# 2010 04 15 FAWG Rochon Notes

.....continued from talk from last week 2010 04 08 FAWG Rochon Notes. (These notes need to be proofread....I seem to have confused some $\chi$'s for $x$.) Colliand 18:21, 15 April 2010 (UTC)

Ricci flow and the determinant of the Laplacian on non-compact surfaces (part 2) (joint work with P. Albin and C. Aldana)

Last time we mostly described the determinant and showed how this is maximized on spaces of constant curvature within a given conformal class. Today, we will discuss the...

## Ricci flow on compact surfaces

Consider a compact surface with some holes $(\Sigma, g)$ with $\chi (\Sigma) < 0$. Let's look at how things conformally transform. If we define $g' = e^\phi g,$ then $$\Delta_{g'} = e^{-\phi} \Delta_g,$$ $$dg' = e^\phi dg,$$ $$R_{g'} = e^{-\phi} ( \Delta_g \phi + R_g).$$ It may be useful to keep these things in mind to understand how various objects transform under conformal change.

### Normalized Ricci flow

Consider the metric flow given by $$\frac{dg}{dt} = (r-R) g$$ where $r = \frac{\int R dg}{\int dg} = \frac{4 \pi \chi (\Sigma)}{Vol(\Sigma, g)}.$ If we consider $g(t) = e^{\phi(t)} g_0$ then $$\frac{d\phi}{dt} = r-R = r - e^{-\phi} (+ \Delta_g \phi + R_g).$$ Short time existence and uniqueness is standard. To extend the local-in-time theory to global-in-time theory, we need an a priori estimate on the curvature. Where do we get such an estimate?

Idea (Hamilton): Introduce a potential function $f \in C^\infty (\Sigma)$ such that $$-\Delta_{g} = R_g - r, ~ \int_\Sigma f dg = 0.$$ This appears to come out of nowhere but it emerges somehow naturally in the Kahler-Ricci flow. In any case, we will soon see that it is useful... Let $$h = -\Delta f + |\nabla f|^2 = R-r + |\nabla f|^2$$. Note that this implies that $R-r \leq h$. The evolution $g_0 \rightarrow g(t)$ induces an evolution $f \rightarrow f(t)$ which in turn induces an evolution on $h$. A calculation shows that $$\frac{\partial h}{\partial t} = - \Delta h - 2 |Z|^2 + rh$$ where $Z_{ij} = \nabla_i \nabla_j f + \frac{1}{2} (\Delta f) g_{ij}.$ Some work leads to $$\frac{\partial h}{\partial t} \leq -\Delta h + rh.$$ By the maximum principle, $h \leq C e^{rt}$ which implies $R-r \leq C e^{rt}$. We can also get a lower bound in a similar way so we find that $$-C e^{rt} \leq R-r \leq C e^{rt}, C>0.$$ Since $\chi (\Sigma) < 0$, we find that $r < 0$. (In the paper of Hamilton, he worked harder on the case where $R=0$ and $R>0$.) From this control, one can get long time existence and convergence of the normalized Ricci flow to constant curvature.

## Non-compact Case

Recall that we were considering a surface $\Sigma \subset \overline{\Sigma} \backslash \partial {\overline{\Sigma}}$ with $\partial {\overline{\Sigma}} = \partial {\overline{\Sigma}}_C \cup \partial {\overline{\Sigma}}_F$. For $\chi \in C^\infty (\overline{\Sigma})$ a boundary defining function, look at $g$ so that

1. near $\partial {\overline{\Sigma}}_C: ~g = e^{\phi} ( \frac{d \chi^2}{\chi^2} + \chi^2 d \theta^2).$
2. near $\partial {\overline{\Sigma}}_F: ~g = e^\phi ( \frac{{d \chi^2} + d \theta^2}{\chi^2} ).$

Here we asume that $\phi \in x^3 C^\infty (\overline{\Sigma}).$

Issues:

1. What happens at infinity when the Ricci flow evolves?
2. Long time existence, convergence?
3. What should be $r$? We will take $r = -2$ and we will replace volume by a renormalized volume....

Consider $\frac{d\phi}{dt} = (r -R), g(t) = e^{\phi(t)} g_0$. Short time existence here is a result of Shi. It was unknown at that time if the flow was unique. In the simple case we are considering with appropriate control at $\infty$, we can prove uniqueness of the flow.

The harder part in the non-compact case is to define the potential function.

Theorem (Ji-Mazzeo-Sesum): If there are only cusps and $\chi (\Sigma) < 0, r= \frac{\int R dg}{Vol(\Sigma, g)}$ then the Ricci flow exists for all time and converges exponentially fast to the hyperbolic metric in the conformal class. You can easily find an $L^2$ solution but this procedure does not give you control on the gradient. J-M-S construct the potential function using a different approach. They make a transformation to make the cusp look asymptotically euclidean. Within this frame, they can prove the gradient control.

When there is also a funnel, there is also the issue of constructing this potential function?

Q (McCann): Are the cone and funnel the only non-compact boundaries to be considered? Is this exhaustive or are there other possibilities? A: This is basically exhaustive in the negative curvature setting. A recent preprint of Isenberg-Mazzeo-Sesum also considers the case with Euclidean ends.

In 2d, because we have this nice formula $\Delta_{g'} = e^{-\phi} \Delta_g$, we can avoid some of the complications used by earlier authors. For the funnel, we do a doubling construction.

Suppose we want to find $f$ such that $\Delta f = h, ~ h \in x^3 C^\infty (\overline{\Sigma}) \cap C^\infty_{g_0} (\Sigma)$. We use the fact that a funnel is asymptotically a Euclidean cone in the conformal class. We make a conformal transformation to $g$ to get a metric which near the funnel end $\partial {\overline{\Sigma}}_F$ is an incomplete cylinder. We get $\Delta_g f = h$ which implies $\Delta_{\tilde{g}} {\tilde{f}} = {\tilde{h}}$ with $\tilde{h} \in x C^\infty (\overline{\Sigma})$.

The trick was then described with pictures. The funnel end is transformed into an incomplete cylinder. We then append a mirror image of the surface off that cylinder end to create a new surface ${\widehat{\Sigma}} = \Sigma \cup_{\partial{\overline{\Sigma}}_F} \Sigma$. On the appended copy of the surface, we replace $h$ by $-h$.

On ${\widehat{\Sigma}}$, we consider $\Delta_{\widehat{g}} {\widehat{f}} = {\widehat{h}}$. By J-M-S, we can find $\widehat{f}$ where we take $f = {\widehat{f}}$ when restricted to the first copy for the new potential function.

What do you want me to do next? Explain how to get the determinant or describe the behavior at infinity? Audience replied.....determinant.

We want to define $$\zeta_{\Delta} (s) = \frac{1}{\Gamma (s)} \int_0^\infty t^{s-1} {overline{Tr}} (e^{-t \Delta_g}) dt.$$ Note that $e^{-t \Delta_g}$ is not trace-class so we need to consider a renormalized trace. The Schwartz kernel $K(t, \sigma, \sigma')$ of $e^{-t \Delta_g}$ $$Tr (e^{-t \Delta_g}) = \int_\Sigma K(t, \sigma, \sigma) dg$$ is not well-defined because the integrand $K$ is not integrable along the diagonal. Instead, we look at $$Tr (\chi^2 e^{-t \Delta_g}) = \int_\Sigma \chi^z K(t, \sigma, \sigma) dg, Re z \gg 0$$ and observe that this has a meromorphic extension.