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Let $K$ be a (locally compact) local field and $G$ be a linear algebraic $K$-group.

Does the topological group $G(K)$ have a cocompact solvable closed subgroup?

If $\mathrm{char}(K)=0$, it is true that $G(K)$ has a cocompact solvable closed subgroup, see Proposition 9.3 in "A. Borel and J. Tits: Groupes reductifs" (Publ. IHES, 1965: link). So the only case left is when the characteristic is positive.

It is also well-known that for any local field $K$, $\mathrm{GL}(n,K)$ has a cocompact solvable closed subgroup.

What about $G=H_{2n+1}\rtimes \mathrm{Sp}_{2n}$, which is the semidirect product of the Heisenberg group $H_{2n+1}$ with the symplectic group $\mathrm{Sp}_{2n}$?

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    $\begingroup$ Every reductive group $G$ over $K$ admits an Iwasawa decomposition $G(K) = P(K) C$ where $P$ is a minimal parabolic $K$-subgroup and $C$ is a maximal compact. If $G$ is quasi-split, then $P$ is just a Borel subgroup, which is solvable and hence satisfies what you want. The above works also for local fields in positive characteristic. $\endgroup$ – user94041 Feb 14 '17 at 0:19
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    $\begingroup$ It's certainly true but so far I haven't seen a rigorous proof. There should indeed be a reduction to the reductive case, but keeping in mind subtleties such as the possibility that the unipotent radical might fail to be defined over $K$. Also one should always be careful: $G\to H$ surjective does not mean that $G(K)\to H(K)$ is surjective, and if one want to claim that the inverse image of a cocompact subgroup of $H(K)$ is cocompact in $G(K)$ one should be careful too. Char $p$ difficulties are probably the reason Borel-Tits restricted to char 0. $\endgroup$ – YCor Feb 14 '17 at 0:33
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    $\begingroup$ For your special case there is no problem since one can take a Borel subgroup $B$ in $Sp(2n)$, defined over $K$, consider $H=H_{2n+1}\rtimes B$; then $H(K)$ is cocompact in $G(K)$. $\endgroup$ – YCor Feb 14 '17 at 0:36
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    $\begingroup$ @YCor: I doubt that there is a simple reduction to the reductive case, for reasons alluded to in my affirmative proof below. (I tried to use the Frobenius isogeny $G\rightarrow G^{(p^n)}$ for $n$ so large that $G^{(p^n)}$ has a split unipotent smooth connected $K$-subgroup $U$ modulo which it is reductive. But I ran into problems trying to pull down to $G$ the result for $G^{(p^n)}/U$, due to various non-smooth groups that arose.) I would be delighted to see such an argument, however. I'm unsure if experience in the reductive case is enough reason to be certain without a rigorous proof. $\endgroup$ – nfdc23 Feb 14 '17 at 7:41
  • $\begingroup$ @nfdc23 Great; I hope you'll publish this somewhere (I would already have had some occasions to quote it). $\endgroup$ – YCor Feb 14 '17 at 14:55
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Yes. By replacing $G$ with its maximal smooth closed subgroup (whose formation commutes with any separable extension on $K$; see Lemma C.4.1 and Remark C.4.2 in the book Pseudo-reductive Groups), we may and do assume $G$ is $K$-smooth. Alternatively, if you prefer, replace $G$ with the Zariski-closure of $G(K)$ to achieve the same end. In such cases we shall describe the solvable group in rather concrete terms.

Let's first treat the case when $G$ is reductive (but possibly not connected), as inspiration for the general case. With $G$ reductive, let $H$ be a maximal $K$-split smooth connected solvable $K$-subgroup. Explicitly, there exists a minimal parabolic $K$-subgroup $P$ of $G^0$ such that $H = S \ltimes U$ where $U = \mathscr{R}_{u,K}(P)$ is the $K$-unipotent radical of $P$ and $S$ is a maximal split $K$-torus in $P$. We claim that $G(K)/H(K)$ is compact.

Since $H$ is smooth, the natural map $G(K)/H(K) \rightarrow (G/H)(K)$ is a homeomorphism onto an open image (via the $K$-analytic implicit function theorem and the Zariski-local structure of smooth morphisms, applied to $G \rightarrow G/H$). But this open image is full since the smooth connected $H$ is $K$-split solvable, so it is the same to show that $(G/H)(K)$ is compact. It suffices to show that $(G^0/H)(K)=G^0(K)/H(K)$ is compact, since $G^0(K)$ is open of finite index in $G(K)$. Thus, we may and do rename $G^0$ as $G$ so that $G$ is connected.

