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.
 A: 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 ;-)
