$\mathrm{SL}_{4}(\mathbf{Z})$ is not purely matricial field

We thank Pierre Deligne for explaining to us the above mentioned fact about representations of $\mathrm{SL}_{3}(\mathbf{Z}/p\mathbf{Z})$. We thank Kevin Boucher, Yves de Cornulier and Olivier Dudas for comments and conversations about this project.

Funding:

M. M. This material is based upon work supported by the National Science Foundation under Grant No. DMS-1926686. This project has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (grant agreement No 949143).

M. S. Research supported by the Charles Simonyi Endowment at the Institute for Advanced Study, and the ANR project ANCG Project-ANR-19-CE40-0002.

Let

$$N=\prod_{\text{$p$ prime}}p^{e(p)}$$

be the prime factorization of $N$. By the Chinese remainder theorem

$$\mathrm{SL}_{4}(\mathbf{Z}/N\mathbf{Z})\cong\prod_{\text{$p$ prime,}e(p)>0}\mathrm{SL}_{4}(\mathbf{Z}/p^{e(p)}\mathbf{Z})$$

and this induces a splitting

$$\phi\cong\bigotimes_{\text{$p$ prime,}e(p)>0}\phi_{p}$$

where $\phi_{p}$ are irreducible representations of $\mathrm{SL}_{4}(\mathbf{Z}/p^{e(p)}\mathbf{Z})$. The assumption that $\phi$ is new implies that each $\phi_{p}$ is new. If we can prove all the $\phi_{p}$ have non-zero $\mathrm{SL}_{2}(\mathbf{Z}/p^{e(p)}\mathbf{Z})$-invariant vectors $v_{p}$, then

$$v=\bigotimes_{\text{$p$ prime,}e(p)>0}v_{p}$$

will be the required non-zero invariant vector for $\mathrm{SL}_{2}(\mathbf{Z}/N\mathbf{Z})\cong\prod_{\text{$p$ prime,}e(p)>0}\mathrm{SL}_{2}(\mathbf{Z}/p^{e(p)}\mathbf{Z})$ — the inclusion of $\mathrm{SL}_{2}$ in $\mathrm{SL}_{4}$ that we use commutes with our applications of the Chinese remainder theorem.

The strategy of the proof is the following:

Step 1:

We prove the representation is non-trivial when restricted to all elementary cyclic subgroups of level $p^{r-1}$.

Step 2:

We use Step 1 to prove that on restriction to a particular copy of the Heisenberg group modulo $p^{r}$, we find a particular type of character, namely, the one described in (2.4).

Step 3:

We take a non-zero vector in the isotypic subspace of the character of the Heisenberg group found in Step 2. By averaging this vector over a copy of $\mathrm{SL}_{2}(\mathbf{Z}/p^{r}\mathbf{Z}$) we find a non-zero $\mathrm{SL}_{2}(\mathbf{Z}/p^{r}\mathbf{Z})$-invariant vector. Here, the form of the Heisenberg group character we found in the previous step is important to make sure that this average is non-zero.

It therefore now suffices to prove Proposition 2.1 when $N=p^{r}$, $r\geq 1$. Let $\phi$ denote the irreducible representation. For $1\leq i\neq j\leq 4$ let $\varepsilon_{ij}$ denote the matrix with one in the $i,j$ entry and zeros elsewhere. The first step is to find a non-trivial subrepresentation of some

$$C_{ij}\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\langle I+p^{r-1}\varepsilon_{ij}\rangle.$$

As $C_{ij}$ is abelian, by further passing to a subrepresentation, we may assume the non-trivial subrepresentation is irreducible and hence a character.

If $r=1$ $\mathrm{SL}_{4}(\mathbf{Z}/p\mathbf{Z})$ is generated by such cyclic subgroups. So suppose for this step that $r>1$.

We could proceed by using a result of Bass–Milnor–Serre [BMS67, Cor. 4.3.b] — stating that the principal congruence subgroup of level $p^{r}$ in $\mathrm{SL}_{4}(\mathbf{Z})$ is normally generated by elementary matrices. For completeness, below we give a simple self-contained proof of what we need.

Let $G(p^{r-1})$ denote the kernel of reduction mod $p^{r-1}$ on $\mathrm{SL}_{4}(\mathbf{Z}/p^{r}\mathbf{Z}$). Since we assume $\phi$ is new, we know $G(p^{r-1})$ is not contained in the kernel of $\phi$. Let $\mathrm{Mat}_{4\times 4}^{0}(\mathbf{Z}/p\mathbf{Z})$ denote the four by four matrices with entries in $\mathbf{Z}/p\mathbf{Z}$ and zero trace. The map

$$A\in\mathrm{Mat}_{4\times 4}^{0}(\mathbf{Z}/p\mathbf{Z})\mapsto I+p^{r-1}A\in G(p^{r-1})$$

(2.1)

is easily seen to be an isomorphism of groups, where the group law on $\mathrm{Mat}_{4\times 4}^{0}(\mathbf{Z}/p\mathbf{Z})$ is addition.