By minimality, $P = Z_{G}(S) \ltimes U$. Note that $H$ is normal in $P$. Since $G/P$ is projective, clearly $(G/P)(K)$ is compact. The map $G/H \rightarrow G/P$ is a right torsor for $P/H = Z_{G}(S)/S$, and $Z_{G}(S)/S$ is a $K$-anisotropic connected reductive group. Thus, $(Z_{G}(S)/S)(K)$ is compact, or in other words $(P/H)(K)$ is compact. It follows that the map $(G/H)(K) \rightarrow (G/P)(K)$ has open image over which its source is a topological torsor for the compact group $(P/H)(K)$. It is therefore enough to show that $(G/H)(K) \rightarrow (G/P)(K)$ is surjective. Even better, the Borel-Tits structure theorem for connected reductive groups over general fields gives that $G(K) \rightarrow (G/Q)(K)$ is surjective for any parabolic $K$-subgroup $Q$ of $G$ (even when ${\rm{H}}^1(K,Q)$ is nontrivial, as can happen).


How to go beyond the reductive case? This is rather trivial when ${\rm{char}}(K)=0$, as then $G$ has a $K$-split unipotent normal smooth connected $K$-subgroup modulo which it is reductive, so let's assume ${\rm{char}}(K)=p>0$. The imperfectness of such $K$ creates many difficulties. I tried some elementary arguments involving Frobenius isogenies to try to justify direct passage to the reductive case over such $K$. Unfortunately, this involves the intervention of various non-smooth group schemes which messed up all of my attempts to try to show various subsets of topological spaces are closed.

So in the end I failed to find an elementary reduction to the reductive case when $K$ is a local function field. All I can find is a proof that involves the full force of the structure theory of pseudo-reductive groups, which you may find somewhat heavy for the task at hand. If someone else can find a way to rigorously justify passage to the reductive case in a more direct manner in positive characteristic then that would be swell. In what follows, [CGP] refers to the book Pseudo-reductive Groups (2nd ed.) by Conrad, Gabber, and Prasad.


Let $H$ be a maximal $K$-split solvable smooth connected $K$-subgroup of $G$. We claim that $G(K)/H(K) = (G/H)(K)$ is compact. Clearly $H$ contains $V := \mathscr{R}_{us,K}(G)$, the maximal $K$-split unipotent smooth connected normal $K$-subgroup of $G$, and $H/V$ is a maximal $K$-split solvable smooth connected $K$-subgroup of $G/V$, with $G/H = (G/V)/(H/V)$. By [CGP, Cor. B.3.5], $G/V$ has $K$-unipotent radical that is $K$-wound (i.e., does not contain $\mathbf{G}_{\rm{a}}$ as a $K$-subgroup); in other words, $G/V$ is "quasi-reductive" in the sense of [CGP, Def. C.2.11]. Thus, we can replace $G$ with $G/V$ to reduce to the case that $G$ is quasi-reductive.

Let $P$ be a minimal pseudo-parabolic $K$-subgroup of $G$ and $S \subset P$ a maximal $K$-split torus. By [CGP, Prop. C.2.4] we have $P = Z_G(S) \cdot \mathscr{R}_{us,K}(P)$; this is a semi-direct product by the first paragraph in [CGP, Rem. C.2.33]. Let $H = S \ltimes \mathscr{R}_{us,K}(P)$, so $H$ is normal in $P$ with $H(K)$ solvable and $G(K)/H(K)=(G/H)(K)$ as in the reductive case. We claim that $G(K)/H(K)$ is compact. As in the reductive case, $G/H \rightarrow G/P$ is a $P/H$-torsor and so $(G/H)(K) \rightarrow (G/P)(K)$ is a topological $(P/H)(K)$-torsor over its open image. This image is full (as in the reductive case) because $G(K) \rightarrow (G/P)(K)$ is surjective [CGP, Lemma C.2.1]. Thus, as in the reductive case it suffices to prove the compactness of $(P/H)(K)$ and $(G/P)(K)$.

