Compact isometries

As seen in this post, any isometry from a compact metric space K to itself is surjective, so the set I(K) of all such isometries is actually a group.

Give I(K) a reasonable metric and prove that it is compact as well. What can one say about the sequence of iterates K, I(K), I(I(K)), \dots?

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Compact isometries

One thought on “Compact isometries

  1. A reasonable metric for I(K) is d(f,g) = sup_{k\in K} d(f(k),g(k)), which in fact defines a metric on K^K. Moreover, the induced topology is the compact open topology.

    Now apply (some version of) Arzelà-Ascoli to deduce that $I(K)$ is compact. One must check that the space is complete, which follows from the fact that K^K is complete and I(K) is closed. It is immediate to check the fact that the maps in I(K) are equicontinuous. Finally the maps are pointwise bounded since K is.

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