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%%%%%%%%%%%% PAPER DATA %%%%%%%%%%%%%%%%%%%
\author{Jose Ricardo Arteaga, Mikhail Malakhaltsev}

\title{Ricci flow of surfaces in three-dimensional Euclidean space}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{document}

\maketitle

\section{Introduction}
\label{sect:intro}

The Ricci flow is a family of Riemannian manifolds $(M,g_t)$ such that
\begin{equation}
\label{eq:1}
\frac{d}{dt} g_t = -2 Ric(g_t).
\end{equation}

In the present paper we study
deformations of surfaces $\Sigma_t$ in the three-dimensional Euclidean
space $E_3$ such that the metric $g_t$ induced on $\Sigma_t$ by the
standard metric of $E_3$ satisfies the Ricci flow equation \fref{eq:1}.
Our considerations are local.


For a two-dimensional Riemannian manifold
$(M,g)$ we have $Ric = K g$, where $K$ is the sectional curvature,
hence in our situation the Ricci-flow equation is written as follows 
\begin{equation}
\label{eq:2}
\frac{d}{dt} g_t = -2 K(g_t) g_t.  
\end{equation}

\begin{stat}
\label{stat:1}
If a family of metrics $g_t$ satisfies differential equation
\emph{\fref{eq:2}}, then $g_t = e^{2\varphi}g_E$, where $g_E$ is the standard
flat metric, and
\begin{equation}
\label{eq:3}
\frac{\pd \varphi}{\pd t} = e^{-4\varphi} \Delta_E \varphi,
\end{equation}
where $\Delta_E$ is the Euclidean Laplacian. 
\end{stat}
\begin{proof}
First let us prove that any solution of \fref{eq:2} has the form
$g_t = e^{2\varphi}g_E$.
Take two vectors $v,w \in T_p M$, $p \in M$, such that $g_0(v,w) = 0$
and consider the function $f(t) = g_t(v,w)$. From \fref{eq:2} we get
the differential equation $\frac{df}{dt} = - 2 K(g_t) f$. Since $f(0)=0$, we get that
$f(t)=0$ is a unique solution to this differential equation. Hence if
two vectors $v$ and $w$ are orthogonal with respect to $g_0$, they are
orthogonal with respect to all $g_t$. It is well known, that in this
case $g_t$ is conformal to $g_0$, and $g_0$ is, in turn, conformal to
the Euclidean metric $g_E$, therefore we can write $g_t =
e^{2\varphi(t,p)}g_E$ (see e.\,g. \cite{DubrovinNovikovFomenko}). 

Now the curvature of $g_0(p) = e^{2\phi(p,t)}g_E(p)$ is 
$
K(g_t)(p) = - e^{-2\phi(t,p)}\Delta_E\phi(t,p)
$ 
(see e.\,g. \cite{DubrovinNovikovFomenko}).  
Hence \fref{eq:2} implies \fref{eq:3}.  
\end{proof}

\comment{ 1) This theorem surely is known, we should find the exact
  reference! Anyway we need to include this proof for ``the reader convenience''. 
2) Here is the subtle question arises: why  cannot we take the metric 
$g_t = e^{2\varphi} g_E$ from the very beginning? We cannot, of course,
since this imposes condition on deformation,
but I feel we should rewrite the proof to make this fact clear. 
 }
\comment{\textit{JR: For 1) Exactly this result I cannot find it, but this is not very strong result. For 2) I am agree with you.}}

It is known that for any analytical positive definite quadratic form
$g = g_{ij}(u^k)$ 
we can find a surface $\Sigma \subset E_3$ (not unique!) such that $g$ is the
metric tensor of $\Sigma$ (see, e.g., \cite{Kagan}, p.191) 
\comment{In fact this is the only book where I could find the exact
  statement and proof of this result}
Therefore, a given surface can be included in the Ricci flow and in
many ways.
\comment{In fact, to make this statement strict we need two things:
1) \fref{eq:3} has solution for any initial conditions (from the
literature it follows that it is true, but we need an exact reference;
\textit{JR: H\"ormander, Lectures on Nonlinar Hyperbolic Differential Equation, Mat\'ematiques \& Applications 26, Springer-Verlag, 1997 , but in this book there is not  exactly the equation \fref{eq:3}.}
2) We need a parametric version of the above mentioned result about
immersion of a quadratic differential form in $E_3$. I think it is
also true, because the proof it based on properties of PDE in normal
form, however it is better to have an exact statement.
\textit{JR: I think so, too. The exact reference I did'nt found} 
}

Therefore, our problem is \emph{to find natural way to immerse the 
family surfaces satisfying the Ricci flow equation in $E_3$ and to
find geometrical properties of surfaces in the family.  
}


\section{Ricci flow of surfaces of revolution}
\label{sect:surf_of_revol}


The most natural class of surfaces which admits complete description
of Ricci flow families is the class of surfaces of revolution.

Let a family $\Sigma_t$ of surfaces of revolution  be given by 
\begin{equation}
  \label{eq:21}
\vec r(u,v) = a(t,u) \vec e(v) + b(t,u) \vec k.  
\end{equation}
We can take the parameter $u$ in such a way that
\begin{equation}
\label{eq:22}
\pd_u a(t,u)^2 + \pd_u b(t,u)^2 = a(t,u)^2.
\end{equation}
Then the metric tensor of $\Sigma_t$ is 
\begin{equation*}
g_t = a(t,u)^2(du^2 + dv^2),  
\end{equation*}
and hence from Statement \ref{stat:1} we get the  equation for
$a(t,u)$:
\begin{equation}
  \label{eq:23}
\frac{\pd a}{\pd t} = \frac{1}{a} \Delta_E \log a.  
\end{equation}
This equation has solution $a(u,t)$ for any initial conditions $a(0,u)=a_0(u)$
\comment{we again need here an exact reference. \textit{JR: This is similar that \fref(3), I do'nt think that we can find the same equation. I know that you think, that It is important the exact reference, but this is the global theory of PDE.}}
and we can find $b(u,t)$ from \fref{eq:22}:
\begin{equation}
  \label{eq:24}
b(t,u) = \int \sqrt{a(t,u)^2 - \pd_u a(t,u)^2} du.  
\end{equation}
Thus we have proved
\begin{stat}
Any surface of revolution can be included into a family of surfaces of
revolution satisfying to Ricci flow equation. This family is given by 
\emph{\fref{eq:21}}, where $a(t,u)$ is a solution of \emph{\fref{eq:23}} and
$b(t,u)$ is given by \emph{\fref{eq:24}}. 
\end{stat}

\section{Derivation equations of infinitesimal deformation}
\label{sect:der_eq}

Let a deformation $\Sigma_t$ of a surface $\Sigma_0$ be given by the
equations $\vec r = \vec r(u^a,t)$. In what follows we denote
the derivative with respect to $t$ by dot. Let us set 
$\vec r_i = \pd_i\vec r$ and let be $\vec n$ be the unit normal vector field.   
Then, for the infinitesimal deformation vector field $\dot{\vec r}$, we
have the expansion:
\begin{equation}
\label{eq:4}
\dot{\vec r} = \lambda^i \vec r_i + \mu \vec n.
\end{equation}
The vector field $\lambda^i \vec r_i$ is the orthogonal projection of
$\dot{\vec r}$ on the tangent plane of $\Sigma_t$. 

\comment{Do terms for $\lambda$ and $\mu$ exist? It would be very good...\textit{JR: They are the components in the basis $\vec{r_i}$, and $\vec{n}$. Our way is to consider surfaces embedded with induced metric.}}


\begin{stat}[Derivation equations of infinitesimal deformations]
\label{stat:2}
\begin{equation}
\label{eq:5}
\begin{aligned}
&\frac{\pd}{\pd t}\vec r_i = (\nabla_i\lambda^s - \mu h^s_i)\vec r_s
  + (\pd_i \mu + h_{is}\lambda^s)\vec n
  \\
&\frac{\pd\vec n}{\pd t} = - (\pd_i \mu + h_{is}\lambda^s) \vec r_s
\end{aligned}
\end{equation}
\end{stat}
\begin{proof}
The derivation equations of the surface are
\begin{equation}
\label{eq:6}
\pd_i \vec r_j = \Gamma^k_{ij} \vec r_k + h_{ij} \vec n,
\end{equation}
where $\Gamma^k_{ij}$ are the coefficients of Riemannian connection of
the induced metric and $h_{ij}$ are the coefficients of the second
fundamental form. We take the derivative of \fref{eq:4} with respect to
$u^j$, then, using \fref{eq:6}, we get the first equation in
\fref{eq:5}.
The second equation is obtained from the equalities $\vec n^2 = 1$ and
$\vec n \vec r_i = 0$ \comment{\textit{(JR: $\vec n \cdot\vec r_i = 0$) }} by differentiating with respect to $t$.  
\end{proof}

\comment{I suspect nobody knows who wrote the derivative equations for
  infinitesimal deformation for the first time. At least, 
  W. Blaschke writes them in ``Einf\"hrung in die
  differerntialgeometrie''
but with respect to an orthonormal frame.
In Russian V.T.Fomenko and his pupils studied infinitesimal
deformations of surfaces and I will try to find references to their works.
\textit{JR: Fomenko can be a better reference.}}


\begin{stat}
\label{stat:3}
Let $\vec r = \vec r(u^i,t)$ be the Ricci flow of surfaces. Then
the vector $\lambda$ and function $\mu$ from \emph{\fref{eq:4}} satisfy
the equation
\begin{equation}
\label{eq:7}
\nabla_i\lambda_j + \nabla_j\lambda_i = -2 K g_{ij} + 2 \mu h_{ij}, 
\end{equation}
where $K$ is the curvature.
\end{stat}
\begin{proof}
Differentiating $g_{ij} = \vec r_i \vec r_j$ with respect to $t$ and using
\fref{eq:5}, we get \fref{eq:7}.
\end{proof}

We will call equations \fref{eq:7} \emph{basic equations of
infinitesimal Ricci flow deformation}.  

\comment{I'm not sure it is the best name. \textit{JR: I beleive that the name is correct, at least it expresses the sense of the  mathematical procedure.}  )}

\subsection{Another form of basic equations  of infinitesimal Ricci
  flow deformations}

The space $S^2(E_2)$ of quadratic forms in the two-dimensional Euclidean space is the
three-dimensional Euclidean space with respect to the scalar product 
\begin{equation}
\label{eq:8}
(u,v) = g^{ik} g^{jm} u_{ij} v_{km} = u^{ij} v_{ij} = u_{ij} v^{ij}.  
\end{equation}
Then the vector product $t=[u,v]$ of two quadratic forms $u_{ij}$,
$v_{ij}$ is given by 
\begin{equation}
\label{eq:9}
t_{ij} = \sqrt{2}
(\varepsilon_i{}^k_\cdot \varepsilon_j{}^m_\cdot +
\varepsilon_j{}^k_\cdot \varepsilon_i{}^m_\cdot) \varepsilon^{pq}
u_{kp} v_{mq},       
\end{equation}
where $\varepsilon_{ij}$ is the area 2-form (``discriminant tensor''
in A.P.\,Norden's terms \cite{Norden}), 
$\varepsilon_i{}^j_\cdot = g^{js}\varepsilon_{is}$, and
$\varepsilon{ij} = g^{ik} g^{jm} \varepsilon_{km}$.

\comment{ I have found out this formula by calculating in an
  orthonormal frame in $S^2(E_2)$ and it is strange but I could not
  find it in classical texts, e.g. in A.P.Norden. Anyway, I think it
  would be interesting to give to a student (possibly to a girl you
  wrote about) a task to prove that this formula really gives vector
  product by tensor calculations, using tensor identities of $e_{ij}$.
\textit{JR: Ha,Ha, Ha. Super!, I like it.}
}


Assume in what follows that the surface has no umbilic points,
therefore at all points $g_{ij}$ and $h_{ij}$ are linearly independent.

A symmetric tensor  $\lambda_{ij}$  lies in the
plane spanned by $g_{ij}$ and $h_{ij}$ if and only if $\lambda_{ij}$
is orthogonal to the symmetric tensor $t = [g,h]$.

Let us write the corresponding equation for $\lambda$.
Using the identity $\varepsilon_{ik}\varepsilon^{jk} = \delta_i^j$, 
from  \fref{eq:9} we immediately get that 
\begin{equation}
\label{eq:12}
t_{ij} = -\sqrt{2}(\varepsilon_{im} h_j^m  + \varepsilon_{jm} h_i^m).  
\end{equation}
and hence 
\begin{equation*}
0 = \lambda_{ij} t^{ij} = \varepsilon^{im} h_m^j \lambda_{ij}.  
\end{equation*}

