Timeline for N-coinvariants of $\mathcal{C}^{sm}_c(N,A)$ (compactly supported smooth functions on $N$) ?

Current License: CC BY-SA 3.0

9 events

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Oct 2, 2012 at 10:39	history	edited	Arkandias	CC BY-SA 3.0	deleted 8 characters in body
Jun 3, 2012 at 10:50	vote	accept	Arkandias
Jun 3, 2012 at 14:32
Jun 2, 2012 at 3:26	answer	added	Filippo Alberto Edoardo		timeline score: 1
May 31, 2012 at 12:14	comment	added	David Loeffler		An $N$-invariant linear functional is the same as a linear functional that factors through the coinvariants.
May 31, 2012 at 11:33	history	edited	Arkandias	CC BY-SA 3.0	added 2 characters in body; edited title
May 31, 2012 at 11:32	comment	added	Arkandias		Thank you for you answer. But I am not sure to understand it... Why would I look for a translation-invariant linear functional ? Maybe my question is a bit messy, I'll try to rewrite it properly.
May 31, 2012 at 7:15	comment	added	David Loeffler		... And if this works, then you are done, since any compact subset of $N$ is contained in a compact open subgroup.
May 31, 2012 at 7:15	comment	added	David Loeffler		Here's a random thought, just off the top of my head. You're looking for a translation-invariant linear functional $\phi$ on $C^{sm}_c(N, A)$. Let $N_0$ be an open compact. By restriction, $\phi$ defines a linear functional on $C^{sm}(N_0, A)$, which is evidently $N_0$-invariant. Now $C^{sm}(N_0, A)$ is just the reduction modulo $\varpi^n$ of $C^{cts}(N_0, \mathcal{O}_E)$, and the continuous linear functionals on this are the Iwasawa algebra of $N_0$, which is a rather nice and explicit ring. By doing this, perhaps you can show that the restriction of $\phi$ to $C^{sm}(N_0, A)$ is zero?
May 31, 2012 at 0:59	history	asked	Arkandias	CC BY-SA 3.0