The space of Riemannian metrics on a given manifold.

Question

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!

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 :). — , Jan 11 '12 at 17:11
@WNY: That seems to assume the existence of a global frame on the manifold, but there are non-parallelisable manifolds. — Zhen Lin, Jan 11 '12 at 17:16
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. — Adam, Jan 12 '12 at 03:24
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. — student, Jan 12 '12 at 05:29
@Zhen: Indeed, my answer was "local." Thanks for pointing that out. — , Jan 13 '12 at 17:44
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. — Ryan Budney, Jan 22 '12 at 21:13

score 16 · Accepted Answer · answered Jan 17 '12 at 00:56

16

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.

answered Jan 17 '12 at 00:56

Yuri Vyatkin

11,679

1

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"? – Alex M. Mar 03 '16 at 12:47
@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 – Yuri Vyatkin Mar 04 '16 at 08:38
Does it known that $\mathcal{Met}(M)$ is (path) connected? – C.F.G Mar 27 '21 at 12:51
@C.F.G I don't know... – Yuri Vyatkin Mar 28 '21 at 04:03

score 10 · Answer 2 · answered Jan 13 '12 at 21:19

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.

score 4 · Answer 3 · answered Feb 05 '15 at 01:54

4

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.

answered Feb 05 '15 at 01:54

mdg

1,152

The space of Riemannian metrics on a given manifold.

3 Answers3

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