# Manifold with no Finsler structure?

Is there a good example for a smooth manifold to which one cannot give a Finsler structure in any meaningful way? Ideal the example should be of low dimension and not too bizarre.

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What does "meaningful" mean? Are Riemannian metrics constructed with partitions of unity not meaningful? –  Lee Mosher Mar 7 '12 at 4:55
Good point. I'm gonna say no, they may not be meaningful, since I vaguely remember that Yau once proved something that states you can modify a given Riemannian metric, under some constraints, and get a constant curvature. –  ssquidd Mar 7 '12 at 8:57
Without a description of what you consider to be "meaningful", this question is not answerable. –  Willie Wong Mar 7 '12 at 9:49
A finite-dimensional manifold admits a Finsler structure if and only if it admits as Riemannian structure (i.e. if and only if it is paracompact). –  alvarezpaiva Mar 7 '12 at 17:36
I see. Too bad I cannot delete my question now. –  ssquidd Mar 8 '12 at 5:23
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## 1 Answer

As Lee Mosher pointed out, every manifold admits a Riemannian metric, by a partition of unity argument. One can easily rescale any Riemannian metric by a positive function to give a complete Riemannian metric. Every Riemannian metric is a Finsler metric. Every Riemannian metric on any compact manifold can be rescaled to have constant scalar curvature (the Yamabe problem, proved by Aubin, Schoen and Yau). But the metric might not be very meaningful. There doesn't appear to be any natural choice of metric on an abstract manifold, say up to isometry, except in low dimension. It might be interesting to study Finsler metrics with some type of constant curvature, or some other natural class of Finsler metrics.

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