Ricci-invariant class of metrics

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

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