9
$\begingroup$

In the mod $p$ local Langlands correspondence for $\mathrm{GL}_{2}(\mathbb{Q}_{p})$, the irreducible supercuspidal representation $\left(\mathrm{ind}^{\mathrm{GL}_{2}(\mathbb{Q}_{p})}_{\mathrm{GL}_{2}(\mathbb{Z}_{p})\mathbb{Q}_{p}^{\times}}\mathrm{Sym}^{r}\overline{\mathbb{F}}_{p}^{2}\right)/T$ of $\mathrm{GL}_{2}(\mathbb{Q}_{p})$ is mapped to the irreducible Galois representation $\mathrm{ind}(\omega_{2}^{r+1})$ for $r\in\lbrace 0,\ldots,p-1\rbrace$. See Definition 1.1 in

https://www.imo.universite-paris-saclay.fr/~breuil/PUBLICATIONS/GL2%28Qp%29II.pdf (with $\eta=1$).

However, I notice that this is not compatible with the local class field theory: on $\mathbb{Z}_{p}^{\times}$, the central character of the supercuspidal representation is the $r$-th power map after going modulo $p$, while the determinant character of the Galois representation is the $(r+1)$-th power map.

So is it defined this way to satisfy some other compatibilities (for example, reduction modulo $p$ of the $p$-adic local Langlands for $\mathrm{GL}_{2}(\mathbb{Q}_{p})$)?

$\endgroup$

1 Answer 1

7
$\begingroup$

You seem to be expecting that mod $p$ local Langlands should satisfy the same compatibilities as "conventional" local Langlands (for smooth representations of $GL_2(\mathbf{Q}_p)$ and $WD(\mathbf{Q}_p)$ with coefficients in $\mathbf{C}$).

However, before you can even talk about reduction mod $p$, you need to check that the coefficients can be descended from $\mathbf{C}$ to a number field. I.e., if $\pi$ is a smooth irred rep of $GL_n(\mathbf{Q}_p)$ on an $L$-vector space, where $L$ is some subfield of $\mathbf{C}$, then is its Langlands parameter $\phi_{\pi}$ an $L$-valued Weil-Deligne rep? The answer, annoyingly, is "no": if $n$ is even, you have to add $\sqrt{p}$ to $L$ in order to get this to work. So it's common to re-normalise by twisting the correspondence for $GL_n$ by $|\det|^{(n-1)/2}$; this makes it compatible with coefficient fields, and also works better for local-global compatibility.

It's this same shift which you are seeing in the mod $p$ theory, and here it's completely impossible to get rid of, even if you extend the coefficient fields as much as you like: the character $|\cdot|^{1/2}$ of $\mathbf{Q}_p^\times$ is not $p$-adically unitary, so there is no way you can reduce it mod $p$.

$\endgroup$
1
  • $\begingroup$ Thanks for this answer. $\endgroup$ Commented Sep 19, 2021 at 12:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.