In the previous notes, we established the *Gleason-Yamabe theorem*:

Theorem 1 (Gleason-Yamabe theorem)Let be a locally compact group. Then, for any open neighbourhood of the identity, there exists an open subgroup of and a compact normal subgroup of in such that is isomorphic to a Lie group.

Roughly speaking, this theorem asserts the “mesoscopic” structure of a locally compact group (after restricting to an open subgroup to remove the macroscopic structure, and quotienting out by to remove the microscopic structure) is always of Lie type.

In this post, we combine the Gleason-Yamabe theorem with some additional tools from point-set topology to improve the description of locally compact groups in various situations.

We first record some easy special cases of this. If the locally compact group has the no small subgroups property, then one can take to be trivial; thus is Lie, which implies that is locally Lie and thus Lie as well. Thus the assertion that all locally compact NSS groups are Lie (Theorem 10 from Notes 4) is a special case of the Gleason-Yamabe theorem.

In a similar spirit, if the locally compact group is connected, then the only open subgroup of is the full group ; in particular, by arguing as in the treatment of the compact case (Exercise 19 of Notes 3), we conclude that any connected locally compact Hausdorff group is the inverse limit of Lie groups.

Now we return to the general case, in which need not be connected or NSS. One slight defect of Theorem 1 is that the group can depend on the open neighbourhood . However, by using a basic result from the theory of totally disconnected groups known as *van Dantzig’s theorem*, one can make independent of :

Theorem 2 (Gleason-Yamabe theorem, stronger version)Let be a locally compact group. Then there exists an open subgoup of such that, for any open neighbourhood of the identity in , there exists a compact normal subgroup of in such that is isomorphic to a Lie group.

We prove this theorem below the fold. As in previous notes, if is Hausdorff, the group is thus an inverse limit of Lie groups (and if (and hence ) is first countable, it is the inverse limit of a *sequence* of Lie groups).

It remains to analyse inverse limits of Lie groups. To do this, it helps to have some control on the dimensions of the Lie groups involved. A basic tool for this purpose is the invariance of domain theorem:

Theorem 3 (Brouwer invariance of domain theorem)Let be an open subset of , and let be a continuous injective map. Then is also open.

We prove this theorem below the fold. It has an important corollary:

Corollary 4 (Topological invariance of dimension)If , and is a non-empty open subset of , then there is no continuous injective mapping from to . In particular, and are not homeomorphic.

Exercise 1 (Uniqueness of dimension)Let be a non-empty topological space. If is a manifold of dimension , and also a manifold of dimension , show that . Thus, we may define the dimension of a non-empty manifold in a well-defined manner.If are non-empty manifolds, and there is a continuous injection from to , show that .

Remark 1Note that the analogue of the above exercise for surjections is false: the existence of a continuous surjection from one non-empty manifold to another doesnotimply that , thanks to the existence of space-filling curves. Thus we see that invariance of domain, while intuitively plausible, is not an entirely trivial observation.

As we shall see, we can use Corollary 4 to bound the dimension of the Lie groups in an inverse limit by the “dimension” of the inverse limit . Among other things, this can be used to obtain a positive resolution to Hilbert’s fifth problem:

Theorem 5 (Hilbert’s fifth problem)Every locally Euclidean group is isomorphic to a Lie group.

Again, this will be shown below the fold.

Another application of this machinery is the following variant of Hilbert’s fifth problem, which was used in Gromov’s original proof of Gromov’s theorem on groups of polynomial growth, although we will not actually need it this course:

Proposition 6Let be a locally compact -compact group that acts transitively, faithfully, and continuously on a connected manifold . Then is isomorphic to a Lie group.

Recall that a continuous action of a topological group on a topological space is a continuous map which obeys the associativity law for and , and the identity law for all . The action is *transitive* if, for every , there is a with , and *faithful* if, whenever are distinct, one has for at least one .

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).

Remark 2It is conjectured that the transitivity hypothesis in Proposition 6 can be dropped; this is known as the Hilbert-Smith conjecture. It remains open; the key difficulty is to figure out a way to eliminate the possibility that is a -adic group . See this previous blog post for further discussion.

** — 1. Van Dantzig’s theorem — **

Recall that a (non-empty) topological space is connected if the only clopen (i.e. closed and open) subsets of are the whole space and the empty set ; a non-empty topological space is *disconnected* if it is not connected. (By convention, the empty set is considered to be neither connected nor disconnected, somewhat analogously to how the natural number is neither considered prime nor composite.)

