Abstract algebra

Section 18.3-19.1.

Today we will discuss finite-dimensional associative algebras and their representations.

Definition 1. Let A be a finite-dimensional associative algebra over a field F . An element a ∈ A is nilpotent if an = 0 for some positive integer n. An algebra is said to be nilpotent if all of its elements are.

Exercise 2. Every subalgebra and factor algebra of a nilpotent algebra are nilpotent. Con- versely, if I ⊂ A is a nilpotent ideal, and the quotient algebra A/I is nilpotent, then so is A.

Example. The algebra of all strictly lower- (or upper-) triangular n×n matrices is nilpotent.

Proposition 3. If the algebra A is nilpotent, then An = 0 for some n ∈ Z+, that is the product of any n elements if the algebra A equals 0.

Proof. Let B ⊂ A be the maximal subspace for which there exists n ∈ Z+ such that Bn = 0. Note that B is closed under multiplication, i.e. B ⊃ Bk for any k ∈ Z+. Assume B 6= A and choose an element a ∈ A\B. Since aBn = 0, there exists k ∈ Z+ such that aBk 6⊂ B. Replacing a with a non-zero element in aBk we obtain aB ⊂ B. Recall that there exists m ∈ Z+ so that am = 0. Now, let us set C = B ⊕〈a〉. Then we have

Cmn = 0

which contradicts the definition of the subspace B. �

Unless A is commutative, the set of all nilpotent elements of A does not have to be an ideal (in general, not even a subspace). On the other hand, if I,J are nilpotent ideals in A, then so is

I + J = {x + y |x ∈ I,y ∈ J} . Therefore, there exists the maximal nilpotent ideal which contains every other nilpotent ideal of A.

Definition 4. The radical of A is the maximal nilpotent ideal of A it is denoted rad(A). The algebra A is semisimple if rad(A) = 0.

If char(F) = 0, there exists an alternative description of semisimple algebras. Consider the regular representation of the algebra A

ρ: A → L(A), ρ(a)(b) = ab, and define a “scalar product” on A via

(a,b) = tr(ρ(ab)) = tr(ρ(a)ρ(b)).

One can see that (·, ·) is a symmetric bilinear function on A satisfying (ab,c) = (a,bc).

Definition 5. For any ideal I ⊂ A, its orthogonal complement I⊥ is defined as I⊥ = {a ∈ A |(a,i) = 0 for all i ∈ I} .

Proposition 6. For any ideal I ⊂ A, its orthogonal complement I⊥ ⊂ A is also an ideal.

Proof. For any a ∈ A, x ∈ I⊥, and y ∈ I we have (xa,y) = (x,ay) = 0 and (ax,y) = (y,ax) = (ya,x) = 0.

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Proposition 7. If A is an algebra over a field F with char(F) = 0, then every element a ∈ A orthogonal to all of its powers is nilpotent.

Proof. Let a ∈ A be such that (a,an) = tr ρ(a)n+1 = 0 for all n ∈ Z+.

Let L ⊃ F be the splitting field of the characteristic polynomial f(x) of the operator ρ(a). Then, over L we have

f(x) = tk0 s∏ i=1

(t−λi)ki,

where λi are distinct for i = 1, . . . ,s, and

tr ρ(a)n+1 =

s∑ i=1

kiλ n+1 i = 0.

Letting n run through the set 1, . . . ,s, the above equation yields a system of s homogeneous linear equations in variables k1, . . . ,ks. The determinant of this system equals

λ21 . . .λ 2 nV (λ1, . . . ,λn),

where V (λ1, . . . ,λn) is the Vandermonde determinant. Therefore, the system is non-degenerate and we derive that kj = 0 in F for all j = 1, . . . ,s. If char(F) = 0 the latter is equivalent to

f(x) = tk0,

which implies that the operator ρ(a) is nilpotent. Therefore, for some m ∈ Z+ we have

am+1 = T(a)m(a) = 0,

and we conclude that a ∈ A is nilpotent. �

Definition 8. The scalar product (·, ·) is non-degenerate if for any a ∈ A there exists a′ ∈ A such that (a,a′) 6= 0. Equivalently, it is nilpotent if A⊥ = 0.

Theorem 9.

(1) If the scalar product (a,b) = tr(ρ(ab)) is non-degenerate, then the algebra A is semisimple.

(2) If the algebra A is semisimple and char(F) = 0, then the scalar product is non- degenerate.

Proof. (1) Let I ⊂ A be a nilpotent ideal. Then for any x ∈ I and any a ∈ A, their product xa is nilpotent and hence

(a,x) = tr ρ(ax) = 0.

