7
$\begingroup$

I have some conceptual questions related to geodesic flows and cuvature.

  1. Suppose you have one parameter group of isometries from your manifold to itself. Since isometry preserves metric then it preserves Levi-Civita connection and curvature. How would one tie this to geodesic flows*. Is there a way to understand whether if a manifold has constant curvature by its geodesics (besides the criteria I gave below). For instance given a point $p$ on $M$, if $p$ can be connected to any other point on the manifold by a geodesic (as in sphere) then does the manifold have constant curvature? I would assume that if you have a "neighbourhood of geodesic flows" then its pullback preserves metric on that nbd. However it is not a global isometry.

*-I know one theorem where if every geodesic circle has constant curvature then the manifold has constant curvature.

  1. My second question is where can I get some information about the set of all isometries of a manifold as a space itself? Is there a good geometry book on this topic as a reference?
$\endgroup$

3 Answers 3

Reset to default
2
$\begingroup$

"For instance given a point $p$ on $M$, if $p$ can be connected to any other point on the manifold by a geodesic (as in sphere) then does the manifold have constant curvature?" I don't think it's hard to construct counterexamples to this. Take a point $p$ in the Euclidean plane. Now pick some region $R$ that doesn't include $p$, and introduce some small change in the metric $g\rightarrow g+\delta g$ that only occurs within $R$, so that the Gaussian curvature no longer vanishes inside $R$. From $p$, you can send out a geodesic at any angle $\theta$. As you increase $\theta$, these geodesics sweep the plane like the beam of a searchlight. It seems pretty clear to me that if $\delta g$ is small, then we will still cover the entire plane with these geodesics.

$\endgroup$
8
  • $\begingroup$ okay your answer seems legitimate from your perspective. But are you sure that after perturbing the metric with that small function you will still get a metric? Also, the geodesic flow preserves Riemannian metric doesn't it? That is given some V at p, and a geodesic flow $F_t$ which takes p to a q in the region you have described so that $g(V,V) = g(D(F_t)V,D(F_t)V)$. Doesn't this mean that if two points (or rather two nbds around those points) are connected by a geodesic flow then the metric at those points should be the same? $\endgroup$
    – Sina
    Sep 12, 2012 at 15:06
  • $\begingroup$ @Sina: "But are you sure that after perturbing the metric with that small function you will still get a metric?" What requirement on the metric do you have in mind that would be violated by changing it? You want its signature to stay the same, but that's no problem for small changes. $\endgroup$
    – user13618
    Sep 12, 2012 at 15:14
  • $\begingroup$ @Sina: "Doesn't this mean that if two points (or rather two nbds around those points) are connected by a geodesic flow then the metric at those points should be the same?" No, I don't think it does. $V$ and $D(F_t)V$ are different vectors. They don't even belong to the same tangent space. The property $g(V,V)=g(D(F_t),D(F_t))$ just says that if a bug starts walking across a surface along a geodesic at constant speed, its speed stays constant. It's a matter of definition of the parametrized geodesic, not a property of the metric. $\endgroup$
    – user13618
    Sep 12, 2012 at 15:23
  • $\begingroup$ Okay thanks, these show that my first idea was wrong (at least in the way it is put). I would still apreaciate some resources on space of isometries of a manifold. I tried searching with the key word "space of isometries" but all I get is very specialized books. I wonder if this topic is done in some basic lecture book on geometry. Thanks $\endgroup$
    – Sina
    Sep 12, 2012 at 17:09
  • $\begingroup$ By the way, I know that killing fields generate isometries but I assume this is only a subset right? Would it help to specialize to space of constant curvatures? $\endgroup$
    – Sina
    Sep 12, 2012 at 17:18
2
$\begingroup$

Okay, I found the answer for the second question in Kobayashi's book: it says isometries of a riemannian manifold (with some additional conditions) forms a lie group whose lie algebra is the space infinitesimal isometries generated by killing fields. Thus this classifies it. Thanks (Kobayashi, Foundations of Differential Geoemtry Vol1. page 236-237).

