Proof. It follows from [Reference Liu and ZhengLZ17, Lemma 3.1.2.(4)] that bdcs maps admitting universal p-codescent are stable under composition. Part (ii) of the claim follows immediately and part (iii) is obvious (see [Reference MannMan22, Example 3.6.11] for the second claim in (iii)). Part (i) follows from the observation that étale maps admit universal p-codescent.

It remains to prove (iv), so let f and g be given as in the claim. Choose a bdcs map $Y' \to Y$ admitting universal p-codescent such that the composed map $Y' \to X$ is bdcs. Denote $g'\colon Z' \to Y'$ the base-change of g along $Y' \to Y$ . Since the map $Z' \to Z$ is bdcs, it follows from (ii) that the composed map $Z' \to X$ is p-fine. We can therefore choose a bdcs map $Z" \to Z'$ admitting universal p-codescent such that $Z" \to X$ is bdcs. Note that the map $Z" \to X$ factors as $Z" \to Y' \to X$ and since $Y' \to X$ is bdcs it follows from [Reference MannMan22, Lemma 3.6.10.(iv)] that $Z" \to Y'$ is bdcs. In particular $Z" \to Y$ is bdcs, so the map $Z" \to Z$ shows that g is p-fine.

Proof. Let $\mathcal C \subset {{\rm vStack}_\Lambda }$ be the full subcategory spanned by those $(X,\Lambda )$ which admit a map (in ${{\rm vStack}_\Lambda }$ ) to some strictly totally disconnected space. Then $\mathcal C$ is a basis of ${{\rm vStack}_\Lambda }$ , hence by [Reference MannMan22, Proposition A.5.16] and [Reference MannMan22, Proposition A.3.11. (i)] it is enough to construct the desired extension of the 6-functor formalism on $\mathcal C$ , i.e. we need to construct a 6-functor formalism extending the one from Theorem 2.2. We now apply [Reference MannMan22, Proposition A.5.14] with $E = bdcs$ , $E' = pfine$ and $S \subset E$ being the subset of those maps which admit universal p-codescent. Condition (a) requires to be presentable, which can be achieved by temporarily restricting to $\kappa $ -condensed objects (as in the proof of [Reference MannMan22, Theorem 3.6.12]). Condition (b) follows from the definition of universal p-codescent, condition (c) follows from the definition of p-fine maps and condition (d) follows from [Reference MannMan22, Lemma 3.6.10] (see [Reference MannMan22, Remark A.5.15.(ii)].

Proof. Let $f\colon Y \to X$ be a bdcs and p-cohomologically smooth map in ${{\rm vStack}_\Lambda }$ with Čech nerve $f_\bullet \colon Y_\bullet \to X$ (i.e. $Y_0 = Y$ , $Y_1 = Y \times _X Y$ and so on); we need to show that the natural functor is an isomorphism. This functor admits a left adjoint

Note that $f_\bullet ^!$ is conservative: This follows from the fact that $f_0^!$ is conservative, which in turn is implied by the fact that $f_0^!$ is just a twist of $f_0^*$ by definition of the p-cohomological smoothness of $f_0$ . Thus in order to show that $f_\bullet ^!$ is an equivalence, it is enough to show that $f_{\bullet !}$ is fully faithful (cf. [Reference LurieLur18, Corollary 03UZ]), i.e. for given we need to see that the natural map $f_\bullet ^! f_{\bullet !} \mathcal M_\bullet \mathbin {\xrightarrow {\sim }} \mathcal M_\bullet $ is an isomorphism. This boils down to showing that the map $f_0^! f_{\bullet !} \mathcal M_\bullet \to \mathcal M_0$ is an isomorphism. Since $f_0$ is p-cohomologically smooth, $f_0^!$ preserves small colimits, so we end up with the claim that the natural map

$$ \begin{align*} \varinjlim_{n\in\Delta} f_0^! f_{n!} \mathcal M_n \mathbin{\xrightarrow{\sim}} \mathcal M_0 \end{align*} $$

is an isomorphism. Let us denote by $f^{\prime }_\bullet \colon Y^{\prime }_\bullet \to Y_0$ the base-change of $Y_\bullet \to X$ along $f_0\colon Y_0 \to X$ . Denote $g_\bullet \colon Y^{\prime }_\bullet \to Y_\bullet $ the natural projections. Then for every $n \ge 0$ the natural morphism $f^{\prime }_{n!} g_n^! \mathcal M_n \mathbin {\xrightarrow {\sim }} f_0^! f_{n!}\mathcal M_n$ is an isomorphism (by p-cohomological smoothness of $f_0$ and $g_n$ this reduces to proper base-change). On the other hand $g_\bullet ^! \mathcal M_\bullet $ defines an object of so the above claimed isomorphism reduces to showing that the natural functor is an equivalence. This means that we can replace $Y_\bullet \to X$ by $Y^{\prime }_\bullet \to Y_0$ . But the latter augmented simplicial object is split, hence the desired equivalence of $\infty $ -categories is formal.

Proof. By [Reference MannMan22, Proposition 3.9.13] the same is true for the 6-functor formalism on bdcs maps. To get this result also on p-fine maps, we note that the same extension procedure as in the proof of Proposition 2.7 applies verbatim to the $\varphi $ -module case so that we get the desired 6-functor formalism. By [Reference MannMan22, Proposition 3.9.13] and functoriality of Kan extensions the forgetful functor preserves small limits and colimits and commutes with all the 6 functors except possibly with $f^!$ . To see that it also commutes with $f^!$ , note first that by the same argument as in the proof of [Reference MannMan22, Lemma 3.9.12. (v)] we can reduce to the case that the target of $f\colon Y \to X$ is a strictly totally disconnected space. Choose also a bdcs cover $g\colon Z \mathrel {\twoheadrightarrow } Y$ such that g admits universal p-codescent and $f \mathbin {\circ } g$ is bdcs. Then by universal p-codescent the functor $g^!$ is conservative, hence to check that $f^!$ commutes with the forgetful functor it is enough to check that $g^! f^!$ commutes with the forgetful functor. This follows from the bdcs case.

Proof. We first note that part (i) is formal if X is strictly totally disconnected and follows for general X once we have shown (ii).(a) (see the proof of [Reference MannMan22, Proposition 3.8.4. (i)]). Moreover, once we have shown (ii).(a), also (iii) is formal (see the proof of [Reference MannMan22, Proposition 3.8.7]) and (ii).(b) is formal (see the proof of [Reference MannMan22, Proposition 3.8.6.(ii)]. Therefore, the whole claim boils down to showing (ii).(a). By the proof of [Reference MannMan22, Proposition 3.8.6.(iii)] the case that g is p-cohomologically smooth follows formally from the case that f is p-cohomologically smooth. Thus from now on we can assume that we are given a cartesian square as in (ii) with f being p-fine and p-cohomologically smooth and we need to show that the morphism $g^{\prime *} f^! \mathbin {\xrightarrow {\sim }} f^{\prime !} g^*$ is an isomorphism. As in the proof of [Reference MannMan22, Proposition 3.8.4.(ii)] we can formally reduce to the case that X and $X'$ are strictly totally disconnected. Then by (i) (which holds unconditionally on (ii).(a) in the strictly totally disconnected case) we are reduced to showing that the natural map $\alpha \colon g^{\prime *} f^! \Lambda ^a \mathbin {\xrightarrow {\sim }} f^{\prime !} \Lambda ^a$ is an isomorphism. We claim that $g^{\prime }_*$ is conservative on bounded objects (in the sense of [Reference MannMan22, Definition 3.3.1. (a)]): Pick any cover $Z' \mathrel {\twoheadrightarrow } X'$ such that the composition $Z \to X$ is bdcs (this can be done by the definition of p-fine maps). Then $Z'$ is a locally spatial diamond, so we can find a pro-étale cover of $Z'$ by totally disconnected spaces $Z_i$ . Letting $W_i := Z_i \times _{X'} Y' = Z_i \times _X X'$ and denoting $g_i'\colon W_i \to Z_i$ the base-change of $g'$ , we note that each $W_i$ is a p-bounded affinoid perfectoid space. Hence by [Reference MannMan22, Theorem 3.5.21] each $g^{\prime }_i$ is a forgetful functor of module categories and in particular conservative. Using also bounded base-change (see [Reference MannMan22, Proposition 3.3.7]) we deduce that $g'$ is conservative on bounded objects. Now note that $f^! \Lambda ^a$ is bounded: If $\Lambda = {\mathcal {O}}^+_X/\pi $ for some $\pi | p$ then this follows from the Riemann-Hilbert correspondence (see [Reference MannMan22, Theorem 3.9.23]); for general $\Lambda $ one can make the usual reductions. Similarly $f^{\prime !}\Lambda ^a$ is bounded. Thus altogether we deduce that it is enough to show that $g^{\prime }_*\alpha \colon g^{\prime }_* g^{\prime *} f^! \Lambda ^a \mathbin {\xrightarrow {\sim }} g^{\prime }_* f^{\prime !} \Lambda ^a$ is an isomorphism. The remaining proof of [Reference MannMan22, Proposition 3.8.4.(ii)] applies verbatim.

