Posted by: matheuscmss | September 10, 2018

## Benoist’s minicourse “Arithmeticity of discrete groups”. Lecture III: From closedness to arithmeticity

Recall that the main goal of this series of posts is the proof of the following result:

Theorem 1 Let ${G}$ be a semisimple algebraic Lie group of real rank ${\textrm{rank}_{\mathbb{R}}(G)\geq 2}$. Denote by ${U\subset G}$ a horospherical subgroup of ${G}$. If ${\Gamma\subset G}$ is a discrete Zariski-dense and irreducible subgroup such that ${\Gamma\cap U}$ is cocompact, then ${\Gamma}$ is commensurable to an arithmetic lattice ${G_{\mathbb{Z}}}$.

Last time, we discussed the first half of the proof of this theorem in the particular case of ${G=SL(2p,\mathbb{R})}$, ${p\geq 2}$, and ${U=\left\{\left( \begin{array}{cc} I & B \\ 0 & I \end{array} \right): B\in M(p,\mathbb{R})\right\}}$. Actually, we saw that this specific form of ${U\subset G}$ is not very important: all results from the previous post hold whenever

• ${U}$ is reflexive: in the context of the example above, this is the fact that ${U}$ is conjugate to the opposite horospherical subgroup ${U^-=\left\{\left( \begin{array}{cc} I & 0 \\ C & I \end{array} \right): C\in M(p,\mathbb{R})\right\}}$;
• ${U}$ is commutative.

Indeed, we observed that ${U}$ reflexive allows to also assume that ${\Gamma\cap U^-}$ is cocompact in ${U^-}$. Then, this property and the commutativity of ${U}$ were exploited to establish the closedness of the ${\textrm{Ad}_L}$-orbit of ${(\Lambda, \Lambda^-)}$ in ${X_{\mathfrak{u}}\times X_{\mathfrak{u}^-}}$, where ${\Lambda:=\log(\Gamma\cap U)\in X_{\mathfrak{u}} = \{\textrm{lattices in }\mathfrak{u}\}}$, ${\Lambda^-:=\log(\Gamma\cap U^-)\in X_{\mathfrak{u}^-} = \{\textrm{lattices in }\mathfrak{u}^-\}}$, and

$\displaystyle L=P\cap P^- = \left\{\left(\begin{array}{cc} A & 0 \\ 0 & D \end{array}\right):\textrm{det}(A)\cdot\textrm{det}(D)=1\right\}$

is the common Levi subgroup of the parabolic subgroups ${P=N_G(U)}$ and ${P^-=N_G(U^-)}$ normalizing ${U}$ and ${U^-}$.

Today, we will discuss the second half of the proof of Theorem 1 in the particular case of ${G=SL(2p,\mathbb{R})}$, ${p\geq 2}$, and ${U=\left\{\left( \begin{array}{cc} I & B \\ 0 & I \end{array} \right): B\in M(p,\mathbb{R})\right\}}$: in other terms, our goal below is to obtain the arithmeticity of ${\Gamma}$ from the closedness of ${\textrm{Ad}_L(\Lambda, \Lambda^-)}$ in the homogenous space ${X_{\mathfrak{u}}\times X_{\mathfrak{u}^-} := G_0/\Gamma_0}$. This step is due to Hee Oh (see Proposition 3.4.4 of her paper).

1. From closedness to infinite stabilizer

Let ${S=[L,L] = \left\{ \left(\begin{array}{cc} A & 0 \\ 0 & D \end{array}\right)\in G: \textrm{det}(A) = \textrm{det}(D) = 1 \right\} \simeq SL(p,\mathbb{R})\times SL(p,\mathbb{R})}$ and ${A=\left\{ \left(\begin{array}{cc} \lambda I & 0 \\ 0 & \lambda^{-1} I \end{array}\right)\in G: \lambda\in\mathbb{R}^*\right\}}$, so that ${L=AS}$. In particular, the closedness of ${\textrm{Ad}_L(\Lambda, \Lambda^-)}$ implies that

${\textrm{Ad}_S(\Lambda, \Lambda^-)}$ is closed in ${X_{\mathfrak{u}}\times X_{\mathfrak{u}^-} = G_0/\Gamma_0}$.