At the opposite extreme to connectedness is the property of being a totally disconnected space. This is a space whose only connected subsets are the singleton sets. Typical examples of totally disconnected spaces include discrete spaces (e.g. the integers with the discrete topology) and Cantor spaces (such as the standard Cantor set).

Most topological spaces are neither connected nor totally disconnected, but some intermediate combination of both. In the case of topological *groups* , this rather vague assertion can be formalised as follows.

Exercise 3

- Define a connected component of a topological space to be a maxmial connected set. Show that the connected components of form a partition of , thus every point in belongs to exactly one connected component.
- Let be a topological group, and let be the connected component of the identity. Show that is a closed normal subgroup of , and that is a totally disconnected subgroup of . Thus, one has a short exact sequence
of topological groups that describes as an extension of a totally disconnected group by a connected group.

- Conversely, if one has a short exact sequence
of topological groups, with connected and totally disconnected, show that is isomorphic to , and is isomorphic to .

- If is locally compact, show that and are also locally compact.

In principle at least, the study of locally compact groups thus splits into the study of connected locally compact groups, and the study of totally disconnected locally compact groups. (Note however that even if one has a complete understanding of the factors of a short exact sequence , it may still be a non-trivial issue to fully understand the combined group , due to the possible presence of non-trivial group cohomology. See for instance this previous blog post for more discussion.)

For totally disconnected locally compact groups, one has the following fundamental theorem of van Dantzig’s:

Theorem 7 (Van Danztig’s theorem)Every totally disconnected locally compact group contains a compact open subgroup (which will of course still be totally disconnected).

Example 1Let be a prime. Then the -adic field (with the usual -adic valuation) is totally disconnected locally compact, and the -adic integers are a compact open subgroup.

Of course, this situation is the polar opposite of what occurs in the connected case, in which the only open subgroup is the whole group.

To prove van Dantzig’s theorem, we first need a lemma from point set topology, which shows that totally disconnected spaces contain enough clopen sets to separate points:

Lemma 8Let be a totally disconnected compact Hausdorff space, and let be distinct points in . Then there exists a clopen set that contains but not .

*Proof:* Let be the intersection of all the clopen sets that contain (note that is obviously clopen). Clearly is closed and contains . Our objective is to show that consists solely of . As is totally disconnected, it will suffice to show that is connected.

Suppose this is not the case, then we can split where are disjoint non-empty closed sets; without loss of generality, we may assume that lies in . As all compact Hausdorff spaces are normal, we can thus enclose in disjoint open subsets of . In particular, the topological boundary is compact and lies outside of . By definition of , we thus see that for every , we can find a clopen neighbourhood of that avoids ; by compactness of (and the fact that finite intersections of clopen sets are clopen), we can thus find a clopen neighbourhood of that is disjoint from . One then verifies that is a clopen neighbourhood of that is disjoint from , contradicting the definition of , and the claim follows.

Now we can prove van Dantzig’s theorem. We will use an argument from the book of Hewitt and Ross. Let be totally disconnected locally compact (and thus Hausdorff). Then we can find a compact neighbourhood of the identity. By Lemma 8, for every , we can find a clopen neighbourhood of the identity that avoids ; by compactness of , we may thus find a clopen neighbourhood of the identity that avoids . By intersecting this neighbourhood with , we may thus find a compact clopen neighbourhood of the identity. As is both compact and open, we may then the continuity of the group operations find a symmetric neighbourhood of the identity such that . In particular, if we let be the group generated by , then is an open subgroup of contained in and is thus compact as required.

Remark 3The same argument shows that a totally disconnected locally compact group contains arbitrarily small compact open subgroups, or in other words the compact open subgroups form a neighbourhood base for the identity.

In view of van Dantzig’s theorem, we see that the “local” behaviour of totally disconnected locally compact groups can be modeled by the compact totally disconnected groups, which are better understood. Thanks to the Gleason-Yamabe theorem for compact groups, such groups are the inverse limits of compact totally disconnected Lie groups. But it is easy to see that a compact totally disconnected Lie group must be finite, and so compact totally disconnected groups are necessarily profinite. The global behaviour however remains more complicated, in part because the compact open subgroup given by van Dantzig’s theorem need not be normal, and so does not necessarily induce a splitting of into compact and discrete factors.