We want to first show that some $C_{ij}$ acts non-trivially in the representation.

Suppose for a contradiction that we do not find a non-trivial irreducible subrepresentation of some $C_{ij}$, so that all $1+p^{r-1}B$ with $B$ zero on the diagonal are in $\ker(\phi)$. Using (2.1), this assumption implies that $\phi$ restricted to $G(p^{r-1})$ is equivalent to a non-trivial representation of

$$\mathrm{Mat}_{4\times 4}^{0}(\mathbf{Z}/p\mathbf{Z})/\{\text{elements of $\mathrm{Mat}_{4\times 4}^{0}(\mathbf{Z}/p\mathbf{Z})$ that are zero on the diagonal\}.}$$

But this is spanned by equivalence classes of diagonal elements. Thus there is necessarily a diagonal matrix $A$ such that $I+p^{r-1}A$ is not in the kernel of $\phi$, without loss of generality (choosing a basis for the diagonal trace zero matrices) $A=\left(\begin{array}[]{cccc}1&0&0&0\\ 0&-1&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right).$

We calculate

	$\displaystyle\left(\begin{array}[]{cccc}1&0&0&0\\ 1&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{array}\right)\left(I+p^{r-1}\left(\begin{array}[]{cccc}0&1&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right)\right)\left(\begin{array}[]{cccc}1&0&0&0\\ -1&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{array}\right)$
	$\displaystyle=I+p^{r-1}\left(\begin{array}[]{cccc}-1&1&0&0\\ -1&1&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right)\in\ker(\phi).$

Then also

	$\displaystyle\left(I+p^{r-1}\left(\begin{array}[]{cccc}-1&1&0&0\\ -1&1&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right)\right)\left(I+p^{r-1}\left(\begin{array}[]{cccc}0&-1&0&0\\ 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right)\right)$
	$\displaystyle=I+p^{r-1}\left(\begin{array}[]{cccc}-1&0&0&0\\ 0&1&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right)\in\ker(\phi),$

a contradiction. The conclusion of this step is no matter $r\geq 1$, we find $i\neq j$ such that $C_{ij}\notin\ker(\phi)$. But in fact, since all $C_{ij}$ are conjugate in $\mathrm{SL}_{4}(\mathbf{Z}/p\mathbf{Z})$, this means that:

No $C_{ij}$ is contained in the kernel of $\phi$.

Let $U_{1}$ denote the group

$$U_{1}\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left\{\Upsilon(u_{1},u_{2},u_{3})\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left(\begin{array}[]{cccc}1&0&0&u_{1}\\ 0&1&0&u_{2}\\ 0&0&1&u_{3}\\ 0&0&0&1\end{array}\right)\right\}\leq\mathrm{SL}_{4}(\mathbf{Z}/p^{r}\mathbf{Z}).$$

The group $U_{1}$ is isomorphic to $(\mathbf{Z}/p^{r}\mathbf{Z},+)^{3}$ so the restriction of $\phi$ to $U_{1}$ breaks into a direct sum of one-dimensional subspaces where $U_{1}$ acts by a character. Moreover, $\mathrm{SL}_{3}(\mathbf{Z}/p^{r}\mathbf{Z})$ normalizes $U_{1}$ so it acts on the characters of $U_{1}$ appearing like this by $g\chi=\chi(g^{-1}\cdot g)$. This action is called the dual action. Every such character is of the form

$$\chi\colon\Upsilon(u_{1},u_{2},u_{3})\mapsto\exp\left(2\pi i\frac{(\xi_{1}u_{1}+\xi_{2}u_{2}+\xi_{3}u_{3})}{p^{r}}\right)$$

