# Prove that the composition map is continuous with respect to the metric topology on $operatorname{Iso}(M)$

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 $$r>0$$.



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)



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) ?



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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