2
$\begingroup$

Is there a long time existence for the Ricci flow on compact negatively curved surfaces? I just read that the normalized Ricci flow has a long time solution converging to a metric of constant negative curvature. I was wondering what happens to the ordinary Ricci flow, or if there is any condition under which one has long time existence? Thanks!

$\endgroup$
  • 1
    $\begingroup$ I guess there is a relationship between the Ricci flow and the normalized Ricci flow. Up to a scaling on each time slice and let $s = \log t$ (and something similar). $\endgroup$ – user99914 Sep 15 '15 at 21:43
2
$\begingroup$

The normalized Ricci flow $\partial_\tau \hat g = - 2 \text{Rc}_{\hat g}+\int \text{Scal}_\hat g d\mu_\hat g$ is related to the standard Ricci flow $\partial_t g = -2 \text{Rc}_g$ by $$\hat g(\tau) = \frac{g(t(\tau))}{\text{Vol}(t(\tau))}$$ where the rescaled time satisfies $t'(\tau) = \text{Vol}(t(\tau))$. Since we are working on a surface $\Sigma$, we have

$$ \frac{\partial}{\partial t} d\mu_g = -\text{Scal}_gd\mu_g = -2\kappa_g d\mu_g$$

and thus Gauss-Bonnet gives us

$$ \text{Vol}'(t) = -4 \pi \chi(\Sigma) \implies \text{Vol}(t) = \text{Vol}(0) - 4 \pi \chi t.$$

Thus if we fix $t(0) = 0$, the rescaled time is $$t(\tau) = \frac{\text{Vol}(0)}{4 \pi \chi}\left(1 - e^{-4\pi \chi \tau}\right).$$

Since $\Sigma$ has negative curvature, we know $\chi < 0$ and thus $t \to +\infty$ as $\tau \to +\infty$; so the long-time existence of the normalized Ricci flow implies that of the standard Ricci flow. Since the volume grows linearly, we don't get convergence of $g(t)$; but we get "convergence of shape", since $\hat g(\tau)$ and $g(t(\tau))$ differ only by a scale factor.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.