Proof. This is formal, see [Reference MannMan22, Lemma 3.8.5].

Proof. The functor r has a left adjoint, given by the (t-exact) functor $\mathrm {ind}_{H}^{G}(-):=-\otimes _{k[H]}k[G]$ of compact induction. We then compute

$$ \begin{align*} \mathrm{Hom}_{{\mathcal{D}}(H)}(C,r{\underline{\operatorname{\mathrm{Hom}}}}_G(A,B)) & \cong\mathrm{Hom}_{{\mathcal{D}}(G)}(\mathrm{ind}_{H}^{G}C,{\underline{\operatorname{\mathrm{Hom}}}}_G(A,B))\\ & \cong\mathrm{Hom}_{{\mathcal{D}}(G)}(A\otimes\mathrm{ind}_{H}^{G}C,B)\\ & \cong\mathrm{Hom}_{{\mathcal{D}}(G)}(\mathrm{ind}_{H}^{G}(rA\otimes C),B)\\ & \cong\mathrm{Hom}_{{\mathcal{D}}(H)}(rA\otimes C,rB)\\ & \cong\mathrm{Hom}_{{\mathcal{D}}(H)}(C,{\underline{\operatorname{\mathrm{Hom}}}}_H(rA,rB)) \end{align*} $$

naturally in $C\in {\mathcal {D}}(H)$ , so the result follows from Yoneda.

Proof. If $A\overset {\sim }{\to }I^{\bullet }$ is a K-injective resolution by injectives, then $R\Gamma _{\mathrm {sm}}(H,A)\cong (I^{\bullet })^{H}$ for any H; here we use that restriction along $\mathrm {Rep}_{k}^{\mathrm {sm}}(G)\to \mathrm {Rep}_{k}^{\mathrm {sm}}(H)$ preserves K-injective complexes of injectives. But then

$$ \begin{align*} \mathrm{colim}_{H\subset G\,\mathrm{open}}R\Gamma_{\mathrm{sm}}(H,A) & \cong\mathrm{colim}_{H\subset G\,\mathrm{open}}(I^{\bullet})^{H}\cong I^{\bullet}\cong A, \end{align*} $$

using the fact that each $I^{n}$ is a smooth representation.

Proof. There are natural isomorphisms

$$ \begin{align*} \mathrm{colim}_{H}R\mathrm{Hom}_{{\mathcal{D}}(H)}(A,B) & \cong\mathrm{colim}_{H}R\Gamma_{\mathrm{sm}}(H,{\underline{\operatorname{\mathrm{Hom}}}}_H(A,B))\\ & \cong\mathrm{colim}_{H}R\Gamma_{\mathrm{sm}}(H,{\underline{\operatorname{\mathrm{Hom}}}}_G(A,B))\\ & \cong {\underline{\operatorname{\mathrm{Hom}}}}_G(A,B), \end{align*} $$

where the second line follows from Lemma 3.2, and the third line follows from Lemma 3.3.

Proof. Easy, using the previous lemma together with the fact that $\mathbf {1}$ is an injective object of $\mathrm {Rep}_{k}^{\mathrm {sm}}(H)$ for any pro-p open compact $H\subset G$ . (Use that any injective map $V\to W$ in $\mathrm {Rep}_{k}^{\mathrm {sm}}(H)$ admits an H-equivariant retraction, by averaging any k-linear retraction against a k-valued Haar measure.)

Proof. As noted above, $\mathcal {S}_{G}(V)$ is defined by the formula

$$\begin{align*}\mathcal{S}_{G}(V)=\mathrm{colim}_{H\subset G\,\mathrm{open}\,\mathrm{compact}}R\mathrm{Hom}_{{\mathbf{F}}_p[[H]]}({\mathbf{F}}_p,V^{\ast}). \end{align*}$$

To actually compute this, choose a uniformly powerful open pro-p subgroup $G_{0}\subset G$ . By work of Lazard [Reference LazardLaz65, Ch. V, Section 2], we can choose a finite length resolution $P_{\bullet }\to {\mathbf {F}}_p\to 0$ of ${\mathbf {F}}_p$ by finite free ${\mathbf {F}}_p[[G_{0}]]$ -modules. Then we can express $\mathcal {S}_{G}(-)$ more concretely as

$$\begin{align*}\mathcal{S}_{G}(V)=\mathrm{colim}_{H\subset G_{0}\,\mathrm{open}}\mathrm{Hom}_{{\mathbf{F}}_p[[H]]}(P_{\bullet},V^{\ast}). \end{align*}$$

Set $I^{n}=\mathrm {Hom}_{{\mathbf {F}}_p}^{\mathrm {cts}}(P_{n},{\mathbf {F}}_p)$ , so $0\to {\mathbf {F}}_p \to I^{\bullet }$ is a finite resolution of $\mathbf {1}$ by injective objects in $\mathrm {Rep}_{{\mathbf {F}}_p}^{\mathrm {sm}}(G_{0})$ . Since Pontryagin duality gives an exact anti-equivalence between smooth H-representations over ${\mathbf {F}}_p$ and pseudocompact ${\mathbf {F}}_p[[H]]$ -modules, we have natural identifications

$$\begin{align*}\mathrm{Hom}_{{\mathbf{F}}_p[[H]]}(P_{\bullet},V^{\ast})\cong\mathrm{Hom}_{\mathrm{Rep}_{{\mathbf{F}}_p}^{\mathrm{sm}}(H)}(V,I^{\bullet})\cong R\mathrm{Hom}_{{\mathcal{D}}(H)}(V,\mathbf{1}). \end{align*}$$

Putting things together, we get

$$ \begin{align*} \mathcal{S}_{G}(V) & \cong\mathrm{colim}_{H\subset G_{0}\,\mathrm{open}}\mathrm{Hom}_{\mathrm{Rep}_{{\mathbf{F}}_p}^{\mathrm{sm}}(H)}(V,I^{\bullet})\\ & \cong\mathrm{colim}_{H\subset G_{0}\,\mathrm{open}}R\mathrm{Hom}_{{\mathcal{D}}(H)}(V,\mathbf{1})\\ & \cong {\underline{\operatorname{\mathrm{Hom}}}}_G(V,\mathbf{1}), \end{align*} $$

where the last line follows by Lemma 3.4.

Proof. The first part is essentially a summary of results from [Reference KohlhaaseKoh17]. The second part is Pontryagin-dual to the biduality discussed in the proof of [Reference KohlhaaseKoh17, Theorem 3.5]; see also [Reference Schneider and SorensenSS22, Proposition 3.4]. The third part is immediate from the first part.

Proof. ‘If’: Given such a $V\to W$ , we get a nonzero map $\mathcal {S}(W)\to \mathcal {S}(V)$ . But the previous proposition shows that $\mathcal {S}(W)=\tau ^{\leq 0}\mathcal {S}(W)=\mathcal {S}^{0}(W)$ , so we get a nonzero map $\mathcal {S}^{0}(W)\to \mathcal {S}(V)$ which necessarily factors over a nonzero map $\mathcal {S}^{0}(W)\to \mathcal {S}^{0}(V)$ .

