While working on my recent paper with Ben Green, I was introduced to the beautiful theorems of Marina Ratner on unipotent flows on homogeneous spaces, and their application to questions in number theory, such as the Oppenheim conjecture (first solved by Margulis, by establishing what can retrospectively be viewed as a special case of Ratner’s theorems). This is a subject that I am still only just beginning to learn, but hope to understand better in the future, especially given that quantitative analogues of Ratner’s theorems should exist, and should have even more applications to number theory (see for instance this recent paper of Einsiedler, Margulis, and Venkatesh). In this post, I will try to describe some of the background for this theorem and its connection with the Oppenheim conjecture; I will not discuss the proof at all, largely because I have not fully understood it myself yet. For a nice introduction to these issues, I recommend Dave Morris’ recent book on the subject (and this post here is drawn in large part from that book).

Ratner’s theorem takes place on a homogeneous space. Informally, a homogeneous space is a space X which looks “the same” when viewed from any point on that space. For instance, a sphere $S^2$ is a homogeneous space, but the surface of a cube is not (the cube looks different when viewed from a corner than from a point on an edge or on a face). More formally, a homogeneous space is a space X equipped with an action $(g,x) \mapsto gx$ of a group G of symmetries which is transitive: given any two points x, y on the space, there is at least one symmetry g that moves x to y, thus y=gx. (For instance the cube has several symmetries, but not enough to be transitive; in contrast, the sphere $S^2$ has the transitive action of the special orthogonal group SO(3) as its symmetry group.) It is not hard to see that a homogeneous space X can always be identified (as a set with an action of G) with a quotient $G/\Gamma := \{ g\Gamma: g \in G \}$, where $\Gamma$ is a subgroup of G; indeed, one can take $\Gamma$ to be the stabiliser $\Gamma := \{ g \in G: gx = x \}$ of an arbitrarily chosen point x in X, and then identify $g\Gamma$ with $g\Gamma x = gx$. For instance, the sphere $S^2$ has an obvious action of the special orthogonal group SO(3), and the stabiliser of (say) the north pole can be identified with SO(2), so that the sphere can be identified with SO(3)/SO(2).  Similarly, the hyperbolic plane ${\Bbb H}$ is isomorphic to $SL(2,{\Bbb R})/SO(2)$. Furthermore, the cosphere bundle $S^* X$ of X – the space of unit (co)tangent vectors on X – is also a homogeneous space with structure group $SL(2,{\Bbb R})$. (For instance, the cosphere bundle $S^* {\Bbb H}$ of the hyperbolic plane ${\Bbb H}$ is isomorphic to $SL(2,{\Bbb R}) / \{ +1, -1 \}$.)

For the purposes of Ratner’s theorem, we only consider homogeneous spaces X in which the symmetry group G is a connected finite-dimensional Lie group, and X is finite volume (or more precisely, it has a finite non-trivial G-invariant measure). Every compact homogeneous space is finite volume, but not conversely; for instance the modular curve $SO(2,{\Bbb R}) \backslash SL(2,{\Bbb R})/SL(2,{\Bbb Z})$ is finite volume but not compact (it has a cusp). (The modular curve has two real dimensions, but just one complex dimension, hence the term “curve”; rather confusingly, it is also referred to as the “modular surface”. As for the term “modular”, observe that the moduli space of unimodular lattices in ${\Bbb R}^2$ has an obvious action of $SL(2,{\Bbb R})$, with the stabiliser of ${\Bbb Z}^2$ being $SL(2,{\Bbb Z})$, as well as an obvious left action of $SO(2,{\Bbb R})$ and so this moduli space can be identified with the modular curve.)

Let U be a subgroup of G. The group U then acts on X, creating an orbit $Ux := \{ gx: g \in U \}$ inside X for every point x in X. Even though X “looks the same” from every point, the orbits of U need not all look alike, basically because we are not assuming U to be a normal subgroup (i.e. $Ug \neq gU$ in general). For instance on the surface of the earth, which we model as a sphere $S^2 = SO(3)/SO(2)$, if we let $U \cong SO(2)$ be the group of rotations around the Earth’s axis, then the orbits Ux are nothing more than the circles of latitude, together with the north and south poles as singleton orbits.

In the above example, the orbits were closed subsets of the space X. But this is not always the case. Consider for instance the 2-torus $X := {\Bbb R}^2/{\Bbb Z}^2$, and let $U \leq {\Bbb R}^2$ be a line $U := \{ (x, \alpha x): x \in {\Bbb R} \}$. Then if the slope $\alpha$ of this line is irrational, the orbit Ux of a point x in the torus will be a dense one-dimensional subset of that two-dimensional torus, and thus definitely not closed. More generally, when considering the orbit of a subspace $U \leq {\Bbb R}^m$ on a torus ${\Bbb R}^m/{\Bbb Z}^m$, the orbit Ux of a point x will always be a dense subset of some subtorus of ${\Bbb R}^m/{\Bbb Z}^m$ (this is essentially Kronecker’s theorem).