The next proposition asserts that the stabilizer of this orbit is large whenever ${S}$ is not compact:

Proposition 2 The stabilizer ${\textrm{Stab}_S(\Lambda,\Lambda^-)=\{s\in S: \textrm{Ad}_s(\Lambda,\Lambda^-) = (\Lambda,\Lambda^-)\}}$ is a lattice in ${S}$.

This proposition is a direct consequence of the closedness of the ${\textrm{Ad}_S(\Lambda, \Lambda^-)}$ in ${G_0/\Gamma_0}$ and the following general fact:

Proposition 3 Let ${G_0}$ be a Lie group, ${\Gamma_0}$ a lattice in ${G_0}$, and ${x_0\in X_0=G_0/\Gamma_0}$. Suppose that ${S_0\subset G_0}$ is a semisimple subgroup with finite center such that ${S_0 x_0}$ is closed, then ${S_0\cap \Gamma_0}$ is a lattice in ${S_0}$.

Proof: The first ingredient of the argument is Howe–Moore’s mixing theorem: it asserts that if ${S_0}$ is a semisimple group with finite center and ${(\mathcal{H}_0, \pi_0)}$ is an unitary representation of ${S_0}$ with ${\{v\in\mathcal{H}_0: \pi_0(S_0)v = v\}=\{0\}}$, then

$\displaystyle \lim\limits_{s\rightarrow\infty}\langle\pi_0(s) v_0, w_0\rangle = 0$

for all ${v_0, w_0\in\mathcal{H}_0}$. (Here, ${s\rightarrow\infty}$ means that the projection of ${s}$ to any simple factor ${S^{(i)}}$ of ${S_0 = \prod S^{(i)}}$ diverges.)

The second ingredient of the argument is Dani–Margulis recurrence theorem: it says that if ${\Gamma_0}$ is a lattice in a Lie group ${G_0}$ and ${u_t}$ is a one-parameter unipotent subgroup of ${G_0}$, then, given ${x_0\in X_0=G_0/\Gamma_0}$ and ${\varepsilon>0}$, there exists a compact subset ${K\subset X_0}$ such that

$\displaystyle \frac{1}{T}\textrm{Leb}(\{t\in [0,T]: u_t x_0\in K\}) \geq 1-\varepsilon$

for all ${T>0}$.

The basic idea to obtain the desired proposition is to apply these ingredients to ${\mathcal{H}_0 = L^2(X_0,\lambda_0)}$, where ${\lambda_0}$ is a ${S_0}$-invariant measure on ${S_0 x_0}$, and ${u_t}$ is a one-parameter unipotent subgroup in the product ${S_0''}$ of non-compact simple factors of ${S_0}$ . Here, we observe that ${\lambda_0}$ is a bona fide Radon measure because we are assuming that ${S_0 x_0}$ is closed, and, if we take ${u_t}$ not contained in proper normal subgroups of ${S_0''}$, then ${u_t\rightarrow\infty}$ as ${t\rightarrow\infty}$ thanks to the absence of compact factors in ${S_0''}$. In this setting, our task is reduced to prove that ${\lambda_0}$ is a finite measure.

In this direction, we apply Dani–Margulis recurrent theorem to get ${A\subset X_0}$ with ${0 < \lambda_0(A) < \infty}$ and a compact subset ${K\subset X_0}$ such that

$\displaystyle \frac{1}{T}\textrm{Leb}(\{t\in [0,T]: u_t x\in K\}) \geq \frac{1}{2}$

for all ${T>0}$ and ${x\in A}$. In this way, we obtain that the characteristic functions ${1_A}$ and ${1_K}$ of ${A}$ and ${K}$ are two elements of ${\mathcal{H}_0}$ with ${\langle \pi_0(u_t) 1_A, 1_K \rangle = \lambda_0(A\cap u_t^{-1}(K))}$, and, hence, by Fubini’s theorem,

