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 kj. 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 rR, 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 ji. 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. ♦

blue-lin

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 ij 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 ij. 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:

chartable_idempotent_s3

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, {e1e2, e3} all belong to the same block.

blue-lin

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 ji. 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 ei, then the idempotent ei acts as the identity on eiP. 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 ei and ej, 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 cde. ♦

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}}.

chartable_block_s4

blue-lin

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 ji. 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 CG, define \alpha_C := \sum_{g\in G} g as a K-linear map VV. Since \alpha_C commutes with all gG, it is a K[G]-linear map VV. 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 MV 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. ♦

Example: S5.

Modular 2, the blocks of the character table are labeled by the circles on the right:

block_s5_mod_2

The corresponding matrices are:

block_s5_mod_2_matrices

Modular 3, the table becomes:

block_s5_mod_3

The corresponding matrices are:

block_s5_mod_3_matrices

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