Mathematics Asked by Abhigyan Saha on December 21, 2020

Let $M$ be a finite dimensional Riemannian manifold and $operatorname{Iso}(M)$ be its set of isometries. It can be shown that $operatorname{Iso}(M)$ is a finite dimensional manifold with a metric as defined below:

Consider $(n + 1)$ points on $M$ 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 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. This distance can be shown to satisfy the usual metric axioms. Here $d_i$ is of course the induced metric on $M$ (Riemannian distance fucntion)

Given $operatorname{Iso}(M)$ is now a metric space with metric $d$ as defined, we thus get a natural metric topology for $operatorname{Iso}(M)$. That is open sets are all subsets that can be realized as the unions of open balls of form $B(f_0, r) = {f in operatorname{Iso}(M): d(f_0,f)< r}$ where $f_0 in operatorname{Iso}(M)$ and $r>0$.

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**I am trying to prove that $$mathscr M: operatorname{Iso}(M) times operatorname{Iso}(M) rightarrow operatorname{Iso}(M), , (f,g) mapsto f circ g$$ is continuous in the metric topology of $operatorname{Iso}(M)$.**

**Attempt:** *Munkres Topology Section 46 Page 287*

Let $Y$ be locally compact Hausdorff, and $X$ and $Z$ general spaces. Also let $mathscr{C}(X,Y),,mathscr{C}(Y,Z),$ and $mathscr{C}(X,Z)$ denote the spaces of continuous functions from the respective spaces with the compact open topology. Then the composition map

$$mathscr M: mathscr{C}(X,Y) timesmathscr{C}(Y,Z)rightarrowmathscr{C}(X,Z)$$

is continuous.

The above is a proven statement, and can be assumed for now. In the statement, $X$ and $Z$ can be replaced with $M$ which has metric topology (and hence manifold topology) and thus is a general space. Further, $Y$ can also be replaced with $M$ as it is locally compact Hausdorff as a manifold. So we end up with $mathscr{C}(M,M)$ for all 3. Furthermore, as isometries are continuous, we get that $operatorname{Iso}(M) subset mathscr{C}(M,M)$. Thus we end up with the following:

$mathscr M: operatorname{Iso}(M) times operatorname{Iso}(M) rightarrow operatorname{Iso}(M)$ is continuous

*Idea*: The compact-open topology and metric topology are the same in case of $operatorname{Iso}(M)$ under these conditions because the topologies of every space involved here comes from same $d_i$ (Riemannian distance function as defined previously)

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**Q) So I’m looking for a proof that the CO Topology and metric topology are the same for $operatorname{Iso}(M)$.**

Alternatively (and preferably)

**Q) Is there a direct way to show continuity of $mathscr M$ in the metric topology of $operatorname{Iso}(M)$ (i.e. showing inverse of an open set in metric topology of $operatorname{Iso}(M)$ is always open in $operatorname{Iso}(M)timesoperatorname{Iso}(M)$, or any of the equivalent definitions of metric continuity) ?**

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In this question you already showed that $operatorname{Iso}(M)$ is a group and the first part of the accepted answer given by Abcde refers to another question to show continuity of the inversion $$iota: operatorname{Iso}(M) to operatorname{Iso}(M), , f mapsto f^{-1},$$ so I will use these two facts in the following answer. Now let $S$ denote the set of the $n+1$ points defining the metric on $operatorname{Iso}(M)$. For $g, f, tilde{f} in operatorname{Iso}(M)$ we have $$d(g circ f, g circ tilde{f}) = max_{x in S} d_i(g(f(x)), g(tilde{f}(x))) = max_{x in S} d_i(f(x), tilde{f}(x)) = d(f, tilde{f})$$ since $g$ is an isometry, so the map $$mathscr{M}(g, -): operatorname{Iso}(M) to operatorname{Iso}(M), , f mapsto g circ f$$ is a $d$-isometry and in particular continuous for any $g in operatorname{Iso}(M).$ Furthermore $$f circ g = left((f circ g)^{-1}right)^{-1} = (g^{-1} circ f^{-1})^{-1}, text{ i.e. } mathscr{M}(f,g) = iota(mathscr{M}(g^{-1}, iota(f)),$$ so the continuity of the inversion and the above shows that $$mathscr{M}(-,g): operatorname{Iso}(M) to operatorname{Iso}(M), , f mapsto f circ g$$ is continuous for $g in operatorname{Iso}(M)$ since $mathscr{M}(-,g) = iota circ mathscr{M}(g^{-1}, -) circ iota$.

Now let $(f_n)_{n in mathbb{N}}, (g_n)_{n in mathbb{N}}$ be sequences in $operatorname{Iso}(M)$ and $f,g in operatorname{Iso}(M)$ such that $$lim_{n to infty} d(f_n, f) = lim_{n to infty} d(g_n, g) = 0.$$ Then the continuity of $mathscr{M}(-,g)$ implies $lim_{n to infty} d(f_n circ g, f circ g) = 0$. Furthermore we can use the triangle inequality of $d$ and the fact that $mathscr{M}(f_n, -)$ is a $d$-isometry for any $n in mathbb{N}$ to obtain $$d(f_n circ g_n, f circ g) leq d(f_n circ g_n, f_n circ g) + d(f_n circ g, f circ g) = d(g_n, g) + d(f_n circ g, f circ g),$$ so $lim_{n to infty} d(f_n circ g_n, f circ g) = 0$. Hence $$mathscr{M}: operatorname{Iso}(M) times operatorname{Iso}(M) to operatorname{Iso}(M), , (f,g) mapsto f circ g$$ is sequentially continuous and therefore continuous on the metric space $operatorname{Iso}(M) times operatorname{Iso}(M)$.

Correct answer by Sebastian Spindler on December 21, 2020

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