Posted by: matheuscmss | September 14, 2018

## Benoist’s minicourse “Arithmeticity of discrete groups”. Lecture IV: Other examples

In this last post of this series, we want to complete the discussion of Oh–Benoist–Miquel theorem by giving a sketch of its proof in the cases not covered in previous posts.

More precisely, let us remind that Oh–Benoist–Miquel theorem (answering a conjecture of Margulis) asserts that:

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 ${G_{\mathbb{Z}}}$.

Moreover, we remind that the proof of this result was worked out in the previous two posts of this series for ${G = SL(4,\mathbb{R})}$ and ${U = \left\{\left(\begin{array}{cccc} 1 & 0 & \ast & \ast \\ 0 & 1 & \ast & \ast \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)\right\}}$. Furthermore, we observed en passant that these arguments can be generalized without too much effort to yield a proof of Theorem 1 when

• ${U}$ is commutative;
• ${U}$ is reflexive (i.e., ${U}$ is conjugate to an opposite horospherical subgroup ${U^-}$);
• ${S}$ is not compact (where ${P=N_G(U)}$, ${P^-=N_G(U^-)}$, ${L=P\cap P^-}$ and ${S=[L,L]}$).

Today, we will divide our discussion below into five sections discussing prototypical examples covering all possible remaining cases for ${U}$ and ${S}$.

Remark 1 The fact that Theorem 1 holds for the examples in Sections 12 and 3 below is originally due to Oh. Similarly, the example in Section 4 was originally treated by Selberg. Finally, the original proof of Theorem 1 for the example in Section 5 is due to Benoist–Oh. Nevertheless, expect for Section 3, the arguments discussed below are some particular examples illustrating the general strategy of Benoist–Miquel and, hence, they provide some proofs which are different from the original ones.

1. ${U}$ is not reflexive

The prototype example of this case is ${G=SL(3,\mathbb{R})}$ and ${U = \left\{\left(\begin{array}{ccc} 1 & \ast & \ast \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)\right\}}$.

The corresponding parabolic subgroup ${P=N_G(U)}$ is the stabilizer of the line ${\mathbb{R}\, e_1}$:

$\displaystyle P=\left\{\left(\begin{array}{ccc} \ast & \ast & \ast \\ 0 & \ast & \ast \\ 0 & \ast & \ast \end{array}\right)\right\} = \{T\in G: T(e_1)\in\mathbb{R}e_1\}.$

Equivalently, ${P}$ is the stabilizer of the flag ${\{0\}\subset\mathbb{R}e_1\subset \mathbb{R}^3}$. Therefore, ${U}$ is not reflexive because its opposite is the stabilizer of a plane.

Since ${\Gamma}$ is Zariski-dense in ${G}$, we can find ${g, h\in\Gamma}$ such that ${\{e_1, g(e_1), h(e_1)\}}$ is a basis of ${\mathbb{R}^3}$. Hence, there is no loss in generality in assuming that ${g(e_1)=e_2}$ and ${h(e_1)=e_3}$. In this setting,

$\displaystyle U'=gUg^{-1} = \left\{\left(\begin{array}{ccc} 1 & 0 & 0 \\ \ast & 1 & \ast \\ 0 & 0 & 1 \end{array}\right)\right\}, \quad U''=hUh^{-1} = \left\{\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \ast & \ast & 1 \end{array}\right)\right\}.$

Also, we know that ${U'/(U'\cap\Gamma)}$ and ${U''/(U''\cap\Gamma)}$ are compact. Moreover, ${\Delta = \langle\Gamma\cap U', \Gamma\cap U''\rangle}$ is a discrete and Zariski dense subgroup of the semi-direct product

$\displaystyle H=\langle U, U'\rangle = \left\{ \left(\begin{array}{ccc} a & b & v_1 \\ c & d & v_2 \\ 0 & 0 & 1 \end{array}\right): \left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\in S, \left(\begin{array}{c} v_1 \\ v_2 \end{array}\right)\in V \right\}$

of ${S=SL(2,\mathbb{R})}$ and ${V=\mathbb{R}^2}$.