We first analyze $G/P$. Recall that $W := \mathscr{R}_{u,K}(G)$ is $K$-wound. By definition of pseudo-parabolicity we have $W \subset P$, and $G/P = (G/W)/(P/W)$ with $P/W$ a minimal pseudo-parabolic $K$-subgroup of the pseudo-reductive $G/W$ (see [CGP, Prop. 2.2.10]). Thus, to prove compactness of $(G/P)(K)$ we may and do assume $G$ is pseudo-reductive. This case is rather subtle since $G/P$ is essentially never proper when $G$ is pseudo-reductive but non-reductive. (The ur-example is $G={\rm{R}}_{k'/k}(G')$ for a non-separable finite extension field $k'/k$ and a nontrivial connected semisimple $k'$-group $G'$, in which case $P = {\rm{R}}_{k'/k}(P')$ for a parabolic $k'$-subgroup $P' \subset G'$. In such cases $G/P={\rm{R}}_{k'/k}(G'/P')$, and this is never proper when $P' \ne G'$ due to [CGP, Example A.5.6].)

In view of the good behavior of pseudo-parabolic $K$-subgroups with respect to quotients of pseudo-reductive groups by central smooth connected $K$-subgroups, the general structure theorem for $G$ (in terms of the standard construction away from characteristics 2 and 3, the "generalized standard" construction in characteristics 2 and 3, and the "totally non-reduced" construction in characteristic 2: see [CGP, Prop. 10.1.4, Theorem 10.2.1(1)]) reduces the compactness of $(G/P)(K)$ to the case when $G = {\rm{R}}_{K'/K}(G')$ for a nonzero finite reduced $K$-algebra $K'$ and a smooth affine $K'$-group $G'$ whose fibers over the factor fields of $G'$ are one of the following: (i) connected semisimple, absolutely simple, and simply connected, (ii) basic exotic in characteristics 2 or 3, (iii) basic non-reduced pseudo-simple in characteristic 2.

By [CGP, Prop. 2.2.13], the pseudo-parabolic $K$-subgroups of such a Weil restriction $G$ are precisely ${\rm{R}}_{K'/K}(Q')$ for fiberwise-pseudo-parabolic $K'$-subgroups $Q' \subset G'$. Thus, $P={\rm{R}}_{K'/K}(P')$ for a $K'$-subgroupp $P' \subset G'$ that is fiberwise minimal pseudo-parabolic. Since $G/P = {\rm{R}}_{K'/K}(G'/P')$, so $(G/P)(K)=(G'/P')(K')$ as topological spaces, we can pass to the fibers over the factor fields of $K'$ separately (renaming such a factor field as $K$) so that $G$ falls into one of the cases (i), (ii), or (iii) above.

Case (i) is easy, as then $G/P$ is projective. For case (ii), [CGP, Prop. 11.4.6(1)] identifies $G/P$ as a closed subscheme of ${\rm{R}}_{K^{1/p}/K}(G'/P')$ for a canonically determined connected semisimple (even absolutely simple and simply connected) $K^{1/p}$-group $G'$ and (minimal) parabolic $K^{1/p}$-subgroup $P'$. Thus, we get the desired compactness in these cases due to the compactness of ${\rm{R}}_{K^{1/p}/K}(G'/P')(K)=(G'/P')(K^{1/p})$ that follows from the projectivity of $G'/P'$ over $K^{1/p}$.

How about case (iii)? In these cases we run into a temporary snag because the natural map $\xi:G/P \rightarrow {\rm{R}}_{K^{1/2}/K}(G'/P')$ analogous to the basic exotic case turns out not to be a closed immersion, and in fact has positive-dimensional fiber over the identity point of the target; see [CGP, Rem. 11.4.5]. But fortunately in these cases a separate miracle is available: the map $\xi$ is bijective on $K$-points. To be precise, topologically this map on $K$-points is identified with $G(K)/P(K)\rightarrow G'(K^{1/2})/P'(K^{1/2})$, and this is bijective by the end of [CGP, Prop. 11.4.4]. But $G(K) \rightarrow G'(K^{1/2})$ is a homeomorphism by [CGP, Prop. 9.9.4(2)], so we win!

It remains to analyze $P/H$. Recall as in the reductive case that $P/H = Z_G(S)/S$. We claim that this group has no $K$-subgroup isomorphic to either of $\mathbf{G}_{\rm{a}}$ or $\mathbf{G}_{\rm{m}}$. The $K$-wound $W$ is a normal $K$-subgroup modulo which its quotient is $Z_{G/W}(S)/S$, so it suffices to show that $Z_{G/W}(S)/S$ has neither $\mathbf{G}_{\rm{a}}$ nor $\mathbf{G}_{\rm{m}}$ as a $K$-subgroup. But $Z_{G/W}(S)$ inherits pseudo-reductivity of $G/W$, so $Z_{G/W}(S)/S$ is a pseudo-reductive group that is $K$-anisotropic in the sense that it does not contain $\mathbf{G}_{\rm{m}}$ as a $K$-subgroup. Hence, $Z_{G/W}(S)/S$ certainly has no proper pseudo-parabolic $K$-subgroup, so it does not contain $\mathbf{G}_{\rm{a}}$ as a $K$-subgroup either, due to [CGP, Thm. C.3.8]!

