An adelic arithmeticity theorem for lattices in products

Abstract

In this paper we prove that, under mild assumptions, a lattice in a product of semi-simple Lie group and a totally disconnected locally compact group is, in a certain sense, arithmetic. We do not assume the lattice to be finitely generated or the ambient group to be compactly generated.

Appendix A. A theorem of Borel revisited

In this appendix we generalise Borel’s Theorem 3(b) from [3]. Our proof is almost identical to Borel’s original proof and we have no doubt the sophisticated reader is well aware of this generalisation. Nevertheless, since the result had not been published elsewhere in the form that we need, we decided to add the short proof for the reader’s convenience.

Theorem A.1

Let K be a field of characteristic 0. Let ${\mathbf {G}}$ be a connected affine K-algebraic group and ${\mathbf {Z}}$ its center. Let $H<{\mathbf {G}}(K)$ be a Zariski dense subgroup. Let A be a K-algebra and let C be the commensurator of H in ${\mathbf {G}}(A)$. Then the image of C in ${\mathbf {G}}/{\mathbf {Z}}(A)$ is contained in ${\mathbf {G}}/{\mathbf {Z}}(K)$.

Before proving the theorem, let us recall some basic facts about the functor of points of a K-algebraic group. For a K-algebraic group ${\mathbf {G}}$ and a K-algebra A we set ${\mathbf {G}}(A)=\mathrm {Hom}_{K\text{-alg }}(K[{\mathbf {G}}],A)$, where $K[{\mathbf {G}}]$ denotes the Hopf algebra of K-regular functions on ${\mathbf {G}}$. Using the coproduct $\Delta : K[{\mathbf {G}}]\rightarrow K[{\mathbf {G}}]\otimes _K K[{\mathbf {G}}]$ we get a product structure on ${\mathbf {G}}(A)$, by sending $x,y\in {\mathbf {G}}(A)$ to the map given by the following composition

$$\begin{aligned} K[{\mathbf {G}}] \overset{\Delta }{\xrightarrow {}} K[{\mathbf {G}}]\otimes _K K[{\mathbf {G}}] \overset{x\otimes y}{\xrightarrow {}} A\otimes _K A \overset{m}{\xrightarrow {}} A. \end{aligned}$$

Further, ${\mathbf {G}}(A)$ has an identity, given by the augmentation $f\mapsto f(e)\in K<A$, and the inverse of $x\in {\mathbf {G}}(A)$ is given by $x\circ S$, where $S:K[{\mathbf {G}}]\rightarrow K[{\mathbf {G}}]$ is the antipode, $Sf(g)=f(g^{-1})$. Thus ${\mathbf {G}}(A)$ is a group. The association $A\leadsto {\mathbf {G}}(A)$ is called the functor of points associated with ${\mathbf {G}}$, see for example Waterhouse’s book [17] for an account on algebraic group theory based on this point of view.

We set $A[{\mathbf {G}}]:=A\otimes _K K[{\mathbf {G}}]$. Consider the conjugation action of ${\mathbf {G}}$ on itself. This is a K-morphic action which gives rise to a K-morphic action of ${\mathbf {G}}/{\mathbf {Z}}$ on ${\mathbf {G}}$. Consider the action map $a:{\mathbf {G}}/{\mathbf {Z}}\times {\mathbf {G}}\rightarrow {\mathbf {G}}$. For every $g\in {\mathbf {G}}/{\mathbf {Z}}(A)$ the map

$$\begin{aligned} K[{\mathbf {G}}] \overset{a^*}{\xrightarrow {}} K[{\mathbf {G}}/{\mathbf {Z}}]\otimes _K K[{\mathbf {G}}] \overset{g\otimes {\text {id}}}{\xrightarrow {}} A\otimes _K K[{\mathbf {G}}] = A[{\mathbf {G}}] \end{aligned}$$

extends uniquely to a A-algebra automorphism $A[{\mathbf {G}}]\rightarrow A[{\mathbf {G}}]$. By this we define an action of ${\mathbf {G}}/\mathbf Z(A)$ on $A[{\mathbf {G}}]$ by A-algebra automorphisms.

