Usually, Berkovich analytic spaces are derived from some Banach rings (or chains of Banach rings) over a completely normed field $k$ through Berkovich spectrum. But when the base field is the complex number $\Bbb{C}$, there is another way of getting analytic spaces, namely gluing open sets in $\Bbb{C}^n$ via biholomorphic transition maps. However, with the "gluing method" we get **more spaces**, for example a non-algebraic complex $2$-torus that admits no non-constant meromorphic functions and thus not the Berkovich spectrum of any $\Bbb{C}$-Banach rings.


What happens if we apply the "gluing method" to open sets of the $n$-dimensional Berkovich affine space $\Bbb{A}^n\_k$? *(For simplicity, only $k=\Bbb{C}\_p$, the $p$-adic complex numbers with $p$-adic metric, are considered.)* Do we also get **compact** "non-algebraic" analytic spaces, in the sense that the function field has a small transcendental degree over $k$, or it can not be realized as the underlying Berkovich space of any $k$-algebraic varieties?


**Edit:** As Wojowu comments, there do exist non-algebraic torus analogues for $\Bbb{C}\_p$-Berkovich spaces. So I'm now asking for a **simply connected** example. (For $\Bbb{C}$-analytic spaces we have non-algebraic $K3$ surfaces.)



"Non-algebraic" Berkovich spaces
Indeed, every complex manifold is locally isomorphic to an open in $\mathbf{C}^n$. More generally, we define complex analytic spaces as those locally isomorphic to a locally ringed space of the form $(Z, i^{-1}\mathcal{O}\_U/I)$ where $U$ is an open in $\mathbf{C}^n$, $I$ is the ideal in the sheaf $\mathcal{O}\_U$ of holomorphic functions defined by finitely many sections $f\_1, \ldots, f\_r$, where $Z\subseteq U$ is the vanishing locus of the $f\_i$, and $i\colon Z\to U$ is the closed embedding.


Following Berkovich, one can define analytic spaces over any Banach ring $A$ in basically the same way. We define $\mathbf{A}^n\_A$ as the set of multiplicative real-valued seminorms on the polynomial ring $A[T\_1, \ldots, T\_n]$ which are bounded on $A$ by the given Banach norm. We give it the coarsest topology making the maps $|\cdot| \mapsto |f|$ ($f\in A$) continuous. We endow it with a structure sheaf $\mathcal{O}$ defined using a certain completion of rational functions (see Berkovich's first book of Lemanissier-Poineau for details). We can then define $A$-analytic spaces as in the previous paragraph.


For $A=\mathbf{C}$ with the Euclidean norm, we recover complex analytic spaces, while for $A$ being a non-Archimedean field, we recover the category of "boundaryless" (or "partially proper") Berkovich spaces. For more general $A$, the study of such spaces is very difficult (see the work of Poineau and Berger for the case $A=\mathbf{Z}$).


However, even for $A=\mathbf{C}\_p$, it is not true that a smooth Berkovich space is locally isomorphic to an open in $\mathbf{A}^n$. For example, the Berkovich analytification of a smooth projective curve over $\mathbf{C}\_p$ of positive genus and good reduction will not have this property: it will have a unique point (of "type 2") which does not admit the open disc as a neighborhood.


Turning to your question. It is indeed easy to produce "non-algebraic" Berkovich spaces over $\mathbf{C}\_p$ by taking a smooth projective variety $X\_0$ over $k=\overline{\mathbf{F}}\_p$ with unobstructed deformations and $H^2(X\_0, \mathcal{O}\_{X\_0})\neq 0$ and taking a "random" formal deformation $\mathfrak{X}$ of $X\_0$ over a complete discrete valuation ring $W$ with residue field $k$, embedding $W$ into $\mathbf{C}\_p$, and taking $X$ to be the Berkovich space associated to the rigid-analytic generic fiber of the formal scheme $\mathfrak{X}$. For a generic choice of the deformation, we will have ${\rm Pic}(X)=0$, so in particular $X$ will not be isomorphic to the analytification of any algebraic variety over $\mathbf{C}\_p$. This applies to abelian varieties of dimension $>1$ and to K3 surfaces.


However, if $X\_0$ is not rational, the $X$ resulting from the above construction will never have the property you imposed that it is locally isomorphic to an open subset of the affine space, because the generic point of $X\_0$ will have a unique preimage in $X$ without such a neighborhood.


To achieve this, we need to be more clever. We pick $X\_0$ to be semistable (or log smooth over the log point $\mathbf{N}\to k$) and maximally degenerate. Think of the product of several Tate curves. In this case, the generic fiber of a "random" smoothing over $W$ will still be nonalgebraic, but will admit a Tate-Raynaud uniformization by $(\mathbf{C}\_p^\*)^g$, and hence will have the property you wanted.


Finally, you asked for a simply connected example. Here are some indications how this could go. Consider the Dwork family of K3 surfaces over $\mathbf{Z}\_p$ given by
$$ p(x^4 + y^4 + z^4 + t^4) = xyzt $$
where $p>5$. This is projective alright. Consider a random smooth formal scheme $\mathfrak{X}$ over $\mathbf{Z}\_p$ which is isomorphic to this one modulo $p^2$, and let $X$ be the associated Berkovich space over $\mathbf{C}\_p$, a non-algebraic rigid-analytic K3 surface. After a suitable blowup (see Section 1 in Harris, Shepherd-Barron, Taylor "A family of Calabi-Yau varieties and potential automorphy"), the model $\mathfrak{X}$ is Zariski-locally isomorphic to
$$ \mathfrak{U} = {\rm Spf}(\mathbf{Z}\_p\langle x, y, z\rangle/(p-xyz)) $$
and hence every point of $X$ admits an open neighborhood (maybe in the weaker sense of an affinoid neighborhood) isomorphic to an open in the generic fiber $U$ of $\mathfrak{U}$. But $U$ itself embeds as an open into $(\mathbf{A}^2)^{\rm an}$, and we are done.


**Postscriptum.** Since you seem to be interested in the homotopy type of the Berkovich space $X$, in the above example it should be homotopy equivalent to $S^2$, as the dual complex of the special fiber is a $2$-sphere. I am sure that the people working on "non-Archimedean SYZ" (following the ideas of Konstevich and Soibelman, Gross and Siebert) have figured it all out in details.


