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Let $G=Gal(\bar{\mathbb Q}/{\mathbb Q})$ be the absolute Galois group of the rationals. Fix two continuous group homomorphisms $\alpha,\beta: G\to {\mathbb Q}_l^\times$, where $l$ is a prime and ${\mathbb Q}_l$ the field of $l$-adic numbers. Let $H(\alpha,\beta)$ be the ${\mathbb Q}_l$ vector space of all continuous maps $\chi:G\to{\mathbb Q}_l$ such that $$ \omega=\left(\begin{array}{cc}\alpha&\chi\\\ \ &\beta\end{array}\right) $$ is a group homomorphism.

My question is: is the space $H(\alpha,\beta)$ one-dimensional?

I can prove this if $\alpha=\beta$ and for the proof one can assume that $\beta=1$. In this case the defining relation for $H(\alpha,\beta)$ is $\chi(xy)=\alpha(x)\chi(y)+\chi(x)$.

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up vote 16 down vote accepted

You're asking if $Ext^1_G(\alpha,\beta)$ is one-dimensional. The short answer is no, yet there are many cases where the answer is yes. Actually, the dimension of this group is not known in all cases, and when it is known, in general it is by a deep theorem, not a trivial computation.

To be more precise, we have $Ext^1_G(\alpha,\beta) = Ext^1_G(1,\alpha^{-1} \beta) = H^1(G,\alpha^{-1} \beta)$ (continuous group cohomology). Setting $\chi=\alpha^{-1} \beta$ one is reduced to compute $H^1(G,\chi)$.

The answer will depend mainly on the sign of $\chi$, that is $\chi(c)$ where $c$ is a complex conjugacy in $G$. We say that $\chi$ is even if $\chi(c)=1$, odd otherwise.

Also, let us assume for simplicity (1) that there is no prime $p$ where you character $\chi$ restricted to $Gal(\overline{ \mathbb Q_p} /\mathbb Q_p)$ is equal to the cyclotomic or to the trivial character.

Then the generic answer is $H^1(G,\chi)$ has dimension $1$ if $\chi$ is odd, $0$ if $\chi$ is even. But there are exceptions. It is always true, and easy to prove, that the dimension of $H^1(G,\chi)$ is at least its expected value.

More precisely:

(a) Fixing a prime number $l$ which is regular in the sense of Kummer, then it is always true that $H^1(G,\chi)$ is of dimension 1 or 0 according to the parity. This follows from Mazur-Wiles's proof of the main conjecture of Iwasawa, and the fact that the $l$-adic zeta function has no zero when $l$ is regular.

(a') For a fixed $l$, which is non-regular, then the same result will be true except for a finite, non-zero, number of exceptions, where $dim \ H^1(G,\chi)$ will be higher. (same argument, using the fact that the $l$-adic zeta function has finitely many zero in any case, being an Iwasawa function, but has at least one zero, since $l$ is not regular).

(b) When $\chi = \chi_{cycl}^n \epsilon$, where $n$ is an integer, $\epsilon$ a finite order character, then it is expected that $H^1(G,\chi)$ has dimension $0$ or $1$ according to the parity. This is even a theorem except if $n$ is negative and $\chi$ even (and hypothesis (1) above is satisfied, that is $n \neq 0,1$). This is a deep result of Soulé combined with some lemmas of Bloch-Kato.

As far as I know, these are the only known results.

In the cases we have left aside (excluded by our hypothesis (1)), the answer may be more complicated. For instance $H^1(G,\chi_{cycl})$ is of infinite dimension, as is easy to see using the Kümmer's exact sequence. For a more thorough discussion, please see my lecture notes on the Bloch-Kato conjecture for the Hawaii 2009's conference, available on my web page.

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@Joel: thanks for this great answer! – Anton Deitmar Jun 10 '13 at 20:00

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