Halloween two-for-one sale

In a previous post, I asked the following question:

Let {F} be a finite extension of {\mathbf{Q}_{p}} with uniformizer {\varpi}, {G=\mathrm{GL}_{n}(F)},{B} (resp. {\overline{B}}) the upper (resp. lower) triangular Borel, {T\cong(F^{\times})^{n}} the diagonal maximal torus. Fix another finite extension {L/\mathbf{Q}_{p}}, and let {\delta:T\rightarrow L^{\times}} be a continuous character, identified with a tuple of characters {\delta_{i}:F^{\times}\rightarrow L^{\times}} in the obvious way. Suppose for all {1\leq i\leq n} we have the inequality {\sum_{1\leq j\leq i}v(\delta_{i}(\varpi))\geq0}, with equality for {i=n}. Does the locally analytic induction

\displaystyle \Pi(\delta)=\mathrm{Ind}_{\overline{B}}^{G}(\delta)^{\mathrm{an}}

admit a {G}-invariant norm?

The answer in this generality is definitely “no”. Here’s a corrected version of this speculation.

Definition. A character {\delta} as above is weakly unitary if

\displaystyle \sum_{j=1}^{i}v_{p}(\delta_{j}(\varpi))\geq0

for all {1\leq i\leq n}, with equality for {i=n}. A character is strongly unitary if every character appearing in {J_{B}(\Pi(\delta))} is weakly unitary.

Here {J_{B}(\Pi(\delta))} denotes Emerton’s locally analytic Jacquet module; the definition of strong unitarity can be unwound (cf. Remark 5.1.8 in Emerton’s second Jacquet paper) into something which doesn’t explicitly mention Jacquet modules. Strong unitary is a necessary condition for {\Pi(\delta)} to admit a {G}-invariant norm, and now we may conjecture the converse.

Conjecture. The representation {\Pi(\delta)} admits a {G}-invariant norm if and only if {\delta} is strongly unitary.

This is closely related to the Breuil-Schneider conjecture. Indeed, if {\delta} is locally algebraic with {B}-dominant algebraic part, then a theorem of Hu (Yongquan Hu, Normes invariantes et existence de filtrations admissibles, Crelle 634) shows that {\delta} is strongly unitary if and only if {\delta} is the parameter of a de Rham and trianguline (“semistabelline”) representation {\rho:G_{F}\rightarrow\mathrm{GL}_{n}(L)} with distinct Hodge-Tate weights, in which case the Breuil-Schneider conjecture predicts that a certain locally algebraic representation {V(\rho)\otimes\pi(\rho)_{\mathrm{cl}}}, which coincides with the space of locally algebraic vectors {\Pi(\delta)_{\mathrm{alg}}\subset\Pi(\delta)} in this case, admits a {G}-invariant norm. So the above conjecture essentially implies the Breuil-Schneider conjecture for semistabelline representations.


Fix a prime {p}, and let {(R,R^{+})} be a perfectoid Tate-Huber pair. Precisely, {R} is a topological ring satisfying the following conditions:

  1. {R} is complete, and the subring {R^{\circ}\subset R} of power-bounded elements is bounded,
  2. {R} contains a topologically nilpotent unit {\varpi} (a pseudouniformizer),
  3. We may choose {\varpi} in 2. so that {\varpi^{p}|p} in {R^{\circ}} and so that the {p}th power map induces an isomorphism {R^{\circ}/\varpi\overset{\sim}{\rightarrow}R^{\circ}/\varpi^{p}}.

Conditions 1 and 2 here simply assert that {R} is a “complete uniform Tate ring”, and condition 3 is the important perfectoid condition. This is Fontaine’s generalization of Scholze’s original definition, and in particular we need not assume that {R} contains a perfectoid field. Note also that when {R} has characteristic {p}, {\varpi^{p}|p} is vacuously true for any {\varpi}.

For many reasons, one would like to define the Fontaine ring {\mathbf{B}_{\mathrm{crys}}^{+}(R,R^{+})}. When {R} has characteristic zero, there is a canonical definition: take the divided power envelope of the canonical surjection {W(R^{+\flat})\rightarrow R^{+}} (or equivalently, of the surjection {W(R^{+\flat})\rightarrow R^{+}/p}), p-adically complete, and then invert {p}.

