Compactness of the automorphic quotient

Let $F$ be a (totally real) number field, and $E$ a (totally imaginary) quadratic extension of $F$. We consider $U$ a unitary group (with respect to a given hermitian form over $E$). The question is:

When is $U(F) \backslash U(\mathbf{A}_F)$ compact?

I found some interesting reflexions about that in Arthur's notes on the trace formula, mainly that generally $U(F) \backslash U(\mathbf{A}_F)$ cannot be compact for it contains some $\mathbf{R}^k$ embedded diagonally, corresponding to the archimedean places. It is what happens for general linear groups, even for the case of $\mathrm{GL}_1$.

So I believe I have to assume that the $U_v$ are compact (that is, the usual $U(n)$) at archimedean places, and then answer the question for $U(F) \backslash U(\mathbf{A}_F)^1$. Is it enough to suppose so? Arthur cites a general characterization due to Borel-Harish-Chandra as well as Mostow-Tamagawa:

$U(F) \backslash U(\mathbf{A}_F)^1$ is compact if and only if $U$ has no proper parabolic subgroup defined over $\mathbf{Q}$

and that seems to be equivalent to $U(F)$ having no unipotent in his radical (is that what we call anisotropic?).

My current interest is in the unitary group in two variables, and I can suppose it to be definite, that is compact at archimedean places. Is it at least true in this case, and is there a simpler and more direct proof of this fact in this case, without invoking such a general result?

• The answer is yes. The quotient $U(\mathbb{A}_F)/U(F)$ is compact if and only if the hermitian form does not represent a zero over the number field $F$. You have assumed that the hermitian form does not represent a zero even over some (in fact, in your case, every) archimedean completion of $F$. The proof is a straightforward consequence of the Mahler criterion . For a reference, see Borel's book on arithmetic groups (it is in French, but since you seem to like Latin,...) – Venkataramana Jan 24 '17 at 11:23
• @Venkataramana Thanks for the answer. However, I do not see why this is such an immediate consequence of the Mahler criterion... moreover, since we are in less than 5 variables (in which case the anisotropic case can only arise at archimedean places, where I suppose it does not), is the anisotropy at archimedean places equivalent to compactness? – Desiderius Severus Jan 24 '17 at 12:21
• @DesideriusSeverus: sorry, I hastily misedited the backslash as setminus, while fixing the F-s in the latex. I know you mean left quotients, my mistake. Feel free to revert it if it bothers you. – M.G. Jan 24 '17 at 13:14
• if the hermitian form is ansotropic at one place of $F$, then it is anisotropic over $F$ as well. If the quotient is not compact, by Mahler criterion, there is a sequence $g_m$ in the quotient, and a sequence $v_m$ of non-zero integral vectors in the vector space, such that $g_m(v_m)$ tends to zero (Mahler). Applying the hermitian form to this, we get $h(v_m)$ tends to zero, but since $h(v_m)$ is an integer in $F$, and it tends to zero in $F\otimes \mathbb{R}$ it follows that $h(v_m)=0$ . That is, $h$ represents a zero (this is a standard argument) – Venkataramana Jan 24 '17 at 13:16
• @Desiderius, no problem! To convert the $\mathbb{R},\mathbb{Z}$ formulation of Borel to the case of integers in a number field, you use the "restriction of scalars" trick; then the unitary group over $F$ (by restricting scalars to $\mathbb{Q}$, becomes a group $G$ over $\mathbb{Q}$, and Mahler's criterion becomes the one given in Borel's book – Venkataramana Jan 24 '17 at 14:01

There are actually several different conditions all of which are equivalent, and one can make various formulations of these conditions depending on whether one sticks to the reductive case and/or global fields of characteristic 0 vs. general global fields. So it may be confusing to sort out exactly what is equivalent to what in which circumstances, and which equivalences are really theorems in number theory rather than instances of general equivalences over arbitrary fields. One thing which should be stressed at the outset, however, is that compactness at archimedean places is very very far from necessary for any such compactness criterion (though it is sufficient in the connected reductive case, as is compactness of the group of points at any single place, due to the various equivalences discussed at the end below).

Below I give a summary of the general situation with references for proofs.

For any connected reductive group $G$ over any field $k$ whatsoever, the absence of proper parabolic $k$-subgroup is equivalent to the condition

$(*)$: all split $k$-tori $T \subset G$ are central (equivalently, the connected semisimple derived group $\mathscr{D}(G)$ is $k$-anisotropic)

(see 20.6(ii) in Borel's textbook on algebraic groups for a proof).

