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Imagine that there is a class of Riemannian metrics $\mathcal{R}$ on 3-dimensional manifolds such that

  1. $\mathcal{R}$ is locally finite dimensional; i.e., there are finite number of real parameters which describe the metric locally at any point;
  2. Any closed 3-dimensional manifold admits a metric from $\mathcal{R}$;
  3. $\mathcal{R}$ is is invariant with respect to Ricci flow.

Likely such $\mathcal{R}$ does not exist. Otherwise there is a good chance to simplify the Perelman's proof.

Can it be proved that such $\mathcal{R}$ does not exist?

UPDATE: You need to assume that metric depend continuously on the real parameters and $\mathcal{R}$ is closed in a reasonable topology.

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Construct ${\mathcal R}$ as follows: For each 3-manifold $M$ pick a Riemannian metric $g$, apply to $g$ Ricci flow. Then the class of metrics R is 1-dimensional. Trivialities aside, there is no known "natural" class of Riemannian metrics which would be locally finite-dimensional up to isometry and would exist on every closed 3-manifold. Proving that such metrics do not exist is, of course, impossibly since the concept of "naturality" is informal. – Misha Aug 26 '12 at 20:52
Well it is 4-dimensional at least, but I am sure it is $\infty$-dimensional. You need to assume that metric depend continuously and $\mathcal{R}$ is closed in a reasonable topology. – ε-δ Aug 27 '12 at 18:25
I would put it this way: You are more or less asking whether every closed 3-manifold admits a natural geometric structure, and thus far nobody has ever found one. So it seems likely such a family does not exist. But it seems quite difficult to prove such a broadly phrased claim rigorously. And of what use would such a theorem be? – Deane Yang Aug 27 '12 at 21:02
@Deane, In the proof, Perelman has to work with an infinite dimensional space (the space of all metrics on all 3-manifolds). This creates many technical problems and it might be also the reason why his proof works. I think it is worth to understand such things, and if the magic class exists why not to simplify the proof. – ε-δ Aug 29 '12 at 17:33

EDIT: As Misha and Deane pointed out, the question is not terribly well posed. So I will interpret it somewhat broadly to better reflect our actual understanding of canonical metrics on 3-manifolds and the Ricci flow.

By Perelman's proof of Thurston's geometrization conjecture, you can decompose any closed 3-manifold into geometric pieces. There are 8 types of geometric pieces, and they behave very simply under Ricci flow (they move just by scaling, etc). So the geometrization gives you essentially the class of metrics that you are asking for, with the constraint that you have to decompose your manifold into pieces first.

I don't think that there is a natural class of metrics, without decomposing the manifold into pieces first. In any case, it seems unlikely that the existence of such a class of metrics would yield to a simplification of Perelman's proof: E.g. a posteriori we know that every simply connected closed 3-manifold admits a positive Einstein metric. But of course, proving that means exactly proving the Poincare conjecture (as Perelman did).

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Robert, th geometrization does not give you $\mathcal{R}$. You need to smooth on the junctions, then after one second in the flow you get an infinite dimensional space of metrics. – ε-δ Aug 28 '12 at 19:18
It gives you $\mathcal{R}$ on the geometric pieces. As said, I don't think that there exists a natural $\mathcal{R}$ that works for manifolds that consist of more pieces, you must decompose them first! – Robert Haslhofer Aug 28 '12 at 21:41
Robert, I also "don't think" that $\mathcal{R}$ exists; I want to know why. – ε-δ Aug 29 '12 at 17:29

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