I'll do my best to formulate everything in the modern language. Let $F$ be a local field. A torus $S$ is an $F$-group scheme of finite type such that $S \times_F \overline{F}$ is isomorphic to a finite product of copies of $\mathbb{G}_{m,\overline{F}}$ as $\overline{F}$-group schemes, where $\mathbb{G}_{m,\overline{F}} = \textrm{Spec } \overline{F}[T,T^{-1}]$. A character $\chi$ of $S \times_F \overline{F}$ is a morphism of $\overline{F}$-group schemes $S \times_F \overline{F} \rightarrow \mathbb{G}_{m,\overline{F}}$.

We say that $\chi$ is defined over $F$ if there exists a morphism of $F$-group schemes $\alpha: S \rightarrow \mathbb{G}_{m,F}$ such that $\chi = \alpha \times 1_{\overline{F}}$. If it exists, $\alpha$ is unique.

The trivial character is the morphism $S \times_F \overline{F} \rightarrow \mathbb{G}_{m,\overline{F}}$ coming from the $\overline{F}$-algebra homomorphism $\overline{F}[T,T^{-1}] \rightarrow \overline{F}[T_1^{\pm 1}, ... T_n^{\pm 1}]$, $T \mapsto 1$. It is always defined over $F$.

We say that $S$ is anisotropic if the trivial character is the only character which is defined over $F$.

There is a natural way of topologizing the $F$-rational points $S(F)$ of $S$, such that it is a locally compact topological group. See for example here (http://math.stanford.edu/~conrad/papers/adelictop.pdf).

I have read that $S(F)$ is compact if and only if $S$ is anisotropic, but I have never seen an explanation of this. What is the most natural way to realize this result?

  • $\begingroup$ @Venkataramana argues the 'if' direction, which is the harder one. The 'only if' is trivial: any rational character has non-compact image. $\endgroup$ – LSpice Feb 25 '17 at 4:26
  • $\begingroup$ Why is that? If $S$ splits over a finite extension $E$ over $F$, then the image of $S(E)$ under a nontrivial character is (most?) of $E$, which is not compact. I'm trying to relate this to the image of $S(F)$. $\endgroup$ – D_S Feb 25 '17 at 4:33
  • $\begingroup$ I'm sorry; I meant any rational cocharacter. Notice the 'rationality' in that description; it means that we have a map $F^\times \to S(F)$, not just $E^\times \to S(E)$. (I am using that the lattices of rational characters and of rational cocharacters are dual, so that one has some non-0 of the former if and only if one has some non-0 of the latter.) $\endgroup$ – LSpice Feb 25 '17 at 4:36
  • $\begingroup$ Although it doesn't seem to contain your particular result, the only place other than Weil (where I fear to look) that I have ever seen the analytic topology discussed is math.stanford.edu/~conrad/vigregroup/vigre02/cosetfinite.pdf . $\endgroup$ – LSpice Feb 25 '17 at 4:37
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    $\begingroup$ Every quotient of ${\rm{GL}}_1$ by a finite $F$-subgroup scheme (which must be some $\mu_n$) is again ${\rm{GL}}_1$, so in fact for the isotropic case one has ${\rm{GL}}_1$ as a closed $F$-subgroup of $S$ (much better than just a nontrivial $F$-homomorphism from ${\rm{GL}}_1$), so $F^{\times}$ with its natural analytic topology is a closed subgroup of $S(F)$, ruling out compactness of the latter. $\endgroup$ – nfdc23 Feb 25 '17 at 5:04