Example: If $(V,h)$ is a non-degenerate finite-dimensional hermitian space over $k$ attached to a separable quadratic extension $k'/k$ then ${\rm{U}}(h)$ satisfies $(*)$ if and only if ${\rm{SU}}(h)$ is $k$-anisotropic, and that in turn is equivalent to the condition that the associated non-degenerate quadratic form $q_h:v \mapsto h(v,v)$ on $V$ is non-vanishing on $V-\{0\}$. This is proved similarly to the analogous fact that if $(W,q)$ is a non-degenerate finite-dimensional quadratic space over $k$ then ${\rm{SO}}(q)$ is $k$-anisotropic if and only if $q$ is non-vanishing on $V-\{0\}$.

Going beyond the reductive case (e.g., smooth connected solvable groups), for an arbitrary smooth connected affine $k$-group $G$ one can also consider the condition

$(*)$': all split $k$-tori $T \subset G$ satisfy $T_{\overline{k}} \subset \mathscr{R}(G_{\overline{k}})$.

In the reductive case, $(*)$' is obviously equivalent to $(*)$ because $\mathscr{R}(G_{\overline{k}})$ is the maximal central $\overline{k}$-torus of $G_{\overline{k}}$ which we know always descends to the maximal central $k$-torus of $G$ (obvious when $k$ is perfect by Galois descent, and part of the structure theory of reductive groups for general $k$). Beyond the reductive case these two conditions are obviously not equivalent (consider solvable $G$).

Although the $\overline{k}$-subgroup $\mathscr{R}(G_{\overline{k}}) \subset G_{\overline{k}}$ does descend to a $k$-subgroup of $G$ when $k$ is perfect (by Galois descent), so for such $k$ we can remove the intervention of $\overline{k}$ in $(*)$', what about in general? Beyond the reductive case whenever $k$ is not perfect there are many examples where $\mathscr{R}(G_{\overline{k}})$ does not descend to a $k$-subgroup of $G$. The theory of pseudo-reductive groups provides zillions of such examples, of which the most concrete are smooth connected affine Weil restrictions $G:= {\rm{R}}_{k'/k}(G')$ for non-separable finite extensions $k'/k$ and non-trivial connected semisimple $k'$-groups $G'$; such $G$ are never reductive.

So if we define the so-called $k$-radical $\mathscr{R}_k(G)$ to be the maximal solvable smooth connected normal $k$-subgroup of $G$ then typically $\mathscr{R}_k(G)_{\overline{k}}$ is a proper $\overline{k}$-subgroup of $\mathscr{R}(G_{\overline{k}})$ (e.g., it is trivial for the non-reductive Weil-restriction examples just mentioned). But despite that, one might wonder nonetheless if for the purpose of containing split tori it may not matter: perhaps $(*)$' is equivalent to the condition

$(*)$'': all split $k$-tori $T \subset G$ satisfy $T \subset \mathscr{R}_k(G)$.

Indeed, it turns out that the equivalence holds in general. The proof requires a digression into the structure of pseudo-reductive groups, which I suspect you don't care about, so I'll refrain from saying more on that here.

Moving on from tori to unipotent groups, recall that a smooth connected unipotent $k$-group $U$ is called $k$-split if it admits a composition series by smooth connected $k$-subgroups such that the successive quotient is $\mathbf{G}_a$; e.g., for perfect $k$ this always holds (see 15.5(ii) in Borel's textbook on linear algebraic groups). There is yet another condition one could consider, on the unipotent side:

(**)': every $k$-split smooth connected unipotent $k$-subgroup $U \subset G$ satisfies $U_{\overline{k}} \subset \mathscr{R}(G_{\overline{k}})$.

Note that in characteristic 0 the group $U(k)$ is Zariski-dense in $U$, and if $u \in G(k)$ is a nontrivial unipotent element then the Zariski closure of the $k$-group it generates is smooth connected unipotent and of course $k$-split. Hence, in characteristic $0$ the condition $(**)$' is equivalent to the condition (mentioned in the question posed) that all unipotent elements of $G(k)$ are contained in the geometric radical. It is asked in the question posed if this is what we call "$k$-anisotropic", which is really defined as the absence of nontrivial split $k$-tori. For connected semisimple groups in characteristic 0 it is true that the absence of nontrivial unipotent $k$-points is equivalent to $k$-anisotropicity, but in characteristic $p>0$ this is a rather more subtle issue since unipotent $k$-points have finite order and so it isn't evident if they must lie inside smooth connected unipotent $k$-subgroups, let alone ones which are $k$-split; let's not dwell on that any further here.

