Modular Representation Theory (II)

We continue our discussion of modular representations; recall that all modules are finitely-generated even if we do not explicitly say so. First, we introduce a new notation: for each projective finitely-generated k[G]-module P, we have a unique projective finitely-generated R[G]-module denoted $\tilde P$ for which $\tilde P/\pi \tilde P \cong P.$

First we have:

Proposition. The matrices for e and d are transpose of each other. Hence, c = de is positive-semidefinite and symmetric.

Proof

We need to show $\left< x, d(y)\right>_k = \left< e(x), y\right>_K$ for all basis elements $x = [P] \in P_k(G)$ and $y = [M] \in R_K(G)$ , i.e. P is projective indecomposable and M is simple. As seen earlier:

By definition $e(x) = [K\otimes_R Q],$ where $Q:=\tilde P.$
Write $M = K\otimes_R N$ for some R[G]-module N, free over R, so that d(y) = N/πN.
We need to show $\dim_k \text{Hom}_{k[G]}(Q/\pi Q, N/\pi N) = \dim_K \text{Hom}_{K[G]}(K\otimes_R Q, K\otimes_R N).$

Consider $X:= \text{Hom}_{R[G]}(Q, N) = \text{Hom}_R(Q, N)^G$ ; this is R-free since it is a submodule of a free R-module. Our desired result would follow if we could show:

$X/\pi X \cong \text{Hom}_{k[G]}(Q/\pi Q, N/\pi N), \qquad K\otimes_R X \cong \text{Hom}_{K[G]}(K\otimes_R Q, K\otimes_R N).$

But since Q is finitely-generated and R[G]-projective, we can write $Q\oplus Q' \cong R[G]^n$ for some R[G]-module Q’ and n>0. Since direct sum commutes with tensor product, it suffices to prove the above two isomorphisms for $Q = R[G].$ Now this is obvious since in this case, X=N, and we have

$\begin{aligned}\text{Hom}_{k[G]}(Q/\pi Q, N/\pi N) &= \text{Hom}_{k[G]}(k[G], N/\pi N) \cong N/\pi N = X/\pi X\\ \text{Hom}_{K[G]}(K\otimes_R Q, K\otimes_R N) &= \text{Hom}_{K[G]}(K[G], K\otimes_R N) = K\otimes_R N = X\otimes_R N\end{aligned}$

as desired. ♦

Note

In fact, we shall see later that c is positive definite, or equivalently, it’s injective. If c(x) = 0, then $0=\left<x, de(x)\right>_k= \left<e(x), e(x)\right>_K$ so e(x) = 0. But a basis of $P_k(G)$ is given by [P] for indecomposable projective P, so it suffices to show that $e([P]) \in R_K(G)$ are linearly independent.

Modular Characters

Recall that a finitely-generated K[G]-module M can be represented by its character $\chi_M : G\to K$ where $\chi_M(g) := \text{tr}(g: M\to M),$ its trace as a K-linear map. Since K[G] is semisimple, standard character theory says $\chi_M$ uniquely determines M. Now, $\chi_M$ is a class function (i.e. it is constant on each conjugancy class) and in fact, forms an orthonormal basis of the space of class functions, where orthonormality follows from Schur’s lemma.

We would like to produce a similar theory for k[G]-modules. The naive approach of taking $G\to k$ where $g\mapsto \text{tr}(g: M\to M)$ does not lead to a satisfactory theory. The better approach is to lift the character to a function G→R.

Definition. An element g of G is said to be p-regular if its order is coprime to p; if it is not p-regular, we say it is p-singular. The collection of p-regular elements of G is denoted $G_{\text{reg}}.$ Note that this is a union of conjugancy classes.

Now assume K contains all n-th roots of 1, where n = |G|. It follows that k contains all m-th roots of 1, where m is the largest factor of n coprime to p. The m-th roots of 1 in k are all distinct and the canonical map R → R/π induces a bijection between the m-th roots of 1 in K and in k. Denote this bijection by λ.

Now we are ready to define modular characters.