First, it should be noted that something much more general is true, and documented with proof in the literature. Suppose $F$ is the fraction field of a rank-1 henselian valuation ring $R$ (for simplicity feel free to assume $R$ is a complete discrete valuation ring, though one only needs that $R$ is henselian rather than complete, which in the rank-1 case is equivalent to the associated absolute value on $F$ admitting a unique extension to every finite extension of $F$: see 2.3.1 and 2.4.3 in Berkovich's paper in IHES 78 for that equivalence). For any affine $F$-scheme $X$ of finite type, it is well-posed to say $X(F)$ is bounded (with respect to $X$) if some some closed immersion $j:X \hookrightarrow \mathbf{A}_F^n$ the closed subset $X(F) \subset F^n$ is bounded; this is independent of the choice of $j$ and if $F$ is locally compact then it is equivalent to compactness of $X(F)$. If $G$ is a connected reductive $F$-group such that $G(F)$ is bounded then obviously $G$ is $F$-anisotropic (i.e., $G$ cannot contain ${\rm{GL}}_1$ as an $F$-subgroup, since boundedness is inherited by closed subschemes and $F^{\times}$ is not bounded (with respect to ${\rm{GL}}_1$). The great fact is that the converse is true: this is due to Rousseau, and a slick proof is given by Gopal Prasad in his paper Elementary proof of a theorem of $\dots$ in Bulletin de la SMF, tome 110 (1982), pp. 197-202.

My recollection is that Prasad's proof treats all such $G$ on equal footing, not requiring preliminary treatment of the case of tori. You might nonetheless consider this to be rather heavy for just treating tori, since Prasad's argument involves a lot of the structure theory of reductive groups (which admittedly simplifies to a triviality for tori).

Here is an argument that is a variant on Venkataramana's answer (comes down to the same fact about integral units being "bounded", but doesn't invoke facts about the semisimplicity of the algebraic representation theory of general tori over fields and includes some details that might make the argument look more complicated than it really is). Over any field $k$, we have an anti-equivalence of categories between $k$-tori $T$ and discrete ${\rm{Gal}}(k_s/k)$-modules $L$ that are finite free as $\mathbf{Z}$-modules (assigning to $T$ the geometric character lattice ${\rm{X}}(T) := {\rm{Hom}}_{k_s}(T_{k_s}, {\rm{GL}}_1)$ equipped with its natural (discrete) Galois action).

Let $\Gamma = {\rm{Gal}}(k'/k)$ for a finite Galois extension $k'/k$ that splits $T$. Then $L := {\rm{X}}(T)$ is a quotient of a finite direct sum of copies of $\mathbf{Z}[\Gamma]$ as a discrete $\Gamma$-lattice. Rationalizing to make the $\Gamma$-action semisimple, the $k$-anisotropicity of $T$ is exactly the condition that ${\rm{X}}(T)_{\mathbf{Q}}$ has no copy of the trivial representation. Equivalently, each map $\mathbf{Q}[\Gamma] \rightarrow {\rm{X}}(T)_{\mathbf{Q}}$ as $\Gamma$-lattices kills the trivial representation $\mathbf{Q}$. Hence, the $\Gamma$-lattice ${\rm{X}}(T)$ is a quotient of a direct sum of copies of $\mathbf{Z}[\Gamma]/\mathbf{Z}$. This latter Galois lattice corresponds to the $k$-subgroup $U^1$ of norm-1 units in ${\rm{R}}_{k'/k}({\rm{GL}}_1)$.

Going from $\Gamma$-lattices back to $k$-tori split by $k'$ turns this quotient presentation into a closed immersion of $k$-groups $T \hookrightarrow \prod_{j=1}^n U^1$, so it suffices to show that if $k$ is a rank-1 henselian valued field then $U^1(k)$ is bounded with respect to $U^1$. But with some review of the construction of Weil restriction through a finite extension of fields, one can show that if $k'/k$ is a finite extension of fields and $X'$ is an affine $k'$-scheme of finite type then the canonical bijection ${\rm{R}}_{k'/k}(X')(k) = X'(k')$ is a topological isomorphism, and that a closed subset if bounded with respect to ${\rm{R}}_{k'/k}(X')$ if and only if it is bounded with respect to $X'$.