Inspired by the success with $(*)$'', one might wonder if there is a unipotent analogue: if we define the $k$-unipotent radical $\mathscr{R}_{u,k}(G)$ to be the maximal unipotent smooth connected normal $k$-subgroup of $G$, one can ask if $(**)$' is equivalent to the condition

$(**)$'': every $k$-split smooth connected unipotent $k$-subgroup $U \subset G$ satisfies $U \subset \mathscr{R}_{u,k}(G)$.

Obviously the equivalence holds when $k$ is perfect or $G$ is reductive. It does hold in general, but again that involves some (quite elementary) considerations in the direction of pseudo-reductive groups, so we'll not say more about it here.

Borel and Tits announced that $(*)$' and $(**)$' are actually equivalent (!) in complete generality. In the connected reductive case this is more familiar via the link to proper parabolic $k$-subgroups, but for imperfect $k$ it is a rather delicate matter with $(**)$' (i.e., showing that the maximal $k$-split unipotent smooth connected $k$-subgroups are precisely the $k$-unipotent radicals of the minimal parabolic $k$-subgroups). It seems that Borel and Tits never published a proof of the general equivalence, but a proof is given as Proposition A.5.1 in the paper Finiteness theorems for algebraic groups over function fields in Compositio 148 (2012) (using in a crucial way the structure theory of pseudo-reductive groups).

Finally, having stated this blizzard of various conditions over arbitrary fields (no number theory yet!), we reach the main point. Let $G$ be an arbitrary smooth connected affine group over a global field $k$. As usual, define $G(\mathbf{A}_k)^1$ to be the subgroup of points $g \in G(\mathbf{A}_k)$ such that $|\!|\chi(g)|\!|_k = 1$ for all $k$-homomorphisms $\chi:G \rightarrow \mathbf{G}_m$. Godement conjectured for global fields $k$ (or at least he did so in characteristic 0) that $G(k)\backslash G(\mathbf{A}_k)^1$ is compact if and only if $G$ satisfies both $(*)$' and $(**)$' that we have noted above are equivalent to each other over general fields.

As is discussed with references early in section A.5 of the Compositio paper mentioned above, the conjecture of Godement was proved over number fields independently in joint work of Borel with Harish-Chandra, and of Mostow with Tamagawa, and over function fields the reductive case was settled by Harder. In case you might wonder about the case of function fields for general $G$ (i.e., not necessarily reductive), the necessity of $(*)$' (or equivalently $(**)$') is due to Oesterle (see 1.4 in Chapter IV of his paper Nombres de Tamagawa et groupes unipotents en caract\'eristique $p$, Inventiones Math. 78 (1984)) but the sufficiency in general again requires the structure theory of pseudo-reductive groups; see Theorem A.5.5 in the Compositio paper mentioned above.

