Let $(M_1,g_1)$ and $(M_2,g_2)$ be two connected Riemannian manifolds and let $f_1$ and $f_2$ be two local isometries such that $f_1(p)=f_1(p)$ and $f_{1*}(v)=f_{2*}(v)$ for some $p\in M$ and $\forall v\in T_pM$. Then $f_1=f_2$ identically.
Can someone help me how can I proceed?
I am stuck on this as we can not define $f_1-f_2$ on a manifold. But I think if we somehow define it and show that $(f_1-f_2)_*(v)=f_{1*}(v)-f_{2*}(v)$, then we are done.
I am not sure choosing a chart $(U,\psi)$ around $f_1(p)$ on $M_2$ and define $(f_1-f_2)(p):= \psi(f_1(p))-\psi(f_2(p))$ would be the write approach. Because then the proof become immediate and no need of $f_1$ or $f_2$ to be local isometry.
Best Answer
Let $A:=\{q\in M_1 \ | d_qf_1=d_qf_2\}$, then it's a closed subset of $M_1$ and it is non-empty (since $p\in A$). We just need to prove that it is open and then conclude by using that $M_1$ is connected. Let's begin with an observation.
Let $q\in M_1$ and $v\in T_qM_1$ such that $\gamma_v(1)$ is defined (where $\gamma_v$ is the unique geodetic such that $\gamma_v(0)=q$ and $\gamma_v'(0)=v$) and $f_1\circ\gamma_v$ is still a geodetic. Notice that this can always be done since the condition of being a geodetic is a local one and it depends only on the metric (recall that $f_1$ is a local isometry). Then, \begin{equation*}d_qf_1(v)=w\in T_{f_1(q)}M_2 \ \Longrightarrow \ \gamma_w=f_1\circ\gamma_v \end{equation*} by uniqueness of the geodetic. Then, $exp(w)=\gamma_w(1)=f_1\circ\gamma_v(1)=f_1\big(exp(v)\big)$, where $exp:\Omega\subset TM_1\to M_1$ is given by $exp(v):=\gamma_v(1)$ and $\Omega:=\{X\in TM_1 \ | \ \gamma_X(1) \ \text{is defined} \ \}$ open subset of $TM_1$. Now we are ready to prove that $A$ is open.
Let $q\in A$ and let $U$ be an open neighborhood of $q$ with normal coordinates centered at $q$. Let $x\in U$, so by definition there exists a unique $v\in T_qM_1$ such that $x=exp_q(v)=\gamma_v(1)$ and \begin{equation*} f_1(x)=f_1(exp(v))=\gamma_{d_qf_1(v)}(1)=\gamma_{d_qf_2(v)}(1)=f_2(\gamma_v(1))=f_2(x) \end{equation*} But this implies that $f_1$ and $f_2$ coincide on $U$, which implies that $d_qf_1=d_qf_2$ for all $q\in U$. Then, $U\subset A$ and $A$ is open.