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!



Vector bundles on adic spaces
$\newcommand{\cO}{\mathcal{O}}\newcommand{\bZ}{\mathbb{Z}}$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.


