Modular Representation Theory (IV)

Continuing our discussion of modular representation theory, we will now discuss block theory. Previously, we saw that in any ring R, there is at most one way to write $1 = e_1 + \ldots + e_r$ where $e_i \in Z(R)$ is a set of orthogonal and centrally primitive idempotents. If such an expression exists, the $e_i$ are called block idempotents of R. For example, block idempotents exist when R is artinian.

We need the following refinement:

Lemma. Let $1 = e_1 + \ldots + e_r$ be block idempotents of R. Suppose $1 = f_1 + \ldots + f_s$ where the $f_j \in Z(R)$ are orthogonal central idempotents. Then there exists a unique map $\phi: \{1, \ldots, r\} \to \{1, \ldots, s\}$ such that:

$f_j = \sum_{i \in \phi^{-1}(j)} e_i.$

Note: if $\phi^{-1}(j) = \emptyset$ the sum is zero

Thus after a suitable renumbering of terms, we have:

$1 = \overbrace{e_1 + \ldots + e_{i_1}}^{f_1} +\overbrace{e_{i_1+1} + \ldots + e_{i_2}}^{f_2}+\ldots + \overbrace{e_{i_{s-1}+1} + \ldots + e_{i_s}}^{f_s}.$

Proof

For each $e_i$ we have $e_i = e_i f_1 + \ldots + e_i f_s$ , where $\{e_i f_1, \ldots, e_i f_s\}$ are orthogonal central idempotents. Since $e_i$ is centrally primitive, we must have $e_i f_j = e_i$ for some unique j and $e_i f_k =0$ for all k≠j. The map $i\mapsto j$ then gives $e_i f_{\phi(i)} = e_i$ and $e_i f_k \ne 0$ for all $k\ne \phi(i).$ And so

$f_j = e_1 f_j + \ldots + e_r f_j = \sum_{i\in \phi^{-1}(j)} e_i.$ ♦

Decomposition of R-Modules

Let M be an R-module and suppose $1 = e_1 + \ldots +e_r$ where the $e_i$ are block idempotents of R. Then $M = \oplus_{i=1}^r e_i M.$

Indeed, $M = \sum_i e_i M$ since each $m = \sum_i e_i m \in \sum_i e_i M.$
On the other hand, if $x \in e_1 M \cap (e_2 M +\ldots + e_r M)$ then we have $x = e_1 m_1 = e_2 m_2 + \ldots + e_r m_r$ and thus $e_1 m_1 = e_1^2 m_1 = e_1 e_2 m_2 + \ldots + e_1 e_r m_r = 0.$

Furthermore, since $e_i$ commutes with every r∈R, $e_i M \subseteq M$ is in fact an R-submodule. The central idempotent $e_i$ acts as the identity on $e_i M$ and zero on $e_j M$ for j≠i. One can thus imagine:

$R \cong R_1 \times R_2 \times \ldots \times R_n, \qquad M \cong M_1 \times M_2 \times \ldots \times M_n$

where each $M_i$ is an $R_i$ -module.

Block Idempotents of Semisimple R

Recall that a semisimple ring R is isomorphic to $\prod_{i=1}^m M_{n_i}(D_i)$ for some division ring $D_i,$ where $M_n(D)$ is the n × n matrix ring with entries in D. Since each matrix ring is a simple ring, we immediately obtain the central idempotents: $e_i$ for i=1,…,m, corresponds to the element whose component in $M_{n_i}(D_i)$ is the identity matrix, and whose component in $M_{n_j}(D_j), j\ne i$ is the zero matrix.

As a module over itself, R is a direct sum of the spaces of column vectors, so $R = \oplus_{i=1}^m S_i^{n_i}$ where $S_1, \ldots, S_r$ runs through a complete collection of simple R-modules, and the component $T_i := S_i^{n_i}$ gives the maximal decomposition $R = \oplus_i T_i$ as a direct sum of ideals.

In particular, this holds for the group ring K[G]. We have:

$K[G] = \oplus V^{\dim_K V}$ , where the direct sum is over all simple V.

Lemma. The block idempotent for $V^{\dim_K V}$ is given by the following formula:

$e_V = \frac{\dim V}{|G|} \sum_{g\in G} \chi_V(g^{-1})g,$

where $\chi_V(g) := \text{tr}(g: V\to V)$ is the character of V.

Proof

Fix V; it suffices to show that $e_V$ induces the identity map on $V^{\dim V}$ and zero on all other components. Now the coefficients of $e_V$ are constant over each conjugancy class, so $e_V\in Z(K[G])$ and $e_V$ is K[G]-linear. If W is simple, $e_V$ induces a scalar map on it, say $\lambda_W\cdot \text{id}_W.$ To compute $\lambda_W$ we take the trace:

$\lambda_W \cdot \dim_K W = \text{tr}(e_V : W\to W) = \frac{\dim V}{|G|} \sum_{g\in G} \chi_V(g^{-1}) \text{tr}(g: W\to W).$

But $\text{tr}(g:W\to W) = \chi_W(g)$ and so the above sum is $\dim V\cdot \left< \chi_V, \chi_W\right> = \dim V\cdot \delta_{V,W}.$ Thus $\lambda_W = \delta_{V,W}$ as desired. ♦

Block Idempotents of R[G]

Since k[G] is artinian, block idempotents are guaranteed to exist. Furthermore these can be lifted to block idempotents of R[G], which is a nice result since R[G] is not artinian.

Lemma. Suppose $e_1 + \ldots + e_n = 1$ are orthogonal central idempotents of Z(k[G]). Then we can find orthogonal central idempotents $\hat{e_1} + \ldots + \hat{e_n} = 1$ of R[G] such that $\hat{e_i} \pmod \pi = e_i.$

Proof

The proof is conceptually similar to an earlier lemma. The main step is to show:

Claim: if S is a commutative ring with ideal I such that $I^2 = 0$ , then any orthogonal idempotents $e_i\in S/I$ summing to 1 can be lifted to orthogonal idempotents $f_i\in S$ summing to 1.

[ If we can show this, then idempotents $e_i \in Z((R/\pi)[G])$ can be lifted to $Z((R/\pi^2)[G]$ , and in turn to $Z((R/\pi^4)[G]$ etc. Since R is complete, this gives idempotents in Z(R[G]). ]

Proof of Claim.

Pick any $x_i \in S$ such that $x_i \pmod I = e_i$ . Thus:

$\sum x_i \equiv 1 \pmod I, \quad x_i x_j \equiv 0 \pmod I \text{ for } i\ne j, \quad x_i^2 \equiv x_i \pmod I.$

As in the earlier proof, let $y_i = 3x_i^2 - 2x_i^3$ and this gives $y_i^2 = y_i.$ Also $y_i y_j \ne 0$ for all i≠j since it is divisible by $(x_i x_j)^2$ (since S is commutative). Finally, we claim that $\sum y_i = 1.$ Indeed, from the factorisation $3T^2 - 2T^3 - 1 = -(T-1)^2(2T+1)$ we obtain:

$\sum_i y_i - 1 = 3\sum_i y_i^2 - 2\sum_i y_i^3 - 1 = -(\sum_i y_i - 1)^2 (2\sum_i y_i + 1)$

where the first equality follows from $y_i^3=y_i^2 = y_i$ and the second follows from $y_i y_j = 0$ for i≠j. Note that $y_i \equiv x_i \pmod I$ so $(\sum y_i - 1)\in I.$ Thus $\sum y_i - 1$ is zero. ♦

Conversely, if $\hat{e_i}\in R[G]$ are orthogonal central idempotents summing to 1, then so are their images $e_i \in k[G].$ Finally, we have:

Lemma. If $\hat{e}, \hat{e}'\in R[G]$ are central idempotents with the same image in k[G], then they are equal.

Proof.

Let $\hat f:=\hat{e} - \hat{e}'$ , which is a central idempotent in πR[G]. But $\hat f \in \pi^m R[G] \implies \hat f = \hat{f}^2\in \pi^{2m} R[G]$ so we must have $\hat f = 0$ and so $\hat{e} = \hat{e}'.$ ♦

Thus, we have shown:

Summary.

There is a 1-1 correspondence between:

orthogonal central idempotents of k[G] summing to 1, and

orthogonal central idempotents of R[G] summing to 1.

In particular, block idempotents for k[G] lift to those for R[G]:

$1 = e_1 + e_2 + \ldots + e_r,\ (e_i \in k[G]) \ \mapsto 1 = \hat{e_1} + \hat{e_2} + \ldots + \hat{e_r},\ (\hat{e_i} \in R[G]).$

Taking each $\hat{e_i} \in K[G]$ , we can write it as a sum of the block idempotents of K[G]. Thus, we can partition the set of simple K[G]-modules as a disjoint union $\cup_i B_i$ , one for each $\hat{e_i}$ , such that:

$\hat{e_i} = \sum_{V\in B_i} e_V.$

For convenience, we also denote $e_V$ by $e_\chi$ where χ is the character of V. This gives the formula $e_\chi = \frac{\chi(1)}{|G|} \sum_{g\in G} \chi(g^{-1})g.$

Example: S₃.

