2
$\begingroup$

Let $\rho_p:G_{\mathbb{Q}_p} \to \text{Gl}_n(\mathbb{Q}_p)$ be semistable representation. In local to global Galois representation, it was asked if one can find a geometric global Galois representation $\rho:G_{\mathbb{Q}}\to \text{Gl}_n(\mathbb{Q}_p)$ such that $\rho\vert_{G_{\mathbb{Q}_p}}=\rho_p$. This doesn't work for one-dimensional representations for cardinality reason, as was pointed out in this answer. In the comments to this answer, BCnrd mentions a concrete obstruction ot liftability

"For a number field $K$ and a (continuous) character $\rho:G_K\to \overline{\mathbb{Q}}_p^\times$ ramified at only finitely many places, the image on inertia at the places away from $p$ is finite. So if unramified at all places dividing $p$ then must have finite image (since inertia subgroups of $G^{ab}_K$ at the non-archimedean places topologically generate a finite-index subgroup, due to finiteness of class groups). So that gives a "concrete" obstruction to globalizing."

Is this obstruction effective? More generally, do we have a good idea which "nice" local representations come from "nice" global representation?

$\endgroup$

1 Answer 1

3
$\begingroup$

A basic necessary condition, following the comment you quote, is that the determinant of Frobenius is finite-order. Is this condition sufficient? It might be, sometimes.

If you look instead at $\ell$-adic representations where the classification of representations is much easier, you're going to run into trouble for cardinality reasons whenever the inertia does not act irreducibly on the representation. We can multiply Frobenius by anything in the centralizer of inertia and get a valid representation, and this centralizer will be bigger than one-dimensional, so will contain uncountably many elements with the same determinant. Again you run into cardinality troubles because there are (or should be...) countably many global geometric representations up to semisimplification.

Said more concretely, in the $n=2$, unramified case, this is like action which numbers are the $p$th Hecke eigenvalue of a modular form of any weight and any level prime to $p$. The answer: quite a lot of them, but not all, and it's hard to say more without just going through the modular forms one-by-one.

In the $\ell$-adic local case, if inertia has finite image, and the determinant of Frobenius has finite order, you're fine to globalize, because then the representation has finite image and those globalize to Artin representations, which are automatically geometric.

Now back to the $p$-adic world, note that $p$-adic geometric representations give, by $p$-adic Hodge theory, Weil-Deligne representations, which behave like the $\ell$-adic setting, plus even more information (a Hodge filtration). So we see there is no straightfoward globalization condition unless inertia acts irreducibly on the Weil-Deligne representation, but maybe one can do something in that case.

$\endgroup$

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.