# Irreducible Subrepresentations of representation of $\operatorname{GL}_{3}(\mathbb{F}_q)$

For a character $\zeta$ of $\mathbb{F}_q^*$, we can construct the representation $\zeta \otimes \zeta \otimes \zeta$ of the diagonal subgroup $L$ of $\operatorname{GL}_{3}(\mathbb{F}_q)$, in the following way:

$$\zeta \otimes \zeta \otimes \zeta \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \\ \end{pmatrix} = \zeta(a)\cdot \zeta(b) \cdot \zeta(c)$$

We can let this representation act on the group $B$ of upper triangular matrices by lettingit act trivial on the group $U$ in the Levi decomposition of $B = U \rtimes L$. I would like to find the decomposition into irreducible representations of $\operatorname{Ind}_B^G(\zeta \otimes \zeta \otimes \zeta)$.

My start: I know from an article of Steinbeck that this should probably give me $3\cdot (q-1)$ irreducible representations. Furthermore we can write $\operatorname{Ind}_B^G(\zeta \otimes \zeta \otimes \zeta) = \operatorname{Ind}_B^G(\zeta \circ \det)$. However, I cannot find the irreducible subrepresentations.

Edit: I should probably add more that I have already found. Let us say $\pi := \zeta \otimes \zeta \otimes \zeta$, then we are trying to compute (by Frobenius Reciprocity $\langle \pi, \operatorname{Ind}_B^G(\operatorname{Ind}_B^G(\pi)) \rangle_B$, which is equal to $$\sum_{w \in W} \langle \pi, \operatorname{Ind}_{B\cap B^w}^{B}(\pi^w) \rangle_B,$$ where $W$ is the Weyl-group (in this case $S_3$). We see that these give us the irreducible components of $\operatorname{Ind}_B^G(\pi)$. For $w = e \in W$, we get $\operatorname{Ind}_{B\cap B^w}^{B}(\pi^w) = \pi$, and of course $\langle \pi, \pi \rangle_B = 1$.

• Just a note on notation (I might return to think about the actual question later if I have time). The group you mean is usually denoted $GL_3(\mathbb{F}_q)$, rather than the other way around. Aug 27, 2015 at 18:51
• Ah yes of course, thank you! Aug 28, 2015 at 11:36

First of all, let me point out that for fixed $\zeta$, this incuded representation will have $3$ distinct irreducible components, not $3(q-1)$; you get the $q-1$ from allowing $\zeta$ to vary. (You probably know that, but just to make sure).

You've got more-or-less the right idea of how to do it. There's a slight issue though, which I'll explain shortly. For simplicity, note that, as $\det$ is a character of $G$, we may write $Ind_B^G \pi=(Ind_B^G 1)\otimes(\zeta\circ\det)$. So we may as well just do this for $\zeta=1$. Now, proceeding just as you did, we get $$\dim End_G(Ind_B^G 1)=\sum_{w\in W}\dim Hom_B(1,Ind_{B\cap B^w}^B 1).$$ Using Frobenius again, this is $\sum_{w\in W}\dim Hom_{B\cap B^w}(1,1)$, which is clearly just $|W|=6$. The reason that we're seeing 6 is that $Ind_B^G 1$ isn't multiplicity-free.

The decomposition is given as follows. Let $M=GL(2)\times GL(1)$ and $M'=GL(1)\times GL(2)$ be the Levi subgroups lying between $L$ and $G$, so that we have a tower of inclusions $L\subset M,M'\subset G$. This gives rise (by transitivity of induction) to a reverse-ordered tower of embeddings $1=Ind_G^G 1\hookrightarrow Ind_P^G 1\simeq Ind_{P'}^G 1\hookrightarrow Ind_B^G 1$, where $P,P'$ are standard parabolics containing $M$ and $M'$, respectively. Clearly 1 is the first irreducible subrepresentation of $Ind_B^G 1$. Next, $1$ admits a complement in $Ind_P^G 1$ (by semisimplicity), which should be irreducible, and similarly $Ind_P^G 1$ has a complement in $Ind_B^G 1$. I'll leave you to check the details; you can basically do it as we did above (i.e. calculate $\dim End_G(Ind_P^G 1)$, which will decompose as a sum over a subgroup of the Weyl group. This will be the set $W_M$ of elements of $W$ which recovers the partition $G=\coprod_{w\in W_M}PwP$; abstractly it will be the Weyl group of $M$, which is $W_M=N_G(M)/M$. You'll also have to look at cosets of the form $BwP$ at some point, again just look at what's needed to recover $G=\coprod BwP$.)

• In another part of this problem I believe I have already calculated $W_M$ as $P\backslash G/P = \{ \{ e, (12) \} , \{ (13), (23), (123), (132) \} \}$. Could I also calculate $B\backslash G/P$ in the same way? Aug 28, 2015 at 13:47
• I don't know; you didn't say how you calculated it! Here's the simplest way to do it though: certainly the full Weyl group $W$ represents every $B\backslash G/P$ coset. Now ask yourself what can possibly lead to two of these cosets being equal. Clearly, it's going to be something coming from $P$. The part of $P$ which isn't accounted for by $W$ is the extra entry coming from the $GL(2)$ in $M$. This means that you can multiply by $\begin{pmatrix} 0& 1&0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{pmatrix}$. Now try to see which cosets $BwP$ with $w\in W$ this eliminates the need for.
– PL.
Aug 28, 2015 at 13:53
• Also, I'd have to check it but your computation of $W_M$ seems to be OK. Certianly, it should be of order 2.
– PL.
Aug 28, 2015 at 13:56
• Although my calculations are too large and sketchy to post here, I have calculated $|B\backslash G/P| = 2$ and $|B\backslash G/P| = 3$ which (together with some proofs of the irreducibility of these reps) gives us the decomposition $\operatorname{Ind}_B^G(1) = 1 + (\operatorname{Ind}_P^G(1) - 1) + 2\cdot (\operatorname{Ind}_{P'}^G(1) - 1)$, which concludes the question, as this is the decomposition into three irreducible representations. Aug 29, 2015 at 18:33
• I should correct the previous comment by me, as I have not so much calculated $|B\backslash G /P|$ or $|B\backslash G /P'|$, but rather the subgroups $W_M$ and $W_{M'}$ as suggested by @PL. and did this according to the following exercise out of my syllabus: imgur.com/76ylEJS , where in our case of course $n = 3$ and $n_1 = 1$ and $n_2 = 2$. Aug 29, 2015 at 18:40