(2.2)

for $(\xi_{1},\xi_{2},\xi_{3})\in(\mathbf{Z}/p^{r}\mathbf{Z})^{3}$ and the dual action corresponds to $(\xi_{1},\xi_{2},\xi_{3})\mapsto(\xi_{1},\xi_{2},\xi_{3})g^{-1}$. If $(\xi_{1},\xi_{2},\xi_{3})\equiv 0\bmod p$ then all $\Upsilon(u_{1},u_{2},u_{3})$ with $p^{r-1}|u_{1},u_{2},u_{3}$ are in the kernel of the character. If all obtained characters satisfy this condition, then $\phi$ restricted to $U_{1}$ has $U_{1}\cap G(p^{r-1})$ in its kernel. But by Step 1, $C_{14}$ is not contained in the kernel of $\phi$. Hence in the restriction of $\phi$ to $U_{1}$ there must be a character of the form (2.2) where $(\xi_{1},\xi_{2},\xi_{3})\not\equiv 0\bmod p$. Since $\mathrm{SL}_{3}(\mathbf{Z}/p^{r}\mathbf{Z})$ acts transitively on the vectors in $(\mathbf{Z}/p^{r}\mathbf{Z})^{3}$ satisfying $(\xi_{1},\xi_{2},\xi_{3})\not\equiv 0\bmod p$, by considering the dual action we may assume

$$(\xi_{1},\xi_{2},\xi_{3})=(0,0,1).$$

Let $V_{\chi}$ be the $\chi$-isotypic space for the restriction of $\phi$ to $U_{1}$, where $\chi$ and $\xi$ are as above.

The group

$$G_{1}\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left\{\left(\begin{array}[]{cccc}*&*&*&0\\ &*&*&0\\ 0&0&1&0\\ 0&0&0&1\end{array}\right)\right\}\leq\mathrm{SL}_{4}(\mathbf{Z}/p^{r}\mathbf{Z})$$

normalizes $U_{1}$ and fixes $\chi$ under the dual action. Hence $V_{\chi}$ is an invariant subspace for $G_{1}$. Now restrict $V_{\chi}$ to the group

$$U_{2}\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left\{[v_{1};v_{2}]\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left(\begin{array}[]{cccc}1&0&v_{1}&0\\ 0&1&v_{2}&0\\ 0&0&1&0\\ 0&0&0&1\end{array}\right)\right\}\leq G_{1}$$

and we will decompose this into characters $\theta$ of $U_{2}$; let $V_{\chi,\theta}$ denote the corresponding isotypic subspace.

Consider now the group

$$H\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left\{[x;y;z]\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left(\begin{array}[]{cccc}1&0&0&0\\ 0&1&x&z\\ 0&0&1&y\\ 0&0&0&1\end{array}\right)\right\}\leq\mathrm{SL}_{4}(\mathbf{Z}/p^{r}\mathbf{Z}).$$

As we already mentioned, $G_{1}$ preserves $V_{\chi}$. Obviously $U_{1}$ fixes all its characters under the dual action induced by conjugation, hence all $\left(\begin{array}[]{cccc}*&*&*&*\\ &*&*&*\\ 0&0&1&*\\ 0&0&0&1\end{array}\right)$ fix our chosen $\chi$ under the dual action, or in other words, leave $V_{\chi}$ invariant. Hence the space $V_{\chi}$ is invariant by $H$.

We have $[0;0;z]=\Upsilon(0,z,0)\in U_{1}$ and for $v\in V_{\chi}$

$$\Upsilon(0,z,0)v=\exp\left(2\pi i\frac{(0\cdot 0+0\cdot z+1\cdot 0)}{p^{r}}\right)v=v.$$

Hence the action of $H$ on $V_{\chi}$ has kernel that contains the subgroup with $x=y=0$, which is isomorphic to $\mathbf{Z}/p^{r}\mathbf{Z}$. Hence the action of $H$ on $V_{\chi}$ factors through an action of

$$H/(\mathbf{Z}/p^{r}\mathbf{Z})\cong(\mathbf{Z}/p^{r}\mathbf{Z})^{2}.$$

We want to find a particular character of $H$ and to do so we split into the following cases.

Case 1. $V_{\chi}$ restricted to $U_{2}$ is trivial. Then obviously $H$ acts on all of $V_{\chi}$ by

$$[x;y;z]\mapsto\exp\left(2\pi i\frac{y}{p^{r}}\right).$$

(2.3)

Case 2. Otherwise, we find a character $\theta$ in $V_{\chi}$ of the form

$$\theta:[v_{1};v_{2}]\mapsto\exp\left(2\pi i\frac{(\zeta_{1}v_{1}+\zeta_{2}v_{2})}{p^{r}}\right)$$

with $(\zeta_{1},\zeta_{2})\not\equiv(0,0)\bmod p^{r}$. Write $(\zeta_{1},\zeta_{2})=p^{R}(z_{1},z_{2})$ with $(z_{1},z_{2})\neq(0,0)\bmod p$. By conjugation in $\mathrm{SL}_{2}(\mathbf{Z}/p^{r}\mathbf{Z})\leq G_{1}$ — which normalizes $U_{2}$ — we can find a new $\theta^{\prime}$ with corresponding $z_{1}=1,z_{2}=0$ so that

$$\theta^{\prime}:[v_{1};v_{2}]\mapsto\exp\left(2\pi i\frac{v_{1}}{p^{r-R}}\right).$$

In particular, on $V_{\chi,\theta^{\prime}}$ $H$ acts by the character (2.3).