To summarize, the proof of compactness of $(P/H)(K)$ is reduced to proving the compactness of $\mathcal{G}(K)$ for any smooth connected affine $K$-group $\mathcal{G}$ that contains neither $\mathbf{G}_{\rm{a}}$ nor $\mathbf{G}_{\rm{m}}$ as a $K$-subgroup. This is Proposition A.5.7 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio 148.

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  • $\begingroup$ Dear nfdc23 Thanks a lot. To be honest, I can not follow your heavy arguments. I suggest you to publish your result. My original motivation is to answer the following question: Let $G$ be a linear algebraic group over a local fields $k$, then the left regular representation of $G(k)$ is Type I? $char(K)=0$ case is known and please see Prop 5.3 and Thm 5.4 in docs.google.com/… $\endgroup$ – m07kl Feb 14 '17 at 12:42
  • $\begingroup$ I do believe the question has positive answer, because it is well-known that if a locally compact separable group G contains a closed cocompact type I subgroup H, then the left regular representation of G is type I. We are more or less (except they used unipotent radicals) done by your answer above and the proofs of Prop 5.3 and Thm 5.4 in docs.google.com/… $\endgroup$ – m07kl Feb 14 '17 at 12:52
  • $\begingroup$ Is the left regular representation of the Jacobi groups over local fields of characteristic $p$ type I? see mathoverflow.net/questions/227551/… $\endgroup$ – m07kl Feb 14 '17 at 13:55
  • $\begingroup$ I don't know anything serious about the notion of type-I group, sorry. $\endgroup$ – nfdc23 Feb 14 '17 at 14:54
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The answer below (is a detailed edit of a previous answer which) provides a (more-or-less) self contained proof of the fact that $\mathbf{G}(K)$ contains a cocompact solvable closed subgroup.

Let $K$ be a local field and $\mathbf{G}$ a linear algebraic group defined over $K$. I denote $G=\mathbf{G}(K)$ and assume as I may that $G$ is Zariski dense in $\mathbf{G}$. Note that by [Borel's "Linear Algebraic Group, AG14.4] the Zariski closure of $G$ in $\mathbf{G}$ is itself a $K$-algebraic group whose $K$-points set is again $G$, thus replacing $\mathbf{G}$ with the Zariski closure of $G$ justifies this assumption. Below topological notions, unless otherwise said, are taken with respect to the Hausdorff (ie $K$-analytic) topology.

Lemma 1: If $G$ is not solvable then $\mathbf{G}$ has an irreducible $K$-representation of dimension $>1$.

Proof: Assuming every $K$-representation is 1-dimensional, fixing an injective $K$-representation $\mathbf{G}\to \text{GL}_n$ the image of $G$ is easily seen to be in a $K$-Borel subgroup of $\text{GL}_n$ (here we use the fact that $G$ is Zariski-dense in $\mathbf{G}$).

Lemma 2: If $G$ is not solvable then there exists a $K$-algebraic variety $\mathbf{X}$ endowed with a $K$-action of $\mathbf{G}$ and a point $x\in X=\mathbf{X}(K)$ which is not $G$-fixed such that $Gx$ is compact.

Proof: Using lemma 1, we fix an irreducible $K$-representation $\mathbf{G}\to \text{GL}_n$ with $n>1$ and consider the associated $K$-action of $\mathbf{G}$ on $\mathbf{X}=\mathbb{P}^{n-1}$. Observe that $X=\mathbf{X}(K)$ is compact. By Zorn Lemma, using compactness, we can find a minimal closed, non-empty, $G$-invariant subset $Y\subset X$. We fix such $Y$ and a point $x\in Y$. By the fact that $n>1$ and the representation is $K$-irreducible we get that $x$ is not $G$-fixed. Note that $\overline{Gx}$ is a non-empty $G$-invariant closed subspace of $Y$, thus by minimality, $Y=\overline{Gx}$. We recall that the action of $G$ on $X$ has locally closed orbits (this is proved in the appendix of http://www.math1.tau.ac.il/~bernstei/Publication_list/publication_texts/B-Zel-RepsGL-Usp.pdf). It follows that $\partial(Gx)$ is a proper closed invariant subset of $Y$, hence by minimality $\partial(Gx)=\emptyset$. We conclude that $Y=Gx$ and indeed $Gx$ is compact.

