The main question is to prove that Isometries of a Riemannian manifold = Iso(M,g) form a finite dimensional Lie group.
I can use one of the following methods to prove it:
Method 1) Showing that that an isometry fixing a point p in M such that df at p is identity, must be the identity map. Now use this to define a map from Iso(M,g) to M x TM x ... x TM and show that it is an embedding
Method 2) For a point p in M, choose a neighbouhood U of zero in $T_p$M on which the exponential map is a diffeomorphism. Choose a basis {$v_1, v_2, ..v_n$} of $T_p$M that lies in U and show that any isometry f:M $\rightarrow$ M is determined by f(p) and f(exp$(v_i)$). This determines a map from Iso(M) to M x (M ... x M). Show this is an embedding
I have been able to solve the first parts of both the methods, but am stuck in the second part(s). My proofs are:-
Method 2 Proof
First we prove that isometry $f: M \rightarrow M$ then $f \circ exp_p = exp_{f(p)} \circ df_p $
Subproof: Let $A_p$ be the domain of $\mathrm{exp}_{p}$, that is, $A_p$ be the open subset of the tangent space $T_{p}M$ such that : $ A_p = \lbrace v \in T_{p}M, \; \gamma_{v}(1) \; \text{exists} \rbrace $ where $\gamma_{v}$ is the unique maximal geodesic of $M$ with initial conditions: $\gamma_{v}(0) = p$ and $\dot{\gamma_{v}}(0) = v$. .
Let $p \in M$ and $v \in A_p$. Since $f$ is an isometry, it is distance-preserving and it sends geodesics of $M$ onto geodesics of $M$. This is true as let $f$ be an isometry between Riemannian manifolds $(M,g)$ and $(N,h).$ Then $f$ also preserves the induced metrics $d_1, d_2$ on $M, N$ from $g, h$ resp. that is, $d_1(x,y)=d_2(f(x),f(y))$ for $x,y \in M.$ Then, $f$ sends geodesics on $M$ to geodesics on $N,$ using the length minimizing property of geodesics and that $f$ is distance-preserving. in this case both M and N are the same, as are the induced metrics.
Hence, $\lambda\ : t \ \mapsto \ f\big( \mathrm{exp}_{p}(tv) \big)$ is a geodesic of M. It satisfies :
$\lambda(0) = f(p)$ and $\dot{\lambda}(0) = df(p)(v)$. The curve $t \mapsto \mathrm{exp}_{f(p)}\big( tdf(p)(v) \big)$ is another geodesic of $M$ which satisfies the same set of initial conditions. we know that geosdesics thus are unique and hence these are the same.
Using the above statement we can prove that local isometries of a connected manifold are completely determined by their values and differentials at a single point, that is, if $\phi, \psi$ : M $\rightarrow$ N are local isometries and p is a point in M such that $d\phi_p$=$d\psi_p$ (and hence $\phi$(p)=$\psi$(p) ), then $\phi = \psi$
Subproof: Let $A_q$ = {q $\epsilon$ M : $d\phi_p$=$d\psi_p$}. By continuity, A is closed in M. Since A is nonempty it suffices to show that A is open. We assert that if q $\epsilon$ $A_q$ then any normal neighborhood $U$ of q is contained in A. If r $\epsilon$ $U$ there is a vector v $\epsilon$ $T_q$M such that $\gamma_v$( 1) = $exp_q$(v) = r. Hence $\phi(r) = \phi(\gamma_v(1)) = \gamma_{d\phi v}(1) = \gamma_{d\psi v}(1) = \psi(\gamma_v(1)) = \psi(r) $. Thus $\phi = \psi$ on $U$ and hence $d\phi_r = \psi_r$ for all r $\epsilon$ $U$
Method 1 proof is relatively simpler and is given on a previous answer, and I would personally prefer to solve the main problem using this method 1: Isometry $f:M\to M$ has a fixed point $p$ with $df_p=\text{id} \Rightarrow f=\text{id}$
So I am stuck in the second part of either method which is to use the proofs to define maps which would be an embedding of Iso(M) in M x TM x.... TM or M x (M x ... M) respectively. As I am unsure as to how to understand the concept of an embedding w.r.t isometry group, a complete proof which shows that the defined map (whatever it may be) is indeed an embedding, would be appreciated.
