## 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. ]

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

## 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.

## 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.

## 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:

• 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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