Lemma A.2

Let K be a field of characteristic 0. Let ${\mathbf {G}}$ be an affine K-algebraic group and ${\mathbf {Z}}$ its center. Let A be a K-algebra. Consider the conjugation representation of ${\mathbf {G}}/{\mathbf {Z}}(A)$ on $A[{\mathbf {G}}]$. Considering $K[{\mathbf {G}}]$ as a subspace of $A[{\mathbf {G}}]$ and ${\mathbf {G}}/{\mathbf {Z}}(K)$ as a subgroup of ${\mathbf {G}}/\mathbf Z(A)$, we have

$$\begin{aligned} {\mathbf {G}}/{\mathbf {Z}}(K)=\{g\in {\mathbf {G}}/\mathbf Z(A)~|~g(K[{\mathbf {G}}])\subset K[{\mathbf {G}}]\}. \end{aligned}$$

Proof

Consider $K[{\mathbf {G}}/{\mathbf {Z}}]$ as a Hopf algebra and consider the comodule structure on $K[{\mathbf {G}}]$ associated with the conjugation representation. By [17, 3.3], $K[{\mathbf {G}}]$ is a direct union of finite dimensional subcomodules. We let $V<K[{\mathbf {G}}]$ be a finite dimensional subcomodule which generates $K[{\mathbf {G}}]$ as a K-algebra ($K[{\mathbf {G}}]$ is a finitely generated K-algebra). By the correspondence between subcomodules and subrepresentations, [17, 3.2], V is a representation of ${\mathbf {G}}/{\mathbf {Z}}$, in the sense that for every K-algebra B, $B\otimes _K V$ is a ${\mathbf {G}}/{\mathbf {Z}}(B)$-invariant subspace of $B[{\mathbf {G}}]$, and we obtain a natural family of group homomorphisms ${\mathbf {G}}/{\mathbf {Z}}(B)\rightarrow {\text {Aut}}_{B}(B\otimes _K V)$. Considering the action of ${\mathbf {G}}/{\mathbf {Z}}(B)$ on $B[{\mathbf {G}}]$, it follows that the left hand side is included in the right hand side in the following equation:

$$\begin{aligned} \{g\in {\mathbf {G}}/{\mathbf {Z}}(B)~|~g(K[{\mathbf {G}}])\subset K[{\mathbf {G}}]\} = \{g\in {\mathbf {G}}/{\mathbf {Z}}(B)~|~gV\subset V\}, \end{aligned}$$

(A.1)

the other inclusion holds as the K-algebra generated in $B[{\mathbf {G}}]$ by V is $K[{\mathbf {G}}]$.

We now take $B={\bar{K}}$, an algebraic closure of K. By definition of ${\mathbf {Z}}$ [4, 1.7], ${\mathbf {G}}/\mathbf Z({\bar{K}})$ acts faithfully on ${\mathbf {G}}({\bar{K}})$. It follows that its action on ${\bar{K}}[{\mathbf {G}}]$ is faithful as well, and by the fact that ${\bar{K}}\otimes _K V$ generates ${\bar{K}}[{\mathbf {G}}]$ as a ${\bar{K}}$-algebra, we get that ${\mathbf {G}}/{\mathbf {Z}}({\bar{K}})$ acts faithfully on ${\bar{K}}\otimes _K V$. Thus the morphism ${\mathbf {G}}/{\mathbf {Z}}({\bar{K}})\rightarrow {{\mathbf {G}}{\mathbf {L}}}({\bar{K}}\otimes _K V)$ is injective. By [4, Cor 1.4(a)] its image is a Zariski closed K-subgroup of ${{\mathbf {G}}{\mathbf {L}}}({\bar{K}}\otimes _K V)$, and by [4, Proposition 6.4] this morphism is a K-isomorphism onto its image since it is separable by the characteristic 0 assumption on K.

As K-vector spaces, $V\cong K^n$. Thus we may identify ${\mathbf {G}}/{\mathbf {Z}}$ with a K-algebraic subgroup ${\mathbf {H}}<{{\mathbf {G}}{\mathbf {L}}}_n$. As subgroups of ${{\mathbf {G}}{\mathbf {L}}}_n(A)$ we have ${\mathbf {H}}(K)={\mathbf {H}}(A)\cap {{\mathbf {G}}{\mathbf {L}}}_n(K)$. We conclude the first equality in

$$\begin{aligned} {\mathbf {H}}(K)=\{g\in {\mathbf {H}}(A)~|~g(K^n)=K^n\}=\{g\in {\mathbf {H}}(A)~|~g(K^n)\subset K^n\}. \end{aligned}$$

The second equality is a triviality since every g is regular. We thus obtain

$$\begin{aligned} {\mathbf {G}}/{\mathbf {Z}}(K)=\{g\in {\mathbf {G}}/{\mathbf {Z}}(A)~|~gV\subset V\}. \end{aligned}$$

The proof is now complete by Equation (A.1). $\square$

Proof of Theorem A.1

Let $g\in C$ be an element. We show that its image in ${\mathbf {G}}/{\mathbf {Z}}(A)$ – denoted also by g – lies in ${\mathbf {G}}/{\mathbf {Z}}(K)$.

