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For a finite-dimensional smooth (Hausdorff, second-countable) manifold $M$, consider the set $$\mathcal{Met}(M) = \{ g : g \text{ is a Riemannian metric on }M \}.$$ I'd like to know about the typical differentiable structures one can place on $\mathcal{Met}(M)$, and how are they constructed. More specifically, is this in general an infinite-dimensional Banach or Hilbert manifold? Or perhaps a Fréchet manifold? What if $M$ is compact?

Also, any references containing a good deal of details, proofs, etc. would be deeply appreciated. Thanks!

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  • $\begingroup$ Maybe the following will help: think about matrix representation of a Riemannian metric, where the matrix entries are smooth functions on the given manifold. That should help :). $\endgroup$
    – user2093
    Jan 11, 2012 at 17:11
  • $\begingroup$ @WNY: That seems to assume the existence of a global frame on the manifold, but there are non-parallelisable manifolds. $\endgroup$
    – Zhen Lin
    Jan 11, 2012 at 17:16
  • $\begingroup$ You'll have to be more specific about what class of Riemannian metrics you mean. As each $g$ is a 2-tensor, they are sections of $T^*M \otimes T^*M$ but do you want $C^0$ sections? $C^\infty$? $L^2$? Once you have decided this, you will have answered your question. $\endgroup$
    – Adam
    Jan 12, 2012 at 3:24
  • $\begingroup$ I want smooth sections, so $C^{\infty}$. Maybe I should've said that explicitly before, but I still don't see how that answers my question. $\endgroup$
    – student
    Jan 12, 2012 at 5:29
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    $\begingroup$ The space of Riemann metrics on a manifold is contractible. It's a simple argument -- the space of inner products on a finite-dimensional vector space is an affine space. $\endgroup$
    – Ryan Budney
    Jan 22, 2012 at 21:13

3 Answers 3

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As far as I understand the situation, the $C^\infty$-topology makes the space of metrics a Fréchet space but if $M$ is not compact the set $\mathcal{Met}(M)$ is not open in it, and a problem of finding a convenient topology arises.

I found that a very good survey of the question is given in the first part of D.E.Blair's "Spaces of Metrics and Curvature Functionals" (Chapter 2 of "Handbook of Differential Geometry", Vol.1, Ed. by F.J.E.Dillen and L.C.A.Verstraelen, Elsevier, 2000).

As for the paper cited by @AlexE it would be convenient to have P.W. Michor's "Manifolds of differentiable mappings" at hand when reading on this subject. It is available on the author's site. One can find there all the necessary foundations including the construction of topologies.

I would love to know more on this subject too.

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    $\begingroup$ No, the space of metrics cannot be a Fréchet space since $0$ is not a metric. Even more, if $g$ is a metric, $-g$ obviously cannot be one. Therefore, the space of metrics is not a vector space. Maybe you meant "a Fréchet manifold"? $\endgroup$
    – Alex M.
    Mar 3, 2016 at 12:47
  • $\begingroup$ @AlexM. You have spotted a sloppy point. Thanks for that. I should have said something like "locally like a Fréchet space". For the details please refer to e.g. David G. Ebin, On the space of Riemannian metrics $\endgroup$
    – Yuri Vyatkin
    Mar 4, 2016 at 8:38
  • $\begingroup$ Does it known that $\mathcal{Met}(M)$ is (path) connected? $\endgroup$
    – C.F.G
    Mar 27, 2021 at 12:51
  • $\begingroup$ @C.F.G I don't know... $\endgroup$
    – Yuri Vyatkin
    Mar 28, 2021 at 4:03
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You may want to have a look at the paper

Gil-Medrano, Michor: The Riemannian Manifold of all Riemannian Metrics, Quarterly Journal of Mathematics (Oxford) 42 (1991), 183-202

and the references therein.

Also available online at http://www.mat.univie.ac.at/~michor/rie-met.pdf.

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Hamilton's 1982 paper The inverse function theorem of Nash and Moser available here and here is also a very good reference.

The space of Riemannian metrics on a compact manifold is a Frechet manifold.

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