# Leopoldt's conjecture and cup-products

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.

-
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
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

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 ;-)
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