Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm looking for "famous" or otherwise well-known 2d Riemannian manifolds which have non-constant curvatures but have a non-trivial Killing vector field. Of course there are tons of spaces like these, for instance if we parametrize the plane (or a subset of it) by $(r,\phi)$ then any conformal rescaling of the flat metric by a conformal factor which only depends on $r$ will be generally good, i.e. have non-constant curvature but the rotations generated by $\partial/\partial\phi$ will still be a symmetry.

Are there special spaces which are somehow famous or well-studied because of some special property? Ideally, I'm looking for deformations of the Poincare disc.

share|improve this question
2  
I was going to say surfaces of revolution, but this looks like what you have written, since in 2 dimensions any two metrics are locally conformal (existence of isothermal coordinates). –  José Figueroa-O'Farrill Apr 28 '11 at 22:16
2  
I think Zoll surfaces are the most "famous", some of them have a lot of symmetry. –  Anton Petrunin Apr 28 '11 at 23:46
    
What is $AdS_2$? –  Deane Yang Apr 29 '11 at 13:30
    
@Deane: I would've assumed that AdS2 is the two-dimensional anti de Sitter space. But that would make it a Lorentzian manifold and not a Riemannian one... –  Willie Wong Apr 29 '11 at 13:49
2  
Could you be more precise by what you mean by a "deformation of the Poincare disk"? Every 2d Riemannian manifold is locally conformally equivalent to the Poincare disk. Are you looking for a complete Riemannian manifold that is conformally equivalent to the Poincare disk? –  Deane Yang Apr 29 '11 at 17:03
show 4 more comments

5 Answers 5

up vote 8 down vote accepted

A Killing field is preserved by the Ricci flow. By a theorem of Daskalopoulos, Hamilton and Sesum (arXiv:0902.1158), on a compact surface an ancient (defined for all negative time) solution to the Ricci flow which is not a shrinking soliton is diffeomorphism equivalent to the Fateev-Onofri-Zamolodchikov-King-Rosenau one-parameter family of metrics. These sausage metrics have the form $$ g(t) = \frac{-4\sinh(2t)(dx^{2} + dy^{2})}{1 + 2\cosh(2t) r^{2} + r^{4}}$$ in which $z = x + iy$ is a standard coordinate on the complement of a point in the complex projective line, and $r = \sqrt{x^{2} + y^{2}}$. There is an obvious rotational symmetry. The scalar curvature $R(t)$ of $g(t)$ satisfies the bounds $$\frac{-2}{\sinh(2t)} = \min_{S^{2}}R(t) \leq R(t) \leq \max_{S^{2}}R(t) = -2\coth(2t).$$

(The long list of names attached to these metrics is explained as follows. For a rotationally symmetric metric on the sphere the Ricci flow can be rewritten as the logarithmic diffusion equation $u_{t} = (\log u)_{zz}$ (not the same $z$ as above). P. Rosenau and J.R. King independently found the above solution in the context of such diffusion equations. In the Ricci flow literature they are usually called the Rosenau metrics or the King-Rosenau metrics. However, Fateev-Onofri-Zamolodchikov found this metric, which they called the sausage metric, earlier, in the context of studying the renormalization group flow for a two-dimensional sigma model, in their paper Integrable deformations of the O(3) sigma model. The sausage model, Nucl. Phys. B 406 (3), 521-565 (1993). (the Ricci flow is the one-loop renormalization group flow).)

On a compact oriented surface of genus at least 2 there is no non-zero Killing field, as follows, for instance, from the classical Bochner argument. Similarly, on a torus, a non-zero Killing field must be parallel. On $S^{2}$ a Killing field generates an isometric $S^1$ action fixing two distinct points; that is, the metric is rotationally symmetric (a proof can be found in the paper of Chen, Lu, and Tian). What this shows is that in the compact case an answer to your question has to be on the sphere or torus. Since the Ricci flow preserves isometries, it is natural to ask what the are interesting rotationally symmetric Ricci flows (A recent survey is arXiv:1103.4669); the theorem above about ancient solutions is one sort of answer.