‘Only if’: Suppose $\mathcal {S}^{0}(V)\neq 0$ . By the previous proposition, this is zero-dimensional as a G-representation, so $\mathcal {S}(\mathcal {S}^{0}(V))\cong \mathcal {S}^{0}(\mathcal {S}^{0}(V))$ . In particular, applying $\mathcal {S}$ to the map $\mathcal {S}^{0}(V)=\tau ^{\leq 0}\mathcal {S}(V)\to \mathcal {S}(V)$ and using biduality, we get a nonzero map $V\cong \mathcal {S}(\mathcal {S}(V))\to \mathcal {S}(\mathcal {S}^{0}(V))\cong \mathcal {S}^{0}(\mathcal {S}^{0}(V))$ , and $\mathcal {S}^{0}(\mathcal {S}^{0}(V))$ is finite-dimensional over ${\mathbf {F}}_p$ .

Proof. This follows from the proof of Theorem V.1.1 in [Reference Fargues and ScholzeFS21], noting that that $[\operatorname {\mathrm {Spd}}(C)/\underline G]_{{\mathrm {\acute {e}t}}}$ locally has uniformly bounded p-cohomological dimension by Lazard-Serre [Reference LazardLaz65, Ch. V, Section 2].

Proof. The $\infty $ -category comes naturally equipped with a symmetric monoidal structure such that the forgetful functor is symmetric monoidal (e.g. one can construct as certain sheaves on the classifying stack $*/G$ in a suitable geometric setup). The forgetful functor $q^*$ has a left adjoint which is given by . It is straightforward to check that $q_\natural $ satisfies the projection formula, i.e. that for all and the natural morphism $q_\natural (N \otimes q^* M) \mathbin {\xrightarrow {\sim }} q_\natural N \otimes M$ is an isomorphism. The symmetric monoidal structure on is closed, i.e. admits internal hom’s ${\underline {\operatorname {\mathrm {Hom}}}}$ (this is clear if we restrict to $\kappa $ -condensed objects for some cardinal $\kappa $ and then follows for unrestricted condensed objects by the following computation). It follows formally from the projection formula for $q_\natural $ that for all the natural morphism

$$ \begin{align*} q^* {\underline{\operatorname{\mathrm{Hom}}}}(M, N) \mathbin{\xrightarrow{\sim}} {\underline{\operatorname{\mathrm{Hom}}}}(q^* M, q^* N) \end{align*} $$

is an isomorphism – in other words, the internal hom in is computed on the underlying solid ${\mathbf {F}}_p$ -vector spaces. In particular, if M and N are static and M is discrete (and thus $q^* M \cong \bigoplus _I {\mathbf {F}}_p$ for some set I) then ${\underline {\operatorname {\mathrm {Hom}}}}(M, N)$ is static. Therefore, from

$$ \begin{align*} \operatorname{\mathrm{Ext}}^i_G(M, N) = \pi_0 R\operatorname{\mathrm{Hom}}(M, N[i]) = \pi_0 \operatorname{\mathrm{Hom}}({\mathbf{F}}_p, {\underline{\operatorname{\mathrm{Hom}}}}(M, N[i])) = H^i(G, {\underline{\operatorname{\mathrm{Hom}}}}(M, N)) \end{align*} $$

we immediately deduce the claim.

Proof. It is clear that the map $* \to */G$ is bdcs (e.g. the pullback along itself is $G \to *$ , which is pro-étale). Now pick any $(X,\Lambda ) \in {{\rm vStack}_\Lambda }$ admitting a map to a strictly totally disconnected space, pick any map $X \to */G$ and denote $q\colon Y \to X$ the base-change of $* \to */G$ . We need to show that q admits p-codescent, i.e. if $q_\bullet \colon Y_\bullet \to X$ denotes the Čech nerve of q then the natural functor

is an equivalence. To see this we employ Lurie’s Beck-Chevalley condition, see [Reference LurieLur, Corollary 4.7.5.3]. Part (2) of this condition is easily satisfied: Since all the maps in the diagram $Y_\bullet \to X$ are proper, the required left adjointability for the upper shriek functors reduces to proper base-change (by passing to left adjoints). Thus in order to apply Lurie’s result it remains to verify the following:

1. (a) The functor is conservative.

2. (b) The functor $q^!$ preserves geometric realizations of $q^!$ -split simplicial objects in .

To prove (a) and (b) we apply ideas by Mathew [Reference MathewMat16] (which were also used extensively in the setup of the p-adic 6-functor formalism, cf. [Reference MannMan22, §2.6]). More concretely, we will prove the following claim:

• (∗) Let be the full subcategory generated by the endofunctor $q_*q^!$ under compositions, retracts and finite (co)limits. Then $\langle q_*q^! \rangle $ contains the identity functor.

Let us first see how ( $\ast $ ) implies (a) and (b), so assume that ( $\ast $ ) holds for the moment. To prove (a), let $\mathcal M \to \mathcal N$ be a given map in such that $q^! \mathcal M \mathbin {\xrightarrow {\sim }} q^! \mathcal N$ is an isomorphism. Then also $q_* q^! \mathcal M \mathbin {\xrightarrow {\sim }} q_* q^! \mathcal N$ is an isomorphism and one deduces easily that $F(\mathcal M) \mathbin {\xrightarrow {\sim }} F(\mathcal N)$ is an isomorphism for all $F \in \langle q_*q^! \rangle $ ; taking $F = \operatorname {\mathrm {id}}$ shows that $\mathcal M \mathbin {\xrightarrow {\sim }} \mathcal N$ is an isomorphism, as desired. To prove (b), let $\mathcal M_\bullet $ be a $q^!$ -split simplicial object in . Then $q_*q^! \mathcal M_\bullet $ is a constant $\operatorname {\mathrm {Ind}}$ -object (i.e. lies in the essential image of ) and it follows immediately that the same is true for $F(\mathcal M_\bullet )$ for all $F \in \langle q_*q^! \rangle $ . Taking $F = \operatorname {\mathrm {id}}$ shows that $\mathcal M_\bullet $ is a constant $\operatorname {\mathrm {Ind}}$ -object, so that its colimit commutes with any exact functor; in particular it commutes with $q^!$ .

It now remains to verify ( $\ast $ ). We compute (using [Reference MannMan22, Corollary 3.6.17. (i)])

$$ \begin{align*} q_*q^! = q_* {\underline{\operatorname{\mathrm{Hom}}}}(\Lambda^a, q^!) = {\underline{\operatorname{\mathrm{Hom}}}}(q_! \Lambda^a, -) = {\underline{\operatorname{\mathrm{Hom}}}}(q_* \Lambda^a, -). \end{align*} $$

Therefore ( $\ast $ ) can be reduced to showing the following: Let be the full subcategory generated by $q_*\Lambda ^a$ under tensor product, retracts and finite (co)limits; then $\langle q_*\Lambda ^a \rangle $ contains the tensor unit $\Lambda ^a$ . By [Reference MathewMat16, Proposition 3.20] this is equivalent to the following claim:

• (∗^′) The algebra object admits descent.

It remains to prove that ( $\ast '$ ) is true. We first perform some reductions. By assumption on $(X,\Lambda )$ there is a map $(X,\Lambda ) \to (Z,\Lambda )$ with $Z = \operatorname {\mathrm {Spa}}(A, A^+)$ being strictly totally disconnected. We thus get an induced map $X \to Z \times */G = Z/G$ such that q is the pullback of the projection $q_Z\colon Z \to Z/G$ . The induced pullback functor is exact, symmetric monoidal and sends $q_{Z*}\Lambda ^a$ to $q_*\Lambda ^a$ (by proper base-change). It is therefore enough to prove that ( $\ast $ ’) holds for $q_Z$ in place of q, so from now on we assume $X = Z/G$ and hence $Y = Z$ .