$\endgroup$
6
  • 2
    $\begingroup$ This is sometimes called the Myers-Steenrod theorem, here's the original paper and here's Palais's simplification of the argument. $\endgroup$
    – t.b.
    Sep 12, 2012 at 17:46
  • $\begingroup$ I'm confused by the interpretation that it's "the space infinitesimal isometries generated by killing fields." In the example of the 2-sphere from the WP article, whose isometries are O(3), this includes reflections. Wouldn't the space generated by the Killing fields not include the reflections? $\endgroup$
    – user13618
    Sep 12, 2012 at 19:37
  • $\begingroup$ Should the statement be instead that the infinitesimal generators of the lie group have the same algebra as the Killing fields? $\endgroup$
    – user13618
    Sep 12, 2012 at 20:02
  • 1
    $\begingroup$ @BenCrowell The isometry group of every semi-Riemannian manifold is a Lie group too, just as in the Riemannian case. More generally, any automorphism group of a $G$-structure of finite type (of which Riemannian and semi-Riemannian isometry groups are particular cases) is a Lie group. You can read about this in Shlomo Sternberg's book, Lectures on Differential Geometry. $\endgroup$
    – student
    Sep 12, 2012 at 22:44
  • 1
    $\begingroup$ @BenCrowell As for Hopf-Rinow, note that even in the Lorentzian case, there are various non-equivalent notions of completeness, so it's hard to establish an analogy. In any case, what is generally understood to be the "equivalent" of Hopf-Rinow in Lorentzian geometry is the theorem that states that in any globally hyperbolic spacetime, two causally related points can always be joined by a maximizing geodesic. Global hyperbolicity here would be the substitute for completeness. $\endgroup$
    – student
    Sep 12, 2012 at 22:52
2
$\begingroup$

Some clarifications:

Every complete Riemannian manifold has the property that for any two points $p, q \in M$, there exists a geodesic $\sigma_{pq}$ connecting $p$ and $q$. Moreover, the geodesic can be chosen to be minimizing, that is, there are no other curves $\alpha : I \rightarrow M$, geodesics or not, with length strictly less than the length of $\sigma_{pq}$. This is part of a basic result known as Hopf-Rinow's Theorem.

Thus, if it were true that being able to connect any $p \in M$ with any other $q \in M$ by a geodesic implies constant curvature, then every complete Riemannian manifold would have constant curvature, which is obviously false.

A related property which does imply constant curvature is homogeneity. A Riemannian manifold $(M, g)$ is homogeneous provided that for any $p, q \in M$, there exists an isometry $\Phi_{pq} : M \rightarrow M$ sending $p$ to $q$. The intuition behind this is that the metric looks the same at every point, and thus everything metric-related (like curvature) must be constant.

About the isometry group: the result is that if $(M, g)$ is Riemannian (finite-dimensional, I'm assuming), the its isometry group is a Lie group. There are no additional conditions. A lot is known about isometry groups of Riemannian manifolds, and since you like Kobayashi, you can take a look at another one of his books, called Transformation groups in Riemannian geometry, which has a nice exposition about this topic and many others.

$\endgroup$
3
  • 1
    $\begingroup$ In Hopf-Rinow, I think you also need the manifold to be connected. (Otherwise you'd have trivial counterexamples in which p and q can't even be connected by any curve.) $\endgroup$
    – user13618
    Sep 12, 2012 at 20:54
  • $\begingroup$ @BenCrowell like most geometers, I always assume the manifold to be connected. $\endgroup$
    – student
    Sep 12, 2012 at 22:40
  • $\begingroup$ Curvature is constant for homogeneous space for surface cases. But for higher dimension, sectional curvature may not be constant. For example, a compact lie group is a symmetric space and can be assigned a bi-invariant metric, i.e. its isometry group is itself. But it has non-constant sectional curvature. Some are zero, some strictly large than zero. $\endgroup$
    – Dai
    Jan 19, 2017 at 6:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .