Download Semisimple algebras and Wedderburn`s theorem

Semisimple algebras and Wedderburn’s theorem
Let G be a finite group. A complex representation of G is a homomorphism ρ :
G → GLn (C), where GLn (C) denotes the group of all invertible n × n matrices over
C. The character afforded by ρ is the map χ : G → C given by χ(g) = Trace ρ(g).
Definition. Let K be any field and let A be a K-vector space. Suppose that there
is a binary operation (x, y) 7→ xy on A and an element 1 = 1A ∈ A satisfying the
following properties:
(i) (xy)z = x(yz) for all x, y, z ∈ A;
(ii) x(y + z) = xy + xz for all x, y, z ∈ A;
(iii) (x + y)z = xz + yz for all x, y, z ∈ A;
(iv) x1 = 1x = x for all x ∈ A;
(v) (cx)y = c(xy) = x(cy) for all x, y ∈ A and c ∈ K.
Then A is called a K-algebra
In other words, a K-algebra is a ring together with a compatible structure of a
vector space over K.
Definition. Let A be a K-algebra and let V be set with an operation A × V → V ,
(x, v) 7→ xv satisfying, for all x, y ∈ A and v, w ∈ V ,
(i) x(v + w) = xv + xw;
(ii) (x + y)v = xv + yv;
(iii) x(yr) = (xy)r;
(iv) 1v = v.
Then V is called a (left) A-module.
Note that each V -module has a structure of the vector space over K: the scalar
multiplication is given by cv = (c1)v, where c ∈ K, v ∈ V , and this structure is
compatible with the algebra: x(cv) = c(xv) = (cx)v for all x ∈ A, v ∈ V and c ∈ K.
Definition. Let V and V ′ be A-modules. A subset W ⊆ V is called a submodule of
V if xw ∈ W for all x ∈ A and w ∈ W . A map f : V → V ′ is called a homomorphism
of A-modules if f (xv) = xf (v) for all x ∈ A and v ∈ V .
Submodules, quotients, direct sums of modules are defined as usual.
The set of all homomorphisms between A-modules V and V ′ is denoted HomA (V, V ′ ).
It has the structure of a vector space over K: if f, g ∈ HomA (V, V ′ ), c ∈ K and
v ∈ V then (cf )(v) = f (cv) and (f + g)(v) = f (v) + g(v).
Definition. Let A and B be K-algebras. A linear transformation f : A → B is
called an algebra homomorphism if
1
f (xy) = f (x)f (y) for all x, y ∈ A;
f (1A ) = 1B . If moreover f is bijective, then f is called an isomorphism of algebras.
From now on, all representations and algebras are assumed to be over the field
C of complex numbers unless otherwise specified. Throughout these notes, A will
denote an algebra. A representation of A is an algebra homomorphism f : A →
Mn (C), where Mn (C) is the algebra of all n × n matrices over C. All algebras,
representations and modules are assumed to be finite-dimensional. To each finite
group G we associate the group algebra


X

CG :=
ag g : ag ∈ C for all g ∈ G ,


g∈G
with multiplication given by the group operation extended by linearity. More precisely,


!
X
X
XX

ag g
bh h =
ag ah gh where ag , bh ∈ C.
g∈G
g∈G h∈G
h∈G
If V is an A-module, we can construct a representation of A as follows. Choose
a basis {v1 , . . . , vn } of M , and, for each x ∈ A, let aij (x) ∈ C (i, j ∈ {1, . . .}) be the
scalars uniquely determined by
xvi =
n
X
aij vj ,
i = 1, . . . , n.
j=1
Let ρ(x) be the n × n matrix (aij (x))1≤i,j≤n . Then ρ : A → Mn (C) is a representation of A. This establishes a one-to-one correspondence between (isomorphism
classes of) A-modules and (isomorphism classes of) representations of A. Also, the
representations of G and those of CG are in one-to-one correspondence: if ρ is a representation of G, the corresponding representation of CG is obtained by extending
ρ linearly.
Definition. A non-zero A-module V is said to be simple (or irreducible) if the only
submodules of V are {0} and V itself.
A group representation ρ : G → GLn (C) and the corresponding character χ are
called irreducible if the corresponding CG-module is simple. We denote by Irr(G)
the set of irreducible characters of G.
Theorem 1. Let V be a module over an algebra A. Then the following are equivalent:
(i) the module V is a direct sum of irreducible submodules;
(ii) the module V is a sum of irreducible submodules;
(iii) for any submodule U of V there exists another submodule W of V such that
V = U + W and U ∩ W = 0. (The submodule W is said to be a complement
to V .)
2
Proof. Clearly, (i) implies (ii). Suppose V = V1 +· · ·+Vk where each Vi is irreducible.
Let
P U be a submodule of V .PLet I be a maximal subset of {1, . . . , k} such that
( i∈I Vi ) ∩ U = 0. Let W = i∈I Vi . We claim that W + U = V . If not, since V is
a sum of the submodules Vi , there exists j ∈ {1, . . . , k} such that Vj is not contained
in U + W . Then Vj ∩ (U + W ) is a submodule of the simple module Vj , distinct
from Vj , so Vj ∩ (U + W ) = 0. However, then (Vj + W ) ∩ U = 0: if u = v + w,
where u ∈ U , v ∈ Vj and w ∈ W , then v = u − w ∈ Vj ∩ (U + W ), so v = 0, whence
u = w = 0 (since U ∩ W = 0). This contradicts the maximality of I and thus proves
the claim. Hence, W is a complement to U in V , so (iii) holds.
Now suppose (iii) holds. Let U be a submodule of V of maximal dimension such
that U is a direct sum of simple modules. Let W be a complement to U . Supposing
W 6= 0, choose a simple submodule Z of W . Then Z + U is a direct sum of simple
modules, contradicting the maximality of dim U . Thus W = 0, whence U is a direct
sum of simple modules.
Definition. An A-module V is said to be semisimple if V satisfies any of the
equivalent properties in the preceding lemma. The algebra A is called semisimple if
it is semisimple as a left module over itself.
Lemma 2. Submodules and quotients of semisimple A-modules are semisimple.
Proof. Let V be a semisimple A-module, and let U be a submodule of V . Suppose
Z is a submodule of U . Since V is semisimple, Z has a complement W in V . We
claim that W ∩ U is a complement to Z in U . Indeed, (W ∩ U ) ∩ Z ⊆ W ∩ Z = 0.
Let u ∈ U . Since V = Z + W , there exist z ∈ Z and w ∈ W such that u = z + w.
Since u, z ∈ U , we have w = u − z ∈ U . Thus, u = z + w ∈ Z + (U ∩ W ). We
conclude that U = Z + (U ∩ W ), proving the claim. Thus U is semisimple.
Now write V = S1 +· · ·+Sk , where each Si is a simple module. Let π : V → V /U
be the canonical homomorphism. Then V /U = π(S1 ) + · · · + π(Sk ). By the first
isomorphism theorem, π(Si ) is isomorphic to a quotient of the simple module Si for
each i, so π(Si ) is either zero or simple. Thus V /U is a sum of simple submodules,
so it is semisimple.
Theorem 3. Let A be a semisimple algebra. Then every A-module is semisimple.
Proof. Let V be an A-module. Choose a basis {v1 , . . . , vk } of V (it suffices for this
set to generate V as an A-module). Denote by Ak the direct sum of k copies of the
left A-module A. Since A is a sum of simple modules, so is Ak . The map Ak →→ V
given by
(x1 , . . . , xk ) 7→ x1 v1 + · · · + xk vk
is an A-module homomorphism. Moreover, it is surjective (since v1 , . . . , vk form a
basis of V ), so by the first isomorphism theorem V is isomorphic to a quotient of
Ak . Thus by Lemma 2 V is semisimple.
Theorem 4 (Maschke). Let G be a finite group. Then the group algebra CG is
semisimple.
3
Proof. Let V be any CG-module, and let U be a submodule of V . Let π : V → U
be any surjective linear map such that π(u) = u for all u ∈ U . (By elementary
linear algebra, one can always find such a map, e.g. via choosing a basis of U and
extending it to a basis of V .) Define π ′ : V → U as follows:
π ′ (v) =
1 X −1
g π(gv).
|G|
g∈G
Clearly, π ′ is linear. We claim that π ′ is a CG-module homomorphism. It is enough
to check that π ′ (hv) = hπ ′ (v) for all h ∈ G, for then the claim would follow by
linearity. However,
X
g−1 π(ghv)
π ′ (hv) = |G|−1
g∈G
= |G|−1
X
hx−1 π(xv)
(we substitute x = gh)
x∈G
= hπ ′ (v),
and the claim follows. Moreover, for all u ∈ U ,
X
g−1 (gu) = u,
π ′ (u) = |G|−1
(1)
g∈G
so U is the image of π ′ . Let W = ker π. Since π ′ is a module homomorphism, W
is a submodule of V . By (1), W ∩ U = 0. Since im π ′ = U , we have dim W =
dim V − dim U , whence V = U + W . So W is a complement to U , and the result
follows.
Let V and W be A-modules, where A is any algebra. We write HomC (V, W ) for
the set of C-linear maps between vector spaces V and W and
HomA (V, W ) = {f ∈ HomC (V, W ) : f (av) = af (v) for all a ∈ A, v ∈ V }.
Also, EndA (V ) = HomA (V, V ).
Lemma 5 (Schur). Let V and W be simple A-modules. Suppose f ∈ HomA (V, W )
and f 6= 0. Then f is an isomorphism. Moreover, if V = W then there exists
λ ∈ C \ {0} such that, for all v ∈ V , f (v) = λv.
Proof. Since f 6= 0, im f is a non-zero submodule of the simple module W . Thus
im f = W . Moreover, ker f is a submodule of V , distinct from V . Thus ker f = 0.
So f is an isomorphism.
Now suppose V = W . Let λ be an eigenvalue of f . Then f − λ idV : V → V is a
module homomorphism. Its kernel is non-trivial, so must be equal to the whole of V .
Thus f = λ idV , as claimed. (Recall that idV denotes the map V → V , v 7→ v.)
Lemma 6. Let T be a simple A-module. Suppose that an A-module V is a sum of
simple submodules Si (i = 1, . . . , n), none of which is isomorphic to T . Then V has
no submodule isomorphic to T .
4
Remark. This (in fact, a stronger result on uniqueness of direct sum decompositions) follows from the Jordan–Hölder theorem for modules.
Proof. Suppose V has a simple submodule T ′ isomorphic to T . Since V is a sum
of simple modules, it is semisimple, so T ′ has a complement in V , say W . Let
π : V /raT be the projection along W : π(w + t) = t for all w ∈ W , t ∈ T . Then π
is a non-zero homomorphism of modules. Since V is the sum of simple submodules
Si , π|Si 6= 0 for some . However, by Schur’s lemma, π|Si is then an isomorphism
between Si and T , which contradicts the hypothesis.
Lemma 7. Suppose A is a semisimple algebra. Then any simple A-module is isomorphic to a submodule of A.
Proof. Let V be a simple A-module, and pick a non-zero element v ∈ V . Then the
map A → V , x 7→ xv is a non-zero module homomorphism, which must be surjective
(since V is simple). Hence, by the first isomorphism theorem, V is isomorphism to
a quotient module of A, say V ∼
= A/U . Since A is semisimple, U has a complement
in the module A, say W . Then V ∼
= W ⊆ A, as required.
If V is an A-module and S is a simple A-module, we denote by S(V ) the sum of
all those submodules of V which are isomorphic to S. We denote by Aop the algebra
with the same underlying vector space as A and with the “opposite” multiplication
(x, y) 7→ yx.
Note that if V is an A-module, then EndA (V ) has the structure of a C-algebra, with
multiplication given by composition:
fg = f ◦ g
f, g ∈ EndA (V )
Lemma 8. For any algebra A, EndA (A) ∼
= Aop .
Proof. Define the map α : Aop → EndA (A) by
α(x)(a) = ax,
a, x ∈ A.
