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

This may be very naive, since I just started trying to learn Ricci flow; but I couldn't really find any answer after looking for a while in all the textbooks and lecture notes I found online...

If $(M,g_t)$ is a solution of the Ricci flow (normalized or not, I don't care), and $i\colon N\hookrightarrow (M,g_0)$ is a submanifold (with the induced metric), what is known about what happens to $(N,i^*g_t)$ in terms of its intrinsic/extrinsic geometry?

This is somewhat vague, so, to be more precise: under what conditions a totally geodesic (resp. minimal) submanifold remains totally geodesic (resp. minimal)? What evolution equation is satisfied by the second fundamental form $B^t_{\xi^t}(X,Y)=g_t(\nabla^t_X Y,\xi^t)$ of $N\subset (M,g_t)$, or shape operator, in the codimension $1$ case? Note that here almost everything depends on $t$: the connection $\nabla^t$, the normal field $\xi^t$ and obviously the metric $g_t$. I tried to take the $t$ derivative using formulas for each of the objects (e.g., the ones found in Topping's notes), but it got incredibly messy very fast -- and there was nothing I could really read off the formulas. I then did some examples, but the only ones I could do all the computations for were somewhat trivial.

I would be interested in any intuition/results related to the above, it could be for hypersurfaces (instead of general submanifolds), only in low dimensions, etc...

share|improve this question
    
I don't know if this has been studied before. You could email Ben Chow and ask him if he knows. Also, you could search "Ricci flow submanifold" on both arxiv and Mathscinet. If you find nothing, I suggest struggling with the calculations more and trying to different approaches. You could fix co-ordinates on $M$ and assume $N$ is given by the vanishing of the last $k$ co-ordinates, where $k$ is the codimension of $N$. I suggest first figuring out what happens to the second fundamental form for an arbitrary deformation of the metric. –  Deane Yang Feb 8 '13 at 1:12
1  
Or you could try to fix geodesic normal co-ordinates relative to the hypersurface. But then the co-ordinates also have to change in time as the metric changes. This sounds more difficult to me but you never know until you try. Anyway, answering questions like this usually require (at least for me) at lot of struggling with the calculations tried different ways. After you've tried all possible angles, what sometimes happens is that suddenly you realize there's an easy way to do it. The main goal is probably to figure out how the second fundamental form evolves. –  Deane Yang Feb 8 '13 at 1:16
2  
I am sure you know it, just in case: If your submanifold is a fixed set of some isometric action then it stay so under Ricci flow. –  Anton Petrunin Feb 8 '13 at 1:55

1 Answer 1

This paragraph doesn't answer the question, but discusses some more elementary related calculations (as in Hamilton's Nonsingular Solutions paper). Consider a closed surface $\Sigma_{t}$ in a $3$-manifold $(M,g(t))$ evolving by Ricci flow, where $\Sigma_{t}$ evolves in the normal direction $N$ with velocity function $V$. Then the induced metric $\operatorname{I}_{t}$ on $\Sigma_{t}$ evolves by $\frac{\partial}{\partial t}\operatorname{I}_{t}=2\operatorname{II} _{t}V-2\operatorname{Ric}|_{T\Sigma_{t}}$, where $\operatorname{II}_{t}$ is the second fundamental form of $\Sigma_{t}$ and $\operatorname{Ric} |_{T\Sigma_{t}}$ is the restriction of the Ricci tensor of $g\left( t\right) $ to $T\Sigma_{t}$. Let $dA_{t}$ denote the area element of $\Sigma_{t}$. Then $\frac{\partial}{\partial t}dA_{t}=\frac{1}{2}\operatorname{trace} _{\operatorname{I}_{t}}(\frac{\partial}{\partial t}\operatorname{I}_{t} )dA_{t}=(H_{t}V-R+\operatorname{Ric}(N,N))dA_{t}$, where $H_{t}$ is the mean curvature of $\Sigma_{t}$. In particular, if $V=0$, then the area $\operatorname{A}_{t}$ of $\Sigma_{t}$ evolves by $\frac{d}{dt} \operatorname{A}_{t}=\int_{\Sigma_{t}}(-R+\operatorname{Ric}(N,N))dA_{t} =\int_{\Sigma_{t}}(-\frac{1}{2}R-\operatorname{sect}(T\Sigma_{t}))dA_{t}$, where $\operatorname{sect}(T\Sigma_{t})$ denotes the sectional curvature of the plane $T\Sigma_{t}$. On the other hand, by the Gauss equations for $\Sigma_{t}\subset M$, the intrinsic Gauss curvature of $(\Sigma _{t},\operatorname{I}_{t})$ is $K_{t}=\operatorname{sect}(T\Sigma_{t} )+\det(\operatorname{II}_{t})$. So $\frac{d}{dt}\operatorname{A}_{t} =\int_{\Sigma_{t}}(-\frac{1}{2}R-K_{t}+\det(\operatorname{II}_{t}))dA_{t}$. This formula is nice at some time, for example, if $\Sigma_{t}$ is a minimal surface and the scalar curvature is bounded from below $R\geq-C_{t}$ (the latter is indeed true for Ricci flow), when and where $\frac{d}{dt} \operatorname{A}_{t}\leq\frac{C_{t}}{2}\operatorname{A}_{t}-2\pi\chi(\Sigma)$ since $\det(\operatorname{II}_{t})\leq0$ follows from $H_{t}=0$ and by the Gauss-Bonnet formula.