Example 2Let be a prime, and let be the semi-direct product , where the integers act on by the map , and we give the product of the discrete topology of and the -adic topology on . One easily verifies that is a totally disconnected locally compact group. It certainly has compact open subgroups, such as . However, it is easy to show that has no non-trivial compact normal subgroups (the problem is that the conjugation action of on has all non-trivial orbits unbounded).

We can pull van Dantzig’s theorem back to more general locally compact groups:

Exercise 4Let be a locally compact group.

- Show that contains an open subgroup which is “compact-by-connected” in the sense that is compact. (
Hint:apply van Dantzig’s theorem to .)- If is compact-by-connected, and is an open neighbourhood of the identity, show that there exists a compact subgroup of in such that is isomorphic to a Lie group. (
Hint:use Theorem 1, and observe that any open subgroup of the compact-by-connected group has finite index and thus has only finitely many conjugates.) Conclude Theorem 2.- Show that any locally compact Hausdorff group contains an open subgroup that is isomorphic to an inverse limit of Lie groups , which each Lie group has at most finitely many connected components. Furthermore, each is isomorphic to for some compact normal subgroup of , with for . If is first countable, show that this inverse limit can be taken to be a sequence (so that the index set is simply the natural numbers with the usual ordering), and the then shrink to zero in the sense that they lie inside any given open neighbourhood of the identity for large enough.

Exercise 5Let be a totally disconnected locally compact group. Show that every compact subgroup of is contained in a compact open subgroup. (Hint:van Dantzig’s theorem provides a compact open subgroup, but it need not contain . But is there a way to modify it so that it is normalised by ? Why would being normalised by be useful?)

** — 2. The invariance of domain theorem — **

In this section we give a proof of the invariance of domain theorem. The main topological tool for this is Brouwer’s famous fixed point theorem:

Theorem 9 (Brouwer fixed point theorem)Let be a continuous function on the unit ball in a Euclidean space . Then has at least one fixed point, thus there exists with .

This theorem has many proofs. We quickly sketch one of these proofs as follows:

Exercise 6For this exercise, suppose for sake of contradiction that Theorem 9 is false, thus there is a continuous map from to with no fixed point.

- Show that there exists a
smoothmap from to with no fixed point.- Show that there exists a smooth map from to the unit sphere , which equals the identity function on .
- Show that there exists a smooth map from to the unit sphere , which equals the map on a neighbourhood of .
- By computing the integral in two different ways (one by using Stokes’ theorem, and the other by using the -dimensional nature of the sphere ), establish a contradiction.

Now we prove Theorem 3. By rescaling and translation invariance, it will suffice to show the following claim:

Theorem 10 (Invariance of domain, again)Let be an continuous injective map. Then lies in the interior of .

Let be as in Theorem 10. The map is a continuous bijection between compact Hausdorff spaces and is thus a homeomorphism. In particular, the inverse map is continuous. Using the Tietze extension theorem, we can find a continuous function that extends .

The function has a zero on , namely at . We can use the Brouwer fixed point theorem to show that this zero is stable:

Lemma 11 (Stability of zero)Let be a continuous function such that for all . Then has at least one zero (i.e. there is a such that ).

*Proof:* Apply Theorem 9 to the function

Now suppose that Theorem 10 failed, so that is not an interior point of . We will use this to locate a small perturbation of that no longer has a zero on , contradicting Lemma 11.

We turn to the details. Let be a small number. By continuity of , we see (if is chosen small enough) that we have whenever and .

On the other hand, since is not an interior point of , there exists a point with that lies outside . By translating if necessary, we may take ; thus avoids zero, , and we have

Let denote the set , where

and

By construction, is compact but does not contain . Crucially, there is a continuous map defined by setting

Note that is continuous and well-defined since avoids zero. Informally, is a perturbation of caused by pushing out a small distance away from the origin (and hence also away from ), with being the “pushing” map.

By construction, is non-zero on ; since is compact, is bounded from below on by some . By shrinking if necessary we may assume that .

By the Weierstrass approximation theorem, we can find a polynomial such that

for all ; in particular, does not vanish on . At present, it is possible that vanishes on . But as is smooth and has measure zero, also has measure zero; so by shifting by a small generic constant we may assume without loss of generality that also does not vanish on . (If one wishes, one can use an algebraic geometry argument here instead of a measure-theoretic one, noting that lies in an algebraic hypersurface and can thus be generically avoided by perturbation. A purely topological way to avoid zeroes in is also given in Kulpa’s paper.)

Now consider the function defined by

This is a continuous function that is never zero. From (3), (2) we have

whenever is such that . On the other hand, if , then from (2), (1) we have

and hence by (3) and the triangle inequality

Thus in all cases we have

for all . But this, combined with the non-vanishing nature of , contradicts Lemma 11.

** — 3. Hilbert’s fifth problem — **

We now establish Theorem 5. Let be a locally Euclidean group. By Exercise 5 of Notes 0, is Hausdorff; it is also locally compact and first countable. Thus, by Exercise 4, such a group contains an open subgroup which is isomorphic to the inverse limit of Lie groups , each of which has only finitely many components. Clearly, is also locally Euclidean. If it is Lie, then is locally Lie and thus Lie. Thus, by replacing with if necessary, we may assume without loss of generality that is the inverse limit , each of which has only finitely many components.

By Exercise 4, each is isomorphic to the quotient of by some compact normal subgroup with . In particular, is isomorphic to the quotient of by a compact normal subgroup . By Cartan’s theorem (Theorem 2 of Notes 2), is also a Lie group. Among other things, this implies that the quotient homomorphism from the Lie algebra of to the Lie algebra of is surjective; indeed, it is the quotient map by the Lie algebra of . This implies that there is a continuous map from to that inverts the quotient map; in other words, we have a continuous map from the one-parameter subgroups of to the one-parameter subgroups of , such that for all .

Exercise 7By iterating these maps and passing to the inverse limit, conclude that for each , there is a continuous map such that for all .

Because is a Lie group, the exponential map is a homeomorphism from a neighbourhood of the origin in to a neighbourhood of the identity in . We can thus obtain a continuous map from a neighbourhood of the identity in to . Since , this map is injective.

Now we use the hypothesis that is locally Euclidean (and in particular, has a well-defined dimension ). By Exercise 1, we have

for all . On the other hand, since each is a quotient of the next Lie group , one has

Since there are only finitely many possible values for the (necessarily integral) dimension between and , we conclude that the dimension must eventually stabilise, i.e. one has

for all sufficiently large . By discarding the first few terms in the sequence and relabeling, we may thus assume that the dimension is constant for *all* . Since , this implies that the Lie groups have dimension zero for all . As the are also compact, they are thus finite. Thus each is a finite extension of . As is the inverse limit of the as , we conclude that is a profinite group, i.e. the inverse limit of finite groups. In particular, is totally disconnected.

We now study the short exact squence

playing off the locally connected nature of the Lie group against the totally disconnected nature of .

As discussed earlier, we have a continuous injective map from a neighbourhood of the identity in to that partially inverts the quotient map. By translation, we may normalise . As is locally connected, we can find a connected neighborhood of the identity in such that .

Now consider the set . On the one hand, this set is contained in and contains ; on the other hand, it is connected. As is totally disconnected, this set must equal , thus for all . A similar argument based on consideration of the set shows that for all . Thus is a homomorphism from the local group to .

Finally, for any , a consideration of the set reveals that commutes with . As a consequence, we see that the preimage of under the quotient map is isomorphic as a local group to , after identifying with for any and .

On the other hand, is locally Euclidean, and hence is locally Euclidean also, and in particular locally connected. This implies that is locally connected; but as is also totally disconnected, it must be discrete. This is now locally isomorphic to and hence to , and is thus locally Lie and hence Lie as required. (Here, we say that two groups are *locally isomorphic* if they have neighbourhoods of the identity which are isomorphic to each other as local groups.)

Exercise 8Let be a locally compact Hausdorff first-countable group which is “finite-dimensional” in the sense that it does not contain continuous injective images of non-trivial open sets of Euclidean spaces of arbitrarily large dimension. Show that is locally isomorphic to the direct product of a Lie group and a totally disconnected compact group . (Note that this local isomorphism does not necessarily extend to a global isomorphism, as the example of the solenoid group shows.)

Remark 4Of course, it is possible for locally compact groups to be infinite-dimensional; a simple example is the infinite-dimensional torus , which is compact, abelian, metrisable, and locally connected, but infinite dimensional. (It will still be an inverse limit of Lie groups, though.)

Exercise 9Show that a topological group is Lie if and only if it is locally compact, Hausdorff, first-countable, locally connected, and finite-dimensional.

