I need to show that the set of isometries of a connected Riemannian manifold is a topological group.
My work so far has been:
Part A: If an isometry $f$ leaves fixed (n + 1) points so close together that n of them lie in an normal neighborhood of the other, and if the points are linearly independent (i.e. not in the same (n-1)-dimensional geodesic hypersurface), then $f$ is the identity
Proof: Suppose $f$ leaves fixed the (n+1) linearly independent points $x_0, x_1,…,x_n$ of which $x_1, … , x_n$ lie in the normal neighborhood of $x_0$. Then the short geodesics $\widehat {x_0x_1}, \widehat {x_ox_2},…, \widehat {x_ox_n}$ are fixed under $f$, as are their initial directions, and transformation of the space of directions at $x_0$ induced by $f$ is the identity. Hence all geodesic arcs issuing from $x_0$ are fixed; since length is preserved, they are pointwise fixed.
Thus any normal neighborhood of $x_0$ is pointwise fixed. If y is a point in an normal neighborhood of $x_0$, any normal neighborhood of y is similarly pointwise fixed. But $x_0$ can be joined to any point z in M by a broken geodesic arc with a finite number of pieces, each corner being in an normal neighborhood of the preceding corner. Thus z is fixed, and since we chose z arbitrarily from M, all of M is fixed, thus making $f$ the identity map.
Corollary: We get that there is at most one isometry which carries (n+1) points $x_i$ of the kind described in the Part A into n + 1 points $y_i$. For if we had two such isometries $f$ and $\tilde{f}$, then $f^{-1} \tilde{f}$ would leave $x_i$ fixed, and so would be the identity, by Part A.
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Comment: The set of isometries is denoted by Iso(M). A sequence of isometries {$f_k$} will be said to converge to an isometry $f$ if for every x in M, $f_k$(x) $\rightarrow$ $f$(x). With this notion of convergence the set satisfies continuity property. It can also be shown that if $f_k$(x) converges to a point x, there exists a subsequence {$f'_k$} of {$f_k$} converging to an isometry $f$ such that $f$(x) = x. If $f_k$(x) converges to x, $f$ leaves x fixed; the sequence $f'_k f^{-1}$ takes x into the same set of points as $f'_k$, and converges to the identity.
Part B We can metrize Iso(M), so that we obtain the above notion of convergence.
Proof: Consider any set of n + 1 points of the kind used in Part A. Then the distance d($f$, $\tilde f$) between two isometries $f$ and $\tilde f$ will be defined as the maximum of the distance $d_i$[$f$(x), $\tilde f$(x)] as x ranges over the given set of n+1 points (here $d_i$ is the induced metric on M). This distance can be shown to satisfy the usual metric axioms. The previous notion of convergence of isometries $f_i$ to $f$ implies d($f_i$, $f$) $\rightarrow$ 0; we must therefore show that d($f_i$, $f$) $\rightarrow$ 0 implies $f_i$(x) $\rightarrow$ $f$(x) for all x on M. This will not only show that convergence in this new topology is equivalent to the previous notion of convergence, but also that the new topology is independent of the particular set of n+1 points used.
So we prove by contradiction. Assume there exists a point y, an $\epsilon$ > 0, and a subsequence {$f'_i$} of {$f_i$} such that $$ d_i[f'_i(y), f(y)] > \epsilon \ \ for\ all\ i \ \ \ … \Omega_2 $$
But we know that $f'_i$(x) $\rightarrow$ $f$(x) on the given set of n+1 points hence {$f'_i$} must contain a subsequence {$f''_i$} such that there exists an isometry $f'$ with $f''_i$(x) $\rightarrow$ $f'$(x) for all x. Now $f'$(x) = $f$(x) for x on the given set of n+1 points; hence by the Corollary, $f'$ = $f$. This contradicts $\Omega_2$, and the proof is complete.
Part C: Set of Iso(M) form a group under composition
Proof: We check the 4 conditions for it to be a group:
Associativity: As compositions of functions is always associative, which is a property inherited from compositions of relations, we get that if a,b,c $\epsilon$ Iso(M) then (ab)c=a(bc)
Closure: Composition of isometries is again an isometry and thus if a,b $\epsilon$ Iso(M) then ab $\epsilon$ Iso(M)
Identity: the identity function is trivially an isometry in itself and thus belongs to Iso(M). For any a $\epsilon$ Iso(M), we have (id)a=a(id)=a
Inverse: By definition, an isometry is a diffeomorphism which means that inverse exists. This inverse is obviously an isometry as well. So if a $\epsilon$ Iso(M) then $a^{-1}$ exists and $\epsilon$ Iso(M)
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Now to finish this I need to do the following and would appreciate proofs of the same:
Q1) Justifiably give Iso(M) a topology
Q2) Show Iso(M) is a topological group with respect to this topology/metric
Now assuming that to solve Q1, I can give Iso(M) the usual metric topology. That is, with the metric d as defined, open sets are all subsets that can be realized as the unions of open balls of form B($f_0$, $r$) = {$f$ $\epsilon$ Iso(M) | d($f_0,f)$< $r$} where $f_0$ $\epsilon$ Iso(M) and $r$>0. Here d($f_0,f)$ is defined as the maximum of the distance $d_i$[$f_0$(x), $f$(x)] as x ranges over the given set of n+1 points as in Part A. So assuming (?) this is the correct topology to give Iso(M) to solve the next part, what will be the proof of Q2 ? Even a possible proof sketch would do.
I keep reading in books that this was shown in a 1928 paper by Danzig and Waerden, but on reading the paper I could understand nothing as the notation used is a century old (not even sure if I got the right paper). Would be helpful if I could get the complete proofs in current notations to study and understand. Thanks!
Best Answer
Links to solutions of both parts are available. Since you asked for a proof sketch:
What is required to prove is that the composition mapping and inverse mapping are continuous. For the inverse mapping to be continuous, you need that {$f_k^{-1}$} $\rightarrow$ $f^{-1}$ if {$f_k$} $\rightarrow$ $f$. Also note that Meyers-Steenrod proved that {$f_k$} $\rightarrow$ $f$ uniformly. Thus the following should take care of this part: Compact convergence of inverse functions
For composition mapping to be continuous it has alreasy been clarified in comments that taking C-O topology on Iso(M) solves the issue, as Iso(M) is a subset of $\mathscr C$(M,M). Here's the link to the solution of the said exercise in Munkres: http://www2.math.ou.edu/~dmccullough/teaching/s05-5863/hw10_soln.pdf