It is clear that α(x) is indeed a (left) A-module homomorphism and that φ is linear.
Moreover, the product of elements x and y in Aop is yx and
α(yx)(a) = ayx = (ay)x = (α(x) ◦ α(y))(a),
so α is an algebra homomorphism (it is clear that α(1) = idA ). If α(x) = 0 for
some x ∈ A then, in particular, 0 = α(x)1 = 1X = x, so ker α = 0. Now suppose
f ∈ EndA (A). Then f (x) = f (x1) = xf (1) for all x ∈ A, so f = α(f (1)). Hence α
is an isomorphism of algebras.
Lemma 9. Let S1 , . . . , Sk be pairwise non-isomorphic simple A-modules. Suppose
V = S1 ⊕ · · · ⊕ S1 ⊕ S2 ⊕ · · · ⊕ S2 ⊕ · · · ⊕ Sk ⊕ · · · ⊕ Sk .
{z
} |
{z
}
|
|
{z
}
n1
n2
Then the algebra EndA (V ) is isomorphic to the direct sum
Mn1 (C) ⊕ · · · ⊕ Mnk (C).
5
nk
Proof. We have V = U1 ⊕ · · · ⊕ Uk where Ui = Si ⊕ · · · ⊕ Si . Informally, the result
{z
}
|
ni
is a consequence of Schur’s lemma: there are no non-zero homomorphisms between
different simple modules, so HomA (Ui , Ui′ ) = 0 for i 6= i′ . On the other hand, an
endomorphism of the module Ui can be represented as an ni × ni matrix with entries
from HomA (Si , Si ) ∼
= C. We now express this idea more formally.
(i)
If 1 ≤ i ≤ k and 1, ≤ j ≤ ni , let ιj : Si → V be the standard embedding
of the j-th component Si of the submodule Ui into V (with respect to the given
(i)
decomposition of V ). Let πj : V → Si be the projection of V onto the j-th
(i)
component of Ui . If 1 ≤ i ≤ k and 1 ≤ j, l ≤ ni , define ejl ∈ EndA (V ) by
(i)
(i)
(i)
(i)
ejl = ιj ◦ πl . (In other words, ejl is obtained by mapping the l-th component Si
isomorphically onto the j-th component and the other simple components to zero.)
Now let C = (C1 , . . . , Ck ) ∈ Mn1 (C) ⊕ · · · ⊕ Mnk (C), with the (j, l) entry of Ci
(i)
equal to cjl ∈ C. Then define α(C) by
α(C) =
ni
ni X
k X
X
(i)
cjl ejl (i).
i=1 j=1 l=1
We claim that α : Mn1 (C) → Mnk (C) is an algebra isomorphism. It is straightforward, though a bit tedious, to check that α is a homomorphism of algebras.
(l)
(i)
(i)
Moreover, if α(C) = 0 then, for all i, j, l, we have 0 = πj ◦ α(C) ◦ ιl = cij idSi :
(i) (i′ )
this follows from the fact that πj ιj ′ = δii′ δjj ′ idSi . (As usual, δab is equal to 1 if
a = b and 0 otherwise.) Then C = 0, so α is injective.
We prove that α is surjective by comparing dimensions. Clearly,
dim(Mn1 (C) ⊕ · · · ⊕ Mnk (C)) = n21 + · · · + n2k .
On the other hand, it is easy to show that EndA (V ) is isomorphic as a vector space
to the direct sum
ni′
ni M
k M
k M
M
HomA (Si , Si′ ).
i=1 j=1 i′ =1 j ′ =1
(The Hom-operation behaves well with respect
P to direct sums.) By Schur’s lemma,
dim HomA (Si , Si′ ) = δii′ , so dim EndA (V ) = ki=1 n2i . The result follows.
Theorem 10 (Wedderburn). Suppose A is semisimple. Up to isomorphism, there
are only finitely many simple A-modules. If we denote these as S1 , . . . , Sk and let
ni = dim Si then
(i) each Si (A) is a two-sided ideal of A;
(ii) the algebra A is the direct sum of the ideals S1 (A), . . . , Sk (A);
(iii) for i = 1, . . . , k, the algebra Si (A) is isomorphic to the matrix algebra Mni (C).
