The classical formulation of Hilbert’s fifth problem asks whether topological groups that have the topological structure of a manifold, are necessarily Lie groups. This is indeed, the case, thanks to following theorem of Gleason and Montgomery-Zippin:
Theorem 1 (Hilbert’s fifth problem) Let
be a topological group which is locally Euclidean. Then
is isomorphic to a Lie group.
We have discussed the proof of this result, and of related results, in previous posts. There is however a generalisation of Hilbert’s fifth problem which remains open, namely the Hilbert-Smith conjecture, in which it is a space acted on by the group which has the manifold structure, rather than the group itself:
Conjecture 2 (Hilbert-Smith conjecture) Let
be a locally compact topological group which acts continuously and faithfully (or effectively) on a connected finite-dimensional manifold
. Then
is isomorphic to a Lie group.
Note that Conjecture 2 easily implies Theorem 1 as one can pass to the connected component of a locally Euclidean group (which is clearly locally compact), and then look at the action of
on itself by left-multiplication.
The hypothesis that the action is faithful (i.e. each non-identity group element acts non-trivially on
) cannot be completely eliminated, as any group
will have a trivial action on any space
. The requirement that
be locally compact is similarly necessary: consider for instance the diffeomorphism group
of, say, the unit circle
, which acts on
but is infinite dimensional and is not locally compact (with, say, the uniform topology). Finally, the connectedness of
is also important: the infinite torus
(with the product topology) acts faithfully on the disconnected manifold
by the action
The conjecture in full generality remains open. However, there are a number of partial results. For instance, it was observed by Montgomery and Zippin that the conjecture is true for transitive actions, by a modification of the argument used to establish Theorem 1. This special case of the Hilbert-Smith conjecture (or more precisely, a generalisation thereof in which “finite-dimensional manifold” was replaced by “locally connected locally compact finite-dimensional”) was used in Gromov’s proof of his famous theorem on groups of polynomial growth. I record the argument of Montgomery and Zippin below the fold.
Another partial result is the reduction of the Hilbert-Smith conjecture to the -adic case. Indeed, it is known that Conjecture 2 is equivalent to
Conjecture 3 (Hilbert-Smith conjecture for
-adic actions) It is not possible for a
-adic group
to act continuously and effectively on a connected finite-dimensional manifold
.
The reduction to the -adic case follows from the structural theory of locally compact groups (specifically, the Gleason-Yamabe theorem discussed in previous posts) and some results of Newman that sharply restrict the ability of periodic actions on a manifold
to be close to the identity. I record this argument (which appears for instance in this paper of Lee) below the fold also.
— 1. Transitive actions —
We now show the transitive case of the Hilbert-Smith conjecture, following Montgomery and Zippin:
Proposition 4 Let
be a locally compact
-compact group that acts transitively, faithfully, and continuously on a connected manifold
. Then
is isomorphic to a Lie group.
The -compact hypothesis is a technical one, and can likely be dropped, but we retain it for this discussion (as in most applications we can reduce to this case).
The advantage of transitivity is that one can now view as a homogeneous space
of
, where
is the stabiliser of a point
(and is thus a closed subgroup of
). Note that a priori, we only know that
and
are identifiable as sets, with the identification map
defined by setting
being continuous; but thanks to the
-compact hypothesis, we can upgrade
to a homeomorphism. Indeed, as
is
-compact,
is also; and so given any compact neighbourhood of the identity
in
,
can be covered by countably many translates of
. By the Baire category theorem, one of these translates
has an image
in
with non-empty interior, which implies that
has
as an interior point. From this it is not hard to see that the map
is open; as it is also a continuous bijection, it is therefore a homeomorphism.
By the Gleason-Yamabe theorem, has an open subgroup
that is the inverse limit of Lie groups. (Note that
is Hausdorff because it acts faithfully on the Hausdorff space
.)
acts transitively on
, which is an open subset of
and thus also a manifold. Thus, we may assume without loss of generality that
is itself the inverse limit of Lie groups
, where
is a sequence of normal compact groups shrinking to the identity.
As is
-compact, the manifold
is also. As
acts faithfully on
, this makes
first countable; and so
is the inverse limit of a sequence of Lie groups
, with each
projecting surjectively onto
.