• Thanks for such a complete and general answer. But after having discovered the existence of the proof of the conjecture of Godement as you say, I am really lost in this level of generality. So I was wondering wether there is a simpler proof in the very specific case I consider, as Venkataramana gave by the means of the Mahler criterium; or how to manage to understand/manipulate better those very general characterizations. For... I do not see why they give the result for the totally definite unitary groups (I am not at ease with some notions of reductive groups theory...). Thanks in advance :) – Desiderius Severus Jan 24 '17 at 14:58
• Let $G$ be a linear algebraic group over a number field $K$ so that $G$ doesn't contain ${\rm{GL}}_1$ as a $K$-subgroup. This holds for ${\rm{SU}}(h)$ if $q_h$ has no nontrivial zero over $K$, and also when some $G(K_v)$ is compact (since ${\rm{GL}}_1(K_v)$ is non-compact and so cannot be a closed subgroup of a compact group). The Weil restriction ${\rm{R}}_{K/\mathbf{Q}}(G)$ has the same adelic points (as a topological group) and the same global points, and doesn't contain ${\rm{GL}}_1$ as a $\mathbf{Q}$-subgroup (use mapping properties), but isn't special unitary when $G$ is. [cont'd] – nfdc23 Jan 24 '17 at 16:12
• So it seems you should be satisfied with a reference that proves for any linear algebraic group $H$ over $\mathbf{Q}$ without ${\rm{GL}}_1$ as a $\mathbf{Q}$-subgroup that $H(\mathbf{Q})\backslash H(\mathbf{A}_{\mathbf{Q}})$ is compact (applying this to ${\rm{R}}_{K/\mathbf{Q}}({\rm{SU}}(h))$ for your initial hermitian space $(V,h)$ over $\mathbf{Q}$ (with $q_h$ having no zero in $V-\{0\}$). Is that correct? The Weil restriction trick avoids a lot of notational headaches, so it is a reason not to focus just on special unitary groups over $K$, but rather to work more broadly over $\mathbf{Q}$. – nfdc23 Jan 24 '17 at 16:16
• Now to be really specific, since you mentioned really only being interested in the case of special unitary groups in 2 variables over $K$, so your group coincides with the (algebraic) group $H_D$ units of reduced norm 1 in a quaternion central division algebra $D$ over $K$, you seem to presently be seeking just a proof that $H_D(K)\backslash H_D(\mathbf{A}_K)$ is compact for such $D$ (or more broadly central division algebras $D$ over $K$, not necessarily rank 4). Is that right? – nfdc23 Jan 24 '17 at 16:22
• @paulgarrett: Might you be thinking of Godement's Bourbaki Expose 257 from 1963? (In 1967 he wrote a Bourbaki report on Langlands' work, which might be why 1968 got conflated with this in your memory?) It is available online through Numdam (as are all of the Seminaire Bourbaki reports): see archive.numdam.org/ARCHIVE/SB/SB_1962-1964__8_/… for the 1963 expose, and archive.numdam.org/ARCHIVE/SB/SB_1966-1968__10_/… for the 1967 one. – nfdc23 Jan 24 '17 at 22:39

This has been answered by @nfdc in great generality. I just wanted to say how Mahler's criterion may be applied. This needs a little bit of work since the group $U(h)$ is "compact at infinity". This should really be an extended comment, but for reasons of space, I write this as an answer.

Mahler's criterion says that if $g_m$ is a sequence in $SL_n(\mathbb{R})/SL_n(\mathbb{Z})$ then it has no convergent subsequence if and only if there exists a sequence $v_m$ of non-zero vectors in $\mathbb{Z}^n$ such that $g_m(v_m)$ tends to zero.

If $\mathbb{A},\mathbb{A}_f$ denote the adeles and finite adles over $\mathbb{Q}$, then $SL_n(\mathbb{Q})$ is dense in $SL_n(\mathbb{A}_f)$ (strong approximation; easy to prove in this case) and hence the quotient $[SL_n(\mathbb{R}\times SL_n(\widehat{\mathbb{Z}})]/SL_n(\mathbb{Z}$ maps isomorphically onto $SL_n(\mathbb{A})/SL_n(\mathbb{Q})$. The former quotient maps onto $SL_n(\mathbb{R}/SL_n(\mathbb{Z})$ with compact fibre (namely $SL_n(\widehat{\mathbb{Z}})$) and hence one may show (with a small amount of work) that a sequence $g_m$ in $SL_n(\mathbb{A})/SL_n(\mathbb{Q})$ tends to infinity if and only if there exists a sequence of rational non-zero vectors $v_m\in \mathbb{Q}^n$ such that $g_m(v_m)$ tends to zero.

If $G=R_{F/\mathbb{Q}}(U(h)$ is the group over $\mathbb{Q}$ obtained by restricting scalars to $\mathbb{Q}$ of the $F$-group $U(h)$, then the inclusion $G(\mathbb{A})/G(\mathbb{Q})$ into $SL_n(\mathbb{A})/SL_n(\mathbb{Q})$ is proper and hence a sequence in the left hand side quotient goes to zero if and only if it goes to zero on the right hand side quotient. By the preceding para, $g_m$ in the left quotient goes to zero if and only if there exists a sequence $v_m\in \mathbb{Q}^n\simeq F^d$ ($h$ is a Hermitian form in $d$ variables, say) such that $g_m(v_m)$ tends to zero. Applying $h$, we get $h(v_m)\in F^d$ and goes to zero. Since $F$ is discrete in $\mathbb{A}\otimes F$, it follows that $h(v_m)$ is zero for large $m$, and hence the Hermitian form represents a zero.