## sheafiness; perversity

$\bullet$ One of the main annoyances in the theory of adic spaces is that, for a given Huber pair $(A,A^+)$, the structure presheaf on $\mathrm{Spa}(A,A^{+})$ is not always a sheaf. One usually remedies this by restricting attention to various classes of Huber rings, e.g. strongly Noetherian Tate rings, perfectoid Tate rings, Noetherian adic rings, etc. However, the following class of rings doesn’t seem to be addressed in the literature:

Definition. Let $A$ be an adic ring with finitely generated ideal of definition $I \subset A$. We say $A$ is strongly Noetherian outside $I$ if, for all $n \geq 0$, the scheme $\mathrm{Spec}\,A\left\langle X_1,\dots,X_n \right\rangle \smallsetminus V(IA\left\langle X_1,\dots,X_n \right\rangle )$ is Noetherian.

Here $A\left\langle X_1,\dots,X_n \right\rangle$ denotes the usual ring of restricted power series. Note that if $A$ is a Tate ring and $(A_0,(\varpi))$ is any couple of definition, then $A$ is strongly Noetherian if and only if $A_0$ is strongly Noetherian outside $(\varpi)$. I should also point out that the condition of being strongly Noetherian outside $I$ is already considered in a very interesting paper of Fujiwara-Gabber-Kato; they use the terminology “topologically universally rigid-Noetherian”, but I prefer my terminology on account of the previous sentence. Anyway, the following conjecture seems reasonable:

Conjecture. If $A$ is strongly Noetherian outside $I$, the structure presheaf on $\mathrm{Spa}(A,A)$ is a sheaf.

This implies that any strongly Noetherian Tate ring is sheafy (which of course is already known), but it also implies e.g. that if $A$ is topologically finitely presented over $\mathcal{O}_K$ for some nonarchimedean field $K$, then $A$ is sheafy. Sheafiness in the latter situation is known when $K$ is discretely valued, but to the best of my knowledge it’s open for general $K$.

I’m sure this conjecture is within reach, and maybe it’s an easy exercise for experts on sheafiness (*cough* Kiran *cough*). Note that FGK already proved some interesting consequences of this definition, which are probably relevant to proving this conjecture. Precisely, they show that if $A$ is strongly Noetherian outside $I$, then:
1. The $I$-power-torsion submodule of any finitely generated $A$-module is killed by a power of $I$.
2. If $N \subset M$ is any inclusion of $A$-modules, with $M$ finitely generated, then the subspace topology on $N$ induced by the $I$-adic topology on $M$ coincides with the $I$-adic topology on $N$.

$\bullet$ Let $j:U \to X$ be some immersion of varieties over a separably closed field. Everyone knows that the intermediate extension functor $j_{!\ast}$ on perverse sheaves (say with coefficients in $\Lambda = \mathbf{Q}_{\ell}$) is pretty great: it’s totally canonical, it commutes with Verdier duality, it preserves irreducibility, it preserves monic and epic maps of perverse sheaves, etc.

Recently I noticed that if $f: Z \to X$ is any map of varieties, with $X$ smooth and $Z$ irreducible, there’s still a natural functor $f^{\ast !}:\mathrm{Perv}(X) \to \mathrm{Perv}(Z)$ which commutes with Verdier duality. To define this functor, note that for any $f$ and any $\mathcal{F} \in D^b_c(X)$, there is a natural map $f^\ast \mathcal{F} \otimes Rf^! \Lambda \to Rf^! \mathcal{F}$, obtained by adjunction from the composite map $Rf_!( f^\ast \mathcal{F} \otimes Rf^! \Lambda) \cong \mathcal{F} \otimes Rf_! Rf^! \Lambda \to \mathcal{F}$ (the first isomorphism here is the projection formula). Since $X$ is smooth, the dualizing complex $\omega_X$ is just $\Lambda[2 \dim X]$, so then $Rf^! \Lambda = Rf^!\omega_X[-2 \dim X] = \omega_Z[-2 \dim X]$. Thus we get a natural map $\alpha: f^\ast \mathcal{F} \otimes \omega_Z[-2 \dim X] \to Rf^! \mathcal{F}$.

Next, note that the complex $\omega_Z$ is concentrated in degrees $[-2 \dim Z,0]$, and in the lowest of these degrees it’s just the constant sheaf, i.e. $\tau^{ \leq -2 \dim Z} \omega_Z \cong \Lambda[2 \dim Z]$. In particular, there is a canonical map $\Lambda[2 \dim Z] \to \omega_Z$. Shifting by $-2 \dim X$ and tensoring with $f^\ast \mathcal{F}$ gives a map $\beta: f^\ast \mathcal{F}[2 \dim Z - 2 \dim X] \to f^\ast \mathcal{F} \otimes \omega_Z[-2 \dim X]$. Putting things together, we get a natural map $\alpha \circ \beta: f^\ast \mathcal{F}[2 \dim Z - 2 \dim X] \to Rf^! \mathcal{F}$. Set $c= \dim X - \dim Z,$ so after shifting this becomes a natural map

$\gamma=\gamma_{\mathcal{F}}: f^{\ast}\mathcal{F}[-c] \to Rf^!\mathcal{F}[c]$.

This shifting has the advantage that Verdier duality exchanges the functors $f^{\ast}[-c]$ and $Rf^![c]$ on $D^b_c$, and one can check that the Verdier dual of $\gamma_{\mathcal{F}}$ identifies with $\gamma_{\mathbf{D}\mathcal{F}}$.

Definition. The functor $f^{\ast !}: \mathrm{Perv}(X) \to \mathrm{Perv}(Z)$ sends any $\mathcal{F}$ to the image of the map $^p\mathcal{H}^0(\gamma):$ $^p\mathcal{H}^0(f^{\ast} \mathcal{F}[-c]) \to$ $^p\mathcal{H}^0(Rf^!\mathcal{F}[c])$.

Here of course $^p\mathcal{H}^0(-)$ denotes the zeroth perverse cohomology sheaf.

Exercise. Show that $f^{\ast !}(\Lambda[\dim X]) \cong \mathcal{IC}_Z$.

It might be interesting to compute this functor in some other examples. Note that it can be quite stupid: if $f$ is a closed immersion (with $c > 0$) and $\mathcal{F} \in \mathrm{Perv}(X)$ is already supported on $Z$, then $f^{ \ast !} \mathcal{F} = 0$. On the other hand, if $f$ is smooth and surjective, then $f^{\ast !} \cong f^{\ast}[-c] \cong Rf^![c]$ is faithful.