5
$\begingroup$

Let $K$ be a finite extension of the field of $p$-adic numbers $\mathbb{Q}_p$ and let $E$ be another such extension, such that all the $\mathbb{Q}_p$ embeddings $K \to \bar{\mathbb{Q}}_p$ are contained in $E$.

Let $V$ be an $n$-dimensional vector space over $E$ which carries a continuous action of $G_K = Gal(\bar{K} / K)$.

Assume the corresponding representation (still denoted $V$) to be crystalline, that is the $K\otimes_{\mathbb{Q}_p}E$-module $D_crys(V) = (B_{crys} \otimes_{\mathbb{Q}_p} V)^{G_K}$ is free of rank $n$. (Here $B_{crys}$ is one of Fontaine's ring of $p$-adic periods).

There is a filtration on $B_{crys}$ (induced by another period ring) and so, we have a filtration on $D_{crys}(V)$ as well and so an induced graduation.

We say that an integer $i$ is a Hodge-Tate weight of $V$ if $gr^{-i} D_{crys}(V) \neq {0}$.

If $\tau : K \to E$ is a $\mathbb{Q}_p$-embedding, say that an integer $i$ is a $\tau$-labeled Hodge-Tate weight of $V$ if $gr^{-i} D_{crys, \tau} (V) \neq {0}$ where $D_{crys,\tau}(V) = (B_{crys} \otimes_{K, \tau}V)^{G_K}$.

Now the question : consider $K$ as before and let $L$ be the unramified extension of $K$ of degree $d$.

/Edit Assume $E$ contains all embeddings $L \to E$. /Edit

Let $W$ be a crystalline representation of $G_L$. Let $V$ be the representation $Ind_{G_L}^{G_K}$(W). It is not hard to see that $V$ is crystalline, but what are the labeled Hodge-Tate weights of $V$ knowing those of $W$ ?

$\endgroup$
  • $\begingroup$ What if there are no embeddings $L \to E$? $\endgroup$ – Will Sawin Apr 19 '15 at 16:52
  • $\begingroup$ I want to avoid this situation so I would allow $E$ to be large enough. $\endgroup$ – JWM Apr 19 '15 at 18:35
5
$\begingroup$

It's simpler (and more general) to merely assume that your representations are de Rham. If $W$ is a de Rham representation of $G_L$, then $D = D_{dR}(W)$ is an $L$-vector space. If $K$ is a subfield of $L$ then $D_{dR}(Ind_L^K W)$ is $D$, now seen as a $K$-vector space via restriction of scalars. The rest should then be an exercise in linear algebra.

Edit: in the above, you can -- and should -- replace "de Rham" by "Hodge-Tate" and $D_{dR}$ by $D_{HT}$, and the resulting exercise is even simpler!

$\endgroup$

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.