Here V is a complex vector space.We define Stab(v) to be the subgroup H of $GL_n(\mathbb Q_p)$ consisting of elements g such that $f(g)v=v$ (preserving v), V':={v\in V|Stab(v) is open in G}. Is it possible to give V some topology such that V' is dense in V?
If $(f,V)$ is a nonzero representation of the reductive group $GL_n(\mathbb Q_p)$, is it true that the smooth subrepresentation $V'$ is dense in $V$?
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1If $V'$ is non-empty, you can take the indiscrete topology. I'm not sure I understand the question. – 2011-08-04
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0@Qiaochu: Dear Qiaochu, There are various theorems that say that for a continuous rep. of $GL_n(\mathbb Q_p)$ on a topological complex vector space $V$, the smooth vectors are dense. (This is true if e.g. $V$ is a topologically irreducible unitary rep. on a Hilbert space, I believe.) The OP wants to know if his abstract $V$ can be placed in such a context. Regards, – 2011-08-04
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1@Matt: thanks for the clarification. As you say in your answer, it seems to me that in practice $V$ will usually come with a topology, so the question seemed a little strange. – 2011-08-04
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0Dear user14242, Given your comments on the two answers below, it might help if you post a more specific question, perhaps with an indication of what you are reading or what precise problem you are trying to solve. Regards, – 2011-08-05
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0@Math E:Thanks for kind reminding.Green hand here,I try to do better next time. – 2011-08-05