Representations of finite groups IV January 10, 2015 We continue to focus on representations of finite groups over C. Although we have proved many interesting theorems on the number of irreducible representations and their dimensions, what's missing is that we have very few ways to actually construct the irreducibles. The main topic of Part IV, induced representations, will provide a powerful new tool in this direction. For this purpose, we first introduce tensor products over an F -algebra. We also take the opportunity to discuss more systematically \pulling back" a representation of G 0 along a homomorphism G −!G, and the resulting right action of Aut G on RepF G. Among other things, this helps to clarify Mackey's double-coset formula and irreducibility criterion. Some of the general constructions below work over an arbitrary field and with any group. So we let F denote any field, G any group, until otherwise specified. We also point out that these constructions have simpler prototypes in the world of G-sets; you may find it helpful to read (or re-read) the G-set notes on \balanced products" and \induced G-sets"; see also the exercises to those notes. 1 Tensor products over F -algebras Let R be an F -algebra, and suppose given a right R-module M and a left R-module N. Then the tensor product over R, denoted M ⊗R N, is the F -module defined by M ⊗R N = (M ⊗F N)=Z, where Z is the F -submodule generated by all elements of the form xr⊗y−x⊗ry (x 2 M, y 2 N, r 2 R). Note that in general there is no natural R-module structure on M ⊗R N; it is only an F -module. If R is commutative, however, then there is no right/left distinction, and moreover M ⊗R N has a natural R-module structure given by r · (x ⊗ y) = (rx) ⊗ y = x ⊗ (ry). This is analogous to the corresponding fact for Hom: If M; N are both left R-modules, say, then in general HomR(M; N) has no natural R-module structure, but as we've seen it does have one when R is commutative. Remark. The tensor product is exactly analogous to the balanced product of G-sets. More- over, the permutation representation functor G-set −! FG-mod given by X 7! FX is compatible with the two constructions: If X is a right G-set and Y a left G-set, then there ∼ is a natural isomorphism of F -modules F (X ×G Y ) = FX ⊗FG FY . Let's note right away that this new tensor product can behave very differently from tensor product over F . In particular, a tensor product of two nonzero modules can be zero. For example, suppose R is a principal ideal domain (e.g. F [x]) and M = R=(a), N = R=(b) with 1 a; b relatively prime. Then by construction M ⊗R N is annihilated by both a and b, hence is zero. We next consider various formal properties, beginning of course with a universal property. If V is any F -module, an F -bilinear map β : M ×N−!V is R-balanced if β(xr; y) = β(x; ry) for all r; x; y. For example the natural map α : M × N−!M ⊗R N given by α(x; y) = x ⊗ y is R-balanced, by the very definition. Note that an R-balanced map is F -bilinear. Proposition 1.1 If β : M × N−!V is R-balanced, then there is a unique F -module homo- morphism β : M ⊗R N−!V such that the following diagram commutes: β M × N - V α p p p p9p !β ? p p p p M ⊗R pNp Proof: By the universal property of tensor product over F , there is a unique γ : M ⊗F N−!V commuting in the appropriate diagram. It is then clear that γ factors uniquely through M ⊗R N, yielding the desired β. As in the case of ⊗F , our tensor product is a functor in both variables. The precise way to say this is that (M; N) 7! M ⊗R N is a functor from the product category mod-R×R-mod to F-mod. Thus given module homomorphisms φ : M1−!M2 and : N1−!N2, we get an induced homomorphism φ ⊗R : M1 ⊗R N1−!M2 ⊗R N2 where (φ ⊗R )(x ⊗ y) = φ(x) ⊗ (y). Proposition 1.2 ⊗R distributes over direct sums in both variables, up to natural isomor- phism. The meaning and the proof of the theorem are left to the reader; compare the corre- sponding fact for tensor products over F . Next we'd like to say that tensor product is associative, but this doesn't even make sense because of the left/right dichotomy. But with some extra structure we can do it, as we digress to explain. 1.1 Bimodules Suppose R; S are rings. An (R; S)-bimodule is an abelian group M equipped with left R- module and right S-module structures (with the same underlying abelian group) such that r(xs) = (rx)s for all r 2 R, s 2 S, x 2 M. When R; S are F -algebras, which is the case we are concerned with here, unless otherwise specified we require further that the F -module 2 structures obtained from R and S are the same. The main example we need at the moment is the following: Example: Let R be an F -algebra. Then R is an (R; R)-bimodule under left and right multiplication in R; the bimodule condition is then equivalent to the associativity of mul- tiplication. More generally if A; B are subalgebras of R, then by restriction R becomes a (A; B)-bimodule. More generally still, suppose A; B are F -algebras and we are given homo- morphisms φ : A−!R, : B−!R. Then R is an (A; B)-bimodule with a · r · b = φ(a)r (b). The utility of this structure comes from: Proposition 1.3 Suppose M is an (R; S)-bimodule and N is a left S-module. Then M ⊗S N has a natural left R-module structure, given by r(x ⊗ y) = (rx) ⊗ y. Proof: One has only to check that the given formula yields a well-defined map R × (M ⊗S N)−!M⊗SN (use universal properties); then the conditions for an R-module follow trivially. Example and definition. Suppose S ⊂ R is a subalgebra. Then we get a functor S-mod −! R-mod by treating R as an (R; S)-bimodule and for any S-module M forming the R-module R ⊗S M. In particular we can take R = FG for a group G and S = FH for a subgroup H. If V is a representation of H, then FG ⊗FH V is the induced representation of G, often G denoted IndH V . These are studied in detail in Serre x7, x8, and below. Induced representations are exactly analogous to induced G-sets. In fact if we consider the permutation representation functor X 7! FX (where X is a G-set or H-set as appropriate), ∼ then for an H-set X we have a natural isomorphism F (G ×H X) = FG ⊗FH FX (check this!). In particular, if X is the trivial one-element H-set, so FH = F with trivial H-action, G we see that IndH F is just the permutation representation defined by G=H. Remark. If M is a left R-module, R ⊗R M is naturally isomorphic to M. The proof is identical to the proof for fields given earlier. Combining this with the fact that tensor products distribute over direct sums, we see that if M is a free right R-module with basis ∼ indexed by a set I, then M ⊗R N = ⊕I N. (A similar result holds for N a free left R-module.) An important example: When H is a subgroup of G, FG is a free right FH-module, with basis given by any set of left coset representatives in G=H. Hence as F -modules an induced representation can be written FG ⊗FH V = ⊕g2G=H gV; where the abusive notation g 2 G=H means that g ranges over a fixed but arbitrary set of coset representatives. We can also give the promised associativity result. Proposition 1.4 Suppose L is a right R-module, M is an (R; S)-bimodule, and N is a left S-module. Then there is a natural isomorphism ∼ (L ⊗R M) ⊗S N = L ⊗R (M ⊗S N): 3 This can be proved similarly to the corresponding statement for F -modules. Now suppose φ : S−!R is an F -algebra homomorphism. If N is a left R-module, we write φ∗N for N regarded as a left S-module via φ. We call this \pullback along φ" (not to be confused with other categorical uses of the term \pullback" that you may encounter). If M is a left S-module, we denote by j the S-module homomorphism M−!R ⊗S M given by j(x) = 1 ⊗ x. Proposition 1.5 Suppose M is a left S-module, N is a left R-module, and f : M−!N is a homomorphism of left S-modules. Then there is a unique homomorphism of R-modules f : R ⊗S M−!N such that the following diagram commutes: f MN- j p 9p!f ? p p p p R ⊗S Mp p Proof: As usual in these situations, the motto \uniqueness yields existence" applies.