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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 $\mathbb{Q}_p$. But the transition maps in this projective system are the identity (simply write down the commutative diagram containing the Kummer sequences for $n$ and $n+1$) so the projective limit is $0$ because there are no infinite $p$-th powers in $F_\infty$. I think this implies the triviality of cohomology also for $\mathbb{Z}_p(-1)$ by duality but I am not sure. If so, it would imply by the above your cup-product is zero also for the cyclotomic character (and hopefully for its powers?).

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 ;-)

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 $\chi=1$.

In the general case, by NSW, 1.5.3, cup-product is compatible with inflation and restriction, and by Inf-Res you are reduced to the case $$ H^1(\Delta,\mathbb{Z}_p(\chi))\times H^1(\Delta,\mathbb{Z}_p(\chi^{-1}))\xrightarrow{\cup} H^2(\Delta,\mathbb{Z}_p) $$ where $\Delta=G_K/\mathrm{Ker}(\chi)$ and now both $H^1$ vanish -- here is where I prefer $\chi$ to be of finite order.

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 $\mathbb{Q}_p$. But the transition maps in this projective system are the identity (simply write down the commutative diagram containing the Kummer sequences for $n$ and $n+1$) so the projective limit is $0$ because there are no infinite $p$-th powers in $F_\infty$. I think this implies the triviality of cohomology also for $\mathbb{Z}_p(-1)$ by duality but I am not sure. If so, it would imply by the above your cup-product is zero also for the cyclotomic character (and hopefully for its powers?).

I think your cup-product is always zero independent of Leopoldt. 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 $\chi=1$.

In the general case, by NSW, 1.5.3, cup-product is compatible with inflation and restriction, and by Inf-Res you are reduced to the case $$ H^1(\Delta,\mathbb{Z}_p(\chi))\times H^1(\Delta,\mathbb{Z}_p(\chi^{-1}))\xrightarrow{\cup} H^2(\Delta,\mathbb{Z}_p) $$ where $\Delta=G_K/\mathrm{Ker}(\chi)$ and now both $H^1$ vanish.

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 $\chi=1$.

In the general case, by NSW, 1.5.3, cup-product is compatible with inflation and restriction, and by Inf-Res you are reduced to the case $$ H^1(\Delta,\mathbb{Z}_p(\chi))\times H^1(\Delta,\mathbb{Z}_p(\chi^{-1}))\xrightarrow{\cup} H^2(\Delta,\mathbb{Z}_p) $$ where $\Delta=G_K/\mathrm{Ker}(\chi)$ and now both $H^1$ vanish -- here is where I prefer $\chi$ to be of finite order.