Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I just started to learn about the Ricci flow and try to understand the Ricci flow evolution equation. It states that a one-parameter family $g_t$, $t\in[0,T)$ of Riemannian metrics on a smooth closed manifold $M$ is a solution of the equation $$ \frac{\partial g_t}{\partial t}=-2 Ric_{g_t}. $$ But, in order to be able to write down the expression $$ \frac{\partial g_t}{\partial t} $$ I need a little bit more than the fact that $g_t$ is a one-parameter family of Riemannian metrics.

There are several ways to understand this expression, which probably are equivalent to each other. I think the most naive idea is to consider $g_t$ as a smooth curve in the infinite-dimensional vector space of all Riemannian metrics on $M$ and $\frac{\partial g_t}{\partial t}$ as tangent vectors to this curve.

Now, my question is: How do I have to understand the left hand side of the Ricci flow equation at the initial value $t=0$?

Does one consider $g_t$ as a smooth family at the open interval $(0,T)$ and requires continuity at the point $t=0$ or has one to deal with differentiability from the right?

Thank you very much!

share|improve this question
6  
Look at a small coordinate neighborhood $U$ on $M$, then $g_{ij}(x,t)$ is assumed to be a smooth matrix-valued function on $U\times [0,T)$, where $x\in U, t\in [0,T)$ and smoothness at $0$ as usual means that you can extend $g_{ij}(x,t)$ to a smooth function on $U\times (-\epsilon,T)$, for some $\epsilon>0$. –  Misha May 8 '12 at 17:13
    
Misha, that argument sounds plausible to me. Thank you very much! –  kassandra May 8 '12 at 18:30
add comment

2 Answers

up vote 7 down vote accepted

Misha's comment could be a bit misleading. In particular, it is not true that the Ricci flow should exist on a slightly bigger interval $(-\epsilon,T)$ with $g(0) = g_0$. One way to see this is by thinking about a theorem of Bando (see http://www.springerlink.com/content/v0764574t4764138/ if you have access) which says that if $(M,g_t)$ is a solution for Ricci Flow on the interval $[0,T)$, then $g_t$ is real analytic with respect to the normal coordinate charts on $M$ for $t>0$. In particular, if $g_0$ was not real analytic, we cannot extend the flow backwards for any $\epsilon>0$ (the comment was only in charts, but by compacntess if we can do it in charts, we can do it on the whole manifold for some small $\epsilon>0$) because then Bando's theorem would imply that $g$ were real analytic.


The correct statement is just like for the heat equation. We say that $f$ is a solution to the heat equation $$ \frac{\partial f}{\partial t} = \Delta f $$ on $[0,T)$ if the above equation is satisfied for $t>0$ and $\lim_{t\searrow 0} f= f_0$. In particular, there is no "meaning" of the equation at $t=0$, only for $t>0$. Do not get confused by trying to apply ODE intuition to the PDE. Parabolic equations are not like ODE's in the sense that you can just "go in the direction of $\Delta f$".


So, for completeness, here is what it means to be a solution to RF on the interval $[0,T)$ with initial data $g_0$:

The metric $g_t$ is smooth for $t\in (0,T)$ and for such $t$, $g_t$ satisfies $$ \frac{\partial g_t}{\partial t} = -2Ric_{g_t}. $$ You can think of this either in local coordinate charts, as Misha does, or just as a coordinate free equation for the symmetric 2-tensor $\frac{\partial}{\partial t} g_t$.

Furthermore, we require that $g$ is continuous up to $t=0$ (because here we're only interested in solving RF with smooth initial data, if we wanted to start with rough data, we'd require a limit in some sense) and $$ g_{t=0} = g_0 $$ at each point in $M$.


If you're confused, you should read up on the heat equation first. Its exactly the same. In particular, after reading about the heat equation, you should read about the De Turk trick, which transforms the RF into a strongly parabolic equation (i.e. heat-type equation) by fixing the diffeomorphism gauge. A quick google suggests the following chapter http://www.springerlink.com/content/0t673151r72133r7/ as a possible reference. Any book on Ricci Flow should have a good description of this.

share|improve this answer
    
Otis, thanks particularly for your valuable references to the literature. –  kassandra May 8 '12 at 20:01
    
No problem! I assume that Terry Tao has something to say about this at his blog, which you might find helpful. –  Otis Chodosh May 8 '12 at 20:05
    
@Otis, can I ask a quick naive question: Is there any other approach to local existence results except Hamilton's original approach and the one using the DeTurck method? In particular, why harmonic coordinates cannot be used? –  timur May 8 '12 at 22:23
    
@timur, I'm no expert, but those are the only two methods I know. Harmonic coordinates might be useful? See Chow-Knopf p 91, where they sort of discuss this, without making any precise claims. It's not clear to me, however, that they work, as I dont know how "far from harmonic" at time $t>0$ coordinates are which are harmonic at $t=0$. –  Otis Chodosh May 9 '12 at 1:08
    
But we can interpret this kind of equations at $t=0$, don't we? One can either use Misha's definition of smoothness (smoothly extendable to a neighborhood of $0$) while remembering that we do not ask the equation to be satisfied for negative times; or define differentiable functions on closed intervals in the natural way (Taylor formula to order $1$). –  Benoît Kloeckner Nov 24 '13 at 9:08
show 2 more comments

Pardon, this is not an answer, but it was too long to put in as a comment, and I wasn't sure how to include this random thought otherwise. Let $g$ be a Riemannian metric on a closed manifold $M$. We can define its temperature $\tau(g)\in\lbrack0,\infty]$ to be the supremum of all $T$ such that there exists a solution to the Ricci flow on $M\times \lbrack0,T]$ with $g(T)=g$. By Brett Kotschwar's backward uniqueness result, there exists a unique solution to the Ricci flow on $M\times(0,\tau(g)]$ with $g\left( \tau\left( g\right) \right) =g$. This solution may or may not be extendable to $t=0$. The temperature of an Einstein metric with nonnegative scalar curvature is infinity (when $R>0$, more generally, a shrinking gradient Ricci soliton). The temperature of an Einstein metric with negative scalar represents its scale. More generally, $\tau (cg) =c\tau (g)$. By Shigetoshi Bando's result, if $\tau(g)>0$, then $g$ is real analytic (as Otis Chodosh wrote above). I'm not sure what this invariant, nontrivial only for real analytic metrics, is good for, though.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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