Recently I read Serre's wonderful book Abelian l-Adic Representations and Elliptic Curves . There is a theorem in section2.2 chapterIV.

Theorem: If $E$ has no complex multiplication, then $g_l=End(V_l)$, i.e. $G_l$ is open in $Aut(V_l)$.

Symbols: Let $E$ be a elliptic curve over a number field $K$. And let $\rho_l:Gal(\bar{K}/K)\rightarrow Aut(V_l(E))$ be the representations of Tate-module, where $V_l(E)=T_l(E)\otimes_{Z_l}Q_l$. Let $G_l=Im(\rho_l)$ and let $g_l\subseteq End(V_l)$ be the Lie algebra of $G_l$.

Question: We know $G_l$ be the closed subgroup of $Aut(V_l(E))$, so by the p-adic version closed subgroup theorem, we have $G_l$ is the embeded Lie subgroup of $Aut(V_l(E))$. But if we have $g_l=End(V_l)$, how can we deduce that $G_l$ is open in $Aut(V_l)$?

In the real Lie groups case, if $H$ is the closed Lie subgroup of a n-dim Lie group $G$, and assume $H$ and $G$ have the same dimension, can we deduce that $H$ is open in $G$?

Thanks for any answers!


I think this is straightforward after noting that for a finite-dimensional vector space $V$, $End(V)$ is the Lie algebra of $Aut(V)$. One has to be a bit careful in the ultrametric case because the exponential map does not converge on the entire Lie algebra, but it does on an open additive subgroup, and its image is an open subgroup (here: of $Aut(V)$); since by the theorem, it's contained in $G_l$, this group is also open in $Aut(V)$.

Cf. Bourbaki, Groupes et algèbres de Lie ch. III, §§ 4 and 7.

Added in response to a comment: It is true, and good to know, that $exp$ is generally not an open map. However:

  1. Actually, what you want already follows from the more general theorem 1 in §4 (and is explicitly stated in its first corollary), whose proof does not use the exponential series, but unfortunately is a bad Bourbaki-style mess referencing a different volume (and to make it worse, only the proof-less Fascicule de résultats of Variétés différentielles et analytiques) in the crucial step. Further, the parts of theorems 2 and 3 (which bring the BCH series into play) in the same section, which again imply openness, rely on theorem 1 in their proofs for that. Likewise, all the statements in §7 which refine this for the ultrametric case again seem to reference that theorem 1 of §4.
  2. Then again, all we need is that the image of $exp$ (if you want, "the image of its entire domain") is open. This is already shown (by explicitly computing radius of convergence for $exp$ and $log$ in the ultrametric case, which gives an open ball around $1$ to which on open ball around $0$ in the Lie algebra exponentiates) in ch. II (!), §8 no. 4, whose setting is that of a complete normed algebra and its unit group, for which $End(V)$ and $Aut(V)$ is a standard example. (Looking through this proof, annoyingly there are again references to that Fascicule, but those seem to be standard convergence results, which anyone other than Bourbaki would take for granted as known from an analysis course.)
  • $\begingroup$ Thanks for your answer. But how do we know the image of the exponential map acting on an open additive subgroup is an open subgroup of $Aut(V)$? As far as I know. the exp map is not an open map in general. $\endgroup$ – Eiang Jul 9 at 7:04

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.