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Let $k$ be a finite extension of the p-adic number field $Q_p$ and G be a connected algebraic (not affine) group over $k$. It is well-known (see e.g. [1] Proposition 3.1) that G decomposes as $1\rightarrow G_{aff}\cap G_{ant}\rightarrow G_{aff}\times G_{ant}\rightarrow G \rightarrow 1,$ where $G_{aff}$ is the smallest normal connected affine subgroup of $G$ such that $G/G_{aff}$ is an abelian variety and $G_{ant}$ is the smallest normal subgroup such that $G/G_{ant}$ is affine.

Then $(G_{ant})_{aff}$ is a connected commutative affine algebraic group, and hence admits a unique decomposition $S\times U$, where $S$ is a torus and $U$ is connected and unipotent. Let $G'$ be a quasi-complement of $S$ in $G_{aff}/R_u$, where $R_u$ is the unipotent radical of $G_{aff}$. If $\pi:G_{aff}\rightarrow G_{aff}/R_u$ is the quotient map, then $\tilde{G}=\pi^{-1}(G')$ is a quasi-complement of $S$ in $G_{aff}$. i.e. $G_{aff}=S\tilde{G}$ with $S \cap\tilde{G}$ is finite. Together with above exact sequence we get an isogeny (see [2] Proposition 2.2)

$\phi:(\tilde{G}\times G_{ant})/U \rightarrow G$ such that the kernel of $\phi$ is $(\tilde{G}\cap G_{ant})/U$. Note that $U$ is the connected component of $\tilde{G}\cap G_{ant}$.

Q1: Is $\phi((\tilde{G}\times G_{ant})/U)(k))=\frac{(\tilde{G}\times G_{ant})/U)(k)}{(\tilde{G}\cap G_{ant})/U)(k)}$ closed in $G(k)$ w.r.t. analytic topology induced from $k$.

In general $\phi$ is not surjective on $k$-rational points. We have the following exact sequence $1 \rightarrow ((\tilde{G}\cap G_{ant})/U)(k)\rightarrow (\tilde{G}\times G_{ant})/U)(k) \rightarrow G(k)\rightarrow H^1(k_s/k, (\tilde{G}\cap G_{ant})/U)$, where $H^1$ is the first Galois cohomology. Since $(\tilde{G}\cap G_{ant})/U$ is a central finite affine algebraic group, $H^1$ is finite abelien group.

Q2: Does there exist a finite Galois extension $K$ of $k$ or $Q_p$ such that $\phi$ is indeed surjective on $K$-rational points? I think this is weaker than $H^1(k_s/k, (\tilde{G}\cap G_{ant})/U)=0$. If it is necessary, we may change the quasi-complement $\tilde{G}$.

[1] Anti-affine algebraic groups, Michel Brion, Journal of Algebra

[2] Principal bundles, quasi-abelian varieties and structure of algebraic groups, Carlos Sancho de Salasa, , Fernando Sancho de Salas, Journal of Algebra

I am very sorry for heavy notation.

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  • $\begingroup$ What is a quasi-complement? Please give a definition. $\endgroup$ – Mikhail Borovoi Feb 22 '15 at 19:03
  • $\begingroup$ @MikhailBorovoi: Given a group scheme $G$, a normal subgroup scheme $H\subseteq G$ and a subgroup scheme $S\subseteq G$, we say that $S$ is a quasi-complement to $H$ in $G$ if $G=HS$ and $H\cap S$ is finite; equivalently, the natural map $S\rightarrow G/H$ is an isogeny. $\endgroup$ – m07kl Feb 22 '15 at 19:22
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YES to Question 1. For an arbitrary homomorphism $\phi\colon G\to F$ of connected algebraic groups over $k$, not necessarily affine, where $k$ is a $p$-adic field or $k=\mathbb{R}$, the image $\phi(G(k))$ is closed in $F(k)$.