To summarize, in any case, there exists a non-zero vector $v\in V_{\chi}$ such that

$$\phi([x;y;z])v=\exp\left(2\pi i\frac{y}{p^{r}}\right)v.$$

(2.4)

Now let

$$G_{2}\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left\{\left(\begin{array}[]{cccc}1&0&0&0\\ 0&a&b&0\\ 0&c&d&0\\ 0&0&0&1\end{array}\right)\right\}\leq\mathrm{SL}_{4}(\mathbf{Z}/p^{r}\mathbf{Z}).$$

From (2.4), $v$ is fixed by the subgroup

$$N\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\left\{\left(\begin{array}[]{cccc}1&0&0&0\\ 0&1&n&0\\ 0&0&1&0\\ 0&0&0&1\end{array}\right)\right\}\leq G_{2}.$$

This implies that if $W$ denotes the representation of $G_{2}\cong\mathrm{SL}_{2}(\mathbf{Z}/p^{r}\mathbf{Z})$ generated by $v$, that $W$ is a quotient of the induced representation

$$\mathrm{Ind}_{N}^{G_{2}}\mathrm{triv}=\mathbb{C}[G_{2}]\otimes_{N}\mathbb{C}.$$

Suppose $g\in G_{2}$, with $a,b,c,d$ as above in $\mathbf{Z}/p^{r}\mathbf{Z}$. We have

	$\displaystyle\phi([0;y;z])\phi(g^{-1})v$	$\displaystyle=\phi(g^{-1})\phi(g[0;y;z]g^{-1})v$
		$\displaystyle=\phi(g^{-1})\phi([0;cz+dy;az+by])v$
		$\displaystyle=\phi(g^{-1})\exp\text{$\left(2\pi i\frac{(dy+cz)}{p^{r}}\right)$}v.$

This means, in this co-adjoint action of $G_{2}$ on characters of the group $\langle[0;y;z]\rangle$, $N$ is precisely the stabilizer of the character of $v$, and hence

$$\dim W=|G_{2}|/|N|=\dim\mathrm{Ind}_{N}^{G_{2}}\mathrm{triv},$$

so in fact, $W\cong\mathrm{Ind}_{N}^{G_{2}}\mathrm{triv}$ as a $G_{2}$ representation. By Frobenius reciprocity, this contains the trivial representation of $G_{2}$. Finally, $G_{2}$ and the upper left copy of $\mathrm{SL}_{2}(\mathbf{Z}/p^{r}\mathbf{Z})$ are conjugate in $\mathrm{SL}_{4}(\mathbf{Z}/p^{r}\mathbf{Z})$. This concludes the proof.

The character tables of $\mathrm{SL}_{3}(\mathbf{F})$ for finite fields $\mathbf{F}$ have been computed in [SF73]. In particular, if we view $\mathrm{SL}_{2}(\mathbf{F})$ as the subgroup of $\mathrm{SL}_{3}(\mathbf{F})$ consisting of matrices of the form $\left(\begin{array}[]{ccc}*&*&0\\ &*&0\\ 0&0&1\\ \end{array}\right)$, we obtain the following example, explained to us by Deligne:

The representations are any of those denoted $\chi_{r^{2}s}(u)$ in [SF73, Table 1b] (that are associated with tori of split rank $0$ in the Deligne-Lusztig theory [Hum81]). The properties of $\mathrm{Tr}(\pi(g))$ follow readily from this table and the description of the conjugacy classes in $\mathrm{SL}_{2}(\mathbf{F}_{q})$ (e.g. [Bon11, Section 1.3]).

Such a representation does not have non-zero $\mathrm{SL}_{2}(\mathbf{F}_{q})$-invariant vectors because, using that there are exactly $q^{2}-1$ unipotent matrices in $\mathrm{SL}_{2}(\mathbf{F}_{q})\setminus\{1\}$ [Bon11, Section 1.3], we can compute that the trace of the projection on the $\mathrm{SL}_{2}(\mathbf{F}_{q})$-invariant vectors is $0$:

$$\mathrm{Tr}\Big{(}\sum_{g\in\mathrm{SL}_{2}(\mathbf{F}_{q})}\pi(g)\Big{)}=1\cdot(q-1)(q^{2}-1)+(q^{2}-1)\cdot(1-q)=0.$$