Zariski closure is semisimple

Suppose that $\Gamma$ is a finitely generated semi-group of $SL(n,\mathbb{Z})$ which acts strongly irreducible on $\mathbb{R}^n$, i.e. there is no finite union of proper nonzero linear sub-spaces of $\mathbb{R}^n$ which is invariant under $\Gamma$. It is easy to see that the zariski closure of $\Gamma$ is a reductive group. Now how to prove that this group is actually semi-simple?

(Actually this is lemma 8.5. in this paper by Benoist-Quint, but I can not understand their proof for finiteness of center!)

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\Qbar{\overline{\mathbb{Q}}}$Let $G$ be the Zariski closure of $\Gamma$. We may (and do) replace $G$ by the connected component of the identity $G_0$ and replace $\Gamma$ by $\Gamma \cap G_0$, in order to assume that $G$ is connected. Note that this is the place where we needed strong irreducibility: An irreducible representation may not stay irreducible when restricted to a finite index subgroup, but a strongly irreducible representation stays strongly irreducible. (For example, the group of matrices of the form $\left( \begin{smallmatrix} z & 0 \\ 0 & z^{-1} \end{smallmatrix} \right)$, $\left( \begin{smallmatrix} 0 & z \\ z^{-1} & 0 \end{smallmatrix} \right)$ acts irreducibly, but becomes reducible when restricted to the index two group of diagonal matrices. But neither action is strongly irreducible.)

Therefore, we now assume $G$ is connected. Let $Z(G)$ be the center of $G$ and let $G^{ab}$ be the abelianization. By structure theory of reductive groups, the composite $Z(G) \to G \to G^{ab}$ is an isogeny. (Not true if $G$ is disconnected! See above example.) Let $H$ be the (finite) kernel of this isogeny, so we have $1 \to H \to Z(G) \to G^{ab} \to 1$. Our goal is to show that $Z(G)$, and equivalently $G^{ab}$, is finite. Suppose for contradiction that $G^{ab}$ is infinite.

Since $\Gamma=G(\ZZ)$ is Zariski dense in $G$, we know that $G^{ab}(\ZZ)$ is Zariski dense in $G^{ab}$. Since we are assuming $G^{ab}$ is infinite, we can find $g \in G^{ab}(\ZZ)$ of infinite order. Replacing $g$ by $g^{|H|}$, we may (and do) assume that $g$ lifts to $Z(G)$.

So, we now have $g \in Z(G)(\ZZ) \subset G(\ZZ) \subset SL_n(\ZZ)$ of infinite order. Let $P(t)$ be the minimal polynomial of $g$, and let $P(t)$ factor over $\mathbb{R}[t]$ as $\prod P_i(t)^{a_i}$. Since $g$ is central in $\Gamma$, we see that $\Gamma$ preserves $\mathrm{Ker}(P_i)$. So, if $P(t)$ is not irreducible (as a polynomial in $\mathbb{R}[t]$) then $\mathbb{R}^n$ is not irreducible (as a $\Gamma$ representation.)

Thus, $P(t)$ is either linear, or a quadratic with imaginary roots, and its constant term is $\pm 1$ since $g \in \mathrm{SL}_n(\ZZ)$. So $P(t)$ is $t \pm 1$, $t^2+1$ or $t^2 \pm t + 1$. But the roots of any of these are torsion, so $g$ is torsion, a contradiction.

The OP asks why I can lift from $G^{ab}$ to $Z(G)$. Let $\phi : T_1 \to T_2$ be an isogeny of algebraic tori with kernel $H$, a group scheme of length $N$. (In case this is a source of confusion, note that the kernel of $z \mapsto z^3$, from $\mathbb{G}_m \to \mathbb{G}_m$ is length 3, not 1, even if our ground field does not contain cube roots of $1$.) Let $g \in T_2(\mathbb{Z})$. The claim is that $g^N$ lifts to $T_1(\mathbb{Z})$.

Since $H$ is a commutative group scheme of order $N$, it is annihilated by $N$. Let $K \subset T_1$ be the kernel of multiplication by $N$. So $H \subseteq K$ as group schemes, and we have $T_1 \overset{\phi}{\longrightarrow} T_2 \overset{\psi}{\longrightarrow} T_1$ where $\psi \circ \phi$ is multiplication by $N$. It must also then be true that $\phi \circ \psi$ is multiplication by $N$. So $\psi(g)$ is a lft of $g^N$.

• +1. I learnt some techniques from your clear answer. Is there a way to relate the dimension of the closure with $\Gamma$.? I am aware $\mathbf{Z}$ can be embedded as a dense subgroup in a torus. Something more on similar lines? – P Vanchinathan Sep 1 '16 at 0:41
• Thanks. I can't think of much to say about relating the dimension of $G$ to properties of $\Gamma$, except that if $\Gamma=\mathbb{Z}$ then $G$ is abelian. (But, as the above proof shows, in that case the action won't be strongly irreducible.) – David E Speyer Sep 1 '16 at 0:59
• Sorry, but I can not understand how do you lift an element of $G^{ab}$ to $Z(G)$. Can you explain it more? For example if the isogeny is actually an isomorphism ($H=\{1\}$), then every element of $G\setminus [G,G]$ is in $Z(G)$? Thank you. – Hesam Sep 14 '16 at 6:40
• I added a proof. Perhaps some examples will help as well? Example 1: If $G = GL_3$, then $Z(G) \cong \mathbb{G}_m \cong G^{ab}$ with maps $z \longrightarrow z \mathrm{Id} \overset{\det}{\longrightarrow} z^3$ and $N=3$. The claim is that, for any $w$, there is a central matrix with determinant $w^3$. – David E Speyer Sep 15 '16 at 3:24
• Example 2: Let $T_1$ and $T_2 \cong \mathbb{G}_m^2$. Map $T_1 \to T_2$ by $(z_1, z_2) \mapsto (z_1 z_2, z_1/z_2)$. The kernel is $\pm (1,1)$, of order $2$, so $N=2$. The claim is that, for any $(w_1, w_2)$, we can solve the equations $(z_1 z_2, z_1/z_2) = (w_1^2, w_2^2)$. Of course, this is ture: take $z_1 = w_1 w_2$ and $z_2 = w_1/w_2$. – David E Speyer Sep 15 '16 at 3:27