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In J. Silverman's book "Arithmetic of Elliptic Curves" on page 40 in the beginning of the proof of Lemma 5.8.1 I have found this statement. Let $K$ be a number field and $\overline{K}$ its algebraic closure. Let $V$ be a $\overline{K}−$vector space and let the absolute Galois group $G_{\overline{K}/K}$ act continuously on $V$. Then, "The fact that $G_{\overline{K}/K}$ acts continuously on $V$ in a compatible way with its action on $\overline{K}$ means precisely that for any $v\in V$ the stabilizer of $v$ as finite index in $G_{\overline{K}/K}$". I do not understand why this has to be true. If $V=\overline{K}$ and more generally if the dimension of $V$ over $\overline{K}$ is finite, then the orbits are always finite (as any algebraic number has only finitely many conjugates), there is no relation with continuity or not. What is the correct meaning of this statement and why should it be true?


This question was asked in math stackexchange with no reaction, see https://math.stackexchange.com/q/1932912.

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    $\begingroup$ I don't know exactly what you mean here and where your confusion is coming from, but your statement about algebraic numbers having finitely many Galois conjugates sounds like you're thinking about first choosing a basis for $V$. That's fine, but it can hold you back a little when you get to the later chapters. If you like, the statement is essentially repeating the maxim that the open subgroups of a profinite group are the ones that are both closed and of finite index, applied to the continuous map $G_K \times V \to V$. Since $V$ has the discrete topology, $\{v\} \subset V$ is open... etc. $\endgroup$
    – stankewicz
    Sep 20, 2016 at 17:05
  • $\begingroup$ I do not oppose if the question is put on hold or removed. But even though it is not easy to see when it is posed with research purposes or not I think the understanding of unclear passages in texts commonly used also for research are legitimate, independently on the "score" of positive votes. I do always ask my questions on StackExchange before doing it here and only if no-one else succeeds in helping me before. They just did not or could not answer, while here Uri Bader and stankewicz kindly did it, helping me a lot. $\endgroup$
    – Hair80
    Sep 22, 2016 at 16:40
  • $\begingroup$ Hiar80, next time when you cross-post you should acknowledge that in both postings. I edited this into your question here, you should do it in MSE. Also, if you do post a question on MSE, please wait some before posting it also in MO, so in case of a need you could honestly say that the MSE community did not react. $\endgroup$
    – Uri Bader
    Sep 23, 2016 at 12:00
  • $\begingroup$ This comment regards the view of this post as "off topic". I disagree (even before realizing it was cross-posted). Truly, the post is about a clarification, not about deep math, but this kind of clarifications are "research level" necessities some times. I have much more sympathy to these than to the many "please prove my lemma, I'm too lazy" posts that I see being approved too often... $\endgroup$
    – Uri Bader
    Sep 23, 2016 at 12:09
  • $\begingroup$ Ok thanks I will do it next time. This is the second time I had no reaction to a question on MSE and still after 3 weeks for the first one. They have the right not to answer and to not knowing the answer as well as I have it to ask somewhere else after 2-3 days. I repeat, no problem if this post is eraced, just someone else may have the same doubt in future and have to ask again. Thanks a lot for your help. $\endgroup$
    – Hair80
    Sep 24, 2016 at 14:06

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To the question in the title: Not in general, e.g the action of (an infinite profinite group) $G$ on itself by left multiplication is continuous with infinite orbits.

However, note that the action of a profinite group on a discrete topological space is continuous iff the stabilizers are open (hence of finite index).

To the question in the body: It is implicitly assumed that $V$ is taken with the discrete topology.

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  • $\begingroup$ Could you tell me why opens of the stabilizers deduces the finiteness of index ? $\endgroup$
    – Duality
    Dec 30, 2021 at 14:09
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    $\begingroup$ Profinite groups are compact. The left cosets of a given open subgroup form an open cover. By compactness, there exists a finite subcover... $\endgroup$
    – Uri Bader
    Dec 30, 2021 at 15:04
  • $\begingroup$ Thank you, clearly understood. $\endgroup$
    – Duality
    Dec 30, 2021 at 15:48

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