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  1. Let $M$ be a complete finite volume Riemannian manifold and $\gamma : \mathbb{R}^{\geq 0} \to M$ a geodesic. Suppose that $\mathrm{im}(\gamma)$ is dense. Is it equidistributed in the Riemannian measure? That is, does $$ \lim_{T \to +\infty} \frac{1}{T} \int_0^T f(\gamma(t)) \, dt = \frac{1}{\mathrm{vol}(M)} \int_M f \, \mathrm{d vol} $$ for every $f \in C_0(M)$? [False in general; true for Nilmanifolds. True a.e. in negative curvature, where the geodesic flow is ergodic. ]

  2. Let now $N \subset M$ be an (immersed) submanifold and $\gamma$ a geodesic of $M$ which is contained densely in $N$. Is the submanifold $N \subset M$ totally geodesic? [False in general, though true for some variants in constant negative curvature. But what if "totally geodesic" is weakened to "minimal"?]

Added. Asaf's answer nonethtless begs a follow-up question to 1:

  1. (Revised). If there is a dense geodesic, must there also be an equidistributed one? Could it in fact be that almost every geodesic is then equidistributed? Does a single dense geodesic imply ergodic geodesic flow? And in particular: does one dense geodesic imply almost all geodesics dense?

[A similar revision of 2 would instead involve the condition that almost every geodesic of $M$ that is tangent to $N$ at some point is contained in $N$; but then it should follow trivially (I think) that $N$ is totally geodesic. ]

Note: There is an analogy with the equidistribution and Manin-Mumford theorems, due to Szpiro, Ullmo, and Zhang, for torsion points in abelian varieties $A/\bar{\mathbb{Q}}$: For a sequence of torsion points which is eventually outside of every torsion translate of an abelian subvariety, the Dirac masses at the Galois orbits converge to the normalized Haar measure on $A(\mathbb{C})$ (where an embedding $\bar{\mathbb{Q}} \hookrightarrow \mathbb{C}$ has been fixed). Here, I would be tempted to think of a geodesic as corresponding to a Galois orbit of torsion points (either minimizes an energy functional -- or a canonical height); and of a totally geodesic subvariety as corresponding to a Galois orbit of a translate of an abelian subvariety by a torsion point (note that in the basic case of a flat torus, the totally geodesic submanifolds are precisely the subtori). The analogy is probably only superficial, but I thought it could be worth pointing out (if only because it led me to asking this question).

Added later. One more (final) question along the line of 2.

In algebraic geometry, we have the following general fact: For $L$ a nef line bundle on a projective variety $X$, if $\deg_LC =0$ for a Zariski-dense set of curves $C \subset X$, then $\deg_LX = 0$. (Nef =non-negative intersection numbers, =non-negative on every curve). For if $\deg_LX > 0$, Riemann-Roch and the almost vanishing of the higher cohomology of powers of nef line bundles imply that $L$ is big, hence a power of $L$ is effective. We may also do this in an arithmetic setting.

In the analogy of the preceding note which led me to consider totally geodesic submanifolds, I was misled by the Manin-Mumford theorem, which is specific to commutative group varieties and fails even for algebraic dynamical systems. Instead, subvarieties of minimal height ought to be analogous to minimal immersed submanifolds: the images of harmonic isometric immersions (which include totally geodesic ones as a particular case, and coincide with the geodesics in dimension one). Considering the previous paragraph, then, does the following question make any sense: If the closure of a minimal submanifold happens to be an immersed submanifold, is this submanifold still minimal?

In the same vein: If we have a sequence of complex algebraic curves in $\mathbb{CP}^n$ (images of non-constant holomorphic maps from compact Riemann surfaces) whose supports converge to a compact real-analytic immersed submanifold $M \subset \mathbb{CP}^n$, must $M$ be a complex (algebraic) submanifold?

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Might be relevant to add a weaker "question 0" about the limit in your question 1 being what it is for test functions... since, e.g., Weyl's equidistribution criterion's proof seems to need more than mere continuity (though I may be mistaken). –  paul garrett Jul 20 at 19:42
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In my opinion, the Andre-Oort situation is more analogous to closed horocycles, as the maps involved are Hecke correspondences, which can be thought somehow as $p$-adic analogues of the horocyclic flow. This gives rise to many powerfull techniques of unipotent flows, such as quantitative recurrence, Ratner's theorems and linearization schemes. In my (not complete) point of view, the Andre-Oort case is more similar to the paper by Nimish you've linked to. You somehow translate closed horocyclic periods. Those periods tend to equidistribute simply by mixing of the geodesic flow. –  Asaf Jul 21 at 20:45

2 Answers 2

up vote 23 down vote accepted

The first question is false as stated. By Artin's encoding, geodesics on $SL_{2}(\mathbb{R})/SL_{2}(\mathbb{Z})$ corresponding to continued fractions, and the geodesic flow corresponds to the shift. It's easy to find one fraction where you'll see any given prefix (hence dense), but you won't be equidistributed (say think about larger and larger blocks composed out of $1$'s).