$\displaystyle \begin{array}{rcl} \frac{1}{T}\int_0^T \langle \pi_0(u_t) 1_A, 1_K \rangle\,dt &=& \frac{1}{T}\int_0^T \lambda_0(A\cap u_t^{-1}(K)) \,dt \\ &=& \int_A \frac{1}{T}\textrm{Leb}(\{t\in [0,T]: u_t x\in K\}) \, d\lambda_0 \\ &\geq& \lambda_0(A)/2 > 0 \end{array}$

for all ${T>0}$. It follows from Howe–Moore’s mixing theorem that

$\displaystyle \{v\in\mathcal{H}_0: \pi_0(S_0'')v=v\}\neq \{0\},$

that is, there exists a non-zero function ${\varphi''\in\mathcal{H}_0}$ which is ${S_0''}$-invariant. By averaging ${\varphi''}$ over the product ${S_0'}$ of the compact factors of ${S_0}$ if necessary, we obtain a non-zero function ${\varphi\in\mathcal{H}_0}$ which is ${S_0}$-invariant. By ergodicity of ${\lambda_0}$, we have that ${\varphi}$ is constant and, a fortiori, ${\lambda_0}$ is a finite measure. $\Box$

2. From infinite stabilizer to ${\Gamma\cap L}$ infinite

Our previous discussions (about ${\textrm{Ad}_L}$-actions) paved the way to understand ${\Gamma\cap L}$. Intuitively, it is important to get some information about ${\Gamma\cap L}$ in our way towards showing the arithmeticity of ${\Gamma}$ because we already know that ${\Gamma\cap U}$ and ${\Gamma\cap U^-}$ are lattices (i.e., ${\Gamma}$ projects to lattices in “other directions”).

The intuition in the previous paragraph is confirmed by Margulis construction of ${\mathbb{Q}}$-forms:

Theorem 4 (Margulis) If ${\Gamma\cap L}$ is infinite, then ${\Gamma}$ is contained in some ${\mathbb{Q}}$-form ${G_{\mathbb{Q}}}$ of ${G}$.

We will discuss the proof of this result in the next section. For now, we want to exploit the information ${\Gamma\subset G_{\mathbb{Q}}}$ in order to derive the arithmeticity of ${\Gamma}$. For this sake, we invoke the following result:

Theorem 5 (Raghunathan-Venkataramana) Assume that ${G}$ is semisimple of ${\textrm{rank}_{\mathbb{R}}(G)\geq 2}$ defined over ${\mathbb{Q}}$, and ${G}$ is ${\mathbb{Q}}$-simple. Suppose that ${U}$ and ${U^-}$ are opposite horospherical subgroups defined over ${\mathbb{Q}}$.If ${\Gamma\subset G_{\mathbb{Z}}}$ is a subgroup such that ${\Gamma\cap U_{\mathbb{Z}}}$, resp. ${\Gamma\cap U_{\mathbb{Z}}^-}$, has finite index in ${U_{\mathbb{Z}}}$, resp. ${U_{\mathbb{Z}}^-}$, then ${\Gamma}$ has finite index in ${G_{\mathbb{Z}}}$.

Remark 1 As it was kindly pointed out to me by David Fisher, Raghunathan-Venkataramana theorem is due to Margulis in the cases covered by Raghunathan (at least).

Remark 2 This result is false when ${\textrm{rank}_{\mathbb{R}}(G)=1}$, e.g., ${G=SL(2,\mathbb{R})}$ (and ${\Gamma=\left\langle\left(\begin{array}{cc}1 & 3 \\ 0 & 1 \end{array}\right), \left(\begin{array}{cc}1 & 0 \\ 3 & 1 \end{array}\right) \right\rangle}$).

Roughly speaking, Raghunathan-Venkataramana theorem essentially establishes the desired Theorem 1 provided we know in advance that ${\Gamma\subset G_{\mathbb{Q}}}$.

In the sequel, we will treat Raghunathan-Venkataramana theorem as a blackbox and we will complete the proof of Theorem 1 in the case ${U}$ reflexive and commutative, and ${S}$ non-compact such as ${G=SL(2p,\mathbb{R})}$ and ${U = \left\{\left(\begin{array}{cc} I & B \\ 0 & I\end{array}\right)\in G: B\in M(p,\mathbb{R})\right\}}$.

