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I'm self studying some group representation theory and I've hit a wall with Exercise 3.10 (Computing the character table of $G = SL_2(\mathbb{Z}/3)$) in Fulton and Harris' book. Any advice or hints would be really appreciated!

Here's what I know so far:

(1) The representatives of the conjugacy classes can be chosen to be:

$I, -I, A_1 = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right), A_2 = \left(\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array} \right), B_1 = \left(\begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array} \right), B_2 = \left(\begin{array}{cc} -1 & -1 \\ 0 & -1 \end{array} \right), \text{ and } C = \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$

Thus there are 7 irreducible representations of $G$.

There are 1, 1, 4, 4, 4, 4, and 6 conjugates for each of these representatives respectively (see this post).

(2) $G/\{\pm I\}$ has a faithful action on the lines in the $\mathbb{Z}/3$ plane, of which there are 4, thus $G/\{\pm I\}$ is a subgroup of $S_4$ of size 12, thus it must be $A_4$. This allows us to pullback the irreducible representations of $A_4$ to get irreducible representations of $G$. By doing this we can obtain 4 rows of the character table:

\begin{equation*} \begin{array}{|c|c|c|c|c|c|c|c|} \hline & 1 & 1 & 4 & 4 & 4 & 4 & 6 \\ \hline SL_2(\mathbb{Z}/3) & I & -I & A_1 & A_2 & B_1 & B_2 & C \\ \hline U & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline U' & 1 & 1 & \omega & \omega^2 & \omega^2 & \omega & 1 \\ \hline U'' & 1 & 1 & \omega^2 & \omega & \omega & \omega^2 & 1 \\ \hline V & 3 & 3 & 0 & 0 & 0 & 0 & -1 \\ \hline W & d_1 & ? & ? & ? & ? & ? & ? \\ \hline W' & d_2 & ? & ? & ? & ? & ? & ? \\ \hline W'' & d_2 & ? & ? & ? & ? & ? & ? \\ \hline \end{array} \end{equation*}

(3) For the remaining three, using the fact that the sum of the squared dimensions of the irreducible representations is $|G| = 24$, we can deduce that $d_1^2+d_2^2+d_3^2 = 12$, implying that we must have $d_1 = d_2 = d_3 = 2$.

My question is, how can we determine the remaining three characters of the 2 dimensional representations? I've tried looking at tensor products of the first 4 representations among other things and nothing seems to work.

Note:

  • This is over $\mathbb{C}$.
  • The person in this post was considering a similar problem, but ran into the same issue.

Thanks in advance!

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I'm rusty in representation theory, but I'll give it a try.

The characters $W,W',W''$ at $A_1,A_2,B_1,B_2$ can't all be zero, since this would break the orthogonality relation (not sure but this feels true).

Therefore one of them must be non-zero, without loss of generality a character $W$. This implies that $W'=W\otimes U'$ and $W''=W \otimes U''$ (modulo permutations). Therefore it is enough to find one irreducible representation of degree 2.

We must have something like this: \begin{equation*} \begin{array}{|c|c|c|c|c|c|c|c|} \hline & 1 & 1 & 4 & 4 & 4 & 4 & 6 \\ \hline SL_2(\mathbb{Z}/3) & I & -I & A_1 & A_2 & B_1 & B_2 & C \\ \hline U & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline U' & 1 & 1 & \omega & \omega^2 & \omega^2 & \omega & 1 \\ \hline U'' & 1 & 1 & \omega^2 & \omega & \omega & \omega^2 & 1 \\ \hline V & 3 & 3 & 0 & 0 & 0 & 0 & -1 \\ \hline W & 2 & a & b & c & d & e & f \\ \hline W' & 2 & a & \omega b & \omega^2 c & \omega^2 d & \omega e & f\\ \hline W'' & 2 & a & \omega^2 b & \omega c & \omega d & \omega^2 e & f\\ \hline \end{array} \end{equation*} Consider the following orthogonality relations:

  • Column $I$ and column $-I$ give $a=-2$.
  • Column $I$ and column $C$ give $f=0$.
  • Column $A_1$ and column $A_1$ give $|b|=1$.
  • Column $A_2$ and column $A_2$ give $|c|=1$.
  • Column $B_1$ and column $B_1$ give $|d|=1$.
  • Column $B_2$ and column $B_2$ give $|e|=1$.
  • Column $A_1$ and column $B_2$ give $b=-e$.
  • Column $A_2$ and column $B_1$ give $c=-d$.

Since $A_1$ has order 3, the eigenvalues of its representations must be $1,\omega$ or $\omega^2$. Therefore the trace (sum of eigenvalues) is in $\{2,2\omega,2\omega^2,-1,-\omega,-\omega^2\}$, but since $|b|=1$ we are only constrained to $\{-1,-\omega,-\omega^2\}$. Up to permuting the $W$'s, we may assume $b=-1$.

\begin{equation*} \begin{array}{|c|c|c|c|c|c|c|c|} \hline & 1 & 1 & 4 & 4 & 4 & 4 & 6 \\ \hline SL_2(\mathbb{Z}/3) & I & -I & A_1 & A_2 & B_1 & B_2 & C \\ \hline U & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline U' & 1 & 1 & \omega & \omega^2 & \omega^2 & \omega & 1 \\ \hline U'' & 1 & 1 & \omega^2 & \omega & \omega & \omega^2 & 1 \\ \hline V & 3 & 3 & 0 & 0 & 0 & 0 & -1 \\ \hline W & 2 & -2 & -1 & c & -c & 1 & 0 \\ \hline W' & 2 & -2 & -\omega & \omega^2 c & -\omega^2 c & \omega & 0\\ \hline W'' & 2 & -2 & -\omega^2 & \omega c & -\omega c & \omega^2 & 0\\ \hline \end{array} \end{equation*}

Since $A_2$ also has order 3, there are three possibilities for the character table of $SL(\mathbb{F}_3)$, depending on the value of $c\in\{-1,-\omega,-\omega^2\}$. If we can show that there is a rational second degree irreducible representation, then $c=1$ and we're finished. But since $W$ is an irreducible representation then $\overline{W}$ is too. Since $\overline{W}$ can't be equal to $W'$ nor $W''$ (look at column of $A_1$), we must have $\overline{W}=W$ and therefore $c$ is real which implies that $c=-1$ thus finishing the proof.

The final table is \begin{equation*} \begin{array}{|c|c|c|c|c|c|c|c|} \hline & 1 & 1 & 4 & 4 & 4 & 4 & 6 \\ \hline SL_2(\mathbb{Z}/3) & I & -I & A_1 & A_2 & B_1 & B_2 & C \\ \hline U & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline U' & 1 & 1 & \omega & \omega^2 & \omega^2 & \omega & 1 \\ \hline U'' & 1 & 1 & \omega^2 & \omega & \omega & \omega^2 & 1 \\ \hline V & 3 & 3 & 0 & 0 & 0 & 0 & -1 \\ \hline W & 2 & -2 & -1 & -1 & 1 & 1 & 0 \\ \hline W' & 2 & -2 & -\omega & -\omega^2 & \omega^2 & \omega & 0\\ \hline W'' & 2 & -2 & -\omega^2 & -\omega & \omega & \omega^2 & 0\\ \hline \end{array} \end{equation*}

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  • $\begingroup$ Some things can be simplified. You can start by multiplying the first row and the third row with two-dimensional reps to obtain values at $-I$ and $C$ (before assuming they're equal), which shows that they are equal, and implies that the remaining part can't be all zero. It's just a different approach. $\endgroup$
    – Kolja
    Jun 23 at 22:05

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