Given a morphism $\Lambda ' \to \Lambda $ of integral torsion coefficients on Z, the induced base-change functor preserves tensor products, retracts and finite (co)limits and sends $q_*\Lambda ^{\prime a}$ to $q_*\Lambda ^a$ (by [Reference MannMan22, Proposition 3.6.16]), hence if ( $\ast '$ ) holds for $\Lambda '$ then it also holds for $\Lambda $ . By definition of integral torsion coefficients we can thus reduce to the case that $\Lambda = {\mathcal {O}}^{+a}_{Z^\sharp }/\pi $ for some untilt $Z^\sharp $ of Z and some pseudouniformizer $\pi $ on $Z^\sharp $ . Pick some pseudouniformizer $\varpi $ on $Z^\sharp $ such that $\varpi | p$ and some integer $n> 0$ such that $\pi | \varpi ^n$ . By [Reference MathewMat16, Proposition 3.35] the map of discrete rings $A^{\sharp +}/\pi \to A^{\sharp +}/\varpi $ admits descent, which after pullback along $Z/G \to Z$ implies that the map ${\mathcal {O}}_{Z^\sharp /G}^{+a}/\pi \to {\mathcal {O}}_{Z^\sharp /G}^{+a}/\varpi $ in admits descent. Therefore, if we can show that ( $\ast '$ ) holds for $\Lambda = {\mathcal {O}}^+_{Z^\sharp /G}/\varpi $ then by [Reference MathewMat16, Proposition 3.24] and [Reference Bhatt and ScholzeBS17, Lemma 11.10] it also follows for $\Lambda = {\mathcal {O}}^+_{Z^\sharp /G}/\pi $ . Thus from now on we can assume that $\pi = \varpi $ , i.e. we now have $\pi | p$ .

Now recall the Riemann-Hilbert functor from [Reference MannMan22, Definition 3.9.21. (a)] (where we ignore the $\varphi $ -module structure). Then $RH$ is symmetric monoidal and exact and satisfies $RH(q_*{\mathbf {F}}_p) = q_*({\mathcal {O}}_{Z^\sharp }^{+a}/\pi )$ . To verify the latter, note that this identity can be checked after applying $q^*$ , where we get (using base-change)

$$ \begin{align*} q^* RH(q_* {\mathbf{F}}_p) &= RH(q^*q_* {\mathbf{F}}_p) = RH({\mathcal{C}}(G,{\mathbf{F}}_p)) = {\mathcal{C}}(G, {\mathbf{F}}_p) \otimes_{{\mathbf{F}}_p} A^{\sharp+,a}/\pi = {\mathcal{C}}(G, A^{\sharp+,a}/\pi) \\&= q^* q_*(A^{\sharp+,a}/\pi) \end{align*} $$

(here we implicitly use the identity ). All in all this means that it is enough to show that $q_*{\mathbf {F}}_p \in {\mathcal {D}}_{\mathrm {\acute {e}t}}(Z/G, {\mathbf {F}}_p)^{\mathrm {oc}}$ is descendable. It follows easily from [Reference MannMan22, Lemma 3.9.18.(iii)] that ${\mathcal {D}}_{\mathrm {\acute {e}t}}(Z/G, {\mathbf {F}}_p)^{\mathrm {oc}} = {\mathcal {D}}^{\mathrm {sm}}(G, {\mathbf {F}}_p(Z))_\omega $ is the derived $\infty $ -category of discrete smooth G-representations on ${\mathbf {F}}_p(Z)$ -modules (use the same argument as in the proof of [Reference MannMan22, Lemma 3.4.26.(iii)]). Our goal is to show that the G-representation ${\mathcal {C}}(G, {\mathbf {F}}_p(Z))$ admits descent in this $\infty $ -category. By considering base-change along ${\mathbf {F}}_p \to {\mathbf {F}}_p(Z)$ we reduce the claim to the following:

• (∗^′′) The algebra object ${\mathcal {C}}(G,{\mathbf {F}}_p) \in {\mathcal {D}}^{\mathrm {sm}}(G, {\mathbf {F}}_p)_\omega $ admits descent.

Note that ${\mathcal {D}}^{\mathrm {sm}}(G,{\mathbf {F}}_p)_\omega $ is equivalent to the full subcategory of spanned by the discrete G-representations (e.g. both $\infty $ -categories can be constructed using the same descent diagram), hence by Proposition 3.10 it follows that for all static $M, N \in {\mathcal {D}}^{\mathrm {sm}}(G,{\mathbf {F}}_p)_\omega $ we have $\pi _0\operatorname {\mathrm {Hom}}(M, N[i]) = 0$ for $i> d := \operatorname {\mathrm {cd}}_p G$ . Now consider the triangle

$$ \begin{align*} {\mathbf{F}}_p \to \operatorname{\mathrm{Tot}}_d({\mathcal{C}}(G,{\mathbf{F}}_p)^{\otimes \bullet+1}) \to X \end{align*} $$

in ${\mathcal {D}}^{\mathrm {sm}}(G,{\mathbf {F}}_p)_\omega $ , where $\operatorname {\mathrm {Tot}}_d$ denotes the d-truncated totalization of a cosimplicial object (i.e. the limit over $\Delta _{\le d}$ ). We know that $\operatorname {\mathrm {Tot}}({\mathcal {C}}(G,{\mathbf {F}}_p)^{\otimes \bullet +1}) = {\mathbf {F}}_p$ (by descent) and by the dual version of [Reference LurieLur, Proposition 1.2.4.5] we have $\pi _i \operatorname {\mathrm {Tot}}_d = \pi _i \operatorname {\mathrm {Tot}}$ for $i> -d$ . It follows that X is concentrated in homological degrees $\le -d$ , so that by the above vanishing of hom’s the connecting map $X \to {\mathbf {F}}_p[1]$ must be zero. It follows that $\operatorname {\mathrm {Tot}}_d({\mathcal {C}}(G,{\mathbf {F}}_p)^{\otimes \bullet +1}) \cong {\mathbf {F}}_p \oplus X$ , which immediately implies that ${\mathcal {C}}(G,{\mathbf {F}}_p)$ admits descent.

Proof. Part (i) follows from Lemma 3.11 by using the factorization $\operatorname {\mathrm {id}} = [* \to */G \to *]$ . To prove (ii) let $(X,\Lambda )$ be given and assume first that $(X,\Lambda )$ admits a map to some strictly totally disconnected space. Let $q\colon X \to X/G$ denote the projection and let $q_\bullet \colon X_\bullet \to X/G$ be the Čech nerve of q. By Lemma 3.11 we have , which means that for all the natural map $\varinjlim _{n\in \Delta } q_{n!} q_n^! \mathcal M \mathbin {\xrightarrow {\sim }} \mathcal M$ is an isomorphism. Since both $f_*$ and $f_!$ preserve small colimits (for $f_*$ this follows from $\operatorname {\mathrm {cd}}_p G < \infty $ ) and all $q_n$ and $f \mathbin {\circ } q_n$ are proper we deduce

$$ \begin{align*} f_* = f_* \varinjlim_n q_{n!} q_n^! = \varinjlim_n (f \mathbin{\circ} q_n)_* q_n^! = \varinjlim_n (f \mathbin{\circ} q_n)_! q_n^! = f_!. \end{align*} $$

The case of general $(X,\Lambda )$ can be deduced via descent (more precisely, do the above computation on sections of the cocartesian fibrations over $\Delta $ ).

Proof. Fix a strictly totally disconnected space $(X,\Lambda ) \in {{\rm vStack}_\Lambda }$ with associated map $f\colon X/G \to X$ . Then is the $\infty $ -category of solid almost $\Lambda (X)$ -modules, while is the $\infty $ -category of smooth G-representations on such modules (see [Reference MannMan22, Lemma 3.4.26]). Moreover, $f_*$ computes smooth G-cohomology and $f^*$ associates to every the trivial G-representation on M.

