Let $K$, $L$ be finite extensions of the $p$-adic numbers. Suppose $\chi:G_K\rightarrow L^{\times}$ is crystalline. It is my understanding that if either $K$ or $L=\mathbb{Q}_p$, then $\chi$ must be a Tate twist of an unramified character. Is there a classification of crystalline characters in general?


3 Answers 3


Hi Kevin,

  1. Go to http://math.stanford.edu/~conrad/
  2. Download "Grunwald--Wang for global character groups"
  3. Read appendix A, especially prop A.3

The answer (note that $K$ and $L$ are switched in Brian's paper) is that once you've identified your character as a character of $K^\times$ via local class field theory, it should be "algebraic" on $O_K^\times$.

  • $\begingroup$ Aha, so the situation is completely analogous to the Hodge-Tate case. Thank you for the reference! $\endgroup$ Apr 17, 2011 at 19:21
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    $\begingroup$ Update: The relevant statements are now in Appendix B of "Lifting global representations with local properties." $\endgroup$ May 13, 2011 at 6:09

[EDIT: I misunderstood the question -- I thought the poster wanted to know why all crystalline characters are unramified twists of powers of cyclotomic when $L$ or $K$ is $\mathbb{Q}_p$, and wrote out a detailed proof. I realise now that the poster already knew this but wanted to understand the classification in the general case. I thought I'd leave this post here anyway in case anyone finds it useful.]

Step 1: Consider the space $\mathbb{D}_{\mathrm{dR}}(\chi)$. This is a free rank 1 module over $K \otimes_{\mathbb{Q}_p} L$, and the steps in the Hodge filtration are $K \otimes_{\mathbb{Q}_p} L$-submodules. By assumption, $K \otimes_{\mathbb{Q}_p} L$ is a field, there is only one Hodge-Tate weight. By twisting by a power of the cyclotomic character, we can assume without loss of generality that the Hodge-Tate weight is 0.

Step 2: Now consider the space $\mathbb{D}_{\mathrm{cris}}(\chi)$. This is a free rank 1 module over $L K_0$, with a Frobenius that is L-linear and $K_0$-semilinear. (Here $K_0$ is the maximal unramified subfield of $K$). If $[K_0 : \mathbb{Q}_p] = d$, then $\varphi^d$ is linear, so acts as multiplication by a scalar $\mu \in L$.

Step 3: Consider the unramified character $G_{K_0} \to L$ mapping geometric Frobenius to $\mu^{-1}$. Tensoring with this character, we may assume that $\mu = 1$.

Step 4: Now choose a basis of $\mathbb{D}_{\mathrm{cris}}(\chi)$ over $LK_0$, and suppose that (with respect to this basis) $\varphi$ acts as multiplication by $\alpha$. The condition that $\varphi^d = 1$ translates to $\alpha \varphi(\alpha) \dots \varphi^{d-1}(\alpha) = 1$. If $K = \mathbb{Q}_p$, then this is just $\alpha = 1$. If $K \ne \mathbb{Q}_p$, then $L = \mathbb{Q}_p$, and $\alpha \in K_0^\times$ satisfies $N_{K_0 / \mathbb{Q}_p} \alpha = 1$. Hence $\alpha = \varphi(x) / x$ for some $x \in K_0^\times$, by Hilbert 90; and after changing basis by $x$ we again have $\alpha = 1$.

So we've shown that $\mathbb{D}_{\mathrm{cris}}(\chi)$ is isomorphic (as a filtered $\varphi$-module) to $\mathbb{D}_{\mathrm{cris}}$ of the trivial character, i.e. $\chi$ is trivial (and hence crystalline!). QED.


By definition, any homomorphism from $G_K$ to an abelian group factorizes through $Gal(K^{ab}/K)$; by the local class field theory the latter group is the completion of $K^\times$. Hence it suffices to calculate $Hom (\overline{K^\times},L^\times)$. Lastly, one can note that up to torsion $K^\times\cong \mathbb{Z}_p^{[K:\mathbb{Q}_p]+1}$ and $L^\times\cong \mathbb{Z}_p^{[L:\mathbb{Q}_p]}\bigoplus \mathbb{Z}$ (via the corresponding logarithms).

  • $\begingroup$ It's not quite obvious (to me) from what you've written how this answers the question: which elements of this hom space correspond to crystalline characters? $\endgroup$ Apr 17, 2011 at 8:33
  • $\begingroup$ (Also, I think you mean $\overline{K^\times} = \mathbb{Z}_p^{[K : \mathbb{Q}_p]} \oplus \overline{\mathbb{Z}}$, where the bar denotes profinite completion.) $\endgroup$ Apr 17, 2011 at 9:43
  • $\begingroup$ Yes, you are right. $\endgroup$ Apr 17, 2011 at 13:51

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