How can I prove that a systole of a close oriented hyperbolic surface S is a simple closed curve?

Question

I looked into all the literature regarding systoles (ex. Katz) and everywhere I only see the statement, but without a proof. Also, if possible can anyone recommend me some articles or books on this topic? Thank you.

Answer 1

From your comment and the link provided there, I take your question to be the following:

Why is the shortest closed geodesic a simple closed geodesic?

Suppose that $\gamma : S^1 \to S$ is a closed geodesic.

I'll prove that if $\gamma$ is not simple then there is a shorter closed geodesic.

Since $\gamma$ is not simple, there exists a proper subarc $[x,y]$ of $S^1$ with $x \ne y$ such that $\gamma(x)=\gamma(y)$. The restriction $\gamma \mid [x,y]$ is therefore a closed curve. Clearly $$\text{Length}(\gamma \mid [x,y]) < \text{Length}(\gamma) $$ Also, $\gamma \mid [x,y]$ is not homotopically trivial, because if it were then $\gamma$ would be freely homotopic to the shorter curve $\gamma \mid S^1 - (x,y)$, contradicting that $\gamma$ is the shortest curve in its free homotopy class.

It follows that $\gamma \mid [x,y]$ is homotopic to a closed geodesic $\delta$. Also, $\delta$ is the shortest curve in its homotopy class, and therefore $$\text{Length}(\delta) \le \text{Length}(\gamma \mid [x,y]) $$ Putting this together, we have found a closed geodesic $\delta$ which is shorter than $\gamma$, using only that $\gamma$ is not simple.