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Let $F$ be a local field of characteristic zero (for simplicity), $\overline{F}$ an algebraic closure of $F$ and $L/F$ a fixed finite Galois extension. If $G$ is a linear algebraic group defined over $F$, then the Galois cohomology group $H^1(F,G)$ can be defined as a direct limit of $H^1(K/F,G)$, where $K$ runs through finite Galois subextensions of $\overline{F}$.

Now the question is: under what conditions is this direct limit just $H^1(L/F,G)$?

I guess there might be restrictions on both $L$ and $G$.

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4 Answers 4

up vote 13 down vote accepted

As Hunter and Sean noted, since the inflation map ${\rm{H}}^1(L/F,G(L)) \rightarrow {\rm{H}}^1(F,G)$ is injective and ${\rm{H}}^1(F,G)$ is always finite (Borel-Serre), such an $L$ always exists. Below we give an explicit sufficient condition on $L$ (often satisfied) when $G$ is connected. (One could probably do better with a closer consideration of the Tate local duality aspects of the argument. I am lazy at this step.) This rests on something deeper than the Borel-Serre result: the Kneser-Bruhat-Tits theorem on vanishing of degree-1 Galois cohomology for simply connected semisimple groups over non-archimedean local fields.

First we set up some notation. Let $U = \mathcal{R}_u(G)$ denote the unipotent radical of $G$ (this is a good notion over $F$ since $F$ is perfect), and let $G' = G/U$ denote the maximal reductive quotient. The identity component $Z'$ of the center of $G'$ is an $F$-torus, and the derived group $\mathcal{D}(G')$ is a semisimple group, say with simply connected central cover $\mathcal{G} \twoheadrightarrow \mathcal{D}(G')$. The preimage $\mu$ of the central subgroup $Z' \cap \mathcal{D}(G')$ is a finite $F$-group of multiplicative type. (It is the kernel of the central covering map if $Z' = 1$.)

Proposition: Assume $G$ is connected and use notation as above. Let $F'/F$ be a finite Galois splitting field for $G'$ (thus for $Z'$ and dual of $\mu$). If $L/F$ is a finite Galois extension containing $F'$ with $[L:F']$ divisible by the order of $\mu$ then ${\rm{H}}^1(F,G) = {\rm{H}}^1(L/F,G(L))$.

In particular, if $G$ is a split connected reductive $F$-group then $L/F$ works provided that $[L:F]$ is divisible by the order of the center of the simply connected central cover of $\mathcal{D}(G)$.

Remark: If $T$ is a maximal $F$-torus in $G$ then it maps isomorphically onto one for $G'$, so could take $F'/F$ to be splitting field for $T$.

Proof: Since $F$ has characteristic 0, the quotient map $G \twoheadrightarrow G'$ admits a section $\sigma$ over $F$, which is to say there exists a connected reductive $F$-subgroup $H \subseteq G$ such that $H \ltimes U \simeq G$ via multiplication (this is a so-called Levi $F$-subgroup of $G$); beware that over any algebraically closed field $k$ with nonzero characteristic Levi subgroups can fail to exist. (A basic natural counterexample is, loosely speaking, ${\rm{SL}} _2(W _2(k))$ as a $k$-group, where $W _2$ denotes length-2 Witt vectors; see Appendix A.6 in the book "Pseudo-reductive groups".)

Using $\sigma$ (or $H$), the natural restriction map ${\rm{H}}^1(F,G) \rightarrow {\rm{H}}^1(F,G')$ is surjective. It is also injective. Indeed, a standard twisting argument (as explained in Serre's book on Galois cohomology) identifies fibers with ${\rm{H}}^1(F,U')$ for various $F$-forms $U'$ of $U$. But since the ground field is perfect, every smooth connected unipotent group is split (i.e., admits a composition series whose successive quotients are $\mathbf{G}_a$) and hence has trivial ${\rm{H}}^1$. Thus, we get the asserted bijectivity. So far we have not used anything about $F$ other than that it has characteristic 0.

We likewise have ${\rm{H}}^1(E/F,U'(E)) = 1$ for any smooth connected unipotent $F$-group $U'$ and any Galois extension $E/F$, so the same argument gives that ${\rm{H}}^1(E/F,G(E)) \rightarrow {\rm{H}}^1(E/F,G'(E))$ is bijective for any Galois extension $E/F$. Thus, the injective inflation map $${\rm{H}}^1(E/F,G(E)) \rightarrow {\rm{H}}^1(F,G)$$ is bijective if and only if the same holds for $G'$ in place of $G$.

