Let $K$ be a $p$-adic field (a finite extension of the field of $p$-adic numbers ${\mathbb Q}_p$).
Let $T$ be a $K$-torus with character group $X={\sf X}^*(T)$ and cocharacter group $Y={\sf X}_*(T)=X^\vee$.
I would like to have explicit cocycles for all Galois cohomology classes in $H^1(K,T)$.
We can compute $H^1(K,T)$ via double duality.
The cup product pairing defines an isomorphism
$$ H^1(K,T)\overset\sim\longrightarrow {\rm Hom}\big(H^1(K,X),{\Bbb Q}/{\Bbb Z}\big),$$
where I write $H^1(K,X)$ for $H^1({\rm Gal}(\overline K/K),X)$. See Milne, Arithmetic Duality Theorems, Corollary I.2.3.
Let $L/K$ be a finite Galois extension of degree $d$ splitting $T$.
We have
$$H^1(K,T)=H^1(\Gamma_{L/K}, Y\otimes_{\Bbb Z} L^\times),$$
where $\Gamma_{L/K}={\rm Gal}(L/K)$.
The finite group $\Gamma_{L/K}$ acts on $X$ and $Y$, and we have
$$H^1(K,X)=H^1(\Gamma_{L/K}, X).$$
The canonical pairing of $\Gamma_{L/K}$-modules
$$ Y\times X\to {\Bbb Z}$$
induces an isomorphism
\begin{align*}
H^1(\Gamma_{L/K}, X)&\overset\sim\longrightarrow
{\rm Hom}\bigg(H^{-2}\big(\Gamma_{L/K},\, Y\otimes_{\Bbb Z} ({\Bbb Q}/{\Bbb Z})\,\big),
\ {\Bbb Q}/{\Bbb Z}\bigg)\\
&\overset\sim\longrightarrow {\rm Hom}\big(H^{-1}(\Gamma_{L/K}, Y),{\Bbb Q}/{\Bbb Z}\big);
\end{align*}
see Brown, Cohomology of Groups, Corollary VI.7.3.
Thus we obtain an isomorphism
\begin{equation*}
\lambda\colon\,(Y_\Gamma)_{\rm tors}= H^{-1}(\Gamma_{L/K}, Y)\overset\sim\longrightarrow H^1(K,T)=H^1(\Gamma_{L/K}, Y\otimes_{\Bbb Z} L^\times),
\end{equation*}
where $\Gamma={\rm Gal}(\overline K/K)$, $\ Y_\Gamma$ denotes the group of coinvariants of $\Gamma$ in $Y$,
and $(\ )_{\rm tors}$ denotes the torsion subgroup of the group in the parentheses.
Question. Let $y\in Y$ be a cocharacter whose image in $Y_\Gamma$ is of finite order (say, of order dividing $d$).
Write explicitly a cocycle
in $Z^1(\Gamma_{L/K}, Y\otimes_{\Bbb Z} L^\times)$ representing $\lambda[y]\in H^1(\Gamma_{L/K}, Y\otimes_{\Bbb Z} L^\times)$.
I think that this is possible, because I know the corresponding formula for $K={\Bbb R}$.
Best Answer
Answer of James S. Milne: Most probably, this homomorphism $$\lambda\colon\, H^{-1}(\Gamma_{L/K}, Y)\overset\sim\longrightarrow H^1(\Gamma_{L/K}, Y\otimes_{\Bbb Z} L^\times)$$ is just the cup-product with the fundamental class in $H^2(\Gamma_{L/K}, L^\times)$.