In this final lecture, we establish a Ratner-type theorem for actions of the special linear group on homogeneous spaces. More precisely, we show:

Theorem 1.Let G be a Lie group, let be a discrete subgroup, and let be a subgroup isomorphic to . Let be an H-invariant probability measure on which is ergodic with respect to H (i.e. all H-invariant sets either have full measure or zero measure). Then ishomogeneousin the sense that there exists a closed connected subgroup and a closed orbit such that is L-invariant and supported on Lx.

This result is a special case of a more general theorem of Ratner, which addresses the case when H is generated by elements which act unipotently on the Lie algebra by conjugation, and when has finite volume. To prove this theorem we shall follow an argument of Einsiedler, which uses many of the same ingredients used in Ratner’s arguments but in a simplified setting (in particular, taking advantage of the fact that H is semisimple with no non-trivial compact factors). These arguments have since been extended and made quantitative by Einsiedler, Margulis, and Venkatesh.

– Representation theory of –

Theorem 1 concerns the action of on a homogeneous space . Before we are able to tackle this result, we must first understand the linear actions of on real or complex vector spaces – in other words, we need to understand the representation theory of the Lie group (and its associated Lie algebra ).

Of course, this theory is very well understood, and by using the machinery of weight spaces, raising and lowering operators, etc. one can completely classify all the finite-dimensional representations of ; in fact, all such representations are isomorphic to direct sums of symmetric powers of the standard representation of on . This classification quickly yields all the necessary facts we will need here. However, we will use only a minimal amount of this machinery here, to obtain as direct and elementary a proof of the results we need as possible.

The first fact we will need is that finite-dimensional representations of are completely reducible.

Lemma 1.(Complete reducibility) Let act linearly (and smoothly) on a finite-dimensional real vector space V, and let W be a -invariant subspace of V. Then there exists a complementary subspace W’ to W which is also -invariant (thus V is isomorphic to the direct sum of W and W’).

**Proof.** We will use Weyl’s unitary trick to create the complement W’, but in order to invoke this trick, we first need to pass from the non-compact group to a compact counterpart. This is done in several stages.

First, we linearise the action of the Lie group by differentiating to create a corresponding linear action of the Lie algebra in the usual manner.

Next, we complexify the action. Let and be the complexifications of V and W respectively. Then the complexified Lie algebra acts on both and , and in particular the special unitary Lie algebra does also.

Since the special unitary group

(1)

Now we can apply the unitary trick. Take any Hermitian form on . This form need not be preserved by the action, but if one defines the averaged form

(2)

where dg is Haar measure on the compact Lie group , then we see that is a Hermitian form which is -invariant; thus this form endows with a Hilbert space structure with respect to which the -action is unitary. If we then define to be the orthogonal complement of in this Hilbert space, then this vector space is invariant under the action, and thus (by differentiation) by the action. But observe that and have the same complex span (namely, ); thus the complex vector space is also -invariant.

The last thing to do is to undo the complexification. If we let W’ be the space of real parts of vectors in which are real modulo , then one easily verifies that W’ is -invariant (hence -invariant, by exponentiation) and is a complementary subspace to W, as required.

**Remark 1.** We can of course iterate the above lemma and conclude that every finite-dimensional representation of is the direct sum of irreducible representations, which explains the term “complete reducibility”. Complete reducibility of finite-dimensional representations of a Lie algebra (over a field of characteristic zero) is equivalent to that Lie algebra being semisimple. The situation is slightly more complicated for Lie groups, though, if such groups are not simply connected.

An important role in our analysis will be played by the one-parameter unipotent subgroup of , where

. (3)

Clearly, the elements of U are unipotent when acting on . It turns out that they are unipotent when acting on all other finite-dimensional representations also:

Lemma 2.Suppose that acts on a finite-dimensional real or complex vector space V. Then the action of any element of U on V is unipotent.

