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A Finsler manifold is defined as a differentiable manifold with a metric defined on so that any well-defined curves of finite arc length is given by a generalized arc length integral of an asymmetric norm over each tangent space defined at a point. This generalizes the Riemannian manifold structure since the norm is no longer required to be induced by an inner product and therefore the Finsler manifold is not necessarily Euclidean in the tangent space structure.

A collegue of mine and I recently got into an arguement over whether or not Finsler manifolds are semi- or psuedo-Reimannian. I say no-by definition, A semi-Reimannian manifold-like an Loretzian manifold in relativity theory-is still required to have a metric tensor as it's normed structure,which is clearly an inner product. We simply weaken the condition of positive definiteness(i.e. the associated quadratic form of the norm is real valued) to nondegeneracy (i.e. the tangent space is isomorphic with its dual space). Both conditions require the distance structure to be induced by an inner product.

My colleague's argument is that the key property of semi-Riemannian manifolds is that they admit local signed coordinate structures that allow the distinction of different kinds of tangent spaces on the manifold. Local isomorphisms can be defined-especially in infinite-dimensional extensions of classical relativistic spaces-that make certain Fisler manifolds eqivilent to relativistic models of space-time.

I honestly don't know enough about the research that's been done on this. Is he right? This seems very bizarre to me, but it may indeed be possible to use specially constructed mappings to convert Finsler spaces to semi-Riemannian ones and vice-versa. I seriously doubt it could be done globally without running into serious topological barriers.

I'd like the geometers to chime in on this,particularly ones who are well-versed in relativistic geometry: Am I right? Can Finsler manifolds be defined in such a manner as to be true semi-Reimannian manifolds? Can local isomorphisms or differentiomorphisms be defined to interconvert them?

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What do you mean by semi-riemannian? Do you allow the metric to be degenerate? I ask this because I have come across at least two uses of semi-riemannian: (1) nondegenerate metric but not necessarily positive-definite (often called pseudo-riemannian), and (2) a possibly degenerate bilinear form on tangent vectors. –  José Figueroa-O'Farrill Jul 15 '10 at 3:43
    
@Jose Well,in the context of this debate,it's definition (1). As for the more general definition (2),is this merely case (1) with the trivial case included or no?Please elaborate a little. –  Andrew L Jul 15 '10 at 3:56
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I would appreciate if the more personal and less professional details of this debate between you and your friend were removed from your question. I also, again, do not understand why this question is CW. It is a legitimate mathematical question. –  Qiaochu Yuan Jul 15 '10 at 4:04
    
@Qiaochu I believe the human side of science is as much a part as the creation itself.As for the community wiki-well,I'm sorry to say I'm still learning the rules of MO,so I apologize. –  Andrew L Jul 15 '10 at 4:55
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Since Andrew L was so kind to make this a CW, I edited the text to remove the editorializing that offends Qiaochu's (and my) sensibilities. Also, I am tempted to say that you and your colleague just don't agree on the definition of a semi-Riemannian manifold. You obviously made sure that you agree on the definition of Finsler manifold. Have you checked that the other definition agrees? –  Willie Wong Jul 15 '10 at 10:12

2 Answers 2

No, a Finsler metric is in general not semiriemannian. As you and José indicate, a semiriemannian metric is always given by a nondegenerate quadratic form on tangent vectors at each point in the manifold. In other words, if you fix a basis of the tangent space for a given point, then the norm of that vector is given by a homogeneous quadratic polynomial in the coefficients of the vector with respect to the basis.

On the other hand, a Finsler metric is given by a norm function on the tangent space of each point in the manifold. This norm function must be convex, and additional regularity and convexity assumptions are often made. However, there is no requirement that the norm function be given by a quadratic form. It could be given by a higher even degree polynomial in the coefficients of a vector with respect to a basis. But it could be an arbitrary sufficiently smooth sufficiently convex function, too.

One way to think about this is to consider the standard flat models. The standard flat semiriemannian model is just $R^n$ with the metric given by a non-degenerate quadratic form. The standard flat Finsler model is $R^n$ with a (sufficiently smooth and convex) Banach norm, i.e. a finite dimensional Banach space. There are obviously a lot more of the latter than the former.

