THE THEORY OF PSEUDOHOLOMORPHIC CURVES
1 1. STABILITY OF PSEUDOHOLOMORPHIC CURVES
Problem 1. Consider the equation ∂zu=∂zufor a pseudoholomorphic curve u:C→Cm,
where z=x+iy with x, y ∈R.
Given the solution u(z)=(z, z2), determine the stability of this solution.
Solution 1. To determine the stability of the given solution u(z) = (z, z2), we need to analyze
the linearization of the equation around this solution.
Let v(z)=(v1(z), v2(z)) be a perturbation around u(z), such that u(z)=(z, z2) + v(z). We
substitute this into the equation ∂zu=∂zuand linearize it.
The linearized equation becomes:
∂zv=∂zu+∂zv
=∂zu+∂zv
=1
2z+∂zv1
∂zv2
To analyze stability, we look at the linearized equation and determine if a small perturbation
v(z)will grow or decay. In this case, the linearized equation shows that the perturbation will grow,
as the term 2zin ∂zvwill lead to an unbounded growth in v2.
Therefore, the given solution u(z)=(z, z2)is unstable.
2 1. BASICS OF PSEUDOHOLOMORPHIC CURVES
Problem 1. Consider the following pseudoholomorphic curve equation on a Riemann surface:
¯
∂Ju= 0
where ¯
∂Jdenotes the Cauchy-Riemann operator associated with a compatible almost complex
structure J.
Given a specific Riemann surface and almost complex structure:
Riemann surface: Σ = C
Almost complex structure: J(z) = iz
a) Find a solution u: Σ →Cto the pseudoholomorphic curve equation.
b) Verify if the solution obtained in part (a) is J-holomorphic.
Solution 1.
a) To find a solution to the pseudoholomorphic curve equation ¯
∂Ju= 0, we need to solve the
Cauchy-Riemann equation ∂u
∂¯z= 0.
Since Σ = Cand J(z) = iz, the Cauchy-Riemann equation simplifies to ∂u
∂¯z= 0. This means
that uis holomorphic with respect to the standard complex structure J0=i.
One possible solution is u(z) = az +b, where aand bare constants.
b) To verify if uis J-holomorphic, we need to check if ¯
∂u = 0. Since ¯
∂u =1
2(∂u
∂x +i∂u
∂y ) =
1
2(a−ib)= 0, the solution u(z) = az +bis not J-holomorphic under the given almost complex
structure.
Therefore, the solution to the pseudoholomorphic curve equation with the specified Riemann
surface and almost complex structure is u(z) = az +b, but it is not J-holomorphic.
3 3. INDEX THEORY FOR PSEUDOHOLOMORPHIC CURVES
Problem 3. Consider a compact, oriented, Riemannian 2-manifold Mwith a compatible almost
complex structure Jand a symplectic form ω. Let u:C→Mbe a nonconstant J-holomorphic
map. Given that the Fredholm index ind(Du)of uis 2, with Dudenoting the linearized Cauchy-
Riemann operator associated to u, determine the total number of positive and negative punctures
of u.
Solution 3.
Since the index of uis 2, we know that the Conley-Zehnder index of each positive puncture of
uis −1and each negative puncture has a Conley-Zehnder index of 1.
Let n+and n−denote the total number of positive and negative punctures of u, respectively.
Then, we have the equation
ind(Du)=2=2−n++n−
Substitute the Conley-Zehnder index values into the equation, we get
2=2−n++n−
0 = −n++n−
n+=n−
Therefore, the total number of positive punctures is equal to the total number of negative punc-
tures in this case.
4 4. EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 4. Consider the following holomorphic curve in C:u:C→Cgiven by u(z) = z2−i.
We want to find the solutions to the Cauchy-Riemann equation ¯
∂Ju= 0.
Solution 4.
a) We start by writing u(z) = u(x1+ix2) = u1(x1, x2) + iu2(x1, x2)where u1, u2:R2→R.
We then compute the Cauchy-Riemann equations:
∂u1
∂x1
=∂u2
∂x2
and ∂u1
∂x2
=−∂u2
∂x1
In this case, u(z) = z2−i= (x1+ix2)2−i= (x2
1−x2
2) + i(2x1x2−1).
Therefore, we have:
u1(x1, x2) = x2
1−x2
2and u2(x1, x2)=2x1x2−1
Calculating the partial derivatives, we find:
∂u1
∂x1
= 2x1and ∂u2
∂x2
= 2x1
∂u1
∂x2
=−2x2and ∂u2
∂x1
= 2x2
Now we plug these into the Cauchy-Riemann equations:
2x1= 2x1and −2x2= 2x2
Solving the above equations, we obtain x1= 0 and x2= 0.
Therefore, the holomorphic curve u(z) = z2−isatisfies the Cauchy-Riemann equations at the
point (0,0).
b) To find other solutions, we can look for constant maps. Let u(z) = cwhere cis a complex
constant. Then, ¯
∂Ju= 0 is automatically satisfied.
c) Another way to find solutions is by looking at the composition of holomorphic functions. If
v:C→Cand u:C→Care holomorphic, then u◦vis also holomorphic.
5 5. FLOER HOMOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 5. Consider the following pseudoholomorphic curve in Cparametrized by z(t) = eit,
where t∈[0,2π]:
u:R×S1→C, u(s, eit) = eiseit.
a) Show that this curve satisfies the Cauchy-Riemann equations.
b) Calculate the energy of the curve.
c) Determine the image of the curve in C.
Solution 5.
a) To show that the curve satisfies the Cauchy-Riemann equations, we need to check if ∂u
∂¯z= 0.
Here, z=x+iy =x+iy(t), so ∂u
∂¯z=1
2(∂u
∂x +i∂u
∂y ) = 1
2(eiseit −ieiseit) = 0, hence the curve satisfies
the Cauchy-Riemann equations.
b) The energy of the curve is calculated using the formula:
E(u) = ZS1||∇u||2dt,
where ||∇u||2is the norm of the gradient of u. In this case, ∇u= (∂su, ∂tu)=(iu(s, eit), eisieit),
so ||∇u||2=|iu(s, eit)|2+|eisieit|2=|i|2+ 1 = 2. Therefore, the energy of the curve is:
E(u) = Z2π
0
2dt = 4π.
c) The image of the curve in Cis given by u(R×S1) = {eiseit |s∈R, t ∈[0,2π]}. Since eis
and eit vary over all complex numbers on the unit circle S1, the image is the unit circle S1itself.
6 6. COMPACTNESS PROPERTIES OF PSEUDOHOLOMORPHIC CURVES
Problem 6. Consider a sequence of pseudoholomorphic curves (uk: Σ →M, Jk)converging
to a constant map, where Σis a Riemann surface and Mis a symplectic manifold.
Let Jkbe a sequence of almost complex structures converging to J∞in the C∞-topology on
Σ. Assume that the energy of the curves E(uk)is uniformly bounded. Show that the sequence uk
converges in the C∞-topology on Σto a constant map u∞.
Solution 6.
Given that the energy of the curves E(uk)is uniformly bounded, we have
E(uk) = ZΣ||duk◦j−Jk◦duk||2dµ
≤C
where Cis a constant independent of k.
By the compactness theorem for Jk, there exists a subsequence (not relabeled) converging to
aJ∞-holomorphic map u∞. This gives us convergence in the C∞-topology for a subsequence.
Now we want to show that the entire sequence converges to u∞. By contradiction, assume
there exists ϵ > 0such that for all Nthere exists kN> N such that ||dukN−du∞||C∞≥ϵ. But this
is a contradiction, as we have a compact set in C∞-topology.
Therefore, the entire sequence ukconverges in the C∞-topology to the constant map u∞.
7 7. GROMOV-WITTEN INVARIANTS AND PSEUDOHOLOMORPHIC CURVES
Problem 7. Consider the moduli space of genus-zero, three-pointed stable pseudoholomor-
phic curves on a Riemann surface of genus one, with the three marked points labeled p1, p2, p3.
Suppose the homology classes of the curves are A= 2α+ 3βand B=−α+ 4β.
a) Calculate the dimension of the moduli space.
b) Determine the number of points in the moduli space that satisfy the conditions given in part
(a).
c) Suppose the intersection number of Aand Bis 1. Find the expected dimension of the moduli
space.
Solution 7.
a) The dimension of the moduli space can be calculated using the Riemann-Roch formula:
dim(M0,3(Σ, A)) = 3d+ 1 −g= 3(2 + 3) + 1 −1 = 10.
b) In this case, the moduli space contains finitely many points due to the constraint on the
homology classes. Hence, there are no points in the moduli space.
c) The expected dimension of the moduli space can be calculated using the formula:
dim(M0,3(Σ, A;B)) = dim(M0,3(Σ, A)) − ⟨A, B⟩= 10 −1 = 9.
Therefore, the expected dimension of the moduli space is 9.
8 8. EVALUATION MAPS FOR PSEUDOHOLOMORPHIC CURVES
Problem 8. Consider the pseudoholomorphic curve equation on a Riemann surface Sgiven by
du +idv = 0, where u, v :S→Care smooth functions. Let z=u+iv be a complex coordinate on
S. Given a pseudoholomorphic curve u=f(z)parametrized by f(z) = z2, evaluate the following:
a) The evaluation map evp:M → S, where Mis the moduli space of solutions to the pseudo-
holomorphic curve equation, at p= 2 + 3i.
b) The derivative d
dz f(z)at z= 1 −i.
c) The image under the map fof the curve g(z) = z3.
Solution 8.
a) To compute the evaluation map at p= 2+3i, we substitute z= 2+3iinto the parametrization
f(z) = z2. Therefore, f(2 + 3i) = (2 + 3i)2= 4 + 12i−9 = −5 + 12i. Hence, ev2+3i(f) = −5 + 12i.
b) To find the derivative d
dz f(z)at z= 1 −i, we differentiate f(z) = z2to get d
dz f(z) = 2z.
Evaluating this derivative at z= 1 −i, we have d
dz f(1 −i) = 2(1 −i)=2−2i.
c) The image under the map fof the curve g(z) = z3is given by f(g(z)) = f(z3)=(z3)2=z6.
Therefore, the image of the curve g(z)under the map fis the curve parametrized by z6.
9 9. MODULI SPACE OF PSEUDOHOLOMORPHIC CURVES
Problem 9. Consider a Riemann surface Mwith genus g= 2 and n= 3 marked points. Let Xbe
a compact oriented Riemann surface of genus 0 with 5 boundary marked points. Denote by Athe
moduli space of smooth maps ¯
M→Xsatisfying certain boundary conditions.
a) Calculate the dimension of the space A.
b) Determine the dimension of the moduli space of pseudo-holomorphic curves in CP 2with
k= 2 marked points.
Solution 9.
a) The dimension of the moduli space Acan be calculated using Riemann-Roch formula. We
know that for a Riemann surface of genus gwith nmarked points, the dimension of the moduli
space is given by:
dim(M)=6g−6+2n
Substitute g= 2 and n= 3 into the formula:
dim(A) = 6 ×2−6+2×3 = 12 −6 + 6 = 12
Therefore, the dimension of the moduli space Ais 12.
b) The dimension of the moduli space of pseudo-holomorphic curves in CP 2with kmarked
points is given by the formula:
dim M= 2k−6+2g
Substitute k= 2 and g= 0 into the formula:
dim M= 2 ×2−6+2×0=4−6 = −2
Therefore, the dimension of the moduli space of pseudo-holomorphic curves in CP 2with 2
marked points is -2.
10 9. THE THEORY OF PSEUDOHOLMORPHIC CURVES
Problem 9. Consider the following pseudoholomorphic curve in CP 1given by the map u:
R×S1→CP 1, where S1is the unit circle in C:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)].
a) Calculate the image of the curve u.
b) Determine the intersection number of the curve uwith the horizontal line [z: 1], where z∈C.
c) Find the set of critical points of the map u.
Solution 9.
a) To find the image of the curve u, we need to substitute the values of sand θinto the map u.
Let’s first express u(s, eiθ)in terms of sand θ:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)] .
Substitute s= 0 and θ=π/2into uto get the image point:
u(0, eiπ/2) = [1 : 2i].
Therefore, the image of the curve uis the point [1 : 2i]in CP 1.
b) To find the intersection number of the curve uwith the horizontal line [z: 1], we need to count
the number of solutions to the equation 2isin(θ)−icos(θ) = z(1 + is). This reduces to finding the
number of solutions to sin(θ)=0:
sin(θ) = 0 =⇒θ=nπ, where n∈Z.
Since θis periodic with period 2π, there are infinitely many intersection points. Hence, the
intersection number is ∞.
c) The critical points of uare the (s, eiθ)such that the derivative of uvanishes. Calculating the
derivative: ∂u
∂s =i[1 : 2i],∂u
∂θ =i[0 : 2 cos(θ) + sin(θ)].
Setting the derivatives equal to zero, we find that sis not involved. Thus, the critical points lie
on the horizontal line.
11 Numeric Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following Riemann surface defined by the equation z=eiθ, where
θ∈[0,2π].
a) Calculate the area of this Riemann surface.
b) Find the length of the curve defined by x=Re(z)and y=Im(z).
Solution 1.
a) To calculate the area of the Riemann surface, we can use the formula for the area of a surface
of revolution, which is given by A= 2πRβ
αf(θ)p1+(f′(θ))2dθ.
In this case, f(θ)=1and f′(θ)=0, so the area simplifies to:
A= 2πR2π
01·√1+0dθ = 2πR2π
01dθ = 2π[2π−0] = 4π2.
Therefore, the area of the Riemann surface is 4π2.
b) The length of the curve is given by the formula L=Rβ
αrdx
dθ 2+dy
dθ 2dθ.
Since x= cos(θ)and y= sin(θ), we have dx
dθ =−sin(θ)and dx
dθ = cos(θ). Therefore, the length
simplifies to:
L=R2π
0p(−sin(θ))2+ (cos(θ))2dθ =R2π
0√1dθ =R2π
01dθ = 2π.
The length of the curve is 2π.
12 12. BUBBLING PHENOMENON IN PSEUDOHOLOMORPHIC CURVES
Problem 12. Consider the sequence of pseudoholomorphic curves defined by un:R×S2→C
given by un(x, z) = eiθnzn, where θn→0as n→ ∞. Determine whether this sequence exhibits
bubbling phenomenon or not.
Solution 12. a) To analyze the behavior of this sequence of pseudoholomorphic curves, let’s
first calculate the energy of each curve. The energy functional for a pseudoholomorphic curve
u: Σ →Cdefined over a Riemann surface Σis given by
E(u) = ZΣ||du||2dA,
where ||du||2is the square of the norm of the differential du and dA is the area element on Σ.
For the curve un(x, z) = eiθnzn, the energy can be calculated as follows:
||dun||2=|eiθnnzn−1|2=n2|z|2n−2,
and integrating over S2(with the standard metric) gives
E(un) = ZS2
n2|z|2n−2dA = 4πn2.
b) Next, let’s examine the behavior of the energy of the sequence as n→ ∞. We have
lim
n→∞ E(un) = lim
n→∞ 4πn2=∞.
c) Since the energy of the sequence of curves diverges as n→ ∞, we conclude that this
sequence exhibits bubbling phenomenon. In this case, the limiting pseudoholomorphic curve will
be a multiply-covered sphere.
Therefore, the sequence of pseudoholomorphic curves defined by un(x, z) = eiθnznexhibits
bubbling phenomenon.
13 13. SYMPLECTIC TOPOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 13. Consider the complex structure Jon R2given by J(x, y)=(−y, x)and the
symplectic form ω=dx ∧dy.
a) Let u(x, y)=(x2−y2,2xy)be a map R2→R2. Determine if uis pseudoholomorphic with
respect to (J, ω).
b) Find the image of the map ufrom part (a).
Solution 13.
a) To determine if uis pseudoholomorphic with respect to (J, ω), we need to check if du +J◦
du ◦J=ω.
Compute du =2x−2y
2y2xand J◦du ◦J=−2y2x
2x2y.
Adding them gives du +J◦du ◦J=0 0
0 0=ω=dx ∧dy.
Since du +J◦du ◦J=ω,uis not pseudoholomorphic.
b) The image of ucan be found by solving the following system of equations:
x2−y2=u1
2xy =u2
Substitute u1=x2−y2into u2= 2xy to get 2x(x2−y2) = u2.
Solving both equations simultaneously gives us x= 0 or x2=y2, which are the equations for
the coordinate axes.
Therefore, the image of the map uis the xand yaxes.
14 14. CAUCHY-RIEMANN EQUATIONS AND PSEUDOHOLOMORPHIC CURVES
Problem 14. Consider the pseudoholomorphic curve u:C→Cdefined by u(z) = z2+i,
where z=x+iy and i=√−1.
a) Find the Cauchy-Riemann equations for the map u.
b) Determine if the map uis pseudoholomorphic.
Solution 14.
a) To find the Cauchy-Riemann equations for u(z) = z2+i, we first express u(z)in terms of x
and y:
u(z)=(x+iy)2+i=x2−y2+ 2ixy +i
Now, we write u(z)as u(x, y) + iv(x, y):
u(x, y) = x2−y2, v(x, y)=2xy + 1
Next, we find the partial derivatives of uand v:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
The Cauchy-Riemann equations are:
∂u
∂x =∂v
∂y and ∂u
∂y =−∂v
∂x
Substitute the partial derivatives we found into these equations to check if they are satisfied.
b) To determine if the map uis pseudoholomorphic, we need to verify if the Cauchy-Riemann
equations hold for u. From part a, we found that:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
As the Cauchy-Riemann equations are satisfied, the map uis pseudoholomorphic.
15 15. GRADIENT FLOW TECHNIQUES FOR PSEUDOHOLOMORPHIC CURVES
Problem 15. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = z2+iz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 1.
Solution 15. The solution to the pseudoholomorphic curve equation is u(t, z) = u(0, z)for all
t, due to the stationary nature of pseudoholomorphic curves. Hence, u(1, z) = u(0, z) = z2+iz.
Therefore, the solution at time t= 1 is u(1, z) = z2+iz for all z∈Σin this case.
Problem 16. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = eiz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 2.
Solution 16. Similar to the previous problem, the solution to the pseudoholomorphic curve
equation is u(t, z) = u(0, z)for all t, due to the stationary nature of pseudoholomorphic curves.
Hence, u(2, z) = u(0, z) = eiz.
Therefore, the solution at time t= 2 is u(2, z) = eiz for all z∈Σin this case.
16 16. TOPOLOGICAL AND GEOMETRIC CONSTRAINTS ON PSEUDOHOLOMORPHIC CURVES
Problem 16. Consider a closed oriented manifold Mof dimension 4, equipped with a compati-
ble almost complex structure J. Let [γ]be a homology class represented by a simple closed curve
γin M. We define the self-intersection number of [γ]as I([γ],[γ]) = γ·γ.
Suppose Σis a pseudoholomorphic curve in Mwith boundary ∂Σ = γ1−γ2, where γ1and γ2
are simple closed curves representing different homology classes in M. Determine the relationship
between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]).
Solution 16. Let’s denote the two simple closed curves γ1and γ2as Aand B, respectively.
Then, we have I([γ1],[γ1]) = A·Aand I([γ2],[γ2]) = B·B, where ·denotes the intersection pairing.
The self-intersection number of the curve Σcan be computed as I(Σ,Σ) = (γ1−γ2)·(γ1−γ2).
Expanding this out, we get:
I(Σ,Σ) = γ1·γ1−γ1·γ2−γ2·γ1+γ2·γ2
=A·A−A·B−B·A+B·B
=A·A−2A·B+B·B
=I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
Therefore, the relationship between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]) is given by:
I(Σ,Σ) = I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
17 Numerical Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in CP 1:
∂u
∂z +λe−u= 0,
where λ∈Ris a constant.
a) Find the general solution u(z, z)to this pseudoholomorphic curve equation.
b) Given λ= 2, solve the equation with the initial condition u(0,0) = 0.
Solution 1.
a) To find the general solution to the pseudoholomorphic curve equation, we first rewrite the
equation in terms of the complex coordinates zand z. Note that ∂u
∂z =1
2∂u
∂x +i∂u
∂y , where z=
x+iy.
Substitute this into the given equation:
1
2∂u
∂x +i∂u
∂y +λe−u= 0.
Separating real and imaginary parts of this equation gives us:
∂u
∂x = 2λe−u
∂u
∂y = 0
Integrating the first equation with respect to xand the second equation with respect to ygives:
u= ln C
2λ+e2λx,
where Cis the constant of integration.
b) Given λ= 2 and the initial condition u(0,0) = 0, we can find the value of Cby substituting
x= 0, u = 0 into the general solution:
0 = ln C
4+ 1.
Solving this equation for Cgives C= 4, so the solution to the equation with λ= 2 and the initial
condition is:
u= ln 4
4+e4x= ln(1 + e4x).
18 The Theory of Pseudoholomorphic Curves
18.1 Problem 1
Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = aeis +bt with a, b ∈C
and |a|= 3,|b|= 2. Determine the image of this pseudoholomorphic curve in C.
Solution: The image of uis given by u(s, t) = aeis +bt. To find the image, we set u(s, t) = z
and solve for sand t:
z=aeis +bt
= 3eis + 2t
= 3(cos s+isin s)+2t.
This implies that the image of the pseudoholomorphic curve uis the set of points z= 3 cos s+
i3 sin s+ 2tin the complex plane C.
18.2 Problem 2
Let u:R×[0,1] →Cbe a pseudoholomorphic curve defined by u(s, t) = s2+it. Determine the
differential of u.
Solution: The differential of uis given by du =∂u
∂s ds +∂u
∂t dt. In this case, we have:
∂u
∂s = 2s,
∂u
∂t =i.
Therefore, the differential of uis du = 2s ds +i dt.
19 19. ENERGY ESTIMATES FOR PSEUDOHOLOMORPHIC CURVES
Problem 19. Consider a pseudoholomorphic curve u:R×S1→Cgiven by u(s, t) = eis sin(t),
where s, t ∈R. Calculate the energy of this pseudoholomorphic curve using the energy functional
E(u) = RR×S1||∂su||2+||∂tu||2dsdt.
Solution 19. The energy of the pseudoholomorphic curve ucan be calculated as follows:
a) First, we calculate the partial derivatives of u:
∂su=ieis sin(t),
∂tu=eis cos(t).
b) Next, we compute the norms of the partial derivatives:
||∂su||2=|ieis sin(t)|2=| − eis sin(t)|2=|eis|2|sin(t)|2= sin2(t),
||∂tu||2=|eis cos(t)|2=|eis||cos(t)|2= 1|cos(t)|2= cos2(t).
c) Now, we substitute the norms into the energy functional and integrate over R×S1:
E(u) = ZR×S1
(sin2(t) + cos2(t)) dsdt
=ZRZS1
1dsdt
=ZR
2π ds
= 2π.
Therefore, the energy of the pseudoholomorphic curve uis 2π.
20 The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in a symplectic mani-
fold:
¯
∂Ju+J(u)◦∂¯zu= 0
where u:C→Mis a holomorphic map, Jis a compatible almost complex structure in M, and
¯
∂Jdenotes the Cauchy-Riemann operator.
Given u(z) = z+i
z−i, determine if uis a pseudoholomorphic curve in the complex plane C.
Solution 1.
To determine if u(z) = z+i
z−iis a pseudoholomorphic curve, we need to check if it satisfies the
pseudoholomorphic curve equation. First, let’s calculate ∂zuand ∂¯zu:
∂zu=1
(z−i)2
∂¯zu= 0
Next, let’s compute ¯
∂Ju:
¯
∂Ju=1
2(∂¯zu−J(u)◦∂zu) = 1
20−J(u)◦1
(z−i)2
Given that Jis the standard almost complex structure on C, i.e., J(z) = iz, we have:
J(u) = iz+i
z−i=i(z+i)
z−i
Therefore, ¯
∂Ju=−i
2(z−i)3. Since ¯
∂Ju+J(u)◦∂¯zu= 0, the function u(z) = z+i
z−iis indeed a
pseudoholomorphic curve in C.
21 21. STABILITY CONDITIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 21. Consider a pseudoholomorphic curve u: Σ →Mwhere Σis a Riemann surface
and Mis a symplectic manifold with symplectic form ω. Let Jbe an almost complex structure on
Mcompatible with ω. Suppose the pseudoholomorphic curve usatisfies the following Cauchy-
Riemann equation:
¯
∂Ju= 0.
a) Show that the linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J.
b) Calculate the Fredholm index of the operator Du.
c) Determine the stability condition for the pseudoholomorphic curve u.
Solution 21.
a) The linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J. This operator acts on sections of the pullback bundle u∗T M,
which is the tangent bundle of the Riemann surface Σpulled back to Mvia the map u.
b) The Fredholm index of the operator Duis given by
index(Du) = dim ker(Du)−dim coker(Du).
Here, ker(Du)represents the kernel of Du(space of solutions to ¯
∂Jv= 0 near u) and coker(Du)
represents the cokernel of Du. The Fredholm index indicates the deformation parameter for the
moduli space of pseudoholomorphic curves.
c) The stability condition for the pseudoholomorphic curve uis satisfied if the Fredholm index
of the linearized operator Duis nonnegative, i.e., index(Du)≥0. This condition ensures that
the moduli space of pseudoholomorphic curves is well-behaved under small perturbations in the
complex structure or metric on M.
22 22. FREDHOLM THEORY AND PSEUDOHOLOMORPHIC CURVES
Problem 22. Consider the following pseudoholomorphic curve equation in C2:
¯
∂u = 0
where u:R×S1→C2is a smooth map.
Given u(t, z) = eit(z, iz)for t∈[0,2π]and z∈S1, compute the Fredholm index of this pseudo-
holomorphic curve.
Solution 22. To compute the Fredholm index, we first need to find the linearized operator asso-
ciated with ¯
∂at u. Let v(t, z)be a small perturbation around u, and write v(t, z) = eit(a(t, z), b(t, z)).
The linearized operator is given by:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb
Plugging in u(t, z)and v(t, z)into the expression and simplifying, we get:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb=eit(∂ta−i∂tb+∂za−i∂zb)
This is equal to 0if and only if both ∂ta−i∂tband ∂za−i∂zbare equal to 0.
From u(t, z), we have a(t, z) = zand b(t, z) = iz. Differentiating with respect to tand z, we get:
∂ta= 0, ∂tb=−z, ∂za= 1, ∂zb=i
So ∂ta−i∂tb= 0 and ∂za−i∂zb= 1 −i= 0. Thus, the Fredholm index is 0−2 = −2.
23 The Theory of Pseudoholomorphic Curves
Problem: Consider the following pseudoholomorphic curve equation in the complex plane C:
¯
∂Ju+i∂tu= 0
where u:C×[0,1] →Cis a smooth map, ∂tdenotes the partial derivative with respect to
t∈[0,1], and ¯
∂Jdenotes the ¯
∂-operator twisted by a compatible almost complex structure Jon C.
Given the initial condition u(x, 0) = x2for x∈C, solve the pseudoholomorphic curve equation.
Solution:
To solve the pseudoholomorphic curve equation, we will first compute the derivatives and then
use the initial condition to determine the solution.
Let u(x, t) = u1(x, t) + iv1(x, t), where u1and v1are real-valued functions. Then the pseudo-
holomorphic curve equation becomes:
∂u1
∂t −∂v1
∂x +i∂v1
∂t +∂u1
∂x = 0
Separating the real and imaginary parts, we get:
∂u1
∂t =∂v1
∂x ,∂v1
∂t =−∂u1
∂x
From the initial condition u(x, 0) = x2, we have u1(x, 0) = x2and v1(x, 0) = 0.
Solving the partial differential equations with these initial conditions gives u1(x, t) = x2and
v1(x, t)=0. Therefore, the solution of the pseudoholomorphic curve equation is u(x, t) = x2+i·0 =
x2.
