REPRESENTATION THEORY, LECTURE 0. BASICS


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1 REPRESENTATION THEORY, LECTURE 0. BASICS IVAN LOSEV Introduction The aim of this lecture is to recall some standard basic things about the representation theory of finite dimensional algebras and finite groups. First, we recall restriction, induction and coinduction functors. Then we recall the Schur lemma and deduce consequences about the action of the center and the structure of completely reducible representations. Then we explain the structure and representation theory of simple finite dimensional algebras over algebraically closed fields. Next, we proceed to semisimple algebras. Finally, we use the latter to recall basics about the representation theory of finite groups. 1. Restriction, induction and coinduction Let A, B be associative unital algebras over a field F with a homomorphism B A. Let M be an Amodule. Of course, we can view M as a Bmodule. On the other hand, A is a left Amodule and a right Bmodule. These two operations commute, one says in this case that A is an ABbimodule. It follows that, for a Bmodule N, the space A B N carries a natural structure of a left Amodule (induced module). Also A is a BAbimodule. So Hom B (A, N) is a left Amodule via (aφ)(a ) := φ(a a), φ Hom B (A, N), a, a A. This is a coinduced module. Lemma 1.1. For an Amodule M and a Bmodule N we have natural isomorphisms Hom B (N, M) = Hom A (A B N, M), Hom B (M, N) = Hom A (M, Hom B (A, N)). Proof. Consider the map Hom B (N, M) Hom A (A B N, M) that sends η to ψ η given ψ η (a n) = aη(n) and the map in the opposite direction that sends ψ to η ψ given by η ψ (n) := ψ(1 n). It is left as an exercise to check that the maps are welldefined (i.e., land in the required Hom spaces) and are mutually inverse. Establishing a natural isomorphism Hom B (M, N) = Hom A (M, Hom B (A, N)) is also left as an exercise. 2. Schur lemma and its consequences 2.1. Schur lemma. The following important result is known as the Schur lemma. Proposition 2.1. Let F be algebraically closed, A be an associative unital Falgebra and let U, V be finite dimensional irreducible Amodules. Then the following is true. (1) If U, V are nonisomorphic, then Hom A (U, V ) = 0. (2) End A (U) consists of constant maps. In particular, dim End A (U) = 1. Under some assumptions, this can be generalized to infinite dimensional irreducible modules. 1
2 2 IVAN LOSEV 2.2. The action of the center. Consider the center of A, Z(A) := {z A za = az, a A}. This is a commutative algebra. The following claim is a corollary of the Schur lemma. Corollary 2.2. Let z Z(A), and let U be a finite dimensional irreducible Amodule. Then z acts on U by a scalar. It follows that there is an algebra homomorphism χ U : Z(A) F (called the central character of U) such that z acts on U as the multiplication by χ U (z) Multiplicities in completely reducible modules. Let A be an associative unital algebra over an algebraically closed field F. Let V i, i I, be the finite dimensional irreducible Amodules, where I is an indexing set. Now let V be a completely reducible finite dimensional Amodule. Since V is completely reducible, there are nonnegative integers m i, i I, with only finitely many nonzero such that V = V m i i. The following lemma is a consequence of the Schur lemma and the additivity of Hom s: Hom A (V i, V V ) = Hom A (V i, V ) Hom A (V i, V ). Lemma 2.3. The number m i coincides with dim Hom A (V i, V ). We call Hom A (V i, V ) the multiplicity space for V i in V. The name is justified by the observation that the natural homomorphism (2.1) V i Hom A (V i, V ) V, v i φ i φ i (v i ) is an isomorphism of Amodules Endomorphisms of completely reducible modules. The isomorphism (2.1) together with the Schur lemma imply the following description of the endomorphism algebra End A (V ): End A (V ) = End(Hom A (V i, V )). Here we assume that the endomorphisms of the zero space are zero (and so we sum over all i such that Hom A (V i, V ) 0, in particular, the sum is finite). The End A (V )module structure on Hom A (V i, V ) is given by the composition: φ ψ := φ ψ, φ End A (V ), ψ Hom A (V i, V ). 3. Simple algebras 3.1. Burnside theorem. Let A be an associative algebra over an algebraically closed field F and let V be a finite dimensional Amodule. So we have an algebra homomorphism A End(V ). Proposition 3.1. If V is irreducible, then the homomorphism A End(V ) is surjective. Proof. The proof is in several steps. Step 1. Consider the Amodule V M, where M is a finite dimensional vector space (and A acts on the first factor). We claim that every Asubmodule U V M has the form V M 0, where M 0 is a subspace in M. Indeed, Hom A (V, U) Hom A (V, V M). By the Schur lemma, the target space is naturally identified with M. The subspace Hom A (V, U) M is M 0 we need: by complete reducibility, if U V M 0, there is a homomorphism φ : V U that does not lie in M 0.