In this context, a key fact is the following result of Auslander (compare with Proposition 4.17 in Benoist–Miquel paper):

Theorem 2 (Auslander) Let ${H}$ be an algebraic subgroup obtained from a semi-direct product of ${S}$ semisimple and ${V}$ solvable, and denote by ${p:H\rightarrow S}$ the natural projection. If ${\Delta\subset H}$ is discrete and Zariski dense, then ${p(\Delta)}$ is also discrete and Zariski dense in ${S}$.

The information about the discreteness of the projection ${p(\Delta)}$ in the previous statement is extremely precious for our purposes. Indeed, Auslander theorem implies that the projections ${p(\Gamma\cap U)}$ and ${p(\Gamma\cap U')}$ are discrete. Using these facts, one checks that

$\displaystyle \Gamma\cap\left\{\left(\begin{array}{ccc} 1 & 0 & \ast \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)\right\} \neq \{\textrm{Id}\} \quad \textrm{ and } \quad \Gamma\cap\left\{\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & \ast \\ 0 & 0 & 1 \end{array}\right)\right\} \neq \{\textrm{Id}\}$

By repeating this argument with ${(U, U'')}$ and ${(U', U'')}$ in the place of ${(U,U')}$, one can “fill all non-diagonal entries”, that is, one essentially gets that ${\Gamma}$ contains finite-index subgroups of

$\displaystyle \left\{ \left(\begin{array}{ccc} 1 & \ast & \ast \\ 0 & 1 & \ast \\ 0 & 0 & 1 \end{array}\right)\in SL(3,\mathbb{Z})\right\} \textrm{ and } \left\{ \left(\begin{array}{ccc} 1 & 0 & 0 \\ \ast & 1 & 0 \\ \ast & \ast & 1 \end{array}\right)\in SL(3,\mathbb{Z})\right\},$

so that Raghunathan–Venkataramana–Oh theorem (stated in the previous post of this series) guarantees that ${\Gamma}$ is commensurable to ${SL(3,\mathbb{Z})}$.

This completes our sketch of proof of Theorem 1 for our prototype of non-reflexive subgroup ${U}$ above.

2. ${U}$ is Heisenberg and ${S}$ is not compact

Heisenberg horospherical subgroup ${U}$ is a ${2}$-step nilpotent whose associated parabolic group ${P=N_G(U)}$ acts by similarities (of some Euclidean norm) on the center of the Lie algebra ${\mathfrak{u}}$ of ${U}$.

A prototypical example of ${U}$ Heisenberg and ${S}$ non-compact is ${G=SL(4,\mathbb{R})}$ and

$\displaystyle U=\left\{ \left( \begin{array}{cccc} 1 & \ast & \ast & \ast \\ 0 & 1 & 0 & \ast \\ 0 & 0 & 1 & \ast \\ 0 & 0 & 0 & 1 \end{array} \right) \right\}$

As it turns out, any Heisenberg ${U}$ is reflexive. Thus, we have that ${U^-=\gamma_0 U \gamma_0^{-1}}$ is opposite to ${U}$ for some adequate choice ${\gamma_0\in\Gamma}$.

In particular, it is tempting to mimmic the arguments from the second and third posts of this series, namely, one introduces the lattices

$\displaystyle \Lambda = \log(\Gamma\cap U)\in X_{\mathfrak{u}}, \quad \Lambda^- = \log(\Gamma\cap U^-)\in X_{\mathfrak{u}^-},$

so that the arithmeticity of ${\Gamma}$ follows from the closedness of the ${\textrm{Ad}_L}$-orbit of ${(\Lambda, \Lambda^-)}$ in ${X_{\mathfrak{u}}\times X_{\mathfrak{u}^-}}$ when ${S}$ is not compact; moreover, the closedness of ${\textrm{Ad}_L(\Lambda, \Lambda^-)}$ is basically a consequence of the closedness and discreteness of ${F(\Lambda)}$ in ${\mathbb{R}}$ for an appropriate choice of polynomial function ${F}$.