**Proof.** By complexifying V if necessary we may assume that V is complex. The action of the Lie group induces a Lie algebra homomorphism . To show that the action of U is unipotent, it suffices to show that is nilpotent, where

(4)

is the infinitesimal generator of U. To show this, we exploit the fact that induces a raising operator. We introduce the diagonal subgroup of , where

. (5)

This group has infinitesimal generator

. (6)

Observe that , and thus (since is a Lie algebra homomorphism)

. (7)

We can rewrite this as

(8)

for any , which on iteration implies that

(9)

for any non-negative integers m, r. But this implies that raises generalised eigenvectors of of eigenvalue to generalised eigenvectors of of eigenvalue . But as V is finite dimensional, there are only finitely many eigenvalues of , and so is nilpotent on each of the generalised eigenvectors of . By the Jordan normal form, these generalised eigenvectors span V, and we are done.

**Exercise 1.** By carrying the above analysis further (and also working with the adjoint of U to create lowering operators) show (for complex V) that is diagonalisable, and the eigenvalues are all integers. For an additional challenge: deduce from this that the representation is isomorphic to a direct sum of the representations of on the symmetric tensor powers of (or, if you wish, on the space of homogeneous polynomials of degree k on 2 variables). Of course, if you are stuck, you can turn to any book on representation theory (I recommend Fulton and Harris).

The group U is merely a subgroup of the group , so it is not a priori evident that any vector (in a space that acts on) which is U-invariant, is also -invariant. But, thanks to the highly non-commutative nature of , this turns out to be the case, even in infinite dimensions, once one restricts attention to *continuous unitary* actions:

Lemma 3(Mautner phenomenon). Let be a continuous unitary action on a Hilbert space V (possibly infinite dimensional). Then any vector which is fixed by U, is also fixed by .

**Proof.** We use an argument of Margulis. We may of course take v to be non-zero. Let be a small number. Then even though the matrix is very close to the identity, the double orbit can stray very far away from U. Indeed, from the algebraic identity

(10)

which is valid for any , we see that this double orbit in fact comes very close to the diagonal group D. Applying (10) to the U-invariant vector v and taking inner products with v, we conclude from unitarity that

. (11)

Taking limits as (taking advantage of the continuity of ) we conclude that . Since has the same length as v, we conclude from the converse Cauchy-Schwarz inequality that , i.e. that v is D-invariant.

As U and D do not quite generate , we have to work a bit more to finish the job. Let be as above. Observe that converges to the identity as , and thus . Using the D-invariance we conclude that , and thus as before v is also invariant with respect to the group U’ generated by the . Since U and U’ (and D, if desired) generate , the claim follows.

**Remark 2.** The key fact about U being used here is that its Lie algebra is not trapped inside any proper ideal of , which, in turn, follows from the fact that this Lie algebra is simple. One can do the same thing for semisimple Lie algebras provided that the unipotent group U is non-degenerate in the sense that it has non-trivial projection onto each simple factor.

This phenomenon has an immediate dynamical corollary:

Corollary 1(Moore ergodic theorem). Suppose that acts in a measure-preserving fashion on a probability space . If this action is ergodic with respect to , then it is also ergodic with respect to U.

**Proof. ** Apply Lemma 3 to .

– Proof of Theorem 1 –

Having completed our representation-theoretic preliminaries, we are now ready to begin the proof of Theorem 1. The key is to prove the following dichotomy:

Proposition 1.(Lack of concentration implies additional symmetry) Let be as in Theorem 1. Suppose there exists a closed connected subgroup such that is L-invariant. Then exactly one of the following statements hold:

- (Concentration) is supported on a closed orbit Lx of L.
- (Additional symmetry) There exists a closed connected subgroup such that is L’-invariant.

Iterating this proposition (noting that the dimension of L’ is strictly greater than that of L) we will obtain Theorem 1. So it suffices to establish the proposition.

We first observe that the ergodicity allows us to obtain the concentration conclusion (1) as soon as assigns any non-zero mass to an orbit of L:

Lemma 4.Let the notation and assumptions be as in Proposition 1. Suppose that for some . Then is closed and is supported on .