From these examples we see that even if an orbit Ux is not closed, its closure $\overline{Ux}$ is fairly “nice” – indeed, in all of the above cases, the closure can be written as a closed orbit Hx of some other group $U \leq H \leq G$ intermediate between U and G.

Unfortunately, this nice state of affairs is not true for arbitrary flows on homogeneous spaces. A classic example is geodesic flow on surfaces M of constant negative curvature (such as the modular curve mentioned earlier). This flow can be viewed as an action of ${\Bbb R}$ (representing time) on the cosphere bundle $S^* M$ (which represents the state space of a particle on M moving at unit speed), which is a homogeneous space with symmetry group $SL(2,{\Bbb R})$. In this example, the subgroup $U \leq SL(2,{\Bbb R})$ is given as

$\displaystyle U := \{ \begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix}: t \in {\Bbb R} \} \cong {\Bbb R}$. (*)

For certain surfaces, this flow is quite chaotic, for instance Morse produced an example of a geodesic flow on a constant negative curvature surface whose closed orbit $\overline{Ux}$ had cross-sections that were homeomorphic to a Cantor set. (For the modular curve, there is an old result of Artin that exhibits an orbit which is dense in the whole curve, but I don’t know if one can obtain Cantor-like behaviour in this curve. There also seems to be some connection between geodesic flow on this curve and continued fractions which I don’t really understand.)  [Edit, Jul 20 2017: these connections are discussed in this paper of Series, and Cantor-like orbits are constructed in this further paper of Series.  Thanks to Caroline Series for the references.]

The reason for the “badness” of the above examples stems from the exponential instabilities present in the action of U, which can already be suspected from the presence of the exponential in (*). (Exponential instability is not a sufficient condition for chaos, but is often a necessary one.) Ratner’s theorems assert, very roughly speaking, that if one eliminates all exponential behaviour from the group U, then the orbits Ux become nicely behaved again; they are either closed, or are dense in larger closed orbits Hx.

What does it mean to eliminate “all exponential behaviour”? Consider a one-dimensional matrix group

$\displaystyle U = \{ A^t: t \in {\Bbb R} \}$

where A is a matrix with some designated logarithm $\log(A)$, and $A^t := \exp( t \log(A) )$. Generally, we expect the coefficients of $A^t$ to contain exponentials (as is the case in (*)), or sines and cosines (which are basically just a complex version of exponentials). However, if A is a unipotent matrix (the only eigenvalue is 1, or equivalently that $A=1+N$ for some nilpotent matrix N), then $A^t$ is a polynomial in t, rather than an exponential or sinusoidal function of t. More generally, we say that an element g of a Lie group G is unipotent if its adjoint action $x \mapsto gxg^{-1}$ on the Lie algebra $\mathfrak{g}$ is unipotent. Thus for instance any element in the centraliser of G is unipotent, and every element of a nilpotent group is unipotent.

We can now state one of Ratner’s theorems.

Ratner’s orbit closure theorem. Let $X = G/\Gamma$ be a homogeneous space of finite volume with a connected finite-dimensional Lie group G as symmetry group, and let U be a connected subgroup of G generated by unipotent elements. Let Ux be an orbit of U in X. Then the closure $\overline{Ux}$ is itself a homogeneous space of finite volume; in particular, there exists a closed subgroup $U \leq H \leq G$ such that $\overline{Ux}=Hx$.

This theorem (first conjectured by Raghunathan, I believe) asserts that the orbit of any unipotent flow is dense in some homogeneous space of finite volume. In the case of algebraic groups, it has a nice corollary: any unipotent orbit in an algebraic homogeneous space which is Zariski dense, is topologically dense as well.

In some applications, density is not enough; we also want equidistribution. Happily, we have this also:

Ratner’s equidistribution theorem. Let X, G, U, x, H be as in the orbit closure theorem. Assume also that U is a one-parameter group, thus $U = \{ g_t: t \in {\Bbb R} \}$ for some homomorphism $t \mapsto g_t$. Then $Ux$ is equidistributed in $Hx$; thus for any continuous function $F: Hx \to {\Bbb R}$ we have

$\displaystyle \lim_{T \to \infty} \frac{1}{T} \int_0^T F(g_t x)\ dt = \int_{Hx} F$
where $\int_{Hx}$ represents integration on the normalised Haar measure on Hx.