Let be the projection of
onto
; this is a closed subgroup of the Lie group
, and each
projects surjectively onto
. Then
are manifolds, and
is the inverse limit of the
. By invariance of domain, the dimensions of the
must be non-decreasing, and bounded above by the dimension of
. Thus, for
large enough, the dimensions of
must be constant; by renumbering, we may assume that all the
have the same dimension. As each
is a cover of
with structure group
, we conclude that the
are zero-dimensional and compact, and thus finite. On the other hand,
is locally connected, which implies that the
are eventually trivial. Indeed, if we pick a simply connected neighbbourhood
of the identity in
, then by local connectedness of
, there exists a connected neighbourhood
of the identity in
whose projection to
is contained in
. Being open,
must contain one of the
. If
is non-trivial for any
, then the projection of
to
will then be disconnected (as this projection will be contained in a neighbourhood with the topological structure of
, and its intersection with the latter fibre is at least as large as
. We conclude that
is trivial for
large enough, and so
is a Lie group as required.
— 2. Periodic actions of prime order —
We now study periodic actions on a manifold
of some prime order
, thus
.
The basic observation to exploit here is that of rigidity: a periodic action (or more precisely, the orbits of this action) cannot be too close to the identity, without actually being the identity. More precisely, we have the following theorem of Newman:
Theorem 5 (Newman’s first theorem) Let
be an open subset of
containing the closed unit ball
, and let
be a homeomorphism of some prime period
. Suppose that for every
, the orbit
has diameter strictly less than
. Then
.
Note that some result like this must be needed in order to establish the Hilbert-Smith conjecture. Suppose for instance that one could find a non-trivial transformation of some period
on the unit ball
that preserved the boundary of that ball. Then by placing infinitely many disjoint copies of that ball into
, and considering maps that are equal to some power of
on each such ball, and on the identity outside all the balls, we can obtain a faithful action of
on
, contradicting the Hilbert-Smith conjecture.
To prove the theorem we will need some basic degree theory. Given a continuously differentiable map , we know from Sard’s theorem that almost every point
in
is a regular point, in that the preimage
is finite and avoids the boundary
of
, with
being non-degenerate at each
. We can define the degree of
at the regular point
to be the number of preimages with
orientation-preserving, minus the number of preimages with
orientation-reversing. One can show that this degree extends to a constant integer-valued function on each connected component
of
; indeed, one can define the degree
on such a component analytically by the formula
for any volume form on
of total mass
(one can show that this definition is independent of the choice of
). This definition is stable under uniform convergence of
, and thus can be used to also define the degree for maps
that are merely continuous rather than continuously differentiable.
Proof: Suppose for contradiction that . We use an averaging argument, combined with degree theory. Let
be the map
Then from construction, is continuous and
-invariant (thus
for all
) and we have
for all
. In particular,
is non-zero on the boundary
of
, and
is contractible to
in
(by taking convex combinations of
and the identity). As such, the degree of
on
near the origin is equal to
.
A similar argument shows that the homeomorphism has degree
on
near the origin, and in particular is orientation-preserving rather than orientation reversing.
On the other hand, it can be shown that the degree of must be divisible by
, leading to a contradiction. This is easiest to see in the case when
is continuously differentiable, for then by Sard’s theorem we may find a regular point
arbitrarily close to
. On the other hand, since
is a positive distance away from
, we see that for
sufficiently close to
,
is not a fixed point of
, and thus (as
has prime order) all elements in the preimage
are not fixed points of
either. Thus
can be partitioned into a finite number of disjoint orbits of
of cardinality
. As
is orientation preserving and
is a regular point of
, each of these orbits contributes
or
to the degree, giving the claim.
The case when is not continuously differentiable is trickier, as the degree is not as easily computed in this case. One way to proceed is to perturb
(or more precisely, the graph
in
) to be piecewise linear near the preimage of
(while preserving the periodicity properties of the graph), so that degree can be computed by hand; this is the approach taken in Newman’s original paper. Another is to use the machinery of singular homology, which is more general and flexible than degree theory; this is the approach taken by Smith and by Dress.
Note that the above argument shows not only that fixes the origin
, but must also fix an open neighbourhood of the origin, by translating
slightly. One can then extend this open neighbourhood to the entire space by the following variant of Theorem 5.
