The fundamental objects of study in higher-order Fourier analysis are nilmanifolds, or in other words spaces given as a quotient of a connected nilpotent Lie group
by a discrete cocompact subgroup
. Starting with Furstenberg’s work on Szemeredi’s theorem and the multiple recurrence theorem, work by Host and Kra, Green and Tao, and several others has gradually established that nilmanifolds control higher-order linear configurations in the same way that the circle, as in the Hardy-Littlewood circle method, controls first-order linear configurations.
Of basic importantance in the study of nilmanifolds is equidistribution: one needs to know when the sequence equidistributes and when it is trapped inside a subnilmanifold. It turns out that this problem was already studied by Leon Green in the 60s. To describe the theorem note first that the abelianisation map
induces a map from
to a torus
which respects the action of
, and recall that equidistribution on tori is well understood by Weyl’s criterion. Leon Green’s beautiful theorem then states that
equidistributes in the nilmanifold if and only if its image in the torus
equidistributes.
Today at our miniseminar, Aled Walker showed us Parry’s nice proof of this theorem, which is more elementary than Green’s original proof. During the talk there was some discussion about the importantance of various hypotheses such as “simply connected” and “Lie”. It turns out that the proof works rather generally for connected locally compact nilpotent groups, so I thought I would record the proof here with minimal hypotheses. The meat of the argument is exactly as in Aled’s talk and, presumably, Parry’s paper.
Let be an arbitrary locally compact connected nilpotent group, say with lower central series
and let be a closed cocompact subgroup. Under these conditions the Haar measure
of
induces a
-invariant probability measure
on
. We say that
is equidistributed if for every
we have
We fix our attention on the sequence
for some and
. As before we have an abelianisation map
from the -space
to the compact abelian group
. We define equidistribution on
similarly. The theorem is then the following.
Theorem 1 (Leon Green’s theorem) For
and
the following are equivalent.
- The sequence
is equidistributed in
.
- The sequence
is equidistributed in
.
- The orbit of
is dense in
.
for every nontrivial character
.
Item 1 above trivially implies every other item. The implication 43 (a generalised Kronecker theorem) follows by pulling back any nontrivial character of
. The implication 3
2 (a generalised Weyl theorem) follows from the observation that every weak* limit point of the sequence of measures
]must be shift-invariant and thus equal to the Haar measure. So the interesting content of the theorem is 21.
A word about the relation to ergodicity: By the ergodic theorem the left shift is ergodic if and only if for almost every
the sequence
equidistributes; on the other hand
is uniquely ergodic, i.e., the only
-invariant measure is the given one, if and only if for every
the sequence
equidistributes. Thus to prove the theorem above we must not only prove that
is ergodic but that it is uniquely ergodic. Fortunately one can prove these two properties are equivalent in this case.
Lemma 2 If
is ergodic then it’s uniquely ergodic.
The following proof is due to Furstenberg.
Proof: By the ergodic theorem the set of
-generic points, in other words points
for which
for every , has
-measure
, and clearly if
and
then
, so
, where
is the projection of
onto
.
Now let be any
-invariant ergodic measure. By induction we may assume that
is uniquely ergodic, so we must have
, so
But by the ergodic theorem the set of -generic points must also have
-measure
, so there must be some point which is both
– and
-generic, and this implies that
.
We need one more preliminary lemma about topological groups before we really get started on the proof.
Lemma 3 If
and
are connected subgroups of some ambient topological group then
is also connected.
Proof: Since is continuous certainly
is connected, so
is also connected, so because
for all
we see that
is connected.
Thus if is connected then every term
in the lower central series of
is connected.
We can now prove Theorem 1. As noted it suffices to prove that acts ergodically on
whenever it acts ergodically on
. By induction we may assume that
acts ergodically on
. So suppose that
is
-invariant. By decomposing
as a
-space we may assume that
obeys
for some character . In particular
is both
-invariant and
-invariant, so it factors through a
-invariant function
, so it must be constant, say
. Moreover for every
the function
is -invariant, and also a
eigenvector:
By integrating this equation we find that either , so
is constant, or
, so either way we have
But since is a continuous function of
and equal to
when
we must have
for all sufficiently small
, and thus for all
by connectedness of
and the identity
Thus setting extends
to a homomorphism
. In fact we can extend
still further to a function
, where
is the unit disc in
, by setting
Now if and
then
and
so
Since we can cancel
, so
Finally observe that is a continuous function of
, and
, so we must have
for all sufficiently small
, and thus by connectedness of
and the identity
we must have identically. But this implies that
vanishes on all
-term commutators and thus on all of
, so in fact
factors through
, so it must be constant. This finishes the proof.
A remark is in order about the possibility that some of the groups and
are not closed. This should not matter. One could either read the above proof as it is written, noting carefully that I never said groups should be Hausdorff, or, what’s similar, instead modify it so that whenever you quotient by a group
you instead quotient by the group
.
Embarrassingly, it’s difficult to come up with a non-Lie group to which this generalised Leon Green’s theorem applies. It seems that many natural candidates have the property that is not connected but
is: for example consider
So it would be interesting to know whether the theorem extends to such a case. Or perhaps there are no interesting non-Lie groups for this theorem, which would be a bit of a let down.