24 24. HOMOLOGICAL ALGEBRA AND PSEUDOHOLOMORPHIC CURVES
Problem 24. Consider the following pseudoholomorphic curve in CP 1given by the equation
u(z) = z2.
a) Determine the differential operator ∂Jassociated with the almost complex structure Jinduced
by the Fubini-Study metric on CP 1.
b) Show that u(z)is a solution to the ∂Jequation on CP 1.
Solution 24.
a) The almost complex structure Jassociated with the Fubini-Study metric on CP 1is given by
J(z) = −iz. The differential operator ∂Jcan be computed as:
∂J=1
2(J+iId) = 1
2(−iz +iId) = 1
2(i(z+ 1)).
b) To show that u(z) = z2is a solution to the ∂Jequation, we need to demonstrate that ∂Ju(z) =
0. Let’s compute it:
∂Ju(z) = 1
2(i(z2+ 1)) = i
2z2+i
2=iz +i
2= 0.
As the result is not zero, the function u(z) = z2is not a solution to the ∂Jequation on CP 1.
25 The Theory of Pseudoholomorphic Curves
Problem 1. Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = eis +t2.
a) Determine the asymptotic behavior of uas s→ −∞.
b) Find the limit lims→∞ |u(s, 1)|.
c) Show that uis a pseudoholomorphic curve.
Solution 1.
a) As s→ −∞, the term eis behaves as oscillatory and does not affect the growth, so we focus
on t2. Thus, the asymptotic behavior of u(s, t)as s→ −∞ is t2.
b) We have |u(s, 1)|=|eis + 1| ≤ |eis|+|1|= 1 + 1 = 2. Thus, lims→∞ |u(s, 1)| ≤ 2.
c) To show that uis a pseudoholomorphic curve, we need to verify that ¯
∂Ju= 0, where Jis the
standard complex structure on C. Compute ∂su=ieis and ∂tu= 2t. Then, ¯
∂Ju=∂su−i∂tu=
ieis −2it.
Since ¯
∂Ju= 0, we have shown that uis a pseudoholomorphic curve.
Therefore, the solution to the pseudoholomorphic curve equation with the specified Riemann
surface and almost complex structure is u(z) = az +b, but it is not J-holomorphic.
3 3. INDEX THEORY FOR PSEUDOHOLOMORPHIC CURVES
Problem 3. Consider a compact, oriented, Riemannian 2-manifold Mwith a compatible almost
complex structure Jand a symplectic form ω. Let u:C→Mbe a nonconstant J-holomorphic
map. Given that the Fredholm index ind(Du)of uis 2, with Dudenoting the linearized Cauchy-
Riemann operator associated to u, determine the total number of positive and negative punctures
of u.
Solution 3.
Since the index of uis 2, we know that the Conley-Zehnder index of each positive puncture of
uis −1and each negative puncture has a Conley-Zehnder index of 1.
Let n+and n−denote the total number of positive and negative punctures of u, respectively.
Then, we have the equation
ind(Du)=2=2−n++n−
Substitute the Conley-Zehnder index values into the equation, we get
2=2−n++n−
0 = −n++n−
n+=n−
Therefore, the total number of positive punctures is equal to the total number of negative punc-
tures in this case.
4 4. EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 4. Consider the following holomorphic curve in C:u:C→Cgiven by u(z) = z2−i.
We want to find the solutions to the Cauchy-Riemann equation ¯
∂Ju= 0.
Solution 4.
a) We start by writing u(z) = u(x1+ix2) = u1(x1, x2) + iu2(x1, x2)where u1, u2:R2→R.
We then compute the Cauchy-Riemann equations:
∂u1
∂x1
=∂u2
∂x2
and ∂u1
∂x2
=−∂u2
∂x1
In this case, u(z) = z2−i= (x1+ix2)2−i= (x2
1−x2
2) + i(2x1x2−1).
Therefore, we have:
u1(x1, x2) = x2
1−x2
2and u2(x1, x2)=2x1x2−1
Calculating the partial derivatives, we find:
∂u1
∂x1
= 2x1and ∂u2
∂x2
= 2x1
∂u1
∂x2
=−2x2and ∂u2
∂x1
= 2x2
Now we plug these into the Cauchy-Riemann equations:
2x1= 2x1and −2x2= 2x2
Solving the above equations, we obtain x1= 0 and x2= 0.
Therefore, the holomorphic curve u(z) = z2−isatisfies the Cauchy-Riemann equations at the
point (0,0).
b) To find other solutions, we can look for constant maps. Let u(z) = cwhere cis a complex
constant. Then, ¯
∂Ju= 0 is automatically satisfied.
c) Another way to find solutions is by looking at the composition of holomorphic functions. If
v:C→Cand u:C→Care holomorphic, then u◦vis also holomorphic.
5 5. FLOER HOMOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 5. Consider the following pseudoholomorphic curve in Cparametrized by z(t) = eit,
where t∈[0,2π]:
u:R×S1→C, u(s, eit) = eiseit.
a) Show that this curve satisfies the Cauchy-Riemann equations.
b) Calculate the energy of the curve.
c) Determine the image of the curve in C.
Solution 5.
a) To show that the curve satisfies the Cauchy-Riemann equations, we need to check if ∂u
∂¯z= 0.
Here, z=x+iy =x+iy(t), so ∂u
∂¯z=1
2(∂u
∂x +i∂u
∂y ) = 1
2(eiseit −ieiseit) = 0, hence the curve satisfies
the Cauchy-Riemann equations.
b) The energy of the curve is calculated using the formula:
E(u) = ZS1||∇u||2dt,
where ||∇u||2is the norm of the gradient of u. In this case, ∇u= (∂su, ∂tu)=(iu(s, eit), eisieit),
so ||∇u||2=|iu(s, eit)|2+|eisieit|2=|i|2+ 1 = 2. Therefore, the energy of the curve is:
E(u) = Z2π
0
2dt = 4π.
c) The image of the curve in Cis given by u(R×S1) = {eiseit |s∈R, t ∈[0,2π]}. Since eis
and eit vary over all complex numbers on the unit circle S1, the image is the unit circle S1itself.
6 6. COMPACTNESS PROPERTIES OF PSEUDOHOLOMORPHIC CURVES
Problem 6. Consider a sequence of pseudoholomorphic curves (uk: Σ →M, Jk)converging
to a constant map, where Σis a Riemann surface and Mis a symplectic manifold.
Let Jkbe a sequence of almost complex structures converging to J∞in the C∞-topology on
Σ. Assume that the energy of the curves E(uk)is uniformly bounded. Show that the sequence uk
converges in the C∞-topology on Σto a constant map u∞.
Solution 6.
Given that the energy of the curves E(uk)is uniformly bounded, we have
E(uk) = ZΣ||duk◦j−Jk◦duk||2dµ
≤C
where Cis a constant independent of k.
By the compactness theorem for Jk, there exists a subsequence (not relabeled) converging to
aJ∞-holomorphic map u∞. This gives us convergence in the C∞-topology for a subsequence.
Now we want to show that the entire sequence converges to u∞. By contradiction, assume
there exists ϵ > 0such that for all Nthere exists kN> N such that ||dukN−du∞||C∞≥ϵ. But this
is a contradiction, as we have a compact set in C∞-topology.
Therefore, the entire sequence ukconverges in the C∞-topology to the constant map u∞.
7 7. GROMOV-WITTEN INVARIANTS AND PSEUDOHOLOMORPHIC CURVES
Problem 7. Consider the moduli space of genus-zero, three-pointed stable pseudoholomor-
phic curves on a Riemann surface of genus one, with the three marked points labeled p1, p2, p3.
Suppose the homology classes of the curves are A= 2α+ 3βand B=−α+ 4β.
a) Calculate the dimension of the moduli space.
b) Determine the number of points in the moduli space that satisfy the conditions given in part
(a).
c) Suppose the intersection number of Aand Bis 1. Find the expected dimension of the moduli
space.
Solution 7.
a) The dimension of the moduli space can be calculated using the Riemann-Roch formula:
dim(M0,3(Σ, A)) = 3d+ 1 −g= 3(2 + 3) + 1 −1 = 10.
b) In this case, the moduli space contains finitely many points due to the constraint on the
homology classes. Hence, there are no points in the moduli space.
c) The expected dimension of the moduli space can be calculated using the formula:
dim(M0,3(Σ, A;B)) = dim(M0,3(Σ, A)) − ⟨A, B⟩= 10 −1 = 9.
Therefore, the expected dimension of the moduli space is 9.
8 8. EVALUATION MAPS FOR PSEUDOHOLOMORPHIC CURVES
Problem 8. Consider the pseudoholomorphic curve equation on a Riemann surface Sgiven by
du +idv = 0, where u, v :S→Care smooth functions. Let z=u+iv be a complex coordinate on
S. Given a pseudoholomorphic curve u=f(z)parametrized by f(z) = z2, evaluate the following:
a) The evaluation map evp:M → S, where Mis the moduli space of solutions to the pseudo-
holomorphic curve equation, at p= 2 + 3i.
b) The derivative d
dz f(z)at z= 1 −i.
c) The image under the map fof the curve g(z) = z3.
Solution 8.
a) To compute the evaluation map at p= 2+3i, we substitute z= 2+3iinto the parametrization
f(z) = z2. Therefore, f(2 + 3i) = (2 + 3i)2= 4 + 12i−9 = −5 + 12i. Hence, ev2+3i(f) = −5 + 12i.
b) To find the derivative d
dz f(z)at z= 1 −i, we differentiate f(z) = z2to get d
dz f(z) = 2z.
Evaluating this derivative at z= 1 −i, we have d
dz f(1 −i) = 2(1 −i)=2−2i.
c) The image under the map fof the curve g(z) = z3is given by f(g(z)) = f(z3)=(z3)2=z6.
Therefore, the image of the curve g(z)under the map fis the curve parametrized by z6.
9 9. MODULI SPACE OF PSEUDOHOLOMORPHIC CURVES
Problem 9. Consider a Riemann surface Mwith genus g= 2 and n= 3 marked points. Let Xbe
a compact oriented Riemann surface of genus 0 with 5 boundary marked points. Denote by Athe
moduli space of smooth maps ¯
M→Xsatisfying certain boundary conditions.
a) Calculate the dimension of the space A.
b) Determine the dimension of the moduli space of pseudo-holomorphic curves in CP 2with
k= 2 marked points.
Solution 9.
a) The dimension of the moduli space Acan be calculated using Riemann-Roch formula. We
know that for a Riemann surface of genus gwith nmarked points, the dimension of the moduli
space is given by:
dim(M)=6g−6+2n
Substitute g= 2 and n= 3 into the formula:
dim(A) = 6 ×2−6+2×3 = 12 −6 + 6 = 12
Therefore, the dimension of the moduli space Ais 12.
b) The dimension of the moduli space of pseudo-holomorphic curves in CP 2with kmarked
points is given by the formula:
dim M= 2k−6+2g
Substitute k= 2 and g= 0 into the formula:
dim M= 2 ×2−6+2×0=4−6 = −2
Therefore, the dimension of the moduli space of pseudo-holomorphic curves in CP 2with 2
marked points is -2.
10 9. THE THEORY OF PSEUDOHOLMORPHIC CURVES
Problem 9. Consider the following pseudoholomorphic curve in CP 1given by the map u:
R×S1→CP 1, where S1is the unit circle in C:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)].
a) Calculate the image of the curve u.
b) Determine the intersection number of the curve uwith the horizontal line [z: 1], where z∈C.
c) Find the set of critical points of the map u.
Solution 9.
a) To find the image of the curve u, we need to substitute the values of sand θinto the map u.
Let’s first express u(s, eiθ)in terms of sand θ:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)] .
Substitute s= 0 and θ=π/2into uto get the image point:
u(0, eiπ/2) = [1 : 2i].
Therefore, the image of the curve uis the point [1 : 2i]in CP 1.
b) To find the intersection number of the curve uwith the horizontal line [z: 1], we need to count
the number of solutions to the equation 2isin(θ)−icos(θ) = z(1 + is). This reduces to finding the
number of solutions to sin(θ)=0:
sin(θ) = 0 =⇒θ=nπ, where n∈Z.
Since θis periodic with period 2π, there are infinitely many intersection points. Hence, the
intersection number is ∞.
c) The critical points of uare the (s, eiθ)such that the derivative of uvanishes. Calculating the
derivative: ∂u
∂s =i[1 : 2i],∂u
∂θ =i[0 : 2 cos(θ) + sin(θ)].
Setting the derivatives equal to zero, we find that sis not involved. Thus, the critical points lie
on the horizontal line.
11 Numeric Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following Riemann surface defined by the equation z=eiθ, where
θ∈[0,2π].
a) Calculate the area of this Riemann surface.
b) Find the length of the curve defined by x=Re(z)and y=Im(z).
Solution 1.
a) To calculate the area of the Riemann surface, we can use the formula for the area of a surface
of revolution, which is given by A= 2πRβ
αf(θ)p1+(f′(θ))2dθ.
In this case, f(θ)=1and f′(θ)=0, so the area simplifies to:
A= 2πR2π
01·√1+0dθ = 2πR2π
01dθ = 2π[2π−0] = 4π2.
Therefore, the area of the Riemann surface is 4π2.
b) The length of the curve is given by the formula L=Rβ
αrdx
dθ 2+dy
dθ 2dθ.
Since x= cos(θ)and y= sin(θ), we have dx
dθ =−sin(θ)and dx
dθ = cos(θ). Therefore, the length
simplifies to:
L=R2π
0p(−sin(θ))2+ (cos(θ))2dθ =R2π
0√1dθ =R2π
01dθ = 2π.
The length of the curve is 2π.
12 12. BUBBLING PHENOMENON IN PSEUDOHOLOMORPHIC CURVES
Problem 12. Consider the sequence of pseudoholomorphic curves defined by un:R×S2→C
given by un(x, z) = eiθnzn, where θn→0as n→ ∞. Determine whether this sequence exhibits
bubbling phenomenon or not.
Solution 12. a) To analyze the behavior of this sequence of pseudoholomorphic curves, let’s
first calculate the energy of each curve. The energy functional for a pseudoholomorphic curve
u: Σ →Cdefined over a Riemann surface Σis given by
E(u) = ZΣ||du||2dA,
where ||du||2is the square of the norm of the differential du and dA is the area element on Σ.
For the curve un(x, z) = eiθnzn, the energy can be calculated as follows:
||dun||2=|eiθnnzn−1|2=n2|z|2n−2,
and integrating over S2(with the standard metric) gives
E(un) = ZS2
n2|z|2n−2dA = 4πn2.
b) Next, let’s examine the behavior of the energy of the sequence as n→ ∞. We have
lim
n→∞ E(un) = lim
n→∞ 4πn2=∞.
c) Since the energy of the sequence of curves diverges as n→ ∞, we conclude that this
sequence exhibits bubbling phenomenon. In this case, the limiting pseudoholomorphic curve will
be a multiply-covered sphere.
Therefore, the sequence of pseudoholomorphic curves defined by un(x, z) = eiθnznexhibits
bubbling phenomenon.
13 13. SYMPLECTIC TOPOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 13. Consider the complex structure Jon R2given by J(x, y)=(−y, x)and the
symplectic form ω=dx ∧dy.
a) Let u(x, y)=(x2−y2,2xy)be a map R2→R2. Determine if uis pseudoholomorphic with
respect to (J, ω).
b) Find the image of the map ufrom part (a).
Solution 13.
a) To determine if uis pseudoholomorphic with respect to (J, ω), we need to check if du +J◦
du ◦J=ω.
Compute du =2x−2y
2y2xand J◦du ◦J=−2y2x
2x2y.
Adding them gives du +J◦du ◦J=0 0
0 0=ω=dx ∧dy.
Since du +J◦du ◦J=ω,uis not pseudoholomorphic.
b) The image of ucan be found by solving the following system of equations:
x2−y2=u1
2xy =u2
Substitute u1=x2−y2into u2= 2xy to get 2x(x2−y2) = u2.
Solving both equations simultaneously gives us x= 0 or x2=y2, which are the equations for
the coordinate axes.
Therefore, the image of the map uis the xand yaxes.
14 14. CAUCHY-RIEMANN EQUATIONS AND PSEUDOHOLOMORPHIC CURVES
Problem 14. Consider the pseudoholomorphic curve u:C→Cdefined by u(z) = z2+i,
where z=x+iy and i=√−1.
a) Find the Cauchy-Riemann equations for the map u.
b) Determine if the map uis pseudoholomorphic.
Solution 14.
a) To find the Cauchy-Riemann equations for u(z) = z2+i, we first express u(z)in terms of x
and y:
u(z)=(x+iy)2+i=x2−y2+ 2ixy +i
Now, we write u(z)as u(x, y) + iv(x, y):
u(x, y) = x2−y2, v(x, y)=2xy + 1
Next, we find the partial derivatives of uand v:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
The Cauchy-Riemann equations are:
∂u
∂x =∂v
∂y and ∂u
∂y =−∂v
∂x
Substitute the partial derivatives we found into these equations to check if they are satisfied.
b) To determine if the map uis pseudoholomorphic, we need to verify if the Cauchy-Riemann
equations hold for u. From part a, we found that:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
As the Cauchy-Riemann equations are satisfied, the map uis pseudoholomorphic.
15 15. GRADIENT FLOW TECHNIQUES FOR PSEUDOHOLOMORPHIC CURVES
Problem 15. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = z2+iz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 1.
Solution 15. The solution to the pseudoholomorphic curve equation is u(t, z) = u(0, z)for all
t, due to the stationary nature of pseudoholomorphic curves. Hence, u(1, z) = u(0, z) = z2+iz.
Therefore, the solution at time t= 1 is u(1, z) = z2+iz for all z∈Σin this case.
Problem 16. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = eiz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 2.
Solution 16. Similar to the previous problem, the solution to the pseudoholomorphic curve
equation is u(t, z) = u(0, z)for all t, due to the stationary nature of pseudoholomorphic curves.
Hence, u(2, z) = u(0, z) = eiz.
Therefore, the solution at time t= 2 is u(2, z) = eiz for all z∈Σin this case.
16 16. TOPOLOGICAL AND GEOMETRIC CONSTRAINTS ON PSEUDOHOLOMORPHIC CURVES
Problem 16. Consider a closed oriented manifold Mof dimension 4, equipped with a compati-
ble almost complex structure J. Let [γ]be a homology class represented by a simple closed curve
γin M. We define the self-intersection number of [γ]as I([γ],[γ]) = γ·γ.
Suppose Σis a pseudoholomorphic curve in Mwith boundary ∂Σ = γ1−γ2, where γ1and γ2
are simple closed curves representing different homology classes in M. Determine the relationship
between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]).
Solution 16. Let’s denote the two simple closed curves γ1and γ2as Aand B, respectively.
Then, we have I([γ1],[γ1]) = A·Aand I([γ2],[γ2]) = B·B, where ·denotes the intersection pairing.
The self-intersection number of the curve Σcan be computed as I(Σ,Σ) = (γ1−γ2)·(γ1−γ2).
Expanding this out, we get:
I(Σ,Σ) = γ1·γ1−γ1·γ2−γ2·γ1+γ2·γ2
=A·A−A·B−B·A+B·B
=A·A−2A·B+B·B
=I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
Therefore, the relationship between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]) is given by:
I(Σ,Σ) = I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
17 Numerical Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in CP 1:
∂u
∂z +λe−u= 0,
where λ∈Ris a constant.
a) Find the general solution u(z, z)to this pseudoholomorphic curve equation.
b) Given λ= 2, solve the equation with the initial condition u(0,0) = 0.
Solution 1.
a) To find the general solution to the pseudoholomorphic curve equation, we first rewrite the
equation in terms of the complex coordinates zand z. Note that ∂u
∂z =1
2∂u
∂x +i∂u
∂y , where z=
x+iy.
Substitute this into the given equation:
1
2∂u
∂x +i∂u
∂y +λe−u= 0.
Separating real and imaginary parts of this equation gives us:
∂u
∂x = 2λe−u
∂u
∂y = 0
Integrating the first equation with respect to xand the second equation with respect to ygives:
u= ln C
2λ+e2λx,
where Cis the constant of integration.
b) Given λ= 2 and the initial condition u(0,0) = 0, we can find the value of Cby substituting
x= 0, u = 0 into the general solution:
0 = ln C
4+ 1.
Solving this equation for Cgives C= 4, so the solution to the equation with λ= 2 and the initial
condition is:
u= ln 4
4+e4x= ln(1 + e4x).
18 The Theory of Pseudoholomorphic Curves
18.1 Problem 1
Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = aeis +bt with a, b ∈C
and |a|= 3,|b|= 2. Determine the image of this pseudoholomorphic curve in C.
Solution: The image of uis given by u(s, t) = aeis +bt. To find the image, we set u(s, t) = z
and solve for sand t:
z=aeis +bt
= 3eis + 2t
= 3(cos s+isin s)+2t.
This implies that the image of the pseudoholomorphic curve uis the set of points z= 3 cos s+
i3 sin s+ 2tin the complex plane C.
18.2 Problem 2
Let u:R×[0,1] →Cbe a pseudoholomorphic curve defined by u(s, t) = s2+it. Determine the
differential of u.
Solution: The differential of uis given by du =∂u
∂s ds +∂u
∂t dt. In this case, we have:
∂u
∂s = 2s,
∂u
∂t =i.
Therefore, the differential of uis du = 2s ds +i dt.
19 19. ENERGY ESTIMATES FOR PSEUDOHOLOMORPHIC CURVES
Problem 19. Consider a pseudoholomorphic curve u:R×S1→Cgiven by u(s, t) = eis sin(t),
where s, t ∈R. Calculate the energy of this pseudoholomorphic curve using the energy functional
E(u) = RR×S1||∂su||2+||∂tu||2dsdt.
Solution 19. The energy of the pseudoholomorphic curve ucan be calculated as follows:
a) First, we calculate the partial derivatives of u:
∂su=ieis sin(t),
∂tu=eis cos(t).
b) Next, we compute the norms of the partial derivatives:
||∂su||2=|ieis sin(t)|2=| − eis sin(t)|2=|eis|2|sin(t)|2= sin2(t),
||∂tu||2=|eis cos(t)|2=|eis||cos(t)|2= 1|cos(t)|2= cos2(t).
c) Now, we substitute the norms into the energy functional and integrate over R×S1:
E(u) = ZR×S1
(sin2(t) + cos2(t)) dsdt
=ZRZS1
1dsdt
=ZR
2π ds
= 2π.
Therefore, the energy of the pseudoholomorphic curve uis 2π.
20 The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in a symplectic mani-
fold:
¯
∂Ju+J(u)◦∂¯zu= 0
where u:C→Mis a holomorphic map, Jis a compatible almost complex structure in M, and
¯
∂Jdenotes the Cauchy-Riemann operator.
Given u(z) = z+i
z−i, determine if uis a pseudoholomorphic curve in the complex plane C.
Solution 1.
To determine if u(z) = z+i
z−iis a pseudoholomorphic curve, we need to check if it satisfies the
pseudoholomorphic curve equation. First, let’s calculate ∂zuand ∂¯zu:
∂zu=1
(z−i)2
∂¯zu= 0
Next, let’s compute ¯
∂Ju:
¯
∂Ju=1
2(∂¯zu−J(u)◦∂zu) = 1
20−J(u)◦1
(z−i)2
Given that Jis the standard almost complex structure on C, i.e., J(z) = iz, we have:
J(u) = iz+i
z−i=i(z+i)
z−i
Therefore, ¯
∂Ju=−i
2(z−i)3. Since ¯
∂Ju+J(u)◦∂¯zu= 0, the function u(z) = z+i
z−iis indeed a
pseudoholomorphic curve in C.
21 21. STABILITY CONDITIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 21. Consider a pseudoholomorphic curve u: Σ →Mwhere Σis a Riemann surface
and Mis a symplectic manifold with symplectic form ω. Let Jbe an almost complex structure on
Mcompatible with ω. Suppose the pseudoholomorphic curve usatisfies the following Cauchy-
Riemann equation:
¯
∂Ju= 0.
a) Show that the linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J.
b) Calculate the Fredholm index of the operator Du.
c) Determine the stability condition for the pseudoholomorphic curve u.
Solution 21.
a) The linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J. This operator acts on sections of the pullback bundle u∗T M,
which is the tangent bundle of the Riemann surface Σpulled back to Mvia the map u.
b) The Fredholm index of the operator Duis given by
index(Du) = dim ker(Du)−dim coker(Du).
Here, ker(Du)represents the kernel of Du(space of solutions to ¯
∂Jv= 0 near u) and coker(Du)
represents the cokernel of Du. The Fredholm index indicates the deformation parameter for the
moduli space of pseudoholomorphic curves.
c) The stability condition for the pseudoholomorphic curve uis satisfied if the Fredholm index
of the linearized operator Duis nonnegative, i.e., index(Du)≥0. This condition ensures that
the moduli space of pseudoholomorphic curves is well-behaved under small perturbations in the
complex structure or metric on M.
22 22. FREDHOLM THEORY AND PSEUDOHOLOMORPHIC CURVES
Problem 22. Consider the following pseudoholomorphic curve equation in C2:
¯
∂u = 0
where u:R×S1→C2is a smooth map.
Given u(t, z) = eit(z, iz)for t∈[0,2π]and z∈S1, compute the Fredholm index of this pseudo-
holomorphic curve.
Solution 22. To compute the Fredholm index, we first need to find the linearized operator asso-
ciated with ¯
∂at u. Let v(t, z)be a small perturbation around u, and write v(t, z) = eit(a(t, z), b(t, z)).
The linearized operator is given by:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb
Plugging in u(t, z)and v(t, z)into the expression and simplifying, we get:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb=eit(∂ta−i∂tb+∂za−i∂zb)
This is equal to 0if and only if both ∂ta−i∂tband ∂za−i∂zbare equal to 0.
From u(t, z), we have a(t, z) = zand b(t, z) = iz. Differentiating with respect to tand z, we get:
∂ta= 0, ∂tb=−z, ∂za= 1, ∂zb=i
So ∂ta−i∂tb= 0 and ∂za−i∂zb= 1 −i= 0. Thus, the Fredholm index is 0−2 = −2.
23 The Theory of Pseudoholomorphic Curves
Problem: Consider the following pseudoholomorphic curve equation in the complex plane C:
¯
∂Ju+i∂tu= 0
where u:C×[0,1] →Cis a smooth map, ∂tdenotes the partial derivative with respect to
t∈[0,1], and ¯
∂Jdenotes the ¯
∂-operator twisted by a compatible almost complex structure Jon C.
Given the initial condition u(x, 0) = x2for x∈C, solve the pseudoholomorphic curve equation.
Solution:
To solve the pseudoholomorphic curve equation, we will first compute the derivatives and then
use the initial condition to determine the solution.
Let u(x, t) = u1(x, t) + iv1(x, t), where u1and v1are real-valued functions. Then the pseudo-
holomorphic curve equation becomes:
∂u1
∂t −∂v1
∂x +i∂v1
∂t +∂u1
∂x = 0
Separating the real and imaginary parts, we get:
∂u1
∂t =∂v1
∂x ,∂v1
∂t =−∂u1
∂x
From the initial condition u(x, 0) = x2, we have u1(x, 0) = x2and v1(x, 0) = 0.
Solving the partial differential equations with these initial conditions gives u1(x, t) = x2and
v1(x, t)=0. Therefore, the solution of the pseudoholomorphic curve equation is u(x, t) = x2+i·0 =
x2.
24 24. HOMOLOGICAL ALGEBRA AND PSEUDOHOLOMORPHIC CURVES
Problem 24. Consider the following pseudoholomorphic curve in CP 1given by the equation
u(z) = z2.
a) Determine the differential operator ∂Jassociated with the almost complex structure Jinduced
by the Fubini-Study metric on CP 1.
b) Show that u(z)is a solution to the ∂Jequation on CP 1.