Consider the central extension structure $$1 \rightarrow \mu \rightarrow Z' \times \mathcal{G} \rightarrow G' \rightarrow 1$$ over $F$, where the second map uses multiplication. By Kneser-Bruhat-Tits, we have an exact sequence of pointed sets $${\rm{H}}^1(F,Z') \rightarrow {\rm{H}}^1(F,G') \rightarrow {\rm{H}}^2(F,\mu)$$ and similarly over $L$, with a commutative diagram using restriction maps.

By local class field theory, the restriction map from ${\rm{H}}^2(F',\mu)$ to ${\rm{H}}^2(L,\mu)$ vanishes, so likewise for ${\rm{H}}^2(F,\mu) \rightarrow {\rm{H}}^2(L,\mu)$. (This is weak; by translating restriction through Tate local duality we can surely give a better sufficient "lower bound" for such an $L$.) For any such $L$, it follows that restriction from ${\rm{H}}^1(F,G')$ to ${\rm{H}}^1(L,G')$ lands in the image of ${\rm{H}}^1(L,Z')$. But this vanishes since the $F$-torus $Z'$ is even $F'$-split, let alone $L$-split. Thus, ${\rm{H}}^1(F,G') \rightarrow {\rm{H}}^1(L,G')$ vanishes, which is to say that $L$ "works" for $G'$, so the same holds for $G$. QED

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Thinking about this sort of question was always a bit tricky for me since going from a larger field (like $E$) to a smaller field (like $L$) reduces the number of cocycles and the number of coboundaries simultaneously. Some cohomology classes over the larger field could disappear entirely, some might merely get smaller, and some might split into multiple distinct classes (two of their cocycles might require a "coboundary" with values in the larger field).

If you wanted to go the route of checking whether the obstruction that occurs in the inflation/restriction sequence is trivial for a particular $L$, it might be helpful to think about this generalization of Tate-Nakayama duality by Kneser and then Kottwitz. It tells you precisely what $H^1(L,G)$ when $L$ is a characteristic $0$ local field but it only works for reductive linear algebraic groups. See the paper "Stable trace formula: cuspidal tempered terms" by Kottwitz (IIRC). The punchline is $H^1(L,G) = (X^{*}(Z(\hat{G}))_{Gal(\bar{L}/L)})_{tor} $ where all those things are character group, center, dual group, coinvariants, and torsion subgroup.

Is it possible that what you really want to do is pick one finite extension $L/F$ and say that every element of $H^1(F, G)$ can be represented by a cocycle that is inflated from a cocycle $Gal(L/F) \rightarrow G(L)$? If so, then you just have to note that $H^1(F, G)$ is a finite set for any linear algebraic group $G$ and take the composite of the fields required to define some arbitrary system of representatives.

As a general reference, the book "Algebraic groups and Number Theory" by Platonov and Rapinchuk has TONS of useful information about Galois cohomology of linear algebraic groups.

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@Brian Conrad: The sequence $$1 \rightarrow \mu \rightarrow Z' \times \mathcal{G} \rightarrow G' \rightarrow 1$$ is not exact in general. Namely, the kernel $\nu$ of the map $Z' \times \mathcal{G} \rightarrow G'$ can be bigger than $\mu$. For example, if $G=G'=GL_n$, then $\mathcal{G}= \mathcal{D}(G')=SL_n$, hence $\mu=1$, while $\nu=\mu_n$, the group of roots of unity of order dividing $n$.

Excuse me for posting this as an answer. As a new user, I am not permitted to post a comment.

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Mikhail, thanks for pointing this out. I have now corrected the definition of \mu to address this. –  BCnrd Feb 21 '10 at 23:12

By the inflation-restriction exact sequence this will be true when $H^1(L, G)^{\text{Gal}\left(L/F\right)}$ is trivial. The superscript "Gal(L/F)" there means to take invariants under an action you define by hand on $H^1(L, G)$. [EDIT: An old version of this post said "precisely when," which is not correct. Thanks!]

This may not be helpful, since it sounds like your point is that you want to avoid working with direct limits. But I'd do that just by requiring that cochains be continuous.

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Thanks Hunter, I would like a way without calculating limits. –  user1832 Jan 11 '10 at 4:50
    
@Hunter: I see that the triviality of H^1(L,G)^{Gal(L/F)} is sufficient, but why is it necessary? –  Bjorn Poonen Jan 11 '10 at 5:54

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