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!



the left hand side of the  Ricci flow equation at the initial value
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.




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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 $\Delta f$".




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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 $M$. 




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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.


