7
$\begingroup$

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 $K={\Bbb R}$.

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.