Proof. The semisimple A-module A can be decomposed as a direct sum of simple
modules. Let
A∼
= T1 ⊕ · · · ⊕ T1 ⊕ · · · ⊕ Tl ⊕ · · · ⊕ Tl ,
{z
}
|
|
{z
}
m1
ml
6
where T1 , . . . , Tl are simple and pairwise non-isomorphic. By Lemmata 8 and 9, we
have the algebra isomorphisms
Aop ∼
= EndA (A) ∼
= Mm1 (C) ⊕ · · · ⊕ Mml (C).
Since Mn (C)op ∼
= Mn (C) (the operation of transposition yields an isomorphism), we
have
A∼
= Mm1 (C) ⊕ · · · ⊕ Mml (C).
Replacing A with an isomorphic algebra, we may assume that A is actually equal
to the direct sums of matrix algebras in question. As we now know the algebra A
explicitly, it will not be hard to identify the simple A-modules.
We shall write elements of A as x = (x1 , . . . , xl ) where xi ∈ Mmi (C). For
i = 1, . . . , l, we may define the structure of an A-module on the vector space C mi
of column vectors via matrix multiplication: xv = xi v for all x ∈ A, v ∈ C mi .
Denote this module Si . We claim that Si is simple. Indeed, if U ⊆ Si is a non-zero
submodule, pick u ∈ U , u 6= 0. Then, by elementary linear algebra, for any vector
v ∈ Si there exists a matrix xi ∈ Mmi (C) such that xi u = v. It follows that v ∈ U ,
so U = Si , and the claim is proved. Let ei = (0, 0, . . . , Ii , 0, . . . , 0) ∈ A. Then
ei Si = Si , but ei Sj = 0 for all j 6= i, whence the modules S1 , . . . , Sl are pairwise not
isomorphic. For i = 1, . . . , l and j = 1, . . . , mi , let
Tij = {(0, . . . , 0, xi , 0, . . . , 0) ∈ A| xi has zero entries outside the j-th column}.
mi
Clearly, each Tij is a submodule of A, Tij ∼
Tij as an A= Si and A = ⊕li=1 ⊕j=1
module. By Lemma 7, the modules S1 , . . . , Sl are all the simple A-modules up to
isomorphism. Note that dim Si = mi .
The theorem will now follow once we show that Si (A) = Mmi (C), the i-th component of the decomposition, for all i. It follows from the direct sum decomposition
of the previous paragraph that Mmi (C) ⊆ Si (A). Suppose that there is a submodule Q ∼
= Si of A that is not contained in Mmi (C). Let π : A → A/Mmi (C)
be the canonical homomorphism of modules. Then π(Q) 6= 0, so π(Q) ∼
= Si .
Thus π(Q) is a submodule of π(A), which is a direct sum of modules isomorphic
to S1 , . . . , Si−1 , Si+1 , . . . , Sl , none of which is isomorphic to Si . This contradicts
Lemma 6, proving the theorem.
The centre of A is
Z(A) = {x ∈ A| xy = yx for all y ∈ A}.
Using the same notation as above, let ei be the identity element of the algebra Si (A).
Under the isomorphism between Si (A) and Mni (C), ei corresponds to the identity
ni × ni matrix. Then Si (A) = Aei A and ei ∈ Z(A). Let Γi : Si (A) → Mni (A) be
an algebra isomorphism. Define the algebra homomorphism ρi : A → Mni (C) by
ρi (x) = Γi (xei ). Then ρi is an irreducible representation of A corresponding to the
module Si . (This can be seen, for example, by observing that ρi (ei ) = Ini and that
Si is the onle simple A-module S with the property that ei T = T .)
Since Z(Si (A)) = Cei , it is easy to see that Z(A) = Ce1 ⊕ · · · ⊕ Cek , that is,
{e1 , . . . , ek } is a basis of Z(CG). The elements e1 , . . . , ek are known as primitive
central idempotents.
7