Solution 24.
a) The almost complex structure Jassociated with the Fubini-Study metric on CP 1is given by
J(z) = −iz. The differential operator ∂Jcan be computed as:
∂J=1
2(J+iId) = 1
2(−iz +iId) = 1
2(i(z+ 1)).
b) To show that u(z) = z2is a solution to the ∂Jequation, we need to demonstrate that ∂Ju(z) =
0. Let’s compute it:
∂Ju(z) = 1
2(i(z2+ 1)) = i
2z2+i
2=iz +i
2= 0.
As the result is not zero, the function u(z) = z2is not a solution to the ∂Jequation on CP 1.
25 The Theory of Pseudoholomorphic Curves
Problem 1. Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = eis +t2.
a) Determine the asymptotic behavior of uas s→ −∞.
b) Find the limit lims→∞ |u(s, 1)|.
c) Show that uis a pseudoholomorphic curve.
Solution 1.
a) As s→ −∞, the term eis behaves as oscillatory and does not affect the growth, so we focus
on t2. Thus, the asymptotic behavior of u(s, t)as s→ −∞ is t2.
b) We have |u(s, 1)|=|eis + 1| ≤ |eis|+|1|= 1 + 1 = 2. Thus, lims→∞ |u(s, 1)| ≤ 2.
c) To show that uis a pseudoholomorphic curve, we need to verify that ¯
∂Ju= 0, where Jis the
standard complex structure on C. Compute ∂su=ieis and ∂tu= 2t. Then, ¯
∂Ju=∂su−i∂tu=
ieis −2it.
Since ¯
∂Ju= 0, we have shown that uis a pseudoholomorphic curve.
Therefore, the solution to the pseudoholomorphic curve equation with the specified Riemann
surface and almost complex structure is u(z) = az +b, but it is not J-holomorphic.
3 3. INDEX THEORY FOR PSEUDOHOLOMORPHIC CURVES
Problem 3. Consider a compact, oriented, Riemannian 2-manifold Mwith a compatible almost
complex structure Jand a symplectic form ω. Let u:C→Mbe a nonconstant J-holomorphic
map. Given that the Fredholm index ind(Du)of uis 2, with Dudenoting the linearized Cauchy-
Riemann operator associated to u, determine the total number of positive and negative punctures
of u.
Solution 3.
Since the index of uis 2, we know that the Conley-Zehnder index of each positive puncture of
uis −1and each negative puncture has a Conley-Zehnder index of 1.
Let n+and n−denote the total number of positive and negative punctures of u, respectively.
Then, we have the equation
ind(Du)=2=2−n++n−
Substitute the Conley-Zehnder index values into the equation, we get
2=2−n++n−
0 = −n++n−
n+=n−
Therefore, the total number of positive punctures is equal to the total number of negative punc-
tures in this case.
4 4. EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 4. Consider the following holomorphic curve in C:u:C→Cgiven by u(z) = z2−i.
We want to find the solutions to the Cauchy-Riemann equation ¯
∂Ju= 0.
Solution 4.
a) We start by writing u(z) = u(x1+ix2) = u1(x1, x2) + iu2(x1, x2)where u1, u2:R2→R.
We then compute the Cauchy-Riemann equations:
∂u1
∂x1
=∂u2
∂x2
and ∂u1
∂x2
=−∂u2
∂x1
In this case, u(z) = z2−i= (x1+ix2)2−i= (x2
1−x2
2) + i(2x1x2−1).
Therefore, we have:
u1(x1, x2) = x2
1−x2
2and u2(x1, x2)=2x1x2−1
Calculating the partial derivatives, we find:
∂u1
∂x1
= 2x1and ∂u2
∂x2
= 2x1
∂u1
∂x2
=−2x2and ∂u2
∂x1
= 2x2
Now we plug these into the Cauchy-Riemann equations:
2x1= 2x1and −2x2= 2x2
Solving the above equations, we obtain x1= 0 and x2= 0.
Therefore, the holomorphic curve u(z) = z2−isatisfies the Cauchy-Riemann equations at the
point (0,0).
b) To find other solutions, we can look for constant maps. Let u(z) = cwhere cis a complex
constant. Then, ¯
∂Ju= 0 is automatically satisfied.
c) Another way to find solutions is by looking at the composition of holomorphic functions. If
v:C→Cand u:C→Care holomorphic, then u◦vis also holomorphic.
5 5. FLOER HOMOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 5. Consider the following pseudoholomorphic curve in Cparametrized by z(t) = eit,
where t∈[0,2π]:
u:R×S1→C, u(s, eit) = eiseit.
a) Show that this curve satisfies the Cauchy-Riemann equations.
b) Calculate the energy of the curve.
c) Determine the image of the curve in C.
Solution 5.
a) To show that the curve satisfies the Cauchy-Riemann equations, we need to check if ∂u
∂¯z= 0.
Here, z=x+iy =x+iy(t), so ∂u
∂¯z=1
2(∂u
∂x +i∂u
∂y ) = 1
2(eiseit −ieiseit) = 0, hence the curve satisfies
the Cauchy-Riemann equations.
b) The energy of the curve is calculated using the formula:
E(u) = ZS1||∇u||2dt,
where ||∇u||2is the norm of the gradient of u. In this case, ∇u= (∂su, ∂tu)=(iu(s, eit), eisieit),
so ||∇u||2=|iu(s, eit)|2+|eisieit|2=|i|2+ 1 = 2. Therefore, the energy of the curve is:
E(u) = Z2π
0
2dt = 4π.
c) The image of the curve in Cis given by u(R×S1) = {eiseit |s∈R, t ∈[0,2π]}. Since eis
and eit vary over all complex numbers on the unit circle S1, the image is the unit circle S1itself.
6 6. COMPACTNESS PROPERTIES OF PSEUDOHOLOMORPHIC CURVES
Problem 6. Consider a sequence of pseudoholomorphic curves (uk: Σ →M, Jk)converging
to a constant map, where Σis a Riemann surface and Mis a symplectic manifold.
Let Jkbe a sequence of almost complex structures converging to J∞in the C∞-topology on
Σ. Assume that the energy of the curves E(uk)is uniformly bounded. Show that the sequence uk
converges in the C∞-topology on Σto a constant map u∞.
Solution 6.
Given that the energy of the curves E(uk)is uniformly bounded, we have
E(uk) = ZΣ||duk◦j−Jk◦duk||2dµ
≤C
where Cis a constant independent of k.
By the compactness theorem for Jk, there exists a subsequence (not relabeled) converging to
aJ∞-holomorphic map u∞. This gives us convergence in the C∞-topology for a subsequence.
Now we want to show that the entire sequence converges to u∞. By contradiction, assume
there exists ϵ > 0such that for all Nthere exists kN> N such that ||dukN−du∞||C∞≥ϵ. But this
is a contradiction, as we have a compact set in C∞-topology.
Therefore, the entire sequence ukconverges in the C∞-topology to the constant map u∞.
7 7. GROMOV-WITTEN INVARIANTS AND PSEUDOHOLOMORPHIC CURVES
Problem 7. Consider the moduli space of genus-zero, three-pointed stable pseudoholomor-
phic curves on a Riemann surface of genus one, with the three marked points labeled p1, p2, p3.
Suppose the homology classes of the curves are A= 2α+ 3βand B=−α+ 4β.
a) Calculate the dimension of the moduli space.
b) Determine the number of points in the moduli space that satisfy the conditions given in part
(a).
c) Suppose the intersection number of Aand Bis 1. Find the expected dimension of the moduli
space.
Solution 7.
a) The dimension of the moduli space can be calculated using the Riemann-Roch formula:
dim(M0,3(Σ, A)) = 3d+ 1 −g= 3(2 + 3) + 1 −1 = 10.
b) In this case, the moduli space contains finitely many points due to the constraint on the
homology classes. Hence, there are no points in the moduli space.
c) The expected dimension of the moduli space can be calculated using the formula:
dim(M0,3(Σ, A;B)) = dim(M0,3(Σ, A)) − ⟨A, B⟩= 10 −1 = 9.
Therefore, the expected dimension of the moduli space is 9.
8 8. EVALUATION MAPS FOR PSEUDOHOLOMORPHIC CURVES
Problem 8. Consider the pseudoholomorphic curve equation on a Riemann surface Sgiven by
du +idv = 0, where u, v :S→Care smooth functions. Let z=u+iv be a complex coordinate on
S. Given a pseudoholomorphic curve u=f(z)parametrized by f(z) = z2, evaluate the following:
a) The evaluation map evp:M → S, where Mis the moduli space of solutions to the pseudo-
holomorphic curve equation, at p= 2 + 3i.
b) The derivative d
dz f(z)at z= 1 −i.
c) The image under the map fof the curve g(z) = z3.
Solution 8.
a) To compute the evaluation map at p= 2+3i, we substitute z= 2+3iinto the parametrization
f(z) = z2. Therefore, f(2 + 3i) = (2 + 3i)2= 4 + 12i−9 = −5 + 12i. Hence, ev2+3i(f) = −5 + 12i.
b) To find the derivative d
dz f(z)at z= 1 −i, we differentiate f(z) = z2to get d
dz f(z) = 2z.
Evaluating this derivative at z= 1 −i, we have d
dz f(1 −i) = 2(1 −i)=2−2i.
c) The image under the map fof the curve g(z) = z3is given by f(g(z)) = f(z3)=(z3)2=z6.
Therefore, the image of the curve g(z)under the map fis the curve parametrized by z6.
9 9. MODULI SPACE OF PSEUDOHOLOMORPHIC CURVES
Problem 9. Consider a Riemann surface Mwith genus g= 2 and n= 3 marked points. Let Xbe
a compact oriented Riemann surface of genus 0 with 5 boundary marked points. Denote by Athe
moduli space of smooth maps ¯
M→Xsatisfying certain boundary conditions.
a) Calculate the dimension of the space A.
b) Determine the dimension of the moduli space of pseudo-holomorphic curves in CP 2with
k= 2 marked points.
Solution 9.
a) The dimension of the moduli space Acan be calculated using Riemann-Roch formula. We
know that for a Riemann surface of genus gwith nmarked points, the dimension of the moduli
space is given by:
dim(M)=6g−6+2n
Substitute g= 2 and n= 3 into the formula:
dim(A) = 6 ×2−6+2×3 = 12 −6 + 6 = 12
Therefore, the dimension of the moduli space Ais 12.
b) The dimension of the moduli space of pseudo-holomorphic curves in CP 2with kmarked
points is given by the formula:
dim M= 2k−6+2g
Substitute k= 2 and g= 0 into the formula:
dim M= 2 ×2−6+2×0=4−6 = −2
Therefore, the dimension of the moduli space of pseudo-holomorphic curves in CP 2with 2
marked points is -2.
10 9. THE THEORY OF PSEUDOHOLMORPHIC CURVES
Problem 9. Consider the following pseudoholomorphic curve in CP 1given by the map u:
R×S1→CP 1, where S1is the unit circle in C:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)].
a) Calculate the image of the curve u.
b) Determine the intersection number of the curve uwith the horizontal line [z: 1], where z∈C.
c) Find the set of critical points of the map u.
Solution 9.
a) To find the image of the curve u, we need to substitute the values of sand θinto the map u.
Let’s first express u(s, eiθ)in terms of sand θ:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)] .
Substitute s= 0 and θ=π/2into uto get the image point:
u(0, eiπ/2) = [1 : 2i].
Therefore, the image of the curve uis the point [1 : 2i]in CP 1.
b) To find the intersection number of the curve uwith the horizontal line [z: 1], we need to count
the number of solutions to the equation 2isin(θ)−icos(θ) = z(1 + is). This reduces to finding the
number of solutions to sin(θ)=0:
sin(θ) = 0 =⇒θ=nπ, where n∈Z.
Since θis periodic with period 2π, there are infinitely many intersection points. Hence, the
intersection number is ∞.
c) The critical points of uare the (s, eiθ)such that the derivative of uvanishes. Calculating the
derivative: ∂u
∂s =i[1 : 2i],∂u
∂θ =i[0 : 2 cos(θ) + sin(θ)].
Setting the derivatives equal to zero, we find that sis not involved. Thus, the critical points lie
on the horizontal line.
11 Numeric Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following Riemann surface defined by the equation z=eiθ, where
θ∈[0,2π].
a) Calculate the area of this Riemann surface.
b) Find the length of the curve defined by x=Re(z)and y=Im(z).
Solution 1.
a) To calculate the area of the Riemann surface, we can use the formula for the area of a surface
of revolution, which is given by A= 2πRβ
αf(θ)p1+(f′(θ))2dθ.
In this case, f(θ)=1and f′(θ)=0, so the area simplifies to:
A= 2πR2π
01·√1+0dθ = 2πR2π
01dθ = 2π[2π−0] = 4π2.
Therefore, the area of the Riemann surface is 4π2.
b) The length of the curve is given by the formula L=Rβ
αrdx
dθ 2+dy
dθ 2dθ.
Since x= cos(θ)and y= sin(θ), we have dx
dθ =−sin(θ)and dx
dθ = cos(θ). Therefore, the length
simplifies to:
L=R2π
0p(−sin(θ))2+ (cos(θ))2dθ =R2π
0√1dθ =R2π
01dθ = 2π.
The length of the curve is 2π.
12 12. BUBBLING PHENOMENON IN PSEUDOHOLOMORPHIC CURVES
Problem 12. Consider the sequence of pseudoholomorphic curves defined by un:R×S2→C
given by un(x, z) = eiθnzn, where θn→0as n→ ∞. Determine whether this sequence exhibits
bubbling phenomenon or not.
Solution 12. a) To analyze the behavior of this sequence of pseudoholomorphic curves, let’s
first calculate the energy of each curve. The energy functional for a pseudoholomorphic curve
u: Σ →Cdefined over a Riemann surface Σis given by
E(u) = ZΣ||du||2dA,
where ||du||2is the square of the norm of the differential du and dA is the area element on Σ.
For the curve un(x, z) = eiθnzn, the energy can be calculated as follows:
||dun||2=|eiθnnzn−1|2=n2|z|2n−2,
and integrating over S2(with the standard metric) gives
E(un) = ZS2
n2|z|2n−2dA = 4πn2.
b) Next, let’s examine the behavior of the energy of the sequence as n→ ∞. We have
lim
n→∞ E(un) = lim
n→∞ 4πn2=∞.
c) Since the energy of the sequence of curves diverges as n→ ∞, we conclude that this
sequence exhibits bubbling phenomenon. In this case, the limiting pseudoholomorphic curve will
be a multiply-covered sphere.
Therefore, the sequence of pseudoholomorphic curves defined by un(x, z) = eiθnznexhibits
bubbling phenomenon.
13 13. SYMPLECTIC TOPOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 13. Consider the complex structure Jon R2given by J(x, y)=(−y, x)and the
symplectic form ω=dx ∧dy.
a) Let u(x, y)=(x2−y2,2xy)be a map R2→R2. Determine if uis pseudoholomorphic with
respect to (J, ω).
b) Find the image of the map ufrom part (a).
Solution 13.
a) To determine if uis pseudoholomorphic with respect to (J, ω), we need to check if du +J◦
du ◦J=ω.
Compute du =2x−2y
2y2xand J◦du ◦J=−2y2x
2x2y.
Adding them gives du +J◦du ◦J=0 0
0 0=ω=dx ∧dy.
Since du +J◦du ◦J=ω,uis not pseudoholomorphic.
b) The image of ucan be found by solving the following system of equations:
x2−y2=u1
2xy =u2
Substitute u1=x2−y2into u2= 2xy to get 2x(x2−y2) = u2.
Solving both equations simultaneously gives us x= 0 or x2=y2, which are the equations for
the coordinate axes.
Therefore, the image of the map uis the xand yaxes.
14 14. CAUCHY-RIEMANN EQUATIONS AND PSEUDOHOLOMORPHIC CURVES
Problem 14. Consider the pseudoholomorphic curve u:C→Cdefined by u(z) = z2+i,
where z=x+iy and i=√−1.
a) Find the Cauchy-Riemann equations for the map u.
b) Determine if the map uis pseudoholomorphic.
Solution 14.
a) To find the Cauchy-Riemann equations for u(z) = z2+i, we first express u(z)in terms of x
and y:
u(z)=(x+iy)2+i=x2−y2+ 2ixy +i
Now, we write u(z)as u(x, y) + iv(x, y):
u(x, y) = x2−y2, v(x, y)=2xy + 1
Next, we find the partial derivatives of uand v:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
The Cauchy-Riemann equations are:
∂u
∂x =∂v
∂y and ∂u
∂y =−∂v
∂x
Substitute the partial derivatives we found into these equations to check if they are satisfied.
b) To determine if the map uis pseudoholomorphic, we need to verify if the Cauchy-Riemann
equations hold for u. From part a, we found that:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
As the Cauchy-Riemann equations are satisfied, the map uis pseudoholomorphic.
15 15. GRADIENT FLOW TECHNIQUES FOR PSEUDOHOLOMORPHIC CURVES
Problem 15. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = z2+iz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 1.
Solution 15. The solution to the pseudoholomorphic curve equation is u(t, z) = u(0, z)for all
t, due to the stationary nature of pseudoholomorphic curves. Hence, u(1, z) = u(0, z) = z2+iz.
Therefore, the solution at time t= 1 is u(1, z) = z2+iz for all z∈Σin this case.
Problem 16. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = eiz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 2.
Solution 16. Similar to the previous problem, the solution to the pseudoholomorphic curve
equation is u(t, z) = u(0, z)for all t, due to the stationary nature of pseudoholomorphic curves.
Hence, u(2, z) = u(0, z) = eiz.
Therefore, the solution at time t= 2 is u(2, z) = eiz for all z∈Σin this case.
16 16. TOPOLOGICAL AND GEOMETRIC CONSTRAINTS ON PSEUDOHOLOMORPHIC CURVES
Problem 16. Consider a closed oriented manifold Mof dimension 4, equipped with a compati-
ble almost complex structure J. Let [γ]be a homology class represented by a simple closed curve
γin M. We define the self-intersection number of [γ]as I([γ],[γ]) = γ·γ.
Suppose Σis a pseudoholomorphic curve in Mwith boundary ∂Σ = γ1−γ2, where γ1and γ2
are simple closed curves representing different homology classes in M. Determine the relationship
between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]).
Solution 16. Let’s denote the two simple closed curves γ1and γ2as Aand B, respectively.
Then, we have I([γ1],[γ1]) = A·Aand I([γ2],[γ2]) = B·B, where ·denotes the intersection pairing.
The self-intersection number of the curve Σcan be computed as I(Σ,Σ) = (γ1−γ2)·(γ1−γ2).
Expanding this out, we get:
I(Σ,Σ) = γ1·γ1−γ1·γ2−γ2·γ1+γ2·γ2
=A·A−A·B−B·A+B·B
=A·A−2A·B+B·B
=I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
Therefore, the relationship between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]) is given by:
I(Σ,Σ) = I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
17 Numerical Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in CP 1:
∂u
∂z +λe−u= 0,
where λ∈Ris a constant.
a) Find the general solution u(z, z)to this pseudoholomorphic curve equation.
b) Given λ= 2, solve the equation with the initial condition u(0,0) = 0.
Solution 1.
a) To find the general solution to the pseudoholomorphic curve equation, we first rewrite the
equation in terms of the complex coordinates zand z. Note that ∂u
∂z =1
2∂u
∂x +i∂u
∂y , where z=
x+iy.
Substitute this into the given equation:
1
2∂u
∂x +i∂u
∂y +λe−u= 0.
Separating real and imaginary parts of this equation gives us:
∂u
∂x = 2λe−u
∂u
∂y = 0
Integrating the first equation with respect to xand the second equation with respect to ygives:
u= ln C
2λ+e2λx,
where Cis the constant of integration.
b) Given λ= 2 and the initial condition u(0,0) = 0, we can find the value of Cby substituting
x= 0, u = 0 into the general solution:
0 = ln C
4+ 1.
Solving this equation for Cgives C= 4, so the solution to the equation with λ= 2 and the initial
condition is:
u= ln 4
4+e4x= ln(1 + e4x).
18 The Theory of Pseudoholomorphic Curves
18.1 Problem 1
Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = aeis +bt with a, b ∈C
and |a|= 3,|b|= 2. Determine the image of this pseudoholomorphic curve in C.
Solution: The image of uis given by u(s, t) = aeis +bt. To find the image, we set u(s, t) = z
and solve for sand t:
z=aeis +bt
= 3eis + 2t
= 3(cos s+isin s)+2t.
This implies that the image of the pseudoholomorphic curve uis the set of points z= 3 cos s+
i3 sin s+ 2tin the complex plane C.
18.2 Problem 2
Let u:R×[0,1] →Cbe a pseudoholomorphic curve defined by u(s, t) = s2+it. Determine the
differential of u.
Solution: The differential of uis given by du =∂u
∂s ds +∂u
∂t dt. In this case, we have:
∂u
∂s = 2s,
∂u
∂t =i.
Therefore, the differential of uis du = 2s ds +i dt.
19 19. ENERGY ESTIMATES FOR PSEUDOHOLOMORPHIC CURVES
Problem 19. Consider a pseudoholomorphic curve u:R×S1→Cgiven by u(s, t) = eis sin(t),
where s, t ∈R. Calculate the energy of this pseudoholomorphic curve using the energy functional
E(u) = RR×S1||∂su||2+||∂tu||2dsdt.
Solution 19. The energy of the pseudoholomorphic curve ucan be calculated as follows:
a) First, we calculate the partial derivatives of u:
∂su=ieis sin(t),
∂tu=eis cos(t).
b) Next, we compute the norms of the partial derivatives:
||∂su||2=|ieis sin(t)|2=| − eis sin(t)|2=|eis|2|sin(t)|2= sin2(t),
||∂tu||2=|eis cos(t)|2=|eis||cos(t)|2= 1|cos(t)|2= cos2(t).
c) Now, we substitute the norms into the energy functional and integrate over R×S1:
E(u) = ZR×S1
(sin2(t) + cos2(t)) dsdt
=ZRZS1
1dsdt
=ZR
2π ds
= 2π.
Therefore, the energy of the pseudoholomorphic curve uis 2π.
20 The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in a symplectic mani-
fold:
¯
∂Ju+J(u)◦∂¯zu= 0
where u:C→Mis a holomorphic map, Jis a compatible almost complex structure in M, and
¯
∂Jdenotes the Cauchy-Riemann operator.
Given u(z) = z+i
z−i, determine if uis a pseudoholomorphic curve in the complex plane C.
Solution 1.
To determine if u(z) = z+i
z−iis a pseudoholomorphic curve, we need to check if it satisfies the
pseudoholomorphic curve equation. First, let’s calculate ∂zuand ∂¯zu:
∂zu=1
(z−i)2
∂¯zu= 0
Next, let’s compute ¯
∂Ju:
¯
∂Ju=1
2(∂¯zu−J(u)◦∂zu) = 1
20−J(u)◦1
(z−i)2
Given that Jis the standard almost complex structure on C, i.e., J(z) = iz, we have:
J(u) = iz+i
z−i=i(z+i)
z−i
Therefore, ¯
∂Ju=−i
2(z−i)3. Since ¯
∂Ju+J(u)◦∂¯zu= 0, the function u(z) = z+i
z−iis indeed a
pseudoholomorphic curve in C.
21 21. STABILITY CONDITIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 21. Consider a pseudoholomorphic curve u: Σ →Mwhere Σis a Riemann surface
and Mis a symplectic manifold with symplectic form ω. Let Jbe an almost complex structure on
Mcompatible with ω. Suppose the pseudoholomorphic curve usatisfies the following Cauchy-
Riemann equation:
¯
∂Ju= 0.
a) Show that the linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J.
b) Calculate the Fredholm index of the operator Du.
c) Determine the stability condition for the pseudoholomorphic curve u.
Solution 21.
a) The linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J. This operator acts on sections of the pullback bundle u∗T M,
which is the tangent bundle of the Riemann surface Σpulled back to Mvia the map u.
b) The Fredholm index of the operator Duis given by
index(Du) = dim ker(Du)−dim coker(Du).
Here, ker(Du)represents the kernel of Du(space of solutions to ¯
∂Jv= 0 near u) and coker(Du)
represents the cokernel of Du. The Fredholm index indicates the deformation parameter for the
moduli space of pseudoholomorphic curves.
c) The stability condition for the pseudoholomorphic curve uis satisfied if the Fredholm index
of the linearized operator Duis nonnegative, i.e., index(Du)≥0. This condition ensures that
the moduli space of pseudoholomorphic curves is well-behaved under small perturbations in the
complex structure or metric on M.
22 22. FREDHOLM THEORY AND PSEUDOHOLOMORPHIC CURVES
Problem 22. Consider the following pseudoholomorphic curve equation in C2:
¯
∂u = 0
where u:R×S1→C2is a smooth map.
Given u(t, z) = eit(z, iz)for t∈[0,2π]and z∈S1, compute the Fredholm index of this pseudo-
holomorphic curve.
Solution 22. To compute the Fredholm index, we first need to find the linearized operator asso-
ciated with ¯
∂at u. Let v(t, z)be a small perturbation around u, and write v(t, z) = eit(a(t, z), b(t, z)).
The linearized operator is given by:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb
Plugging in u(t, z)and v(t, z)into the expression and simplifying, we get:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb=eit(∂ta−i∂tb+∂za−i∂zb)
This is equal to 0if and only if both ∂ta−i∂tband ∂za−i∂zbare equal to 0.
From u(t, z), we have a(t, z) = zand b(t, z) = iz. Differentiating with respect to tand z, we get:
∂ta= 0, ∂tb=−z, ∂za= 1, ∂zb=i
So ∂ta−i∂tb= 0 and ∂za−i∂zb= 1 −i= 0. Thus, the Fredholm index is 0−2 = −2.
23 The Theory of Pseudoholomorphic Curves
Problem: Consider the following pseudoholomorphic curve equation in the complex plane C:
¯
∂Ju+i∂tu= 0
where u:C×[0,1] →Cis a smooth map, ∂tdenotes the partial derivative with respect to
t∈[0,1], and ¯
∂Jdenotes the ¯
∂-operator twisted by a compatible almost complex structure Jon C.
Given the initial condition u(x, 0) = x2for x∈C, solve the pseudoholomorphic curve equation.
Solution:
To solve the pseudoholomorphic curve equation, we will first compute the derivatives and then
use the initial condition to determine the solution.
Let u(x, t) = u1(x, t) + iv1(x, t), where u1and v1are real-valued functions. Then the pseudo-
holomorphic curve equation becomes:
∂u1
∂t −∂v1
∂x +i∂v1
∂t +∂u1
∂x = 0
Separating the real and imaginary parts, we get:
∂u1
∂t =∂v1
∂x ,∂v1
∂t =−∂u1
∂x
From the initial condition u(x, 0) = x2, we have u1(x, 0) = x2and v1(x, 0) = 0.
Solving the partial differential equations with these initial conditions gives u1(x, t) = x2and
v1(x, t)=0. Therefore, the solution of the pseudoholomorphic curve equation is u(x, t) = x2+i·0 =
x2.
24 24. HOMOLOGICAL ALGEBRA AND PSEUDOHOLOMORPHIC CURVES
Problem 24. Consider the following pseudoholomorphic curve in CP 1given by the equation
u(z) = z2.
a) Determine the differential operator ∂Jassociated with the almost complex structure Jinduced
by the Fubini-Study metric on CP 1.
b) Show that u(z)is a solution to the ∂Jequation on CP 1.
Solution 24.
a) The almost complex structure Jassociated with the Fubini-Study metric on CP 1is given by
J(z) = −iz. The differential operator ∂Jcan be computed as:
∂J=1
2(J+iId) = 1
2(−iz +iId) = 1
2(i(z+ 1)).
b) To show that u(z) = z2is a solution to the ∂Jequation, we need to demonstrate that ∂Ju(z) =
0. Let’s compute it:
∂Ju(z) = 1
2(i(z2+ 1)) = i
2z2+i
2=iz +i
2= 0.
As the result is not zero, the function u(z) = z2is not a solution to the ∂Jequation on CP 1.