The following is added to my original answer in response to comments by the original questioner asking for metrics related to the hyperbolic metric on the disk. Rescaling the sausage metrics $g(t)$ to have constant volume (in this case this means multiplying by a constant multiple of $t^{-1}$), they solve the volume normalized Ricci flow, and by a theorem of Chow-Hamilton as $t\to 0$ these rescaled metrics converge (as here can be checked directly) to a round metric on the sphere. Consider the metrics $$ \tilde{g}(t) = \frac{-4\sin(2t)(dx^{2} + dy^{2})}{1 + 2\cos(2t)r^{2} + r^{4}}$$ which are rotationally symmetric and solve the Ricci flow for $t \in (-\pi/2, 0)$. (Formally they are related to the sausage metrics by a complex rotation $t \to it$). Their scalar curvatures $R_{\tilde{g}(t)}$ are bounded as follows $$-2\cot(2t) = \min_{S^{2}}R(t) \leq R(t) \leq \max_{S^{2}}R(t) \leq -2\csc(t2).$$ $R(t)$ is everywhere positive for $t \in (-\pi/4, 0)$, but for $t \in (-\pi/2, -\pi/4)$ it is both positive and negative. More precisely, it is positive in an equatorial band, and negative on the complementary disks. As $t \to -\pi/2$ these disks expand to fill the complement of the equator. The homothetic metrics $$h(t) = -\frac{1}{\sin(2t)}\tilde{g}(t) = \frac{4(dx^{2} + dy^{2})}{1 + 2\cos(4t)r^{2} + r^{4}}$$ are determined by the normalization $\max_{S^{2}}R_{h(t)} = 2$, and satisfy the lower bound $-2 \leq 2\cos(2t)\leq \min_{S^{2}}R_{h(t)}$. As $t \to 0$ these metrics converge pointwise to the round metric on the sphere of volume $4\pi$, while as $t \to -\pi/2$ they converge pointwise on either of the disks complementary to the equator to the hyperbolic metric $4(1 - r^{2})^{-1}(dx^{2} + dy^{2})$ of scalar curvature $-2$. Thus the family $h(t)$ interpolates between the hyperbolic metric and the round metric. There is a similar rescaling of the sausage metrics which interpolates in a similar way between the flat metric on the punctured plane and the round metric on the sphere. There are similar families of metrics on the torus, though I am not going to write them down here. I came across the metrics $h(t)$ thinking about Einstein-Weyl structures on surfaces (this is explained in the arXiv:1011.5723, although there the relations with the Ricci flow and sausage metrics are not mentioned, as I was not then aware of them). I don't know what the characterization of the $\tilde{g}(t)$ in Ricci flow terms is, though I suspect they've been described somewhere in the literature on 2-d sigma models and RG flows.

share|improve this answer
    
Thanks a lot, this was very helpful. One question: since I'm primarily interested in the non-compact case (unit disc or the entire plane) can I stick a minus sign in front of the cosh(2t) in the metric or switch the $\cosh$ and $\sinh$ to $\cos$ and $\sin$ in order to have something which looks like the one parameter deformation of the Poincare disc? –  Daniel Apr 29 '11 at 16:33
add comment

There aren't many 2D surfaces with a Killing vector field.

  • If the Killing field is non-trivial, its singularities must be isolated. And locally near the singularity the symmetry is a rotation.
  • Away from singularities, consider unit speed geodesics orthogonal to the Killing vector field. Using the symmetry this implies locally there exists an orthogonal coordinate system $(r,\theta)$ such that $\partial_r$ has unit length and $\partial_\theta$ is Killing.

This implies that your surface is everywhere locally isometric to surfaces of revolution.

If your surface is geodesically complete and connected, then necessarily the number of singularities of the Killing vector field must be 0, 1, or 2. (Else consider the geodesics joining the singularities.) In the case of 2 singularities, your surface must be closed, and homeomorphic to a sphere. In the case of 1 singularity, your surface is a non-compact surface diffeomorphic to $\mathbb{R}^2$ with a polar coordinate chart in which $\partial_\theta$ is Killing. In both these cases you have bona fide surfaces of revolutions.

The case of 0 singularities is more delicate. The orbits of the symmetry doesn't have to be a closed curve (consider the torus with an "irrational translation", as well as just the flat or hyperbolic planes). If the surface is compact, Poincare-Hopf implies that it must be a topological torus.

share|improve this answer
    
Willie, I don't believe that a Killing field implies a surface of revolution. In particular, doesn't a helicoid have a nontrivial Killing field? –  Deane Yang Apr 29 '11 at 1:20
3  
True, but the helicoid is locally isometric to the catenoid. The screw-motion symmetry of the helicoid is taken to the rotational symmetry of the catenoid. –  Matt Noonan Apr 29 '11 at 2:05
    
Ah, I missed the "locally isometric". That is correct. –  Deane Yang Apr 29 '11 at 2:34
add comment

As Willie and José point out, surfaces of revolution provide an infinite set of examples, but I thought it would be worthwhile pointing out some of the most "famous" examples of non-constant curvature:

a) A standard (non-flat) torus obtained by rotating a circle disjoint from the origin about the $y$-axis

b) A paraboloid

c) A catenoid

As for "famous" surfaces that are not surfaces of revolution, I would add "famous" examples of minimal surfaces:

1) helicoid

2) Enneper surface

share|improve this answer
    
I didn't know about the Enneper surface. Thanks for showing me something new. –  Willie Wong Apr 29 '11 at 13:45
    
