How submanifolds evolve under Ricci flow? 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...
 A: 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.
