Modular Representation Theory (I)

Let K be a field and G a finite group. We know that when char(K) does not divide |G|, the group algebra K[G] is semisimple. Conversely we have:

Proposition. If char(K) divides |G|, then K[G] is not semisimple.

Proof

Let I := \{ \sum_{g\in G} a_g\cdot g: a_g\in K, \text{ such that } \sum a_g =0\}, a two-sided ideal of K[G]. If K[G] = IJ for some left ideal J of K[G], then dim J = 1 so it is spanned by, say, \alpha = \sum_g b_g\cdot g. Since αg is a scalar multiple of α for each gG, all the bg‘s must be equal to some bK. But this means \alpha = b(\sum_ g g) lies in I since its sum of coefficients is |G|b = 0. ♦

For the case where char(K) divides |G|, we have modular representationt theory. Here, we’ll require some knowledge of complete discrete valuation rings; throughout this article, we adopt the following notations and assumptions.

  • R is a complete discrete valuation ring with maximal ideal generated by π∈R.
  • K is the field of fractions of R.
  • k := R/(π) is the residue field.
  • p := char(k) divides |G| and char(K) = 0.

Representations of G over k are studied by first lifting them to R, then projecting to K. Let R_k(G), R_K(G) be the Grothendieck group of finitely-generated modules over k[G], K[G] respectively. Since char(K) = 0, K[G] is semisimple. Since k[G] is of finite dimension over k, it is artinian; we let J be its Jacobian radical. Finally, let P_k(G) be the Grothendieck group of finitely-generated projective modules over k[G]. From earlier results:

  • R_k(G), R_K(G), P_k(G) are free abelian groups, with the following bases, which we will fix from now on.
    • Basis of R_k(G) or R_K(G): [M] for simple modules M.
    • Basis of P_k(G): [P] for finitely-generated indecomposable projective modules P.
  • We have a map c:P_k(G) \to R_k(G) taking [P] to [P].
  • We have an isomorphism P_k(G) \cong R_k(G), taking [P] to [P/JP]; this corresponds to identity matrix (under the above bases).

Note: if [M] = [N] \in R_k(G), then M and N have identical composition factors but M and N may not be isomorphic. On the other hand, if [M] = [N] \in P_k(G), then the decomposition factors of M and N are the same and thus M ≅ N.

Next, we introduce the following pairings:

  • The pairing \left<-, -\right>_K : R_K(G) \times R_K(G) \to\mathbf{Z} taking [M], [N] \mapsto \dim_K \text{Hom}_{K[G]}(M, N) is a bi-additive map. The above basis for R_K(G) is orthogonal by Schur’s lemma. Extending K if necessary, we may assume it is orthonormal.
  • The pairing \left<-, -\right>_k : P_k(G) \times R_k(G) \to \mathbf{Z} taking [P], [M] \mapsto \dim_k \text{Hom}_{k[G]}(P, M) is a bi-additive map. The given bases for P_k(G) and R_k(G) are dual with respect to this pairing. [ Indeed, for [M] \in R_k(G) with M simple, we have Hom(PM) = Hom(P/JPM) since JM = 0; if P is projective indecomposable, then P/JP is simple and the result follows from Schur’s lemma. Assuming k is large enough, we may assume that \text{Hom}(P, M) = k if P/JP ≅ M. ]

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The next step is to define the maps d and e in the following diagram, such that cde:

modular_rep_diagram

Definition of d

To define d:R_K(G) \to R_k(G), let M be a finitely-generated K[G]-module.

Proposition. There is an R[G]-module N such that K\otimes_R N \cong M. The class of this module [N/\pi N] \in R_k(G) depends only on M.

Proof

Step 1. First prove the existence of N.

Let N’ ⊂ M be a free R-module with a basis spanning M, so K\otimes_R N' \cong M as K-vector spaces. Now let N := \sum_{g\in G} gN' which is also a free R-module since it is torsion-free, and has a basis spanning M. Thus K\otimes_R N \cong M. Since N is G-invariant it is an R[G]-module.

Step 2. Proof of uniqueness: initial step.

Suppose N_1, N_2 are R[G]-modules such that K\otimes_R N_1 \cong K\otimes_R N_2. Replacing N_1 by a scalar multiple (this does not affect [N_1]), we may assume N_2 \subseteq N_1. Now for each x\in N_1 we have \pi^r x\in N_2 for some r; since N_1 has a finite basis, we have \pi^r N_1 \subseteq N_2\subseteq N_1 for some r>0.

Step 3. Now we show: if \pi^r N_1 \subseteq N_2\subseteq N_1, then [N_1/\pi N_1]= [N_2/\pi N_2] in R_k(G).

This is by induction on r. When r=1, let T:= N_1/N_2; we have an exact sequence of k[G]-modules:

0\to \pi N_1 / \pi N_2 \to N_2 /\pi N_2 \to N_1/\pi N_1 \to N_1/N_2 \to 0.

Since \pi N_1/\pi N_2 \cong N_1/N_2 we have [N_2/\pi N_2] = [N_1/\pi N_1] in the group R_k(G). For r>1, let N_3 := \pi^{r-1}N_1 + N_2 we have:

\pi^{r-1} N_1\subseteq N_3\subseteq N_1, \qquad \pi N_3\subseteq N_2\subseteq N_3.

By induction hypothesis, [N_1/\pi N_1] = [N_3/\pi N_3] = [N_2/\pi N_2] and we’re done. ♦

We thus define d:R_K(G) \to R_k(G) via: given [M]\in R_K(G) pick an R[G]-module N such that K\otimes_R N \cong M. We then define d([M]) := [N/\pi N].

Note

An exact sequence of K[G]-modules 0\to M'\to M\to M''\to 0 splits as M ≅ M’ ⊕ M”; picking modules N’ and N” for M’ and M” via the above proposition, N := N’ ⊕ N” also satisfies the proposition for M, and we obtain d([M]) = d([M’]) + d([M”]) as desired, so d is well-defined.

Example

Suppose G = {eg} is of order 2. Let RZ2, KQ2 and kF2. Take MK[G] itself. One can obtain the R[G]-module N (of rank 2 over R) in different ways, e.g.:

g \mapsto \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}, \qquad g\mapsto \begin{pmatrix} 1 & 1 \\ 0 & -1\end{pmatrix}.

Indeed, they’re isomorphic over K[G] since the above two matrices are diagonalizable with eigenvalues -1, +1. However, reducing modulo 2 gives non-isomorphic k[G]-modules:

g\mapsto \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \pmod 2, \qquad g\mapsto \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \pmod 2.

Note that they have the same composition factors, although one is semisimple and the other is not.

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Definition of e

Next, we will define e: P_k(G) \to R_K(G). This is given in two steps: first we consider the map \psi: P_R(G) \to P_k(G), [P] \mapsto [P/\pi P] where P_R(G) is the Grothendieck group of finitely-generated projective R[G]-modules. [ Warning: since R[G] is not artinian, we have to avoid using results from the previous two articles. ] Note that Ψ is well-defined:

  • If P is projective over R[G], then P\oplus Q\cong R[G]^n for some n so (P/\pi P)\oplus (Q/\pi Q)\cong k[G]^n and so PP is projective over k[G].
  • An exact sequence of projective modules must split, so Ψ is a well-defined homomorphism of abelian groups.

The next step is to show that Ψ is an isomorphism. First, injectivity:

Lemma. If P, Q are finitely-generated and projective such that P/\pi P\cong Q/\pi Q as k[G]-modules, then P\cong Q as R[G]-modules.

Proof

Let f_0 : P/\pi P \to Q/\pi Q be an isomorphism of k[G]-modules. Since P is projective, the map P\to P/\pi P \to Q/\pi Q lifts to a map f_1: P\to Q of R[G]-modules such that the reduction of f1 mod π is f0. To show that f1 is bijective, it suffices to show it is an isomorphism of R-modules. But PQ are finite R-free since they are projective over R[G] and thus torsion-free. So we only need to show that det(f1) is invertible in R, i.e. ≠0 modulo π. But det(f1) mod π = det(f0) ≠ 0. ♦

Hence, \psi: P_R(G) \to P_k(G) is injective. Finally, we have the following:

Lemma. For any finitely-generated projective k[G]-module N, there is a finitely-generated projective R[G]-module M such that M/\pi M\cong N.

Proof

We may assume N is indecomposable, so N is a direct summand of k[G], i.e. k[G] = N\oplus N'. The image of 1∈k[G] in N then gives an idempotent xN (i.e. x2=x). We wish to lift x to an idempotent yR[G] such that xy mod π.