Definition. Let M be a finitely-generated k[G]-module and $g\in G_{\text{reg}}.$ Consider the k-linear map $g:M\to M.$ Let r be the order of g; since r|m, k contains all the eigenvalues of g, denoted $\zeta_1(g), \ldots, \zeta_r(g).$

Now the modular character of M is defined as follows:

$\varphi_M : G_{\text{reg}} \to K, \quad g \mapsto \sum_i \lambda^{-1}(\zeta_i(g)).$

We have the following properties:

If N ⊆ M is a submodule, then $\varphi_M = \varphi_N + \varphi_{M/N}.$ This is easily seen by taking a k-basis of N, then extending to give a k-basis of M. Hence $\varphi_M =: \varphi_{[M]}$ is independent of our choice of [M] in the Grothendieck group $R_k(G).$
For any k[G]-module M, consider its k-dual M* with G acting on it via $g\cdot f := f\circ g^{-1} : M\to M.$ Then $\varphi_{M^*}(g) = \varphi_M(g^{-1}).$
For any two k[G]-modules M, N, we have $\varphi_{M\otimes_k N}(g) = \varphi_M(g) \varphi_N(g)$ for all p-regular g. [ Note that the tensor product is over k and not k[G]. ]

Next we would like to relate modular characters and standard ones, so suppose N is an R[G]-module. Then we have:

$N_K := K\otimes_R N, \qquad N_k := N/\pi N,$

a K[G]-module and k[G]-module. The first one gives a standard character $\chi_N := \chi_{N_K} : G\to K$ and the second one gives a modular character $\varphi_N := \varphi_{N_k} : G_{\text{reg}} \to K.$ From the definition, we have:

$\varphi_N = \chi_N |_{G_{\text{reg}}} : G_{\text{reg}} \to K.$

Hence, in our c–d–e diagram:

we have: $\varphi_{d(N)} = \chi_N|_{G_\text{reg}}.$

On the other hand, let us compare $\varphi_P$ and $\chi_{e(P)}.$ These clearly give the same values for all p-regular g. Furthermore, the following shows that $\varphi_P(g) = 0$ when g is p-singular.

Proposition. Let M be a finitely-generated projective R[G]-module. If g is p-singular then $\chi_M(g) = 0.$

Proof

First, note that M is projective over R[H] for any subgroup H of G since R[G] is free over R[H]. Hence, replacing G with the subgroup generated by g, we may assume G is cyclic and generated by g. This gives:

$R[G] = R[T]/\left< T^{m\cdot p^r} - 1\right>$

where m is coprime to p and r>0 since g is p-singular. We need to show that multiplication-by-T has trace zero for every indecomposable projective R[G]-module M. But this corresponds to an indecomposable projective k[G]-module N := M/πM and we may classify all N by decomposing k[G]:

$k[G] = k[T]/\left<T^{m\cdot p^r}-1\right> \cong \oplus_i k[T]/\left<(T - \zeta_i)^{p^r}\right>$

where each direct-sum factor is indecomposable since its radical $\left< T-\zeta_i\right>$ is of codimension 1. The corresponding R[G]-module is then $M_i=R[T]/\left<T^{p^r} - \zeta_i'\right>.$ Since r>0, multiplication by T has trace zero in each $M_i$ : this is readily seen by taking the basis $\{1, T, \ldots, T^{p^r-1}\}$ which maps to $\{T, \ldots, T^{p^r-1}, \zeta_i'\}.$ ♦

Summary. In the c-d-e diagram above, we have:

$\varphi_{d(N)} = \chi_N|_{G_{\text{reg}}}, \qquad \chi_{e(P)} = \varphi_P$

where in the second equality, $\varphi_P$ is extended to a function G→K by mapping all p-singular elements to zero.

Computing Pairing over k

Our next job is to find a formula for the pairing:

$\left< P, M\right>_k := \dim_k \text{Hom}_{k[G]}(P, M)$

for a projective k[G]-module P and general k[G]-module M. First some preliminaries:

Lemma. If M is a projective k[G]-module, then so are $M^*$ and $M\otimes_k N$ for any k[G]-module N.

Proof

Since M is projective, it is a direct summand of some k[G]ⁿ. Since dual and tensor product commute with finite direct sums, it suffices to prove this for M = k[G]. In this case, $M^*$ and $M\otimes_k N$ are in fact free over k[G].

To check that a general k[G]-module V is free, it suffices to find a k-subspace W ⊆ V such that $V=\oplus_{g\in G} gW.$ For the dual we pick $k\cdot f \subseteq k[G]^*$ where f : k[G]→k takes e to 1 and all other g to 0; and for the tensor product, we pick $k\cdot e\otimes_k N \subseteq k[G]\otimes_k N.$ ♦

Now we compute the pairing:

$\left< P, M\right>_k = \dim_k (\overbrace{P^* \otimes_k M}^{:=Q})^G = \dim_k Q^G = \dim_R (\tilde{Q})^G = \dim_K (K\otimes_R \tilde{Q})^G$

The third equality follows from this:

If Q is a projective k[G]-module, then $(\tilde Q)^G$ is a free R-module and $\dim_R (\tilde Q)^G = \dim_k Q^G.$
[ To prove this, note that $(\tilde Q)^G \subseteq \tilde Q$ is a R-submodule of a free R-module, so it must be R-free. Now write $\tilde Q \oplus \tilde M \cong R[G]^n$ for some R[G]-module $\tilde M$ and so $(\tilde Q)^G \oplus (\tilde M)^G\cong (R\cdot \sum_g g)^n.$ On the other hand, we have $Q\oplus M\cong k[G]^n$ so $Q^G \oplus M^G \cong (k\cdot \sum_g g)^n.$ Comparing terms then gives $(\tilde Q)^G/\pi (\tilde Q)^G \cong Q^G.$ ]

Since $\tilde Q$ is R[G]-projective, its character $\chi_{\tilde Q}$ vanishes on the p-singular elements and the above equals:

$\frac 1 {|G|} \sum_{g\in G_{\text{reg}}} \chi_{\tilde Q}(g) =\frac 1{|G|} \sum_{g\in G_{\text{reg}}} \varphi_Q(g) = \frac 1 {|G|} \sum_{g\in G_{\text{reg}}} \varphi_P(g) \varphi_M(g^{-1}).$

Thus, we define the following:

Definition. A class function $f:G_{\text{reg}} \to K$ is said to be a p-regular class function. An inner product is defined on the space of all such functions:

$\left<\phi, \psi\right>_k := \frac 1 {|G|} \sum_{g\in G_{\text{reg}}} \phi(g) \psi(g^{-1}).$

Hence, the basis $[P]\in P_k(G)$ and $[M]\in R_k(G)$ for projective indecomposable P and simple M give rise to dual bases $\{\varphi_P\}$ and $\{\varphi_M\}$ under the above inner product.

We thus have:

Theorem. The map $e:P_k(G) \to R_K(G)$ is injective. Hence $r=\text{rank} R_k(G) = \text{rank} P_k(G)$ is the number of p-regular conjugancy classes of G.

Proof

We will show that for projective indecomposable P, the collection of $\varphi_P$ is linearly independent over K and hence over Z. Indeed, if $\sum_P c_P \varphi_P = 0$ then taking the inner product with $\varphi_M$ for each simple M gives:

$0 = \sum_P c_P \left< \varphi_P, \varphi_M\right>_k\ \forall\ M\implies c_P = 0\ \forall\ P$

since {[M]} forms a dual basis for {[P]}.

Note

This also shows that d, being the transpose of e, is surjective when extended to K. In fact, it is even true that $d: R_K(G)\to R_k(G)$ is surjective but the proof of this is rather involved and we may revisit this again some day.

Grothendieck Rings

Given finitely-generated k[G]-modules M and N, their tensor product $M\otimes_k N$ over k is also finitely-generated. By linear algebra, tensor product over a field is always exact for both $M\otimes_k -$ and $-\otimes_k M.$ So if $0 \to M' \to M \to M'' \to 0$ is an exact sequence of k[G]-modules, then so is:

$0\to M'\otimes_k N \to M\otimes_k N\to M''\otimes_k N \to 0.$

This gives a pairing $R_k(G) \times R_k(G) \to R_k(G)$ which is bi-additive. Since tensor product is associative, we obtain a ring structure on $R_k(G).$ Furthermore if M is projective, then so is $M\otimes_k N$ as we saw above. Thus, identifying $P_k(G)$ with its image $c(P_k(G)) \subseteq R_k(G),$ we see that $P_k(G)$ is an ideal of $R_k(G).$

Example

Let us revisit the example from the previous article. The character table of S₃ is given by:

For p=3, we have $D = \begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 1\end{pmatrix}$ and $E = \begin{pmatrix} 1& 0 \\ 0 & 1\\ 1 & 1\end{pmatrix}.$ So the modular character tables for $P_k(G)$ and $R_k(G)$ are given by:

Note that $e(\varphi_1'') = \chi_{\text{triv}} + \chi$ and $e(\varphi_2'') = \chi_{\text{alt}} + \chi$ and both vanish at the 3-cycle as expected. Furthermore one can verify by direct computation that $\left< \varphi_i'', \varphi_j'\right>_k = \delta_{ij}$ taking into account (1,2) has class size of 3.

The Grothendieck ring $R_k(G)$ is isomorphic to $\mathbf{Z}[T]/\left<T^2 - 1\right>$ where {1, T} maps to $\{\varphi_1', \varphi_2'\}$ respectively. Then $P_k(G)$ is the ideal generated by T+2 and $R_k(G)/P_k(G) \cong \mathbf{F}_3.$