In this way we are reduced to showing that the kernel of ${\rm{N}}_{k'/k}: {k'}^{\times} \rightarrow k^{\times}$ is bounded inside the "hyperbola" $uv=1$ within $k' \times k'$. But that kernel is a closed subgroup of $O_{k'}^{\times}$, so the desired boundedness is clear.


  • $\begingroup$ +1 for the "details that might make the argument look more complicated than it really is" ! $\endgroup$ – ACL Feb 25 '17 at 11:39
  • $\begingroup$ you seem to be taking $k'/k$ to be a quadratic extension; why? $\endgroup$ – Venkataramana Feb 25 '17 at 15:26
  • $\begingroup$ @Venkataramana: I think you're misunderstanding the reason for the "hyperbola" at the end, which is just describing how $\mathbf{G}_m$ is Zariski-closed in the affine plane (over $k'$) in accordance with the general procedure to identify an affine scheme of finite type as closed in an affine space; this has nothing to do with the computation of ${\rm{N}}_{k'/k}$ and definitely $k'/k$ is just a finite Galois extension splitting the original torus (not quadratic in general). $\endgroup$ – nfdc23 Feb 25 '17 at 15:32
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    $\begingroup$ @UriBader: I don't think such an argument is possible. The map you have in mind is surjective as varieties but likely not on rational points. I will be very surprised that the result for anisotropic connected reductive groups can really proved without anything deeper than the case of tori and the "soft" input that a connected reductive group is the image under multiplication from a direct product (as varieties) of finitely many tori over the ground field. Note that over local function fields there are "anisotropic" unipotent smooth connected groups admitting $\mathbf{G}_a$ as a quotient. $\endgroup$ – nfdc23 Feb 27 '17 at 5:45
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    $\begingroup$ @UriBader: Nice. I'm only able to make arguments that work in all characteristics. :) I find the "boundedness" formulation somehow very appealing since it applies for henselian fields which are not necessarily complete, though I concede the original question didn't ask for that generality. $\endgroup$ – nfdc23 Feb 27 '17 at 17:19

Sorry for adding a second proof, but I think I have now a much clearer one.

Let $k$ be a local field. Let $\mathbf{T}$ be a $k$-anisotropic torus, that is $\mathbf{T}$ contains no $k$-subgroup $k$-isomorphic to $\mathbf{G}_m$. It is standard that $T$ also doesn't have a non-trivial $k$-homorphism to $\mathbf{G}_m$. We argue to show that $T=\mathbf{T}(k)$ is compact. We assume as we may that $T$ is Zariski dense in $\mathbf{T}$. We fix a non-trivial irreducible $k$-representation $\mathbf{T}\to\text{GL}_n$. As mentioned above, $n>1$. We use the fact that the $T$-orbits on $\mathbb{P}^{n-1}(k)$ are locally closed to find a point $x\in \mathbb{P}^{n-1}(k)$ with a $T$-closed orbit. We let $S$ be the stabilizer of $x$ in $T$ and $\mathbf{S}$ be the Zariski closure of $S$ in $\mathbf{T}$. $\mathbf{S}$ is a proper $k$-subtorus. By an induction argument we may assume that $S=\mathbf{S}(k)$ is compact. We deduce that $T$ is compact.

For completeness, let me add the remark that in charactersitic 0 the same proof yields the fact that $\mathbf{G}(k)$ is compact when $\mathbf{G}$ is a $k$-anisotropic reductive group. The only thing to observe is that every $k$-subgroup of $\mathbf{G}$ is again reductive, what enables one to make an inductive argument as above. This follows from Jacobson-Morozov theorem: $\mathbf{G}$ has no unipotent $k$-subgroup.


This is true and is well known. I don't remember a specific reference but the proof is easy (I do not know the "most natural way" for which you have asked). I will assume that $Char (F)=0$.

