p-adic Langlands is weird

Parabolic induction is one of the most basic operations in the representation theory of p-adic groups. In the classical setting of complex coefficients, parabolic induction has a left adjoint given by the Jacquet module, and also a right adjoint given – miraculously – by the (twisted) Jacquet module for the opposite parabolic. Here all three functors are exact. One then defines supercuspidal representations as those which are killed by all Jacquet module functors, or equivalently as those which don’t occur in any parabolic induction.

With mod p coefficients, parabolic induction is still exact, so it easily passes to a functor on derived categories. More precisely, fix G a p-adic reductive group, and let D(G) be the derived category of the category of smooth \overline{\mathbf{F}_p} representations of G. Let P=MU \subset G be any parabolic subgroup. Then the usual operation of parabolic induction upgrades to a t-exact functor \mathrm{Ind}_P^G:D(M) \to D(G) which preserves D^b_{\mathrm{adm}}. By general nonsense, \mathrm{Ind}_P^G commutes with all direct sums, and hence admits a right adjoint \mathbf{R}_{G}^{P}: D(G) \to D(M). Much less obviously, a recent theorem of Heyer shows that \mathrm{Ind}_P^G commutes with direct products, and hence admits a left adjoint \mathbf{L}_{G}^{P}: D(G) \to D(M). Moreover, Heyer also shows that \mathbf{L}_{G}^{P} preserves D^{b}_{\mathrm{adm}}, and computes its values in some examples.

Exercise. Show that \mathbf{R}_{G}^{P} restricted to D^{b}_{\mathrm{adm}}(G) satisfies the isomorphism \mathbf{R}_{G}^{P} \cong (\mathcal{S}_M \circ \mathbf{L}_{G}^{P} \circ \mathcal{S}_G)[-\dim U] \otimes \chi_P, where \mathcal{S}_G is Kohlhaase’s derived duality funtor and \chi_P:M \to \mathbf{F}_p^\times is the integral modulus character. Deduce that \mathbf{R}_{G}^{P} preserves D^{b}_{\mathrm{adm}}.

Now, if you start with an irreducible representation \pi in degree zero, it is formal that \mathbf{R}_{G}^{P}(\pi) resp. \mathbf{L}_{G}^{P}(\pi) will be concentrated in nonnegative resp. nonpositive degrees, and H^0 of it is something explicit: H^0 \mathbf{L}_{G}^{P}(\pi) is basically the (naive) Jacquet module, and H^0 \mathbf{R}_{G}^{P}(\pi) is Emerton’s functor \mathrm{Ord}_{\overline{P}}(\pi) of ordinary parts. In particular, when \pi is supersingular, both of these things vanish in degree zero. But of course, they might be nonzero in other degrees, since \mathbf{L}_{G}^{P}(\pi) and \mathbf{R}_{G}^{P}(\pi) are not t-exact.

In the special case where G=\mathrm{GL}_2(\mathbf{Q}_p) and P=B is the Borel, Heyer showed that \mathbf{L}_{G}^{B}(\pi) vanishes identically for any irreducible supersingular representation \pi, and the above exercise then implies that also \mathbf{R}_{G}^{B}(\pi) vanishes identically. However, if there’s one thing we’ve learned in recent years, it’s that p-adic Langlands is only simple for \mathrm{GL}_2(\mathbf{Q}_p) – for every other group, the whole story is completely different.

Theorem (Yongquan Hu). If G=\mathrm{GL}_2(\mathbf{Q}_{p^2}), there are plenty of irreducible admissible supersingular representations \pi such that \mathbf{L}_{G}^{B}(\pi) and \mathbf{R}_{G}^{B}(\pi) are both nonzero!

This is actually immediate from Corollary 1.2 here and basic adjunctions.

On further reflection, it is probably true that “most” supersingular representations of a given group have the property that some \mathbf{L}_{G}^{P}(-) or \mathbf{R}_{G}^{P}(-) is nonzero. If you believe in some version of the mod p Langlands correspondence, this is reflected in the fact that “most” mod p Galois representations are reducible (e.g., they are Zariski-dense in the Emerton-Gee stack).

Question. Is it true that “second adjointness” holds in this setting, in the sense that \mathbf{L}_{G}^{P} \cong \mathbf{R}_{G}^{\overline{P}}[\dim U] \otimes \chi_P as functors on on D^b_{\mathrm{adm}}(G), or even on all of D(G)?

One can check by hand that this isomorphism is OK on irreps of \mathrm{GL}_2(\mathbf{Q}_{p}) using the calculations in Heyer’s paper and the exercise above. If this question has an affirmative answer, then \mathbf{L}_{G}^{P}(\pi) is necessarily concentrated in degrees [1-\dim U,-1] for any supersingular \pi, and similarly for \mathbf{R}_{G}^{P}(\pi). In particular, in the setting of Hu’s example above, we would get that \mathbf{L}_{G}^{B}(\pi) is concentrated in degree -1, and \mathbf{R}_{G}^{B}(\pi) is concentrated in degree 1.