Hilbert’s fifth problem concerns the minimal hypotheses one needs to place on a topological group to ensure that it is actually a Lie group. In the previous set of notes, we saw that one could reduce the regularity hypothesis imposed on to a “” condition, namely that there was an open neighbourhood of that was isomorphic (as a local group) to an open subset of a Euclidean space with identity element , and with group operation obeying the asymptotic
for sufficiently small . We will call such local groups local groups.
We now reduce the regularity hypothesis further, to one in which there is no explicit Euclidean space that is initially attached to . Of course, Lie groups are still locally Euclidean, so if the hypotheses on do not involve any explicit Euclidean spaces, then one must somehow build such spaces from other structures. One way to do so is to exploit an ambient space with Euclidean or Lie structure that is embedded or immersed in. A trivial example of this is provided by the following basic fact from linear algebra:
Lemma 1 If is a finite-dimensional vector space (i.e. it is isomorphic to for some ), and is a linear subspace of , then is also a finite-dimensional vector space.
We will establish a non-linear version of this statement, known as Cartan’s theorem. Recall that a subset of a -dimensional smooth manifold is a -dimensional smooth (embedded) submanifold of for some if for every point there is a smooth coordinate chart of a neighbourhood of in that maps to , such that , where we identify with a subspace of . Informally, locally sits inside the same way that sits inside .
Note that the hypothesis that is closed is essential; for instance, the rationals are a subgroup of the (additive) group of reals , but the former is not a Lie group even though the latter is.
Exercise 1 Let be a subgroup of a locally compact group . Show that is closed in if and only if it is locally compact.
A variant of the above results is provided by using (faithful) representations instead of embeddings. Again, the linear version is trivial:
Lemma 3 If is a finite-dimensional vector space, and is another vector space with an injective linear transformation from to , then is also a finite-dimensional vector space.
Here is the non-linear version:
Actually, it will suffice for the homomorphism to be locally injective rather than injective; related to this, von Neumann’s theorem localises to the case when is a local group rather a group. The requirement that be locally compact is necessary, for much the same reason that the requirement that be closed was necessary in Cartan’s theorem.
Example 1 Let be the two-dimensional torus, let , and let be the map , where is a fixed real number. Then is a continuous homomorphism which is locally injective, and is even globally injective if is irrational, and so Theorem 4 is consistent with the fact that is a Lie group. On the other hand, note that when is irrational, then is not closed; and so Theorem 4 does not follow immediately from Theorem 2 in this case. (We will see, though, that Theorem 4 follows from a local version of Theorem 2.)
As a corollary of Theorem 4, we observe that any locally compact Hausdorff group with a faithful linear representation, i.e. a continuous injective homomorphism from into a linear group such as or , is necessarily a Lie group. This suggests a representation-theoretic approach to Hilbert’s fifth problem. While this approach does not seem to readily solve the entire problem, it can be used to establish a number of important special cases with a well-understood representation theory, such as the compact case or the abelian case (for which the requisite representation theory is given by the Peter-Weyl theorem and Pontryagin duality respectively). We will discuss these cases further in later notes.
In all of these cases, one is not really building up Euclidean or Lie structure completely from scratch, because there is already a Euclidean or Lie structure present in another object in the hypotheses. Now we turn to results that can create such structure assuming only what is ostensibly a weaker amount of structure. In the linear case, one example of this is is the following classical result in the theory of topological vector spaces.
Remark 1 The Banach-Alaoglu theorem asserts that in a normed vector space , the closed unit ball in the dual space is always compact in the weak-* topology. Of course, this dual space may be infinite-dimensional. This however does not contradict the above theorem, because the closed unit ball is not a neighbourhood of the origin in the weak-* topology (it is only a neighbourhood with respect to the strong topology).
The full non-linear analogue of this theorem would be the Gleason-Yamabe theorem, which we are not yet ready to prove in this set of notes. However, by using methods similar to that used to prove Cartan’s theorem and von Neumann’s theorem, one can obtain a partial non-linear analogue which requires an additional hypothesis of a special type of metric, which we will call a Gleason metric:
Definition 6 Let be a topological group. A Gleason metric on is a left-invariant metric which generates the topology on and obeys the following properties for some constant , writing for :
- (Escape property) If and is such that , then .
- (Commutator estimate) If are such that , then
where is the commutator of and .
Exercise 2 Let be a topological group that contains a neighbourhood of the identity isomorphic to a local group. Show that admits at least one Gleason metric.