The situation is the same even for cocompact (hyperbolic) homogeneous spaces, and relays on the fact that the corresponding dynamical system is a Bernoulli system, see for example the survey by Katok in the Clay Pisa proceedings for more information about the encoding.

In the case where the manifold is a Nilmanifold, the answer is indeed true, which follows from say Furstenberg's theorem about skew-products (when you use both the topological version and the ergodic version). Finer (quantitative) results are probably attained by Green-Tao (see Tao's post about the Nilmanifold version of Ratner's theorem). In the toral case, this boils done to merely Fourier series computations and Weyl's equidistribution criterion or so.

In the higher rank (semisimple) case, things get more complicated, as one might think about multi-parameter actions, and then the measure-classification theorem by Lindenstrauss kicks in, but it was observed by Furstenberg in the $60$'s (and maybe before that) that even for multi-parameter actions, there might be dense but not equidistributed orbits. Maybe the easiest toy model to think of is to think about the multiplicative action of $<2,3>$ as a semi-group on the torus $\mathbb{R}/\mathbb{Z}$, and start at say a Liouville number for base $6$. This action is some $S$-adic analouge for a higher-rank multi-parameter diagonalizable action.

Edit - to address the revised question, here the geometrical settings are being addressed more intimately. In the case of homogeneous spaces ($G/\Gamma$, or you can take the appropriate locally symmetric space as well), where $G$ is semi-simple say, then the geodesic flow is ergodic (it follows for example from the Howe-Moore theorem, or from the Bernoullicity theorem I've mentioned above). As a result, a simple application of the pointwise ergodic theorem will tell you that for almost every point and every direction (the approperiate measures here will be the Liouville measure on the unit tangent bundle, which is really where the geodesic flow "lives"), the orbit is equidistributed. For the variable curvature case, as long as some natural conditions are met (say an upper bound on the sectional curvature making it negative everywhere), the dynamical picture is pretty much the same (but the proofs are significantly more involved, as you don't have rep. theory at hand). Again in the Nilmanifold case, the situation is much more simple, the toy model for that is tori, where the question of rationality implies both density and equidistribution.

I will address the Andre-Oort question in the comments, as I'm not an expert on this subject.

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Thank you very much! For the case of Nilmanifolds, I would wonder, then, if the closure of a geodesic is always (a) an immersed submanifold, which (b) is totally geodesic. –  Vesselin Dimitrov Jul 21 at 12:12

The answer to question 2 is negative.

Let $M$ be the suspension of the round sphere $S^2$ by an irrational rotation $\phi$ (whose fixed point should be called the poles); i.e. $M$ is the quotient of $S^2\times \mathbb{R}$ by the relation $(x,t+1)\sim (\phi(x),t)$. Since $\phi$ is an isometry of $S^2$, $M$ inherits a well-defined Riemannian metric, which is locally isometric to a Riemannian product of $S^2$ by a small interval.In particular, the "vertical lines" $\{x\}\times\mathbb{R}/\sim$ are geodesics of $M$. Now, consider the vertical line issued from a point $p=(x,0)$; its closure is the (quotient of the) product of a circle by the $\mathbb{R}$ factor. If $x$ is neither a pole nor lying on the equator, this submanifold is clearly not totally geodesic.

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Thank you! Meanwhile, since asking the question, I found an article by N. Shah ("Closures of totally geodesic immersions in manifolds of constant negative curvature") which proves that for compact manifolds of constant negative curvature, the closure of a totally geodesic immersed submanifold of dimension at least $2$ is still a totally geodesic immersed submanifold. –  Vesselin Dimitrov Jul 21 at 11:52
    
Excuse my ignorance, but is this submanifold minimal? If so, are there still counterexamples with "totally geodesic" replaced by "minimal immersed submanifold" (=image of a harmonic isometric immersion)? Or could it be that the closure of a minimal immersed submanifold, when it happens to be a submanifold, is still a minimal immersed submanifold? –  Vesselin Dimitrov Jul 25 at 6:21
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@VesselinDimitrov: no, the submanifold is the Riemannian product of an embedded strictly convex circle in the sphere, by a straight line, so it is strictly mean convex. So the latest version of question 2 is still false. –  Benoît Kloeckner Jul 25 at 16:20
    
I see. Thanks! It would seem, then, that there is no hope of characterizing the submanifolds containing a dense geodesic of $M$, or a dense minimal submanifold of $M$ -- not in this generality. Perhaps, as in N. Shah's paper, one could say something for the case of constant negative curvature; whether or not this is possible, I have no idea. I am still curious, though, about the complex projective variant, which I formulated in the linked question. –  Vesselin Dimitrov Jul 25 at 17:37

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