We consider the conjugation action of ${\mathbf {G}}(A)$ on $A[{\mathbf {G}}]$ and claim that $gK[{\mathbf {G}}]\subset K[{\mathbf {G}}]$. By Lemma A.2 this suffices. Let $x\in K[{\mathbf {G}}]$. We write

$$\begin{aligned} gx=x_0+\alpha _1x_1+\cdots +\alpha _nx_n, \quad \alpha _i\in A,~x_i\in K[{\mathbf {G}}], \end{aligned}$$

where $1,\alpha _1,\ldots ,\alpha _n$ are independent over K and $x_i\ne 0$. If $n=0$ then indeed $gx=x_0\in K[{\mathbf {G}}]$.

We assume $n\ge 1$ and argue by contradiction. Note that $H\cap gHg^{-1}<H$ is of finite index, thus $H\cap gHg^{-1}$ is Zariski dense in ${\mathbf {G}}(K)$ since ${\mathbf {G}}$ is connected. Thus we can find $h\in H\cap gHg^{-1}<{\mathbf {G}}(K)=\mathrm {Hom}_{K\text{-alg }}(K[{\mathbf {G}}],K)$ such that $h(x_1)\ne 0$. From

$$\begin{aligned} h(x_0)+\alpha _1h(x_1)+\cdots +\alpha _nh(x_n)=h(gx)=(g^{-1}hg)(x) \in K \end{aligned}$$

we get a non-trivial linear combination of $1,\alpha _1,\ldots ,\alpha _n$ over K. This gives the desired contradiction. $\square$

Remark A.3

Neither Theorem A.1 nor Lemma A.2 are correct as stated if K has positive characteristic. Take for example K to be the algebraic closure of ${\mathbb {F}}_2$, ${\mathbf {G}}={{{\mathbf {S}}}{{\mathbf {L}}}}_2$, and $A=K[t]/(t^2)$. Then ${\mathbf {G}}(K)={{{\mathbf {S}}}{{\mathbf {L}}}}_2(K)$ has trivial center ${\mathbf {Z}}$, thus ${\mathbf {G}}/{\mathbf {Z}}={\mathbf {G}}$. However, ${\mathbf {G}}/{\mathbf {Z}}(A)= {\mathbf {G}}(A)\cong {{{\mathbf {S}}}{{\mathbf {L}}}}_2(K)\ltimes {{\mathfrak {s}}}{{\mathfrak {l}}}_2(K)$ has a non-trivial center L corresponding to the scalar matrices in ${{\mathfrak {s}}}{{\mathfrak {l}}}_2(K)$. Then L is not contained in ${\mathbf {G}}/{\mathbf {Z}}(K)$, but

$$\begin{aligned} L<\{g\in {\mathbf {G}}/{\mathbf {Z}}(A)~|~g(K[{\mathbf {G}}])<K[{\mathbf {G}}]\}, \end{aligned}$$

and for any H in ${\mathbf {G}}(K)$, L commensurates H in ${\mathbf {G}}/{\mathbf {Z}}(A)$.

Correct statements are obtained by replacing the smooth subgroup ${\mathbf {Z}}$ with a scheme theoretically defined center ${\mathcal {Z}}$. For example, for ${\mathbf {G}}={{{\mathbf {S}}}{{\mathbf {L}}}}_2$, regardless of the field of definition, ${\mathbf {G}}/{\mathcal {Z}}\cong {\mathbf {PGL}}_2$. In characteristic 0 we always have ${\mathcal {Z}}={\mathbf {Z}}$. The proof of the corrected statement of the theorem follows formally from the corrected lemma. We will not give the proof of the corrected lemma but merely remark that it is verbatim the proof above, upon replacing the notion of “faithfulness of an action" by a scheme theoretical analogue. Moreover, there is no need to pass to an algebraic closure anymore.

Working with the scheme theoretical center has advantages even if one is concerned only with the classical setting. For example, if K is reduced then ${\mathbf {G}}/{\mathcal {Z}}(K)={\mathbf {G}}/\mathbf Z(K)$. To see this one observes that the algebra representing ${\mathbf {G}}/{\mathcal {Z}}$, modulo its nil-radical, equals the algebra representing ${\mathbf {G}}/{\mathbf {Z}}$.

The following application of Theorem A.1 is needed in Sect. 3.

Theorem A.4

We retain the setup in Sect. 1.2. Let ${\mathbf {H}}$ be a connected adjoint semi-simple K-algebraic group. Assume in addition that ${\mathbf {H}}({\mathcal {O}}[S])$ is Zariski dense in ${\mathbf {H}}$. Then the commensurator of ${\mathbf {H}}({\mathcal {O}}[S])$ in ${\mathbf {H}}_S$ (via the diagonal embedding) is exactly ${\mathbf {H}}(K)$.

