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.

  • $\begingroup$ What is a quasi-complement? Please give a definition. $\endgroup$ 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

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 $K/k$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ The paper of Tits on abstract homomorphisms is not relevant here: you deal with algebraic homomorphisms. $\endgroup$ Feb 23 '15 at 6:01
  • 1
    $\begingroup$ Since you write "Thank you for your answer", consider upvoting the answer. $\endgroup$ Feb 23 '15 at 6:05
  • 1
    $\begingroup$ In characteristic 0 any morphism is separable and any isogeny is central. $\endgroup$ Feb 25 '15 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.