Therefore, I ⊂ A⊥ = 0. (2) Conversely, if char(F) = 0 the previous two propositions show that A⊥ is a nilpotent

ideal, therefore A⊥ = 0. �

Definition 10. An algebra is simple if it has no proper ideals.

Note that a simple algebra is semi-simple unless it is nilpotent. However, in that case A2 ⊂ A is a proper ideal of A, hence A2 = 0. Therefore, a simple algebra is semi-simple, unless A is 1-dimensional with 0 multiplication.

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Theorem 11. Any semisimple associative algebra A can be decomposed into a direct sum of simple algebras

A = A1 ⊕···⊕As, moreover, any ideal of A is a sum of some direct summands in the above decomposition.

Proof. We will prove this theorem in the assumption char(F) = 0. If A is simple, we have s = 1 and the statement is trivial. Otherwise, let A1 ⊂ A be a minimal ideal of A. Then either

A = A1 ⊕A⊥1 , or A1 ⊂ A⊥1 . In the second case, A1 has to be nilpotent, which would contradict the semisim- plicity of A. Then, indeed A is a direct sum of A1 and A

⊥ 1 , in which case every ideal of A1 or

A⊥1 is also an ideal of A. This allows us to conclude that A1 is simple and A ⊥ 1 is semisimple.

We know apply the same reasoning to the algebra A⊥1 . Let now I ⊂ A be any ideal. Denote by πk : A → Ak the canonical projections, and set

Ik = πk(I). Then Ik is an ideal of the algebra Ak. If Ik 6= 0 we have Ik = Ak which yields

Ak = A 2 k = AkIk = AkI ⊂ I,

which means that I is the sum of several summands Aj. �

Theorem 12. Any non-trivial simple associative algebra A over an algebraically closed field F is isomorphic to L(V ) for some vector space V over F . Moreover, any non-trivial irre- ducible representation of A is isomorphic to the tautological representation of L(V ).

Proof. Let ρ: A → L(A) be the regular representation, and V be a minimal invariant subspace, equivalently a minimal left ideal in A. Denote φ = ρ|V : A → L(V ). Using the fact that ker(φ) ⊂ A is an ideal, and one of the results from the previous lecture, we get either

A ' φ(A) = L(V ) or dim(V ) = 1 and φ(A) = 0. In the latter case, we see that AV = 0 and therefore the subspace

A0 = {x ∈ A |Ax = 0} is nonzero. Since A0 ⊂ A is an ideal, we must have A = A0 which would imply that A is trivial.

Therefore, φ(A) = L(V ) for some vector space V , and φ is a tautological representation of A in V . In that case we have previously shown that the regular representation ρ: A → L(A) is isomorphic to the sum of φ with itself n times:

ρ = nφ.

Now, let τ : A → L(U) be any irreducible representation of A, and u0 ∈ U be any nonzero vector. Then the map

T : A → U, T(a) = τ(a)(u0) is a morphism between the regular representation (A,ρ) and the representation (U,τ), indeed for any x ∈ A we have

T(ρ(a)(x)) = T(ax) = τ(ax)(u0) = τ(a)τ(x)(u0) = τ(a)T(x).

If the representation (U,τ) is non-trivial, we have im(T) = U, therefore U is a factor repre- sentation of the regular representation ρ. Since U is irreducible, we get (U,τ) ' (V,φ). �

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Definition 13. Subalgebra Z(A) ⊂ A of an associative algebra A defined as

Z(A) = {z ∈ A |za = az ∀a ∈ A}

is called the center of A.

Corollary 14. Every semisimple associative algebra over an algebraically closed field F is isomorphic to the algebra

A = L(V1) ⊕···⊕L(Vs) where s = dim Z(A)

for some vector spaces V1, . . . ,Vs over F . If dim(Vi) = ni for i = 1, . . . ,s, we get

dim(A) = n21 + · · · + n 2 s.

Proof. The only statement that have not been yet proven is that the number of direct summands is equal to dim Z(A). It follows from the fact that the center of each summand is 1-dimensional and consists of scalar matrices. The latter statement is left as an exercise. �

Corollary 15. Let ρj for j = 1, . . . ,s be the irreducible representation of the algebra

A ' L(V1) ⊕···⊕L(Vs)

in the space Vj obtained via projection πj : A → L(Vj). Then any non-trivial irreducible representation of A is isomorphic to one of the representations ρj.

Remark 16. Note that representations ρj are pairwise non-isomorphic since their kernels are distinct.

