Let $X$ be an n-dimensional polyhedral space with, say, $n\geq 3.$ Let also $p\in X$ be a vertex on a triangulation $\tau$ of $X,$ so a vertex on the polyhedral space.


The tangent cone (as a metric space) of $X$ at $p$ is given by the limit $$\lim\_{\lambda \to \infty} (X,\lambda d\_X, p),$$ where $d\_X$ is the distance on $X.$ Lebedeva and Petrunin (<https://arxiv.org/pdf/1402.6670.pdf>) have shown that a compact length space $X$ is polyhedral if and only if a neighbourhood of each point on $X$ admits an open isometric embedding to a Euclidean cone which sends said point to the tip of the point. I am trying to figure out what the cone on a vertex point would be.


Say, in particular, that $n=3.$ So we have that the vertex point $p\in X$ is where 3 edges meet. Any intuition for how the cone should look like? In particular, the space of directions or the link of this cone.



Tangent cone on polyhedral spaces
Your question isn't completely clear, but I think you mean that $p$ is the vertex of a simplex in the triangulation of $X$. In that case, take all the simplices containing $p$, viewed as simplices embedded in $\mathbb R^3$, and note how they are glued together inside $X$. Now take the *cone* over each simplex *based at* $p$; this is set of all geodesic rays emanating from $p$ which pass through another point of the simplex. This gives you a bunch of convex cones in $\mathbb R^3$, which you now glue together using the same gluing scheme as for the original simplices. The resulting space is the Euclidean cone you want—it is the *tangent cone* of $X$ at $p$, denoted by $T\_pX$.


The unit sphere about $p$ in $T\_pX$, equipped with its intrinsic metric, is what I would call the *link* of $p$ in $X$, but I don't know how standard this is. To get the space of directions, take the link as I just defined it, and (as suggested by Anton) replace any distance greater than $\pi$ by $\pi$.


