Modular Representation Theory (III)

Let’s work out some explicit examples of modular characters. First, we have a summary of the main results.

  • Let \varphi_1', \varphi_2', \ldots, \varphi_r' be the modular characters of the simple k[G]-modules; they form a basis of R_k(G).
  • Let \varphi_1'', \varphi_2'', \ldots, \varphi_r'' be those of the projective indecomposable k[G]-modules; they form a basis of P_k(G).
  • We have r = \dim R_k(G) = \dim P_k(G), the number of p-regular conjugancy classes of G.
  • The \varphi_i' and \varphi_i'' form a dual basis under the inner product \left<\varphi'', \varphi'\right>_k := \frac 1 {|G|} \sum_{g\in G_{\text{reg}}} \varphi''(g) \varphi'(g^{-1}) so that \left<\varphi_i'', \varphi_j'\right>_k = \delta_{ij}.

These relate to ordinary characters as follows: let \chi_1, \chi_2, \ldots,\chi_s be the standard irreducible characters of K[G], so they form an orthonormal basis of R_K(G).

  • The map d:P_k(G) \to R_K(G) satisfies: for each \varphi''\in P_k(G), the function d(\varphi'') is zero on the p-singular conjugancy classes of G.
  • The map e:R_K(G) \to R_k(G) is the transpose of d and is injective.
  • Thus, cde is symmetric and positive definite.

Group S4 with p=2.

Let’s consider the usual character table for S4chartable_s4_2

For the ring R_k(G), we leave only the columns for e and (1,2,3), since the remaining conjugancy classes are 2-singular. Immediately, we obtain some linear relations:

  • d(\chi_{\text{alt}}) = d(\chi_{\text{triv}});
  • d(\chi_1) = d(\chi_2) + d(\chi_{\text{triv}}) = d(\chi_1 \chi_{\text{alt}});

So it remains to consider if d2) is simple. If it weren’t it would be the sum of two 1-dimensional representations. But these are easily classifiable for Sn.

Lemma. There are at most two 1-dimensional representations of Sn over any field: the trivial and the alternating.

Proof

Indeed, these correspond to S_n \to k^* and since the image is abelian, it factors through the commutator [S_n, S_n] = A_n and we obtain S_n/A_n \to k^*. So we are left with the trivial and alternating representations. ♦

So d2) is simple since otherwise it would be d(2χ1) which it clearly is not. Hence:

D = \begin{pmatrix} 1 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1\end{pmatrix}\implies E = D^T = \begin{pmatrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \\ 1 & 1 \\ 1 & 1\end{pmatrix}, \ C = DE=\begin{pmatrix} 4 & 2 \\ 2 & 3\end{pmatrix}.

The basis elements of P_k(G) and R_k(G) are thus:

mod_char_s4_mod_2

Note that we do have \left< \varphi_i'', \varphi_j'\right>_k = \delta_{ij} as expected. E.g.

\left< \varphi_1'', \varphi_1'\right>_k = \frac 1 {24}(1\cdot(8\cdot 1) + 8\cdot(2\cdot 1)) = 1.

Also e(\varphi_1'') = \chi_{\text{triv}}+ \chi_{\text{alt}}+ \chi_1 + \chi_1 \chi_{\text{alt}} and e(\varphi_2'') = \chi_2 + \chi_1 + \chi_1\chi_{\text{alt}} take all the 2-singular classes to 0.

Exercise

Let φ be the regular representation. Find the multiplicity of \varphi_1'' and \varphi_2'' in the decomposition of φ, and the multiplicty of \varphi_1' and \varphi_2' among its composition factors.

blue-lin

Group S4 with p=3.

We remove the column (1,2,3) and keep the remaining four. Clearly, d(\chi_{\text{triv}}) and d(\chi_{\text{alt}}) are simple since they’re of dimension 1. Next, we have d(\chi_2) = d(\chi_{\text{triv}}) + d(\chi_{\text{alt}}). It remains to see if d(\chi_1) and d(\chi_1 \chi_{\text{alt}}) are simple. Consider d(\chi_1). If it weren’t simple it must contain a submodule of dimension 1, which we saw is either d(\chi_{\text{triv}}) or d(\chi_{\text{alt}}).

  • In the former case, d(\chi_1) = d(\chi_{\text{triv}})+\varphi where \varphi((1,2,3,4)) = -2. Since dim \varphi = 2 this means both eigenvalues for (1,2,3,4) are -1, and so those for its square (1,3)(2,4) are +1. But this contradicts \varphi((1,2)(3,4)) = -2.
  • In the latter case, d(\chi_1) = d(\chi_{\text{alt}})+\varphi where \varphi((1,2)) = +2 so both eigenvalues for (ab) are +1. This means all elements of S4 have both eigenvalues equal to +1, which is absurd.

