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?

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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