25 The Theory of Pseudoholomorphic Curves
Problem 1. Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = eis +t2.
a) Determine the asymptotic behavior of uas s→ −∞.
b) Find the limit lims→∞ |u(s, 1)|.
c) Show that uis a pseudoholomorphic curve.
Solution 1.
a) As s→ −∞, the term eis behaves as oscillatory and does not affect the growth, so we focus
on t2. Thus, the asymptotic behavior of u(s, t)as s→ −∞ is t2.
b) We have |u(s, 1)|=|eis + 1| ≤ |eis|+|1|= 1 + 1 = 2. Thus, lims→∞ |u(s, 1)| ≤ 2.
c) To show that uis a pseudoholomorphic curve, we need to verify that ¯
∂Ju= 0, where Jis the
standard complex structure on C. Compute ∂su=ieis and ∂tu= 2t. Then, ¯
∂Ju=∂su−i∂tu=
ieis −2it.
Since ¯
∂Ju= 0, we have shown that uis a pseudoholomorphic curve.
Therefore, the solution to the pseudoholomorphic curve equation with the specified Riemann
surface and almost complex structure is u(z) = az +b, but it is not J-holomorphic.
3 3. INDEX THEORY FOR PSEUDOHOLOMORPHIC CURVES
Problem 3. Consider a compact, oriented, Riemannian 2-manifold Mwith a compatible almost
complex structure Jand a symplectic form ω. Let u:C→Mbe a nonconstant J-holomorphic
map. Given that the Fredholm index ind(Du)of uis 2, with Dudenoting the linearized Cauchy-
Riemann operator associated to u, determine the total number of positive and negative punctures
of u.
Solution 3.
Since the index of uis 2, we know that the Conley-Zehnder index of each positive puncture of
uis −1and each negative puncture has a Conley-Zehnder index of 1.
Let n+and n−denote the total number of positive and negative punctures of u, respectively.
Then, we have the equation
ind(Du)=2=2−n++n−
Substitute the Conley-Zehnder index values into the equation, we get
2=2−n++n−
0 = −n++n−
n+=n−
Therefore, the total number of positive punctures is equal to the total number of negative punc-
tures in this case.
4 4. EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 4. Consider the following holomorphic curve in C:u:C→Cgiven by u(z) = z2−i.
We want to find the solutions to the Cauchy-Riemann equation ¯
∂Ju= 0.
Solution 4.
a) We start by writing u(z) = u(x1+ix2) = u1(x1, x2) + iu2(x1, x2)where u1, u2:R2→R.
We then compute the Cauchy-Riemann equations:
∂u1
∂x1
=∂u2
∂x2
and ∂u1
∂x2
=−∂u2
∂x1
In this case, u(z) = z2−i= (x1+ix2)2−i= (x2
1−x2
2) + i(2x1x2−1).
Therefore, we have:
u1(x1, x2) = x2
1−x2
2and u2(x1, x2)=2x1x2−1
Calculating the partial derivatives, we find:
∂u1
∂x1
= 2x1and ∂u2
∂x2
= 2x1
∂u1
∂x2
=−2x2and ∂u2
∂x1
= 2x2
Now we plug these into the Cauchy-Riemann equations:
2x1= 2x1and −2x2= 2x2
Solving the above equations, we obtain x1= 0 and x2= 0.
Therefore, the holomorphic curve u(z) = z2−isatisfies the Cauchy-Riemann equations at the
point (0,0).
b) To find other solutions, we can look for constant maps. Let u(z) = cwhere cis a complex
constant. Then, ¯
∂Ju= 0 is automatically satisfied.
c) Another way to find solutions is by looking at the composition of holomorphic functions. If
v:C→Cand u:C→Care holomorphic, then u◦vis also holomorphic.
5 5. FLOER HOMOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 5. Consider the following pseudoholomorphic curve in Cparametrized by z(t) = eit,
where t∈[0,2π]:
u:R×S1→C, u(s, eit) = eiseit.
a) Show that this curve satisfies the Cauchy-Riemann equations.
b) Calculate the energy of the curve.
c) Determine the image of the curve in C.
Solution 5.
a) To show that the curve satisfies the Cauchy-Riemann equations, we need to check if ∂u
∂¯z= 0.
Here, z=x+iy =x+iy(t), so ∂u
∂¯z=1
2(∂u
∂x +i∂u
∂y ) = 1
2(eiseit −ieiseit) = 0, hence the curve satisfies
the Cauchy-Riemann equations.
b) The energy of the curve is calculated using the formula:
E(u) = ZS1||∇u||2dt,
where ||∇u||2is the norm of the gradient of u. In this case, ∇u= (∂su, ∂tu)=(iu(s, eit), eisieit),
so ||∇u||2=|iu(s, eit)|2+|eisieit|2=|i|2+ 1 = 2. Therefore, the energy of the curve is:
E(u) = Z2π
0
2dt = 4π.
c) The image of the curve in Cis given by u(R×S1) = {eiseit |s∈R, t ∈[0,2π]}. Since eis
and eit vary over all complex numbers on the unit circle S1, the image is the unit circle S1itself.
6 6. COMPACTNESS PROPERTIES OF PSEUDOHOLOMORPHIC CURVES
Problem 6. Consider a sequence of pseudoholomorphic curves (uk: Σ →M, Jk)converging
to a constant map, where Σis a Riemann surface and Mis a symplectic manifold.
Let Jkbe a sequence of almost complex structures converging to J∞in the C∞-topology on
Σ. Assume that the energy of the curves E(uk)is uniformly bounded. Show that the sequence uk
converges in the C∞-topology on Σto a constant map u∞.
Solution 6.
Given that the energy of the curves E(uk)is uniformly bounded, we have
E(uk) = ZΣ||duk◦j−Jk◦duk||2dµ
≤C
where Cis a constant independent of k.
By the compactness theorem for Jk, there exists a subsequence (not relabeled) converging to
aJ∞-holomorphic map u∞. This gives us convergence in the C∞-topology for a subsequence.
Now we want to show that the entire sequence converges to u∞. By contradiction, assume
there exists ϵ > 0such that for all Nthere exists kN> N such that ||dukN−du∞||C∞≥ϵ. But this
is a contradiction, as we have a compact set in C∞-topology.
Therefore, the entire sequence ukconverges in the C∞-topology to the constant map u∞.
7 7. GROMOV-WITTEN INVARIANTS AND PSEUDOHOLOMORPHIC CURVES
Problem 7. Consider the moduli space of genus-zero, three-pointed stable pseudoholomor-
phic curves on a Riemann surface of genus one, with the three marked points labeled p1, p2, p3.
Suppose the homology classes of the curves are A= 2α+ 3βand B=−α+ 4β.
a) Calculate the dimension of the moduli space.
b) Determine the number of points in the moduli space that satisfy the conditions given in part
(a).
c) Suppose the intersection number of Aand Bis 1. Find the expected dimension of the moduli
space.
Solution 7.
a) The dimension of the moduli space can be calculated using the Riemann-Roch formula:
dim(M0,3(Σ, A)) = 3d+ 1 −g= 3(2 + 3) + 1 −1 = 10.
b) In this case, the moduli space contains finitely many points due to the constraint on the
homology classes. Hence, there are no points in the moduli space.
c) The expected dimension of the moduli space can be calculated using the formula:
dim(M0,3(Σ, A;B)) = dim(M0,3(Σ, A)) − ⟨A, B⟩= 10 −1 = 9.
Therefore, the expected dimension of the moduli space is 9.
8 8. EVALUATION MAPS FOR PSEUDOHOLOMORPHIC CURVES
Problem 8. Consider the pseudoholomorphic curve equation on a Riemann surface Sgiven by
du +idv = 0, where u, v :S→Care smooth functions. Let z=u+iv be a complex coordinate on
S. Given a pseudoholomorphic curve u=f(z)parametrized by f(z) = z2, evaluate the following:
a) The evaluation map evp:M → S, where Mis the moduli space of solutions to the pseudo-
holomorphic curve equation, at p= 2 + 3i.
b) The derivative d
dz f(z)at z= 1 −i.
c) The image under the map fof the curve g(z) = z3.
Solution 8.
a) To compute the evaluation map at p= 2+3i, we substitute z= 2+3iinto the parametrization
f(z) = z2. Therefore, f(2 + 3i) = (2 + 3i)2= 4 + 12i−9 = −5 + 12i. Hence, ev2+3i(f) = −5 + 12i.
b) To find the derivative d
dz f(z)at z= 1 −i, we differentiate f(z) = z2to get d
dz f(z) = 2z.
Evaluating this derivative at z= 1 −i, we have d
dz f(1 −i) = 2(1 −i)=2−2i.
c) The image under the map fof the curve g(z) = z3is given by f(g(z)) = f(z3)=(z3)2=z6.
Therefore, the image of the curve g(z)under the map fis the curve parametrized by z6.
9 9. MODULI SPACE OF PSEUDOHOLOMORPHIC CURVES
Problem 9. Consider a Riemann surface Mwith genus g= 2 and n= 3 marked points. Let Xbe
a compact oriented Riemann surface of genus 0 with 5 boundary marked points. Denote by Athe
moduli space of smooth maps ¯
M→Xsatisfying certain boundary conditions.
a) Calculate the dimension of the space A.
b) Determine the dimension of the moduli space of pseudo-holomorphic curves in CP 2with
k= 2 marked points.
Solution 9.
a) The dimension of the moduli space Acan be calculated using Riemann-Roch formula. We
know that for a Riemann surface of genus gwith nmarked points, the dimension of the moduli
space is given by:
dim(M)=6g−6+2n
Substitute g= 2 and n= 3 into the formula:
dim(A) = 6 ×2−6+2×3 = 12 −6 + 6 = 12
Therefore, the dimension of the moduli space Ais 12.
b) The dimension of the moduli space of pseudo-holomorphic curves in CP 2with kmarked
points is given by the formula:
dim M= 2k−6+2g
Substitute k= 2 and g= 0 into the formula:
dim M= 2 ×2−6+2×0=4−6 = −2
Therefore, the dimension of the moduli space of pseudo-holomorphic curves in CP 2with 2
marked points is -2.
10 9. THE THEORY OF PSEUDOHOLMORPHIC CURVES
Problem 9. Consider the following pseudoholomorphic curve in CP 1given by the map u:
R×S1→CP 1, where S1is the unit circle in C:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)].
a) Calculate the image of the curve u.
b) Determine the intersection number of the curve uwith the horizontal line [z: 1], where z∈C.
c) Find the set of critical points of the map u.
Solution 9.
a) To find the image of the curve u, we need to substitute the values of sand θinto the map u.
Let’s first express u(s, eiθ)in terms of sand θ:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)] .
Substitute s= 0 and θ=π/2into uto get the image point:
u(0, eiπ/2) = [1 : 2i].
Therefore, the image of the curve uis the point [1 : 2i]in CP 1.
b) To find the intersection number of the curve uwith the horizontal line [z: 1], we need to count
the number of solutions to the equation 2isin(θ)−icos(θ) = z(1 + is). This reduces to finding the
number of solutions to sin(θ)=0:
sin(θ) = 0 =⇒θ=nπ, where n∈Z.
Since θis periodic with period 2π, there are infinitely many intersection points. Hence, the
intersection number is ∞.
c) The critical points of uare the (s, eiθ)such that the derivative of uvanishes. Calculating the
derivative: ∂u
∂s =i[1 : 2i],∂u
∂θ =i[0 : 2 cos(θ) + sin(θ)].
Setting the derivatives equal to zero, we find that sis not involved. Thus, the critical points lie
on the horizontal line.
11 Numeric Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following Riemann surface defined by the equation z=eiθ, where
θ∈[0,2π].
a) Calculate the area of this Riemann surface.
b) Find the length of the curve defined by x=Re(z)and y=Im(z).
Solution 1.
a) To calculate the area of the Riemann surface, we can use the formula for the area of a surface
of revolution, which is given by A= 2πRβ
αf(θ)p1+(f′(θ))2dθ.
In this case, f(θ)=1and f′(θ)=0, so the area simplifies to:
A= 2πR2π
01·√1+0dθ = 2πR2π
01dθ = 2π[2π−0] = 4π2.
Therefore, the area of the Riemann surface is 4π2.
b) The length of the curve is given by the formula L=Rβ
αrdx
dθ 2+dy
dθ 2dθ.
Since x= cos(θ)and y= sin(θ), we have dx
dθ =−sin(θ)and dx
dθ = cos(θ). Therefore, the length
simplifies to:
L=R2π
0p(−sin(θ))2+ (cos(θ))2dθ =R2π
0√1dθ =R2π
01dθ = 2π.
The length of the curve is 2π.
12 12. BUBBLING PHENOMENON IN PSEUDOHOLOMORPHIC CURVES
Problem 12. Consider the sequence of pseudoholomorphic curves defined by un:R×S2→C
given by un(x, z) = eiθnzn, where θn→0as n→ ∞. Determine whether this sequence exhibits
bubbling phenomenon or not.
Solution 12. a) To analyze the behavior of this sequence of pseudoholomorphic curves, let’s
first calculate the energy of each curve. The energy functional for a pseudoholomorphic curve
u: Σ →Cdefined over a Riemann surface Σis given by
E(u) = ZΣ||du||2dA,
where ||du||2is the square of the norm of the differential du and dA is the area element on Σ.
For the curve un(x, z) = eiθnzn, the energy can be calculated as follows:
||dun||2=|eiθnnzn−1|2=n2|z|2n−2,
and integrating over S2(with the standard metric) gives
E(un) = ZS2
n2|z|2n−2dA = 4πn2.
b) Next, let’s examine the behavior of the energy of the sequence as n→ ∞. We have
lim
n→∞ E(un) = lim
n→∞ 4πn2=∞.
c) Since the energy of the sequence of curves diverges as n→ ∞, we conclude that this
sequence exhibits bubbling phenomenon. In this case, the limiting pseudoholomorphic curve will
be a multiply-covered sphere.
Therefore, the sequence of pseudoholomorphic curves defined by un(x, z) = eiθnznexhibits
bubbling phenomenon.
13 13. SYMPLECTIC TOPOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 13. Consider the complex structure Jon R2given by J(x, y)=(−y, x)and the
symplectic form ω=dx ∧dy.
a) Let u(x, y)=(x2−y2,2xy)be a map R2→R2. Determine if uis pseudoholomorphic with
respect to (J, ω).
b) Find the image of the map ufrom part (a).
Solution 13.
a) To determine if uis pseudoholomorphic with respect to (J, ω), we need to check if du +J◦
du ◦J=ω.
Compute du =2x−2y
2y2xand J◦du ◦J=−2y2x
2x2y.
Adding them gives du +J◦du ◦J=0 0
0 0=ω=dx ∧dy.
Since du +J◦du ◦J=ω,uis not pseudoholomorphic.
b) The image of ucan be found by solving the following system of equations:
x2−y2=u1
2xy =u2
Substitute u1=x2−y2into u2= 2xy to get 2x(x2−y2) = u2.
Solving both equations simultaneously gives us x= 0 or x2=y2, which are the equations for
the coordinate axes.
Therefore, the image of the map uis the xand yaxes.
14 14. CAUCHY-RIEMANN EQUATIONS AND PSEUDOHOLOMORPHIC CURVES
Problem 14. Consider the pseudoholomorphic curve u:C→Cdefined by u(z) = z2+i,
where z=x+iy and i=√−1.
a) Find the Cauchy-Riemann equations for the map u.
b) Determine if the map uis pseudoholomorphic.
Solution 14.
a) To find the Cauchy-Riemann equations for u(z) = z2+i, we first express u(z)in terms of x
and y:
u(z)=(x+iy)2+i=x2−y2+ 2ixy +i
Now, we write u(z)as u(x, y) + iv(x, y):
u(x, y) = x2−y2, v(x, y)=2xy + 1
Next, we find the partial derivatives of uand v:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
The Cauchy-Riemann equations are:
∂u
∂x =∂v
∂y and ∂u
∂y =−∂v
∂x
Substitute the partial derivatives we found into these equations to check if they are satisfied.
b) To determine if the map uis pseudoholomorphic, we need to verify if the Cauchy-Riemann
equations hold for u. From part a, we found that:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
As the Cauchy-Riemann equations are satisfied, the map uis pseudoholomorphic.
15 15. GRADIENT FLOW TECHNIQUES FOR PSEUDOHOLOMORPHIC CURVES
Problem 15. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = z2+iz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 1.
Solution 15. The solution to the pseudoholomorphic curve equation is u(t, z) = u(0, z)for all
t, due to the stationary nature of pseudoholomorphic curves. Hence, u(1, z) = u(0, z) = z2+iz.
Therefore, the solution at time t= 1 is u(1, z) = z2+iz for all z∈Σin this case.
Problem 16. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = eiz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 2.
Solution 16. Similar to the previous problem, the solution to the pseudoholomorphic curve
equation is u(t, z) = u(0, z)for all t, due to the stationary nature of pseudoholomorphic curves.
Hence, u(2, z) = u(0, z) = eiz.
Therefore, the solution at time t= 2 is u(2, z) = eiz for all z∈Σin this case.
16 16. TOPOLOGICAL AND GEOMETRIC CONSTRAINTS ON PSEUDOHOLOMORPHIC CURVES
Problem 16. Consider a closed oriented manifold Mof dimension 4, equipped with a compati-
ble almost complex structure J. Let [γ]be a homology class represented by a simple closed curve
γin M. We define the self-intersection number of [γ]as I([γ],[γ]) = γ·γ.
Suppose Σis a pseudoholomorphic curve in Mwith boundary ∂Σ = γ1−γ2, where γ1and γ2
are simple closed curves representing different homology classes in M. Determine the relationship
between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]).
Solution 16. Let’s denote the two simple closed curves γ1and γ2as Aand B, respectively.
Then, we have I([γ1],[γ1]) = A·Aand I([γ2],[γ2]) = B·B, where ·denotes the intersection pairing.
The self-intersection number of the curve Σcan be computed as I(Σ,Σ) = (γ1−γ2)·(γ1−γ2).
Expanding this out, we get:
I(Σ,Σ) = γ1·γ1−γ1·γ2−γ2·γ1+γ2·γ2
=A·A−A·B−B·A+B·B
=A·A−2A·B+B·B
=I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
Therefore, the relationship between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]) is given by:
I(Σ,Σ) = I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
17 Numerical Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in CP 1:
∂u
∂z +λe−u= 0,
where λ∈Ris a constant.
a) Find the general solution u(z, z)to this pseudoholomorphic curve equation.
b) Given λ= 2, solve the equation with the initial condition u(0,0) = 0.
Solution 1.
a) To find the general solution to the pseudoholomorphic curve equation, we first rewrite the
equation in terms of the complex coordinates zand z. Note that ∂u
∂z =1
2∂u
∂x +i∂u
∂y , where z=
x+iy.
Substitute this into the given equation:
1
2∂u
∂x +i∂u
∂y +λe−u= 0.
Separating real and imaginary parts of this equation gives us:
∂u
∂x = 2λe−u
∂u
∂y = 0
Integrating the first equation with respect to xand the second equation with respect to ygives:
u= ln C
2λ+e2λx,
where Cis the constant of integration.
b) Given λ= 2 and the initial condition u(0,0) = 0, we can find the value of Cby substituting
x= 0, u = 0 into the general solution:
0 = ln C
4+ 1.
Solving this equation for Cgives C= 4, so the solution to the equation with λ= 2 and the initial
condition is:
u= ln 4
4+e4x= ln(1 + e4x).
18 The Theory of Pseudoholomorphic Curves
18.1 Problem 1
Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = aeis +bt with a, b ∈C
and |a|= 3,|b|= 2. Determine the image of this pseudoholomorphic curve in C.
Solution: The image of uis given by u(s, t) = aeis +bt. To find the image, we set u(s, t) = z
and solve for sand t:
z=aeis +bt
= 3eis + 2t
= 3(cos s+isin s)+2t.
This implies that the image of the pseudoholomorphic curve uis the set of points z= 3 cos s+
i3 sin s+ 2tin the complex plane C.
18.2 Problem 2
Let u:R×[0,1] →Cbe a pseudoholomorphic curve defined by u(s, t) = s2+it. Determine the
differential of u.
Solution: The differential of uis given by du =∂u
∂s ds +∂u
∂t dt. In this case, we have:
∂u
∂s = 2s,
∂u
∂t =i.
Therefore, the differential of uis du = 2s ds +i dt.
19 19. ENERGY ESTIMATES FOR PSEUDOHOLOMORPHIC CURVES
Problem 19. Consider a pseudoholomorphic curve u:R×S1→Cgiven by u(s, t) = eis sin(t),
where s, t ∈R. Calculate the energy of this pseudoholomorphic curve using the energy functional
E(u) = RR×S1||∂su||2+||∂tu||2dsdt.
Solution 19. The energy of the pseudoholomorphic curve ucan be calculated as follows:
a) First, we calculate the partial derivatives of u:
∂su=ieis sin(t),
∂tu=eis cos(t).
b) Next, we compute the norms of the partial derivatives:
||∂su||2=|ieis sin(t)|2=| − eis sin(t)|2=|eis|2|sin(t)|2= sin2(t),
||∂tu||2=|eis cos(t)|2=|eis||cos(t)|2= 1|cos(t)|2= cos2(t).
c) Now, we substitute the norms into the energy functional and integrate over R×S1:
E(u) = ZR×S1
(sin2(t) + cos2(t)) dsdt
=ZRZS1
1dsdt
=ZR
2π ds
= 2π.
Therefore, the energy of the pseudoholomorphic curve uis 2π.
20 The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in a symplectic mani-
fold:
¯
∂Ju+J(u)◦∂¯zu= 0
where u:C→Mis a holomorphic map, Jis a compatible almost complex structure in M, and
¯
∂Jdenotes the Cauchy-Riemann operator.
Given u(z) = z+i
z−i, determine if uis a pseudoholomorphic curve in the complex plane C.
Solution 1.
To determine if u(z) = z+i
z−iis a pseudoholomorphic curve, we need to check if it satisfies the
pseudoholomorphic curve equation. First, let’s calculate ∂zuand ∂¯zu:
∂zu=1
(z−i)2
∂¯zu= 0
Next, let’s compute ¯
∂Ju:
¯
∂Ju=1
2(∂¯zu−J(u)◦∂zu) = 1
20−J(u)◦1
(z−i)2
Given that Jis the standard almost complex structure on C, i.e., J(z) = iz, we have:
J(u) = iz+i
z−i=i(z+i)
z−i
Therefore, ¯
∂Ju=−i
2(z−i)3. Since ¯
∂Ju+J(u)◦∂¯zu= 0, the function u(z) = z+i
z−iis indeed a
pseudoholomorphic curve in C.
21 21. STABILITY CONDITIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 21. Consider a pseudoholomorphic curve u: Σ →Mwhere Σis a Riemann surface
and Mis a symplectic manifold with symplectic form ω. Let Jbe an almost complex structure on
Mcompatible with ω. Suppose the pseudoholomorphic curve usatisfies the following Cauchy-
Riemann equation:
¯
∂Ju= 0.
a) Show that the linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J.
b) Calculate the Fredholm index of the operator Du.
c) Determine the stability condition for the pseudoholomorphic curve u.
Solution 21.
a) The linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J. This operator acts on sections of the pullback bundle u∗T M,
which is the tangent bundle of the Riemann surface Σpulled back to Mvia the map u.
b) The Fredholm index of the operator Duis given by
index(Du) = dim ker(Du)−dim coker(Du).
Here, ker(Du)represents the kernel of Du(space of solutions to ¯
∂Jv= 0 near u) and coker(Du)
represents the cokernel of Du. The Fredholm index indicates the deformation parameter for the
moduli space of pseudoholomorphic curves.
c) The stability condition for the pseudoholomorphic curve uis satisfied if the Fredholm index
of the linearized operator Duis nonnegative, i.e., index(Du)≥0. This condition ensures that
the moduli space of pseudoholomorphic curves is well-behaved under small perturbations in the
complex structure or metric on M.
22 22. FREDHOLM THEORY AND PSEUDOHOLOMORPHIC CURVES
Problem 22. Consider the following pseudoholomorphic curve equation in C2:
¯
∂u = 0
where u:R×S1→C2is a smooth map.
Given u(t, z) = eit(z, iz)for t∈[0,2π]and z∈S1, compute the Fredholm index of this pseudo-
holomorphic curve.
Solution 22. To compute the Fredholm index, we first need to find the linearized operator asso-
ciated with ¯
∂at u. Let v(t, z)be a small perturbation around u, and write v(t, z) = eit(a(t, z), b(t, z)).
The linearized operator is given by:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb
Plugging in u(t, z)and v(t, z)into the expression and simplifying, we get:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb=eit(∂ta−i∂tb+∂za−i∂zb)
This is equal to 0if and only if both ∂ta−i∂tband ∂za−i∂zbare equal to 0.
From u(t, z), we have a(t, z) = zand b(t, z) = iz. Differentiating with respect to tand z, we get:
∂ta= 0, ∂tb=−z, ∂za= 1, ∂zb=i
So ∂ta−i∂tb= 0 and ∂za−i∂zb= 1 −i= 0. Thus, the Fredholm index is 0−2 = −2.
23 The Theory of Pseudoholomorphic Curves
Problem: Consider the following pseudoholomorphic curve equation in the complex plane C:
¯
∂Ju+i∂tu= 0
where u:C×[0,1] →Cis a smooth map, ∂tdenotes the partial derivative with respect to
t∈[0,1], and ¯
∂Jdenotes the ¯
∂-operator twisted by a compatible almost complex structure Jon C.
Given the initial condition u(x, 0) = x2for x∈C, solve the pseudoholomorphic curve equation.
Solution:
To solve the pseudoholomorphic curve equation, we will first compute the derivatives and then
use the initial condition to determine the solution.
Let u(x, t) = u1(x, t) + iv1(x, t), where u1and v1are real-valued functions. Then the pseudo-
holomorphic curve equation becomes:
∂u1
∂t −∂v1
∂x +i∂v1
∂t +∂u1
∂x = 0
Separating the real and imaginary parts, we get:
∂u1
∂t =∂v1
∂x ,∂v1
∂t =−∂u1
∂x
From the initial condition u(x, 0) = x2, we have u1(x, 0) = x2and v1(x, 0) = 0.
Solving the partial differential equations with these initial conditions gives u1(x, t) = x2and
v1(x, t)=0. Therefore, the solution of the pseudoholomorphic curve equation is u(x, t) = x2+i·0 =
x2.
24 24. HOMOLOGICAL ALGEBRA AND PSEUDOHOLOMORPHIC CURVES
Problem 24. Consider the following pseudoholomorphic curve in CP 1given by the equation
u(z) = z2.
a) Determine the differential operator ∂Jassociated with the almost complex structure Jinduced
by the Fubini-Study metric on CP 1.
b) Show that u(z)is a solution to the ∂Jequation on CP 1.
Solution 24.
a) The almost complex structure Jassociated with the Fubini-Study metric on CP 1is given by
J(z) = −iz. The differential operator ∂Jcan be computed as:
∂J=1
2(J+iId) = 1
2(−iz +iId) = 1
2(i(z+ 1)).
b) To show that u(z) = z2is a solution to the ∂Jequation, we need to demonstrate that ∂Ju(z) =
0. Let’s compute it:
∂Ju(z) = 1
2(i(z2+ 1)) = i
2z2+i
2=iz +i
2= 0.
As the result is not zero, the function u(z) = z2is not a solution to the ∂Jequation on CP 1.
25 The Theory of Pseudoholomorphic Curves
Problem 1. Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = eis +t2.
a) Determine the asymptotic behavior of uas s→ −∞.
b) Find the limit lims→∞ |u(s, 1)|.
c) Show that uis a pseudoholomorphic curve.
