Let $P$ be a non-negatively curved (in the Alexandrov sense) polyhedral space (of dimension 3, say), $p,q\in P$ be vertices, and let $e$ be an edge connecting $p$ and $q$. Assume $e$ has cone angle $0< \alpha \leq 2\pi$. We know that the space of directions of $p,$ $\Sigma_p P,$ is a polyhedral surface of curvature bounded below by 1 and there is a conical point $x\in \Sigma_p P$ corresponding to the edge $e$ 'coming into $p$'. In this case, we also know that our surface is homeomorphic to $\mathbb{S}^2$ by Gauss-Bonnet.
By definition, there is a small $r_0>0$ such that the ball $B(p,r_0)$ is isometric to the cone over $\Sigma_p P,$ $C(\Sigma_pP).$ The same is true for a tubular neighbourhood of the edge $e$. If the length of $e$ is $l>0$, then there exists $r_1>0$ such that a tubular neighbourhood of $e$, denoted here by $U_{r_1}(e)$, is isometric to $(0,l)\times C(S^1_{\alpha})$, where $S^1_{\alpha}$ is just $S^1$ with the metric rescaled so that its length is equal to $\alpha.$ Finally, on $\Sigma_p P$ there is a small ball centred at $x$ which is isometric to $C(S^1_{\alpha}).$
First of all, I would appreciate it if someone could tell me if this is all correct. I have been reading about polyhedral spaces in quite a few books and papers, so I might have mixed things up (any reference recommendation is also very welcome).
Assuming all of the above is fine, my question is the following. Do $B(p,r_0)$ and $U_{r_1}(e)$ intersect? In the picture I have in mind, they do, but I'm not sure. Also, if they do intersect, what happens to the distance metric at the intersection? On one hand, it has to be isometric to a 3-dimensional cone over a spherical polyhedral surface (the space of directions at $p$). On the other hand, it has to be isometric to $(0,l)\times C(S^1_{\alpha})$. Am I missing something obvious here? Thank you in advance!