Let us now get back to the representation theory of finite groups.

Theorem 17. Let G be a finite group of order n. Then the algebra FG is semisimple provided char(F) does not divide n.

Proof. Let ρ: FG → FG be the regular representation. Then it is easy to see that

tr ρ(g) =

{ n, g = e,

0, g 6= e.

Therefore, for any g,h ∈ G we have

(g,h) =

{ n, gh = e,

0, gh 6= e.

If char(F) does not divide n, this scalar product is non0degenerate, therefore FG is semisim- ple. �

In what follows we will assume that F is an algebraically closed field whose characteristic does not divide |G|. By the results of the previous lecture, FG is isomorphic to a direct sum of matrix algebras L(Vi).

Theorem 18. For any finite group G with |G| = n, there are only finitely many non- isomorphic irreducible representations of G over the field F . Dimensions n1, . . . ,ns of these representations satisfy

n21 + · · · + n 2 s = n,

and s is the number of conjugacy classes in G.

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Proof. The first statement and the equality

n21 + · · · + n 2 s = n

follow from the fact that FG is a semisimple associative algebra, and a correspondence between representations of FG and G. We also know that s = dim Z(FG). An element a =

∑ g∈G agg belongs to the center Z(FG) if and only if

hah−1 = ∑ g∈G

aghgh −1 =

∑ g∈G

ah−1ghg = a

for all h ∈ G. The latter is equivalent to the statement that the coefficients ag are constant on conjugacy classes of G:

ag = ahgh−1

for all g ∈ G. Therefore, Z(FG) is a linear span of elements of the form ∑

g∈C g, where

C ⊂ G is a conjugacy class, and dim Z(FG) equals the number of conjugacy classes of G. � Example. (1) Any irreducible representation of an abelian group is 1-dimensional, there-

fore a finite abelian group of order n has precisely n non-isomorphic irreducible rep- resentations.

(2) The permutation group S3 has a trivial representation, a 1-dimensional sign repre- sentation, where σ 7→ sgn(σ), and a 2-dimensional irreducible representation, where S3 acts by permuting vertices of a regular triangle. Up to isomorphism, these are all irreducible representations of S3 because

6 = 12 + 12 + 22.

(3) In a similar fashion, one can find the following non-isomorphic irreducible represen- tations of S4: • the trivial one; • sign representation; • a composition of the projection S4/K4 ' S3 with the irreducible 2-dimensional

representation of S3 considered above; • an isomorphism with the group of rotations of a cube (composed with its natural

3-dimensional representation); • an isomorphism with the permutation group of the vertices of a regular tetrahe-

dron composed (composed with its natural 3-dimensional representation). Checking dimensions we get

24 = 12 + 12 + 22 + 32 + 32,

and conclude that these are all irreducible representations of S4 up to isomorphism.

Let us summarize, what we have proved so far. Let G be a finite group, and ρi : V → Vi, i = 1, . . . ,s be all distinct (up to isomorphism) irreducible representations of G over an algebraically closed field F , whose characteristic does not divide |G|. Then the group algebra FG can be written as

FG ' L(V1) ⊕···⊕L(Vs), and ρi is simply a projection onto the i-th summand. Then the subspaces L(Vi) are the isotypic components of the regular representation ρ of the algebra FG, and the restriction of ρ onto L(Vi) is isomorphic to niρi where ni = dim(Vi). Therefore, for any a,b ∈ FG, we have

(a,b) = s∑ i=1

ni tr ρi(a)ρi(b).

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Now, let us consider the (finite-dimensional) space F [G] of F-valued functions on G. Every function f ∈ F[G] can be uniquely continued to a linear function on FG by

f( ∑ g∈G

agg) = ∑ g∈G

agf(g),

which allows us to identify F[G] with the space of linear functionals on FG, dual to FG. On the other hand, the non-degenerate scalar product

(·, ·) : FG×FG −→ F provides a bijection between FG and its dual via∑

g∈G agg 7→

∑ g∈G

agfg where fg(h) = (g,h).

Since

(g,h) =

{ n, gh = e,

0, gh 6= e, we see that

fg = nδg−1,

where δg−1 is the Kronecker delta. Above isomorphisms allow us to drag the scalar product (·, ·) from FG onto F[G]. In terms

of δ-functions we obtain

(δg,δh) = 1

n2 (g−1,h−1) =

{ 1 n , gh = e,

0, gh 6= e,

and for any functions f1,f2 ∈ F[G] we have

(f1,f2) = 1

n

∑ g∈G

f1(g)f2(g −1).

