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We consider the unipotent radical $N$ of the Borel subgroup $B$ of $\operatorname{GL}_2(F)$ where $F$ is a local field. Let $\phi$ be a character of the maximal split torus $T$. We inflate $\phi$ to $B$ and consider its smooth induction $\operatorname{Ind}_B^G \phi = (\Sigma, X)$ to $G$ (here $X$ is the space of $G$-smooth functions $f:G \rightarrow \mathbb{C}$ satisfying $f(bg)=\phi(b)f(g)$ for all $b \in B$, $g \in G$, and $\Sigma$ acts on this space via right translation).

Let $V=\ker \alpha_\phi$, where $\alpha_\phi:X \rightarrow \mathbb{C}$ is the surjective map given by $f \mapsto f(1)$. Since $V$ is a representation of $B$, it is also a representation of $N$. We denote the space of locally constant functions $\varphi:N \rightarrow \mathbb{C}$ having compact support by $C_c^\infty(N)$. This space carries a representation of $N$ given by right translation. My question is the following:

Given $f \in V$, define $f_N \in C_c^\infty (N)$ by $f_N(n)=f(\omega n)$. Show that the map $f \mapsto f_N$ is an $N$-isomorphism.

So far, I have shown that indeed $f_N$ as defined above is in $C_c^\infty(N)$, that the map is linear and commutes with the $N$-action. However I've been unable to construct an inverse. How would one do this, or can the bijection be shown in another way?

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  • $\begingroup$ What does $G$-smooth mean (as opposed to smooth)? What is $\omega$? $\endgroup$
    – LSpice
    Mar 23, 2022 at 1:47
  • $\begingroup$ (on mobile, apologies for lack of format) a function is G-smooth if there is a compact open subgroup K of G such that f(gk)=f(g) for all g in G, k in K. Omega is the 2×2 permutation matrix (0,1;1,0). $\endgroup$
    – carraig
    Mar 23, 2022 at 1:50
  • $\begingroup$ I think that would usually just be called just "smooth" ($G$ is already there, as the domain), but maybe you would like to have this terminology to accommodate a general space which we might equip with actions of several different groups. $\endgroup$
    – LSpice
    Mar 23, 2022 at 1:52

1 Answer 1

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The key point is that $\operatorname{GL}_2(F)$ equals $B \sqcup B\omega N$, and the multiplication map $B \times N \to B\omega N$ is a homeomorphism in the analytic topology. Thus, given $\varphi \in \operatorname C_\text c^\infty(N)$, we can define $\varphi^N \in V$ by extending $b\omega n \mapsto \phi(b)\varphi(n)$ by $0$ to all of $\operatorname{GL}_2(F)$. The resulting map $\varphi \mapsto \varphi^N$ is inverse to $f \mapsto f_N$.

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  • $\begingroup$ I'm not sure what is meant by U here, could you clarify that? And could you elaborate on "analytic isomorphism", I'm not familiar with that in this context. $\endgroup$
    – carraig
    Mar 23, 2022 at 2:00
  • $\begingroup$ $U$ should have been your $N$; sorry, fixed. The analytic topology is the word I'm used to for the topology coming from the valued-field topology on $F$, hence on $\operatorname{GL}_2(F)$ (and then on its subgroups) as an open subset of $\operatorname M_{2\times2}(F) = F^4$; and I used "analytic isomorphism" somewhat archaically to mean a homeomorphism in that topology, but it's confusing since it's not an isomorphism of groups, so I changed it to just "homeomorphism in the analytic topology". $\endgroup$
    – LSpice
    Mar 23, 2022 at 2:14
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    $\begingroup$ I'm not entirely sure I understand your construction. Most of all, how would $f_N(b)$ be defined, as $b$ is not necessarily in $N$. Besides this, do I understand correctly that given $\varphi \in C_c^\infty(N)$ (you say "given $f_N$", by this do you mean an arbitrary element, or the image of some $f \in V$?) and define its extension $\varphi'$ to $G$ by $\varphi'(g)=0$ if $g \in B$ and $\varphi'(g)=\varphi(b)$ if $g \in N\omega B$. Then the claim is that this provides an inverse to the map $f \mapsto f_N$? $\endgroup$
    – carraig
    Mar 24, 2022 at 12:49
  • $\begingroup$ Your objection can be handled by extending $f_N$ appropriately to $B$, but a simpler way is to present the big cell as $B\omega N$ rather than $N\omega B$; I have done so. Yes, you are right that I am writing $f_N$ for an arbitrary element of $C_c^\infty(N)$, and so that it would be clearer to denote it by some other symbol $\varphi$; I have changed this. Then the map $\varphi \mapsto \varphi^N$, say, is inverse to $f \mapsto f^N$. $\endgroup$
    – LSpice
    Mar 24, 2022 at 13:01
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    $\begingroup$ I can verify that this provides an inverse, but I run into trouble showing $\varphi^N \in V$, particularly, that $\varphi^N(bg)=\phi(b)\varphi^N(g)$ holds for all $b \in B, g \in G.$ If $g \in B$, then both sides of the above equality equal 0. If $g \in B\omega N$ so $g=b'\omega n,$ then I find $\varphi^N(bg)=\varphi^N(bb'\omega n)=\varphi(n)$ whereas $\phi(b)\varphi^N(g)=\phi(b)\varphi(n)$. Since $\phi$ may be taken to be a non-trivial character, $\phi(b) \neq 1$ for some $b$. This would imply $\varphi^N \not\in X$, so I assume I've erred somewhere. Can you see where? $\endgroup$
    – carraig
    Mar 24, 2022 at 14:23

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