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Among the many equivalent formulations of Leopoldt's conjecture, this one is probably the shortest: For any number field $K$, prime number $p$, finite set $S$ of primes of $K$ containing the primes above $p$, one has

Leopoldt's conjecture: $H^2(G_{K,S},\mathbb{Q}_p)=0$.

Here $G_{K,S}$ is as usual the Galois group of the maximal algebraic extension of $K$ un ramified outside $S$ and places at infinity, the $H^2$ is continuous cohomology.

Now, one of the most natural way to get a class in an $H^2$ is as a cup-product of two classes in an $H^1$. For example, if $\chi : G_{K,S} \rightarrow Q_p^\ast $ is a continuous character, then there is a cup-product map $$H^1(G_{K,S},\chi) \times H^1(G_{K,S},\chi^{-1}) \rightarrow H^2(G_{K,S},\mathbb{Q}_p),$$ which, according to Leopoldt's conjecture, should be zero.

Is it any easier to prove that the above morphism is zero than to prove Leopoldt's conjecture itself ?

I would also be interested to know the answer in special cases (of $K$, $\chi$, $p$) where Leopoldt's conjecture is not known.

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Shouldn't the $\mathbb{Q}_p$ be replaced by $\mathbb{Q}_p/\mathbb{Z}_p$ in the statement of Leopoldt's conjecture? So that the statement should be $H^{2}(G_{K,S},\mathbb{Q}_p/\mathbb{Z}_p)=0$? –  Henri Johnston Feb 6 '12 at 20:56
    
@Henri: Hum, I do think the statement I wrote is equivalent to Leopoldt's conjecture. Actually, your statement and mine are equivalent if $p>2$, or if $K$ is totally complex. This is easy to see since the the $p$-cohomological dimension of $G_{K,S}$ is at most 2 under these hypotheses (Cf e.g. Jannsen, On the l-adic cohomology of varieties over number field and its Galois cohomilogy, in Galois groups over Q, Lemma 1). For the somewhat anecdotical reminding cases, I am not sure. –  Joël Feb 6 '12 at 21:43
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According to Neukirch–Schmidt–Wingberg Corollary 10.3.10, the $\mathbf{Z}_p$-rank of $H^2(G_{K,S},\mathbf{Z}_p)$ equals the Leopoldt defect. This rank is the $\mathbf{Q}_p$-dimension of $H^2(G_{K,S},\mathbf{Q}_p)$, so Joël's formulation is correct (at least away from $p=13$, right Joël?). Shouldn't this be the same as $H^2(G_{K,S},\mathbf{Q}_p/\mathbf{Z}_p)$ being finite? –  Rob Harron Feb 7 '12 at 0:59
    
@Rob: Yes. The point is that $H^2(G_{K,S},\mathbb{Q}_p/\mathbb{Z}_p)$ is the dual of $H_2(G_{K,S},\mathbb{Z}_p)$ and the $\mathbb{Z}_p$-Rank of the later is precisely the Leopoldt defect. –  Guillermo Mantilla Feb 7 '12 at 3:25
    
@Rob: Right. And this is also equivalent to $H^2(G_{K,S},\mathbb{Q}_p/\mathbb{Z}_p)=0$ when $p>2$ as I have said, and actually, even when $p=2$. Look at the long exact sequence of cohomology attached to the short exact sequence $0 \rightarrow \mathbb{Z}_p \rightarrow \mathbb{Q}_p \rightarrow \mathbb{Q}_p / \mathbb{Z}_p \rightarrow 0$. One gets a surjection $H^2(G_{K,S},\mathbb{Q}_p) \rightarrow H^2(G_{K,S},\mathbb{Q}_p/\mathbb{Z}_p) \rightarrow H^3(G_{K,S},\mathbb{Z}_p)$ but the latter is $0$ in any case, even when $p=2$. So my formulation and Henri's are equivalent. –  Joël Feb 7 '12 at 3:59

