# Hodge-Tate weights of induced representation

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 $W$ ?

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

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!