  • Claim: if S is any ring and I ⊂ S is an ideal satisfying I2=0, then any idempotent y ∈ S/I lifts to an idempotent x ∈ S.
  • Proof: let z ∈ S be any lift of y. Then x := 3z2-2z3 satisfies, modulo Ix ≡ 3z2-2z3 ≡ 3z – 2zz so it is another lift of y. Now x2x = 0 since it is divisible by (z2z)2 ∈ I= 0. ♦

Now consider the ring S := R[G]/\pi^2 R[G] with ideal I:=\pi R[G]/\pi^2 R[G]. This satisfies I= 0 and S/I ≅ k[G] so by the claim, x lifts to an idempotent y’S.

Repeat the process with the ring S' := R[G]/\pi^3 R[G] with ideal I' := \pi^2 R[G]/\pi^3 R[G]. Again I'^2 = 0 and S’/I’ ≅ S so y’ lifts to an idempotent y”S’. Repeating this process, since R is a complete discrete valuation ring, we obtain an idempotent yR[G] such that xy mod π.

Then R[G] is the direct sum of left modules M := R[G]y and M’ := R[G](1-y), and M/\pi M\cong N. ♦

Thus, Ψ is an isomorphism and we can define:

e : P_k(G) \stackrel{\psi^{-1}}{\longrightarrow} P_R(G) \longrightarrow R_K(G)

where the second map takes [M]\in P_R(G) to [K\otimes_R M] \in R_K(G).

Relations Between d and e

Finally we wish to prove that cde. Indeed, let [P]\in P_k(G) and [Q] = \psi^{-1}([P]) \in P_R(G) so that Q/\pi Q\cong P. Then the element e([P]) is given by [K\otimes_R Q] \in R_K(G). To obtain d([K\otimes_R Q]) we recover Q as the R[G]-module in the definition of d and take: d([K\otimes_R Q]) = [Q/\pi Q] = [P] \in R_k(G) as desired.

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Concrete Example

Let’s compute an explicit example for GS3RZ3K = Q3k = F3. It’s easy to calculate R_K(G) since K[G] is semisimple. From elementary character theory, there are three simple modules:

rep_of_s3

  • trivial rep. : g acts as trivially on K;
  • alternating rep.g acts via sgn(g) on K;
  • other : take the subspace of K3 given by (xyz) satisfying x+y+z = 0; G acts by permuting the coordinates.

This gives a basis of R_K(G). On the other hand, let’s consider all simple k[G]-modules. We still have the trivial and alternating representations, clearly simple. But now the third representation has a submodule:

V = \{(x,y,z) \in k^3 : x+y+z = 0\} \implies k\cdot (1,1,1) \in V

since char(k) = 3. This gives a subspace which is the trivial representation, and the quotient is alternating. Hence, R_k(G) has two simple modules, given by the trivial and alternating modules, and the matrix for d:R_K(G) \to R_k(G) is:

D = \begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 1\end{pmatrix}

Next we compute e; to do that we need to find the indecomposable projective k[G]-modules. These can be found by decomposing k[G] itself. From above, we see that the key is to find idempotents of k[G] (specifically indecomposable projectives correspond to primitive idempotents, but we won’t use that here). Clearly x = \frac 1 2 (e + (1,2)) is idempotent, and we obtain:

\begin{aligned} W := k[G]x &= \left< e + (1,2), (1,3) + (1,2,3), (2,3)+(1,3,2)\right>, \\ W' :=k[G](1-x) &= \left< e - (1,2), (1,3) - (1,2,3), (2,3) - (1,3,2)\right>.\end{aligned}

We leave it to the reader to prove that these are indecomposable. To compute e([W]) and e([W’]), we lift them to projective R[G]-modules then tensor with K: this still gives

\begin{aligned} K[G]x &= \left< e + (1,2), (1,3) + (1,2,3), (2,3)+(1,3,2)\right>, \\ K[G](1-x) &= \left< e - (1,2), (1,3) - (1,2,3), (2,3) - (1,3,2)\right>.\end{aligned}

which has character values (3, 1, 0) and (3, -1, 0), i.e. \chi + \chi_{\text{triv}} and \chi + \chi_{\text{alt}}. So the matrix for e is:

E = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1\end{pmatrix}\implies C=DE =\begin{pmatrix} 2&1 \\ 1&2\end{pmatrix}.

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