Thus d(\chi_1) and d(\chi_1 \chi_{\text{alt}}) are simple and we have:

D = \begin{pmatrix} 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1\end{pmatrix},\ E =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix},\ C =\begin{pmatrix} 2 & 1 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}.

The basis elements of P_k(G) and R_k(G) are thus:

mod_char_s4_mod_3

blue-lin

Group S5 with p=2.

First, we look at the usual character table for S5.

chartable_s5_2

Removing the 2-singular conjugancy classes leaves us with the columns for e, 3-cyc and 5-cyc. Note that d(\chi) = d(\chi\chi_{\text{alt}}) for any character χ so we are left with 4 rows. Next d(\chi_3) = d(\chi_4) + d(\chi_{\text{triv}}) so we are left with d(\chi_{\text{triv}}), d(\chi_1), d(\chi_4), which are clearly linearly independent. Writing them as linear combinations of simple modular characters:

\begin{pmatrix} 1 & 0 & 0 \\ * & * & * \\ * & * & *\end{pmatrix} \begin{pmatrix} \varphi_1' \\ \varphi_2' \\ \varphi_3'\end{pmatrix} = \begin{pmatrix} d(\chi_{\text{triv}}) \\ d(\chi_1) \\ d(\chi_4)\end{pmatrix}.

where the matrix entries are all non-negative integers. It is not hard, albeit rather tedious, to list all possible 2 × 3 matrices. After solving for \varphi_2', \varphi_3', we are further reduced to 12 possibilities (corresponding to their values at e, the 3-cycle and 5-cycle):

((2, 5, 2), (2, -4, -3)),  ((2, -1, -3), (2, -1, 2)), ((2, -1, -3), (2, -4, -3)),

((2, -1, -3), (3, 0, 3)), ((2, -1, -3), (3, -3, -2)), ((2, -1, -3), (4, -2, -1)),

((2, -1, -3), (5, -1, 0)), ((3, 0, -2), (3, -3, -2)), ((3, 0, -2), (4, -2, -1)),

((3, 0, -2), (5, -1, 0)), ((4, 1, -1), (4, -2, -1)), ((4, 1, -1), (5, -1, 0)).

Since |\varphi(g)| \le |\varphi(e)| for any g, this immediately removes the first 7 possibilities. Next \varphi = (3, 0, -2) is invalid since for \varphi((1,2,3,4,5)) = -2 we must have 3 fifth roots of unity summing up to -2, which is impossible. So we’re left with two choices. It turns out \varphi_3' = (4, -2, -1) is the right choice, which we shall show below.

Construction

We need to show that modulo 2, the modular representation d(\chi_4) contains the trivial representation. Recall that \chi_4 is found in the representation W :=\text{Sym}^2 V where V is a 4-dimensional representation given by:

V := \{(x_1, x_2, x_3, x_4, x_5) \in k^5 : \sum_i x_i = 0\}

and G=S_5 acts on V by permuting the coordinates. Another way of expressing this is: v_1, \ldots, v_5 \in V where \sum_i v_i = \mathbf{0} and gG acts on \{v_i\} by taking v_i \mapsto v_{g(i)}. Now W = \text{Sym}^2 V is spanned by v_i v_j where multiplication commutes. A basis of W is given by v_i v_j where 1 \le i \le j \le 4 and

v_5^2 = \sum_{j=1}^4 v_j^2 \pmod 2,\ v_i v_5 =\sum_{j=1}^4 v_i v_j \pmod 2.

In the semisimple case, W contains exactly one copy of the trivial representation but modulo 2, we can find two copies.

  • First, take the subspace X spanned by \sum_{1\le i<j\le 5} v_i v_j\in W, which is G-invariant. Note that this vector is non-zero (simplify via v_5 = v_1 + v_2 + v_3 + v_4).
  • Next, consider the map fWk which takes v_i^2\mapsto 0, v_i v_j \mapsto 1 for all 1≤ij≤4. Note that f(v_5^2) = f(v_1^2 + \ldots + v_4^2) = 0 and f(v_i v_5) = f(v_i(v_1 + \ldots + v_4)) =1 and so we see that f is G-equivariant, where G acts trivially on k. Note that f(X)=0 so we have at least two copies of the trivial representation among the composition factors.

Thus, d(\chi_4) has at least one copy of the trivial representation, and we get:

\varphi_1' = (1,1,1), \quad \varphi_2' = (4, 1, -1), \quad \varphi_3' = (4, -2, -1).

An explicit representation of φ3‘ is given by:

(1,2)\mapsto \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{pmatrix}, (1,2,3,4,5)\mapsto \begin{pmatrix} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1 \end{pmatrix},\text{ so }(1,2,3)\mapsto \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1\\ 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\end{pmatrix}.

