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.


1 Answer 1


Let $n$ be the dimension of $M$. Given what you've already done in Method 1, defining the embedding $$F : \text{Iso}(M,g) \to M \times (TM)^n $$ is pretty straight forward: choose $p \in M$, choose a basis of its tangent space $v_1,...,v_n \in T_p M$, and define $$F(p) = (f(p), D_p f(v_1), \ldots, D_p f(v_n)) $$ You can then use that to derive the formula for the embedding in Method 2, however you should to make sure that you choose $v_1,...,v_n$ to be contained in a small enough ball around the origin of the vector space $T_p M$ so that the exponential map is defined on that ball.

Added: You have emphasized your interest in the issue of $F$ being an "embedding". This can be a vague term, and your post does not resolve the vagueness, so there's not too much to be said definitively.

On the set theoretic level, one would certainly start with

$F$ is an injective function...

That is the only level on which I have addressed your question, although I leave the proof of injectivity for you.

But, perhaps one would want to continue on to topological level with

...which is a homeomorphism onto its image...

For that, one needs a topology on the domain which is $\text{Iso}(M,g)$, which has not been specified. Hopefully that's not too much of a problem, for instance perhaps that can be achieved using the compact open topology on $\text{Iso}(M,g)$, so let me assume that has been accomplished.

Next, perhaps one would want to continue on to the smooth level with

... and is a diffeomorphism onto its image...

However, serious trouble is beginning to rear its ugly head. For instance, how do we even know that $\text{Iso}(M,g)$ is a smooth manifold? This, of course, is a key issue of your "main question".

At this stage, with so many vaguenesses yet unresolved, I would be afraid to go on to the next level, i.e. the Riemannian level:

... and is an isometry onto its image...

Here I honestly do not know what to suggest, because there are quite large unresolved issues. Assuming $\text{Iso}(M,g)$ is indeed a smooth manifold, what Riemannian metric on it do you wish to use? And is there some kind of Riemannian metric you want to use on $M \times (TM)^n$?

  • $\begingroup$ I'm having a bit of confusion with regards to the "embedding" aspect of the group of Isometries. As it must be that metric is preserved, for any two isometries lets say f and h, we must have that <dF(f),dF(h)> in the riemannian metric g, is the same as.... (?) $\endgroup$
    – Bizwhiz
    May 11, 2020 at 13:49
  • $\begingroup$ I have added some comments to address the "embedding" aspect. $\endgroup$
    – Lee Mosher
    May 11, 2020 at 15:29
  • $\begingroup$ The smooth manifold aspect is indeed not easy. The shortest proof I know is in the following paper: people.mpim-bonn.mpg.de/hwbllmnn/archiv/autmor00.pdf $\endgroup$
    – Maik Pickl
    May 11, 2020 at 15:34

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