# $G(k)/H(k)$ as a submanifold of $G/H(k)$

Let $$k$$ be a local field (if necessary, assume characteristic zero). In general, if $$X$$ is a smooth variety of finite type over $$k$$ of dimension $$n$$, then the set of $$k$$-rational points $$X(k)$$ is an analytic manifold over $$k$$ of dimension $$n$$. I was thinking about the passage $$X \mapsto X(k)$$ from smooth varieties to manifolds in the context of coset spaces, and had a couple of questions.

Let $$H$$ be a closed subgroup of a linear algebraic group $$G$$ over $$k$$. Assume $$H$$ is defined over $$k$$. Then the coset space $$G/H$$ has the structure of a smooth quasi-projective variety over $$k$$, which tells us that $$G/H(k)$$ is an analytic manifold over $$k$$ of dimension $$n = \operatorname{Dim}G - \operatorname{Dim}H$$.

On the other hand, $$H(k)$$ is a closed subgroup of the analytic Lie group $$G(k)$$, and the quotient $$G(k)/H(k)$$ has the structure of an $$n$$ dimensional analytic manifold (the quotient map $$G(k) \rightarrow G(k)/H(k)$$ is a principal fiber bundle with $$H(k)$$ as a fiber). This just follows from the general theory of Lie groups.

Of course, $$G(k)/H(k)$$ injects into $$G/H(k)$$, but they don't have to be equal. They are the same if $$H^1(\operatorname{Gal}(k_s/k),H)$$ is trivial.

What can we say about the inclusion $$G(k)/H(k) \rightarrow G/H(k)$$ from the perspective of manifolds? Is the inclusion an analytic map? Does it make $$G(k)/H(k)$$ into an open submanifold? How different can these two manifolds be?

• Would you specify what you call a local field? Everybody agrees that $\mathbf{Q}_p$ is a local field, but some define it as an arbitrary nondiscrete locally compact field (which is a reasonable setting for your question), while others would exclude $\mathbf{R}$ but would allow some non-locally compact fields such as Laurent Series over infinite discrete fields.
– YCor
Commented Nov 10, 2018 at 22:29
• Borel-Serre (CMH 1964) obtains the finiteness of the relevant cohomology groups when one has a perfect field with finitely many extensions of each degree. This probably yields the openness result in this case. The openness result fails for the non-perfect locally compact field $K=\mathbf{F}_p((t))$ and $G\to G/H$ for $G=\mathrm{SL}_p$, $H=\mathrm{PGL}_p$.
– YCor
Commented Nov 10, 2018 at 22:33
• I just mean $\mathbb R, \mathbb C$, or a finite extension of $\mathbb Q_p$ or $\mathbf F_q((t))$
– D_S
Commented Nov 10, 2018 at 23:48
• OK: these are precisely the nondiscrete locally compact fields.
– YCor
Commented Nov 11, 2018 at 1:57

If $$k$$ has characteristic zero, then $$H^1(k,H)$$ is finite (Borel-Serre: https://mathscinet.ams.org/mathscinet-getitem?mr=181643) and the map $$G(k)/H(k) \rightarrow (G/H)(k)$$ has open image (can also be proved by the implicit function theorem). Moreover, a Baire category argument shows that this map is a homeomorphism onto the image. In positive characteristic, there is an old paper by Bernstein Zelevinsky (https://mathscinet.ams.org/mathscinet-getitem?mr=425030; see the appendix) which says (I think) that the map is a homeo onto the image, and that the image is closed.

• Could add references please ? Commented Nov 11, 2018 at 7:46
• I guess the implicit function argument (and hence openness) works over arbitrary complete normed fields of characteristic zero, even without the finiteness condition.
– YCor
Commented Nov 11, 2018 at 8:08
• @YCor: I think that is correct; you need not use Borel-Serre for openness. Commented Nov 11, 2018 at 8:32
• mathoverflow.net/a/176011/23291 Commented Nov 11, 2018 at 8:44
• For more general results, over henselian valued fields, see content.algebraicgeometry.nl/2014-5/2014-5-025.pdf Commented Nov 12, 2018 at 7:13

From the number of votes, it seems that the useful comment of Laurent Moret-Bailly has been under appreciated, so I thought it would be useful to explicitly record what the main theorem in the linked paper says (in community wiki mode, since this is really his answer).

Theorem 1.2 of GGMB14 (in the special case of $$k$$ a non-archimedean local field) The map $$G(k)/H(k)\to (G/H)(k)$$

$$\bullet$$ is always a homeomorphism onto its image,

$$\bullet$$ has closed image if $$H$$ satisfies the condition ($$\ast$$),

$$\bullet$$ has an open image if $$H$$ is smooth.

The condition ($$\ast$$) appearing in this theorem is technical (see Definition 2.4.3 of the paper), but $$H$$ satisfies the condition ($$\ast$$) if it has either one of the following properties: smooth, unipotent, commutative, or being a normal subgroup of a smooth group. Interestingly, there are examples of $$H$$ not satisfying $$(\ast)$$ for which the map $$G(k)/H(k)\to (G/H)(k)$$ has non-closed image (see example 7.1 of the paper, taking for example $$k=\mathbf{F}_p(\!(T)\!)$$).

Note that in characteristic $$0$$, (affine) group schemes (of finite type) are always smooth by a result of Cartier. Also, this puts in perspective the comment of YCor on the non-oppeness of $$\text{SL}_p(K)\to \text{PGL}_p(K)$$ for $$K = \mathbf{F}_p(\!(T)\!)$$. Finally, let me remark that the implicit function theorem should prove in all characteristic the openness of $$G(k)/H(k)\to (G/H)(k)$$ when $$H$$ is smooth (as suggested in Venkataramana's answer in characteristic $$0$$).