In the case of ${U}$ commutative, we took ${F(X) = \Phi(\exp(X)\gamma_0)}$, where ${\Phi(g) = \textrm{det}_{\mathfrak{u}}(M(g))}$, ${M(g)=\pi\circ \textrm{Ad}(g)\circ \pi}$ and ${\pi:\mathfrak{g}\rightarrow \mathfrak{u}}$ was the natural projection with respect to the decomposition ${\mathfrak{g} = \mathfrak{u}\oplus \mathfrak{l} \oplus \mathfrak{u}^-}$.

As it turns out, the case of ${U}$ Heisenberg can be dealt with by slightly modifying the construction in the previous paragraph. More precisely, one considers a natural graduation

$\displaystyle \mathfrak{g} = \underbrace{\mathfrak{g}_{-2}\oplus \mathfrak{g}_{-1}}_{=\mathfrak{u}^-} \oplus \underbrace{\mathfrak{g}_0}_{=\mathfrak{l}} \oplus \underbrace{\mathfrak{g}_1 \oplus \overbrace{\mathfrak{g}_2}^{=\textrm{ center of }\mathfrak{u}}}_{=\mathfrak{u}}$

and one sets ${F(X)=\Phi(\exp(X)\gamma_0)}$, ${\Phi(g) = \textrm{det}(M(g))}$, ${M(g)=\pi\circ\textrm{Ad}(g)\circ \pi}$, and ${\pi}$ is the natural projection ${\pi:\mathfrak{g}\rightarrow \mathfrak{g}_2}$. In our prototypical example, the polynomial function ${F(X) = \Phi(\exp(X)\gamma_0)}$ is very explicit:

$\displaystyle F(\left(\begin{array}{cccc} 0 & x_1 & x_2 & z \\ 0 & 0 & 0 & y_1 \\ 0 & 0 & 0 & y_2 \\ 0 & 0 & 0 & 0 \end{array}\right)) = (x_1 y_1 + x_2 y_2)^2 - z^2$

This completes our sketch of proof of Theorem 1 when ${U}$ is Heisenberg and ${S}$ is not compact.

3. ${U}$ is not commutative and ${U}$ is not Heisenberg

Our prototype of non-commutative and non-Heisenberg ${U}$ is the subgroup

$\displaystyle U=\left\{ \left(\begin{array}{cccc} 1 & \ast & \ast & \ast \\ 0 & 1 & \ast & \ast \\ 0 & 0 & 1 & \ast \\ 0 & 0 & 0 & 1 \end{array}\right) \right\}$

of ${G=SL(4,\mathbb{R})}$.

In this context, we will explore some well-known results from the theory of lattices in nilpotent groups to reduce our task to the case of ${U}$ commutative and reflexive.

More concretely, the properties of nilpotent groups together with our hypothesis that ${\Delta=\Gamma\cap U}$ is a lattice in ${U}$ allow to conclude that ${\Delta':=[\Delta,\Delta]}$ is a lattice in

$\displaystyle U':=[U, U] = \left\{ \left( \begin{array}{cccc} 1 & 0 & \ast & \ast \\ 0 & 1 & 0 & \ast \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \right\},$

and, consequently, the centralizer ${\Delta_0}$ of ${\Delta'}$ in ${\Delta}$ is a lattice in the centralizer

$\displaystyle U_0= \left\{ \left( \begin{array}{cccc} 1 & 0 & \ast & \ast \\ 0 & 1 & \ast & \ast \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \right\}$

of ${U'}$ in ${U}$. Therefore, we reduced matters to the case of ${U_0}$ commutative and reflexive which was discussed in the previous two posts of this series.