Remark 3 In some natural situations (e.g., subgroups generated by the matrices of the so-called Kontsevich–Zorich cocycle) we have that ${\Gamma\subset G_{\mathbb{Z}}}$. In particular, it is a pity that the lack of time made that Yves Benoist could not explain to me the proof of Raghunathan-Venkataramana theorem. Anyhow, I hope to come back to discuss this point in more details in the future.

Proof of Theorem 1:  We consider the subgroup ${\Gamma':=\langle \Gamma\cap U, \Gamma\cap U^-\rangle}$ of ${\Gamma}$. It is discrete and Zariski dense in ${G}$. Therefore, the normalizer ${\Gamma''=N_G(\Gamma')}$ is Zariski dense in ${G}$, and it is not hard to check that it is also discrete.

Observe that Proposition 2 says that ${\Gamma''\cap S}$ is a lattice in ${S}$ (because ${\Gamma''=N_G(\Gamma')\supset \textrm{Stab}_S(\Lambda, \Lambda^-)}$). Since ${S}$ is non-compact, we have that ${\Gamma'\cap L}$ is infinite. Thus, Margulis’ Theorem 4 implies that ${\Gamma'\subset\Gamma''\subset G_{\mathbb{Q}}}$ for some ${\mathbb{Q}}$-form of ${G}$.

Note also that ${U}$ and ${U^-}$ are defined over ${\mathbb{Q}}$: in fact, if ${H\subset GL(n,\mathbb{R})}$ is an algebraic subgroup such that ${H\cap GL(n,\mathbb{Q})}$ is Zariski dense in ${H}$, then ${H}$ is defined over ${\mathbb{Q}}$.

Hence, we can apply Raghunathan-Venkataramana Theorem 5 to get that ${\Gamma'}$ is commensurable to ${G_{\mathbb{Z}}}$. Since ${\Gamma'\subset\Gamma}$, this proves the arithmeticity of ${\Gamma}$. $\Box$

Remark 4 As we noticed above, the arguments presented so far allows to prove Theorem 1 when ${U}$ is reflexive and commutative, and ${S}$ is non-compact.

3. From ${\Gamma\cap L}$ infinite to arithmeticity

In this (final) section (of this post), we discuss some steps in the proof of Margulis  Theorem 4 stated above.

We write ${\mathfrak{g} = \mathfrak{u}^-\oplus \underbrace{\mathfrak{s}\oplus\mathfrak{a}}_{=\mathfrak{l}} \oplus\mathfrak{u}}$, i.e., we decompose the Lie algebra of ${G}$ in terms of the Lie algebras of ${U^-}$, ${U}$ and ${L=AS}$.

Our goal is to find a ${\mathbb{Q}}$-form of ${G}$ containing ${\Gamma}$. For this sake, let us do some “reverse engineering”: assuming that we found ${G_{\mathbb{Q}}\supset\Gamma}$, what are the constraints satisfied by the ${\mathbb{Q}}$-structure ${\mathfrak{g}_{\mathbb{Q}}}$ on its Lie algebra?

First, we note that we dispose of lattices ${\Lambda\subset \mathfrak{u}}$ and ${\Lambda^-\subset\mathfrak{u}^-}$. Hence, we are “forced” to define ${\mathfrak{u}_{\mathbb{Q}}}$ and ${\mathfrak{u}^-_{\mathbb{Q}}}$ as the ${\mathbb{Q}}$-vector spaces spanned by ${\Lambda}$ and ${\Lambda^-}$.

Next, we observe that the choice of ${\mathfrak{u}_{\mathbb{Q}}}$ above imposes a natural ${\mathbb{Q}}$-structure ${\mathfrak{s}_{\mathbb{Q}}}$ on ${\mathfrak{s}}$ via the adjoint map ${\textrm{Ad}:S\rightarrow SL(\mathfrak{u})}$. In fact, ${S}$ is defined over ${\mathbb{Q}}$ because we know that ${\Gamma\cap S}$ is a lattice (and, hence, Zariski-dense) in ${S}$ and ${\textrm{Ad}(S\cap\Gamma)\subset SL(\mathfrak{u}_{\mathbb{Q}})}$ (by definition).