The relevant computations must be somewhere in the literature, but I don't know where; so the following is off the top of my head and needs to be checked. About the normal $N_{t}$ in the case of a static hypersurface, consider a family of inner products $g_{t}$ on a vector space $E$ (e.g., $T_{x}M$) and a fixed hyperplane $P$ (e.g., $T_{x}\Sigma$). By $g_{t} (X,N_{t})\equiv0$ for each $X\in P$, we have $\frac{\partial g_{t}}{\partial t}(X,N_{t})+g_{t}(X,\frac{\partial N_{t}}{\partial t})=0$. So $\frac{\partial g_{t}}{\partial t}(N_{t})+g_{t}(\frac{\partial N_{t}}{\partial t})=cN_{t}$ for some $c\in\mathbb{R}$, identifying $T_{x}M$ and $T_{x}^{\ast}M$ by $g_{t}$ here and below. Dotting with $N_{t}$ yields $c=\frac{1}{2}\frac{\partial g_{t}}{\partial t}(N_{t},N_{t})$ since $g_{t}(\frac{\partial N_{t}}{\partial t},N_{t})=-\frac{1}{2}\frac{\partial g_{t}}{\partial t}(N_{t},N_{t})$ from $g_{t}(N_{t},N_{t})\equiv1$. We obtain $\frac{\partial N_{t}}{\partial t}=\frac{1}{2}\frac{\partial g_{t}}{\partial t}(N_{t},N_{t})N_{t} -\frac{\partial g_{t}}{\partial t}(N_{t})$. Combining this with some other formulas of this nature, such as the standard $\frac{\partial}{\partial t}\nabla_{t}$, where $\nabla_{t}$ denotes the Levi-Civita connection of $g_{t}$, one should be able to compute the evolution of $\operatorname{II} _{t}$, etc.

[Dec 3, 2013] In response to Chris Gerig's question:

Let $F:N\times(0,T)\rightarrow M$ be a parametrized hypersurface in a Riemannian manifold $(M^{n},g)$. The first fundamental form (induced metric) at time $t$ is $\operatorname{I}_{t}(X,Y)=g(dF_{t}(X),dF_{t}(Y))$, where $F_{t}(x)=F(x,t)$. The unit normal $\nu$ and the velocity $dF_{t} (\frac{\partial}{\partial t})=\frac{\partial F}{\partial t}=V\nu$ are vector fields along the map $F$. We compute that $$ \frac{\partial}{\partial t}\operatorname{I}_{t}(X,Y)=g(\frac{D}{dt} dF_{t}(X),dF_{t}(Y))+g(dF_{t}(X),\frac{D}{dt}dF_{t}(Y)), $$ where $\frac{D}{dt}$ is covariant differentiation along the path $\alpha _{x}(t)=F(x,t)$. Basically, since $[\frac{\partial}{\partial t},X]=0$ in $N^{n-1}\times(0,T)$ and by pushing this forward by $F$, we have $\frac{D} {dt}dF_{t}(X)=\frac{D}{dX}\left( dF_{t}(\frac{\partial}{\partial t})\right) =\frac{D}{dX}\left( V\nu\right) $, where $\frac{D}{dX}$ is covariant differentiation along $F$ restricted to a path in $N\times\{t\}$ tangent to $X$ (heuristically, $\nabla_{dF_{t}(\frac{\partial}{\partial t})} dF_{t}(X)-\nabla_{dF_{t}(X)}dF_{t}(\frac{\partial}{\partial t})=[dF_{t} (\frac{\partial}{\partial t}),dF_{t}(X)]=dF_{t}([\frac{\partial}{\partial t},X])=0$). Using $\langle X,\nu\rangle=0$ and the product rule for $\frac {D}{dX}$, we obtain $$ \frac{\partial}{\partial t}\operatorname{I}_{t}(X,Y)=V\left( g(\frac{D} {dX}\nu,dF_{t}(Y))+g(dF_{t}(X),\frac{D}{dY}\nu)\right) =2V\operatorname{II} {}_{t}(X,Y) $$ by the definition of the second fundamental form in terms of the derivative of the unit normal.

share|improve this answer
    
Could you briefly explain how $2II_tV$ arises in the induced metric evolution? –  Chris Gerig Dec 1 '13 at 20:19

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.