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
.
Theorem 2 (Cartan’s theorem) If
is a (topologically) closed subgroup of a Lie group
, then
is a smooth submanifold of
, and is thus also a Lie group.
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:
Theorem 4 (von Neumann’s theorem) If
is a Lie group, and
is a locally compact group with an injective continuous homomorphism
, then
also has the structure of a Lie group.
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.
Theorem 5 Let
be a locally compact Hausdorff topological vector space. Then
is isomorphic (as a topological vector space) to
for some finite
.
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.
Theorem 7 (Building Lie structure from Gleason metrics) Let
be a locally compact group that has a Gleason metric. Then
is isomorphic to a Lie group.
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.
Lemma 10 Suppose that the identity
is not an isolated point of
(i.e.
is not discrete). Then
is non-trivial (i.e. it does not consist solely of
).
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:
Lemma 11 There exists a neighbourhood
of the identity in
, and a neighbourhood
of the origin in
, such that
is a homeomorphism.
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 —
We will now turn to the study of topological vector spaces, which we will need to establish Theorem 7. We begin by recalling the definition of a topological vector space.
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
Theorem 14 Every finite-dimensional Hausdorff topological vector space has the usual topology.
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:
Corollary 15 In a Hausdorff topological space
, every finite-dimensional subspace
is closed.
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:
and
Proof: We begin with the first inequality. By the triangle inequality, it suffices to show that
. By left-invariance and approximate right-invariance, the left-hand side is comparable to
which by (2) is bounded above by
as required.
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
is totally bounded. By the Arzelá-Ascoli theorem, it suffices to show that the family of functions in is equicontinuous.
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
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
and
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
and thus
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).
Proposition 21 For any neighbourhood
of the origin in
,
is a neighbourhood of the identity in
.
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
and thus
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
.
Exercise 9 If we identify
with
for some
, show that the exponential map
is bilipschitz.
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
).
Combining this proposition with Lemma 16 from the previous notes, we obtain Theorem 7.
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.)

11 comments
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11 September, 2011 at 3:07 pm
David Roberts
The last sentence of the first paragraph should be ‘we call such local groups…’.
Also, the sentence ‘the one-parameter subgroups are in one-to-one correspondence with the Lie algebra g’ doesn’t make sense. Do you mean ‘elements of the Lie algebra g’?
[Corrected, thanks - T.]
27 September, 2011 at 3:30 pm
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4 October, 2011 at 12:58 pm
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9 October, 2011 at 7:03 am
pavel zorin
Dear Prof. Tao,
In the proof of theorem 14 one can use the continuity of scalar multiplication at (0,0) to make U bounded. This is easier and also shows that the continuity of addition is not needed in this step.
best regards,
pavel
[Good point; I've made the change. -T.]
16 October, 2011 at 10:26 am
Ben Hayes
In Lemma 17 I don’t quite see how you apply the escape property. To apply the escape property, you need to already know that
whereas it appears that at this stage you only know 
16 October, 2011 at 2:51 pm
Terence Tao
If
exceeds 1/C, then one works instead with the largest positive integer m for which
; this will be an integer between 1 and n, and for
small enough one can show that one must have
to avoid a contradiction.
(More generally, one often encounters “almost circular reasoning” issues in these sorts of problems, in which one almost has to assume the result one is trying to prove as a hypothesis. Often, one has to deal with the apparent circularity by an induction argument, a bootstrap argument, a continuity argument, an infinite descent argument, or passing to the “first counterexample” or “last example”. The above is an instance of the latter strategy.)
16 October, 2011 at 4:43 pm
Ben Hayes
Ah, when I go through the estimates I see the contradiction now. Thanks.
27 October, 2011 at 8:37 pm
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1 December, 2011 at 7:58 am
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