# On the definition on the Ricci flow [closed]

I'm trying to understand how one can define the Ricci flow equation.

First you have to parametrized the set of all Riemannian metrics. Then you have to define the derivative on this parametrized family.

HERE IS MY PROBLEM.

The set of all metrics has the structure of a vector space (2-tensor fields). But this is not enough, We need to be this a normed vector space.

Is the set of all Riemannian metrics a Normed vector space? If not, How we can define the derivative ?

## closed as off-topic by Deane Yang, Willie Wong, Joonas Ilmavirta, coudy, abxApr 30 '15 at 12:05

This question appears to be off-topic. The users who voted to close gave these specific reasons:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Joonas Ilmavirta, abx
• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Deane Yang, Willie Wong, coudy
If this question can be reworded to fit the rules in the help center, please edit the question.

• There seems to be some fundamental misunderstanding about the Ricci flow on your part. In terms of "how can we define the derivative": it is basically no different on how you can define the Lie derivative of an arbitrary tensor field. In particular, the Ricci flow is not an ODE with values in some infinite dimensional vector space, it is a PDE with values in a finite dimensional vector bundle. If this doesn't answer your question, please edit to clarify what you actually mean. – Willie Wong Apr 30 '15 at 11:12
• @WillieWong Wait, now I am confused. I do think of Ricci flow as an ODE in an open subset of an infinite dimensional vector space. Let $X$ be a smooth $n$-fold. Let $V$ be the vector space of sections of the vector bundle $\mathrm{Sym}^2 TX$ and let $U$ be the open subset of positive definite metrics. Then Ricci flow is an ODE in $U$: $dg/dt = -2 R(g)$. What am I missing? – David E Speyer Apr 30 '15 at 11:43
• Oh, okay, I think I get it. $R(g)$ is made up of various derivatives of $g$. So we can view this instead as a (very nonlinear) PDE on $X \times \mathbb{R}$. – David E Speyer Apr 30 '15 at 11:49
• @DavidSpeyer: My point is that the Ricci flow is not merely an ODE in an infinite dimensional vector space. The equations the comes from a PDE are more special and some generalities one has to deal with when dealing with infinite dimensional ODEs can be avoided. – Willie Wong Apr 30 '15 at 11:49
• Just to give an example: suppose you want to think about the linear wave (or heat) equation as a second order ODE on the Sobolev space $H^1$, you have the problem that the Laplacian is not a bounded operator and so you cannot appeal to the usual theory of infinite dimensional ODEs with continuous right hand sides. The local existence theory for these equations come in many flavours (energy estimates, fundamental solutions, Fourier representations), but they all make use of details of the equation that are beyond simply "ODE in Banach space" properties. – Willie Wong Apr 30 '15 at 11:58