1 Answer 1

I think your cup-product is always zero independently of Leopoldt (at least if $\chi$ is of finite order). Consider $\mathbb{Z}_p$-coefficients instead (enough by Neukirch-Schmidt-Wingberg, 2.3.10). If $\chi=1=\chi^{-1}$, then both $H^1$ degenerate to $H^1(\tilde{K}/K,\mathbb{Z}_p)$ where $\tilde{K}$ is the compositum of all $\mathbb{Z}_p$ extensions, because cocyles are Hom's. In particular, your cup-product factors through $$ H^1(\tilde{K}/K,\mathbb{Z}_p)\times H^1(\tilde{K}/K,\mathbb{Z}_p)\xrightarrow{\cup}H^2(\tilde{K}/K,\mathbb{Z}_p) $$ but the $H^2$ is trivial because a free $p$-group has $p$-cohomological dimension $1$ (and NSW, 2.3.5 tells you that cohomology with $\mathbb{Z}_p$-coefficients is the projective limit of those with $\mathbb{Z}/p^n$-coefficients). So, your cup-product is zero if $\chi=1$.

Now suppose $\chi$ is of non-trivial and set $\Delta=G_K/\mathrm{Ker}(\chi)$.
Assume $H^i(\Delta,\mathbb{Z}_p(\chi^{\pm 1}))=0$ for $i=1,2$
Then $H^1(G_{K,S},\mathbb{Z}_p(\chi^{\pm 1}))=H^1(G_{L,S},\mathbb{Z}_p)^{\Delta}$ by Hochschild-Serre, where $L$ is the field attached to $\chi$. Since cup-product is compatible with restriction by NSW 1.5.3, it is compatible with the above identification and coincides with $$ H^1(\tilde{L}/L,\mathbb{Z}_p)^\Delta\times H^1(\tilde{L}/L,\mathbb{Z}_p)^\Delta\xrightarrow{\cup}H^2(\tilde{L}/L,\mathbb{Z}_p)^\Delta=0 $$ by the $\chi=1$ case.
Now when is $\Delta$-cohomology trivial? If $\chi$ is of finite order, its order is prime to $p$ and we win. If $\chi$ is the cyclotomic character, then $H^2(\Delta,\mathbb{Z}_p(\pm 1))=0$ because $cd_p(\Delta)=1$. For $i=1$, by Kummer theory we see that $H^1(\Delta,\mathbb{Z}_p(1))=\varprojlim F_\infty^{\times}/(F_\infty^{\times})^{p^n}$, where $F_\infty$ is the cyclotomic extension of $\mathbb{Q}_p$...and I do not know what to do ;-)

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Dear Filippo, thank you very much. Actually I had arrived at the same conclusion a few weeks ago (with my student Yu Fang) that for a finite order character the cup product was always zero (with a proof very close to yours). But the method does not seem to generalize to infinite-order characters. For example, what about $\chi=$ cyclotomic character ? I don't know how to do it, nor any infinite-order character for that matter. –  Joël Apr 8 '12 at 16:51
    
Dear Joël, for the cyclotomic character you get 0 again, I guess. My argument shows that you reduce to Gamma-cohomology, and H^2 is zero bacuase cd(Gamma)=1. I actually think this restriction is not severe, you might always write the Galois group on which chi acts faithfully as product of something free (of dim. 1) and something finite. Then both H^2 with Z_p-coefficients are trivial. Or am I wrong? –  Filippo Alberto Edoardo Apr 9 '12 at 0:27
    
Dear Filippo, I am not sure I completely get the argument yet. Already in your answer, second part, how do you prove that you can reduce to the case of Delta ? –  Joël Apr 9 '12 at 13:34
    
Dear Joël, I see the issue with Inf-Res that I thought would have been trivial – I am sorry. I will try to write down a proper proof, if I can and come back if I have it. –  Filippo Alberto Edoardo Apr 9 '12 at 16:20

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