Solution 1.
a) As s→ −∞, the term eis behaves as oscillatory and does not affect the growth, so we focus
on t2. Thus, the asymptotic behavior of u(s, t)as s→ −∞ is t2.
b) We have |u(s, 1)|=|eis + 1| ≤ |eis|+|1|= 1 + 1 = 2. Thus, lims→∞ |u(s, 1)| ≤ 2.
c) To show that uis a pseudoholomorphic curve, we need to verify that ¯
∂Ju= 0, where Jis the
standard complex structure on C. Compute ∂su=ieis and ∂tu= 2t. Then, ¯
∂Ju=∂su−i∂tu=
ieis −2it.
Since ¯
∂Ju= 0, we have shown that uis a pseudoholomorphic curve.
Therefore, the solution to the pseudoholomorphic curve equation with the specified Riemann
surface and almost complex structure is u(z) = az +b, but it is not J-holomorphic.
3 3. INDEX THEORY FOR PSEUDOHOLOMORPHIC CURVES
Problem 3. Consider a compact, oriented, Riemannian 2-manifold Mwith a compatible almost
complex structure Jand a symplectic form ω. Let u:C→Mbe a nonconstant J-holomorphic
map. Given that the Fredholm index ind(Du)of uis 2, with Dudenoting the linearized Cauchy-
Riemann operator associated to u, determine the total number of positive and negative punctures
of u.
Solution 3.
Since the index of uis 2, we know that the Conley-Zehnder index of each positive puncture of
uis −1and each negative puncture has a Conley-Zehnder index of 1.
Let n+and n−denote the total number of positive and negative punctures of u, respectively.
Then, we have the equation
ind(Du)=2=2−n++n−
Substitute the Conley-Zehnder index values into the equation, we get
2=2−n++n−
0 = −n++n−
n+=n−
Therefore, the total number of positive punctures is equal to the total number of negative punc-
tures in this case.
4 4. EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 4. Consider the following holomorphic curve in C:u:C→Cgiven by u(z) = z2−i.
We want to find the solutions to the Cauchy-Riemann equation ¯
∂Ju= 0.
Solution 4.
a) We start by writing u(z) = u(x1+ix2) = u1(x1, x2) + iu2(x1, x2)where u1, u2:R2→R.
We then compute the Cauchy-Riemann equations:
∂u1
∂x1
=∂u2
∂x2
and ∂u1
∂x2
=−∂u2
∂x1
In this case, u(z) = z2−i= (x1+ix2)2−i= (x2
1−x2
2) + i(2x1x2−1).
Therefore, we have:
u1(x1, x2) = x2
1−x2
2and u2(x1, x2)=2x1x2−1
Calculating the partial derivatives, we find:
∂u1
∂x1
= 2x1and ∂u2
∂x2
= 2x1
∂u1
∂x2
=−2x2and ∂u2
∂x1
= 2x2
Now we plug these into the Cauchy-Riemann equations:
2x1= 2x1and −2x2= 2x2
Solving the above equations, we obtain x1= 0 and x2= 0.
Therefore, the holomorphic curve u(z) = z2−isatisfies the Cauchy-Riemann equations at the
point (0,0).
b) To find other solutions, we can look for constant maps. Let u(z) = cwhere cis a complex
constant. Then, ¯
∂Ju= 0 is automatically satisfied.
c) Another way to find solutions is by looking at the composition of holomorphic functions. If
v:C→Cand u:C→Care holomorphic, then u◦vis also holomorphic.
5 5. FLOER HOMOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 5. Consider the following pseudoholomorphic curve in Cparametrized by z(t) = eit,
where t∈[0,2π]:
u:R×S1→C, u(s, eit) = eiseit.
a) Show that this curve satisfies the Cauchy-Riemann equations.
b) Calculate the energy of the curve.
c) Determine the image of the curve in C.
Solution 5.
a) To show that the curve satisfies the Cauchy-Riemann equations, we need to check if ∂u
∂¯z= 0.
Here, z=x+iy =x+iy(t), so ∂u
∂¯z=1
2(∂u
∂x +i∂u
∂y ) = 1
2(eiseit −ieiseit) = 0, hence the curve satisfies
the Cauchy-Riemann equations.
b) The energy of the curve is calculated using the formula:
E(u) = ZS1||∇u||2dt,
where ||∇u||2is the norm of the gradient of u. In this case, ∇u= (∂su, ∂tu)=(iu(s, eit), eisieit),
so ||∇u||2=|iu(s, eit)|2+|eisieit|2=|i|2+ 1 = 2. Therefore, the energy of the curve is:
E(u) = Z2π
0
2dt = 4π.
c) The image of the curve in Cis given by u(R×S1) = {eiseit |s∈R, t ∈[0,2π]}. Since eis
and eit vary over all complex numbers on the unit circle S1, the image is the unit circle S1itself.
6 6. COMPACTNESS PROPERTIES OF PSEUDOHOLOMORPHIC CURVES
Problem 6. Consider a sequence of pseudoholomorphic curves (uk: Σ →M, Jk)converging
to a constant map, where Σis a Riemann surface and Mis a symplectic manifold.
Let Jkbe a sequence of almost complex structures converging to J∞in the C∞-topology on
Σ. Assume that the energy of the curves E(uk)is uniformly bounded. Show that the sequence uk
converges in the C∞-topology on Σto a constant map u∞.
Solution 6.
Given that the energy of the curves E(uk)is uniformly bounded, we have
E(uk) = ZΣ||duk◦j−Jk◦duk||2dµ
≤C
where Cis a constant independent of k.
By the compactness theorem for Jk, there exists a subsequence (not relabeled) converging to
aJ∞-holomorphic map u∞. This gives us convergence in the C∞-topology for a subsequence.
Now we want to show that the entire sequence converges to u∞. By contradiction, assume
there exists ϵ > 0such that for all Nthere exists kN> N such that ||dukN−du∞||C∞≥ϵ. But this
is a contradiction, as we have a compact set in C∞-topology.
Therefore, the entire sequence ukconverges in the C∞-topology to the constant map u∞.
7 7. GROMOV-WITTEN INVARIANTS AND PSEUDOHOLOMORPHIC CURVES
Problem 7. Consider the moduli space of genus-zero, three-pointed stable pseudoholomor-
phic curves on a Riemann surface of genus one, with the three marked points labeled p1, p2, p3.
Suppose the homology classes of the curves are A= 2α+ 3βand B=−α+ 4β.
a) Calculate the dimension of the moduli space.
b) Determine the number of points in the moduli space that satisfy the conditions given in part
(a).
c) Suppose the intersection number of Aand Bis 1. Find the expected dimension of the moduli
space.
Solution 7.
a) The dimension of the moduli space can be calculated using the Riemann-Roch formula:
dim(M0,3(Σ, A)) = 3d+ 1 −g= 3(2 + 3) + 1 −1 = 10.
b) In this case, the moduli space contains finitely many points due to the constraint on the
homology classes. Hence, there are no points in the moduli space.
c) The expected dimension of the moduli space can be calculated using the formula:
dim(M0,3(Σ, A;B)) = dim(M0,3(Σ, A)) − ⟨A, B⟩= 10 −1 = 9.
Therefore, the expected dimension of the moduli space is 9.
8 8. EVALUATION MAPS FOR PSEUDOHOLOMORPHIC CURVES
Problem 8. Consider the pseudoholomorphic curve equation on a Riemann surface Sgiven by
du +idv = 0, where u, v :S→Care smooth functions. Let z=u+iv be a complex coordinate on
S. Given a pseudoholomorphic curve u=f(z)parametrized by f(z) = z2, evaluate the following:
a) The evaluation map evp:M → S, where Mis the moduli space of solutions to the pseudo-
holomorphic curve equation, at p= 2 + 3i.
b) The derivative d
dz f(z)at z= 1 −i.
c) The image under the map fof the curve g(z) = z3.
Solution 8.
a) To compute the evaluation map at p= 2+3i, we substitute z= 2+3iinto the parametrization
f(z) = z2. Therefore, f(2 + 3i) = (2 + 3i)2= 4 + 12i−9 = −5 + 12i. Hence, ev2+3i(f) = −5 + 12i.
b) To find the derivative d
dz f(z)at z= 1 −i, we differentiate f(z) = z2to get d
dz f(z) = 2z.
Evaluating this derivative at z= 1 −i, we have d
dz f(1 −i) = 2(1 −i)=2−2i.
c) The image under the map fof the curve g(z) = z3is given by f(g(z)) = f(z3)=(z3)2=z6.
Therefore, the image of the curve g(z)under the map fis the curve parametrized by z6.
9 9. MODULI SPACE OF PSEUDOHOLOMORPHIC CURVES
Problem 9. Consider a Riemann surface Mwith genus g= 2 and n= 3 marked points. Let Xbe
a compact oriented Riemann surface of genus 0 with 5 boundary marked points. Denote by Athe
moduli space of smooth maps ¯
M→Xsatisfying certain boundary conditions.
a) Calculate the dimension of the space A.
b) Determine the dimension of the moduli space of pseudo-holomorphic curves in CP 2with
k= 2 marked points.
Solution 9.
a) The dimension of the moduli space Acan be calculated using Riemann-Roch formula. We
know that for a Riemann surface of genus gwith nmarked points, the dimension of the moduli
space is given by:
dim(M)=6g−6+2n
Substitute g= 2 and n= 3 into the formula:
dim(A) = 6 ×2−6+2×3 = 12 −6 + 6 = 12
Therefore, the dimension of the moduli space Ais 12.
b) The dimension of the moduli space of pseudo-holomorphic curves in CP 2with kmarked
points is given by the formula:
dim M= 2k−6+2g
Substitute k= 2 and g= 0 into the formula:
dim M= 2 ×2−6+2×0=4−6 = −2
Therefore, the dimension of the moduli space of pseudo-holomorphic curves in CP 2with 2
marked points is -2.
10 9. THE THEORY OF PSEUDOHOLMORPHIC CURVES
Problem 9. Consider the following pseudoholomorphic curve in CP 1given by the map u:
R×S1→CP 1, where S1is the unit circle in C:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)].
a) Calculate the image of the curve u.
b) Determine the intersection number of the curve uwith the horizontal line [z: 1], where z∈C.
c) Find the set of critical points of the map u.
Solution 9.
a) To find the image of the curve u, we need to substitute the values of sand θinto the map u.
Let’s first express u(s, eiθ)in terms of sand θ:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)] .
Substitute s= 0 and θ=π/2into uto get the image point:
u(0, eiπ/2) = [1 : 2i].
Therefore, the image of the curve uis the point [1 : 2i]in CP 1.
b) To find the intersection number of the curve uwith the horizontal line [z: 1], we need to count
the number of solutions to the equation 2isin(θ)−icos(θ) = z(1 + is). This reduces to finding the
number of solutions to sin(θ)=0:
sin(θ) = 0 =⇒θ=nπ, where n∈Z.
Since θis periodic with period 2π, there are infinitely many intersection points. Hence, the
intersection number is ∞.
c) The critical points of uare the (s, eiθ)such that the derivative of uvanishes. Calculating the
derivative: ∂u
∂s =i[1 : 2i],∂u
∂θ =i[0 : 2 cos(θ) + sin(θ)].
Setting the derivatives equal to zero, we find that sis not involved. Thus, the critical points lie
on the horizontal line.
11 Numeric Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following Riemann surface defined by the equation z=eiθ, where
θ∈[0,2π].
a) Calculate the area of this Riemann surface.
b) Find the length of the curve defined by x=Re(z)and y=Im(z).
Solution 1.
a) To calculate the area of the Riemann surface, we can use the formula for the area of a surface
of revolution, which is given by A= 2πRβ
αf(θ)p1+(f′(θ))2dθ.
In this case, f(θ)=1and f′(θ)=0, so the area simplifies to:
A= 2πR2π
01·√1+0dθ = 2πR2π
01dθ = 2π[2π−0] = 4π2.
Therefore, the area of the Riemann surface is 4π2.
b) The length of the curve is given by the formula L=Rβ
αrdx
dθ 2+dy
dθ 2dθ.
Since x= cos(θ)and y= sin(θ), we have dx
dθ =−sin(θ)and dx
dθ = cos(θ). Therefore, the length
simplifies to:
L=R2π
0p(−sin(θ))2+ (cos(θ))2dθ =R2π
0√1dθ =R2π
01dθ = 2π.
The length of the curve is 2π.
12 12. BUBBLING PHENOMENON IN PSEUDOHOLOMORPHIC CURVES
Problem 12. Consider the sequence of pseudoholomorphic curves defined by un:R×S2→C
given by un(x, z) = eiθnzn, where θn→0as n→ ∞. Determine whether this sequence exhibits
bubbling phenomenon or not.
Solution 12. a) To analyze the behavior of this sequence of pseudoholomorphic curves, let’s
first calculate the energy of each curve. The energy functional for a pseudoholomorphic curve
u: Σ →Cdefined over a Riemann surface Σis given by
E(u) = ZΣ||du||2dA,
where ||du||2is the square of the norm of the differential du and dA is the area element on Σ.
For the curve un(x, z) = eiθnzn, the energy can be calculated as follows:
||dun||2=|eiθnnzn−1|2=n2|z|2n−2,
and integrating over S2(with the standard metric) gives
E(un) = ZS2
n2|z|2n−2dA = 4πn2.
b) Next, let’s examine the behavior of the energy of the sequence as n→ ∞. We have
lim
n→∞ E(un) = lim
n→∞ 4πn2=∞.
c) Since the energy of the sequence of curves diverges as n→ ∞, we conclude that this
sequence exhibits bubbling phenomenon. In this case, the limiting pseudoholomorphic curve will
be a multiply-covered sphere.
Therefore, the sequence of pseudoholomorphic curves defined by un(x, z) = eiθnznexhibits
bubbling phenomenon.
13 13. SYMPLECTIC TOPOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 13. Consider the complex structure Jon R2given by J(x, y)=(−y, x)and the
symplectic form ω=dx ∧dy.
a) Let u(x, y)=(x2−y2,2xy)be a map R2→R2. Determine if uis pseudoholomorphic with
respect to (J, ω).
b) Find the image of the map ufrom part (a).
Solution 13.
a) To determine if uis pseudoholomorphic with respect to (J, ω), we need to check if du +J◦
du ◦J=ω.
Compute du =2x−2y
2y2xand J◦du ◦J=−2y2x
2x2y.
Adding them gives du +J◦du ◦J=0 0
0 0=ω=dx ∧dy.
Since du +J◦du ◦J=ω,uis not pseudoholomorphic.
b) The image of ucan be found by solving the following system of equations:
x2−y2=u1
2xy =u2
Substitute u1=x2−y2into u2= 2xy to get 2x(x2−y2) = u2.
Solving both equations simultaneously gives us x= 0 or x2=y2, which are the equations for
the coordinate axes.
Therefore, the image of the map uis the xand yaxes.
14 14. CAUCHY-RIEMANN EQUATIONS AND PSEUDOHOLOMORPHIC CURVES
Problem 14. Consider the pseudoholomorphic curve u:C→Cdefined by u(z) = z2+i,
where z=x+iy and i=√−1.
a) Find the Cauchy-Riemann equations for the map u.
b) Determine if the map uis pseudoholomorphic.
Solution 14.
a) To find the Cauchy-Riemann equations for u(z) = z2+i, we first express u(z)in terms of x
and y:
u(z)=(x+iy)2+i=x2−y2+ 2ixy +i
Now, we write u(z)as u(x, y) + iv(x, y):
u(x, y) = x2−y2, v(x, y)=2xy + 1
Next, we find the partial derivatives of uand v:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
The Cauchy-Riemann equations are:
∂u
∂x =∂v
∂y and ∂u
∂y =−∂v
∂x
Substitute the partial derivatives we found into these equations to check if they are satisfied.
b) To determine if the map uis pseudoholomorphic, we need to verify if the Cauchy-Riemann
equations hold for u. From part a, we found that:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
As the Cauchy-Riemann equations are satisfied, the map uis pseudoholomorphic.
15 15. GRADIENT FLOW TECHNIQUES FOR PSEUDOHOLOMORPHIC CURVES
Problem 15. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = z2+iz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 1.
Solution 15. The solution to the pseudoholomorphic curve equation is u(t, z) = u(0, z)for all
t, due to the stationary nature of pseudoholomorphic curves. Hence, u(1, z) = u(0, z) = z2+iz.
Therefore, the solution at time t= 1 is u(1, z) = z2+iz for all z∈Σin this case.
Problem 16. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = eiz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 2.
Solution 16. Similar to the previous problem, the solution to the pseudoholomorphic curve
equation is u(t, z) = u(0, z)for all t, due to the stationary nature of pseudoholomorphic curves.
Hence, u(2, z) = u(0, z) = eiz.
Therefore, the solution at time t= 2 is u(2, z) = eiz for all z∈Σin this case.
16 16. TOPOLOGICAL AND GEOMETRIC CONSTRAINTS ON PSEUDOHOLOMORPHIC CURVES
Problem 16. Consider a closed oriented manifold Mof dimension 4, equipped with a compati-
ble almost complex structure J. Let [γ]be a homology class represented by a simple closed curve
γin M. We define the self-intersection number of [γ]as I([γ],[γ]) = γ·γ.
Suppose Σis a pseudoholomorphic curve in Mwith boundary ∂Σ = γ1−γ2, where γ1and γ2
are simple closed curves representing different homology classes in M. Determine the relationship
between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]).
Solution 16. Let’s denote the two simple closed curves γ1and γ2as Aand B, respectively.
Then, we have I([γ1],[γ1]) = A·Aand I([γ2],[γ2]) = B·B, where ·denotes the intersection pairing.
The self-intersection number of the curve Σcan be computed as I(Σ,Σ) = (γ1−γ2)·(γ1−γ2).
Expanding this out, we get:
I(Σ,Σ) = γ1·γ1−γ1·γ2−γ2·γ1+γ2·γ2
=A·A−A·B−B·A+B·B
=A·A−2A·B+B·B
=I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
Therefore, the relationship between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]) is given by:
I(Σ,Σ) = I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
17 Numerical Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in CP 1:
∂u
∂z +λe−u= 0,
where λ∈Ris a constant.
a) Find the general solution u(z, z)to this pseudoholomorphic curve equation.
b) Given λ= 2, solve the equation with the initial condition u(0,0) = 0.
Solution 1.
a) To find the general solution to the pseudoholomorphic curve equation, we first rewrite the
equation in terms of the complex coordinates zand z. Note that ∂u
∂z =1
2∂u
∂x +i∂u
∂y , where z=
x+iy.
Substitute this into the given equation:
1
2∂u
∂x +i∂u
∂y +λe−u= 0.
Separating real and imaginary parts of this equation gives us:
∂u
∂x = 2λe−u
∂u
∂y = 0
Integrating the first equation with respect to xand the second equation with respect to ygives:
u= ln C
2λ+e2λx,
where Cis the constant of integration.
b) Given λ= 2 and the initial condition u(0,0) = 0, we can find the value of Cby substituting
x= 0, u = 0 into the general solution:
0 = ln C
4+ 1.
Solving this equation for Cgives C= 4, so the solution to the equation with λ= 2 and the initial
condition is:
u= ln 4
4+e4x= ln(1 + e4x).
18 The Theory of Pseudoholomorphic Curves
18.1 Problem 1
Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = aeis +bt with a, b ∈C
and |a|= 3,|b|= 2. Determine the image of this pseudoholomorphic curve in C.
Solution: The image of uis given by u(s, t) = aeis +bt. To find the image, we set u(s, t) = z
and solve for sand t:
z=aeis +bt
= 3eis + 2t
= 3(cos s+isin s)+2t.
This implies that the image of the pseudoholomorphic curve uis the set of points z= 3 cos s+
i3 sin s+ 2tin the complex plane C.
18.2 Problem 2
Let u:R×[0,1] →Cbe a pseudoholomorphic curve defined by u(s, t) = s2+it. Determine the
differential of u.
Solution: The differential of uis given by du =∂u
∂s ds +∂u
∂t dt. In this case, we have:
∂u
∂s = 2s,
∂u
∂t =i.
Therefore, the differential of uis du = 2s ds +i dt.
19 19. ENERGY ESTIMATES FOR PSEUDOHOLOMORPHIC CURVES
Problem 19. Consider a pseudoholomorphic curve u:R×S1→Cgiven by u(s, t) = eis sin(t),
where s, t ∈R. Calculate the energy of this pseudoholomorphic curve using the energy functional
E(u) = RR×S1||∂su||2+||∂tu||2dsdt.
Solution 19. The energy of the pseudoholomorphic curve ucan be calculated as follows:
a) First, we calculate the partial derivatives of u:
∂su=ieis sin(t),
∂tu=eis cos(t).
b) Next, we compute the norms of the partial derivatives:
||∂su||2=|ieis sin(t)|2=| − eis sin(t)|2=|eis|2|sin(t)|2= sin2(t),
||∂tu||2=|eis cos(t)|2=|eis||cos(t)|2= 1|cos(t)|2= cos2(t).
c) Now, we substitute the norms into the energy functional and integrate over R×S1:
E(u) = ZR×S1
(sin2(t) + cos2(t)) dsdt
=ZRZS1
1dsdt
=ZR
2π ds
= 2π.
Therefore, the energy of the pseudoholomorphic curve uis 2π.
20 The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in a symplectic mani-
fold:
¯
∂Ju+J(u)◦∂¯zu= 0
where u:C→Mis a holomorphic map, Jis a compatible almost complex structure in M, and
¯
∂Jdenotes the Cauchy-Riemann operator.
Given u(z) = z+i
z−i, determine if uis a pseudoholomorphic curve in the complex plane C.
Solution 1.
To determine if u(z) = z+i
z−iis a pseudoholomorphic curve, we need to check if it satisfies the
pseudoholomorphic curve equation. First, let’s calculate ∂zuand ∂¯zu:
∂zu=1
(z−i)2
∂¯zu= 0
Next, let’s compute ¯
∂Ju:
¯
∂Ju=1
2(∂¯zu−J(u)◦∂zu) = 1
20−J(u)◦1
(z−i)2
Given that Jis the standard almost complex structure on C, i.e., J(z) = iz, we have:
J(u) = iz+i
z−i=i(z+i)
z−i
Therefore, ¯
∂Ju=−i
2(z−i)3. Since ¯
∂Ju+J(u)◦∂¯zu= 0, the function u(z) = z+i
z−iis indeed a
pseudoholomorphic curve in C.
21 21. STABILITY CONDITIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 21. Consider a pseudoholomorphic curve u: Σ →Mwhere Σis a Riemann surface
and Mis a symplectic manifold with symplectic form ω. Let Jbe an almost complex structure on
Mcompatible with ω. Suppose the pseudoholomorphic curve usatisfies the following Cauchy-
Riemann equation:
¯
∂Ju= 0.
a) Show that the linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J.
b) Calculate the Fredholm index of the operator Du.
c) Determine the stability condition for the pseudoholomorphic curve u.
Solution 21.
a) The linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J. This operator acts on sections of the pullback bundle u∗T M,
which is the tangent bundle of the Riemann surface Σpulled back to Mvia the map u.
b) The Fredholm index of the operator Duis given by
index(Du) = dim ker(Du)−dim coker(Du).
Here, ker(Du)represents the kernel of Du(space of solutions to ¯
∂Jv= 0 near u) and coker(Du)
represents the cokernel of Du. The Fredholm index indicates the deformation parameter for the
moduli space of pseudoholomorphic curves.
c) The stability condition for the pseudoholomorphic curve uis satisfied if the Fredholm index
of the linearized operator Duis nonnegative, i.e., index(Du)≥0. This condition ensures that
the moduli space of pseudoholomorphic curves is well-behaved under small perturbations in the
complex structure or metric on M.
22 22. FREDHOLM THEORY AND PSEUDOHOLOMORPHIC CURVES
Problem 22. Consider the following pseudoholomorphic curve equation in C2:
¯
∂u = 0
where u:R×S1→C2is a smooth map.
Given u(t, z) = eit(z, iz)for t∈[0,2π]and z∈S1, compute the Fredholm index of this pseudo-
holomorphic curve.
Solution 22. To compute the Fredholm index, we first need to find the linearized operator asso-
ciated with ¯
∂at u. Let v(t, z)be a small perturbation around u, and write v(t, z) = eit(a(t, z), b(t, z)).
The linearized operator is given by:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb
Plugging in u(t, z)and v(t, z)into the expression and simplifying, we get:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb=eit(∂ta−i∂tb+∂za−i∂zb)
This is equal to 0if and only if both ∂ta−i∂tband ∂za−i∂zbare equal to 0.
From u(t, z), we have a(t, z) = zand b(t, z) = iz. Differentiating with respect to tand z, we get:
∂ta= 0, ∂tb=−z, ∂za= 1, ∂zb=i
So ∂ta−i∂tb= 0 and ∂za−i∂zb= 1 −i= 0. Thus, the Fredholm index is 0−2 = −2.
23 The Theory of Pseudoholomorphic Curves
Problem: Consider the following pseudoholomorphic curve equation in the complex plane C:
¯
∂Ju+i∂tu= 0
where u:C×[0,1] →Cis a smooth map, ∂tdenotes the partial derivative with respect to
t∈[0,1], and ¯
∂Jdenotes the ¯
∂-operator twisted by a compatible almost complex structure Jon C.
Given the initial condition u(x, 0) = x2for x∈C, solve the pseudoholomorphic curve equation.
Solution:
To solve the pseudoholomorphic curve equation, we will first compute the derivatives and then
use the initial condition to determine the solution.
Let u(x, t) = u1(x, t) + iv1(x, t), where u1and v1are real-valued functions. Then the pseudo-
holomorphic curve equation becomes:
∂u1
∂t −∂v1
∂x +i∂v1
∂t +∂u1
∂x = 0
Separating the real and imaginary parts, we get:
∂u1
∂t =∂v1
∂x ,∂v1
∂t =−∂u1
∂x
From the initial condition u(x, 0) = x2, we have u1(x, 0) = x2and v1(x, 0) = 0.
Solving the partial differential equations with these initial conditions gives u1(x, t) = x2and
v1(x, t)=0. Therefore, the solution of the pseudoholomorphic curve equation is u(x, t) = x2+i·0 =
x2.
24 24. HOMOLOGICAL ALGEBRA AND PSEUDOHOLOMORPHIC CURVES
Problem 24. Consider the following pseudoholomorphic curve in CP 1given by the equation
u(z) = z2.
a) Determine the differential operator ∂Jassociated with the almost complex structure Jinduced
by the Fubini-Study metric on CP 1.
b) Show that u(z)is a solution to the ∂Jequation on CP 1.
Solution 24.
a) The almost complex structure Jassociated with the Fubini-Study metric on CP 1is given by
J(z) = −iz. The differential operator ∂Jcan be computed as:
∂J=1
2(J+iId) = 1
2(−iz +iId) = 1
2(i(z+ 1)).
b) To show that u(z) = z2is a solution to the ∂Jequation, we need to demonstrate that ∂Ju(z) =
0. Let’s compute it:
∂Ju(z) = 1
2(i(z2+ 1)) = i
2z2+i
2=iz +i
2= 0.
As the result is not zero, the function u(z) = z2is not a solution to the ∂Jequation on CP 1.
25 The Theory of Pseudoholomorphic Curves
Problem 1. Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = eis +t2.
a) Determine the asymptotic behavior of uas s→ −∞.
b) Find the limit lims→∞ |u(s, 1)|.
c) Show that uis a pseudoholomorphic curve.
Solution 1.
a) As s→ −∞, the term eis behaves as oscillatory and does not affect the growth, so we focus
on t2. Thus, the asymptotic behavior of u(s, t)as s→ −∞ is t2.
b) We have |u(s, 1)|=|eis + 1| ≤ |eis|+|1|= 1 + 1 = 2. Thus, lims→∞ |u(s, 1)| ≤ 2.
c) To show that uis a pseudoholomorphic curve, we need to verify that ¯
∂Ju= 0, where Jis the
standard complex structure on C. Compute ∂su=ieis and ∂tu= 2t. Then, ¯
∂Ju=∂su−i∂tu=
ieis −2it.