Proof. If $G$ is a connected $k$-group, and $X=G/H$ is a homogeneous space of $G$, then every orbit of $G(k)$ in $X(k)$ is open. Indeed, for any point $x\in X(k)$, the differential at any element $g\in G(k)$ of the map $\psi_x\colon G(k)\to X(k)\colon\ g\mapsto g\cdot x$ is surjective (we are in characteristic 0), and we can apply the Implicit Function Theorem over $k$. Since every orbit is open, we see that every orbit is closed.

Now if $\phi\colon G\to F$ is an epimorphism of connected $k$-groups, then $F$ is a homogeneous space of $G$ with respect to the action $g*f=\phi(g)f$, hence the orbit $\phi(G(k))$ of $1\in F(k)$ is open and closed in $F(k)$. If $\phi\colon G\to F$ is an arbitrary homomorphism of connected $k$-groups (not necessarily epimorphism), then the image ${\rm im}\,\phi$ is Zariski closed in $F$, see Borel's book, Cor. 1.4(a), and $G\to{\rm im}\,\phi$ is an epimorphism. We see that $\phi(G(k))$ is open and closed in the closed subgroup $({\rm im}\,\phi)(k)\subset F(k)$, hence $\phi(G(k))$ is closed in $F(k)$.

NO? to Question 2. Set $H={\rm SL}_{2,k}\times_k {\rm SL}_{2,k}$, $C=\{\pm 1 \}\times \{ \pm1\} $ $=Z/2Z\oplus Z/2Z$. Let $E$ be an elliptic curve over $k$ that has all the points of order 2 defined over $k$. We choose central embeddings $C\hookrightarrow H$ and $C\hookrightarrow E$ and set $G=(H\times_k E)/C$ with respect to the diagonal embedding. Then $G_{\rm ant}=E$, $G_{\rm aff}=H$, $S=1$, $\bar G=H$, $U=1$. In this particular case Question 2 is whether the map $$ \phi\colon H(K)\times E(K)\to G(K) $$ is surjective for some finite extension $K/k$. Since $H^1(K,H)=1,$ an easy cohomological argument reduces Question 2 in this case to the question whether for any elliptic curve $E$ the map $$ E(K)\to E(K)\colon\ x\mapsto 2x $$ can become surjective after passing to some finite extension $K/k$.

I am not an expert on elliptic curves. In a separate question you can ask to construct an elliptic curve $E$ over $k=\mathbb{Q}_p$ such that the homomorphism $$ E(K)\to E(K), \quad x\mapsto 2x $$ is not surjective for any finite extension $K/k$.

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  • $\begingroup$ Dear Mikhail. Thank you for your answer and I will reply soon. BTW, the algebraic groups, I talk about, are not necessarily affine. $\endgroup$ – m07kl Feb 22 '15 at 19:28
  • $\begingroup$ Dear Mikhail: why (im$\phi$)(k) is closed in $F(k)$? I think im$\phi$(k) is not Zariski closed in F(k) in general, in my case im$\phi$(k) is in fact Zariski dense in F(k), because it has finite index. It follows from Proposition 3.19 of [3] we need homomorphism to be central and surjective. However, I don't know whether they only talk about affine algebraic groups and whether the isogeny $\phi:(\tilde{G}\times G_{ant})/U \rightarrow G$ is central and separable? [3] jstor.org/stable/1970833?seq=1#page_scan_tab_contents $\endgroup$ – m07kl Feb 22 '15 at 22:23
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    $\begingroup$ The paper of Tits on abstract homomorphisms is not relevant here: you deal with algebraic homomorphisms. $\endgroup$ – Mikhail Borovoi Feb 23 '15 at 6:01
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    $\begingroup$ Since you write "Thank you for your answer", consider upvoting the answer. $\endgroup$ – Mikhail Borovoi Feb 23 '15 at 6:05
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    $\begingroup$ In characteristic 0 any morphism is separable and any isogeny is central. $\endgroup$ – Mikhail Borovoi Feb 25 '15 at 16:39

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