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I am reading John Lee's book Riemannian Manifolds. On page 91, he begins a chapter called "Geodesics and Distance," which is I think the first chapter that seriously addresses geodesics.

I was very surprised when I came across the following sentence:

Most of the results of this chapter do not apply to pseudo-Riemmanian metrics, at least not without substantial modification.

I thought the only real difference between the two was about the positive-definite constraint. But this makes it sound like there's a whole host of properties that don't apply to pseudo-Riemannian metrics but that do apply to Riemannian metrics.

Can someone clarify this for me? Are the things we can do on Riemannian manifolds that can't be done on pseudo-Riemannian ones?

Stan Shunpike
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    The author of the book that you mentioned is an active participant of math.stackexchange: http://math.stackexchange.com/users/1421/jack-lee Possibly he can answer your question when he reads your question. – Paul Jan 14 '15 at 03:10

2 Answers2

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OK, I'll accept the challenge...

The biggest difference in the pseudo-Riemannian case is that curves can have zero length, and the "Riemannian distance function" (the supremum of the lengths of curves between two points) is not a metric in the sense of metric spaces. Thus most of the results in Chapter 6 of my book don't make sense if the metric isn't positive definite.

Nonetheless, there is still a lot that can be said, at least in the Lorentz case (pseudo-Riemannian metrics of index $1$). In that case, one has to distinguish curves according to the nature of their velocity vectors: a curve $\gamma$ is spacelike if $g(\dot\gamma,\dot\gamma)>0$, lightlike if $g(\dot\gamma,\dot\gamma)\equiv 0$, and timelike if $g(\dot\gamma,\dot\gamma)<0$. The resulting geometric properties, as you might imagine, are intimately connected with the physics of space-time.

But this is a subject that is quite different from Riemannian geometry, which is why I didn't treat it in my book.

Jack Lee
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    Why is the distance function not a metric space? You say on 94 "With the distance function $d$ defined above, any connected Riemannian manifold is a metric space whose induced topology is the same as the given manifold topology" . doesn't that right there suggest that the distance function is a metric in the sense of metric space? By the way, I love all your books and thank you for writing them. Despite i am a somewhat average mathematician, I can still understand them they are so well written. – Stan Shunpike Jan 14 '15 at 05:03
  • Notice that the lemma on page 94 assumes the metric is Riemannian, and in that case the distance function does turn the manifold into a metric space. But if the metric is merely pseudo-Riemannian, there will be admissible curves that have zero length, and thus the "distance" between two points can be zero. This violates the definition of metric space. – Jack Lee Jan 14 '15 at 05:08
  • Ahhhh, I get it. Thank you so much. I was so confused. – Stan Shunpike Jan 14 '15 at 05:27
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For one thing, a Lorentz-signature metric on a compact manifold can fail to be geodesically complete. If memory serves, Chapter 3 of Einstein Manifolds by Besse contains an example of a metric on a torus where a finite-length geodesic "winds" infinitely many times.

Generally, the "unit sphere" in a tangent space is non-compact for a metric of indefinite signature (e.g., it's a hyperbola on a Lorentz-signature surface), which can cause all manner of fun.

Andrew D. Hwang
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