Corollary: If $G$ is not solvable then there exists a proper $K$-algebraic group $\mathbf{H}<\mathbf{G}$ such that $H=\mathbf{H}(K)$ is cocompact in $G$.

Proof: Fix $\mathbf{X}$ and $x\in X$ as in Lemma 2 and let $H$ be the stabilizer of $x$ in $G$ and $\mathbf{H}$ be the Zariski-closure of $H$ in $\mathbf{G}$. By [Borel's "Linear Algebraic Group, AG14.4], $\mathbf{H}$ is a $K$-group and clearly $H=\mathbf{H}(K)$ (note however that $\mathbf{H}$ might not be the stabilizer of $x$ in $\mathbf{G}$). We get a continuous bijective $G$-map $G/H\to Gx$. By a standard argument in the theory of locally compact groups (using Baire category) this map is open, thus $H<G$ is cocompact. By the fact that $x$ is not $G$ fixed, $H<G$ is proper. Thus also $\mathbf{H}<\mathbf{G}$ is proper.

Theorem: There exists a $K$-subgroup $\mathbf{S}<\mathbf{G}$ such that $S=\mathbf{S}(K)$ is solvable and cocompact in $G$.

Proof: The collection $$ \{\mathbf{H}\mid \mathbf{H}<\mathbf{G} \text{ is a $K$-algebraic subgrop such that $\mathbf{G}(K)/\mathbf{H}(K)$ is compact} \} $$ is not empty, as it contains $\mathbf{G}$, hence it contains a minimal element by Noetherianity. Let $\mathbf{S}$ be such a minimal element. Denote $S=\mathbf{S}(K)$ and note that by minimality $S$ is Zariski dense in $\mathbf{S}$. If $S$ is not solvable, then by the corollary there exists $\mathbf{H}<\mathbf{S}$ such that $H=\mathbf{H}(K)$ is cocompact in $S$, hence also in $G$, contradicting the minimality of $\mathbf{S}$. Thus $S$ is indeed solvable.

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  • $\begingroup$ In positive characteristic how do you show the orbit on $K$-points is locally closed and has its subspace topology from the ambient projective space coinciding with the quotient topology from $\mathbf{G}(K)$? In characteristic 0 orbit maps are smooth onto their Zariski-locally closed image, so passing to $K$-points gives a $K$-analytic submersion, but in positive characteristic it seems more work is needed for topological aspects of orbits. The main result of the sort-of-recent paper arxiv.org/abs/1309.6430 should take care of that (maybe it simplifies for complete valued fields?) $\endgroup$ – nfdc23 Feb 14 '17 at 22:15
  • $\begingroup$ For a locally compact group $G$ and a closed subgroup $H<G$ there is a unique $G$ invariant locally compact topology on $G/H$. Eventhough the map $G/H\to$ Orbit might not be the identity (algebraically), it is a homeomorphism for the hausdorff topology. $\endgroup$ – Uri Bader Feb 14 '17 at 22:18
  • $\begingroup$ Ah, so local compactness helps for this. But we have to know that orbits are locally closed to know that they're locally compact. How are you seeing that the orbits on $K$-points are locally closed in positive characteristic? Is this aspect of the result of that arxiv paper relatively easy to prove for locally compact $K$? $\endgroup$ – nfdc23 Feb 14 '17 at 22:21
  • $\begingroup$ One minor point is that this argument appears to tacitly use that $\mathbf{G}(K)=\mathbf{G}^{\rm{sm}}(K)$ for the maximal smooth closed $K$-subgroup $\mathbf{G}^{\rm{sm}}$ of $\mathbf{G}$, so that we may restrict attention to $\mathbf{G}$ and $\mathbf{H}$ that are $K$-smooth. This ensures (by $K$-analytic considerations) that $\mathbf{G}(K)$ is (schematically) dense in $\mathbf{G}$, so more specifically solvability as a $K$-group agrees with solvability on $K$-points, and that irreducibility as a representation of an algebraic group agrees with irreducibility on $K$-points. $\endgroup$ – nfdc23 Feb 14 '17 at 22:25
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    $\begingroup$ @m07kl, yes, it is also true for projective varieties. $\endgroup$ – Uri Bader Feb 14 '17 at 22:44

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