We first prove (i). By [Reference MannMan22, Lemma 3.4.19] the collection of functors for open compact subgroups H is conservative. Therefore the natural map $f^* \otimes f^! \Lambda ^a \to f^!$ is an equivalence if and only if for all compact open subgroups $H \subset G$ the map $R\Gamma _{\mathrm {sm}}(H, f^* \otimes f^! \Lambda ^a) \to R\Gamma _{\mathrm {sm}}(H, f^!)$ is an equivalence of endofunctors of . Let us denote $f_H\colon */H \to *$ the projection, which factors over the finite étale projection $*/H \to */G$ . Then we see that $f^* \otimes f^! \Lambda ^a \to f^!$ is an equivalence if and only if for all compact open subgroups $H \subset G$ the map

$$ \begin{align*} f_{H*} (f_H^* \otimes f_H^! \Lambda^a) \to f_{H*} f_H^! \end{align*} $$

is an equivalence. By Corollary 3.12 we have $f_{H*} = f_{H!}$ , so in particular $f_{H*}$ satisfies the projection formula. We compute

$$ \begin{align*} f_{H*} (f_H^* \otimes f_H^! \Lambda^a) &= - \otimes f_{H*} f_H^! \Lambda^a = - \otimes {\underline{\operatorname{\mathrm{Hom}}}}(f_{H*}\Lambda^a, \Lambda^a),\\ f_{H*} f_H^! &= f_{H*} {\underline{\operatorname{\mathrm{Hom}}}}(\Lambda^a, f_H^!) = {\underline{\operatorname{\mathrm{Hom}}}}(f_{H!}\Lambda^a, -) = {\underline{\operatorname{\mathrm{Hom}}}}(f_{H*}\Lambda^a, -). \end{align*} $$

It follows that $f^* \otimes f^! \Lambda ^a \to f^!$ is an equivalence if and only if $R\Gamma _{\mathrm {sm}}(H, \Lambda (X)^a) = f_{H*} \Lambda ^a$ is dualizable in for all open compact $H \subset G$ . Now assume that G satisfies condition (i) of Definition 3.13. Then $R\Gamma _{\mathrm {sm}}(H, {\mathbf {F}}_p)$ is dualizable in for all H. If $p = 0$ in $\Lambda (X)$ then the same follows for $R\Gamma _{\mathrm {sm}}(H, \Lambda (X)^a) = R\Gamma _{\mathrm {sm}}(H, {\mathbf {F}}_p) \otimes _{{\mathbf {F}}_p} \Lambda (X)^a$ . For general $\Lambda $ we can argue as in the proof of [Reference MannMan22, Lemma 3.8.4. (i)]. Conversely, assume that $f^* \otimes f^! \Lambda ^a \mathbin {\xrightarrow {\sim }} f^!$ is an isomorphism for all $(X,\Lambda )$ . Pick $X = \operatorname {\mathrm {Spa}} C$ for some algebraically closed field over ${\mathbf {Q}}_p$ and $\Lambda = {\mathcal {O}}_C/p$ . Then the Riemann-Hilbert correspondence from [Reference MannMan22, Theorem 3.9.23] implies that the base-change functor $RH\colon {\mathcal {D}}^{\mathrm {sm}}(G,{\mathbf {F}}_p) \mathrel {\hookrightarrow } {\mathcal {D}}^{\mathrm {sm}}(G,{\mathcal {O}}^a_C/p)^\varphi $ is fully faithful and induces an equivalence of dualizable objects. On the other hand, this base-change functor sends $R\Gamma _{\mathrm {sm}}(H,{\mathbf {F}}_p)$ to $R\Gamma _{\mathrm {sm}}(H,{\mathcal {O}}^a_C/p)$ , hence it follows from the above that $R\Gamma _{\mathrm {sm}}(H,{\mathbf {F}}_p)$ is dualizable for all open compact subgroups $H \subset G$ . But every G-representation M on a finite static ${\mathbf {F}}_p$ -vector space is a finite colimit of the generating representations ${\mathbf {F}}_p[G/H]$ , hence also $R\Gamma _{\mathrm {sm}}(G,M)$ is dualizable. In particular all $H^i(G,M)$ are finite (here we implicitly use $R\Gamma (G,M) = R\Gamma _{\mathrm {sm}}(G,M)$ since M is discrete).

We now prove (ii), so from now on we can assume that G satisfies the equivalent conditions of (i). We then need to see that G is p-Poincaré if and only if for every strictly totally disconnected $(X,\Lambda ) \in {{\rm vStack}_\Lambda }$ with $\Lambda = {\mathcal {O}}^+_{X^\flat }/\pi $ for some pseudouniformizer $\pi $ on $X^\flat $ the object is invertible. Given $(X,\Lambda )$ , let us denote $q\colon X \to X/G$ the projection. Since invertibility satisfies descent, $f^! \Lambda ^a$ is invertible if and only if $q^* f^! \Lambda ^a$ is invertible. For every compact open subgroup $H \subset G$ we denote $f_H\colon */H \to *$ and $q_H\colon */H \to */G$ the natural maps and we compute (using [Reference MannMan22, Proposition 3.3.9.(iii)] and Corollary 3.12)

$$ \begin{align*} q^* f^! \Lambda^a = f_* q_* q^* f^! \Lambda^a = \varinjlim_H f_* q_{H*} q_H^* f^! \Lambda^a = \varinjlim_H f_{H*} f_H^! \Lambda^a = \varinjlim_H {\underline{\operatorname{\mathrm{Hom}}}}(f_{H*} \Lambda^a, \Lambda^a). \end{align*} $$

By the proof of (i) we know that $f_{H*} \Lambda ^a = R\Gamma (H, \Lambda (X)^a)$ is dualizable in and similarly $R\Gamma (H,{\mathbf {F}}_p)$ is dualizable in . It follows that

$$ \begin{align*} {\underline{\operatorname{\mathrm{Hom}}}}(f_{H*} \Lambda^a, \Lambda^a) = {\underline{\operatorname{\mathrm{Hom}}}}(\Gamma(H,{\mathbf{F}}_p), {\mathbf{F}}_p) \otimes_{{\mathbf{F}}_p} \Lambda(X)^a. \end{align*} $$

Therefore we obtain

$$ \begin{align*} q^* f^! \Lambda^a = \varinjlim_H {\underline{\operatorname{\mathrm{Hom}}}}(f_{H*} \Lambda^a, \Lambda^a) = (\varinjlim_H {\underline{\operatorname{\mathrm{Hom}}}}(\Gamma(H,{\mathbf{F}}_p), {\mathbf{F}}_p)) \otimes_{{\mathbf{F}}_p} \Lambda(X)^a \end{align*} $$

We have $R\Gamma (H,{\mathbf {F}}_p) = \bigoplus _i H^i(H, {\mathbf {F}}_p)[-i]$ and since there are no higher $\operatorname {\mathrm {Ext}}$ -groups between discrete ${\mathbf {F}}_p$ -vector spaces we even have a decomposition of diagrams $(R\Gamma (H,{\mathbf {F}}_p)_H = \bigoplus _i (H^i(H,{\mathbf {F}}_p))_H$ . On the other hand, we know that if $(M_i)_i$ is any cofiltered system of finite discrete ${\mathbf {F}}_p$ -vector spaces then $\varprojlim _i M_i$ is concentrated in degree $0$ and ${\underline {\operatorname {\mathrm {Hom}}}}(\varprojlim _i M_i, {\mathbf {F}}_p) = \varinjlim _i {\underline {\operatorname {\mathrm {Hom}}}}(M_i, {\mathbf {F}}_p)$ . Altogether this implies

We deduce:

In particular, if G is p-Poincaré then $q^* f^! \Lambda ^a$ is invertible. Conversely, assume that $*/G \to *$ is p-cohomologically smooth. Choosing $X = \operatorname {\mathrm {Spa}} C$ for some algebraically closed field C, the Riemann-Hilbert correspondence (see [Reference MannMan22, Theorem 3.9.23]) tells us that the invertibility of $q^* f^! \Lambda ^a$ implies that is invertible. But then also is invertible and thus G is p-Poincaré.