Since ¯
∂Ju= 0, we have shown that uis a pseudoholomorphic curve.
Therefore, the solution to the pseudoholomorphic curve equation with the specified Riemann
surface and almost complex structure is u(z) = az +b, but it is not J-holomorphic.
3 3. INDEX THEORY FOR PSEUDOHOLOMORPHIC CURVES
Problem 3. Consider a compact, oriented, Riemannian 2-manifold Mwith a compatible almost
complex structure Jand a symplectic form ω. Let u:C→Mbe a nonconstant J-holomorphic
map. Given that the Fredholm index ind(Du)of uis 2, with Dudenoting the linearized Cauchy-
Riemann operator associated to u, determine the total number of positive and negative punctures
of u.
Solution 3.
Since the index of uis 2, we know that the Conley-Zehnder index of each positive puncture of
uis −1and each negative puncture has a Conley-Zehnder index of 1.
Let n+and n−denote the total number of positive and negative punctures of u, respectively.
Then, we have the equation
ind(Du)=2=2−n++n−
Substitute the Conley-Zehnder index values into the equation, we get
2=2−n++n−
0 = −n++n−
n+=n−
Therefore, the total number of positive punctures is equal to the total number of negative punc-
tures in this case.
4 4. EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 4. Consider the following holomorphic curve in C:u:C→Cgiven by u(z) = z2−i.
We want to find the solutions to the Cauchy-Riemann equation ¯
∂Ju= 0.
Solution 4.
a) We start by writing u(z) = u(x1+ix2) = u1(x1, x2) + iu2(x1, x2)where u1, u2:R2→R.
We then compute the Cauchy-Riemann equations:
∂u1
∂x1
=∂u2
∂x2
and ∂u1
∂x2
=−∂u2
∂x1
In this case, u(z) = z2−i= (x1+ix2)2−i= (x2
1−x2
2) + i(2x1x2−1).
Therefore, we have:
u1(x1, x2) = x2
1−x2
2and u2(x1, x2)=2x1x2−1
Calculating the partial derivatives, we find:
∂u1
∂x1
= 2x1and ∂u2
∂x2
= 2x1
∂u1
∂x2
=−2x2and ∂u2
∂x1
= 2x2
Now we plug these into the Cauchy-Riemann equations:
2x1= 2x1and −2x2= 2x2
Solving the above equations, we obtain x1= 0 and x2= 0.
Therefore, the holomorphic curve u(z) = z2−isatisfies the Cauchy-Riemann equations at the
point (0,0).
b) To find other solutions, we can look for constant maps. Let u(z) = cwhere cis a complex
constant. Then, ¯
∂Ju= 0 is automatically satisfied.
c) Another way to find solutions is by looking at the composition of holomorphic functions. If
v:C→Cand u:C→Care holomorphic, then u◦vis also holomorphic.
5 5. FLOER HOMOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 5. Consider the following pseudoholomorphic curve in Cparametrized by z(t) = eit,
where t∈[0,2π]:
u:R×S1→C, u(s, eit) = eiseit.
a) Show that this curve satisfies the Cauchy-Riemann equations.
b) Calculate the energy of the curve.
c) Determine the image of the curve in C.
Solution 5.
a) To show that the curve satisfies the Cauchy-Riemann equations, we need to check if ∂u
∂¯z= 0.
Here, z=x+iy =x+iy(t), so ∂u
∂¯z=1
2(∂u
∂x +i∂u
∂y ) = 1
2(eiseit −ieiseit) = 0, hence the curve satisfies
the Cauchy-Riemann equations.
b) The energy of the curve is calculated using the formula:
E(u) = ZS1||∇u||2dt,
where ||∇u||2is the norm of the gradient of u. In this case, ∇u= (∂su, ∂tu)=(iu(s, eit), eisieit),
so ||∇u||2=|iu(s, eit)|2+|eisieit|2=|i|2+ 1 = 2. Therefore, the energy of the curve is:
E(u) = Z2π
0
2dt = 4π.
c) The image of the curve in Cis given by u(R×S1) = {eiseit |s∈R, t ∈[0,2π]}. Since eis
and eit vary over all complex numbers on the unit circle S1, the image is the unit circle S1itself.
6 6. COMPACTNESS PROPERTIES OF PSEUDOHOLOMORPHIC CURVES
Problem 6. Consider a sequence of pseudoholomorphic curves (uk: Σ →M, Jk)converging
to a constant map, where Σis a Riemann surface and Mis a symplectic manifold.
Let Jkbe a sequence of almost complex structures converging to J∞in the C∞-topology on
Σ. Assume that the energy of the curves E(uk)is uniformly bounded. Show that the sequence uk
converges in the C∞-topology on Σto a constant map u∞.
Solution 6.
Given that the energy of the curves E(uk)is uniformly bounded, we have
E(uk) = ZΣ||duk◦j−Jk◦duk||2dµ
≤C
where Cis a constant independent of k.
By the compactness theorem for Jk, there exists a subsequence (not relabeled) converging to
aJ∞-holomorphic map u∞. This gives us convergence in the C∞-topology for a subsequence.
Now we want to show that the entire sequence converges to u∞. By contradiction, assume
there exists ϵ > 0such that for all Nthere exists kN> N such that ||dukN−du∞||C∞≥ϵ. But this
is a contradiction, as we have a compact set in C∞-topology.
Therefore, the entire sequence ukconverges in the C∞-topology to the constant map u∞.
7 7. GROMOV-WITTEN INVARIANTS AND PSEUDOHOLOMORPHIC CURVES
Problem 7. Consider the moduli space of genus-zero, three-pointed stable pseudoholomor-
phic curves on a Riemann surface of genus one, with the three marked points labeled p1, p2, p3.
Suppose the homology classes of the curves are A= 2α+ 3βand B=−α+ 4β.
a) Calculate the dimension of the moduli space.
b) Determine the number of points in the moduli space that satisfy the conditions given in part
(a).
c) Suppose the intersection number of Aand Bis 1. Find the expected dimension of the moduli
space.
Solution 7.
a) The dimension of the moduli space can be calculated using the Riemann-Roch formula:
dim(M0,3(Σ, A)) = 3d+ 1 −g= 3(2 + 3) + 1 −1 = 10.
b) In this case, the moduli space contains finitely many points due to the constraint on the
homology classes. Hence, there are no points in the moduli space.
c) The expected dimension of the moduli space can be calculated using the formula:
dim(M0,3(Σ, A;B)) = dim(M0,3(Σ, A)) − ⟨A, B⟩= 10 −1 = 9.
Therefore, the expected dimension of the moduli space is 9.
8 8. EVALUATION MAPS FOR PSEUDOHOLOMORPHIC CURVES
Problem 8. Consider the pseudoholomorphic curve equation on a Riemann surface Sgiven by
du +idv = 0, where u, v :S→Care smooth functions. Let z=u+iv be a complex coordinate on
S. Given a pseudoholomorphic curve u=f(z)parametrized by f(z) = z2, evaluate the following:
a) The evaluation map evp:M → S, where Mis the moduli space of solutions to the pseudo-
holomorphic curve equation, at p= 2 + 3i.
b) The derivative d
dz f(z)at z= 1 −i.
c) The image under the map fof the curve g(z) = z3.
Solution 8.
a) To compute the evaluation map at p= 2+3i, we substitute z= 2+3iinto the parametrization
f(z) = z2. Therefore, f(2 + 3i) = (2 + 3i)2= 4 + 12i−9 = −5 + 12i. Hence, ev2+3i(f) = −5 + 12i.
b) To find the derivative d
dz f(z)at z= 1 −i, we differentiate f(z) = z2to get d
dz f(z) = 2z.
Evaluating this derivative at z= 1 −i, we have d
dz f(1 −i) = 2(1 −i)=2−2i.
c) The image under the map fof the curve g(z) = z3is given by f(g(z)) = f(z3)=(z3)2=z6.
Therefore, the image of the curve g(z)under the map fis the curve parametrized by z6.
9 9. MODULI SPACE OF PSEUDOHOLOMORPHIC CURVES
Problem 9. Consider a Riemann surface Mwith genus g= 2 and n= 3 marked points. Let Xbe
a compact oriented Riemann surface of genus 0 with 5 boundary marked points. Denote by Athe
moduli space of smooth maps ¯
M→Xsatisfying certain boundary conditions.
a) Calculate the dimension of the space A.
b) Determine the dimension of the moduli space of pseudo-holomorphic curves in CP 2with
k= 2 marked points.
Solution 9.
a) The dimension of the moduli space Acan be calculated using Riemann-Roch formula. We
know that for a Riemann surface of genus gwith nmarked points, the dimension of the moduli
space is given by:
dim(M)=6g−6+2n
Substitute g= 2 and n= 3 into the formula:
dim(A) = 6 ×2−6+2×3 = 12 −6 + 6 = 12
Therefore, the dimension of the moduli space Ais 12.
b) The dimension of the moduli space of pseudo-holomorphic curves in CP 2with kmarked
points is given by the formula:
dim M= 2k−6+2g
Substitute k= 2 and g= 0 into the formula:
dim M= 2 ×2−6+2×0=4−6 = −2
Therefore, the dimension of the moduli space of pseudo-holomorphic curves in CP 2with 2
marked points is -2.
10 9. THE THEORY OF PSEUDOHOLMORPHIC CURVES
Problem 9. Consider the following pseudoholomorphic curve in CP 1given by the map u:
R×S1→CP 1, where S1is the unit circle in C:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)].
a) Calculate the image of the curve u.
b) Determine the intersection number of the curve uwith the horizontal line [z: 1], where z∈C.
c) Find the set of critical points of the map u.
Solution 9.
a) To find the image of the curve u, we need to substitute the values of sand θinto the map u.
Let’s first express u(s, eiθ)in terms of sand θ:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)] .
Substitute s= 0 and θ=π/2into uto get the image point:
u(0, eiπ/2) = [1 : 2i].
Therefore, the image of the curve uis the point [1 : 2i]in CP 1.
b) To find the intersection number of the curve uwith the horizontal line [z: 1], we need to count
the number of solutions to the equation 2isin(θ)−icos(θ) = z(1 + is). This reduces to finding the
number of solutions to sin(θ)=0:
sin(θ) = 0 =⇒θ=nπ, where n∈Z.
Since θis periodic with period 2π, there are infinitely many intersection points. Hence, the
intersection number is ∞.
c) The critical points of uare the (s, eiθ)such that the derivative of uvanishes. Calculating the
derivative: ∂u
∂s =i[1 : 2i],∂u
∂θ =i[0 : 2 cos(θ) + sin(θ)].
Setting the derivatives equal to zero, we find that sis not involved. Thus, the critical points lie
on the horizontal line.
11 Numeric Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following Riemann surface defined by the equation z=eiθ, where
θ∈[0,2π].
a) Calculate the area of this Riemann surface.
b) Find the length of the curve defined by x=Re(z)and y=Im(z).
Solution 1.
a) To calculate the area of the Riemann surface, we can use the formula for the area of a surface
of revolution, which is given by A= 2πRβ
αf(θ)p1+(f′(θ))2dθ.
In this case, f(θ)=1and f′(θ)=0, so the area simplifies to:
A= 2πR2π
01·√1+0dθ = 2πR2π
01dθ = 2π[2π−0] = 4π2.
Therefore, the area of the Riemann surface is 4π2.
b) The length of the curve is given by the formula L=Rβ
αrdx
dθ 2+dy
dθ 2dθ.
Since x= cos(θ)and y= sin(θ), we have dx
dθ =−sin(θ)and dx
dθ = cos(θ). Therefore, the length
simplifies to:
L=R2π
0p(−sin(θ))2+ (cos(θ))2dθ =R2π
0√1dθ =R2π
01dθ = 2π.
The length of the curve is 2π.
12 12. BUBBLING PHENOMENON IN PSEUDOHOLOMORPHIC CURVES
Problem 12. Consider the sequence of pseudoholomorphic curves defined by un:R×S2→C
given by un(x, z) = eiθnzn, where θn→0as n→ ∞. Determine whether this sequence exhibits
bubbling phenomenon or not.
Solution 12. a) To analyze the behavior of this sequence of pseudoholomorphic curves, let’s
first calculate the energy of each curve. The energy functional for a pseudoholomorphic curve
u: Σ →Cdefined over a Riemann surface Σis given by
E(u) = ZΣ||du||2dA,
where ||du||2is the square of the norm of the differential du and dA is the area element on Σ.
For the curve un(x, z) = eiθnzn, the energy can be calculated as follows:
||dun||2=|eiθnnzn−1|2=n2|z|2n−2,
and integrating over S2(with the standard metric) gives
E(un) = ZS2
n2|z|2n−2dA = 4πn2.
b) Next, let’s examine the behavior of the energy of the sequence as n→ ∞. We have
lim
n→∞ E(un) = lim
n→∞ 4πn2=∞.
c) Since the energy of the sequence of curves diverges as n→ ∞, we conclude that this
sequence exhibits bubbling phenomenon. In this case, the limiting pseudoholomorphic curve will
be a multiply-covered sphere.
Therefore, the sequence of pseudoholomorphic curves defined by un(x, z) = eiθnznexhibits
bubbling phenomenon.
13 13. SYMPLECTIC TOPOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 13. Consider the complex structure Jon R2given by J(x, y)=(−y, x)and the
symplectic form ω=dx ∧dy.
a) Let u(x, y)=(x2−y2,2xy)be a map R2→R2. Determine if uis pseudoholomorphic with
respect to (J, ω).
b) Find the image of the map ufrom part (a).
Solution 13.
a) To determine if uis pseudoholomorphic with respect to (J, ω), we need to check if du +J◦
du ◦J=ω.
Compute du =2x−2y
2y2xand J◦du ◦J=−2y2x
2x2y.
Adding them gives du +J◦du ◦J=0 0
0 0=ω=dx ∧dy.
Since du +J◦du ◦J=ω,uis not pseudoholomorphic.
b) The image of ucan be found by solving the following system of equations:
x2−y2=u1
2xy =u2
Substitute u1=x2−y2into u2= 2xy to get 2x(x2−y2) = u2.
Solving both equations simultaneously gives us x= 0 or x2=y2, which are the equations for
the coordinate axes.
Therefore, the image of the map uis the xand yaxes.
14 14. CAUCHY-RIEMANN EQUATIONS AND PSEUDOHOLOMORPHIC CURVES
Problem 14. Consider the pseudoholomorphic curve u:C→Cdefined by u(z) = z2+i,
where z=x+iy and i=√−1.
a) Find the Cauchy-Riemann equations for the map u.
b) Determine if the map uis pseudoholomorphic.
Solution 14.
a) To find the Cauchy-Riemann equations for u(z) = z2+i, we first express u(z)in terms of x
and y:
u(z)=(x+iy)2+i=x2−y2+ 2ixy +i
Now, we write u(z)as u(x, y) + iv(x, y):
u(x, y) = x2−y2, v(x, y)=2xy + 1
Next, we find the partial derivatives of uand v:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
The Cauchy-Riemann equations are:
∂u
∂x =∂v
∂y and ∂u
∂y =−∂v
∂x
Substitute the partial derivatives we found into these equations to check if they are satisfied.
b) To determine if the map uis pseudoholomorphic, we need to verify if the Cauchy-Riemann
equations hold for u. From part a, we found that:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
As the Cauchy-Riemann equations are satisfied, the map uis pseudoholomorphic.
15 15. GRADIENT FLOW TECHNIQUES FOR PSEUDOHOLOMORPHIC CURVES
Problem 15. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = z2+iz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 1.
Solution 15. The solution to the pseudoholomorphic curve equation is u(t, z) = u(0, z)for all
t, due to the stationary nature of pseudoholomorphic curves. Hence, u(1, z) = u(0, z) = z2+iz.
Therefore, the solution at time t= 1 is u(1, z) = z2+iz for all z∈Σin this case.
Problem 16. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = eiz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 2.
Solution 16. Similar to the previous problem, the solution to the pseudoholomorphic curve
equation is u(t, z) = u(0, z)for all t, due to the stationary nature of pseudoholomorphic curves.
Hence, u(2, z) = u(0, z) = eiz.
Therefore, the solution at time t= 2 is u(2, z) = eiz for all z∈Σin this case.
16 16. TOPOLOGICAL AND GEOMETRIC CONSTRAINTS ON PSEUDOHOLOMORPHIC CURVES
Problem 16. Consider a closed oriented manifold Mof dimension 4, equipped with a compati-
ble almost complex structure J. Let [γ]be a homology class represented by a simple closed curve
γin M. We define the self-intersection number of [γ]as I([γ],[γ]) = γ·γ.
Suppose Σis a pseudoholomorphic curve in Mwith boundary ∂Σ = γ1−γ2, where γ1and γ2
are simple closed curves representing different homology classes in M. Determine the relationship
between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]).
Solution 16. Let’s denote the two simple closed curves γ1and γ2as Aand B, respectively.
Then, we have I([γ1],[γ1]) = A·Aand I([γ2],[γ2]) = B·B, where ·denotes the intersection pairing.
The self-intersection number of the curve Σcan be computed as I(Σ,Σ) = (γ1−γ2)·(γ1−γ2).
Expanding this out, we get:
I(Σ,Σ) = γ1·γ1−γ1·γ2−γ2·γ1+γ2·γ2
=A·A−A·B−B·A+B·B
=A·A−2A·B+B·B
=I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
Therefore, the relationship between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]) is given by:
I(Σ,Σ) = I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
17 Numerical Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in CP 1:
∂u
∂z +λe−u= 0,
where λ∈Ris a constant.
a) Find the general solution u(z, z)to this pseudoholomorphic curve equation.
b) Given λ= 2, solve the equation with the initial condition u(0,0) = 0.
Solution 1.
a) To find the general solution to the pseudoholomorphic curve equation, we first rewrite the
equation in terms of the complex coordinates zand z. Note that ∂u
∂z =1
2∂u
∂x +i∂u
∂y , where z=
x+iy.
Substitute this into the given equation:
1
2∂u
∂x +i∂u
∂y +λe−u= 0.
Separating real and imaginary parts of this equation gives us:
∂u
∂x = 2λe−u
∂u
∂y = 0
Integrating the first equation with respect to xand the second equation with respect to ygives:
u= ln C
2λ+e2λx,
where Cis the constant of integration.
b) Given λ= 2 and the initial condition u(0,0) = 0, we can find the value of Cby substituting
x= 0, u = 0 into the general solution:
0 = ln C
4+ 1.
Solving this equation for Cgives C= 4, so the solution to the equation with λ= 2 and the initial
condition is:
u= ln 4
4+e4x= ln(1 + e4x).
18 The Theory of Pseudoholomorphic Curves
18.1 Problem 1
Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = aeis +bt with a, b ∈C
and |a|= 3,|b|= 2. Determine the image of this pseudoholomorphic curve in C.
Solution: The image of uis given by u(s, t) = aeis +bt. To find the image, we set u(s, t) = z
and solve for sand t:
z=aeis +bt
= 3eis + 2t
= 3(cos s+isin s)+2t.
This implies that the image of the pseudoholomorphic curve uis the set of points z= 3 cos s+
i3 sin s+ 2tin the complex plane C.
18.2 Problem 2
Let u:R×[0,1] →Cbe a pseudoholomorphic curve defined by u(s, t) = s2+it. Determine the
differential of u.
Solution: The differential of uis given by du =∂u
∂s ds +∂u
∂t dt. In this case, we have:
∂u
∂s = 2s,
∂u
∂t =i.
Therefore, the differential of uis du = 2s ds +i dt.
19 19. ENERGY ESTIMATES FOR PSEUDOHOLOMORPHIC CURVES
Problem 19. Consider a pseudoholomorphic curve u:R×S1→Cgiven by u(s, t) = eis sin(t),
where s, t ∈R. Calculate the energy of this pseudoholomorphic curve using the energy functional
E(u) = RR×S1||∂su||2+||∂tu||2dsdt.
Solution 19. The energy of the pseudoholomorphic curve ucan be calculated as follows:
a) First, we calculate the partial derivatives of u:
∂su=ieis sin(t),
∂tu=eis cos(t).
b) Next, we compute the norms of the partial derivatives:
||∂su||2=|ieis sin(t)|2=| − eis sin(t)|2=|eis|2|sin(t)|2= sin2(t),
||∂tu||2=|eis cos(t)|2=|eis||cos(t)|2= 1|cos(t)|2= cos2(t).
c) Now, we substitute the norms into the energy functional and integrate over R×S1:
E(u) = ZR×S1
(sin2(t) + cos2(t)) dsdt
=ZRZS1
1dsdt
=ZR
2π ds
= 2π.
Therefore, the energy of the pseudoholomorphic curve uis 2π.
20 The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in a symplectic mani-
fold:
¯
∂Ju+J(u)◦∂¯zu= 0
where u:C→Mis a holomorphic map, Jis a compatible almost complex structure in M, and
¯
∂Jdenotes the Cauchy-Riemann operator.
Given u(z) = z+i
z−i, determine if uis a pseudoholomorphic curve in the complex plane C.
Solution 1.
To determine if u(z) = z+i
z−iis a pseudoholomorphic curve, we need to check if it satisfies the
pseudoholomorphic curve equation. First, let’s calculate ∂zuand ∂¯zu:
∂zu=1
(z−i)2
∂¯zu= 0
Next, let’s compute ¯
∂Ju:
¯
∂Ju=1
2(∂¯zu−J(u)◦∂zu) = 1
20−J(u)◦1
(z−i)2
Given that Jis the standard almost complex structure on C, i.e., J(z) = iz, we have:
J(u) = iz+i
z−i=i(z+i)
z−i
Therefore, ¯
∂Ju=−i
2(z−i)3. Since ¯
∂Ju+J(u)◦∂¯zu= 0, the function u(z) = z+i
z−iis indeed a
pseudoholomorphic curve in C.
21 21. STABILITY CONDITIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 21. Consider a pseudoholomorphic curve u: Σ →Mwhere Σis a Riemann surface
and Mis a symplectic manifold with symplectic form ω. Let Jbe an almost complex structure on
Mcompatible with ω. Suppose the pseudoholomorphic curve usatisfies the following Cauchy-
Riemann equation:
¯
∂Ju= 0.
a) Show that the linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J.
b) Calculate the Fredholm index of the operator Du.
c) Determine the stability condition for the pseudoholomorphic curve u.
Solution 21.
a) The linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J. This operator acts on sections of the pullback bundle u∗T M,
which is the tangent bundle of the Riemann surface Σpulled back to Mvia the map u.
b) The Fredholm index of the operator Duis given by
index(Du) = dim ker(Du)−dim coker(Du).
Here, ker(Du)represents the kernel of Du(space of solutions to ¯
∂Jv= 0 near u) and coker(Du)
represents the cokernel of Du. The Fredholm index indicates the deformation parameter for the
moduli space of pseudoholomorphic curves.
c) The stability condition for the pseudoholomorphic curve uis satisfied if the Fredholm index
of the linearized operator Duis nonnegative, i.e., index(Du)≥0. This condition ensures that
the moduli space of pseudoholomorphic curves is well-behaved under small perturbations in the
complex structure or metric on M.
22 22. FREDHOLM THEORY AND PSEUDOHOLOMORPHIC CURVES
Problem 22. Consider the following pseudoholomorphic curve equation in C2:
¯
∂u = 0
where u:R×S1→C2is a smooth map.
Given u(t, z) = eit(z, iz)for t∈[0,2π]and z∈S1, compute the Fredholm index of this pseudo-
holomorphic curve.
Solution 22. To compute the Fredholm index, we first need to find the linearized operator asso-
ciated with ¯
∂at u. Let v(t, z)be a small perturbation around u, and write v(t, z) = eit(a(t, z), b(t, z)).
The linearized operator is given by:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb
Plugging in u(t, z)and v(t, z)into the expression and simplifying, we get:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb=eit(∂ta−i∂tb+∂za−i∂zb)
This is equal to 0if and only if both ∂ta−i∂tband ∂za−i∂zbare equal to 0.
From u(t, z), we have a(t, z) = zand b(t, z) = iz. Differentiating with respect to tand z, we get:
∂ta= 0, ∂tb=−z, ∂za= 1, ∂zb=i
So ∂ta−i∂tb= 0 and ∂za−i∂zb= 1 −i= 0. Thus, the Fredholm index is 0−2 = −2.
23 The Theory of Pseudoholomorphic Curves
Problem: Consider the following pseudoholomorphic curve equation in the complex plane C:
¯
∂Ju+i∂tu= 0
where u:C×[0,1] →Cis a smooth map, ∂tdenotes the partial derivative with respect to
t∈[0,1], and ¯
∂Jdenotes the ¯
∂-operator twisted by a compatible almost complex structure Jon C.
Given the initial condition u(x, 0) = x2for x∈C, solve the pseudoholomorphic curve equation.
Solution:
To solve the pseudoholomorphic curve equation, we will first compute the derivatives and then
use the initial condition to determine the solution.
Let u(x, t) = u1(x, t) + iv1(x, t), where u1and v1are real-valued functions. Then the pseudo-
holomorphic curve equation becomes:
∂u1
∂t −∂v1
∂x +i∂v1
∂t +∂u1
∂x = 0
Separating the real and imaginary parts, we get:
∂u1
∂t =∂v1
∂x ,∂v1
∂t =−∂u1
∂x
From the initial condition u(x, 0) = x2, we have u1(x, 0) = x2and v1(x, 0) = 0.
Solving the partial differential equations with these initial conditions gives u1(x, t) = x2and
v1(x, t)=0. Therefore, the solution of the pseudoholomorphic curve equation is u(x, t) = x2+i·0 =
x2.
24 24. HOMOLOGICAL ALGEBRA AND PSEUDOHOLOMORPHIC CURVES
Problem 24. Consider the following pseudoholomorphic curve in CP 1given by the equation
u(z) = z2.
a) Determine the differential operator ∂Jassociated with the almost complex structure Jinduced
by the Fubini-Study metric on CP 1.
b) Show that u(z)is a solution to the ∂Jequation on CP 1.
Solution 24.
a) The almost complex structure Jassociated with the Fubini-Study metric on CP 1is given by
J(z) = −iz. The differential operator ∂Jcan be computed as:
∂J=1
2(J+iId) = 1
2(−iz +iId) = 1
2(i(z+ 1)).
b) To show that u(z) = z2is a solution to the ∂Jequation, we need to demonstrate that ∂Ju(z) =
0. Let’s compute it:
∂Ju(z) = 1
2(i(z2+ 1)) = i
2z2+i
2=iz +i
2= 0.
As the result is not zero, the function u(z) = z2is not a solution to the ∂Jequation on CP 1.
25 The Theory of Pseudoholomorphic Curves
Problem 1. Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = eis +t2.
a) Determine the asymptotic behavior of uas s→ −∞.
b) Find the limit lims→∞ |u(s, 1)|.
c) Show that uis a pseudoholomorphic curve.
Solution 1.
a) As s→ −∞, the term eis behaves as oscillatory and does not affect the growth, so we focus
on t2. Thus, the asymptotic behavior of u(s, t)as s→ −∞ is t2.
b) We have |u(s, 1)|=|eis + 1| ≤ |eis|+|1|= 1 + 1 = 2. Thus, lims→∞ |u(s, 1)| ≤ 2.
c) To show that uis a pseudoholomorphic curve, we need to verify that ¯
∂Ju= 0, where Jis the
standard complex structure on C. Compute ∂su=ieis and ∂tu= 2t. Then, ¯
∂Ju=∂su−i∂tu=
ieis −2it.
Since ¯
∂Ju= 0, we have shown that uis a pseudoholomorphic curve.
Therefore, the solution to the pseudoholomorphic curve equation with the specified Riemann
surface and almost complex structure is u(z) = az +b, but it is not J-holomorphic.
