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Let $M$ be a connected finite dimensional topological manifold and $g$ be any metric on it that induces the topology of $M$ ($g$ is not a Riemannian metric). How to prove that the group of isometries of $(M,g)$ is a finite dimensional Lie group?

PS. Note that in case $M$ is a Riemannian manifold, this statement is classical.

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    $\begingroup$ This seems relevant: Karube, Takashi, The group of isometries of a metric manifold, Sci. Bull. Fac. Educ., Nagasaki Univ. 40, 1-3 (1989). ZBL0674.54021 = MR90e:57064. $\endgroup$
    – Francois Ziegler
    Commented Nov 2, 2020 at 23:43
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    $\begingroup$ @FrancoisZiegler If my answer is correct, this reference proves the Hilbert-Smith conjecture. The MR review roughly says: *The author shows that the isometry group I(M) of any connected metric manifold M is a Lie group wrt to any topology finer than the pointwise topology. (...) This follows from I(M) being a locally compact transformation group, together with a 1966 theorem of the author which states that a locally compact, locally Lipschitzian, faithful transformation group of a connected metric manifold is necessarily a Lie group. (...) $\endgroup$
    – YCor
    Commented Nov 3, 2020 at 0:01
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    $\begingroup$ (...) and this 1989 paper (which apparently has no detectable quotation) sounds suspicious to me. Indeed the 1966 paper (which is available) seems to use manifolds with charts in the Euclidean space. So locally Lipschitz action is a condition for a metric that is locally Lipschitz to the Euclidean space, not an arbitrary distance. $\endgroup$
    – YCor
    Commented Nov 3, 2020 at 0:04
  • $\begingroup$ @YCor I meant to point to possible hypotheses (admittedly unclear from the MR) under which something like the original question (itself yet unclear) could have a positive answer. In fact the question still seems unclear in that the group’s topology is unspecified. E.g. any group “is Lie” in the discrete topology, which is not excluded. $\endgroup$
    – Francois Ziegler
    Commented Nov 3, 2020 at 10:47
  • $\begingroup$ @FrancoisZiegler In any case the reference purports to apply for the compact-open topology, which is very natural in this context (notably it's compact when $M$ is compact). The 1989 reference is perfectly relevant! But my claim is that it is probably seriously flawed (or the MR summary misses a hypothesis, such as assuming that the metric is locally Lipschitz to the Euclidean space). $\endgroup$
    – YCor
    Commented Nov 3, 2020 at 14:08

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The assertion (for connected topological manifolds) is equivalent to the Hilbert–Smith conjecture. The latter (in one of its equivalent versions) asserts that every continuous action of every profinite group on every topological metrizable manifold has an open kernel.

Indeed, let $G$ be a profinite group acting continuously on a connected topological metrizable manifold $X$. Let $d$ be a proper distance on $X$ (inducing the topology). Then $d'(x,y)=\int_G d(gx,gy)$, integrating with respect to the Haar measure, defines another proper distance, continuous on $X^2$, and which in addition is $G$-invariant. Hence it defines the topology and is complete. If OP's assertion holds, then the isometry group being a Lie group, the map has an open kernel.

The Hilbert–Smith conjecture is known in dimension $\le 3$ ($\le 2$ is classical, and $3$ is more recent work of John Pardon).

The assertion is also known to hold when the isometry group acts transitively.

Added remarks on the Hilbert–Smith conjecture:

If "green" qualifies topological groups, say that a topological group $G$ has "no small green subgroups" if there is a neighborhood of the identity $V$ in $G$ such that the only green subgroup of $G$ contained in $V$ is $\{1\}$.

For instance, the solution to Hilbert's 5th problem yields that a locally compact group has no small subgroups iff it is isomorphic (as topological group) to some Lie group.

One reformulation of the Hilbert–Smith conjecture is:

For every connected topological (metrizable) manifold $M$, the topological group $\mathrm{Homeo}(M)$ (with the compact-open topology) has no small compact subgroups.

Newman's theorems (see T. Tao's blog) implies that it has no small finite subgroups. The too optimistic hope would be that $\mathrm{Homeo}(M)$ has no small subgroups, but this is false: as soon as $M$ has positive dimension, it is easy to check that $\mathrm{Homeo}(M)$ has small discrete infinite cyclic subgroups.

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  • $\begingroup$ Thanks, that's very interesting. I guess, this is a classical fact that a profinite group has a Haar measure, is it easy to prove? Also, when you say "open kernel" you mean that the kernel of the action is an open subset of the group $G$? $\endgroup$
    – aglearner
    Commented Nov 2, 2020 at 23:57
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    $\begingroup$ Actually for the HS conjecture it's enough to check it for $\mathbf{Z}_p$ for all $p$, for which the Haar measure is explicit (the various equivalences between versions of the HS conjecture are far harder than the existence of Haar measure on a compact group, anyway). $\endgroup$
    – YCor
    Commented Nov 3, 2020 at 0:00
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    $\begingroup$ Yes open kernel = kernel is open (= an open subgroup, which in a profinite group implies finite index). $\endgroup$
    – YCor
    Commented Nov 3, 2020 at 0:05
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    $\begingroup$ @aglearner Yes. Because every continuous homomorphism of a profinite group into any Lie group has an open kernel. $\endgroup$
    – YCor
    Commented Nov 3, 2020 at 0:21
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    $\begingroup$ The best I'm aware of is at Tao's blog $\endgroup$
    – YCor
    Commented Nov 3, 2020 at 0:30

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