Finally, since ${U}$, ${U^-}$ and ${S}$ are already defined over ${\mathbb{Q}}$ and we want a ${\mathbb{Q}}$-structure on ${\mathfrak{g} = \mathfrak{u}^-\oplus \mathfrak{s}\oplus\mathfrak{a} \oplus\mathfrak{u}}$, it remains to put a ${\mathbb{Q}}$-structure on ${\mathfrak{a}}$. In general this is not difficult: for instance, we can take

$\displaystyle \mathfrak{a}_{\mathbb{Q}} = \left\{ \left( \begin{array}{cc} \mu I & 0 \\ 0 & -\mu I\end{array}\right):\mu\in\mathbb{Q} \right\}$

in the context of the example ${G=SL(2p,\mathbb{R})}$ and ${U = \left\{\left(\begin{array}{cc} I & B \\ 0 & I\end{array}\right)\in G: B\in M(p,\mathbb{R})\right\}}$.

Once we understood the constraints on a ${\mathbb{Q}}$-form of ${G}$ containing ${\Gamma}$, we can work backwards and set

$\displaystyle \mathfrak{g}_\mathbb{Q} = \mathfrak{u}^-_\mathbb{Q}\oplus \mathfrak{s}_\mathbb{Q}\oplus\mathfrak{a}_\mathbb{Q} \oplus\mathfrak{u}_\mathbb{Q}$

where ${\mathfrak{u}_\mathbb{Q}}$, ${\mathfrak{u}^-_\mathbb{Q}}$, ${\mathfrak{s}_\mathbb{Q}}$ and ${\mathfrak{a}_\mathbb{Q}}$ are the ${\mathbb{Q}}$-structures from the previous paragraphs.

At this point, the proof of Theorem 4 is complete once we show the following facts:

• ${\mathfrak{g}_{\mathbb{Q}}}$ doesn’t depend on the choices (of ${U^-}$, etc.);
• ${\textrm{Ad}(\gamma) \mathfrak{g}_{\mathbb{Q}}\subset \mathfrak{g}_{\mathbb{Q}}}$ for all ${\gamma\in\Gamma}$;
• ${\mathfrak{g}_{\mathbb{Q}}}$ is a Lie algebra.

The proof of these statements is described in the proof of Proposition 4.11 of Benoist–Miquel paper. Closing this post, let us just make some comments on the independence of ${\mathfrak{g}_{\mathbb{Q}}}$ on the choices. For this sake, suppose that ${U'}$ is another choice of horospherical subgroup. Denote by ${\mathfrak{u}'}$ its Lie algebra, and let ${P'=N_G(U')}$ be the associated parabolic subgroup. Our task is to verify that the Lie algebra ${\mathfrak{p}' = \textrm{Lie}(P')}$ is defined over ${\mathbb{Q}}$. In this direction, the basic strategy is to reduce to the case when ${\mathfrak{u}'}$ is opposite to ${\mathfrak{u}}$ and ${\mathfrak{u}^-}$ in order to get ${\mathfrak{p}' = (\mathfrak{p}'\cap\mathfrak{p})\oplus(\mathfrak{p}'\cap\mathfrak{p}^-)}$. Finally, during the implementation of this strategy, one relies on the following properties discussed in Lemma 4.8 of Benoist–Miquel article of the action of the unimodular normalizers ${Q=\{g\in P: \textrm{det}_{\mathfrak{u}}\textrm{Ad}(g) = 1\}}$ of horospherical subgroups ${U\subset P}$ on the space ${X=G/\Gamma}$ with basepoint ${x_0=\Gamma/\Gamma}$:

• If ${Ux_0}$ is compact, then ${Qx_0}$ is closed;
• If ${Qx_0}$ and ${Q^-x_0}$ are closed, then ${Sx_0 = (Q\cap Q^-) x_0}$ is closed;
• If ${Ux_0}$ is compact and ${S x_0}$ is closed, then ${(\Gamma\cap S)(\Gamma\cap U)}$ has finite index in ${\Gamma\cap SU = \Gamma\cap Q}$.

In any event, this completes our discussion of Theorem 4. In particular, we gave a (sketch of) proof of Theorem 1 when ${U}$ is commutative and reflexive, and ${S}$ is non-compact (cf. Remark 4).

Next time, we will establish Theorem 1 in the remaining cases of ${U}$ and ${S}$.