We will rely on Theorem 7 to solve Hilbert’s fifth problem; this theorem reduces the task of establishing Lie structure on a locally compact group to that of building a metric with suitable properties. Thus, much of the remainder of the solution of Hilbert’s fifth problem will now be focused on the problem of how to construct good metrics on a locally compact group.
In all of the above results, a key idea is to use one-parameter subgroups to convert from the nonlinear setting to the linear setting. Recall from the previous notes that in a Lie group , the one-parameter subgroups are in one-to-one correspondence with the elements of the Lie algebra , which is a vector space. In a general topological group , the concept of a one-parameter subgroup (i.e. a continuous homomorphism from to ) still makes sense; the main difficulties are then to show that the space of such subgroups continues to form a vector space, and that the associated exponential map is still a local homeomorphism near the origin.
Exercise 3 The purpose of this exercise is to illustrate the perspective that a topological group can be viewed as a non-linear analogue of a vector space. Let be locally compact groups. For technical reasons we assume that are both -compact and metrisable.
- (i) (Open mapping theorem) Show that if is a continuous homomorphism which is surjective, then it is open (i.e. the image of open sets is open). (Hint: mimic the proof of the open mapping theorem for Banach spaces, as discussed for instance in these notes. In particular, take advantage of the Baire category theorem.)
- (ii) (Closed graph theorem) Show that if a homomorphism is closed (i.e. its graph is a closed subset of ), then it is continuous. (Hint: mimic the derivation of the closed graph theorem from the open mapping theorem in the Banach space case, as again discussed in these notes.)
- (iii) Let be a homomorphism, and let be a continuous injective homomorphism into another Hausdorff topological group . Show that is continuous if and only if is continuous.
- (iv) Relax the condition of metrisability to that of being Hausdorff. (Hint: Now one cannot use the Baire category theorem for metric spaces; but there is an analogue of this theorem for locally compact Hausdorff spaces.)
— 1. The theorems of Cartan and von Neumann —
We now turn to the proof of Cartan’s theorem. As indicated in the introduction, the fundamental concept here will be that of a one-parameter subgroup:
Definition 8 (One-parameter subgroups) Let be a topological group. A one-parameter subgroup of is a continuous homomorphism . The space of all such one-parameter subgroups is denoted .
Remark 2 Strictly speaking, the terminology “one-parameter subgroup” is a misnomer, because it is the image of which is a subgroup of , rather than itself. Note that we consider reparameterisations of a one-parameter subgroup , where is a non-zero real number, to be distinct from when , even though both one-parameter subgroups have the same image.
We recall Exercise 12 from the previous set of notes, which we reformulate here as a lemma:
Lemma 9 (Classification of one-parameter subgroups) Let be a Lie group, with Lie algebra . Then if is an element of , then is a one-parameter subgroup; conversely, if is a one-parameter subgroup, then there is a unique such that for all . Thus we have a canonical one-to-one correspondence between and .
Now let be a closed subgroup of a Lie group . Every one-parameter subgroup of is clearly also a one-parameter subgroup of , which by the above lemma can be viewed as an element of :
Thus we can think of as a subset of :
We claim that is in fact a linear subspace of . Indeed, it contains the zero element of (which corresponds to the trivial one-parameter subgroup ), and from reparameterisation we see that if , then for all . Finally, if , then by definition we have for all . But recall from Exercise 11(ii) of the previous notes that
Since is a group, we see that lies in . Since is closed, we conclude that for all , which implies that . Thus is closed under both addition and scalar multiplication, and so it is a vector space. (It turns out that is in fact a Lie algebra, but we will not need this fact yet.)
The next step is to show that is “large” enough to serve as the “Lie algebra” of . To illustrate this type of fact, let us first establish a simple special case.
Proof: As is not isolated, there exists a sequence of elements of that converge to . As is a local homeomorphism near the identity, we may thus find a sequence of elements of converging to zero such that for all sufficiently large .
Let us arbitrarily endow the finite-dimensional vector space with a norm (it will not matter which norm we select). Then the sequence lies on the unit sphere with respect to this norm, and thus by the Heine-Borel theorem (and passing to a subsequence) we may assume that converges to some element of norm .
Let be any positive real number. Then converges to , and so converges to . As lies in , so does ; as is closed, we conclude that for all positive , and hence for all . We conclude that , and the fclaim follows.
