# Explicit cocycles for the first Galois cohomology of a $p$-adic torus

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 $$K={\Bbb R}$$.

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 $$H^2(\Gamma_{L/K}, L^\times)$$.