2
$\begingroup$

Let $\breve{\mathbb Q}_p$ denote the completion of the maximal unramified extension of $\mathbb Q_p$. I‘d like to compute Galois cohomology groups and sets related to $\breve{\mathbb Q}_p \otimes_{\mathbb Q_p} \breve{\mathbb Q}_p =: R_2$ such as $H^i(\Gamma_{\mathbb Q_p} \times \Gamma_{\mathbb Q_p}, \mathrm{GL}_n(R_2))$, where the $i$-th copy of $\Gamma_{\mathbb Q_p}$ (denoting the absolute Galois group of $\mathbb Q_p$) acts on the $i$-th copy of $\breve{\mathbb Q}_p$ in $R_2$.

$R_2$ seems to be an unwieldy object. Bourbaki (Algebra V.10.4. Cor.) proves that for any possibly infinite Galois extension $L/K$ there is a natural injection $L \otimes_K L \hookrightarrow \prod_{\mathrm{Gal}(L/K)} L$. For a finite extension, this is an isomorphism of $K$-algebras. $\breve{\mathbb Q}_p$ however is only the completion of a Galois extension, so the result doesn‘t apply. Even if there is an analogous result here, it seems that it would be of limited use without a good description of the quotient.

I would be grateful for either a more workable description of $R_2$ or ideas as to how to compute its cohomology. I haven‘t had much luck considering the obvious restriction and inflation maps. For what it‘s worth, I’m interested more generally in the $n$-fold tensor product $R_n$ and in arbitrary reductive algebraic groups $G$.

$\endgroup$
3
  • 6
    $\begingroup$ My gut feeling is that you don't actually want to consider this tensor product. I don't know what your motivation is, so it's hard to say, but if I would run into this tensor product, I'd try to see if I can either replace $\breve{\mathbb Q}_p$ by its uncompleted version, or the tensor product by its completion. $\endgroup$ Commented Mar 20, 2023 at 13:41
  • $\begingroup$ @PeterScholze Thank you for the intuition! How does the situation change when we consider the completed tensor product instead? $\endgroup$
    – bsbb4
    Commented Mar 20, 2023 at 20:40
  • 1
    $\begingroup$ The completed tensor product is isomorphic to continuous functions from $\widehat{\mathbb Z}$ to $\breve{\mathbb Q}_p$. $\endgroup$ Commented Mar 21, 2023 at 13:10

0

You must log in to answer this question.