3 REPRESENTATION THEORY, LECTURE 0. BASICS 3 Step 2. The space V is a right Amodule via (φ a)(v) := φ(a v), φ V, v V, a A. Note that V is irreducible (if U V is a proper submodule, then the annihilator of U is a proper Asubmodule in V ). Step 3. Recall that End(V ) is naturally identified with V V. Both End(V ) and V V are Abimodules and the isomorphism End(V ) = V V is that of Abimodules. Replacing A with its image in End(V ), we may assume that A End(V ). Clearly, A V V is a subbimodule. Apply Step 1 to A viewed as a left Amodule. We get that A = V M, where M V. Applying (the obvious analog of) Step 1 to the right Amodule V V, we get A = M V, where M V. Since V M = M V, we see that M = V, M = V Simple algebras over algebraically closed fields. Let A be an associative unital algebra over F. We say that A is simple if it has no proper twosided ideals. For example, Mat n (F) is a simple algebra, this can be deduced similarly to Step 3 of the proof of the Burnside theorem. Proposition 3.2. Let F be algebraically closed and A be a finite dimensional simple A algebra. Then A = Mat n (F) for some n. Proof. Consider a minimal (w.r.t. inclusion) left ideal I A (just take a left ideal of minimal dimension). It is an irreducible left Amodule. So we get a homomorphism A End(I). Its injective because A is simple. It is surjective by the Burnside theorem. So A End(I) Representations of the matrix algebra. Let V be a finite dimensional vector space over F. We are going to understand the representation theory of the algebra A = End(V ). Proposition 3.3. Every finite dimensional Amodule U is completely reducible and the only irreducible module is V itself. Proof. There are u 1,..., u k U such that U = Au Au k. This gives an Amodule epimorphism A k U, (a 1,..., a k ) a 1 u a k u k. Moreover, A = V n, where n = dim V and so V nk U. Being a quotient of a completely reducible module, U is completely reducible itself. If U is irreducible, Hom A (V nk, U) 0. Therefore Hom A (V, U) 0. By the Schur lemma, V = U. 4. Semisimple algebras Let A be a finite dimensional algebra over F (still assumed to be algebraically closed). We say that A is semisimple, if it is a direct sum of simple algebras Criteria for semisimplicity. We are going to explain some criteria for semisimplicity. Lemma 4.1. Let A be a finite dimensional associative unital algebra. Let I, J be twosided ideals in A consisting of nilpotent elements (we say that a A is nilpotent if a n = 0 for some n > 0). Then I + J is a twosided ideal consisting of nilpotent elements. The proof is left as an exercise. So A has the unique maximal twosided ideal consisting of nilpotent elements, it is called the radical of A and is denoted by Rad(A). Also we can define a distinguished element tr A A. It sends a A to the the trace of m a, the operator A A given by m a (b) = ab. Note that tr A (ab) = tr A (ba). So (a, b) A := tr A (ab) is a symmetric bilinear form.