Now we establish a stronger version of the above lemma:
Proof: Let be a neighbourhood of the origin in such that is a homeomorphism (this exists since is a local homeomorphism in a neighbourhood of the origin in . Clearly lies in and contains . If contains a neighbourhood of the in then we are done, so suppose that this is not the case. Then we can find a sequence of elements in that converge to . We may write for some converging to zero in .
As is a subspace of the finite-dimensional vector space , we may write for some vector space transverse to (i.e. ). (We do not require to be a Lie algebra.) From the inverse function theorem, the map from to is a local homeomorphism near the identity. Thus we may write for sufficiently large , where and both go to zero as . Since , we see that is non-zero for sufficiently large.
We arbitrarily place a norm on . As before, we may pass to a subsequence and assume that converges to some limit in the unit sphere of ; in particular, .
Since and both lie in , does also. By arguing as in the proof of Lemma 10 we conclude that lies in for all , and so , yielding the desired contradiction.
From the above lemma we see that locally agrees with near the identity, and thus locally agrees with near for every . This implies that is a smooth submanifold of ; since it is also a topological group, it is thus a Lie group. This establishes Cartan’s theorem.
Remark 3 Observe a posteriori that is the Lie algebra of , and in particular is closed with respect to Lie brackets. This fact can also be established directly using Exercise 22 from the previous notes.
There is a local version of Cartan’s theorem, in which groups are replaced by local groups:
Theorem 12 (Local Cartan’s theorem) If is a locally compact local subgroup of a local Lie group , then there is an open neighbourhood of the identity in that is a smooth submanifold of , and is thus also a local Lie group.
The proof of this theorem follows the lines of the global Cartan’s theorem, with some minor technical changes, and we set this proof out in the following exercise.
Exercise 4 Define a local one-parameter subgroup of a local group to be a continuous homomorphism from the (additive) local group to . Call two local one-parameter subgroups equivalent if they agree on a neighbourhood of the origin, and let be the set of all equivalence classes of local one-parameter subgroups. Establish the following claims:
- (i) If is a global group, then there is a canonical one-to-one correspondence that identifies this definition of with the definition of given previously.
- (ii) In the situation of Theorem 12, show that can be identified with a linear subspace of , namely
- (iii) Let the notation and assumptions be as in (ii). For any neighbourhood of the identity in , there is a neighbourhood of the origin in such that .
- (iv) Let the notation and assumptions be as in (ii). There exists a neighbourhood of the identity in , and a neighbourhood of the origin in , such that is a homeomorphism.
- (v) Prove Theorem 12.
One can then use Theorem 12 to establish von Neumann’s theorem, as follows. Suppose that is a locally compact group with an injective continuous homomorphism into a Lie group . As is locally compact, there is an open neighbourhood of the origin in whose closure is compact. The map from to is a continuous bijection from a compact set to a Hausdorff set, and is therefore a homeomorphism (since it maps closed (and hence compact) subsets of to compact (and hence closed) subsets of ). The set is then a locally compact local subgroup of and thus has a neighbourhood of the identity which is a local Lie group, by Theorem 12. Pulling this back by , we see that some neighbourhood of the identity in is a local Lie group, and thus is a global Lie group by Exercise 15 of the previous notes.
Remark 4 State and prove a local version of von Neumann’s theorem, in which and are local groups rather than global groups, and the global injectivity condition is similarly replaced by local injectivity.
— 2. Locally compact vector spaces —
Definition 13 (Topological vector space) A topological vector space is a (real) vector space equipped with a topology that makes the vector space operations and (jointly) continuous. (In particular, is necessarily a topological group.)
One can also consider complex topological vector spaces, but the theory for such spaces is almost identical to the real case, and we will only need the real case for what follows. In the literature, it is often common to restrict attention to Hausdorff topological vector spaces, although this is not a severe restriction in practice, as the following exercise shows:
Exercise 5 Let be a topological vector space. Show that the closure of the origin is a closed subspace of , and the quotient space is a Hausdorff topological vector space. Furthermore, show that a set is open in if and only if it is the preimage of an open set in under the quotient map .
An important class of topological vector spaces are the normed vector spaces, in which the topology is generated by a norm on the vector space. However, not every topological vector space is generated by a norm. See these notes for some further discussion.
