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I have read that Riemannian manifolds have the structure of a metric space. In this sense, they have a distance function and it satisfies the definition of metric space.

However, I have recently learned that pseudo-Riemannian metrics do not have this property and pseudo-Riemannian manifolds are not considered metric spaces. In this sense then, they do not have a distance function.

Two question:

(1) if a pseudo-Riemannian manifold isn't a metric space, why is it called a metric?

(2) without a distance function, how can we have any notion of length on a manifold?

Remark:

I thought to be a metric space, by definition, for two points $x,y$ in the space, $d(x,y)\geq 0$ and if $d(x,y)=0$, then $x=y$. So any notion of zero or negative length distances would be impossible then, correct? Thus, relaxing the positive definite constraint of the Riemannian metric gives us a pseudo-Riemannian metric, but it also then violates those definitions of metric space and so, if I understand correctly, then pseudo-Riemannian metrics can't be metric spaces.

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    $\begingroup$ You do have the notion of length in a pseudo Riemannian manifold, you just have to accept that it could be negative. As for the terminology, you just have to accept it. Riemannian metrics are also not metrics (they are not distance functions but can be converted to such). This leads to the confusion with the notion of an isometry: A Riemannian isometry is not the same as an isometry in the sense of distances. $\endgroup$ – Moishe Kohan Jan 15 '15 at 21:48
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    $\begingroup$ I am confused and would like to clarify your comment. If you are saying that Rimannian manifolds aren't metric spaces, then look here en.wikipedia.org/wiki/Riemannian_manifold and explain why it says they can be metric spaces with distance functions. On the other hand, if you are saying Riemannian manifolds aren't necessarily metric spaces but can be converted to them (including having a distance function), then my question is "Is it true that pseudo-Riemannian manfiolds cannot be converted to metric spaces because you can have curves of zero or negative length?" $\endgroup$ – Stan Shunpike Jan 15 '15 at 21:52
  • $\begingroup$ I have heard the term Riemannian isometry but do not really understand what it is. Can you explain why that is relevant here? $\endgroup$ – Stan Shunpike Jan 15 '15 at 21:53
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    $\begingroup$ In linear algebra, non-degenerate symmetric bilinear forms are often called "metrics" as well, even though they do not give rise to a metric space in the usual sense. It's just an overuse of the term "metric". $\endgroup$ – Santiago Canez Jan 15 '15 at 22:23
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As an American president once said "it all depends on what the meaning of the word is is". If you look at the definition, it is clear that Riemannian metrics, as defined, are not distance functions. When you say that a (connected) Riemannian manifold $(M,g)$ is a metric space, you are actually saying that there exists a certain functor from the category of Riemannian manifolds to the category of metric spaces: $$ \Phi: (M,g)\mapsto (M, d_g) $$ where $d_g$ is the Riemannian distance function. (If you do not know what functors and categories are, just think of functors as maps and categories as sets, even though this is, strictly speaking, false.) This functor is like a conversion procedure from, say, kilograms to pounds (except, it is not invertible). The functor $\Phi$ is defined by first assigning a quantity, which we call length $L_g(p)$ to each (rectifiable) path $p$ in $M$ and then minimizing.

The step $g\mapsto L_g$ makes sense even for pseudo-Riemannian manifolds, just you get some paths of negative length. It is the 2nd step which fails: If you try to minimize, you will (frequently) get $-\infty$, which is not particularly useful.

Remark. What I said above about isometries: Both categories of Riemannian manifolds and metric spaces have their own notions of isometric maps and these two notions do not correspond to each other under the functor $\Phi$. For instance, a Riemannian geometer would think of the circle of length $2\pi$ isometrically embedded into $R^2$ (with the image of the embedding being the standard unit circle). But this embedding does not preserve distances between points! This is another reason I do not like to use the word is for the relation between Riemannian manifolds and metric spaces. However, assuming that you know what functors are, otherwise, just ignore this: $\Phi$ does become a functor if we use 1-Lipschitz maps between spaces as morphisms for both Riemannian manifolds and metric spaces.

Now, to your question of why do we call pseudo-Riemannian metrics metrics, it is all matter of habit and tradition. You can think of three different worlds:

  1. Metric geometry

  2. Riemannian geometry.

  3. Pseudo-Riemannian geometry.

They all have their notions of metrics (and isometries), but these notions have different meanings. It is as if people who speak different languages can occasionally use the same word, but it has different meaning in these languages. My favorite example is the word application, which has different meaning in English and in French.

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  • $\begingroup$ Thanks for answering. That's a much more clear definition of the specific meaning when we call a connected Riemannian manifold a metric space than the definition I saw on Wikipedia. $\endgroup$ – Stan Shunpike Jan 15 '15 at 22:24
  • $\begingroup$ @StanShunpike: Consider then accepting the answer in order to remove this question from the "unanswered list"; the same applies to other questions you asked and which were answered. $\endgroup$ – Moishe Kohan Jan 17 '15 at 3:59
  • $\begingroup$ Done. I was told to wait last time I accepted immediately so now I wait a few days unless someone with enough rep asks me to $\endgroup$ – Stan Shunpike Jan 17 '15 at 4:04
  • $\begingroup$ Would you mind explaining your last remark about "application"? $\endgroup$ – Geoffrey Sangston May 25 '17 at 1:38
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    $\begingroup$ @GeoffreySangston: "Application" in French means "a map" or "a function". $\endgroup$ – Moishe Kohan May 25 '17 at 2:42
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A Riemannian metric is not a metric, in the same way that an alleged criminal is not (necessarily) a criminal. A Riemannian metric is a smoothly varying positive-definite bilinear form on the tangent spaces. By extension, we call any non-degenerate --- not necessarily positive-definite --- bilinear form on the tangent spaces a pseudo-Riemannian metric. In this sense, it is not strange to call the object you get when you drop positive definiteness a (pseudo) metric.

To turn a Riemannian metric into a metric, you have to integrate.

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Well in a pseudo Riemannian manifold, at least in a Lorentz manifold, you can define a "distance" between points that are connected with causal curves, the diference between this distance and some Riemannian distance is that there could be some points p,q with p not equal to q such that d'(p,q)=0 (I am using the "distance" here defined for a Lorentz manifold) ... Check this out

http://www.math.miami.edu/~galloway/vienna-course-notes.pdf

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