Are any of these deformations of the Poincare disc? (I should have added that ideally I'm looking for "famous" one or two parameter deformations of the Poincare disc such that the deformation introduces the non-constant curvature.) –  Daniel Apr 29 '11 at 16:34
    
The torus is unlikely to be conformal to the Poincare disk, and I doubt the paraboloid is, either. However, the helicoid (which is the universal cover of the catenoid) and the Enneper surface probably are. If I had to guess, any complete immersed minimal surface with Gauss curvature bounded from above by a negative constant is conformally equivalent to the Poincare disk. But that's just a guess. –  Deane Yang Apr 29 '11 at 17:06
add comment

My favorite 2D metrics of nonconstant curvature admitting a Killing vector field are the so-called Darboux-superintegrable metrics. Their definition is: the space of Killing tensors of degree two (i.e., integrals of the geodesic flow that are quadratic in velocities) is 4-dimensional. These metrics necessary admit Killing vector field. Their local description was known to Koenigs (Note II from Darboux "Sur la theorie generale des surfaces, Vol. IV, 1896) and S. Lie (1882), one can get a list of these metrics from arXiv:math-ph/0307039v1 or, alternatively from the paper arXiv:0705.3592 (the metrics 2a,2b,2c from Theorem 1). The lists are equivalent modulo coordinate transformation and contain also semi-riemannian metrics.

Let me explain why I love these metrics:

(1) They appear in differential geometry and physics. In differential geometry, they appear in studying of projective connections with big group of symmetries and their analysis was one of the main ingredients of the solution of two problems explicitly stated by S. Lie which is one of my bests results. They also appeared in mathoverflow: for example the metrics suggested in the Bryant's answer on Riemannian surfaces with an explicit distance function? are Darboux-superintegrable.

(2) They appear in physics: many physical phenomena lead to such metrics. There is a big activity about it in the "superintegrability" community (consult, for example, the papers of Winternitz et al in ArXive).

(3) For these metrics, one can answer many natural questions that normally require solving systems of ODE or PDE (such that description of geodesics or eigenvalues of the Laplacian) by algebraic methods, using the Killing tensors.

My personal opinion is that the metrics appeared naturally in differential geometry, because they are "the third best" metrics: the best metric is flat, the second best is of constant curvature, and the Darboux-superintegrable are the next simplest choice. The metrics appeared in physics, since physicists love solvable models, and models described by these metrics are sometimes solvable.

share|improve this answer
add comment

This is not an answer to the question but could be interesting as well, regarding non-constant curvature manifolds and Killing fields. If you take the tensor product of two Killing vectors, you obtain a second-degree (contravariant) Killing tensor. However, not all the second-degree Killing tensors can be obtained in this way, depending on the (pseudo-)Riemannian manifold. For example, in all constant curvature (pseudo-)Riemannian manifolds all the second-degree Killing tensors are linear combinations of symmetric tensor products of Killing vectors (and similarly for higher degree Killing tensors), Killing tensors of this type are called reducible. This is not true for manifolds with non constant curvature. The (very classical) example is the skew ellipsoid, which obviously does not admit Killing vectors but admits a second-degree Killing tensor. The existence of this tensor is connected with the integrability of the geodesic flow on the skew ellipsoid and the celebrated Jacobi's ellipsoidal coordinates. A kind of connection that holds for any (pseudo-)Riemannian manifold under certain conditions. As far I know, there is no rule to determine if a manifold of non-constant curvature admits or not non-reducible Killing tensors of some degree.

share|improve this answer
    
Dear Giovanni, The Killing equations is a an overdetermined linear system of PDE of finite type and in theory there exists an algorithm that decides whether a given metric admits a Killing tensor of a given degree and determine the dimension of the space of Killing tensors. Then, this algorithm also answers whether a metric admits a non-reducible Killing tensors of some degree. To be precise, the algorithm is local and gives the answer in a neighborhood of almost every point. The algorithm is computationally very hard even in simple situations. –  Vladimir S Matveev May 3 '11 at 10:21
    
Dear Vladimir, it seems that this algorithm connects the dimension of the space of Killing tensors of any degree with their reducibility or not, at least locally. That procedure seems very much intriguing. Could you give precise references? –  Giovanni Rastelli May 3 '11 at 18:53
    
Dear Giovanni, I did not understand what statement should I confirm by a reference. I claimed that for an explicitly given metric one can algorithmically understand the dimension space of Killing tensors of a given degree. This is a folklore but I do not know a good reference. Now, once we have the dimensions of the spaces of Killing tensors of degree less or equal to $k$, we know whether there exists a Killing tensor of degree k by counting the dimensions. Does it help. –  Vladimir S Matveev May 4 '11 at 19:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.