Let’s compute the central idempotents for K[S₃] using the above formula. Let a = (1,2) + (2,3) + (3,1) and b = (1,2,3) + (1,3,2). We recover the example at the end of the previous article:

Now let’s consider the case p=2. We have $1 = (e_1 + e_2) + e_3 = \frac 1 3(1 + b) + \frac 2 3 (2-b)$ as block idempotents of R[G] (and also of k[G], after reduction mod 2), so the blocks are {e₁, e₂}, {e₃}. For p=3, {e₁, e₂, e₃} all belong to the same block.

If M is an indecomposable k[G]-module, then $M = \oplus_i e_i M$ and thus there is exactly one i for which $M = e_i M$ , and $e_j M = 0$ for all j≠i. As a result, the basis elements $[P] \in P_k(G)$ and $[M] \in R_k(G)$ can be classified into blocks, where P (resp. M) belongs to block $e_i$ if and only if $e_i P = P$ (resp. $e_i M = M$ ). Note that if P belongs to block e_i, then the idempotent e_i acts as the identity on e_iP. The same holds for M.

Similarly, for a simple K[G]-module V, there is a unique i for which $\hat{e_i}V = V.$ In summary, we can think of a block as a collection of:

indecomposable finitely-generated projective k[G]-modules P;
simple k[G]-modules M;
simple K[G]-modules V.

Lemma. Suppose basis elements $[P]\in P_k(G), [V]\in R_K(G)$ belong to distinct blocks e_i and e_j, where i≠j. Then the matrix entry of $e: P_k(G) \to R_K(G)$ corresponding to [P], [V] is zero.

Proof

Indeed, we have $e_i P = P, e_j P = 0$ and $\hat{e_i} V = 0, \hat{e_j}V = V.$ If $[\hat P] \in P_R(G)$ is the lift of [P], we have $\hat{e_i} \hat P \ne 0$ and thus $\hat{e_i} \hat P = \hat P$ since $\hat P$ is an indecomposable R[G]-module. Hence $\hat{e_i}(K \otimes_R \hat P) = K\otimes_R \hat P$ and we must have $\hat{e_i} W = W$ for any irreducible component W of $K\otimes_R \hat P.$ This shows that V cannot be a component of $K\otimes_R \hat P.$ ♦

Corollary. Let $[P] \in P_k(G), [M] \in R_k(G), [V]\in R_K(G)$ be basis elements.

If [M] and [V] belong to different blocks, the matrix entry of $d:R_K(G) \to R_k(G)$ is zero.

If [P] and [V] belong to different blocks, the matrix entry of $c:P_k(G) \to R_k(G)$ is zero.

Proof

The first statement follows from the fact that the matrix for d is the transpose of that for e; the second statement follows from c = de. ♦

Thus, the matrices for $c:P_k(G) \to R_k(G), d:R_K(G) \to R_k(G), e:P_k(G) \to R_K(G)$ can be broken up as block matrices, one for each block idempotent of k[G].

Example: S₄.

The character table of S₄ gives us: letting a = (sum of 2-cycles), b = (sum of 3-cycles), c = (sum of 4-cycles), d = (sum of (2+2)-cycles), the central idempotents of K[S₄] are:

$\begin{aligned} e_{\text{triv}} = \frac 1 {24} (1 + a + b + c + d),&\quad e_{\text{alt}} = \frac 1 {24} (1 - a + b - c + d)\\ e_2 = \frac 1 {12}(2 - b + 2d),&\quad e_1 = \frac 1 8 (3 + a - c - d)\\ e_{1,\text{alt}} = \frac 1 8 (3 - a + c -d).&\end{aligned}$

For p = 2, all five simple characters belong to the same block. For p = 3, we have:

$1 = \overbrace{e_{\text{triv}} + e_{\text{alt}} + e_2}^{\text{Block 1}} + \overbrace{e_1}^{\text{Block 2}} + \overbrace{e_{1, \text{alt}}}^{\text{Block 3}}.$

Finally, we can check when two characters lie within the same block. First, we need the following lemma:

Lemma. Let R be a commutative k-algebra of finite dimension over k and $1 = e_1 + \ldots + e_r$ be its block idempotents. Then any k-algebra homomorphism $\phi : R\to k$ is uniquely determined by the image of the block idempotents $\phi(e_1), \ldots, \phi(e_r).$

Proof

Corresponding to the block idempotents, write $R = R_1\times \ldots \times R_r$ as a product of commutative k-algebras. Since $R_i$ is artinian, $R_i/J(R_i)$ is semisimple and hence a product of matrix algebras. Since $R_i$ has no idempotent except 0 and 1, $R_i/J(R_i)$ itself is a matrix algebra, i.e. $M_n(D)$ for some division ring D/k. But R is commutative, so n=1 and D is a field extension k’ of k.

