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, c = de is symmetric and positive definite.

Group S₄ with p=2.

Let’s consider the usual character table for S₄.

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 d(χ₂) is simple. If it weren’t it would be the sum of two 1-dimensional representations. But these are easily classifiable for S_n.

Lemma. There are at most two 1-dimensional representations of S_n 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 d(χ₂) 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:

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.

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 (a, b) 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:

Group S₅ with p=2.

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

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 g∈G 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 f : W→k which takes $v_i^2\mapsto 0, v_i v_j \mapsto 1$ for all 1≤i≤j≤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:

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

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:

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:

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