Two isometries that have same value and differential at some point are the same.

Question

I also have trouble in this problem:

Let $f, g$ be two isometries of a connected Riemannian manifold $(M, g)$. If $f(p)=g(p)$, $df_p = dg_p$, show that $f=g$.

Any comment is expected. I know it is a well-known rigidity result of Riemannian manifold. Could anyone give some advice?

Answer 1

Let $q\in{M}$ and $\alpha(s)$ geodesic from $p$ to $q$ with $0\leq{s}\leq{l}$, also $\alpha(0)=p$ and $\alpha(l)=q$.

If $f, g$ are isomteries imply $f(\alpha(s))$ and $g(\alpha(s))$ are geodesics from $f(p)$ to $f(q)$ and $g(p)$ to $g(q)$ respectively.

By hypothesis $f(p)=g(p)$ and $df(\alpha'(0))=dg(\alpha'(0))$, these geodesics coincide and

$$f(q)=f(\alpha(l))=g(\alpha(l))=g(q),$$

therefore $f=g$.