Remark A.5

The subgroup ${\mathbf {H}}({\mathcal {O}}[S])<{\mathbf {H}}(K)$ depends on a choice of embedding ${\mathbf {H}}<{{\mathbf {G}}{\mathbf {L}}}_n$ but not its commensurability class. Since ${\mathbf {H}}$ is connected its Zariski-density is unambiguous.

Proof

Consider the locally compact totally disconnected group given by the restricted product

$$\begin{aligned} {\prod }^{'}_{\nu \in {\mathcal {V}}-{\mathcal {V}}^\infty -S} {\mathbf {H}}(K_\nu ), \end{aligned}$$

which is taken with respect to the compact open subgroups ${\mathbf {H}}({\mathcal {O}}_\nu )<{\mathbf {H}}(K_\nu )$. Then

$$\begin{aligned} {\prod }_{\nu \in {\mathcal {V}}-{\mathcal {V}}^\infty -S} {\mathbf {H}}({\mathcal {O}}_\nu ) \end{aligned}$$

is a compact open subgroup, hence it is commensurated. Its preimage under the diagonal embedding

$$\begin{aligned} {\mathbf {H}}(K)\rightarrow {\prod }^{'}_{\nu \in {\mathcal {V}}-{\mathcal {V}}^\infty -S} {\mathbf {H}}(K_\nu ) \end{aligned}$$

is ${\mathbf {H}}({\mathcal {O}}[S])$. By Lemma 5.2, ${\mathbf {H}}({\mathcal {O}}[S])$ is commensurated by ${\mathbf {H}}(K)$.

We need to show that the commensurator group of ${\mathbf {H}}({\mathcal {O}}[S])$ inside ${\mathbf {H}}_S$ is contained in the diagonal image of ${\mathbf {H}}(K)$. Let A be the K-algebra of S-adeles, that is, the restricted product $\prod '_{\nu \in S} K_\nu$ taken with respect to the subgroups ${\mathcal {O}}_\nu <K_\nu$. In [12, pp.  249–250] the identification ${\mathbf {H}}(A)\cong {\mathbf {H}}_S$ is explained. Under this identification, ${\mathbf {H}}(K)$ is identified with its diagonal image. Now apply Theorem A.1 and use the assumption that ${\mathbf {H}}$ is adjoint, hence has a trivial center. $\square$

For the special case where S is a set of archimedean places we get the following.

Theorem A.6

We retain the setup in Sect. 1.2. Let $S\subset {\mathcal {V}}^{\infty }$ be a set of archimedean places. Let ${\mathbf {H}}$ be a connected adjoint semi-simple K-algebraic group. Assume that there exists $\nu \in {\mathcal {V}}^\infty$ such that the Lie group ${\mathbf {H}}(K_\nu )$ is not compact. Then the commensurator of ${\mathbf {H}}({\mathcal {O}})$ in $\prod _{\nu \in S}{\mathbf {H}}(K_\nu )$ is exactly ${\mathbf {H}}(K)$.

Proof

This follows from Theorem A.4, once we show that ${\mathbf {H}}({\mathcal {O}})$ is Zariski dense in ${\mathbf {H}}$. By [3, Theorem 3(a)] ${\mathbf {H}}({\mathcal {O}})$ is Zariski dense if and only if $R_{K/{{\mathbb {Q}}}}{\mathbf {H}}({{\mathbb {R}}})$ is non-compact where $R_{K/{{\mathbb {Q}}}}$ denotes the restriction of scalars from K to ${{\mathbb {Q}}}$. By applying [11, Theorem I.3.1.4(i)] to the real place,

$$\begin{aligned} R_{K/{{\mathbb {Q}}}}{\mathbf {H}}({{\mathbb {R}}}) \cong \prod _{\nu \in V^\infty } {\mathbf {G}}(K_\nu ). \end{aligned}$$

The right hand side is non-compact by assumption. $\square$

In view of Remark A.3, Theorem A.4 has the following analogue in positive characteristic. The proof is the same as of Theorem A.4, taking into account that for adjoint groups the scheme theoretical center is trivial.

Theorem A.7

Let K be field of rational functions over a finite field, ${\mathcal {O}}$ its ring of polynomials and ${\mathcal {V}}$ the set of places of k. We fix a subset $S\subset {\mathcal {V}}$. Let ${\mathbf {H}}$ be a connected adjoint semi-simple K-algebraic group. We assume that ${\mathbf {H}}({\mathcal {O}}[S])$ is Zariski dense in ${\mathbf {H}}$. Then the commensurator of ${\mathbf {H}}({\mathcal {O}}[S])$ in $\prod '_{\nu \in S}{\mathbf {H}}(K_\nu )$ is exactly ${\mathbf {H}}(K)$.