# Computing the character table of $SL_2(\mathbb{Z}/3)$

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.

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

• 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. Jun 23 at 22:05