rationality question while dealing with an isogeny

Question

I don't think that the following is known, but before going to other things, I would like to know what can be said about it. Thanks in advance for any relevant comment !

So here is the situation. Let $k'\subset k$ be a finite field extension. Take affine algebraic group schemes defined over $k'$ together with a central isogeny $\pi : \tilde{G}\rightarrow G$ (i.e. $\pi $ is surjective on underlying topological space, and is a finite flat morphism such that $ker$ $\pi\subset Z(G)$).

Given a subgroup $\Gamma \leq \tilde{G}(k)$, assume that $\Gamma _{down}=\pi_{k}(\Gamma )$ lies in fact already in $G(k')$. Can we conclude that $\pi_{k'}^{-1}(\Gamma_{down})$ is big enough compared to $\Gamma $ (more precisely, that its intersection with $\Gamma $ is of finite index in $\Gamma $ ?)

Ok, now more precision about assumption (but I would be interested to have counter-examples if not in that situation, especially for the first assumption) :
1) $k$ is a local field, and $k'$ a closed subfield
2) $\tilde{G}$ (resp. $G$) is absolutey simple simply connected (resp. adjoint)

Also, let me stress that I am mainly interetsed in the positive characteristic case, i.e. $k$ a finite extension of $\mathbb{F}_{p}$((T)), and that I do not assume that the extension is Galois (but I would be interested to know what Galois cohomology can bring to the matter, I'm not at all familiar with that theory).

EDIT (prompted by the answer of user 76758)

I'm very happy to have forgotten to mention the crucial assumption on $\Gamma $ since I got this very illuminating answers without it. But now, here it is : $\Gamma $ is in fact assumed to be an open compact subgroup of $\tilde{G}(k)$ (in the topology given by the local field $k$).

Also, the purpose of all that was to prove the following claim : if $\Gamma $ is an open compact subgroup of $\tilde{G}(k)$ avoiding the center (as above, $\tilde{G}$ absolutely simple simply connected), then any closed normal subgroup of $\Gamma $ is of finite index. I finally found a Paper of Carl Riehm, "The congruence subgroup problem over local fields", which completely settles the question.

But still, I would be interesting to know what happens in the above situation when $\Gamma $ is open compact (and if the finite index fact is true in this case, do we need the compactness assumption ?).

Answer 1

The answer is negative (even under all of the given hypotheses); this expresses a standard difficulty in the arithmetic aspects of connected semisimple groups over local function fields in contrast with $p$-adic fields.

As S.Carnahan notes, it is equivalent to ask that the subgroup $\Gamma_{\rm{down}}$ of $G(k')$ contains the image of its preimage on $k'$-points with finite index. This in turn is equivalent to the image of $\Gamma_{\rm{down}}$ under the connecting map $\delta:G(k') \rightarrow {\rm{H}}^1(k', \ker \pi)$ having finite image.

The answer is generally negative because ${\rm{H}}^1(k',\mu_p) = {k'}^{\times}/({k'}^{\times})^p$ is infinite whenever $k'$ is a local function field of characteristic $p$ even though $[{k'}^{1/p}:k'] = p < \infty$ in such cases.

To be specific, consider the central isogeny $\pi_0: {\rm{SL}}_p \rightarrow {\rm{PGL}}_p$ between absolutely simple and connected semisimple groups over $\mathbf{F}_p$ (with ${\rm{SL}}_p$ simply connected and ${\rm{PGL}}_p$ of adjoint type), so the induced map $D \rightarrow \overline{D}$ between split diagonal tori is a direct product of the $p$-power endomorphism of ${\rm{GL}}_1$ and the identity map on ${\rm{GL}}_1^{p-2}$. So if we let $k'$ be any local function field of characteristic $p > 0$, $k = {k'}^{1/p}$ (a finite extension!), and $\pi$ the base change of $\pi_0$ over $k'$ then for the group $\Gamma = k^{\times}$ of $k$-points in a 1-parameter subgroup ${\rm{GL}}_1 \subset D$ multiplied by $p$ under $\pi_0$ we have $\Gamma_{\rm{down}} = {k'}^{\times}$ and the image of the preimage of $\Gamma_{\rm{down}}$ on $k'$-points is $({k'}^{\times})^p$, so the quotient is the group ${k'}^{\times}/({k'}^{\times})^p$ is infinite. (In this case the connecting map $\delta$ carrie $\Gamma_{\rm{down}}$ onto the group ${\rm{H}}^1(k', \ker \pi) = {\rm{H}}^1(k', \mu_p)$ that is infinite.) Equivalently, the preimage of $\Gamma_{\rm{down}}$ on $k'$-points is the subgroup ${k'}^{\times} = (k^{\times})^p \subset k^{\times} = \Gamma$ that has infinite index in $\Gamma$.

If some motivation were given for the question then it might be clearer if there is an appropriate salvage to the question. But it is hard to tell what (if anything) to "fix" since the OP doesn't indicate where the question is coming from. (There are ways around the above difficulty in practice, depending on what the aim is.)