## 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, 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.)