Let $\mathbb{H}^{2}$ be hyperbolic $2$-space and $G$ a discrete subgroup of $PSL_{2}(\mathbb{R})$. Then the quotient $\mathbb{H}^{2} / G$ is a surface. Can this be extended to $2$-complexes? In other words, are negatively curved $2$-complexes homeomorphic to quotients of the form $\mathbb{H}^{2}/G$, where $G$ is some group?
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$\begingroup$ Strictly speaking, graphs are negatively curved (locally $CAT(-1)$, and indeed locally CAT($-\infty$)) and are the most obvious counterexamples. $\endgroup$– YCorCommented Mar 26, 2017 at 20:11
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In this vague form your question has an easy negative answer: take two copies of the hyperbolic plane and identify them along a half-space, this gives you an example of a negatively curved space which is not a surface.
In fact there are more "natural" (and more complicated) objects which answer your question in the negative, called hyperbolic buildings. As in the second example they are contractible polygonal complexes locally modeled the hyperbolic plane but the branching pattern is along the 1-skeleton of a regular tesselation (for example that by right-angled pentagons). Some of these spaces also admit discrete cocompact actions. You can look at this paper of Frédéric Haglund and Frédéric Paulin and its references for more information : https://arxiv.org/abs/math/9812167.
(1st paragraph edited for clarity following Ycor's comment)
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1$\begingroup$ Actually gluing infinitely copies of $H^2$ in this way, in an arborescent way, with singularities at all points with integer Busemann function, yields "natural" spaces with cocompact (nondiscrete) isometry groups, and called millefeuille spaces. Groups such as the semidirect product $(\mathbf{R}\times\mathbf{Q}_p)\rtimes\mathbf{Z}$ with contracting action, e.g., by multiplication by $(1/2,p)$, admit isometric cocompact proper action on such spaces (branched $(1+p)$ times at every point of integral Busemann function). $\endgroup$– YCorCommented Mar 26, 2017 at 15:49