In particular, our sketch of proof of Theorem 1 when ${U}$ is non-commutative and non-Heisenberg is complete.

4. ${U}$ commutative and ${S}$ is compact

The basic example of ${U}$ commutative and ${S}$ compact is ${G=SL(2,\mathbb{R})\times SL(2,\mathbb{R})}$ and

$\displaystyle U= \left\{ (\left(\begin{array}{cc} 1 & \ast \\ 0 & 1 \end{array}\right), \left(\begin{array}{cc} 1 & \ast \\ 0 & 1 \end{array}\right)) \right\}.$

In this setting, we consider

$\displaystyle L = \left\{ (\left(\begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_1^{-1} \end{array}\right), \left(\begin{array}{cc} \lambda_2 & 0 \\ 0 & \lambda_2^{-1} \end{array}\right)): \lambda_1, \lambda_2\in \mathbb{R}^*\right\} = P\cap P^-$

the common Levi subgroup of ${P=N_G(U)}$ and the parabolic subgroup ${P^-}$ normalizing an opposite of ${U}$, and the “unimodular Levi subgroup”

$\displaystyle \begin{array}{rcl} L_0 &=& \{\ell\in L: \textrm{det}_{\mathfrak{u}} \textrm{Ad}(\ell)=1\} \\ &=& \left\{ (\left(\begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_1^{-1} \end{array}\right), \left(\begin{array}{cc} \lambda_2 & 0 \\ 0 & \lambda_2^{-1} \end{array}\right)): \lambda_1 \lambda_2=\pm1\right\} \\ &\simeq& \mathbb{R}^*\times\{\pm1\} \end{array}$

The discussion in the second post of this series ensures that the ${\textrm{Ad}_{L_0}}$-orbit of ${(\Lambda, \Lambda^-)}$ is closed in ${X_{\mathfrak{u}}\times X_{\mathfrak{u}^-}}$.

We affirm that ${\textrm{Ad}_{L_0}(\Lambda, \Lambda^-)}$ is compact. Indeed, this fact can be proved via Mahler’s compactness criterion: more concretely, recall from the second post of this series that the proof of the closedness of ${\textrm{Ad}_L(\Lambda, \Lambda')}$ produced a polynomial ${F}$ on ${\mathfrak{u}}$ which is ${\textrm{Ad}_{L_0}}$-invariant and whose values ${F(\Lambda)}$ on ${\Lambda}$ form a closed and discrete subset of ${\mathbb{R}}$; in our prototypical example, a direct computation shows that

$\displaystyle F(\left(\begin{array}{cc} 0 & x_1 \\ 0 & 0 \end{array}\right), \left(\begin{array}{cc} 0 & x_2 \\ 0 & 0 \end{array}\right)) = x_1^2 x_2^2;$

in particular, ${\inf\limits_{\ell\in L_0} \inf\limits_{X\in\Lambda\setminus\{0\}} \|\textrm{Ad}(\ell) X\|^4 > \inf\limits_{\ell\in L_0} \inf\limits_{X\in\Lambda\setminus\{0\}} F(\textrm{Ad}(\ell) X)}$; therefore, the ${\textrm{Ad}_{L_0}}$-invariance of ${F}$ together with the closedness and discreteness of ${F(\Lambda)}$ imply that

$\displaystyle \begin{array}{rcl} \inf\limits_{\ell\in L_0} \inf\limits_{X\in\Lambda\setminus\{0\}} \|\textrm{Ad}(\ell) X\|^4 &>& \inf\limits_{\ell\in L_0} \inf\limits_{X\in\Lambda\setminus\{0\}} F(\textrm{Ad}(\ell) X) \\ &=& \inf\limits_{X\in\Lambda\setminus\{0\}} F(X) =\min\limits_{X\in\Lambda\setminus\{0\}} F(X); \end{array}$

since ${\Lambda}$ is irreducible, ${\min\limits_{X\in\Lambda\setminus\{0\}} F(X)>0}$, and, a fortiori, there are no arbitrarily short non-trivial vectors in the closed family of lattices ${\textrm{Ad}_{L_0}(\Lambda)}$; hence, we can apply Mahler’s compactness criterion to complete the proof of our affirmation.

