Local geodesics

Question

Suppose I have a $K$-local geodesic in a metric space $M$, $\alpha(t) : [0,L] \rightarrow M$, meaning $$\forall t ,t' \in [0,L] \quad \text{s.t.} \quad |t-t'|\leq K, \quad |\alpha(t)-\alpha(t')|=|t-t'| $$
and suppose I take two points $t,t' \in [0,L]$ such that $|t-t'|=2K$ , assuming t'>t and I divide the segment in the following way: $$[\alpha(t),\alpha(t')]=[\alpha(t),\alpha(t+K)]\, \cup \,[\alpha(t+K),\alpha(t')]$$ so every part is locally geodesic. Now suppose I want to calculate $d(\alpha(t),\alpha(t'))$ I get the following: $$d(\alpha(t),\alpha(t'))= d(\alpha(t),\alpha(t+K))+d(\alpha(t+K),\alpha(t'))=(t+K-t)+ (t'-t-K)=t'-t.$$ But the original $t,t'$ were in a distance that doesn't ensure that. In that way I shall get the same result for every $ t,t'$ such that $|t-t'|>K$. So what am I missing?

Answer 1

Pretty much everything.

Start with the faulty notation $[\alpha(t),\alpha(t')]$. What can it possibly mean? Do you mean the image of a geodesic (there might be many!) connecting $\alpha(t), \alpha(t')$?

More importantly, the equality $$d(\alpha(t),\alpha(t'))=d(\alpha(t),\alpha(t+K))+d(\alpha(t+K),\alpha(t'))$$ comes out of nowhere and is, in general false.

I suggest, you think of the map $$ \alpha: [0, 2\pi]\to M=S^1\subset {\mathbb C}, $$ $$ \alpha(t)=e^{it}, $$ where $S^1$ is equipped with the angular metric. Prove that $\alpha$ is a $\pi$-local geodesic. See where your argument breaks down in this example.