3 3. INDEX THEORY FOR PSEUDOHOLOMORPHIC CURVES
Problem 3. Consider a compact, oriented, Riemannian 2-manifold Mwith a compatible almost
complex structure Jand a symplectic form ω. Let u:C→Mbe a nonconstant J-holomorphic
map. Given that the Fredholm index ind(Du)of uis 2, with Dudenoting the linearized Cauchy-
Riemann operator associated to u, determine the total number of positive and negative punctures
of u.
Solution 3.
Since the index of uis 2, we know that the Conley-Zehnder index of each positive puncture of
uis −1and each negative puncture has a Conley-Zehnder index of 1.
Let n+and n−denote the total number of positive and negative punctures of u, respectively.
Then, we have the equation
ind(Du)=2=2−n++n−
Substitute the Conley-Zehnder index values into the equation, we get
2=2−n++n−
0 = −n++n−
n+=n−
Therefore, the total number of positive punctures is equal to the total number of negative punc-
tures in this case.
4 4. EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 4. Consider the following holomorphic curve in C:u:C→Cgiven by u(z) = z2−i.
We want to find the solutions to the Cauchy-Riemann equation ¯
∂Ju= 0.
Solution 4.
a) We start by writing u(z) = u(x1+ix2) = u1(x1, x2) + iu2(x1, x2)where u1, u2:R2→R.
We then compute the Cauchy-Riemann equations:
∂u1
∂x1
=∂u2
∂x2
and ∂u1
∂x2
=−∂u2
∂x1
In this case, u(z) = z2−i= (x1+ix2)2−i= (x2
1−x2
2) + i(2x1x2−1).
Therefore, we have:
u1(x1, x2) = x2
1−x2
2and u2(x1, x2)=2x1x2−1
Calculating the partial derivatives, we find:
∂u1
∂x1
= 2x1and ∂u2
∂x2
= 2x1
∂u1
∂x2
=−2x2and ∂u2
∂x1
= 2x2
Now we plug these into the Cauchy-Riemann equations:
2x1= 2x1and −2x2= 2x2
Solving the above equations, we obtain x1= 0 and x2= 0.
Therefore, the holomorphic curve u(z) = z2−isatisfies the Cauchy-Riemann equations at the
point (0,0).
b) To find other solutions, we can look for constant maps. Let u(z) = cwhere cis a complex
constant. Then, ¯
∂Ju= 0 is automatically satisfied.
c) Another way to find solutions is by looking at the composition of holomorphic functions. If
v:C→Cand u:C→Care holomorphic, then u◦vis also holomorphic.
5 5. FLOER HOMOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 5. Consider the following pseudoholomorphic curve in Cparametrized by z(t) = eit,
where t∈[0,2π]:
u:R×S1→C, u(s, eit) = eiseit.
a) Show that this curve satisfies the Cauchy-Riemann equations.
b) Calculate the energy of the curve.
c) Determine the image of the curve in C.
Solution 5.
a) To show that the curve satisfies the Cauchy-Riemann equations, we need to check if ∂u
∂¯z= 0.
Here, z=x+iy =x+iy(t), so ∂u
∂¯z=1
2(∂u
∂x +i∂u
∂y ) = 1
2(eiseit −ieiseit) = 0, hence the curve satisfies
the Cauchy-Riemann equations.
b) The energy of the curve is calculated using the formula:
E(u) = ZS1||∇u||2dt,
where ||∇u||2is the norm of the gradient of u. In this case, ∇u= (∂su, ∂tu)=(iu(s, eit), eisieit),
so ||∇u||2=|iu(s, eit)|2+|eisieit|2=|i|2+ 1 = 2. Therefore, the energy of the curve is:
E(u) = Z2π
0
2dt = 4π.
c) The image of the curve in Cis given by u(R×S1) = {eiseit |s∈R, t ∈[0,2π]}. Since eis
and eit vary over all complex numbers on the unit circle S1, the image is the unit circle S1itself.
6 6. COMPACTNESS PROPERTIES OF PSEUDOHOLOMORPHIC CURVES
Problem 6. Consider a sequence of pseudoholomorphic curves (uk: Σ →M, Jk)converging
to a constant map, where Σis a Riemann surface and Mis a symplectic manifold.
Let Jkbe a sequence of almost complex structures converging to J∞in the C∞-topology on
Σ. Assume that the energy of the curves E(uk)is uniformly bounded. Show that the sequence uk
converges in the C∞-topology on Σto a constant map u∞.
Solution 6.
Given that the energy of the curves E(uk)is uniformly bounded, we have
E(uk) = ZΣ||duk◦j−Jk◦duk||2dµ
≤C
where Cis a constant independent of k.
By the compactness theorem for Jk, there exists a subsequence (not relabeled) converging to
aJ∞-holomorphic map u∞. This gives us convergence in the C∞-topology for a subsequence.
Now we want to show that the entire sequence converges to u∞. By contradiction, assume
there exists ϵ > 0such that for all Nthere exists kN> N such that ||dukN−du∞||C∞≥ϵ. But this
is a contradiction, as we have a compact set in C∞-topology.
Therefore, the entire sequence ukconverges in the C∞-topology to the constant map u∞.
7 7. GROMOV-WITTEN INVARIANTS AND PSEUDOHOLOMORPHIC CURVES
Problem 7. Consider the moduli space of genus-zero, three-pointed stable pseudoholomor-
phic curves on a Riemann surface of genus one, with the three marked points labeled p1, p2, p3.
Suppose the homology classes of the curves are A= 2α+ 3βand B=−α+ 4β.
a) Calculate the dimension of the moduli space.
b) Determine the number of points in the moduli space that satisfy the conditions given in part
(a).
c) Suppose the intersection number of Aand Bis 1. Find the expected dimension of the moduli
space.
Solution 7.
a) The dimension of the moduli space can be calculated using the Riemann-Roch formula:
dim(M0,3(Σ, A)) = 3d+ 1 −g= 3(2 + 3) + 1 −1 = 10.
b) In this case, the moduli space contains finitely many points due to the constraint on the
homology classes. Hence, there are no points in the moduli space.
c) The expected dimension of the moduli space can be calculated using the formula:
dim(M0,3(Σ, A;B)) = dim(M0,3(Σ, A)) − ⟨A, B⟩= 10 −1 = 9.
Therefore, the expected dimension of the moduli space is 9.
8 8. EVALUATION MAPS FOR PSEUDOHOLOMORPHIC CURVES
Problem 8. Consider the pseudoholomorphic curve equation on a Riemann surface Sgiven by
du +idv = 0, where u, v :S→Care smooth functions. Let z=u+iv be a complex coordinate on
S. Given a pseudoholomorphic curve u=f(z)parametrized by f(z) = z2, evaluate the following:
a) The evaluation map evp:M → S, where Mis the moduli space of solutions to the pseudo-
holomorphic curve equation, at p= 2 + 3i.
b) The derivative d
dz f(z)at z= 1 −i.
c) The image under the map fof the curve g(z) = z3.
Solution 8.
a) To compute the evaluation map at p= 2+3i, we substitute z= 2+3iinto the parametrization
f(z) = z2. Therefore, f(2 + 3i) = (2 + 3i)2= 4 + 12i−9 = −5 + 12i. Hence, ev2+3i(f) = −5 + 12i.
b) To find the derivative d
dz f(z)at z= 1 −i, we differentiate f(z) = z2to get d
dz f(z) = 2z.
Evaluating this derivative at z= 1 −i, we have d
dz f(1 −i) = 2(1 −i)=2−2i.
c) The image under the map fof the curve g(z) = z3is given by f(g(z)) = f(z3)=(z3)2=z6.
Therefore, the image of the curve g(z)under the map fis the curve parametrized by z6.
9 9. MODULI SPACE OF PSEUDOHOLOMORPHIC CURVES
Problem 9. Consider a Riemann surface Mwith genus g= 2 and n= 3 marked points. Let Xbe
a compact oriented Riemann surface of genus 0 with 5 boundary marked points. Denote by Athe
moduli space of smooth maps ¯
M→Xsatisfying certain boundary conditions.
a) Calculate the dimension of the space A.
b) Determine the dimension of the moduli space of pseudo-holomorphic curves in CP 2with
k= 2 marked points.
Solution 9.
a) The dimension of the moduli space Acan be calculated using Riemann-Roch formula. We
know that for a Riemann surface of genus gwith nmarked points, the dimension of the moduli
space is given by:
dim(M)=6g−6+2n
Substitute g= 2 and n= 3 into the formula:
dim(A) = 6 ×2−6+2×3 = 12 −6 + 6 = 12
Therefore, the dimension of the moduli space Ais 12.
b) The dimension of the moduli space of pseudo-holomorphic curves in CP 2with kmarked
points is given by the formula:
dim M= 2k−6+2g
Substitute k= 2 and g= 0 into the formula:
dim M= 2 ×2−6+2×0=4−6 = −2
Therefore, the dimension of the moduli space of pseudo-holomorphic curves in CP 2with 2
marked points is -2.
10 9. THE THEORY OF PSEUDOHOLMORPHIC CURVES
Problem 9. Consider the following pseudoholomorphic curve in CP 1given by the map u:
R×S1→CP 1, where S1is the unit circle in C:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)].
a) Calculate the image of the curve u.
b) Determine the intersection number of the curve uwith the horizontal line [z: 1], where z∈C.
c) Find the set of critical points of the map u.
Solution 9.
a) To find the image of the curve u, we need to substitute the values of sand θinto the map u.
Let’s first express u(s, eiθ)in terms of sand θ:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)] .
Substitute s= 0 and θ=π/2into uto get the image point:
u(0, eiπ/2) = [1 : 2i].
Therefore, the image of the curve uis the point [1 : 2i]in CP 1.
b) To find the intersection number of the curve uwith the horizontal line [z: 1], we need to count
the number of solutions to the equation 2isin(θ)−icos(θ) = z(1 + is). This reduces to finding the
number of solutions to sin(θ)=0:
sin(θ) = 0 =⇒θ=nπ, where n∈Z.
Since θis periodic with period 2π, there are infinitely many intersection points. Hence, the
intersection number is ∞.
c) The critical points of uare the (s, eiθ)such that the derivative of uvanishes. Calculating the
derivative: ∂u
∂s =i[1 : 2i],∂u
∂θ =i[0 : 2 cos(θ) + sin(θ)].
Setting the derivatives equal to zero, we find that sis not involved. Thus, the critical points lie
on the horizontal line.
11 Numeric Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following Riemann surface defined by the equation z=eiθ, where
θ∈[0,2π].
a) Calculate the area of this Riemann surface.
b) Find the length of the curve defined by x=Re(z)and y=Im(z).
Solution 1.
a) To calculate the area of the Riemann surface, we can use the formula for the area of a surface
of revolution, which is given by A= 2πRβ
αf(θ)p1+(f′(θ))2dθ.
In this case, f(θ)=1and f′(θ)=0, so the area simplifies to:
A= 2πR2π
01·√1+0dθ = 2πR2π
01dθ = 2π[2π−0] = 4π2.
Therefore, the area of the Riemann surface is 4π2.
b) The length of the curve is given by the formula L=Rβ
αrdx
dθ 2+dy
dθ 2dθ.
Since x= cos(θ)and y= sin(θ), we have dx
dθ =−sin(θ)and dx
dθ = cos(θ). Therefore, the length
simplifies to:
L=R2π
0p(−sin(θ))2+ (cos(θ))2dθ =R2π
0√1dθ =R2π
01dθ = 2π.
The length of the curve is 2π.
12 12. BUBBLING PHENOMENON IN PSEUDOHOLOMORPHIC CURVES
Problem 12. Consider the sequence of pseudoholomorphic curves defined by un:R×S2→C
given by un(x, z) = eiθnzn, where θn→0as n→ ∞. Determine whether this sequence exhibits
bubbling phenomenon or not.
Solution 12. a) To analyze the behavior of this sequence of pseudoholomorphic curves, let’s
first calculate the energy of each curve. The energy functional for a pseudoholomorphic curve
u: Σ →Cdefined over a Riemann surface Σis given by
E(u) = ZΣ||du||2dA,
where ||du||2is the square of the norm of the differential du and dA is the area element on Σ.
For the curve un(x, z) = eiθnzn, the energy can be calculated as follows:
||dun||2=|eiθnnzn−1|2=n2|z|2n−2,
and integrating over S2(with the standard metric) gives
E(un) = ZS2
n2|z|2n−2dA = 4πn2.
b) Next, let’s examine the behavior of the energy of the sequence as n→ ∞. We have
lim
n→∞ E(un) = lim
n→∞ 4πn2=∞.
c) Since the energy of the sequence of curves diverges as n→ ∞, we conclude that this
sequence exhibits bubbling phenomenon. In this case, the limiting pseudoholomorphic curve will
be a multiply-covered sphere.
Therefore, the sequence of pseudoholomorphic curves defined by un(x, z) = eiθnznexhibits
bubbling phenomenon.
13 13. SYMPLECTIC TOPOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 13. Consider the complex structure Jon R2given by J(x, y)=(−y, x)and the
symplectic form ω=dx ∧dy.
a) Let u(x, y)=(x2−y2,2xy)be a map R2→R2. Determine if uis pseudoholomorphic with
respect to (J, ω).
b) Find the image of the map ufrom part (a).
Solution 13.
a) To determine if uis pseudoholomorphic with respect to (J, ω), we need to check if du +J◦
du ◦J=ω.
Compute du =2x−2y
2y2xand J◦du ◦J=−2y2x
2x2y.
Adding them gives du +J◦du ◦J=0 0
0 0=ω=dx ∧dy.
Since du +J◦du ◦J=ω,uis not pseudoholomorphic.
b) The image of ucan be found by solving the following system of equations:
x2−y2=u1
2xy =u2
Substitute u1=x2−y2into u2= 2xy to get 2x(x2−y2) = u2.
Solving both equations simultaneously gives us x= 0 or x2=y2, which are the equations for
the coordinate axes.
Therefore, the image of the map uis the xand yaxes.
14 14. CAUCHY-RIEMANN EQUATIONS AND PSEUDOHOLOMORPHIC CURVES
Problem 14. Consider the pseudoholomorphic curve u:C→Cdefined by u(z) = z2+i,
where z=x+iy and i=√−1.
a) Find the Cauchy-Riemann equations for the map u.
b) Determine if the map uis pseudoholomorphic.
Solution 14.
a) To find the Cauchy-Riemann equations for u(z) = z2+i, we first express u(z)in terms of x
and y:
u(z)=(x+iy)2+i=x2−y2+ 2ixy +i
Now, we write u(z)as u(x, y) + iv(x, y):
u(x, y) = x2−y2, v(x, y)=2xy + 1
Next, we find the partial derivatives of uand v:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
The Cauchy-Riemann equations are:
∂u
∂x =∂v
∂y and ∂u
∂y =−∂v
∂x
Substitute the partial derivatives we found into these equations to check if they are satisfied.
b) To determine if the map uis pseudoholomorphic, we need to verify if the Cauchy-Riemann
equations hold for u. From part a, we found that:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
As the Cauchy-Riemann equations are satisfied, the map uis pseudoholomorphic.
15 15. GRADIENT FLOW TECHNIQUES FOR PSEUDOHOLOMORPHIC CURVES
Problem 15. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = z2+iz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 1.
Solution 15. The solution to the pseudoholomorphic curve equation is u(t, z) = u(0, z)for all
t, due to the stationary nature of pseudoholomorphic curves. Hence, u(1, z) = u(0, z) = z2+iz.
Therefore, the solution at time t= 1 is u(1, z) = z2+iz for all z∈Σin this case.
Problem 16. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = eiz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 2.
Solution 16. Similar to the previous problem, the solution to the pseudoholomorphic curve
equation is u(t, z) = u(0, z)for all t, due to the stationary nature of pseudoholomorphic curves.
Hence, u(2, z) = u(0, z) = eiz.
Therefore, the solution at time t= 2 is u(2, z) = eiz for all z∈Σin this case.
16 16. TOPOLOGICAL AND GEOMETRIC CONSTRAINTS ON PSEUDOHOLOMORPHIC CURVES
Problem 16. Consider a closed oriented manifold Mof dimension 4, equipped with a compati-
ble almost complex structure J. Let [γ]be a homology class represented by a simple closed curve
γin M. We define the self-intersection number of [γ]as I([γ],[γ]) = γ·γ.
Suppose Σis a pseudoholomorphic curve in Mwith boundary ∂Σ = γ1−γ2, where γ1and γ2
are simple closed curves representing different homology classes in M. Determine the relationship
between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]).
Solution 16. Let’s denote the two simple closed curves γ1and γ2as Aand B, respectively.
Then, we have I([γ1],[γ1]) = A·Aand I([γ2],[γ2]) = B·B, where ·denotes the intersection pairing.
The self-intersection number of the curve Σcan be computed as I(Σ,Σ) = (γ1−γ2)·(γ1−γ2).
Expanding this out, we get:
I(Σ,Σ) = γ1·γ1−γ1·γ2−γ2·γ1+γ2·γ2
=A·A−A·B−B·A+B·B
=A·A−2A·B+B·B
=I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
Therefore, the relationship between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]) is given by:
I(Σ,Σ) = I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
17 Numerical Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in CP 1:
∂u
∂z +λe−u= 0,
where λ∈Ris a constant.
a) Find the general solution u(z, z)to this pseudoholomorphic curve equation.
b) Given λ= 2, solve the equation with the initial condition u(0,0) = 0.
Solution 1.
a) To find the general solution to the pseudoholomorphic curve equation, we first rewrite the
equation in terms of the complex coordinates zand z. Note that ∂u
∂z =1
2∂u
∂x +i∂u
∂y , where z=
x+iy.
Substitute this into the given equation:
1
2∂u
∂x +i∂u
∂y +λe−u= 0.
Separating real and imaginary parts of this equation gives us:
∂u
∂x = 2λe−u
∂u
∂y = 0
Integrating the first equation with respect to xand the second equation with respect to ygives:
u= ln C
2λ+e2λx,
where Cis the constant of integration.
b) Given λ= 2 and the initial condition u(0,0) = 0, we can find the value of Cby substituting
x= 0, u = 0 into the general solution:
0 = ln C
4+ 1.
Solving this equation for Cgives C= 4, so the solution to the equation with λ= 2 and the initial
condition is:
u= ln 4
4+e4x= ln(1 + e4x).
18 The Theory of Pseudoholomorphic Curves
18.1 Problem 1
Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = aeis +bt with a, b ∈C
and |a|= 3,|b|= 2. Determine the image of this pseudoholomorphic curve in C.
Solution: The image of uis given by u(s, t) = aeis +bt. To find the image, we set u(s, t) = z
and solve for sand t:
z=aeis +bt
= 3eis + 2t
= 3(cos s+isin s)+2t.
This implies that the image of the pseudoholomorphic curve uis the set of points z= 3 cos s+
i3 sin s+ 2tin the complex plane C.
18.2 Problem 2
Let u:R×[0,1] →Cbe a pseudoholomorphic curve defined by u(s, t) = s2+it. Determine the
differential of u.
Solution: The differential of uis given by du =∂u
∂s ds +∂u
∂t dt. In this case, we have:
∂u
∂s = 2s,
∂u
∂t =i.
Therefore, the differential of uis du = 2s ds +i dt.
19 19. ENERGY ESTIMATES FOR PSEUDOHOLOMORPHIC CURVES
Problem 19. Consider a pseudoholomorphic curve u:R×S1→Cgiven by u(s, t) = eis sin(t),
where s, t ∈R. Calculate the energy of this pseudoholomorphic curve using the energy functional
E(u) = RR×S1||∂su||2+||∂tu||2dsdt.
Solution 19. The energy of the pseudoholomorphic curve ucan be calculated as follows:
a) First, we calculate the partial derivatives of u:
∂su=ieis sin(t),
∂tu=eis cos(t).
b) Next, we compute the norms of the partial derivatives:
||∂su||2=|ieis sin(t)|2=| − eis sin(t)|2=|eis|2|sin(t)|2= sin2(t),
||∂tu||2=|eis cos(t)|2=|eis||cos(t)|2= 1|cos(t)|2= cos2(t).
c) Now, we substitute the norms into the energy functional and integrate over R×S1:
E(u) = ZR×S1
(sin2(t) + cos2(t)) dsdt
=ZRZS1
1dsdt
=ZR
2π ds
= 2π.
Therefore, the energy of the pseudoholomorphic curve uis 2π.
20 The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in a symplectic mani-
fold:
¯
∂Ju+J(u)◦∂¯zu= 0
where u:C→Mis a holomorphic map, Jis a compatible almost complex structure in M, and
¯
∂Jdenotes the Cauchy-Riemann operator.
Given u(z) = z+i
z−i, determine if uis a pseudoholomorphic curve in the complex plane C.
Solution 1.
To determine if u(z) = z+i
z−iis a pseudoholomorphic curve, we need to check if it satisfies the
pseudoholomorphic curve equation. First, let’s calculate ∂zuand ∂¯zu:
∂zu=1
(z−i)2
∂¯zu= 0
Next, let’s compute ¯
∂Ju:
¯
∂Ju=1
2(∂¯zu−J(u)◦∂zu) = 1
20−J(u)◦1
(z−i)2
Given that Jis the standard almost complex structure on C, i.e., J(z) = iz, we have:
J(u) = iz+i
z−i=i(z+i)
z−i
Therefore, ¯
∂Ju=−i
2(z−i)3. Since ¯
∂Ju+J(u)◦∂¯zu= 0, the function u(z) = z+i
z−iis indeed a
pseudoholomorphic curve in C.
21 21. STABILITY CONDITIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 21. Consider a pseudoholomorphic curve u: Σ →Mwhere Σis a Riemann surface
and Mis a symplectic manifold with symplectic form ω. Let Jbe an almost complex structure on
Mcompatible with ω. Suppose the pseudoholomorphic curve usatisfies the following Cauchy-
Riemann equation:
¯
∂Ju= 0.
a) Show that the linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J.
b) Calculate the Fredholm index of the operator Du.
c) Determine the stability condition for the pseudoholomorphic curve u.
Solution 21.
a) The linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J. This operator acts on sections of the pullback bundle u∗T M,
which is the tangent bundle of the Riemann surface Σpulled back to Mvia the map u.
b) The Fredholm index of the operator Duis given by
index(Du) = dim ker(Du)−dim coker(Du).
Here, ker(Du)represents the kernel of Du(space of solutions to ¯
∂Jv= 0 near u) and coker(Du)
represents the cokernel of Du. The Fredholm index indicates the deformation parameter for the
moduli space of pseudoholomorphic curves.
c) The stability condition for the pseudoholomorphic curve uis satisfied if the Fredholm index
of the linearized operator Duis nonnegative, i.e., index(Du)≥0. This condition ensures that
the moduli space of pseudoholomorphic curves is well-behaved under small perturbations in the
complex structure or metric on M.
22 22. FREDHOLM THEORY AND PSEUDOHOLOMORPHIC CURVES
Problem 22. Consider the following pseudoholomorphic curve equation in C2:
¯
∂u = 0
where u:R×S1→C2is a smooth map.
Given u(t, z) = eit(z, iz)for t∈[0,2π]and z∈S1, compute the Fredholm index of this pseudo-
holomorphic curve.
Solution 22. To compute the Fredholm index, we first need to find the linearized operator asso-
ciated with ¯
∂at u. Let v(t, z)be a small perturbation around u, and write v(t, z) = eit(a(t, z), b(t, z)).
The linearized operator is given by:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb
Plugging in u(t, z)and v(t, z)into the expression and simplifying, we get:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb=eit(∂ta−i∂tb+∂za−i∂zb)
This is equal to 0if and only if both ∂ta−i∂tband ∂za−i∂zbare equal to 0.
From u(t, z), we have a(t, z) = zand b(t, z) = iz. Differentiating with respect to tand z, we get:
∂ta= 0, ∂tb=−z, ∂za= 1, ∂zb=i
So ∂ta−i∂tb= 0 and ∂za−i∂zb= 1 −i= 0. Thus, the Fredholm index is 0−2 = −2.
23 The Theory of Pseudoholomorphic Curves
Problem: Consider the following pseudoholomorphic curve equation in the complex plane C:
¯
∂Ju+i∂tu= 0
where u:C×[0,1] →Cis a smooth map, ∂tdenotes the partial derivative with respect to
t∈[0,1], and ¯
∂Jdenotes the ¯
∂-operator twisted by a compatible almost complex structure Jon C.
Given the initial condition u(x, 0) = x2for x∈C, solve the pseudoholomorphic curve equation.
Solution:
To solve the pseudoholomorphic curve equation, we will first compute the derivatives and then
use the initial condition to determine the solution.
Let u(x, t) = u1(x, t) + iv1(x, t), where u1and v1are real-valued functions. Then the pseudo-
holomorphic curve equation becomes:
∂u1
∂t −∂v1
∂x +i∂v1
∂t +∂u1
∂x = 0
Separating the real and imaginary parts, we get:
∂u1
∂t =∂v1
∂x ,∂v1
∂t =−∂u1
∂x
From the initial condition u(x, 0) = x2, we have u1(x, 0) = x2and v1(x, 0) = 0.
Solving the partial differential equations with these initial conditions gives u1(x, t) = x2and
v1(x, t)=0. Therefore, the solution of the pseudoholomorphic curve equation is u(x, t) = x2+i·0 =
x2.
24 24. HOMOLOGICAL ALGEBRA AND PSEUDOHOLOMORPHIC CURVES
Problem 24. Consider the following pseudoholomorphic curve in CP 1given by the equation
u(z) = z2.
a) Determine the differential operator ∂Jassociated with the almost complex structure Jinduced
by the Fubini-Study metric on CP 1.
b) Show that u(z)is a solution to the ∂Jequation on CP 1.
Solution 24.
a) The almost complex structure Jassociated with the Fubini-Study metric on CP 1is given by
J(z) = −iz. The differential operator ∂Jcan be computed as:
∂J=1
2(J+iId) = 1
2(−iz +iId) = 1
2(i(z+ 1)).
b) To show that u(z) = z2is a solution to the ∂Jequation, we need to demonstrate that ∂Ju(z) =
0. Let’s compute it:
∂Ju(z) = 1
2(i(z2+ 1)) = i
2z2+i
2=iz +i
2= 0.
As the result is not zero, the function u(z) = z2is not a solution to the ∂Jequation on CP 1.
25 The Theory of Pseudoholomorphic Curves
Problem 1. Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = eis +t2.
a) Determine the asymptotic behavior of uas s→ −∞.
b) Find the limit lims→∞ |u(s, 1)|.
c) Show that uis a pseudoholomorphic curve.
Solution 1.
a) As s→ −∞, the term eis behaves as oscillatory and does not affect the growth, so we focus
on t2. Thus, the asymptotic behavior of u(s, t)as s→ −∞ is t2.
b) We have |u(s, 1)|=|eis + 1| ≤ |eis|+|1|= 1 + 1 = 2. Thus, lims→∞ |u(s, 1)| ≤ 2.
c) To show that uis a pseudoholomorphic curve, we need to verify that ¯
∂Ju= 0, where Jis the
standard complex structure on C. Compute ∂su=ieis and ∂tu= 2t. Then, ¯
∂Ju=∂su−i∂tu=
ieis −2it.
Since ¯
∂Ju= 0, we have shown that uis a pseudoholomorphic curve.
Therefore, the solution to the pseudoholomorphic curve equation with the specified Riemann
surface and almost complex structure is u(z) = az +b, but it is not J-holomorphic.
3 3. INDEX THEORY FOR PSEUDOHOLOMORPHIC CURVES
Problem 3. Consider a compact, oriented, Riemannian 2-manifold Mwith a compatible almost
complex structure Jand a symplectic form ω. Let u:C→Mbe a nonconstant J-holomorphic
map. Given that the Fredholm index ind(Du)of uis 2, with Dudenoting the linearized Cauchy-
Riemann operator associated to u, determine the total number of positive and negative punctures
of u.
Solution 3.
Since the index of uis 2, we know that the Conley-Zehnder index of each positive puncture of
uis −1and each negative puncture has a Conley-Zehnder index of 1.
