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Let $W_p$ be the Weil group of $\mathbf{Q}_p$. What is the Galois cohomology group $H^2(W_p,\mathbf{C}^{\times} )$ (with trivial action)? Is it zero, or something huge and complicated?

(This group comes up, at least for me, when you want to compare two Weil group representations whose projectivizations agree.)

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Just to nitpick a bit: the Weil group $W_p$ is not a profinite group, so "the Galois cohomology group $H^2(W_p,\mathbb{C}^{\times})$" is not well-defined. I believe though that this group can be made sense of in some reasonable way, and Marty's answer addresses this. –  Pete L. Clark Jan 18 '11 at 9:20

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It is known that $H^2(W, C^\times)$ is trivial, when $W$ is the Weil group of a global or local field, with the trivial action on $C^\times$, and the cohomology is taken in the sense of Moore (measurable cochains). This is the main result of C.S. Rajan, "On the vanishing of the measurable cohomology groups of Weil groups", Compositio Math. 140 (2004) 84-98 (also easy to find online).

Is this well-known? I don't know. I only found this paper recently, and hopefully now it will become better-known!

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Beautiful, thanks! –  David Hansen Jan 17 '11 at 2:57
    
As is explained in the introduction to Rajan's paper, Tate had proved the analogous result for Galois groups instead of Weil groups. –  Chandan Singh Dalawat Jan 18 '11 at 4:56

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