**Proof. **Since is H-invariant and is H-ergodic, the set must either have full measure or zero measure. It cannot have zero measure by hypothesis, thus . Thus, if we show that is closed, we automatically have that is supported on .

As is a homogeneous space, we may assume without loss of generality (conjugating L if necessary) that is at the origin, then . The measure on this set can then be pulled back to a measure m on L by the formula

. (12)

By construction, m is left L-invariant (i.e. a left Haar measure) and right -invariant. From uniqueness of left Haar measure up to constants, we see that for any g in L there is a constant such that for all measurable E. It is not hard to see that is a character, i.e. it is continuous and multiplicative, thus for all g, h in L. Also, it is the identity on and thus descends to a continuous function on .

Now let be a compact subset of with positive -measure, and let be arbitrary. By the Poincare recurrence theorem, is non-empty for arbitrarily large , and thus is non-empty for arbitrary large . Since is bounded above and below, we conclude that for all (i.e. L is unimodular). Thus m is right-invariant, which implies that obeys the right-invariance property for any g in L with sufficiently close to and any sufficiently small compact set (small enough to fit inside a single fundamental domain of ).

Suppose that is not closed; then one can find a sequence in that converges to but with the staying bounded away from the identity for . For a sufficiently small compact neighbourhood of the identity in , the sets then are disjoint and all have the same measure for large enough; since , this forces these sets to be null. But then the invariant measure annihilates and is thus null as well, a contradiction.

We return to the proof of Proposition 1. In view of Lemma 4, we may assume that is totally non-concentrated on L-orbits in the sense that

. (14)

In particular, for -almost every x and y, y does not lie in the orbit Lx of x and vice versa; informally, the group elements in G that are used to move from x to y should be somehow “transverse” to L.

On the other hand, we are given that is ergodic with respect to H, and thus (by Corollary 1) ergodic with respect to U. This implies (cf. Proposition 2 from Lecture 9) that -almost every point x in is *generic* (with respect to U) in the sense that

. (15)

for all continuous compactly supported .

**Exercise 2**. Prove this claim. (Hint: obtain continuous analogues of the theory from Lecture 8 and Lecture 9.)

The equation (15) (and the Riesz representation theorem) lets us describe the measure in terms of the U-orbit of a generic point. On the other hand, from (14) and the ensuing discussion we see that any two generic points are likely to be separated from each other by some group element “transverse” to L. It is the interaction between these two facts which is going to generate the additional symmetry needed for Proposition 1. We illustrate this with a model case, in which the group element centralises U:

Proposition 2(central case). Let the notation and assumptions be as in Proposition 1. Suppose that x, y are generic points such that for some that centralises U (i.e. it commutes with every element of u). Then is invariant under the action of g.

**Proof. ** Let be continuous and compactly supported. Applying (15) with x replaced by y=gx we obtain

. (16)

Commuting g with and using (15) again, we conclude

(17)

and the claim follows from the Riesz representation theorem.

Of course, we don’t just want invariance under one group element g; we want a whole group L’ of symmetries for which one has invariance. But it is not hard to leverage the former to the latter, provided one has enough group elements:

Lemma 5.Let the notation and assumptions be as in Proposition 1. Suppose one has a sequence of group elements tending to the identity, such that the action of each of the preserve , and such that none of the lie in L. Then there exists a closed connected subgroup such that is L-invariant.

**Proof.** Let S be the stabiliser of , i.e. the set of all group elements g whose action preserves . This is clearly a closed subgroup of G which contains L. If we let be the identity connected component of S, then L’ is a closed connected subgroup containing L which will contain for all sufficiently large n, and in particular is not equal to L. The claim follows.