One can also formulate this theorem (first conjectured by Dani, I believe) for groups U that have more than one parameter, but it is a bit technical to do so and we shall omit it. My paper with Ben Green concerns a quantitative version of this theorem in the special case when X is a nilmanifold, and where the continuous orbit Ux is replaced by a discrete polynomial sequence. (There is an extensive literature on generalising Ratner’s theorems from continuous U to discrete U, which I will not discuss here.)

From the equidistribution theorem and a little bit of ergodic theory one has a measure-theoretic corollary, which describes ergodic measures of a group generated by unipotent elements:

Ratner’s measure classification theorem. Let X be a finite volume homogeneous space for a connected Lie group G, and let U be a connected subgroup of G generated by unipotent elements. Let $\mu$ be a probability measure on X which is ergodic under the action of U. Then $\mu$ is the Haar measure of some closed finite volume orbit Hx for some $U \leq H \leq G$.

— The Oppenheim conjecture —

To illustrate the power of Ratner’s orbit closure theorem, we discuss the first major application of this theorem, namely to solve the Oppenheim conjecture. (Margulis’ solution of the Oppenheim conjecture predates Ratner’s papers by a year or two, but Margulis solved the conjecture by establishing a special case of the orbit closure theorem.) I will not discuss applications of the other two theorems of Ratner here.

The Oppenheim conjecture concerns the possible value of quadratic forms in more than one variable, when all the variables are restricted to be integer. For instance, the famous four squares theorem of Lagrange asserts that the set of possible values of the quadratic form

$Q(n_1,n_2,n_3,n_4) := n_1^2 + n_2^2 + n_3^2 + n_4^2,$

where $n_1,n_2,n_3,n_4$ range over the integers, are precisely the natural numbers $\{0,1,2,\ldots\}$. More generally, if Q is a positive definite quadratic form in m variables, possibly with irrational coefficients, then the set $Q({\Bbb Z}^m)$ of possible values of Q can be easily seen to be a discrete subset of the positive real axis. I can’t resist mentioning here a beautiful theorem of Jon Hanke and my friend Manjul Bhargava: if a positive-definite quadratic form with integer coefficients represents all positive integers up to 290, then it in fact represents all positive integers. If the off-diagonal coefficients are even, one only needs to represent the integers up to 15; this was done by John Conway and my classmate from Princeton, Will Schneeberger.

What about if Q is indefinite? Then a number of things can happen. If Q has integer coefficients, then clearly $Q({\Bbb Z}^m)$ must take integer values, and can take arbitrarily large positive or negative such values, but can have interesting gaps in the representation. For instance, the question of which integers are represented by $Q(n_1,n_2) := n_1^2 - d n_2^2$ for some integer d already involves a little bit of class field theory of ${\Bbb Q}(\sqrt{-d})$, and was first worked out by Gauss.

Similar things can happen of course if Q has commensurate coefficients, i.e. Q has integer coefficients after dividing out by a constant. What if Q has incommensurate coefficients? In the two-variable case, we can still have some discreteness in the representation. For instance, if $\phi := \frac{1+\sqrt{5}}{2}$ is the golden ratio, then the quadratic form

$Q(n_1,n_2) = n_1^2 - \phi^2 n_2^2 = (n_1-\phi n_2) (n_1 + \phi n_2)$

cannot get arbitrarily close to 0, basically because the golden ratio is very hard to approximate by a rational a/b (the best approximants being given, of course, by the Fibonacci numbers).

However, for indefinite quadratic forms Q of three or more variables $m \geq 3$ with incommensurate coefficients, Oppenheim conjectured in 1929 that there was no discreteness whatsoever – the set $Q({\Bbb Z}^m)$ was dense in ${\Bbb R}$. There was much partial progress on this problem in the case of many variables (in large part due to the power of the Hardy-Littlewood circle method in this setting), but the hardest case of just three variables was only solved by Margulis in 1989.

Nowadays we can obtain Margulis’ result rather easily from Ratner’s theorem as follows. It is straightforward to reduce to the most difficult case, namely when m=3. We need to show that the image of ${\Bbb Z}^3$ under the quadratic form $Q: {\Bbb R}^3 \to {\Bbb R}$ is dense in ${\Bbb R}$. Now, every quadratic form comes with a special orthogonal group SO(Q), defined as the orientation-preserving linear transformations that preserve Q; for instance, the Euclidean form $x_1^2 + x_2^2 + x_3^2$ in ${\Bbb R^3}$ has the rotation group SO(3), the Minkowski form $x_1^2 + x_2^2 + x_3^2 - x_4^2$ has the Lorentz group SO(3,1), and so forth. The image of ${\Bbb Z}^3$ under Q is the same as that of the larger set $SO(Q) {\Bbb Z}^3$. [We may as well make our domain as large as possible, as this can only make our job easier, in principle at least.] Since Q is indefinite, $Q({\Bbb R}^3) = {\Bbb R}$, and so it will suffice to show that $SO(Q) {\Bbb Z}^3$ is dense in ${\Bbb R}^3$. Actually, for minor technical reasons it is convenient to just work with the identity component $SO(Q)^+$ of SO(Q) (which has two connected components).