Now suppose {R} is of characteristic {p}, so {R^{+}\cong R^{+\flat}}. Here one must make an auxiliary choice in defining {\mathbf{B}_{\mathrm{crys}}^{+}(R,R^{+})} – in order to get something interesting, we need an auxiliary surjection {W(R^{+\flat})\rightarrow S} where {S} is a nonperfect characteristic {p} ring. Choosing {\varpi} as above does the trick: we may define {\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})} by applying the same recipe to the surjection {W(R^{+\flat})\rightarrow R^{+}/\varpi}. This certainly depends on the choice of {\varpi}. However, I claim:

Proposition. For any reduced rational {a/h\in\mathbf{Q}_{\geq0}}, the {\mathbf{Q}_{p}}-vector space {\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})^{\varphi^{h}=p^{a}}\subset\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})} is functorially independent of {\varpi}, and the assignment {\mathrm{Spa}(R,R^{+})\rightsquigarrow\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})^{\varphi^{h}=p^{a}}} defines a (pre)sheaf {\mathbf{B}_{a/h}} of {\mathbf{Q}_{p}}-vector spaces with Frobenius action on the category {\mathrm{Perf}} of perfectoid spaces in characteristic {p}.

The idea here is very simple: if {\varpi|\varpi'} in {R^{\circ}}, then there is a natural inclusion {\mathbf{B}_{\mathrm{crys},\varpi'}^{+}(R,R^{+})\subset\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})}. (More generally, given a commutative diagram of homomorphisms of characteristic {p} rings

\displaystyle \begin{array}{rcl} S & \rightarrow & S/I\\ \downarrow & & \downarrow\\ T & \rightarrow & T/J\end{array}

with {S,T} perfect, there is a ring map {\mathbf{B}_{\mathrm{crys}}^{+}(W(S)\rightarrow S/I)\rightarrow\mathbf{B}_{\mathrm{crys}}^{+}(W(T)\rightarrow T/J)}, with the evident meaning. In general this isn’t an injection, but it is in our case, i.e. with {S=T=R^{+\flat}=R^{+}} and {I=(\varpi'),J=(\varpi)}.) In particular, given any {j} large enough that {\varpi'|\varpi^{p^{j}}}, we have

\displaystyle \varphi^{j}(\mathbf{B}_{\mathrm{crys},\varpi}^{+})\subset\mathbf{B}_{\mathrm{crys},\varpi'}^{+}\subset\mathbf{B}_{\mathrm{crys},\varpi}^{+},

so { }any {x=p^{-aj}\varphi^{hj}(x)\in\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})^{\varphi^{h}=p^{a}}} factors through the inclusion {\mathbf{B}_{\mathrm{crys},\varpi'}^{+}(R,R^{+})^{\varphi^{h}=p^{a}}\subset\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})^{\varphi^{h}=p^{a}}}, and we’re basically done. The assignment {(R,R^{+})\rightsquigarrow\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})^{\varphi^{h}=p^{a}}} is a little irritating since some choice of uniformizer is still present, but we can get something totally canonical by defining the inverse limit

\displaystyle \mathbf{B}_{a/h}(R,R^{+})=\lim_{\varpi\in\mathcal{P}}\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+})^{\varphi^{h}=p^{a}},

where {\mathcal{P}} denotes the set of {\varpi}‘s for {R} with the partial ordering given by divisibility in {R^{\circ}} (since any two {\varpi}‘s divide a common third {\varpi}, this is clearly a cofiltered limit). In fact we can throw away the eigenspace condition first and define the ring

\displaystyle \tilde{\mathbf{B}}_{\mathrm{rig}}^{+}(R,R^{+})=\lim_{\varpi\in\mathcal{P}}\mathbf{B}_{\mathrm{crys},\varpi}^{+}(R,R^{+}),

which is totally canonical and functorial in characteristic {p} perfectoid Tate-Huber pairs. When {(R,R^{+})=(\mathbf{C}_{p}^{\flat},\mathcal{O}_{\mathbf{C}_{p}}^{\flat})} this recovers the usual {\tilde{\mathbf{B}}_{\mathrm{rig}}^{+}}.

It is a remarkable fact that for {0<a/h\leq1}, the functor {\mathrm{Spa}(R,R^{+})\rightsquigarrow\mathbf{B}_{a/h}(R,R^{+})} is representable by a perfectoid space, and in fact by the universal cover of a suitable p-divisible group. In particular,

\displaystyle \mathbf{B}_{1}(R,R^{+})=\mathrm{Hom}_{\mathrm{Perf}}(\mathrm{Spa}(R,R^{+}),\tilde{\mathbf{G}}_{m});

by perfectness it’s enough to give an element of {\mathrm{Hom}(\mathrm{Spa}(R,R^{+}),\mathbf{G}_{m})} ({\mathbf{G}_{m}=\mathrm{Spf}\mathbf{F}_{p}[[T]]}), which is to say any element {r\in1+R^{\circ\circ}} whatsoever ({T\mapsto r-1}), and then this maps to {\log[r]}. For {a/h>1}, it seems likely that {\mathbf{B}_{a/h}} is no longer representable by any kind of adic space. This is actually an example of a diamond – it is “proetale under a perfectoid space” in a suitable sense. (This probably isn’t written down anywhere; JW kindly explained to me why it’s a diamond in the case {a/h=2}, and the construction he provided pretty clearly generalizes.)

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