From Proposition 2 and Lemma 5 we see that we are done if we can find pairs of nearby generic points with going to the identity such that and that centralises U. Now we need to consider the non-central case; thus suppose for instance that we have two generic points x, y=gx in which g is close to the identity but does not centralise U. The key observation here is that we can use the U-invariance of the situation to pull x and y slowly apart from each other. More precisely, since x and y are generic, we observe that and are also generic for any t, and that these two points differ by the conjugated group element . Taking logarithms (which are well-defined as long as stays close to the identity), we can write

(18)

where is the adjoint representation. From Lemma 2, we know that is nilpotent, and so (by Taylor expansion of the exponential) depends polynomially on t. In particular, if g does not centralise U, then is non-constant and thus must diverge to infinity as . In particular, given some small ball B around the origin in (with respect to some arbitrary norm), then whenever lies inside B around the origin and is not central, there must be a first time such that reaches the boundary of this ball. We write for the location of g when it escapes. We now have the following variant of Proposition 2:

Proposition 3(non-central case). Let the notation and assumptions be as in Proposition 1. Suppose that are generic points such that for some which do not centralise u, but such that converge to the identity (in particular, for all sufficiently large n). Suppose furthermore that areuniformly genericin the sense that for any continuous compactly supported , the convergence of (15) (with x replaced by or ) is uniform in n. Then is invariant under the action of any limit point of the .

** Proof.** By passing to a subsequence if necessary we may assume that converges to . For each sufficiently large n, we write , thus for all , and . We rescale this by defining the functions by . From the unipotent nature of U, these functions are polynomial (with bounded degree), and also bounded (as they live in ), and are thus equicontinuous (since all norms are equivalent on finite dimensional spaces). Thus, by the Arzelà-Ascoli theorem, we can assume (after passing to another subsequence) that is uniformly convergent to some limit f, which is another polynomial. Since we already have converging to , this implies that for any there exists such that for all and all sufficiently large n. In other words, we have

(19)

for sufficiently large n, whenever .

This is good enough to apply a variant of the Proposition 2 argument. Namely, if is continuous and compactly supported, then by uniform genericity we have for T sufficiently large that

(20)

for all n. Applying (19) we can write on the support of f, and so by uniform continuity of f

(21)

where o(1) goes to zero as , uniformly in n. Using (15) again and then letting , we obtain the -invariance of as desired.

Now we have all the ingredients to prove Proposition 1, and thus Theorem 1.

**Proof of Proposition 1.** We know that -almost every point is generic. Applying Egoroff’s theorem, we can find sets of measure arbitrarily close to 1 (e.g. ) on which the points are uniformly generic.

Now let V be a small neighbourhood the origin in L. Observe from the Fubini-Tonelli theorem that

(22)

where m is the Haar measure on the unimodular group L, from which one can find a set of positive measure such that for all ; one can view E’ as “points of density” of E in some approximate sense (and with regard to the L action).

Since E’ has positive measure, and using (14), it is not hard to find sequences with for any n and with (using some reasonable metric on ).

**Exercise 3.** Verify this. (*Hint*: can be covered by countably many balls of a fixed radius.)

Next, recall that acts by conjugation on the Lie algebra of G, and also leaves the Lie algebra of L invariant. By Lemma 1, this implies there is a complementary subspace W of in which is also H-invariant (and in particular, U-invariant). From the inverse function theorem, we conclude that for any group element g in G sufficiently close to the identity, we can factor where is also close to the identity, and is small (in fact this factorisation is unique). We let be the map from g to l; this is well-defined and smooth near the identity.

Let n be sufficiently large, and write where goes to the identity as n goes to infinity. Pick at random (using the measure m conditioned to V). Using the inverse function theorem and continuity, we see that the random variable is supported in a small neighbourhood of V, and that its distribution converges to the uniform distribution of V (in, say, total variation norm) as . In particular, we see that with probability at least 0.7 and with probability at least 0.6 (say) if n is large enough. In particular we can find an such that both lie in E. Also by construction we see that for some ; since , we see that is non-zero. On the other hand, since W is transverse to and the distance between go to zero, we see that goes to zero.