We emphasise that in order to be a topological vector space, the vector space operations need to be jointly continuous; merely being continuous in the individual variables is not sufficient to qualify for being a topological vector space. We illustrate this with some non-examples of topological vector spaces:
Example 2 Consider the one-dimensional vector space with the co-compact topology (a non-empty set is open iff its complement is compact in the usual topology). In this topology, the space is a space (though not Hausdorff), the scalar multiplication map is jointly continuous, and the addition map is continuous in each coordinate (i.e. translations are continuous), but not jointly continuous; for instance, the set does not contain a non-trivial Cartesian product of two sets that are open in the co-compact topology. So this is not a topological vector space. Similarly for the cocountable or cofinite topologies on (the latter topology, incidentally, is the same as the Zariski topology on ).
Example 3 Consider the topology of inherited by pulling back the usual topology on the unit circle . This pullback topology is not quite Hausdorff, but the addition map is jointly continuous (so that this gives the structure of a topological group). On the other hand, the scalar multiplication map is not continuous at all. A slight variant of this topology comes from pulling back the usual topology on the torus under the map for some irrational ; this restores the Hausdorff property, and addition is still jointly continuous, but multiplication remains discontinuous.
Example 4 Consider with the discrete topology; here, the topology is Hausdorff, addition is jointly continuous, and every dilation is continuous, but multiplication is not jointly continuous. If one instead gives the half-open topology, then again the topology is Hausdorff and addition is jointly continuous, but scalar multiplication is only jointly continuous once one restricts the scalar to be non-negative.
These examples illustrate that a vector space such as can have many topologies on it (and many topological group structures), but only one topological vector space structure. More precisely, we have
Proof: Let be a finite-dimensional Hausdorff topological space, with topology . We need to show that every set which is open in the usual topology, is open in , and conversely.
Let be a basis for the finite-dimensional space . From the continuity of the vector space operations, we easily verify that the linear map given by
is continuous. From this, we see that any set which is open in , is also open in the usual topology.
Now we show conversely that every set which is open in the usual topology, is open in . It suffices to show that there is a bounded open neighbourhood of the origin in , since one can then translate and dilate this open neighbourhood to obtain a (sub-)base for the usual topology. (Here, “bounded” refers to the usual sense of the term, for instance with respect to an arbitrarily selected norm on (note that on a finite-dimensional space, all norms are equivalent).)
We use to identify (as a vector space) with . As is continuous, every set which is compact in the usual topology, is compact in . In particular, the unit sphere (in, say, the Euclidean norm on ) is compact in . Using this and the Hausdorff assumption on , we can find an open neighbourhood of the origin in which is disjoint from .
At present, need not be bounded (note that we are not assuming to be locally connected a priori). However, we can fix this as follows. Using the joint continuity of the scalar multiplication map, one can find another open neighbourhood of the origin and an open interval around such that the product set is contained in . In particular, since avoids the unit sphere , must avoid the region and is thus bounded, as required.
We isolate one important consequence of the above theorem:
Proof: It suffices to show that every vector is in the exterior of . But this follows from Theorem 14 after restricting to the finite-dimensional space spanned by and .
We can now prove Theorem 5. Let be a locally compact Hausdorff space, thus there exists a compact neighbourhood of the origin. Then the dilate is also a neighbourhood of the origin, and so by compactness can be covered by finitely many translates of , thus
for some finite set . If we let be the finite-dimensional vector space generated by , we conclude that
Iterating this we have
for any . On the other hand, if is a neighbourhood of the origin, then for every we see that for sufficiently large . By compactness of (and continuity of the scalar multiplication map at zero), we conclude that for some sufficiently large , and thus
for any neighbourhood of the origin; thus is in the closure of . By Corollary 15, we conclude that
But is a neighbourhood of the origin, thus for every we have for all sufficiently large , and thus . Thus , and the claim follows.
Exercise 6 Establish the Riesz lemma: if is a normed vector space, is a proper closed subspace of , and , then there exists a vector in with and . (Hint: pick an element of not in , and then pick that nearly minimises . Use these two vectors to construct a suitable .) Using this lemma and the Heine-Borel theorem, give an alternate proof of Theorem 5 in the case when is a normed vector space.
— 3. From Gleason metrics to Lie groups —
Now we prove Theorem 7. The argument will broadly follow the lines of Cartan’s theorem, but we will have to work harder in many stages of the argument in order to compensate for the lack of an obvious ambient Lie structure in the initial hypotheses. In particular, the Gleason metric hypothesis will substitute for the type structure enjoyed by Lie groups, which as we saw in the previous set of notes was needed to obtain good control on the exponential map.