Remark 5It is interesting to note that this characterisation barely uses the real numbers , which are of course fundamental in defining the smooth structure of a Lie group; the only remaining reference to comes through the notion of finite dimensionality. It is also possible, using dimension theory, to obtain alternate characterisations of finite dimensionality (e.g. finite Lebesgue covering dimension) that avoid explicit mention of the real line, thus capturing the concept of a Lie group using only the concepts of point-set topology (and the concept of a group, of course).

** — 4. Transitive actions — **

We now prove Proposition 6. As this is a stronger statement than Theorem 5, it will not be surprising that we will be using a very similar argument to prove the result.

Let be a locally compact -compact group that acts transitively, faithfully, and continuously on a connected manifold . 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 (Theorem 2), 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.

As is -compact, the manifold is also. As acts faithfully on , this makes first countable; and so (by Exercise 4) is the inverse limit of a *sequence* of Lie groups , with each projecting surjectively onto , and with the shrinking to the identity.

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 .

Exercise 10Show that the dimensions of the are be non-decreasing, and bounded above by the dimension of . (Hint:repeat the arguments of the previous section. The need no longer be compact, but they are still closed, and this still suffices to make the preceding arguments go through.)

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.

## 9 comments

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10 October, 2011 at 10:43 am

Marius BuligaA comment and a questions:

1. Re Remark 5, I would say that, according to the definition of “finite dimensional” from Exercise 7, there is a lot of structure entering by the back door, namely something akin to rectifiability (statements about continuous images of euclidean spaces). Maybe related, the problem of defining rectifiability in sub-riemannian spaces (which are main examples of spaces non euclidean at any scale) is open, in the sense that there is no satisfactory definition of rectifiability for these spaces.

2. Question: In your Theorem 12 (Freiman’s type theorem) from your Notes 0, if you take to be the dyadic integers (or some noncommutative nilpotent like, but dyadic version), what could be the locally compact group generated by the said sequence of approximate groups, etc ?

Thanks!

10 October, 2011 at 11:35 am

Terence TaoI’m not sure what the “group of dyadic integers” is in Q2; also, to specify an approximate group, one needs to not just specify the ambient group G, but also a subset A of that group G obeying the approximate group axioms. So I don’t have enough information to answer your question.

If you mean G to be the additive group of rationals whose denominator is a power of two, and let be the multiples of between -1 and 1 for some large , then this is an approximate group, and in the limit , the associated ultraproduct can be modeled by the interval in the real line. But I don’t know if this is the type of example you have in mind.

11 October, 2011 at 12:04 am

Marius BuligaThank you for the reply. I don’t know your proof, but I am thinking about taking a group G which is not locally euclidean (say Q2 or nilpotent versions, easy to construct, or even some strange group a la Grigorchuk). Inside this group, maybe, you can choose a (sequence of?) approximate group(s) with two properties: first, the approximate group is big enough to feel (??) all the group G, second, you can tell exactly what is the locally euclidean group (say G’) which has inside a progression which is roughly equivalent with the approximate group.

The question is: is there any relation between G and G’? (like approximate morphism, or a functorial correspondence G-G’…)

10 October, 2011 at 11:11 am

Neville“Roughly speaking, this theorem … of a locally group …”

locally compact group?

[Corrected, thanks - T.]27 October, 2011 at 8:37 pm

254A, Notes 7: Models of ultra approximate groups « What’s new[...] precompact neighbourhood of the identity such that . By the Gleason-Yamabe theorem (Theorem 1 of Notes 5), there is an open subgroup of , and a compact normal subgroup of contained in , such that is [...]

6 November, 2011 at 11:45 am

Ben HayesIn Exercise 8, I’m a little confused by the concept of “local isomorphism.” Do you mean that there is an open subset of ,a Lie Group , a compact totally disconnected group , and a homeomorphism such that , are defined for all and equal , ? Or do you mean there are open subset , of , and a homeomorphism which respects multiplication and inversion (when it is defined).

6 November, 2011 at 11:56 am

Terence TaoI mean the latter; I’ll reword to clarify. (The example of the solenoid group from Notes 0 shows that the former is not true.)

6 November, 2011 at 12:11 pm

Ben HayesOkay, thanks It became clearer when I read the surrounding text.

4 October, 2014 at 8:43 am

Tomek KaniaLet me point out a minor typo: “theorem of van Dantzig’s theorem”.

[Corrected, thanks - T.]