[An analogy with the Euclidean case $Q(x_1,x_2,x_3) = x_1^2+x_2^2+x_3^2$ might be enlightening here. If one spins around the lattice ${\Bbb Z}^3$ by the Euclidean orthogonal group SO(Q)=SO(3), one traces out a union of spheres around the origin, where the radii of the spheres are precisely those numbers whose square can be expressed as the sum of three squares. In this case, $SO(Q) {\Bbb Z}^3$ is not dense, and this is reflected in the fact that not every number is the sum of three perfect squares. The Oppenheim conjecture asserts instead that if you spin a lattice by an irrational Lorentz group, one traces out a dense set.]

In order to apply Ratner’s theorem, we will view $SO(Q)^+ {\Bbb Z}^3$ as an orbit Ux in a symmetric space $G/\Gamma$. Clearly, U should be the group $SO(Q)^+$, but what to do about the set ${\Bbb Z}^3$? We have to turn it somehow into a point in a symmetric space. The obvious thing to do is to view ${\Bbb Z}^3$ as the zero coset (i.e. the origin) in the torus ${\Bbb R}^3/{\Bbb Z}^3$, but this doesn’t work, because $SO(Q)^+$ does not act on this torus (it is not a subgroup of ${\Bbb R}^3$). So we need to lift up to a larger symmetric space $G/\Gamma$, with a symmetry group G which is large enough to accommodate $SO(Q)^+$.

The problem is that the torus is the moduli space for translations of the lattice ${\Bbb Z}^3$, but $SO(Q)^+$ is not a group of translations; it is instead a group of unimodular linear transformations, i.e. a subgroup of the special linear group $SL(3,{\Bbb R})$. This group acts on lattices, and the stabiliser of ${\Bbb Z}^3$ is $SL(3,{\Bbb Z})$. Thus the right homogeneous space to use here is $X := SL(3,{\Bbb R})/SL(3,{\Bbb Z})$, which has a geometric interpretation as the moduli space of unimodular lattices in ${\Bbb R}^3$ (i.e. a higher-dimensional version of the modular curve); X is not compact, but one can verify that X has finite volume, which is good enough for Ratner’s theorem to apply. Since the group $G = SL(3,{\Bbb R})$ contains $U=SO(Q)^+$, U acts on X. Let $x = SL(3,{\Bbb Z})$ be the origin in X (under the moduli space interpretation, x is just the standard lattice ${\Bbb Z}^3$). If Ux is dense in X, this implies that the set of matrices $SO(Q)^+ SL(3,{\Bbb Z})$ is dense in $SL(3,{\Bbb R})$; applying this to, say, the unit vector (1,0,0), we conclude that $SO(Q)^+ {\Bbb Z}^3$ is dense in ${\Bbb R}^3$ as required. (These reductions are due to Raghunathan. Note that the new claim is actually a bit stronger than the original Oppenheim conjecture; not only are we asserting now that $SO(Q)^+$ applied to the standard lattice ${\Bbb Z}^3$ sweeps out a dense subset of Euclidean space, we are saying the stronger statement that one can use $SO(Q)^+$ to bring the standard lattice “arbitrarily close” to any given unimodular lattice one pleases, using the topology induced from $SL(3,{\Bbb R})$.)

How do we show that Ux is dense in X? We use Ratner’s orbit closure theorem! This theorem tells us that if Ux is not dense in X, it must be much smaller – it must be contained in a closed finite volume orbit Hx for some proper closed connected subgroup H of $SL(3,{\Bbb R})$ which still contains $SO(Q)^+$. [To apply this theorem, we need to check that U is generated by unipotent elements, which can be done by hand; here is where we need to assume $m \geq 3$.] An inspection of the Lie algebras of $SL(3,{\Bbb R})$ and $SO(Q)^+$ shows in fact that the only such candidate for H is $SO(Q)^+$ itself (here is where we really use the hypothesis m=3!). Thus $SO(Q)^+ x$ is closed and finite volume in X, which implies that $SO(Q)^+ \cap SL(3,{\Bbb Z})$ is a lattice in $SO(Q)^+$. Some algebraic group theory (specifically, the Borel density theorem) then shows that $SO(Q)^+$ lies in the Zariski closure of $SO(Q)^+ \cap SL(3,{\Bbb Z})$, and in particular is definable over ${\Bbb Q}$. It is then not difficult to see that the only way this can happen is if Q has rational coefficients (up to scalar multiplication), and the Oppenheim conjecture follows.

[Update, September 29: Link to Dave Morris’ book repaired slightly.]