Every curve is a geodesic??

Question

I've been reading up on how isometries send geodesics to geodesics. I recently saw a proof of another theorem that used the fact:

The set of fixed points of an isometry is a geodesic.

But isnt the Identity always an isometry, which would then imply every curve, in say the Poincare half plane, is a geodesic. Whats wrong with my reasoning?

Thanks

As stated, that clearly can't be true because a geodesic is a curve, whereas the fixed set of the identity (which is the entire manifold, as you point out) can have dimension $>1$. The correct statement of the theorem you're referring to is here. — Zev Chonoles, May 16 '13 at 00:25

score 5 · Accepted Answer · answered May 16 '13 at 00:27

You're probably recalling or interpreting the fact they used incorrectly. The relevant fact is the following:

If $M$ is a Riemannian manifold and $f: M \longrightarrow M$ an isometry, then each connected component of $\mathrm{Fix}(f)$ is a closed totally geodesic submanifold of $M$.

Note that a submanifold $N \subset M$ being totally geodesic doesn't mean that $N$ is a geodesic in $M$, it means that every geodesic in $N$ is a geodesic in $M$ as well.

Every curve is a geodesic??

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