# 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 $V$?

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?

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If $V'$ is non-empty, you can take the indiscrete topology. I'm not sure I understand the question. – Qiaochu Yuan Aug 4 '11 at 15:11
@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, – Matt E Aug 4 '11 at 15:56
@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. – Qiaochu Yuan Aug 4 '11 at 16:05
Dear 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, – Matt E Aug 5 '11 at 0:53
@Math E:Thanks for kind reminding.Green hand here,I try to do better next time. – user14242 Aug 5 '11 at 4:37

Suppose that $V$ is a complete locally convex $\mathbb C$-vector space with a continuous action of $G := GL_n(\mathbb Q_p)$. Then I believe that the smooth vectors $V'$ are dense in $V$.

On the other hand, suppose that we take $V = \mathbb C[GL_n(\mathbb Q_p)]$ (i.e. the group ring of $G$ as an abstract group), with its left regular action. Then $V'$ is quite a bit smaller than $V$, and I don't see that there is a sensible choice of topology to make $V'$ dense in $V$.

In the abstract, I don't see anything sensible to say. My experience in practice, though, is that $V$ will normally come equipped with a complete locally convex structure (e.g. it might be $L^p$-functions on $G$, or some kind of distributions on $G$), and then the result of my first paragraph applies. Do you have some particular context in mind when $V$ arises without any natural topology given?

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 Thanks for help.I think (I am not shure of it) that V may be a subspace of the space of locally constant,cptly supported functions from G to $\mathbb{C}$.I am not very clear what repn we should consider here,either. – user14242 Aug 4 '11 at 23:34 @user14242: If $V$ is contained in the compactly supported locally constant functions (the latter being equipped with its left regular action, say) then $V = V'$. Regards, – Matt E Aug 5 '11 at 0:52

In brief, "yes", in a repn of a totally disconnected topological group on a locally convex, quasi-complete (complex) topological vector space, smooth vectors are dense.

(This includes unitary repns of p-adic reductive groups, and many other natural repns on spaces of functions.)

For some reason, it is unfortunately common for sources to say that repn spaces for p-adic groups "have no topology" or "have the discrete topology". NB, the latter is not a topological vector space topology. The apparent fact that this hasn't wrecked everything is that admissible smooth repns in fact have "the finest" TVS topology, namely, the (locally convex) colimit of finite-dimensional subspaces. However, larger repn spaces won't have this topology.

But when we've recovered from that misdirection, just as Garding's argument, smooth vectors are dense, because we make an approximate identity $u_i$, and the integrated action $u_i\cdot v=\int_G u_i(g)\cdot v\;dg$ gives smooth vectors which approach $v$ in the original topology. (The integral, being compactly supported and continuous, converges nicely, e.g., in a Gelfand-Pettis/"weak" sense.)

Edit: in response to request for saying more about the topologies of (presumably, naturally-arising) spaces. For example, for $G=GL_n(\mathbb Q_p)$ or such, we could look at $V=L^2(G)$. This has a natural Hilbert-space topology, and Urysohn's lemma proves that $C^o_c(G)$ is dense, which by a two-or-three epsilon argument proves that $G\times V\rightarrow V$ by (e.g., right) translation is continuous, fulfilling topological conditions for a repn.

One should be aware that one should not find oneself needing to "define" a topology on a space of functions that appears in nature, but, rather, be able to see that there is at least one natural topology that makes sense. If one fails to see that a given function-space has a natural topology it means that one fails to understand that class of functions...

So, in general, for any topological group $G$, whenever $V$ is a reasonable TVS (locally convex, quasi-complete), and $G\times V\rightarrow V$ is continuous, whatever class of functions one has on $G$ that can give "approximate identities" ... and so on... produces "smooth/whatever-adjective" functions in the repn space, by the argument indicated.

Square-integrable spaces of functions on "physical" spaces are the most frequent real-life examples. Spaces of functions with additional conditions are just variants... For example, for a discrete subgroup $\Gamma$ of $G=SL_n(\mathbb R)\times SL_n(\mathbb Q_p)$, such as $\Gamma=SL_n(\mathbb Z[1/p]$, the space of "automorphic" functions $V=L^2(\Gamma\backslash G)$ is a reasonable thing to consider. Or "rapidly-decreasing" on $\Gamma\backslash G$. Or "Schwartz functions", ...

Really, all the usual intuitive conditions-on-functions.

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 Thanks for help.Can you say some more about the spaces V we consider in the repn of redutive groups over $\mathbb{Q}p$ and the topology of them? I am not very clear about this. – user14242 Aug 4 '11 at 23:38 Great!It really helps me. – user14242 Aug 5 '11 at 4:38