To calculate the modular character values at (1,2,3) and (1,2,3,4,5), we compute their characteristic polynomials, giving x^4 + x^2 + 1 = (x^2 + x + 1)^2 and x^4 + x^3 + x^2 + x + 1 respectively. Lifting the roots of unity to K gives us \omega, \omega, \omega^2, \omega^2 for the first case and \zeta, \zeta^2, \zeta^3, \zeta^4 for the second, where \omega = e^{2\pi i/3} and \zeta = e^{2\pi i/5} so the values are 2(\omega + \omega^2) = -2 and \sum_{j=1}^4 \zeta^j = -1 respectively.

Conclusion

This gives:

D = \begin{pmatrix} 1 & 1 & 0 & 0 & 1 & 1 & 2\\ 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1\end{pmatrix}, E = \begin{pmatrix} 1 & 0 & 0\\ 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \\ 2 & 0 & 1\end{pmatrix} \implies C=\begin{pmatrix}8 & 0 & 4 \\ 0 & 2 & 0 \\ 4 & 0 & 3\end{pmatrix}

mod_char_s5_mod_2

blue-lin

Group S5 with p=3

From the character table of S5, we remove the two columns for 3-cyc and (2+3)cyc. The resulting modular characters satisfy the following:

d(\chi_4) = d(\chi_1) + d(\chi_{\text{alt}}), \quad d(\chi_4 \chi_{\text{alt}}) = d(\chi_1 \chi_{\text{alt}}) + d(\chi_{\text{triv}}).

The remaining 5 modular characters are linearly independent:

chartable_s5_mod_3

On the other hand, denote the 5 simple modular characters of R_k(G) by:

\varphi_1' := d(\chi_{\text{triv}}),\ \varphi_2' := d(\chi_{\text{alt}}), \ \varphi_3',\ \varphi_4',\ \varphi_5'.

Let us say \varphi\in R_k(G) is even if \varphi \cdot \varphi_2' = \varphi. Note that this holds if and only if φ is zero on the odd permutations: 2-cyc and 4-cyc (we’re ignoring the (2+3)cyc column for modular characters mod 3). Since φ is simple if and only if \varphi\cdot \varphi_2' is, we see that \varphi_3', \varphi_4', \varphi_5' are either all even, or exactly one of them is even. The former case is impossible since that would imply \varphi_1', \ldots, \varphi_5' all take the same values for 2-cyc and 4-cyc. Hence, we may assume \varphi_4' = \varphi_3' \cdot\varphi_2' and that \varphi_5' is even.

  • First, d(\chi_1) cannot contain \varphi_1' or \varphi_2'. E.g. if d(\chi_1) = \varphi_1' + \varphi we would have φ(e) = 3 and φ(5-cyc) = -2, which is impossible since we cannot have 3 fifth roots of unity summing to -2.
  • Clearly, d(\chi_1) = \varphi_3' + \varphi_4' = \varphi_3'(1 + \varphi_2') is impossible.
  • Now suppose d(\chi_1) = \varphi_3' + \varphi_5'. This means \varphi_3', \varphi_4', \varphi_5' are all of dimension 2. Since d(\chi_3) is even, it must be \varphi_3' + \varphi_4' + \varphi_5'. Hence we have \varphi_4' = d(\chi_3) - d(\chi_1) which takes 2-cyc to -2. Hence \varphi_3' = \varphi_4' \varphi_2' takes 2-cyc to +2 and so it must be the identity (contradiction).

Hence, we have shown that d(\chi_1) and d(\chi_1 \chi_{\text{alt}}) are both simple. Finally d(\chi_3) is either simple, or contains \varphi_1' + \varphi_2'. The latter would imply d(\chi_3) = \varphi_1' + \varphi_2' + \varphi where \varphi = (4, 0, 0, -1, -4). Hence the eigenvalues for (2+2)cyc are all -1 and since its order is coprime to p=3, the matrix for (2+2)cyc is –I. But then we have (1,3)(2,4)·(1,2,3,4,5) = (1,4,5,3,2) and we cannot have both M^5 = I and (-M)^5 = I.

Thus we may write \varphi_3' = d(\chi_1), \varphi_4' = d(\chi_1 \chi_{\text{alt}}), \varphi_5' = d(\chi_3) and we have:

D = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 &1 & 0 \\ 0 & 1 & 0 & 0 & 1 &0 & 0 \\ 0 & 0 & 1 &0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1&0 & 1 & 0 \\ 0 & 0 & 0 & 0& 0 & 0 & 1 \end{pmatrix}, E = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0& 0 & 1\end{pmatrix}, C = \begin{pmatrix} 2 & 0 & 0 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 1 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1\end{pmatrix}.

 

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