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!

$$\newcommand{\cO}{\mathcal{O}}\newcommand{\bZ}{\mathbb{Z}}$$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.

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.