On the other hand, let $\phi_i := \phi|_{R_i} : R_i \to k.$ Since $\phi(e_1 + \ldots + e_r) = 1$ and the $\phi(e_i)$ are orthogonal idempotents, exactly one $\phi_i$ is a k-algebra homomorphism while the remaining $\phi_j$ are zero maps for j≠i. Now $\phi_j : R_j \to k$ must factor through the nilpotent ideal $J(R_j)$ and so it is determined by $\phi_j' : k' \to k$ which is either the zero map or the identity, depending on whether $\phi(e_j)$ is 0 or 1. ♦

Theorem. Let V, W be simple K[G]-modules; the following are equivalent.

V and W are in the same block.

For any conjugancy class C⊆G and g∈C, we have:

$\frac{|C| \chi_V(g)}{\dim_K V} \equiv \frac{|C| \chi_W(g)}{\dim_K W} \pmod \pi.$

Note

In the course of the proof, we will see that $\frac{|C| \chi_V(g)}{\dim_K V}\in R$ for any simple V, so the congruence is well-defined.

Proof

Step 1. For each simple V, we will define a ring homomorphism $\lambda_V : Z(K[G]) \to K.$

Given any conjugancy class C⊆G, define $\alpha_C := \sum_{g\in G} g$ as a K-linear map V→V. Since $\alpha_C$ commutes with all g∈G, it is a K[G]-linear map V→V. But V is simple, hence such a map is a scalar; thus we get a ring homomorphism $\lambda_V : Z(K[G]) \to K.$

Step 2. Show that $\lambda_V(\alpha_C)$ is the LHS of the congruence and that this lies in R.

Taking the trace of $\alpha_C : V\to V$ , we get:

$\lambda_V(\alpha_C) \dim V = \sum_{g\in C} \text{tr}(g: V\to V) = |C| \chi_V(g).$

So $\lambda_V(\alpha_C) = \frac{|C| \chi_V(g_C)}{\dim V}$ for any representative $g_C \in C.$ To prove that this lies in R, recall that we can pick an R-lattice M⊂V which is an R[G]-module. Thus $\alpha_C (M) \subseteq M.$ Picking a basis for M, we have $\det(\alpha_C) \in R \implies \lambda_V(\alpha_C) \in R.$

Step 3. Compute $\lambda_V(e_W)$ where $e_W$ is the K[G] block idempotent for W.

Let us rewrite

$e_W = \frac{\dim W}{|G|} \sum_{g\in G} \chi_W(g^{-1})g = \frac{\dim W}{|G|} \sum_{\text{conj. cl. }C} \chi_W(g_C^{-1}) \alpha_C$

and so $\lambda_V(e_W) = \frac{\dim W}{\dim V\cdot |G|} \sum_C |C| \chi_V(g_C)\chi_W(g_C^{-1}) = \frac{\dim W}{\dim V} \left< \chi_V, \chi_W\right> = \delta_{V,W}.$

Step 4. Complete the proof.

Since $\hat{e_i}= \sum_{V\in B_i} e_V$ , step 3 tells us V and W belong to the same block $B_i$ if and only if $\lambda_V(\hat{e_i}) = \lambda_W(\hat{e_i})$ for all i. But this value is either 0 or 1, so it holds if and only if

$\lambda_V(\hat{e_i}) \equiv \lambda_W(\hat{e_i}) \pmod \pi$ for all i.

On the other hand, $\lambda_V(Z(R[G])) \subseteq R$ so we also obtain a ring homomorphism $\lambda_V : Z(R[G]) \to R.$ Reduction mod π then gives us a k-algebra homomorphism $\lambda_V' : Z(k[G]) \to k.$ By the above, V and W belong to the same block if and only if $\lambda_V'(e_i) = \lambda_W'(e_i) \in k$ for all i. Since Z(k[G]) is a commutative k-algebra of finite dimension over k, the above lemma says this holds if and only if $\lambda_V' = \lambda_W' : Z(k[G]) \to k$ which is equivalent to $\lambda_V(\alpha_C) \equiv \lambda_W(\alpha_C) \pmod \pi$ for all C. ♦