REPRESENTATIONS OF GROUP ALGEBRAS. Contents. 1. Maschke s theorem: every submodule is a direct summand. References 6

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1 REPRESENTATIONS OF GROUP ALGEBRAS ZAJJ DAUGHERTY MARCH 27, 28 Contents. Maschke s theorem: every submodule is a direct summand References 6 See [DF, Ch. 8] for proofs and more examples. Jumping back into chapter 8, we ll mostly be considering group algebras for a while. Before we do, we consider a broader definition. A ring R is semisimple if every R module is isomorphic to the direct sum of simple modules. Equivalently, R is semisimple if it is a direct sum of simple rings. For example, we ve seen that not every Z module is projective. So not every SES splits. And so there are non-decomposable reducible Z modules. In particular, the left regular module Z has submodules nz for every n. So no submodule of Z is simple! In contrast, we will show in this section that for any finite group G and field F whose characteristic of F does not divide the order of G, we have that F G is semisimple. So for example, CS n is semisimple because C has characteristic. On the other hand, F 3 has characteristic 3, which divides S 3 6; and indeed F 3 S 3 has indecomposible reducible modules (e.g. the permutation module).. Maschke s theorem: every submodule is a direct summand Theorem. (Maschke s theorem [DF, Thm. 8..]). Let G be a finite group and let F be a field whose characteristic does not divide. Let V be an F G-module, and U be a submodule. Then there is also a submodule W satisfying V U W. Example.2. Let G be a group of order n, F a field of characteristic not dividing n, and consider the left regular module F G over itself. Define the symmetric bilinear form x, y on F G by xg, hy δ g,h for all g, h P G, and extend linearly. () x, y is non-degenerate: If a ř α gg satisfies So for all h P G. So a. xa, F Gy, then xa, hy for all h P G. C G ÿ α g g, h ÿ α g xg, hy α h

2 2 ZAJJ DAUGHERTY MARCH 27, 28 (2) x, y is invariant: Fix an order on G, and let that determine an ordered basis for F G as a vector space. Then for g P G, since g acts on G by a permutation, the associated matrix M MG G pρpgqq for the action of g on F G is a permutation matrix (Ch4: a group acts on itself by left multiplication, which induces a homomorphism into S G ). In particular, M M t. So, identifying a P F G with MB paq, we have for all a, b P F G. Now let U be a submodule of F G, and define xga, by pmaq t b a t M t b a t M b xa, g by U K tv P F G xu, vy u. Since x, y is nondegenerate, we have dimpu K q dimpf Gq dimpuq. Since x, y is bilinear, U K is closed under subtraction: If xu, vy xu, v y, then xu, v v y xu, vy xu, v y. So since P U K, U K is at least a subgroup. Also by linearity, U K is a subspace: for all α P F, we have xu, αvy αxu, vy whenever xu, vy. Moreover, since x, y is G-invariant, U K is also closed under the action by F G: If xu, vy for all u P U, then for all g P G, xu, gvy xg u, vy, since U is closed under the G-action. So gv P U K for all g P G, and thus ř αg v P U K. Finally, we must show that U ` U K F G and U X U K. The first implies second by a dimension count: since x, y is non-degenerate, we have dimpuq ` dimpu K q dimpf Gq. If they had non-trivial overlap, then we would have dimpuq`dimpu K q ă dimpf Gq, a contradiction. But the fact that U ` U K F G is a special artifact of dot product. Namely, in any vector space V F n with subspace U, any vector v can be decomposed as v u`u K where u P U and u K P U K by projecting v onto U to get u, and then letting u K ` v u. For example, if U is one-dimensional: For any a, b P V, the vector b K a xa,by xb,by b is orthogonal to b, since and xb K, by xa, by xa, by xb, by ; xb, by () a b {{ ` b K, where b {{ xa, by xb, by b. (This should look very familiar, from vector calculus and/or linear algebra.) For larger dimensional subspaces: For all v P V, the goal is to find u P U and u K P U K such that v u ` u K. Now let B tu,..., u m u be a basis of U, and write u α u ` ` α m u m for some α i P F. Finding such a u so that v u P U K is the same as solving xu i, v uy xu i, vy ` α xu i, u y ` ` α m xu, u m y for all i,..., m.