Theorem 6 (Newman’s second theorem) Let
be a connected manifold, and let
be a homeomorphism of some prime order
that fixes a non-empty open set
. Then
is the identity.
Proof: We need to show that fixes all points in
, and not just
. Suppose for sake of contradiction that
just fixes some of the points in
and not others. By a continuity argument, and applying a homeomorphic change of variables if necessary, we can find a coordinate chart containing the ball
, where
fixes all
with
, but does not fix all
with
. By shrinking
we may assume that the entire orbit
stays inside the coordinate chart (and can thus be viewed as a subset of
). Then the map
can be defined as before. This map is the identity on the left hemisphere
. On the right hemisphere
, one observes for
small enough that
must all stay on the right-hemisphere (as they must lie in
and cannot enter the left-hemisphere, where
is the identity) and so
stays on the right. This implies that
has degree
near
on a small ball around the origin, but as before one can argue that the degree must in fact be divisible by
, leading again to a contradiction.
— 3. Reduction to the -adic case —
We are now ready to prove Conjecture 2 assuming Conjecture 3. Let be a locally compact group acting continuously and faithfully on a connected manifold
; we wish to show that
is Lie.
We first make some basic reductions. From the Gleason-Yamabe theorem (as discussed in this post) every locally compact group contains an open subgroup which is an extension of a Lie group by a compact subgroup of
. Since a group with a Lie group as an open subgroup is again Lie (because all outer automorphisms of Lie groups are smooth), and an extension of a Lie group by a Lie group is again Lie (this latter result, first proved by Gleason, follows for instance from the fact that a locally compact group is Lie if and only if it is NSS), it thus suffices to prove the claim when
is compact. In particular, all orbits of
on
are also compact.
Let be a small ball in chart of
around some origin
. By continuity, there is some neighbourhood
of the identity in
such that
for all
. By the Peter-Weyl theorem, there is a compact normal subgroup
of
in
with
linear (and hence Lie). The set
is then a
-invariant manifold, which is precompact and connected (because all the shifts
are connected and share a common point). If we let
be the subgroup of
that fixes
, then
is a compact normal subgroup of
, and
acts faithfully on
. By Newman’s first theorem (Theorem 5), we see that if
is small enough, then
cannot contain any elements of prime order, and hence cannot contain any non-trivial periodic elements whatsoever; by Conjecture 3, it also cannot contain a continuously embedded copy of
for any
. We claim that this forces
to be trivial.
As is precompact, the space of
of continuous maps from
to itself (with the compact-open topology) is first countable, which makes
first-countable as well (since
is homeomorphic to a subspace of
). The claim now follows from
Lemma 7 Let
be a compact first-countable group which does not contain any non-trivial periodic elements or a continuously embedded copy of
for any
. Then
is trivial.
Proof: Every element of
is contained in a compact abelian subgroup of
, namely the closed group
generated by
. Thus we may assume without loss of generality that
is abelian.
As is both compact and first countable, it can be written (using the Peter-Weyl theorem) as the inverse limit
of a countable sequence of compact abelian Lie groups
, with surjective continuous projection homomorphisms
between these Lie groups. (Note that without first countability, one might only be an inverse limit of a net of Lie groups rather than a sequence, as one can see for instance with the example
.)
Suppose for contradiction that at least one of the , say
, is non-trivial. It is a standard fact that every compact abelian Lie group is isomorphic to the direct product of a torus and a finite group. (Indeed, in the connected case one can inspect the kernel of the exponential map, and then one can extend to the general case by viewing a compact Lie group as an extension of a finite group by a connected compact Lie group.) in particular, the periodic points (i.e. points of finite order) are dense, and so there exists an element
of
of finite non-trivial order. By raising
to a suitable power, we may assume that
has some prime order
.
We now claim inductively that for each ,
can be lifted to an element
of some order
, where
are a non-decreasing set of integers. Indeed, suppose inductively that we have already lifted
up to
with order
. The preimage of
in
is then a dense subset of the preimage of
, which is a compact abelian Lie group, and thus contains an element
of some finite order. As
has order
,
must have an order divisible by
, and thus of the form
for some
coprime to
and some
. By raising
to a multiple of
that equals
mod
, we may eliminate
and obtain a preimage
of order
, and the claim follows.
