Vector bundles on adic spaces

Question

Let $X = \mathrm{Spa}(A,A^+)$ be an analytic sheafy adic space. Let $\mathcal{E}$ be a locally finite free $\mathcal{O}_X$ sheaf. Does $\mathcal{E}$ correspond to a geometric vector bundle over $X$? In other words, does there exist an analytic adic space $E$ with a morphism to $X$, unique up to isomorphism as an analytic adic space over $X$, such that for every open immersion of analytic adic spaces $S\rightarrow X$ there is a natural isomorphism $\mathcal{E}(S) \cong \mathrm{Hom}_X(S,E)$?

By a theorem of Kedlaya-Liu, we do know that there is a correspondence between $\mathcal{E}$'s and finite projective $A$-modules, which is achieved by mapping $\mathcal{E}$ to the global sections $M=\mathrm{H}^0(X,\mathcal{E})$. So a natural guess for $E$ would be to take $\mathrm{Spa}(B,B^+)$ with $B=\mathrm{Sym}_A(M^\vee)$; however, I do not know what to take for $B^+$. (Is there some sort of $M^+$?)

Thanks!

Answer 1

$\newcommand{\cO}{\mathcal{O}}\newcommand{\bZ}{\mathbb{Z}}$Let's first work out the case $\mathcal{E}=\mathcal{O}_X$. We want a space $E\to X$ such that $Hom_X(S, E)=\cO_S(S)=Hom(S,\mathbb{A}^1)$. Here $\mathbb{A}^1$ is the adic space representing the functor $Y\mapsto \cO_Y(Y)$ on all adic spaces, it is given by $\mathbb{A}^1=\mathrm{Spa}(\bZ[x],\bZ)$ where the topology on $\bZ[x]$ is discrete, see e.g. 4.1 in Scholze's lectures https://www.math.uni-bonn.de/people/scholze/Berkeley.pdf

Therefore in the case of $\mathcal{E}=\cO_X$ the total space $E$ is $X\times \mathbb{A}^1$ where the product is taken in the category of adic spaces. If $X$ lives over some $\mathrm{Spa}(K,\cO_K)$ for a non-archimedean field $K$ then $E$ is equal to $X\times_K \mathbb{A}^1_K$ where $\mathbb{A}^1_K$ is the analitifycation of the algebraic scheme $\mathbb{A}^1_K$. In particular the total space $E$ is never quasi-compact, hence it can't be affinoid. Roughly speaking, the issue with the guess you give is that it does not account for the topology on the module $M$.

To get to the general case from here we simply need to glue together total spaces over trivializing charts. To make sure that the non-sheafiness issue does not get in out way, let's assume that $X$ lives over $K$. Then, if $X=\bigcup U_i$ is a cover trivializing $\mathcal{E}$, the total space $E$ is obtained by gluing together the spaces $U_i\times \mathbb{A}^{rk\,\mathcal{E}}$.

I'm afraid I can't supply a reputable source, but hopefully somebody else will.

Answer 2

The question is local on $X$, so we may assume that $\mathcal E$ is finite free, of rank $n$, say. In that case, as also SashaP points out, the question amounts to the question whether $\mathbb A^n_X$ is an adic space. As this is covered by an increasing union of balls $\mathbb B^n_X=\mathrm{Spa}(A\langle T_1,\ldots,T_n\rangle,A^+\langle T_1,\ldots,T_n\rangle)$, it is also equivalent to ask whether $A\langle T_1,\ldots,T_n\rangle$ is sheafy. (Note that your choice of $B$ is not correct, as your $B$ is not a Huber ring. You have to complete $B$ in a certain way -- this is like the difference between $A[T_1,\ldots,T_n]$ and $A\langle T_1,\ldots,T_n\rangle$ -- and then take a union over all possible such choices of completions.)

Now I'm not up to speed about the known counterexamples to sheafyness, but I think there ought to be examples where $A$ is sheafy but $A\langle T\rangle$ is not, so in this sense the answer to your question would be no.

On the other hand, any practical condition guaranteeing sheafyness of $A$ usually also implies sheafyness of $A\langle T\rangle$ -- e.g., strongly noetherian, or sousperfectoid, or ... .

Finally, let me also add again the advertisement that in the setting of Analytic Geometry, it is possible to generalize Huber's theory of adic spaces to incorporate non-sheafy Huber rings (in particular, by allowing the structure sheaf to be a sheaf of animated condensed rings). So in this generalized setting, the answer to the question would also be Yes.