4 4 IVAN LOSEV Proposition 4.2. Let A be a finite dimensional algebra over an algebraically closed field F. The following conditions are equivalent. (i) The algebra A is semisimple. (ii) Rad(A) = {0}. (iii) A is completely reducible as a left Amodule. (iv) Every finite dimensional representation of A is completely reducible. If the characteristic of F is zero, then (i)(iv) are equivalent to the following condition. (v) The form (, ) A is nondegenerate. Proof. Proof of (i) (ii). Note that the radical of a simple algebra is zero. Also the radical of the direct sum is the direct sum of radicals. This proves the required implication. Proof of (ii) (iii). We have an Amodule filtration A = A 0 A 1... A n A n+1 = 0 such that A i /A i+1 is irreducible for all i = 0,..., n. Consider the corresponding algebra homomorphism φ : A n i=0 End(A i/a i+1 ). The inclusion a ker φ is equivalent to aa i A i+1 for all i. So for any a ker φ we have a n+1 = 0 and hence ker φ Rad(A). Therefore ker φ = {0} and we have an embedding A n i=0 End(A i/a i+1 ) of algebras, so, in particular, of left Amodules. The Amodule n i=0 End(A i/a i+1 ) = n i=0 (A i/a i+1 ) (A i /A i+1 ) is completely reducible. Being a submodule in a completely reducible module, A is completely reducible. Proof of (iii) (iv) repeats (a part of) the proof of Proposition 3.3. Proof of (iv) (i). We just need that A is completely reducible. Let A = k i=1 V m i i, where all V i are irreducible and all m i are positive. Since A is a faithful Amodule (only zero acts by zero), the same is true for k i=1 V i. So we get an algebra embedding A k In particular, this is a left Amodule embedding. It follows that m i dim V i, as the right hand side is the multiplicity of V i in the left Amodule k On the other hand, by the Burnside theorem, the composition of the embedding A k i=1 End(V i) with the projection to End(V i ) is surjective. It follows that m i dim V i. We conclude that m i = dim V i and A k Proof of (ii) (v). It is enough to show that the radical of A coincides with ker(, ) A when char F = 0. Indeed, if a is in the radical, then ab is nilpotent for all b A, and (a, b) A = 0. So Rad(A) = ker(, ) A. On the other hand, ker(, ) A is a twosided ideal because (ab, c) A = (a, bc) A for all a, b, c A. Also if a ker(, ) A, then tr A (a n ) = 0 for all n > 0. It follows that a is nilpotent (here we use that char F = 0). We see that ker(, ) A is contained in the radical. This finishes the proof of (ii) (v) Representations of semisimple algebras. Lemma 4.3. Let A = k Then the set of irreducible Amodules, to be denoted by Irr(A), coincides with {V 1,..., V k }. Proof. Let A 1, A 2 be associative algebras. Then any A 1 A 2 module V canonically decomposes as V 1 V 2, where V i is an A i module. Namely, if e i is the unit in A i, then V i = e i V. In particular, Irr(A 1 A 2 ) = Irr(A 1 ) Irr(A 2 ). The claim of the lemma follows from here (and trivial induction). Often one wants to compute the number of irreducible representations of A without referring to the decomposition A = k Lemma 4.4. Let A be a semisimple algebra. Then Irr(A) = dim Z(A).
5 REPRESENTATION THEORY, LECTURE 0. BASICS 5 Proof. This follows from the observation that the center of End(V i ) consists of scalar operators combined with Z(A 1 A 2 ) = Z(A 1 ) Z(A 2 ) Irreducible representations of arbitrary finite dimensional algebras. Now let A be a finite dimensional Falgebra. By (ii) of Proposition 4.2, the algebra A/ Rad(A) is semisimple. Besides Rad(A) acts by 0 on all irreducible representations. We conclude that pulling back a representation from A/ Rad(A) to A gives rise to a bijection between Irr(A/ Rad(A)) and Irr(A). 5. Finite groups 5.1. Group algebra and its semisimplicity. Let G be a finite group and F be a field. We can form the group algebra FG of G, a vector space with basis G, where the basis elements multiply as in G. A representation of G is the same thing as a representation of FG. Proposition 5.1. Let F be algebraically closed and of characteristic 0. Then FG is a semisimple algebra. In particular, any finite dimensional representation of G over F is completely reducible. Proof. We will check (v) of Proposition 4.2. On the basis elements, we have tr FG (g) = δ g1 G, where δ g1 is the Kroneker symbol. So (g, h) FG = δ g,h 1 G. Clearly, this form is nondegenerate. We remark that this proposition is no longer true when the characteristic of F is positive The number of irreducible representations. By Lemma 4.4, the number of the irreducible representations of G coincides with the dimension of the center. So let us investigate the structure of Z(FG) as a vector space. Proposition 5.2. There is a basis b C Z(FG), where C runs over the set of conjugacy classes in G. It is given by b C := g C g. Proof. The inclusion g G c gg Z(FG) is equivalent to h g G c gg = g G c ggh, which in its turn is equivalent to g G c g(hgh 1 ) = g G c gg. In other words, g G c gg Z(FG) if and only if the function g c g is constant on conjugacy classes. This implies the claim of the proposition. 6. What happens when F is not algebraically closed First, we need to modify the second part of the Schur lemma: the endomorphism algebra End A (V ) is a skewfield. An analog of the Burnside theorem still works: the image of A in End(V ) is End S (V ), where S is the skewfield End A (V ). The proof is somewhat more involved. The simple algebras are precisely Mat n (S), where S is a finite dimensional skewfield over F, the proof repeats that of Proposition 3.2. An analog of Proposition 3.3 holds for Mat n (S). An analog of Proposition 4.2 holds too. The details are left to the reader.
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