Henceforth, is a locally compact group with a Gleason metric (and an associated “norm” ). In particular, by the Heine-Borel theorem, is complete with this metric.
We use the asymptotic notation in place of for some constant that can vary from line to line (in particular, need not be the constant appearing in the definition of a Gleason metric), and write for . We also let be a sufficiently small constant (depending only on the constant in the definition of a Gleason metric) to be chosen later.
Note that the left-invariant metric properties of give the symmetry property
and the triangle inequality
From the commutator estimate (1) and the triangle inequality we also obtain a conjugation estimate
whenever . Since left-invariance gives
we then conclude an approximate right invariance
whenever . In a similar spirit, the commutator estimate (1) also gives
This has the following useful consequence, which asserts that the power maps behave like dilations:
Proof: We begin with the first inequality. By the triangle inequality, it suffices to show that
uniformly for all . By left-invariance and approximate right-invariance, the left-hand side is comparable to
uniformly for all . By left-invariance and approximate right-invariance, the left-hand side is comparable to
which by (2) is bounded above by
Now we prove the second estimate. Write , then . We have
thanks to the escape property (shrinking if necessary). On the other hand, from the first inequality, we have
If is small enough, the claim now follows from the triangle inequality.
Remark 5 Lemma 16 implies (by a standard covering argument) that the group is locally of bounded doubling, though we will not use this fact here. The bounds above should be compared with the bounds in Exercise 9 of the previous notes. Indeed, just as the bounds in that exercise were used in the previous notes to build the exponential map for Lie groups, the bounds in Lemma 16 are crucial for controlling the exponential function on the locally compact group equipped with the Gleason metric .
Now we bring in the space of one-parameter subgroups. We give this space the compact-open topology, thus the topology is generated by balls of the form
for , , and compact . Actually, using the homomorphism property, one can use a single compact interval , such as , to generate the topology if desired, thus making a metric space.
Given that is eventually going to be shown to be a Lie group, must be isomorphic to a Euclidean space. We now move towards this goal by establishing various properties of that Euclidean spaces enjoy.
Lemma 17 is locally compact.
Proof: It is easy to see that is complete. Let . As is continuous, we can find an interval small enough that for all . By the Heine-Borel theorem, it will suffice to show that the set
By construction, we have whenever . By the escape property, this implies (for small enough, of course) that for all and , thus whenever . From the homomorphism property, we conclude that whenever , which gives uniform Lipschitz control and hence equicontinuity as desired.
We observe for future reference that the proof of the above lemma also shows that all one-parameter subgroups are locally Lipschitz.
Now we put a vector space structure on , which we define by analogy with the Lie group case, in which each tangent vector generates a one-parameter subgroup . From this analogy, the scalar multiplication operation has an obvious definition: if and , we define to be the one-parameter subgroup
which is easily seen to actually be a one-parameter subgroup.
which is easily seen to actually be a one-parameter subgroup.
Now we turn to the addition operation. In the Lie group case, one can express the one-parameter subgroup in terms of the one-parameter subgroups , by the limiting formula
cf. Exercise 15 from the previous notes. In view of this, we would like to define the sum of two one-parameter subgroups by the formula
Proof: To show well-definedness, it suffices to show that for each , the sequence is a Cauchy sequence. It suffices to show that
as . We will in fact prove the slightly stronger claim
Observe from continuity of multiplication that to prove this claim for a given , it suffices to do so for ; thus we may assume without loss of generality that is small.
Let be a small number to be chosen later. Since are locally Lipschitz, we see (if is sufficiently small depending on ) that
for all . From Lemma 16, we conclude that
if and is sufficiently large. Another application of Lemma 16 then gives
if , is sufficiently large, and . The claim follows.
The above argument in fact shows that is uniformly Cauchy for in a compact interval, and so the pointwise limit is in fact a uniform limit of continuous functions and is thus continuous. To prove that is a homomorphism, it suffices by density of the rationals to show that
for all and all positive integers . To prove the first claim, we observe that
and similarly for and , whence the claim. To prove the second claim, we see that
but is conjugated by , which goes to the identity; and the claim follows.
also has an obvious zero element, namely the trivial one-parameter subgroup .
Lemma 19 is a topological vector space.
