3
$\begingroup$

I'm doing some reading into Riemannian geometry and PDEs and I have the following question. When we evolve a Riemannian metric (by say the Ricci flow) we are evolving a bilinear form on a manifold $\mathcal{M}$. I have seen diagrams that demonstrate this phenomenon as say $S^2$ embedded in $\mathbb{R}^3$ with the sphere undergoing a process of deformation as the metric evolves in time.

However, deforming a metric does not move any points, it just changes the notion of distances between points. So my question is this. If we denote the Euclidean metric in $\mathbb{R}^3$ by $g$ and consider the standard sphere $(S^2,g_{|S^2})$ embedded in $\mathbb{R}^3$ and we have another metric $g'$ on $S^2$ arrived at from $g_{|S^2}$ through some smooth evolution, then does there exist a map $f:S^2\to\mathbb{R}^3$ (to a deformed sphere) such that the "distance" between any two points $f(x),f(y)$ on $(f(S^2),g)$ is the same as the distance between $x,y$ on $(S^2,g')$?

If not then why do we visualise such geometric flows as deformations of submanifolds in $\mathbb{R}^3$, which as an ambient space always has the same Euclidean metric (in the literal sense of the diagram because we reside, ourselves, in such a flat space)?

Clarification: I realise that when given a family of diffeomorphisms $f_t:S^2\to \mathbb{R}^3$ that evolve smoothly in time $t$ we obtain a metric satisfying the reqiurements above through the pullback. What I'm looking for is the other direction, i.e. given an evolving metric, can we always find an $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.