3 REPRESENTATIONS OF GROUP ALGEBRAS 3 But since x, y is the dot product on V, this is precisely equivalent to finding α pα,..., α m q t such that M t pv Mαq, where M u u m P M n,m pf q. Note that M is not square, but M t M P M m pf q is. independent, M t M is invertible. So Then And since the columns of M are linearly M t pv Mαq iff M t Mα M t v iff α pm t Mq M t v. u α u ` ` α m u m Mα MpM t Mq M t v P U, and u K v u P U K. Sanity check : Equation () is the case when m ; so M b and MpM t Mq M t a bpb t bq b t a bpxb, byq xb, ay xa, by xb, by b, as desired. Sanity check 2: If you started with a basis that was orthonormal (pairwise orthogonal and xb i, b i y ), and considered the space spanned by the first m basis vectors, then M m.... So M t M I m and MpM t Mq M t MM t I m n m... m m..., which is indeed the projection onto the space spanned by the first m basis vectors as desired. Finally, note that π MpM t Mq M t : V Ñ U is a F G homomorphism: we have π U id u which is a homomorphism on U, and for any v P V, writing v u ` u K, we have gu K P U K (since x, y is G-invariant), so that πpgvq πpgu ` gu K q gu gπpvq. Which parts of this example were particular to F G V on the ring side?? And which parts were particular to F G R on the ring side? Certainly, most of it depended on there being a bilinear form on the module, and at certain steps we used the fact that it was non-degenerate or invariant

4 4 ZAJJ DAUGHERTY MARCH 27, 28 at different times, or the fact that we had an orthonormal basis (something that made the form into a dot product). On the side of the ring, F G is an algebra over a field, so every finite-dimensional F G module V is a vector space isomorphic to F d for d dimpv q. Taking the standard dot product on F d, our projection process in Example.2 will still produce an orthogonal complement U K to any subspace U Ď V, and show that V U U K as vector spaces. The only missing piece is to show that U K is a submodule (is closed under the action of F G). Specifically, in our example above, invariance came from the fact that under the representation ρ associated to the module F G, for every g P G we had ρpgq ρpgq t.this is certainly not always the case for all modules/representations; in particular, it requires tools specific to the module. And in fact, it is not always the orthogonal complement that we want anyway! Even if you have a nondegenerate bilinear form, but which is not G-invariant, the orthogonal complement to a submodule is usually not itself a submodule. Example.3. Consider the action of CS 2 on itself. Fix the basis B tv, v 2 u, where v and v 2 p2q, and use it to identify CS 2 with C 2 and EndpCS 2 q with M 2 pcq. Consider the bilinear form ˆ x, y x, y J where J. Since J is symmetric and invertible, we know x, y is non-degenerate and symmetric. However, and ˆ ˆ ˆ xv, v y p, q p, q and ˆ ˆ xv 2, v 2 y p, q ˆ p, q, so that xp2qv, p2qv y xv 2, v 2 y xv, v y. Thus x, y is not S 2 -invariant. Now, let s look at the decomposition of CS 2. We already know that T Ce` is isomorphic to the trivial module, where e` v ` v 2. Further, T K tv P CS 2 xv ` v 2, vy u So v αv ` αv 2 P T K whenever ˆ ˆ xv, v ` v 2 y pα, α 2 q But that implies that T K T, so that T ` T K T V. ˆ pα, α 2 q α α 2. Example.4. Consider the action of CS 2 on itself again with the same basis, but this time with ˆ x, y x, y J where J, 2 which again gives a non-degenerate and symmetric bilinear form. Again, ˆ ˆ xv, v y p, q 2 and ˆ ˆ xv 2, v 2 y p, q 2, 2