Proof. This follows immediately from Proposition 3.14 by observing that being p-fine and being p-cohomologically smooth can both be checked étale locally on the source (note that if $H \subseteq G$ is an open subgroup then the map $*/H \mathrel {\twoheadrightarrow } */G$ is an étale cover).

Proof. Factor $f'$ as $Y/G \to X/G \to X$ . We claim that the first map, i.e. $f/G\colon Y/G \to X/G$ , is bdcs: Base-changing this map along the pro-étale cover $X \mathrel {\twoheadrightarrow } X/G$ gives f, hence the map $Y/G \to X/G$ is separated by [Reference ScholzeSch18a, Proposition 10.11.(ii)], compactifiable by [Reference ScholzeSch18a, Proposition 22.3.(iii)] (note that f is automatically compactifiable by [Reference ScholzeSch18a, Proposition 22.3. (v)]), p-bounded by [Reference MannMan22, Lemma 3.5.10.(ii)] and representable in locally spatial diamonds by [Reference ScholzeSch18a, Proposition 13.4.(iv)]. Since $X/G \to X$ is p-fine by Corollary 3.12 we deduce that $f'$ is p-fine. This proves (i).

To prove (ii), assume in addition that f is p-cohomologically smooth and G is virtually p-Poincaré. Then it follows from [Reference MannMan22, Lemma 3.8.5.(ii)] that $f/G$ is p-cohomologically smooth, and it follows from Theorem 3.16 that the map $X/G \to X$ is p-cohomologically smooth. Together this implies that $f'$ is p-cohomologically smooth, as desired.

Proof of Theorem 3.19

Let G be as in the statement of the theorem. Then condition (i) in Definition 3.13 is classical, see, e.g. [Reference Symonds and WeigelSW00, Theorems 5.1.1 and 5.1.5]. For condition (ii), recall that $R\Gamma _{\mathrm {cts}}(G,{\mathbf {F}}_p[[G]]) \simeq {\mathbf {F}}_p[-\dim G]$ by [Reference LazardLaz65, Ch. V, Sec. 2] and [Reference Symonds and WeigelSW00, Theorem 5.1.9]. Since , the conclusion now follows from Lemma 3.20.

Proof of Lemma 3.20

Arguing as in the proof of [Reference MannMan22, Proposition 3.4.6], one sees that $R\Gamma (G,\underline {V})$ is computed as the cohomology of the complex

$$\begin{align*}0 \to \underline{V} \to \underline{\operatorname{\mathrm{Hom}}}(\underline{G},\underline{V}) \to \underline{\operatorname{\mathrm{Hom}}}(\underline{G}^2,\underline{V}) \to \cdots\end{align*}$$

where the internal homs are taken in condensed sets. By the definition of $\underline {R\Gamma _{\mathrm {cts}}(G,V)}$ , it now suffices to show that $\underline {\operatorname {\mathrm {Hom}}}(\underline {S},\underline {V})$ identifies with $\underline {{\mathcal {C}}(S,V)}$ functorially in locally profinite S. This is a special case of the next lemma.

Proof. The claim regarding the topology on ${\mathcal {C}}(X,Y)$ is easy. For the remaining claim, let T be any profinite space. We then compute that

$$ \begin{align*} \operatorname{\mathrm{Hom}}_{\mathrm{Cond}}(\underline{T}, \underline{\operatorname{\mathrm{Hom}}}(\underline{X},\underline{Y})) &= \operatorname{\mathrm{Hom}}_{\mathrm{Cond}}(\underline{T} \times \underline{X},\underline{Y}) \\ &= \operatorname{\mathrm{Hom}}_{\mathrm{Top}}(T \times X, Y) \\ &= \operatorname{\mathrm{Hom}}_{\mathrm{Top}}(T, {\mathcal{C}}(X,Y)) \\ &= \operatorname{\mathrm{Hom}}_{\mathrm{Cond}}(\underline{T}, \underline{{\mathcal{C}}(X,Y)}) \end{align*} $$

naturally in $T,X,Y$ . The first equality is immediate from the universal property of $\underline {\operatorname {\mathrm {Hom}}}$ . The second equality follows from the fact that on compactly generated T1 spaces, $S \mapsto \underline {S}$ is fully faithful and commutes with finite products. The fourth equality follows similarly, using the first claim in the lemma. The third equality follows from the assumption that X is locally compact Hausdorff, which is exactly the condition needed to ensure that ${\mathcal {C}}(X,-)$ with its compact-open topology is right-adjoint to $- \times X$ . Since T is arbitrary, the result now follows from Yoneda.

Proof. First assume that M is dualizable. Then by [Reference MannMan22, Proposition 3.7.5] both M and $M^\vee $ are discrete. It follows from [Reference MannMan22, Lemmas 3.9.2, 3.9.3] that M is bounded. The isomorphism $M \mathbin {\xrightarrow {\sim }} M^{\vee \vee }$ is clear. This proves the ‘only if’ part of the claim.

We now prove the ‘if’ part, so assume that M is bounded, M and $M^\vee $ are discrete and $M \mathbin {\xrightarrow {\sim }} M^{\vee \vee }$ is an isomorphism. We want to show that M is dualizable, for which we can assume that M is concentrated in homological degree $\ge 0$ . Now consider $M_0 := \pi _0 M$ , which is a static discrete ${\mathcal {O}}^a_K$ -module. From the fact that $M^\vee $ is discrete it follows that $\pi _0 M_0^\vee = \pi _0 M^\vee $ is discrete. After choosing a surjection $\bigoplus _I {\mathcal {O}}^a_K/\pi \mathrel {\twoheadrightarrow } M_0$ we get an injection $\pi _0 M_0^\vee \mathrel {\hookrightarrow } \prod _I {\mathcal {O}}^a_K/\pi $ . Now apply $\pi _0 (-)_*$ to get a submodule

$$ \begin{align*} M_0' := \pi_0 (\pi_0 M_0^\vee)_* \subseteq \prod_I A, \end{align*} $$

where we abbreviate $A := \pi _0 ({\mathcal {O}}^a_K/\pi )_*$ . Note that $M_0'$ is equipped with the subspace topology since it is a kernel of a map $\prod _I A \to \prod _J A$ . Moreover, the fact that $\pi _0 M_0^\vee $ is discrete implies that for every $\varepsilon \in {\mathfrak {m}}_K$ the submodule $T_\varepsilon \subset M_0'$ of $\varepsilon $ -torsion in $M_0'$ is open. In particular there is an open subset $U_\varepsilon \subset \prod _I A$ containing $0$ such that $U_\varepsilon \mathbin {\cap } M_0' \subseteq T_\varepsilon $ . In fact we can choose $U_\varepsilon $ to be an element of the standard basis for the product topology, i.e. $U_\varepsilon = U_\varepsilon ' \times \prod _{I \setminus I_\varepsilon } A$ for some finite subset $I_\varepsilon \subset I$ and some open subset $U_\varepsilon ' \subset \prod _{I_\varepsilon } A$ . This means that if $(a_i)_i, (b_i)_i \in \prod _I A$ lie in $M_0'$ and $a_i = b_i$ for $i \in I_\varepsilon $ then $\varepsilon (a_i)_i = \varepsilon (b_i)_i$ . Now let $M_{0,\varepsilon }' \subset \prod _{I_\varepsilon } A$ be the image of $M_0'$ under the projection $\prod _I A \mathrel {\twoheadrightarrow } \prod _{I_\varepsilon } A$ (equipped with the subspace topology). Then there is a natural projection map $f^{\prime }_\varepsilon \colon M_0' \to M_{0,\varepsilon }'$ . We can also construct a map $g^{\prime }_\varepsilon \colon M_{0,\varepsilon }' \to M_0'$ as follows: Given $a \in M_{0,\varepsilon }'$ we pick any $a' \in M_0'$ such that $f^{\prime }_\varepsilon (a') = a$ . Then we set $g^{\prime }_\varepsilon (a) := \varepsilon a'$ , which by the above observation is independent of the choice of $a'$ . Note that $f^{\prime }_\varepsilon g^{\prime }_\varepsilon = \varepsilon $ and $g^{\prime }_\varepsilon f^{\prime }_\varepsilon = \varepsilon $ , hence we have $M^{\prime }_0 \approx _\varepsilon M^{\prime }_{0,\varepsilon }$ . Going back to the almost world we now see that

$$ \begin{align*} \pi_0 M_0^\vee \approx_\varepsilon M^{\prime a}_{0,\varepsilon}, \qquad \text{with }M^{\prime a}_{0,\varepsilon} \subseteq \prod_{i=1}^{n_\varepsilon} {\mathcal{O}}^a_K/\pi \text{ a discrete submodule}, \end{align*} $$

where we took the liberty of naming $I_\varepsilon = \{ 1, \dots , n_\varepsilon \}$ . By [Reference MannMan22, Lemma 3.7.15] the discrete dualization functor $N \mapsto (N^\vee )_\omega $ is t-exact on discrete N. By applying this functor to $M^\vee $ it follows from the isomorphism $M \mathbin {\xrightarrow {\sim }} M^{\vee \vee }$ that also $M_0 \mathbin {\xrightarrow {\sim }} ((\pi _0 M_0^\vee )^\vee )_\omega $ is an isomorphism. Letting $M_{0,\varepsilon } := ((M^{\prime a}_{0,\varepsilon })^\vee )_\omega $ we deduce that $M_0 \approx _{\varepsilon } M_{0,\varepsilon }$ and there is a surjection $\bigoplus _{i=1}^{n_\varepsilon } {\mathcal {O}}^a_K/\pi \mathrel {\twoheadrightarrow } M_{0,\varepsilon }$ . As this works for all $\varepsilon $ , we deduce that $M_0$ is almost finitely generated as an ${\mathcal {O}}^a_K/\pi $ -module. Then it is also almost finitely generated as an ${\mathcal {O}}^a_K$ -module (where we equip ${\mathcal {O}}^a_K$ with the discrete topology in this proof), hence by [Reference ScholzeSch13, Theorem 2.5] we have

$$ \begin{align*} M_0 \approx ({\mathcal{O}}^a_K/\pi)^r \oplus {\mathcal{O}}^a_K/I_{\gamma_1} \oplus {\mathcal{O}}^a_K/I_{\gamma_2} \oplus \dots \end{align*} $$

for a series of real numbers $1> \gamma _1 \ge \gamma _2 \ge \dots \ge 0$ converging to $0$ . Here for a real number $\gamma $ we denote by $I_\gamma \subset {\mathcal {O}}_K$ the ideal generated by $\pi ^{\gamma '}$ for all $\gamma '> \gamma $ (see [Reference ScholzeSch13, Example 2.4. (i)]). The $\varphi $ -module structure $\varphi ^* M \mathbin {\xrightarrow {\sim }} M$ induces an isomorphism $\pi _0 \varphi ^* M_0 \mathbin {\xrightarrow {\sim }} M_0$ . On the other hand for $\gamma < 1$ we have

$$ \begin{align*} \pi_0 \varphi^*({\mathcal{O}}^a_K/I_\gamma) = \begin{cases} {\mathcal{O}}^a_K/\pi, \qquad &\text{if }\gamma \ge 1/p,\\ {\mathcal{O}}^a_K/I_{p\gamma}, \qquad &\text{else}, \end{cases} \end{align*} $$

as ${\mathcal {O}}^a_K/\pi $ -modules. This forces all $\gamma _i = 0$ , i.e. we have $M_0 \approx ({\mathcal {O}}^a_K/\pi )^r$ . It follows that $M_0$ is dualizable: We need to check that the natural map $\alpha \colon M_0 \otimes M_0^\vee \to {\underline {\operatorname {\mathrm {Hom}}}}(M_0, M_0)$ is an isomorphism but since the same statement holds for $({\mathcal {O}}^a_K/\pi )^r$ in place of $M_0$ one deduces easily that both kernel and cokernel of $\pi _i \alpha $ are killed by $\varepsilon $ for all $i \in {\mathbf {Z}}$ and $\varepsilon \in {\mathfrak {m}}_K$ (cf. the proof of [Reference MannMan22, Lemma 3.7.17]). In particular $M_0$ is flat (which can also be seen directly), so that $\pi _0 \varphi ^* M_0 = \varphi ^* M$ and therefore the $\varphi $ -module structure on $M_0$ restricts to a $\varphi $ -module structure on $M_0$ . Since the conditions on M are preserved under finite limits, we reduce the claimed dualizability of M to that of $\tau _{\ge 1} M$ , which satisfies the same assumptions as M. Since M is bounded, we can conclude by induction.

Proof. Note first that (i) implies (ii) because $R\Gamma _{\mathrm {sm}}(H,V) \otimes _{{\mathbf {F}}_p} {\mathcal {O}}^a_C/\pi = R\Gamma _{\mathrm {sm}}(H, V \otimes _{{\mathbf {F}}_p} {\mathcal {O}}^a_C/\pi )$ for all $V \in {\mathcal {D}}^{\mathrm {sm}}(G,{\mathbf {F}}_p)$ (e.g. by the explicit computation of smooth group cohomology in [Reference MannMan22, Corollary 3.4.17]). Conversely, assume that (ii) is satisfied. Then M is discrete because $M = \varinjlim _H R\Gamma _{\mathrm {sm}}(H,M)$ , where on the right-hand side H ranges through all compact open subgroups of G (see [Reference MannMan22, Lemma 3.4.19]). We claim that M lies in ${\mathcal {D}}^{\mathrm {sm}}(G,{\mathbf {F}}_p)$ , i.e. that the natural morphism $M^\varphi \otimes _{{\mathbf {F}}_p} {\mathcal {O}}^a_C \mathbin {\xrightarrow {\sim }} M$ is an isomorphism. By the colimit property we just used the collection of functors $R\Gamma _{\mathrm {sm}}(H,-)$ , where H ranges through a basis of compact open subgroups of G, is conservative, hence it is enough to show that $R\Gamma _{\mathrm {sm}}(H,M^\varphi \otimes _{{\mathbf {F}}_p} {\mathcal {O}}^a_C) = R\Gamma _{\mathrm {sm}}(H,M)$ for all H in the basis. Noting that that $R\Gamma _{\mathrm {sm}}(H,-)$ commutes with $(-)^\varphi $ (which is formal) and with $- \otimes _{{\mathbf {F}}_p} {\mathcal {O}}^a_C/\pi $ (as used above), the claim boils down to showing that for $M_H := R\Gamma _{\mathrm {sm}}(H,M)$ we have $M_H^\varphi \otimes _{{\mathbf {F}}_p} {\mathcal {O}}^a_C/\pi \mathbin {\xrightarrow {\sim }} M_H$ . But by assumption $M_H$ is dualizable, hence the desired equivalence follows from the Riemann-Hilbert correspondence (see [Reference MannMan22, Theorem 3.9.23]). From the Riemann-Hilbert correspondence we also deduce that $R\Gamma _{\mathrm {sm}}(H, M^\varphi )$ is dualizable, hence $M^\varphi $ is admissible, as desired.