There are now two cases, depending on whether goes to infinity or not. If the
stay bounded, then they converge to a limit
, and the inverse limit of the
is then an element
of
of finite order
, a contradiction. But if the
are unbounded, then
contains a continuously embedded copy of the inverse limit of the
, which is
, and again we have a contradiction.
Since and
are both Lie groups,
is Lie too. Thus it suffices to show that
is Lie. Without loss of generality, we may therefore replace
by
and assume that
is fixed by
.
Now let be the closure of the interior of the set of fixed points of
. Then
is non-empty, and is also clearly closed. We claim that
is open; by connectedness of
, this implies that
, which by faithfulness of
implies that
is trivial, giving the claim. Indeed, let
, and let
be a small ball around
. As before,
is then a connected
-invariant
-compact manifold, and so if
is the subgroup of
that fixes
, then
is a compact Lie group that acts faithfully on
fixing a nontrivial open subset of
. By Newman’s theorem (Theorem 6),
cannot contain any periodic elements; by Conjecture 3, it also cannot contain any copy of
. By Lemma 7,
is trivial, and so
fixes all of
. Thus gives the desired openness of
as required.
Remark 1 Whereas actions of the finite group
on manifolds can be analysed by degree theory, it appears that actions of
-adic groups
require more sophisticated homological tools; the
-adic analogues of the averaged map
now typically have infinite preimages and it is no longer obvious how to compute the degree of such maps. Nevertheless, some progress has been made along these lines under some additional regularity hypotheses on the action, such as Lipschitz continuity; see for instance this paper of Repovs and Scepin. Note that if
is the action of the generator of
, then the powers
will converge to the identity locally uniformly as
. This is already enough to rule out generating maps
that are smooth but non-trivial by use of Taylor expansion; see the text of Montgomery and Zippin. (In fact, this type of argument even works for
actions.)
8 comments
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27 August, 2011 at 11:35 am
254A, Notes 0 – Hilbert’s fifth problem and related topics « What’s new
[…] fifth problem for group actions, known as the Hilbert-Smith conjecture, which was discussed at this blog post.) To answer this question, we need to relax the notion of a Lie group to that of a topological […]
30 August, 2011 at 3:46 am
handaoyuan
Hello, Professor Tao, Louis F. McAuley has published his preprint in arxiv about the proof of Hilbert-Smith conjecture.
20 September, 2011 at 8:06 am
More on Hilbert’s tenth problem | cartesian product
[…] The Hilbert-Smith conjecture (terrytao.wordpress.com) […]
8 October, 2011 at 12:57 pm
254A, Notes 5: The structure of locally compact groups, and Hilbert’s fifth problem « What’s new
[…] difficulty is to figure out a way to eliminate the possibility that is a -adic group . See this previous blog post for further […]
6 May, 2012 at 4:28 pm
On Tao’s talk and the 3-dimensional Hilbert-Smith conjecture « Area 777
[…] mentioned in Tao’s post, in fact is the only possible bad case! i.e. it is sufficient to […]
29 April, 2013 at 3:58 am
IPAM’s program “Interactions between Analysis and Geometry” and John Pardon’s talk on Hilbert-Smith conjecture for 3-manifolds | Disquisitiones Mathematicae
[…] recommend consulting Terence Tao’s notes on Hilbert’s 5th problem (as well as his post here on Hilbert-Smith […]
7 March, 2016 at 4:37 am
Hilbert’s Problems: Part 1 | mathsbyagirl
[…] seen to be resolved by Andrew Gleason in 1954, however if it is seen to be equivalent to the Hilbert-Smith Conjecture, then it remains […]
21 November, 2016 at 7:38 pm
anonymous
Dear Prof. Tao,
In your statement of Hilbert’s 5th, needn’t we assume G is second-countable? For example, the real line with the discrete topology is locally Euclidean and is not isomorphic to a Lie group.
Thanks.
[I am not requiring manifolds (or Lie groups) here to be second countable, though one can certainly impose this condition if desired (it doesn’t change the difficulty of Hilbert’s fifth much) -T.]