At this point, we observe that ${L_0\simeq \mathbb{R}^*\times \{\pm1\}}$ is not compact (because ${\textrm{rank}_{\mathbb{R}}(G)\geq 2}$), so that the compactness of ${\textrm{Ad}_{L_0}(\Lambda,\Lambda^-)}$ means that the stabilizer of this orbit is infinite. Consequently, ${\Gamma\cap L}$ is infinite, and a quick inspection of the previous post reveals that this is precisely the information needed to apply Margulis’ construction of ${\mathbb{Q}}$-forms and Raghunathan–Venkataramana theorem in order to derive the arithmeticity of ${\Gamma}$. This completes our sketch of proof of Theorem 1 when ${U}$ is commutative and ${S}$ is compact (and the reader is invited to consult Section 4.6 of Benoist–Miquel paper for more details).

5. ${U}$ is Heisenberg and ${S}$ compact

Closing this series of post, let us discuss the remaining case of ${U}$ Heisenberg and ${S}$ compact. A concrete example of this situation is ${G=SL(3,\mathbb{R})}$ and

$\displaystyle U=\left\{ \left(\begin{array}{ccc} 1 & \ast & \ast \\ 0 & 1 & \ast \\ 0 & 0 & 1 \end{array}\right) \right\}.$

In this context, ${L=\left\{ \left(\begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right): abc=1 \right\}}$ and an unimodular Levi subgroup is

$\displaystyle L_0 = \left\{ \left(\begin{array}{ccc} a & 0 & 0 \\ 0 & 1/a^2 & 0 \\ 0 & 0 & a \end{array}\right): a\in\mathbb{R}^* \right\}$

Once again, let us recall that we know that ${\textrm{Ad}_{L_0}(\Lambda, \Lambda^-)}$ is closed, where

$\displaystyle \Lambda= \left\{ u=u(x,y,z)=\left(\begin{array}{ccc} 0 & x & z \\ 0 & 0 & y \\ 0 & 0 & 0 \end{array}\right)\in \mathfrak{u} \right\}$

We affirm that there is no loss of generality in assuming that ${x\neq 0}$ and ${y\neq 0}$ for all ${u=u(x,y,z)\in \Lambda\setminus (\Lambda\cap [U, U])}$. Indeed, if this is not the case (say ${y=0}$ for some ${u\in\Lambda\setminus (\Lambda\cap [U,U])}$), then we are back to the setting of Section 1 above (of the horospherical subgroup ${\left\{\left(\begin{array}{ccc} 1 & \ast & \ast \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)\right\}}$).

Here, we can derive the arithmeticity of ${\Gamma}$ along the same lines in Section 4 above (where it sufficed to study an appropriate polynomial ${F(\left(\begin{array}{cc} 0 & x_1 \\ 0 & 0 \end{array}\right), \left(\begin{array}{cc} 0 & x_2 \\ 0 & 0 \end{array}\right))=x_1^2 x_2^2}$ to employ Mahler’s compactness criterion). More precisely, one uses the fact that ${x\neq0}$ and ${y\neq0}$ for all ${u\in\Lambda\setminus (\Lambda\cap [U,U])}$ to prove that ${\textrm{Ad}_{L_0}(\Lambda, \Lambda')}$ is compact, so that ${\Gamma\cap L}$ is infinite and, thus, by Margulis’ construction of ${\mathbb{Q}}$-forms and Raghunathan–Venkataramana theorem, ${\Gamma}$ is arithmetic.

This site uses Akismet to reduce spam. Learn how your comment data is processed.