3
$\begingroup$

Let $G$ be the absolute Galois group of a local field $K$ (i.e $K$ is a finitely extension of $\mathbb{Q}_{\ell}$.) Let $T$ be a finitely generated $\mathbb{Z}_{p}$ module with a continuous $G$ action. Here for simplicity we assume that $\ell \neq p, \infty$ though both these cases are equally important.

As usual we consider $V=T \otimes_{\mathbb{Z}_{p}} \mathbb{Q}_{p}$ and $W=T \otimes_{\mathbb{Z}_{p}} \mathbb{Q}_{p}/ \mathbb{Z}_{p}$. We then define the finite part of $H_{f}^1(K,V)$ as the kernel of the map $H^1(K,V) \to H^{1}(I,V)$ where $I$ is the inertial group of $K$. By the Hochschild-Serre spectral sequence this is the same as $H^1(G_{ur}, V^{I})$ where $G_{ur}=\text{Gal}(K_{ur}/K)=G/I$. We then define $H_{f}^{1}(G,W)$ and $H_{f}^{1}(G,V)$ to be the image and pre-image of $H_{f}^1(K,T)$ respectively under the following maps $$H^1(K,T) \to H^{1}(K,V) \to H^{1}(K,W).$$ These groups are studied extensively in the book "Euler systems" of Prof. Rubin.

I wonder there is a natural way to define $H_{f}^{2}(K,A)$ for $A=T,V,W$. My naive guess is the following. As usual, we would start with defining $H^{2}_{ur}(K,V)$ as the kernel of the map $H^2(K,V) \to H^2(I,V)$. However, $I$ has $p$-cohomological dimension $1$ so the kernel is just $H^2(K,V)$ (which in turn equals to $H^1(G_{ur}, H^{1}(I,V))$ by the Hochschild-Serre spectral sequence). However, this definition does not seem to be as satisfying as that of $H^1$.

My question is the following: is there a better definition?

$\endgroup$
2
  • $\begingroup$ The "natural way to define" it to do what? Given the duality, it seems to me only $H^1$ is really of interest here. $\endgroup$ Sep 5, 2016 at 19:55
  • $\begingroup$ Dear Prof. Wuthrich, thank you for your reply. The situation that leads me to the above question is follow. I have some elements in $H^2(K,T)$ where $K$ is a number field (more precisely $K=Q(mp^{i})$) and $T$ is some finitely generated $\mathbb{Z}_{p}$. I try to see whether these classes have similar (local) properties as generalized Beilinson-Flach classes constructed by Lei-Loeffler-Zerbes. From your response, the above question seems not the right one to ask. $\endgroup$
    – user97464
    Sep 6, 2016 at 2:43

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.