We have now finished the proof that (i) and (ii) are equivalent. To prove that they are also equivalent to (iii) we make the following computation. Let $X = \operatorname {\mathrm {Spa}} C$ , $\Lambda = {\mathcal {O}}_C/\pi $ , $\omega _G := f^! \Lambda ^a$ and for every compact open subgroup $H \subset G$ let $f_H\colon X/H \to X$ be the natural map. Then from Corollary 3.12 and Proposition 3.14 we get:

$$ \begin{align*} f_{H*} M^\vee &= f_{H*} {\underline{\operatorname{\mathrm{Hom}}}}(M, \Lambda^a) = f_{H*} {\underline{\operatorname{\mathrm{Hom}}}}(M \otimes \omega_G, \omega_G) = {\underline{\operatorname{\mathrm{Hom}}}}(f_{H*}(M \otimes \omega_G), \Lambda^a)\\ &= (f_{H*}(M \otimes \omega_G))^\vee. \end{align*} $$

Note also that for small enough H the action of H on $\omega _G$ is trivial, so that $- \otimes \omega _G$ is just a shift in the $\infty $ -category of H-representations. Consequently for small enough H the functor $R\Gamma _{\mathrm {sm}}(H,-)$ preserves duals up to shift. Since the functors $R\Gamma _{\mathrm {sm}}(H,-)$ are also conservative (for H varying in a basis of compact open subgroups of G) it follows that (iii) is equivalent to the claim that M and $M^\vee $ are bounded and for a basis H of compact open subgroups of G we have that both $M_H := R\Gamma _{\mathrm {sm}}(H,M)$ and $M_H^\vee $ are discrete and the natural morphism $M_H \mathbin {\xrightarrow {\sim }} M_H^{\vee \vee }$ is an isomorphism. By Lemma 3.25 this implies (ii). Moreover, if (ii) is satisfied then all the conditions in (iii) follow immediately except possibly that $M^\vee $ is also bounded. To see that $M^\vee $ is indeed bounded, note first that $M^\vee = \varinjlim _H R\Gamma _{\mathrm {sm}}(H, M^\vee )$ (see [Reference MannMan22, Lemma 3.4.19]) and by the above computation the right-hand side is (up to a constant shift) isomorphic to $\varinjlim _H M_H^\vee $ . On the other hand, the discrete dualization functor $N \mapsto (N^\vee )_\omega $ is t-exact on discrete objects in (see [Reference MannMan22, Lemma 3.7.15]). In particular, if $M_H$ is concentrated in homological degrees $[a,b]$ then $M_H^\vee $ is concentrated in homological degrees $[-b,-a]$ . The fact that M is bounded implies that we can choose a and b independently of H, which then implies that $M^\vee $ is also bounded.

It remains to prove the last part of the claim, so assume that M is admissible and bounded. It follows immediately from (iii) that also $M^\vee $ (computed in ) is admissible (and bounded), so in particular it is discrete and comes from an ${\mathbf {F}}_p$ -representation via the Riemann-Hilbert correspondence. It is thus formal that this dual can be computed in any of the claimed $\infty $ -categories.

Proof. For every we get the following chain of natural equivalences (using that g is proper and smooth and that f is smooth):

$$ \begin{align*} {\underline{\operatorname{\mathrm{Hom}}}}(g_* f^* A, {\mathcal{O}}^{+a}_{B\underline{D^\times}}/p) &= {\underline{\operatorname{\mathrm{Hom}}}}(g_! f^* A, {\mathcal{O}}^{+a}_{B\underline{D^\times}}/p)\\ &= g_* {\underline{\operatorname{\mathrm{Hom}}}}(f^* A, g^!({\mathcal{O}}^{+a}_{B\underline{D^\times}}/p))\\ &= g_* {\underline{\operatorname{\mathrm{Hom}}}}(f^* A, {\mathcal{O}}^{+a}_X/p)[2n-2](n-1)\\ &= g_* f^* {\underline{\operatorname{\mathrm{Hom}}}}(A, {\mathcal{O}}^{+a}_{B\underline G}/p)[2n-2](n-1). \end{align*} $$

In the last step we used that by the smoothness of f, $f^*$ commutes with internal hom (see [Reference MannMan22, Proposition 3.8.7].

Proof of Theorem 4.2

Fix an admissible $A \in {\mathcal {D}}^{\mathrm {sm}}(G,{\mathbf {F}}_p)$ as in the claim and let us denote $A' := A \otimes \mathcal {O}_{B\underline {G}}^{+a}/p$ the corresponding object in given by the Riemann-Hilbert embedding. Then $A'$ is admissible (by definition) and hence by Proposition 3.26 both $A'$ and $A^{\prime \vee }$ are bounded and discrete and the natural map $A' \mathbin {\xrightarrow {\sim }} A^{\prime \vee \vee }$ is an isomorphism. Clearly preserves bounded and discrete objects (this is true for $f^*$ like for any pullback and it follows for $g_*$ from the fact that g is proper and p-bounded). Thus it follows from Lemma 4.3 that both and are bounded and discrete. Moreover, Lemma 4.3 also implies that there is an isomorphism , so altogether it follows from Proposition 3.26 (and Remark 3.27) that is admissible. This immediately implies (i). It also implies (ii) because we now know that lies in the image of the Riemann-Hilbert functor. Now (iii) follows from Lemma 4.3 and the last part of Proposition 3.26.

Proof. Evaluate both sides of the duality isomorphism $\mathcal {S}_{D^\times } \circ \mathcal {J}) \cong (\mathcal {J} \circ \mathcal {S}_G)[2n-2](n-1)$ on $\pi $ . Since $\mathcal {J}(\pi )=\mathcal {J}^{c}(\pi )[-c]$ by assumption, the left-hand side evaluates to $\mathcal {S}_{D^\times }(\mathcal {J}^{c}(\pi )[-c]) \cong \mathcal {S}_{D^\times }(\mathcal {J}^{c}(\pi ))[c]$ . Next, since $\mathcal {S}_G(\pi )=\check {\pi }[-\dim _G \pi ]$ by assumption, the right-hand side simplifies to $\mathcal {J}(\check {\pi }[-\dim _G \pi ])[2n-2](n-1) \cong \mathcal {J}(\check {\pi })[2n-2-\dim _G \pi ](n-1)$ . Equating both sides and rearranging the shifts, we get $\mathcal {S}_{D^\times }(\mathcal {J}^{c}(\pi )) = \mathcal {J}(\check {\pi })[2n-2-\dim _G \pi -c](n-1)$ . Passing to cohomology on both sides (and ignoring the Tate twists), we get the result.

Proof of Theorem 1.6

Let $d=\dim _{G}(\pi )$ , so $\mathcal {S}_{G}(\pi )\in {\mathcal {D}}^{\mathrm {sm}}(G)^{[0,d]}$ . Since $\mathcal {J}(-)$ carries ${\mathcal {D}}^{\mathrm {sm}}(G)^{[a,b]}$ into ${\mathcal {D}}^{\mathrm {sm}}(D^\times )^{[a,b+2n-2]}$ , we deduce that

$$\begin{align*}(\mathcal{J}\circ\mathcal{S}_{G})(\pi)[2n-2]\in {\mathcal{D}}^{\mathrm{sm}}(D^\times)^{ \leq d}. \end{align*}$$

By the duality isomorphism $(\mathcal {S}_{D^\times }\circ \mathcal {J})(\pi )\cong (\mathcal {J}\circ \mathcal {S}_{G})(\pi )[2n-2]$ , we get that $M=(\mathcal {S}_{D^\times }\circ \mathcal {J})(\pi )\in {\mathcal {D}}^{\mathrm {sm}}(D^\times )^{ \leq d}$ . Now look at the fourth quadrant spectral sequence

$$\begin{align*}E_{2}^{i,j}=\mathcal{S}_{D^\times}^{i}(\mathcal{J}^{-j}(\pi))\Rightarrow H^{i+j}(M). \end{align*}$$

Let $c=\max _{i}\dim _{D^\times }\mathcal {J}^{i}(\pi )$ , and let $i_{0}$ be the least i realizing this maximum. Then $E_{2}^{c,-i_{0}}$ is c-dimensional as an $D^\times $ -representation, and as we turn the pages of the spectral sequence it only interacts with representations of dimension $<c$ via the incoming differentials. Therefore a nonzero subquotient survives to the $E_{\infty }$ -page, giving that $H^{c-i_{0}}(M)\neq 0$ . But $M\in {\mathcal {D}}^{\mathrm {sm}}(D^\times )^{ \leq d}$ , so this implies that $c-i_{0}\leq d$ , and therefore $c\leq d+i_{0}\leq d+N_{\pi }$ , as desired.