5 REPRESENTATIONS OF GROUP ALGEBRAS 5 so that x, y is not S 2 -invariant. Again, let T Ce` and consider T K : v αv ` αv 2 P T K whenever ˆ ˆ ˆ xv, v ` v 2 y pα, α 2 q pα 2, α 2 q α 2 ` 2α 2. So T K Cp2v v 2 q. This time, we do get that T X T K and T ` T K V, so that V T T K as vector space. However, this is still the wrong decomposition as CS 2 -modules because T K is not closed under the S 2 action: p2qp2v v 2 q v ` 2v 2 R T K. So V fl T T K. The last two examples do not imply that V doesn t decompose as a CS 2 -module. We ve seen already that, as CS 2 -modules, V T S, where S Ce Cpv v 2 q. The issue was that we couldn t use the chosen forms to do the decomposition for us. Namely, in Example.2, we had an F G-module homomorphism π MpM t Mq M t : V Ñ U satisfying π U id. The goal in the following proof is to compute a similar F G homomorphism, without the use of an inner-product. Proof of Maschke s theorem. Let G be a finite group and let F be a field whose characteristic does not divide. Let V be an F G-module, and U be a submodule. The idea is to construct a module homomorphism π : V Ñ U such that π U id U. Namely, since πpvq P U Ď V for all v P V, we have π 2 pvq πpπpvqq πpvq (π is an idempotent in EndpV q). This homomorphism will be the analogue to the projections we saw at the end of Example.2 (see using π below). Existence of π: Since V is a vector space and U is a subspace, there is a complement U to U (take a basis of U, extend it to a basis of V, and then let U be the subspace spanned by the basis vectors of V that aren t in U). By definition, V U ` U, and certainly V U U as vector spaces, but we do not yet know anything about whether U K is closed under the F G-action, so we can t say anything yet about the decomposition as F G-modules. However, we do know that U X U, so for all v P V, there are unique u P U and w P U such that v u ` w. Let π be the projection onto U, so that πpu ` wq u. Since the choice of u and w are unique, we know π is well-defined; it is also linear because u and w are closed as subspaces. It may not be an F G homomorphism yet, though, because for some w P U, we may have gw R W so that gπpwq π pgwq. To fix this, we ll average over all twists of π by elements of G as follows. First, consider gπ g restricted to U: since π U id U, we have gπ U g gid U g gg id U id U. You might want to call U the orthogonal complement at this moment, but by those examples above, we know that s not always what we re talking about here.

6 6 ZAJJ DAUGHERTY MARCH 27, 28 So for u P U, ÿ u u. gπ g puq ÿ If charpf qˇˇ, then this is ; otherwise, we can solve for u. Namely, if π ÿ gπ g, then πpuq u. And still, since π is linear and the action of F G is linear, we have π is linear. It remains to show that π commutes with the action of G, showing that an F G-homomorphism on V. To that end, let h P G and v P V. Then πphvq ÿ gπ g phvq ÿ phh qgπ g phvq ÿ hph gqπ ph gq v ÿ hkπ k v kpg ÿ h kπ k pvq hπpvq, kpg since G acts on itself as a permutation by left multiplication (so we can reindex over k h g). // Using π: Now that we have π, let W kerpπq. Since π is a F G-module homomorphism, W is a submodule of V. Now suppose v P U X W. Then since v P U, we have πpvq v; and since u P W, we have πpvq. So v, implying U X W. Finally, for any v P V, we have Therefore, Thus U ` W V. So πpv πpvqq πpvq π 2 pvq ; so v πpvq P W. v u ` w, where u πpvq P U and v v πpvq P W. V U W imgpπq kerpπq. References [DF] D. Dummit and R. Foote, Abstract Algebra, Third edition. John Wiley & Sons, Inc., Hoboken, NJ, 24. xii+932 pp. ISBN: