I want to find an explicit geodesic between two points on the geometric realization of a connected graph.
It is intuitively clear that the geometric realization of a connected graph is a geodesic space, but I don't know how to find an explicit geodesic between two arbitrary elements on it.
Thanks!
Here is my proof: The geometric realization of a connected graph is obtained by identifying each edge with a unit interval. Assume Y is the geometric realization of the connected graph X=(V,E). $\left ((u,v),t \right )$ is the point on Y between the vertices u and v at the time $t\in [0,1]$ where {u,v} is in E. Let $((u,v),t),((a,b),s)\in Y$. Assume $d_{X}(u,b)=n$= minmal length of a path P connecting u and b in X. Say $P=u=u_{0},....,u_{n}=b$. Let I=[0,n], then $I= \bigcup_{i=1}^{n}I_{i}$ where $I_{i}$ is the unit interval $[i-1,i], i=1,...,n.$ Define: $ \theta_{i} :[i-1,i]\rightarrow Y$ by $\theta_{i}(t)=\frac{i-t}{i}u_{i-1}+\frac{t}{i}u_{i}$ and then define $\theta:[0,n]\rightarrow Y$ by $\theta (t)=\theta _{i}(t)$ if t is in [i-1,i]. Then $\theta:[0,n]\rightarrow Y$ is (continuous by the pasting lemma) a geodesic between u and b and hence $\theta |_{[t,s]}$ is a geodesic between $((u,v),t),((a,b),s)\in Y$.
Is there anything wrong with my proof?