Fix a faithful representation $T\rightarrow GL(V)$ defined over $F$, with $V$ a finite dimensional $F$-vector space. Write $V=\oplus V_i$ as a direct sum of $F$ irreducible representations $V_i$. Then $End _T(V_i)$ is a division ring with centre $F_i$ say. Then $F_i$ is a finite extension of $F$. Taking the "reduced determinant" in $GL(V_i)$ we get a morphism from $T$ into $\mathbb{G}_m$, which is trivial by assumption. Hence $T$ is realised as a closed subgroup of the product $\prod R^1_{F_i/F}(\mathbb{G}_m)$ where $R$ is the Weil restriction of scalars and $R^1$ is the subgroup of relative norm one elements in $F_i$.

It is therefore enough to prove that the group $R^1_{F_i/F}(\mathbb{G}_m)(F)$ is compact. But this is really the kernel to the norm map from $F_i^*\rightarrow F^*$. This kernel is contained in the group of units of $F_i^*$ and is hence compact.

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    $\begingroup$ (1) How are you using characteristic 0? (2) Isn't the 'reduced determinant' here really just the determinant? $\endgroup$ – LSpice Feb 25 '17 at 4:27
  • $\begingroup$ I used semi-simplicity of the representation. Yes, reduced determinant is determinant, which, when restricted to the division ring, is the reduced determinant. $\endgroup$ – Venkataramana Feb 25 '17 at 7:33
  • $\begingroup$ @L Spice : I now understand why you asked. I wanted to get a faithful semi-simple representation, where the centres $F_i$ of the division rings are separable (this can actually be got, but that becomes a separate issue. I did not want to get into that). $\endgroup$ – Venkataramana Feb 25 '17 at 8:21

Here is a proof of a different flavor. One can rightly argue that it is less straight forward, but in fact it gives a bit more (eg when regarding a non-split unipotent group). Below we fix a complete valued field $k$ (eg a local field) of any characteristic.

The following theorem is a reminiscent of "Furstenberg's Lemma", related to "Borel Density Theorem". For a complete proof of the stated form, see proposition 3.10 in https://arxiv.org/pdf/1411.5327.pdf (sorry for self-referencing). On demand I can sketch a self contained proof.

Theorem: Let $\mathbf{G}$ be a $k$-algebraic group and $\mathbf{X}$ a $k$-variety endowed with a $k$-action of $\mathbf{G}$. Let $\mu$ be a probability measure on $X=\mathbf{X}(k)$. Then the stabilizer of $\mu$ in $G=\mathbf{G}(k)$ has a cocompact normal subgroup which fixes all points in the support of $\mu$. This subgroup is the $k$-point of a $k$-subgroup of $\mathbf{G}$.

Corollary: Let $\mathbf{S}$ be a solvable $k$-algebraic group. Let $\mathbf{V}$ be a $k$-irreducible finite dimensional $k$-representation of $\mathbf{S}$. Then the image of $S=\mathbf{S}(k)$ in $\text{PGL}(\mathbf{V})$ is bounded.

For a general complete $k$ there is an extra step which I will give on demand. Below is the

Proof for $k$ local: We let $\mathbf{X}=\mathbb{P}(\mathbf{V})$ amd $\mathbf{G}=\text{PGL}(\mathbf{V})$. Using the compactness of $X$ and the amenability of $S$, we find an $S$-invariant probability measure $\mu$ on $X$. By irreducibility the support of $\mu$ spans $V$, hence it has a trivial fixator. By the previous theorem the stabilizer of $\mu$ in $G$ is bounded modulo the fixator of the support. It follows that the image of $S$ is bounded.

Now we are ready to specialize to the case where $\mathbf{G}$ is a $k$-anisotropic torus. We argue to show that $G$ is bounded. We assume as we may that this holds true for every proper $k$-subgroup of $\mathbf{G}$. Fixing a $k$-irreducible representation $\mathbf{V}$, the result now follows from the corollary above.


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