Proof: We first show that is a vector space. It is clear that the zero element of is an additive and scalar multiplication identity, and that scalar multiplication is associative. To show that addition is commutative, we again use the observation that is conjugated by an element that goes to the identity. A similar argument shows that , and a change of variables argument shows that for all positive integers , hence for all rational , and hence by continuity for all real . The only remaining thing to show is that addition is associative, thus if , that for all . By the homomorphism property, it suffices to show this for all sufficiently small .
An inspection of the argument used to establish (18) reveals that there is a constant such that
for all small and all large , and hence also that
(thanks to Lemma 16). Similarly we have (after adjusting if necessary)
From Lemma 16 we have
Similarly for . By the triangle inequality we conclude that
sending to zero, the claim follows.
Finally, we need to show that the vector space operations are continuous. It is easy to see that scalar multiplication is continuous, as are the translation operations; the only remaining thing to verify is that addition is continuous at the origin. Thus, for every we need to find a such that whenever and . But if are as above, then by the escape property (assuming small enough) we conclude that for , and then from the triangle inequality we conclude that for , giving the claim.
Exercise 7 Show that for any , the quantity
exists and defines a norm on that generates the topology on .
As is both locally compact, metrisable, and a topological vector space, it must be isomorphic to a finite-dimensional vector space with the usual topology, thanks to Theorem 14.
In analogy with the Lie algebra setting, we define the exponential map by setting . Given the topology on , it is clear that this is a continuous map.
Exercise 8 Show that the exponential map is locally injective near the origin. (Hint: from Lemma 16, obtain the unique square roots property: if are sufficiently close to the identity and , then .)
We have proved a number of useful things about , but at present we have not established that is large in any substantial sense; indeed, at present, could be completely trivial even if was large. In particular, the image of the exponential map could conceivably be quite small. We now address this issue. As a warmup, we show that is at least non-trivial if is non-discrete (cf. Lemma 10):
Proposition 20 Suppose that is not a discrete group. Then is non-trivial.
Of course, the converse is obvious; discrete groups do not admit any non-trivial one-parameter subgroups.
Proof: As is not discrete, there is a sequence of non-identity elements of such that as . Writing for the integer part of , then as , and we conclude from the escape property that for all .
We define the approximate one-parameter subgroups by setting
Then we have for , and we have the approximate homomorphism property
uniformly whenever . As a consequence, is asymptotically equicontinuous on , and so by (a slight generalisation of) the Arzéla-Ascoli theorem, we may pass to a subsequence in which converges uniformly to a limit , which is a genuine homomorphism that is genuinely continuous, and is thus can be extended to a one-parameter subgroup. Also, for all , and thus ; in particular, is non-trivial, and the claim follows.
We now generalise the above proposition to a more useful result (cf. Lemma 11).
Proof: We use an argument of Hirschfeld (communicated to me by van den Dries and Goldbring). By shrinking if necessary, we may assume that is a compact star-shaped neighbourhood, with contained in the ball of radius around the origin. As is compact, is compact also.
Suppose for contradiction that is not a neighbourhood of the identity, then there is a sequence of elements of such that as . By the compactness of , we can find an element of that minimises the distance . If we then write , then
and hence as .
Let be the integer part of , then as , and for all .
Let be the approximate one-parameter subgroups defined as
As before, we may pass to a subsequence such that converges uniformly to a limit , which extends to a one-parameter subgroup .
In a similar vein, since , we can find such that , which by the escape property (and the smallness of implies that for . In particular, goes to zero in .
We now claim that is close to . Indeed, from Lemma 16 we see that
Since , we conclude from the triangle inequality and left-invariance that
But from Lemma 16 again, one has
But for large enough, lies in , and so the distance from to is . But this contradicts the minimality of for large enough, and the claim follows.
If is a sufficiently small compact neighbourhood of the identity in , then is bijective by Lemma 8; since it is also continuous, is compact, and is Hausdorff, we conclude that is a homeomorphism. The local group structure on then pulls back to a local group structure on .
Proposition 22 is a radially homogeneous local group (as defined in Definition 7 and (11) from the previous notes), after identifying with for some finite .
Proof: The radial homogeneity is clear from (4) and the homomorphism property, so the main task is to establish the property
for the local group law on . By Exercise 9, this is equivalent to the assertion that
for sufficiently close to the identity in . By definition of , it suffices to show that
for all ; but this follows from Lemma 16 (and the observation, from the escape property, that and ).
Exercise 10 State and prove a version of Theorem 7 for local groups. (In order to do this, you must first decide how to define an analogue of a Gleason metric on a local group.)