Let n+and n−denote the total number of positive and negative punctures of u, respectively.
Then, we have the equation
ind(Du)=2=2−n++n−
Substitute the Conley-Zehnder index values into the equation, we get
2=2−n++n−
0 = −n++n−
n+=n−
Therefore, the total number of positive punctures is equal to the total number of negative punc-
tures in this case.
4 4. EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 4. Consider the following holomorphic curve in C:u:C→Cgiven by u(z) = z2−i.
We want to find the solutions to the Cauchy-Riemann equation ¯
∂Ju= 0.
Solution 4.
a) We start by writing u(z) = u(x1+ix2) = u1(x1, x2) + iu2(x1, x2)where u1, u2:R2→R.
We then compute the Cauchy-Riemann equations:
∂u1
∂x1
=∂u2
∂x2
and ∂u1
∂x2
=−∂u2
∂x1
In this case, u(z) = z2−i= (x1+ix2)2−i= (x2
1−x2
2) + i(2x1x2−1).
Therefore, we have:
u1(x1, x2) = x2
1−x2
2and u2(x1, x2)=2x1x2−1
Calculating the partial derivatives, we find:
∂u1
∂x1
= 2x1and ∂u2
∂x2
= 2x1
∂u1
∂x2
=−2x2and ∂u2
∂x1
= 2x2
Now we plug these into the Cauchy-Riemann equations:
2x1= 2x1and −2x2= 2x2
Solving the above equations, we obtain x1= 0 and x2= 0.
Therefore, the holomorphic curve u(z) = z2−isatisfies the Cauchy-Riemann equations at the
point (0,0).
b) To find other solutions, we can look for constant maps. Let u(z) = cwhere cis a complex
constant. Then, ¯
∂Ju= 0 is automatically satisfied.
c) Another way to find solutions is by looking at the composition of holomorphic functions. If
v:C→Cand u:C→Care holomorphic, then u◦vis also holomorphic.
5 5. FLOER HOMOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 5. Consider the following pseudoholomorphic curve in Cparametrized by z(t) = eit,
where t∈[0,2π]:
u:R×S1→C, u(s, eit) = eiseit.
a) Show that this curve satisfies the Cauchy-Riemann equations.
b) Calculate the energy of the curve.
c) Determine the image of the curve in C.
Solution 5.
a) To show that the curve satisfies the Cauchy-Riemann equations, we need to check if ∂u
∂¯z= 0.
Here, z=x+iy =x+iy(t), so ∂u
∂¯z=1
2(∂u
∂x +i∂u
∂y ) = 1
2(eiseit −ieiseit) = 0, hence the curve satisfies
the Cauchy-Riemann equations.
b) The energy of the curve is calculated using the formula:
E(u) = ZS1||∇u||2dt,
where ||∇u||2is the norm of the gradient of u. In this case, ∇u= (∂su, ∂tu)=(iu(s, eit), eisieit),
so ||∇u||2=|iu(s, eit)|2+|eisieit|2=|i|2+ 1 = 2. Therefore, the energy of the curve is:
E(u) = Z2π
0
2dt = 4π.
c) The image of the curve in Cis given by u(R×S1) = {eiseit |s∈R, t ∈[0,2π]}. Since eis
and eit vary over all complex numbers on the unit circle S1, the image is the unit circle S1itself.
6 6. COMPACTNESS PROPERTIES OF PSEUDOHOLOMORPHIC CURVES
Problem 6. Consider a sequence of pseudoholomorphic curves (uk: Σ →M, Jk)converging
to a constant map, where Σis a Riemann surface and Mis a symplectic manifold.
Let Jkbe a sequence of almost complex structures converging to J∞in the C∞-topology on
Σ. Assume that the energy of the curves E(uk)is uniformly bounded. Show that the sequence uk
converges in the C∞-topology on Σto a constant map u∞.
Solution 6.
Given that the energy of the curves E(uk)is uniformly bounded, we have
E(uk) = ZΣ||duk◦j−Jk◦duk||2dµ
≤C
where Cis a constant independent of k.
By the compactness theorem for Jk, there exists a subsequence (not relabeled) converging to
aJ∞-holomorphic map u∞. This gives us convergence in the C∞-topology for a subsequence.
Now we want to show that the entire sequence converges to u∞. By contradiction, assume
there exists ϵ > 0such that for all Nthere exists kN> N such that ||dukN−du∞||C∞≥ϵ. But this
is a contradiction, as we have a compact set in C∞-topology.
Therefore, the entire sequence ukconverges in the C∞-topology to the constant map u∞.
7 7. GROMOV-WITTEN INVARIANTS AND PSEUDOHOLOMORPHIC CURVES
Problem 7. Consider the moduli space of genus-zero, three-pointed stable pseudoholomor-
phic curves on a Riemann surface of genus one, with the three marked points labeled p1, p2, p3.
Suppose the homology classes of the curves are A= 2α+ 3βand B=−α+ 4β.
a) Calculate the dimension of the moduli space.
b) Determine the number of points in the moduli space that satisfy the conditions given in part
(a).
c) Suppose the intersection number of Aand Bis 1. Find the expected dimension of the moduli
space.
Solution 7.
a) The dimension of the moduli space can be calculated using the Riemann-Roch formula:
dim(M0,3(Σ, A)) = 3d+ 1 −g= 3(2 + 3) + 1 −1 = 10.
b) In this case, the moduli space contains finitely many points due to the constraint on the
homology classes. Hence, there are no points in the moduli space.
c) The expected dimension of the moduli space can be calculated using the formula:
dim(M0,3(Σ, A;B)) = dim(M0,3(Σ, A)) − ⟨A, B⟩= 10 −1 = 9.
Therefore, the expected dimension of the moduli space is 9.
8 8. EVALUATION MAPS FOR PSEUDOHOLOMORPHIC CURVES
Problem 8. Consider the pseudoholomorphic curve equation on a Riemann surface Sgiven by
du +idv = 0, where u, v :S→Care smooth functions. Let z=u+iv be a complex coordinate on
S. Given a pseudoholomorphic curve u=f(z)parametrized by f(z) = z2, evaluate the following:
a) The evaluation map evp:M → S, where Mis the moduli space of solutions to the pseudo-
holomorphic curve equation, at p= 2 + 3i.
b) The derivative d
dz f(z)at z= 1 −i.
c) The image under the map fof the curve g(z) = z3.
Solution 8.
a) To compute the evaluation map at p= 2+3i, we substitute z= 2+3iinto the parametrization
f(z) = z2. Therefore, f(2 + 3i) = (2 + 3i)2= 4 + 12i−9 = −5 + 12i. Hence, ev2+3i(f) = −5 + 12i.
b) To find the derivative d
dz f(z)at z= 1 −i, we differentiate f(z) = z2to get d
dz f(z) = 2z.
Evaluating this derivative at z= 1 −i, we have d
dz f(1 −i) = 2(1 −i)=2−2i.
c) The image under the map fof the curve g(z) = z3is given by f(g(z)) = f(z3)=(z3)2=z6.
Therefore, the image of the curve g(z)under the map fis the curve parametrized by z6.
9 9. MODULI SPACE OF PSEUDOHOLOMORPHIC CURVES
Problem 9. Consider a Riemann surface Mwith genus g= 2 and n= 3 marked points. Let Xbe
a compact oriented Riemann surface of genus 0 with 5 boundary marked points. Denote by Athe
moduli space of smooth maps ¯
M→Xsatisfying certain boundary conditions.
a) Calculate the dimension of the space A.
b) Determine the dimension of the moduli space of pseudo-holomorphic curves in CP 2with
k= 2 marked points.
Solution 9.
a) The dimension of the moduli space Acan be calculated using Riemann-Roch formula. We
know that for a Riemann surface of genus gwith nmarked points, the dimension of the moduli
space is given by:
dim(M)=6g−6+2n
Substitute g= 2 and n= 3 into the formula:
dim(A) = 6 ×2−6+2×3 = 12 −6 + 6 = 12
Therefore, the dimension of the moduli space Ais 12.
b) The dimension of the moduli space of pseudo-holomorphic curves in CP 2with kmarked
points is given by the formula:
dim M= 2k−6+2g
Substitute k= 2 and g= 0 into the formula:
dim M= 2 ×2−6+2×0=4−6 = −2
Therefore, the dimension of the moduli space of pseudo-holomorphic curves in CP 2with 2
marked points is -2.
10 9. THE THEORY OF PSEUDOHOLMORPHIC CURVES
Problem 9. Consider the following pseudoholomorphic curve in CP 1given by the map u:
R×S1→CP 1, where S1is the unit circle in C:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)].
a) Calculate the image of the curve u.
b) Determine the intersection number of the curve uwith the horizontal line [z: 1], where z∈C.
c) Find the set of critical points of the map u.
Solution 9.
a) To find the image of the curve u, we need to substitute the values of sand θinto the map u.
Let’s first express u(s, eiθ)in terms of sand θ:
u(s, eiθ) = [1 + is : 2isin(θ)−icos(θ)] .
Substitute s= 0 and θ=π/2into uto get the image point:
u(0, eiπ/2) = [1 : 2i].
Therefore, the image of the curve uis the point [1 : 2i]in CP 1.
b) To find the intersection number of the curve uwith the horizontal line [z: 1], we need to count
the number of solutions to the equation 2isin(θ)−icos(θ) = z(1 + is). This reduces to finding the
number of solutions to sin(θ)=0:
sin(θ) = 0 =⇒θ=nπ, where n∈Z.
Since θis periodic with period 2π, there are infinitely many intersection points. Hence, the
intersection number is ∞.
c) The critical points of uare the (s, eiθ)such that the derivative of uvanishes. Calculating the
derivative: ∂u
∂s =i[1 : 2i],∂u
∂θ =i[0 : 2 cos(θ) + sin(θ)].
Setting the derivatives equal to zero, we find that sis not involved. Thus, the critical points lie
on the horizontal line.
11 Numeric Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following Riemann surface defined by the equation z=eiθ, where
θ∈[0,2π].
a) Calculate the area of this Riemann surface.
b) Find the length of the curve defined by x=Re(z)and y=Im(z).
Solution 1.
a) To calculate the area of the Riemann surface, we can use the formula for the area of a surface
of revolution, which is given by A= 2πRβ
αf(θ)p1+(f′(θ))2dθ.
In this case, f(θ)=1and f′(θ)=0, so the area simplifies to:
A= 2πR2π
01·√1+0dθ = 2πR2π
01dθ = 2π[2π−0] = 4π2.
Therefore, the area of the Riemann surface is 4π2.
b) The length of the curve is given by the formula L=Rβ
αrdx
dθ 2+dy
dθ 2dθ.
Since x= cos(θ)and y= sin(θ), we have dx
dθ =−sin(θ)and dx
dθ = cos(θ). Therefore, the length
simplifies to:
L=R2π
0p(−sin(θ))2+ (cos(θ))2dθ =R2π
0√1dθ =R2π
01dθ = 2π.
The length of the curve is 2π.
12 12. BUBBLING PHENOMENON IN PSEUDOHOLOMORPHIC CURVES
Problem 12. Consider the sequence of pseudoholomorphic curves defined by un:R×S2→C
given by un(x, z) = eiθnzn, where θn→0as n→ ∞. Determine whether this sequence exhibits
bubbling phenomenon or not.
Solution 12. a) To analyze the behavior of this sequence of pseudoholomorphic curves, let’s
first calculate the energy of each curve. The energy functional for a pseudoholomorphic curve
u: Σ →Cdefined over a Riemann surface Σis given by
E(u) = ZΣ||du||2dA,
where ||du||2is the square of the norm of the differential du and dA is the area element on Σ.
For the curve un(x, z) = eiθnzn, the energy can be calculated as follows:
||dun||2=|eiθnnzn−1|2=n2|z|2n−2,
and integrating over S2(with the standard metric) gives
E(un) = ZS2
n2|z|2n−2dA = 4πn2.
b) Next, let’s examine the behavior of the energy of the sequence as n→ ∞. We have
lim
n→∞ E(un) = lim
n→∞ 4πn2=∞.
c) Since the energy of the sequence of curves diverges as n→ ∞, we conclude that this
sequence exhibits bubbling phenomenon. In this case, the limiting pseudoholomorphic curve will
be a multiply-covered sphere.
Therefore, the sequence of pseudoholomorphic curves defined by un(x, z) = eiθnznexhibits
bubbling phenomenon.
13 13. SYMPLECTIC TOPOLOGY AND PSEUDOHOLOMORPHIC CURVES
Problem 13. Consider the complex structure Jon R2given by J(x, y)=(−y, x)and the
symplectic form ω=dx ∧dy.
a) Let u(x, y)=(x2−y2,2xy)be a map R2→R2. Determine if uis pseudoholomorphic with
respect to (J, ω).
b) Find the image of the map ufrom part (a).
Solution 13.
a) To determine if uis pseudoholomorphic with respect to (J, ω), we need to check if du +J◦
du ◦J=ω.
Compute du =2x−2y
2y2xand J◦du ◦J=−2y2x
2x2y.
Adding them gives du +J◦du ◦J=0 0
0 0=ω=dx ∧dy.
Since du +J◦du ◦J=ω,uis not pseudoholomorphic.
b) The image of ucan be found by solving the following system of equations:
x2−y2=u1
2xy =u2
Substitute u1=x2−y2into u2= 2xy to get 2x(x2−y2) = u2.
Solving both equations simultaneously gives us x= 0 or x2=y2, which are the equations for
the coordinate axes.
Therefore, the image of the map uis the xand yaxes.
14 14. CAUCHY-RIEMANN EQUATIONS AND PSEUDOHOLOMORPHIC CURVES
Problem 14. Consider the pseudoholomorphic curve u:C→Cdefined by u(z) = z2+i,
where z=x+iy and i=√−1.
a) Find the Cauchy-Riemann equations for the map u.
b) Determine if the map uis pseudoholomorphic.
Solution 14.
a) To find the Cauchy-Riemann equations for u(z) = z2+i, we first express u(z)in terms of x
and y:
u(z)=(x+iy)2+i=x2−y2+ 2ixy +i
Now, we write u(z)as u(x, y) + iv(x, y):
u(x, y) = x2−y2, v(x, y)=2xy + 1
Next, we find the partial derivatives of uand v:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
The Cauchy-Riemann equations are:
∂u
∂x =∂v
∂y and ∂u
∂y =−∂v
∂x
Substitute the partial derivatives we found into these equations to check if they are satisfied.
b) To determine if the map uis pseudoholomorphic, we need to verify if the Cauchy-Riemann
equations hold for u. From part a, we found that:
∂u
∂x = 2x, ∂u
∂y =−2y
∂v
∂x = 2y, ∂v
∂y = 2x
As the Cauchy-Riemann equations are satisfied, the map uis pseudoholomorphic.
15 15. GRADIENT FLOW TECHNIQUES FOR PSEUDOHOLOMORPHIC CURVES
Problem 15. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = z2+iz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 1.
Solution 15. The solution to the pseudoholomorphic curve equation is u(t, z) = u(0, z)for all
t, due to the stationary nature of pseudoholomorphic curves. Hence, u(1, z) = u(0, z) = z2+iz.
Therefore, the solution at time t= 1 is u(1, z) = z2+iz for all z∈Σin this case.
Problem 16. Consider the pseudoholomorphic curve equation on a Riemann surface Σgiven
by ∂J(u) = 0, where Jis a compatible almost complex structure. Let u:R×Σ→Mbe a solution.
Given that u(0, z) = eiz for all z∈Σ, and M=C. Find the solution u(t, z)at time t= 2.
Solution 16. Similar to the previous problem, the solution to the pseudoholomorphic curve
equation is u(t, z) = u(0, z)for all t, due to the stationary nature of pseudoholomorphic curves.
Hence, u(2, z) = u(0, z) = eiz.
Therefore, the solution at time t= 2 is u(2, z) = eiz for all z∈Σin this case.
16 16. TOPOLOGICAL AND GEOMETRIC CONSTRAINTS ON PSEUDOHOLOMORPHIC CURVES
Problem 16. Consider a closed oriented manifold Mof dimension 4, equipped with a compati-
ble almost complex structure J. Let [γ]be a homology class represented by a simple closed curve
γin M. We define the self-intersection number of [γ]as I([γ],[γ]) = γ·γ.
Suppose Σis a pseudoholomorphic curve in Mwith boundary ∂Σ = γ1−γ2, where γ1and γ2
are simple closed curves representing different homology classes in M. Determine the relationship
between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]).
Solution 16. Let’s denote the two simple closed curves γ1and γ2as Aand B, respectively.
Then, we have I([γ1],[γ1]) = A·Aand I([γ2],[γ2]) = B·B, where ·denotes the intersection pairing.
The self-intersection number of the curve Σcan be computed as I(Σ,Σ) = (γ1−γ2)·(γ1−γ2).
Expanding this out, we get:
I(Σ,Σ) = γ1·γ1−γ1·γ2−γ2·γ1+γ2·γ2
=A·A−A·B−B·A+B·B
=A·A−2A·B+B·B
=I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
Therefore, the relationship between I([γ1],[γ1]),I([γ2],[γ2]), and I([γ1],[γ2]) is given by:
I(Σ,Σ) = I([γ1],[γ1]) −2I([γ1],[γ2]) + I([γ2],[γ2]).
17 Numerical Problems on The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in CP 1:
∂u
∂z +λe−u= 0,
where λ∈Ris a constant.
a) Find the general solution u(z, z)to this pseudoholomorphic curve equation.
b) Given λ= 2, solve the equation with the initial condition u(0,0) = 0.
Solution 1.
a) To find the general solution to the pseudoholomorphic curve equation, we first rewrite the
equation in terms of the complex coordinates zand z. Note that ∂u
∂z =1
2∂u
∂x +i∂u
∂y , where z=
x+iy.
Substitute this into the given equation:
1
2∂u
∂x +i∂u
∂y +λe−u= 0.
Separating real and imaginary parts of this equation gives us:
∂u
∂x = 2λe−u
∂u
∂y = 0
Integrating the first equation with respect to xand the second equation with respect to ygives:
u= ln C
2λ+e2λx,
where Cis the constant of integration.
b) Given λ= 2 and the initial condition u(0,0) = 0, we can find the value of Cby substituting
x= 0, u = 0 into the general solution:
0 = ln C
4+ 1.
Solving this equation for Cgives C= 4, so the solution to the equation with λ= 2 and the initial
condition is:
u= ln 4
4+e4x= ln(1 + e4x).
18 The Theory of Pseudoholomorphic Curves
18.1 Problem 1
Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = aeis +bt with a, b ∈C
and |a|= 3,|b|= 2. Determine the image of this pseudoholomorphic curve in C.
Solution: The image of uis given by u(s, t) = aeis +bt. To find the image, we set u(s, t) = z
and solve for sand t:
z=aeis +bt
= 3eis + 2t
= 3(cos s+isin s)+2t.
This implies that the image of the pseudoholomorphic curve uis the set of points z= 3 cos s+
i3 sin s+ 2tin the complex plane C.
18.2 Problem 2
Let u:R×[0,1] →Cbe a pseudoholomorphic curve defined by u(s, t) = s2+it. Determine the
differential of u.
Solution: The differential of uis given by du =∂u
∂s ds +∂u
∂t dt. In this case, we have:
∂u
∂s = 2s,
∂u
∂t =i.
Therefore, the differential of uis du = 2s ds +i dt.
19 19. ENERGY ESTIMATES FOR PSEUDOHOLOMORPHIC CURVES
Problem 19. Consider a pseudoholomorphic curve u:R×S1→Cgiven by u(s, t) = eis sin(t),
where s, t ∈R. Calculate the energy of this pseudoholomorphic curve using the energy functional
E(u) = RR×S1||∂su||2+||∂tu||2dsdt.
Solution 19. The energy of the pseudoholomorphic curve ucan be calculated as follows:
a) First, we calculate the partial derivatives of u:
∂su=ieis sin(t),
∂tu=eis cos(t).
b) Next, we compute the norms of the partial derivatives:
||∂su||2=|ieis sin(t)|2=| − eis sin(t)|2=|eis|2|sin(t)|2= sin2(t),
||∂tu||2=|eis cos(t)|2=|eis||cos(t)|2= 1|cos(t)|2= cos2(t).
c) Now, we substitute the norms into the energy functional and integrate over R×S1:
E(u) = ZR×S1
(sin2(t) + cos2(t)) dsdt
=ZRZS1
1dsdt
=ZR
2π ds
= 2π.
Therefore, the energy of the pseudoholomorphic curve uis 2π.
20 The Theory of Pseudoholomorphic Curves
Problem 1. Consider the following pseudoholomorphic curve equation in a symplectic mani-
fold:
¯
∂Ju+J(u)◦∂¯zu= 0
where u:C→Mis a holomorphic map, Jis a compatible almost complex structure in M, and
¯
∂Jdenotes the Cauchy-Riemann operator.
Given u(z) = z+i
z−i, determine if uis a pseudoholomorphic curve in the complex plane C.
Solution 1.
To determine if u(z) = z+i
z−iis a pseudoholomorphic curve, we need to check if it satisfies the
pseudoholomorphic curve equation. First, let’s calculate ∂zuand ∂¯zu:
∂zu=1
(z−i)2
∂¯zu= 0
Next, let’s compute ¯
∂Ju:
¯
∂Ju=1
2(∂¯zu−J(u)◦∂zu) = 1
20−J(u)◦1
(z−i)2
Given that Jis the standard almost complex structure on C, i.e., J(z) = iz, we have:
J(u) = iz+i
z−i=i(z+i)
z−i
Therefore, ¯
∂Ju=−i
2(z−i)3. Since ¯
∂Ju+J(u)◦∂¯zu= 0, the function u(z) = z+i
z−iis indeed a
pseudoholomorphic curve in C.
21 21. STABILITY CONDITIONS FOR PSEUDOHOLOMORPHIC CURVES
Problem 21. Consider a pseudoholomorphic curve u: Σ →Mwhere Σis a Riemann surface
and Mis a symplectic manifold with symplectic form ω. Let Jbe an almost complex structure on
Mcompatible with ω. Suppose the pseudoholomorphic curve usatisfies the following Cauchy-
Riemann equation:
¯
∂Ju= 0.
a) Show that the linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J.
b) Calculate the Fredholm index of the operator Du.
c) Determine the stability condition for the pseudoholomorphic curve u.
Solution 21.
a) The linearized operator for the Cauchy-Riemann equation is given by
Du=D¯
∂Ju: Γ(u∗T M)→Γ(u∗T M )
where D¯
∂Jis the linearization of ¯
∂J. This operator acts on sections of the pullback bundle u∗T M,
which is the tangent bundle of the Riemann surface Σpulled back to Mvia the map u.
b) The Fredholm index of the operator Duis given by
index(Du) = dim ker(Du)−dim coker(Du).
Here, ker(Du)represents the kernel of Du(space of solutions to ¯
∂Jv= 0 near u) and coker(Du)
represents the cokernel of Du. The Fredholm index indicates the deformation parameter for the
moduli space of pseudoholomorphic curves.
c) The stability condition for the pseudoholomorphic curve uis satisfied if the Fredholm index
of the linearized operator Duis nonnegative, i.e., index(Du)≥0. This condition ensures that
the moduli space of pseudoholomorphic curves is well-behaved under small perturbations in the
complex structure or metric on M.
22 22. FREDHOLM THEORY AND PSEUDOHOLOMORPHIC CURVES
Problem 22. Consider the following pseudoholomorphic curve equation in C2:
¯
∂u = 0
where u:R×S1→C2is a smooth map.
Given u(t, z) = eit(z, iz)for t∈[0,2π]and z∈S1, compute the Fredholm index of this pseudo-
holomorphic curve.
Solution 22. To compute the Fredholm index, we first need to find the linearized operator asso-
ciated with ¯
∂at u. Let v(t, z)be a small perturbation around u, and write v(t, z) = eit(a(t, z), b(t, z)).
The linearized operator is given by:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb
Plugging in u(t, z)and v(t, z)into the expression and simplifying, we get:
Du¯
∂(v) = ∂ta−i∂tb+∂za−i∂zb=eit(∂ta−i∂tb+∂za−i∂zb)
This is equal to 0if and only if both ∂ta−i∂tband ∂za−i∂zbare equal to 0.
From u(t, z), we have a(t, z) = zand b(t, z) = iz. Differentiating with respect to tand z, we get:
∂ta= 0, ∂tb=−z, ∂za= 1, ∂zb=i
So ∂ta−i∂tb= 0 and ∂za−i∂zb= 1 −i= 0. Thus, the Fredholm index is 0−2 = −2.
23 The Theory of Pseudoholomorphic Curves
Problem: Consider the following pseudoholomorphic curve equation in the complex plane C:
¯
∂Ju+i∂tu= 0
where u:C×[0,1] →Cis a smooth map, ∂tdenotes the partial derivative with respect to
t∈[0,1], and ¯
∂Jdenotes the ¯
∂-operator twisted by a compatible almost complex structure Jon C.
Given the initial condition u(x, 0) = x2for x∈C, solve the pseudoholomorphic curve equation.
Solution:
To solve the pseudoholomorphic curve equation, we will first compute the derivatives and then
use the initial condition to determine the solution.
Let u(x, t) = u1(x, t) + iv1(x, t), where u1and v1are real-valued functions. Then the pseudo-
holomorphic curve equation becomes:
∂u1
∂t −∂v1
∂x +i∂v1
∂t +∂u1
∂x = 0
Separating the real and imaginary parts, we get:
∂u1
∂t =∂v1
∂x ,∂v1
∂t =−∂u1
∂x
From the initial condition u(x, 0) = x2, we have u1(x, 0) = x2and v1(x, 0) = 0.
Solving the partial differential equations with these initial conditions gives u1(x, t) = x2and
v1(x, t)=0. Therefore, the solution of the pseudoholomorphic curve equation is u(x, t) = x2+i·0 =
x2.
24 24. HOMOLOGICAL ALGEBRA AND PSEUDOHOLOMORPHIC CURVES
Problem 24. Consider the following pseudoholomorphic curve in CP 1given by the equation
u(z) = z2.
a) Determine the differential operator ∂Jassociated with the almost complex structure Jinduced
by the Fubini-Study metric on CP 1.
b) Show that u(z)is a solution to the ∂Jequation on CP 1.
Solution 24.
a) The almost complex structure Jassociated with the Fubini-Study metric on CP 1is given by
J(z) = −iz. The differential operator ∂Jcan be computed as:
∂J=1
2(J+iId) = 1
2(−iz +iId) = 1
2(i(z+ 1)).
b) To show that u(z) = z2is a solution to the ∂Jequation, we need to demonstrate that ∂Ju(z) =
0. Let’s compute it:
∂Ju(z) = 1
2(i(z2+ 1)) = i
2z2+i
2=iz +i
2= 0.
As the result is not zero, the function u(z) = z2is not a solution to the ∂Jequation on CP 1.
25 The Theory of Pseudoholomorphic Curves
Problem 1. Consider a pseudoholomorphic curve u:R×[0,1] →Cgiven by u(s, t) = eis +t2.
a) Determine the asymptotic behavior of uas s→ −∞.
b) Find the limit lims→∞ |u(s, 1)|.
c) Show that uis a pseudoholomorphic curve.
Solution 1.
a) As s→ −∞, the term eis behaves as oscillatory and does not affect the growth, so we focus
on t2. Thus, the asymptotic behavior of u(s, t)as s→ −∞ is t2.
b) We have |u(s, 1)|=|eis + 1| ≤ |eis|+|1|= 1 + 1 = 2. Thus, lims→∞ |u(s, 1)| ≤ 2.
c) To show that uis a pseudoholomorphic curve, we need to verify that ¯
∂Ju= 0, where Jis the
standard complex structure on C. Compute ∂su=ieis and ∂tu= 2t. Then, ¯
∂Ju=∂su−i∂tu=
ieis −2it.
Since ¯
∂Ju= 0, we have shown that uis a pseudoholomorphic curve.