Now, recall that FG = L(V1) ⊕···⊕L(Vs).

Let us now choose a basis 〈ei〉, i = 1, . . . ,ni in the representation Vi, and denote by fijk ∈ F[G] the (j,k) matrix coefficient in the representation Vi, that is

fijk(g) = 〈ej,ρi(g)ek〉 . Then one can see that

(fijk,f i′ j′k′) =

1

ni δii′δjk′δj′k.

Definition 19. A character χφ ∈ F [G] of the representation φ: G → GL(V ) is defined by χ(g) = tr φ(g)

for any g ∈ G.

It is clear that he character of a sum of two representations equals the sum of the characters:

χφ+ψ = χφ + χψ,

and that a character is a class function:

χ(g) = χ(hgh−1)

for any g,h ∈ G. The following theorem is a direct corollary of above calculations:

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Theorem 20. Let χi, i = 1, . . . ,s be the characters of irreducible representations ρi of the group G. Then

(χi,χj) = δij,

equivalently, characters χi form an orthonormal basis in the subspace of class functions in F[G].

Corollary 21. Let χ be the character of a representation φ: G → GL(V ). Then,

V ' s∑ i=1

(χ,χi)Vi.

In other words, (χ,χi) is the multiplicity of the irreducible representation Vi in the decompo- sition of the representation V .

Proof. Representation V can be written as

V =

s∑ i=1

kiVi

for some ki ∈ Z>0. Therefore

χ = s∑ i=1

kiχi and (χ,χi) = ki.

�

Corollary 22. A representation φ: G → GL(V ) with character χ is irreducible if and only if (χ,χ) = 1.

Proof. By above,

(χ,χ) =

s∑ i=1

k2i

which is equal to 1 if and only if one of the ki is 1, and all the other are 0. �

Remark 23. Above corollaries show that a character, which is clearly less data than a representation, since it only knows about traces of operators rather than operators themselves, is enough to recover the whole representation.

Finally, if F = C it is more convenient for computations to replace the bilinear form (·, ·) with the following Hermitian scalar product:

(f1 |f2) = 1

n

∑ g∈G

f1(g)f2(g),

where x denotes the complex conjugate to x. If now in each of the spaces Vi, one chooses a basis orthonormal with respect to the Hermitian inner product, they would get

fikj(g −1) = fijk(g),

which yields

(fijk |fi ′ j′k′) = (f

i jk,f

i′ k′j′) =

1

ni δii′δjj′δkk′.

Example.

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(1) A character of a 1-dimensional representation coincides with the representation. If G = 〈g |gn = e〉 is a cyclic group of order n, its irreducible representations are defined by

ρk(g) = ω k−1,

where ω = e2πi/n. Therefore, the table of characters of G takes the following form

χ0 χ1 . . . χn−1 e 1 1 . . . 1 g 1 ω . . . ωn−1

g2 1 ω2 . . . ω2(n−1)

... ...

... . . .

...

gn−1 1 ωn−1 . . . ω(n−1) 2

(2) Using the description of irreducible representations of the group S4, we can write out its table of characters as well

χ1 χ ′ 1 χ2 χ3 χ

′ 3

e 1 1 2 3 3 1 (12) 1 -1 0 -1 1 6

(12)(34) 1 1 2 -1 -1 3 (123) 1 1 -1 0 0 8 (1234) 1 -1 0 1 -1 6

where the first column shows representatives of conjugacy classes in S4, and the last column shows number of elements in the corresponding conjugacy class. For example one gets

(χ2,χ3) = (χ2 |χ3) = 1

24 (2 · 3 · 1 + 0 · (−1) · 6 + 2 · (−1) · 3 + (−1) · 0 · 8 + 0 · 1 · 6) = 0.

(3) Let V be a 6-dimensional space of functions on the set of faces of a cube. An iso- morphism between S4 and the rotation group of a cube defines a representation of φ: S4 → GL(V ), let χ be its character. Every element g ∈ S4 permutes the faces of a cube, and therefore permutes δ-functions of these faces accordingly. Therefore, χ(g) = tr φ(g) equals the number of faces fixed by g. This allows us to compute the character χ:

e (12) (12)(34) (123) (1234) χ 6 0 2 0 2

Computing scalar products of χ with those of irreducible representations of S4 we find

(χ |χ1) = 1, (χ |χ′1) = 0, (χ |χ2) = 1, (χ |χ3) = 1, (χ |χ ′ 3) = 0.

Therefore, φ ' ρ1 ⊕ρ2 ⊕ρ3.