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Thanks to both you and Will,Deane. Can either of you recommend some books and/or papers to study this question further?The thing that confused me the most was the fact this guy claims there's been applications of Finsler manifolds to modeling space-time. If that was correct,how could Minkowski space-time (or it's variants) be both semi-Riemannian and Finsler at the same time? It's trivially clear any semi-Riemannian manifold is Finsler,but the converse is false. Unless that triviality is what he meant and he was just trying to confuse me. –  Andrew L Jul 15 '10 at 5:01
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You can have a distinguished time component (hence space-time) without a Minkowski structure. Take $R^4$ with the standard coordinates $x_0,\ldots,x_3$ which induces the coordinates on the tangent space $\xi_0,\ldots,\xi_3$. Now take the Finsler norm to be Globally $-\xi_0^4 + \xi_1^4 + \xi_2^4 + \xi_3^4$. The characteristic cone is not factorizable as the characteristic of a quadratic form; yet the "future-time-like" vectors still form a convex cone, so you can still do Causal geometry. Lastly, just because papers have been written on it doesn't mean most people take it seriously. –  Willie Wong Jul 15 '10 at 10:07
    
Andrew, you can look at articles and lecture notes available here: math.iupui.edu/~zshen/Finsler/index.html math.poly.edu/research/finsler/intro/one.html –  Deane Yang Jul 16 '10 at 2:45

The answer is no. Any smooth manifold admits a Riemannian metric using paracompactness and partitions of unity: in short, a convex sum of positive definite symmetric matrices is positive definite symmetric. So any manifold has such a structure. But there are topological obstructions to the existence of global pseudo-Riemmannian metrics of other prescribed signatures. http://en.wikipedia.org/wiki/Semi-Riemannian_manifold#Properties_of_pseudo-Riemannian_manifolds

EDIT, 16 July: I was looking for an example of said topological obstructions, and with an assist fom Willie Wong it has worked out: the ordinary sphere $$ \mathbb S^2 \subseteq \mathbb R^3 $$ does not possess a signature $(+,-) $ metric which I suppose ought to be called Lorentzian for this dimension. The topological obstruction is that $ \mathbb S^2$ cannot have a smooth (tangent) "line field," just as it cannot have a smooth nonzero tangent vector field by Brouwer. Now, a Lorentzian metric would give (pointwise) null cones, in this case a pair of distinct but intersecting lines in each tangent plane. As we are using $\mathbb R^3, $ we can cheat and define a line field from the angle bisector of the $+$ part of the cone.

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I've read some of the wonderful book by O'Niel and it's very sad it's out of print. To further study this question,I'll need to spend a little more time reading it in depth.Agree? –  Andrew L Jul 15 '10 at 4:57
    
Let me just say that O'Neill (two Ls, by the way) does not answer your question per se. It is nevertheless a very good book and the result that Will mentioned in his answer is, if I remember correctly, an exercise in the book. –  Willie Wong Jul 15 '10 at 10:01
    
@Will: just to nitpick on your argument. Any Riemannian manifold is also automatically semi-Riemannian, so in particular any smooth manifold admits a semi-Riemannian metric. To complete your argument one needs something to the effect that there always exist Finsler structure of arbitrary causal structure. The topological obstruction that I know of (vanishing Euler class for Lorentzian signature) also implies that there cannot be a Finsler structure with causality. Of course, there are probably other topological obstructions that I am not aware of. –  Willie Wong Jul 15 '10 at 10:19
    
Hello, Willie. Right now I am fiddling with trying to show that $\mathbb S^2$ does not possess a signature $(+,-)$ metric, by elementary methods. I think it would lead to a line field, which I then think ought to be impossible, although at the moment I am only sure about vector fields by Brouwer. –  Will Jagy Jul 15 '10 at 17:14
    
To finish your argument you just need that $\mathbb{S}^2$ is simply connected, and that by taking the double-cover a line-field becomes a vector field. So any line-field on the sphere (should it exist) would be equivalent to a vector field. –  Willie Wong Jul 16 '10 at 10:24

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