Symmetric Groups and Iwahori–Hecke Algebras
4.5 Semisimple algebras and modules
4.5.1 Semisimple modules
LetA be an algebra. By an A-module, we mean a left A-module, that is, a K-vector spaceM together with aK-bilinear mapA×M →M,(a, m)→am such that a(bm) = (ab)m and 1m = m for all a, b ∈ A, and m ∈ M. The map a → (m → am) (a ∈ A, m ∈ M) defines an algebra homomorphism A→EndK(M) with values in the algebra ofK-linear endomorphisms ofM. Conversely, any algebra homomorphismχ :A →EndK(M) gives rise to an A-module structure onM byam=χ(a)(m) fora∈Aandm∈M.
By a finite-dimensional A-module we mean an A-module that is finite- dimensional as a vector space overK.
A homomorphism of A-modules f : M → M is a K-linear map such thatf(am) =af(m) for all a∈A andm∈M. We write HomA(M, M) for the vector space of all homomorphisms ofA-modulesM →M. We also set EndA(M) = HomA(M, M).
IfM is a linear subspace of an A-moduleM such thatam ∈M for all a∈A and m ∈ M, then we say that M is a A-submodule or, for short, a submodule ofM. In this case, the embeddingM →M is a homomorphism ofA-modules.
Definition 4.27.(a) An A-module M is simple if M has no A-submodules except0 andM.
(b) An A-module is semisimple if it is isomorphic to a direct sum of a finite number of simpleA-modules.
(c) An A-module M is completely reducible if for any A-submodule M ofM there is anA-submodule M such thatM =M⊕M.
Note that a simple A-module is semisimple, and if an A-module M is completely reducible, then any short exact sequence ofA-modules
0→M→M →M→0 splits, i.e.,M ∼=M ⊕M.
Proposition 4.28.Let M be a finite-dimensional A-module. The following assertions are equivalent.
(i)M is semisimple.
(ii)M is completely reducible.
(iii)M =
i∈I Mi is a sum of simple submodules Mi.
Proof. We first prove the implication (ii) ⇒(iii). Assume that M is nonzero and completely reducible. SinceM is finite-dimensional overK, it must have nonzero submodules of minimal dimension as vector spaces overK; such sub- modules are necessarily simple. Consider the sumM⊂M of all simple sub- modules ofM. This is a nonzero submodule ofM. We are done ifM=M. If not, sinceM is completely reducible, there is a nonzero submoduleM⊂M such thatM =M⊕M. A nonzero submodule ofM of minimal dimension is a simple submodule ofM that is not inM. This contradicts the definition ofM. Therefore,M=M.
We next prove the implication (iii)⇒(i). Suppose thatM =
i∈I Mi is a sum of simple submodulesMi. LetI ⊂I be a maximal subset such that Mi = 0 for i ∈ I and the sum
i∈I Mi is direct. Such a subsetI exists and is finite becauseM is finite-dimensional. LetM =
i∈I Mi. We claim that M = M, which implies that M is a direct sum of a finite number of simple submodules. To prove the claim, it suffices to check that Mk ⊂ M for any k ∈ I−I. Clearly, Mk ∩M is a submodule of Mk. Since Mk is simple, either Mk∩M = 0 or Mk∩M = Mk. If Mk∩M = 0, then the
sum
i∈I∪{k}Miis direct. This contradicts the maximality ofI. Therefore, Mk∩M =Mk, which implies thatMk⊂M.
We finally prove the implication (i)⇒ (ii). Suppose thatM =
i∈I Mi is a direct sum of simple submodules, where I is a finite indexing set. Let M be a submodule ofM. Consider a maximal subset I ⊂I such that the sumM+
i∈I Mi is direct. Reasoning analogous to that in the previous paragraph shows that M +
i∈I Mi = M. Set M =
i∈I Mi. Then M⊕M=M, which proves thatM is completely reducible.
Proposition 4.29.LetM be a finite-dimensional semisimpleA-module. Any A-submodule and any quotientA-module of M is semisimple.
Proof. Let M0 be a submodule of M. Let M0 be the sum of all simple submodules of M0. Since by Proposition 4.28, M is completely reducible, M = M0 ⊕M for some submodule M of M. Together with M0 ⊂ M0, this implies thatM0=M0 ⊕(M0∩M). If M0∩M= 0, then this module contains a nonzero simple submodule, which is then contained in M0. This is impossible. Therefore, M0∩M = 0 and M0 = M0 is a sum of simple submodules. Using Proposition 4.28, we conclude thatM0 is semisimple.
Consider the quotient of M by a submodule M. By Proposition 4.29, there is a submodule M ⊂M such that M = M⊕M. By the previous paragraph,M is semisimple; hence so isM/M∼=M.
Recall that a division ring is a ring in which each nonzero element is invertible. A left module over a division ringDis called a leftD-vector space.
Any left D-vector space V has a basis, and two bases of V have the same cardinality, so that the concept of the dimension dimDV of V makes sense (these results can be proved in the same way as the corresponding ones for vector spaces over a field).
The following proposition is calledSchur’s lemma.
Proposition 4.30.(a) Let M and M be simple A-modules. If M and M are not isomorphic asA-modules, thenHomA(M, M) = 0.
(b) The ring EndA(M) of A-module endomorphisms of a nonzero simple A-moduleM is a division ring.
(c) If the ground fieldK is algebraically closed andM is a nonzero finite- dimensional simple A-module, thendimKEndA(M) = 1.
Proof. (a) Let f ∈ HomA(M, M). The kernel Ker(f) of f is a submodule ofM. SinceM is simple, Ker(f) =M or Ker(f) = 0. In the first case,f = 0.
In the second case,f is injective. Its image f(M) is a submodule ofM. By the simplicity of the latter, f(M) = M or f(M) = 0. If f(M) =M, then f is an isomorphism M → M. Thus, f = 0 or f is an isomorphism. The latter contradicts the assumptions. Hence,f = 0.
(b) By the proof of (a), any nonzerof ∈EndA(M) is bijective. It is easy to check that the inversef−1 off is a homomorphism ofA-modules. Hence, f is invertible in EndA(M).
(c) For any scalar λ∈K, the endomorphism m→λmlies in EndA(M).
Conversely, let f ∈ EndA(M). Since K is algebraically closed and M is a finite-dimensionalK-vector space,f has a nonzero eigenspace for some eigen- valueλ∈K. The eigenspace Ker(f−λidM), being a nonzero submodule of the simple moduleM, must be equal to M. Hence,f =λidM. In conclusion,
EndA(M)∼=K.
Corollary 4.31.If K is algebraically closed and M, M are isomorphic nonzero finite-dimensional simpleA-modules, thendimKHomA(M, M) = 1.
LetΛbe the set of isomorphism classes of nonzero finite-dimensional sim- pleA-modules. For eachλ∈Λ, fix a simpleA-moduleVλ in the isomorphism class λ. For any integer d≥1, we denote by Vλd the direct sum of d copies ofVλ. We agree that Vλd = 0 if d = 0. The following proposition, known as the Krull–Schmidt theorem, asserts that the decomposition of a semisimple module into a direct sum of simple modules is unique.
Proposition 4.32.If for some families{d(λ)}λ∈Λ,{e(λ)}λ∈Λ of nonnegative integers, there is anA-module isomorphism
λ∈Λ
Vλd(λ) ∼=
λ∈Λ
Vλe(λ), thend(λ) =e(λ) for allλ∈Λ.
Proof. Pickλ0∈Λ and setD= EndA(Vλ0). By Proposition 4.30 (a),
Hom
λ∈Λ
Vλd(λ), Vλ0
∼=
λ∈Λ
HomA(Vλd(λ), Vλ0)
∼=
λ∈Λ
HomA(Vλ, Vλ0)d(λ)
∼= HomA(Vλ0, Vλ0)d(λ0)
∼=Dd(λ0). Similarly,
Hom
λ∈Λ
Vλe(λ), Vλ0
∼=De(λ0).
The assumptions imply thatDd(λ0)∼=De(λ0). By Proposition 4.30 (b),Dis a division ring. Therefore, taking the dimensions overD, we obtain
d(λ0) = dimDDd(λ0)= dimDDe(λ0)=e(λ0). 4.5.2 Simple algebras
Definition 4.33.An algebra A is simple if A is finite-dimensional and the only two-sided ideals ofA are0 andA.
We give a typical example of a simple algebra.
Proposition 4.34.Let V be a finite-dimensional left vector space over a di- vision ringD. Then the algebra EndD(V)is simple.
Proof. Pick a basis {v1, . . . , vd} of V. We have V = Dv1⊕ · · · ⊕Dvd. For i, j ∈ {1, . . . , d}, define fi,j ∈ A = EndD(V) by fi,j(vk) = δj,kvi for all k= 1, . . . , d(hereδj,k is the Kronecker symbol whose value is 1 ifj =k, and 0 otherwise). One checks easily that{fi,j}i,j∈{1,...,d}is a basis ofAconsidered as a vector space overD, and thatfi,j◦fk,=δj,kfi,for alli,j,k,.
LetIbe a nonzero two-sided ideal ofAand letf ∈Ibe a nonzero element.
Writef =
i,j ai,jfi,j, whereai,j ∈D for alli, j∈ {1, . . . , d}. Suppose that ak, = 0 for somek, ∈ {1, . . . , d}. Then
fk,k◦f◦f,= d i, j=1
ai,jfk,k◦fi,j◦f,=ak,fk,
belongs toI. It follows thatfk,∈I. The relationfi,j=fi,k◦fk,◦f,jimplies thatfi,j ∈Ifor alli, j= 1, . . . , d. Consequently,I=A.
The following is a converse to the previous proposition. It is a version of Wedderburn’s theorem.
Proposition 4.35.For any simple algebra A, there is a division ringD and a finite-dimensionalD-vector space V such that A∼= EndD(V).
Proof. Pick a left ideal V ⊂A of A of minimal positive dimension overK (possibly,V =A). The ideal V is an A-module, and by the minimality con- dition, it is a simple module. By Proposition 4.30 (b), D = EndA(V) is a division ring. We conclude using the next lemma withI=V. Lemma 4.36.Let A be an algebra having no two-sided ideals besides 0 and A. For any nonzero left ideal I ⊂ A, there is an algebra isomorphism A∼= EndD(I), where D = EndA(I)and I is viewed as a (left)D-module via the action ofD onI defined by(f, x)→f(x) for allf ∈D andx∈I.
In this lemma we impose no conditions on the dimension ofA, which may be finite or infinite.
Proof. For a∈A, define La (resp. Ra) to be the left (resp. the right) multi- plication byainA. By definition,
La(b) =ab and Ra(b) =ba (4.25) for allb∈A. We have
La◦Lb = Lab and Ra◦Rb= Rba (4.26) for alla,b∈A. Since I is a left ideal ofA, we have La(I)⊂I for alla∈A, which implies that La∈EndK(I). Since
La(f(x)) =af(x) =f(ax) =f(La(x))
for allf ∈D andx∈I, the endomorphism La belongs to EndD(I).
Let L : A → EndD(I) be the map sending a ∈ A to La ∈ EndD(I).
Since La◦Lb = Lab for alla, b∈ A and L1 = idI, the map L is an algebra homomorphism. Let us show that L is an isomorphism. The kernel of L is a two-sided ideal ofA. Since L = 0, the assumptions of the lemma imply that the kernel of L must be zero. This proves the injectivity of L.
The proof of the surjectivity of L is a little bit more complicated; it goes as follows. Ifx∈I, then Rx(I)⊂I. We claim that Rx∈D= EndA(I). Indeed, for alla∈A,x,y∈I,
aRx(y) =a(yx) = (ay)x= Rx(ay). Ifu∈EndD(I), then for allx,y∈I,
u(yx) =u(Rx(y)) = Rxu(y) =u(y)x . In particular, for anya∈A,x,y∈I,
u(yax) =u(y(ax)) =u(y)ax .
In other words, for alla∈Aandy∈I,
u◦Lya= Lu(y)a. (4.27)
Now,IAis a nonzero two-sided ideal ofA. By the assumptions of the lemma, IA = A. Equation (4.27) then implies that u◦Lb ∈ L(A) ⊂ EndD(I) for all u∈ EndD(I) and b ∈ A. This shows that the image of L is a left ideal of EndD(I). Since idI = L1 is in the image, the latter is equal to the whole algebra EndD(I), and the map L is surjective.