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I guess the following is well-known (and probably follows from Chebotarev's density theorem, but I'm not very comfortable with it):

Define some notation:

  • $K$ a global field,
  • $G$ the absolute Galois group $\mathrm{Gal}(\bar{K}/K)$,
  • $G_{v}$ the decomposition group at a place $v$ of $K$,
  • $U,W$ two $\mathbb{Q}_{\ell}$-representations of $G$,
  • $V$ an extension: $0 \to U \to V \to W \to 0$ of $G$-rep's.

Suppose that for almost all (say: cofinitely many) places $v$ of $K$ the extension $V$ is split as representation of $G_{v}$. Can we deduce that $V$ is split as representation of $G$?

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up vote 16 down vote accepted

Definitely no. To see why, consider just the case where $W=\mathbb Q_\ell$ is the trivial representation (this is not really a loss of generality, because extensions of $W$ by $V$ are "the same thing" as extensions of $\mathbb Q_\ell$ by $W^\ast \otimes V$). Then extensions (as $G$-representations) of $\mathbb Q_\ell$ by $V$ are parametrized by $H^1(G,V)$. You're asking if such an extension which is split at almost all places $w$ (say all places $w$ not in a finite set $S$) is split. In other words, you're asking if the kernel $H^1_S(G,V)$ of $H^1(G,V) \mapsto \prod_{w \not \in S} H^1(G_w,V)$ is trivial.

The spaces $H^1_S(G,V)$ are examples of Selmer groups and while they are known to be finite dimensional, they are not always 0. In fact, computing their dimensions in general is still an open problem, essentially the characteristic zero part of the Bloch-Kato conjecture.

To give an example where $H^1_S(G,V)$ is not zero, let $K=\mathbb Q$, $V=\mathbb Q_\ell(n)$ with $n \neq 1$, $n$ odd (that is $V$ is the one-dimensional rep. where $G$ acts as the cyclotomic character to the power $n$) and $S$ a finite set of places containing $\ell$. Then it is known that $H^1(G,V)$ has dimension 1 in this case (to prove that it has dimension at least 1, which is enough for the argument, use Global Poitou-Tate Euler-Poincaré. For at most 1, you need deep result of Soulé). On the other hand, $H^1(G_w,V)$ has dimension $0$, as follows easily from local Euler-characteristic formula. Hence $H^1_S(G,V)$ has dimension 1. In other words, there is one (and only one) non-split extension of $1$ by $\mathbb Q_\ell(n)$, but it is split at all places $w \neq \ell$. That's a counter-example to your question. (It is interesting to note that the question whether this extension is split as $\ell$ or not is still open for positive $n$).

Let me add (paragraph removed after Kevin's comment)

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Thank you very much for your answer! It was not what I expected, and there is quite some mathematics in your answer that I do not know much about yet. So that is cool, because I get to learn new stuff! Thanks! – jmc May 1 '14 at 17:31
I'm confused about the last paragraph. If $L_p(\chi,s)$ has a zero away from $s=0$, doesn't this correspond exactly to a global extension which is split everywhere locally? – Kevin Ventullo May 1 '14 at 19:21
Hmm, Kevin, you're right. I was thinking, I believe, to motivic representation, where this phenomenon is (IIRC) not supposed to happen. Let me remove the last paragraph, which is not related to the initial question anyway. – Joël May 1 '14 at 19:46
@Joël — Could you please point me to some literature (Google turns up a lot of stuff about Selmer groups of ab.var.'s)? In particular, is anything/more known in the special case where $V$ is of weight $0$? – jmc May 26 '14 at 9:40
You can find related stuff in my expository paper I wrote for a graduate summer school in Hawaii, here: – Joël May 26 '14 at 12:47

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