There are now two cases. If centralises U for infinitely many n, then from Proposition 2 followed by Lemma 5 we obtain conclusion 2 of Proposition 1 as required. Otherwise, we may pass to a subsequence and assume that none of the centralise U. Since W is preserved by U, we see that the group elements also lie in for some compact set K in W, and also on the boundary of B. This space is compact, and so by Proposition 3 we see that is invariant under some group element , which cannot lie in L. Since the ball B can be chosen arbitrarily small, we can thus apply Lemma 5 to again obtain conclusion 2 of Proposition 1 as required.

[*Update*, Oct 18 2011: A gap in the proof of Lemma 3 has been repaired, following a suggestion of Hee Oh.]

## 12 comments

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17 March, 2008 at 5:30 am

www.smasra.comlooks nice thank you .. waiting for more maths operations :)

18 March, 2008 at 5:16 am

This w33k in the arXivs… « It’s Equal, but It’s Different…[...] 254A, Lecture 17: A Ratner-type theorem for SL_2(R) orbits [...]

22 January, 2012 at 5:15 pm

RexDear Terry,

How do we know that the integrals in line 13 in the proof of lemma 4 converge?

A priori, it seems that the function c(g) need not be bounded, since the domain L.x_0 is not known to be compact.

22 January, 2012 at 6:59 pm

Terence TaoOops, you’re right. I’m reinstating Einseidler’s original argument here (based on the Poincare recurrence theorem) instead.

1 February, 2012 at 11:55 pm

RexCould you elaborate a little more on how one makes the equicontinuity deduction

“…these functions are polynomial (with bounded degree), and also bounded (as they live in B), and are thus equicontinuous (since all norms are equivalent on finite dimensional spaces)”

as found in the proof of Prop 3?

2 February, 2012 at 8:08 am

Terence TaoAs all norms are equivalent on finite dimensional spaces, the sup norm on the space of bounded degree polynomials is equivalent to, say, the Lipschitz norm, which certainly controls equicontinuity.

4 June, 2012 at 7:36 pm

RexHow much of the proof of lemma 4 carries over when we assume L to be, not a Lie group, but only a locally compact group?

I am mostly worried about the spots where we have to “choose a sufficiently small compact set K”.

8 June, 2012 at 1:42 am

RexFirst, a few typos: I believe in the statement of Proposition 3, you want the to be points in rather than . Also, in the proof of Proposition 3, it seems that has suddenly turned into in a few spots. I am assuming that , anyway.

Now, a question: I’m confused about the notion of uniform genericity mentioned in Proposition 3. It seems to be a property of a sequence of points rather than a single point (since the uniformity demanded is over ).

However, in the beginning of the proof of Proposition 1, it is mentioned that Egoroff’s theorem can be used to show that a large subset of is uniformly generic. But what does it mean for individual points in to be uniformly generic?

8 June, 2012 at 7:03 am

Terence TaoThanks for the corrections!

To be uniformly generic in a subset E of means that the convergence of (15) is uniform in that subset E. (In the proposition, E was a countable set , but when applying Egoroff’s theorem, E becomes the complement of a set of small measure.)

8 June, 2012 at 2:56 pm

RexAh, okay. This makes sense.

In this case, maybe you shouldn’t use the phrase “…on which the points are uniformly generic.” as you do in the beginning of the proof of Proposition 1. It gave me the impression that there is a meaningful notion of uniform genericity for single points.

20 June, 2012 at 3:30 am

RexIn equations (20) and (21), is taken to be a large time (large enough that the RHS comes close to the LHS as promised by the Birkhoff ergodic theorem).

However, we are using equation (19) to deduce (21), and (19) only seems to apply for within the range .

How do we reconcile these two things? It seems that we need to set for some to get the equation (21), but a priori we could have for all , and this might not be large enough for Birkhoff’s ergodic theorem to make the two sides of (20) close to equal.

20 June, 2012 at 12:22 pm

Terence TaoThe are converging to the identity, and so the must necessarily go to infinity (otherwise cannot reach the boundary of the unit ball).