Now if  $\lambda = a g + b h$ in $S^2(E_2)$, we obtain by standard
argument that 
\begin{equation*}
a = \frac{(\lambda,g)(h,h)-(h,g)(\lambda,h)}{(g,g)(h,h)-(g,h)^2},  
\end{equation*}
where $(\ ,\ )$ stands for the scalar product \fref{eq:8}.
Then 
\begin{equation*}
(\lambda,g) = g^{ij}\lambda_{ij}, \quad
(\lambda,h) = h^{ij}\lambda_{ij}, \quad  
(h,g) = g^{ij}h_{ij} = H, \quad 
(h,h)=h^{ij}h_{ij}=k_1^2+k_2^2=H^2-2K,
\end{equation*}
where $K$ is the total curvature, $H$ is the mean curvature, and
$k_1$, $k_2$ are principal curvatures.

Now, taking $\lambda_{ij} = \nabla_i \lambda_j + \nabla_j \lambda_i$,
we get 

\begin{stat}[Another form of basic equations]
$\lambda^i$ is the tangent component of infinitesimal Ricci flow
deformation if and only if
\begin{equation}
\label{eq:10}
\begin{aligned}
&(\varepsilon^{im}h_m^j +   \varepsilon^{jm}h_m^i)\nabla_i\lambda_j =
0,
\\
&(H^2-2K)(\operatorname{div}\lambda) - H h^{ij}\nabla_i\lambda_j = K(H^2-4K).
\end{aligned}
\end{equation}
\end{stat}

An infinitesimal  deformation is called normal if $\pd_t\vec r$ is
collinear to $\vec n$. 
\comment{I have found this term in a thesis of V.T.Fomenko's pupil,
  and she refers to Kagan. But I did not find this term in Kagan.}
From \fref{eq:10} (as well as from \fref{eq:7}. \textit{JR: Then it is better to write as reference Fomenko.}) it follows  
\begin{cor}
If the infinitesimal Ricci flow  is normal, then the Ricci flow
consists of spheres.   
\end{cor}
\begin{proof}
If the deformation is normal, then $\lambda^i=0$, and from
\fref{eq:10} (as well as from \fref{eq:7}) it follows that all points
of $\Sigma_t$ are umbilic.   
\end{proof}

\begin{rem}
The corollary statement is true for any conformal deformation.  
\end{rem}


%---------------------------------

\subsection{Calculation for minimal surfaces}

Let $\Sigma$ be a minimal surface. In \cite{Palais_Terng} it is proved
that at a neighborhood of any point of $\Sigma$
there exists a coordinate system $(u^1,u^2)$ such that 
\begin{equation}
\label{eq:11}
\begin{aligned}
& g_{11} = g_{22} = \frac{1}{\kappa}, \quad g_{12}=0
\\
& h_{11}=1 \quad h_{12}=0 \quad h_{22}=-1   
\end{aligned}
\end{equation}
\comment{I take this from Palais, but I'm sure it can be found in many
other classical books. Possibly, it is better to refer to other sources. \textit{JR: I did'nt want to say you that the hard disk of my computer in the office was damaged. Mauricio, our Mauricio gave me another disc of 20 GB only. He promised me  will change to me soon my PC (Excuse me, I returned to Windows, I know that you did'nt speak about this). On the other hand some files of the my discs  either want to open for example, I don't knok, but I cannot open A. D. Aleksandrov and V. A. Zalgaller. This book is free in the AMS but I prefer download tomorow in the office.}}
Then $\varepsilon^{12}=\frac{1}{\kappa}$, and for $t^{ij}$ defined by
\fref{eq:12} we have 
\begin{equation*}
t^{11}=t^{22}0 \quad t^{12} = t^{21} = -2\kappa.  
\end{equation*}
Then the first equation in  \fref{eq:10} is written in the form 
$\nabla_1 \lambda_2 + \nabla_2 \lambda_1 = 0$, and, as $H=0$ and
$K=-\kappa^2$, from the second equation in \fref{eq:10} we get that 
$\nabla_1\lambda_1 + \nabla_2\lambda_2 = - 2 \kappa$. 

Now with respect to this coordinate system we have 
\begin{equation*}
\begin{aligned}
\Gamma^1_{11} = - \frac{\pd_1 \kappa}{2\kappa} &
\Gamma^1_{12} = - \frac{\pd_2 \kappa}{2\kappa} & 
\Gamma^1_{22} =    \frac{\pd_1 \kappa}{2\kappa}
\\     
\Gamma^2_{11} =  \frac{\pd_2 \kappa}{2\kappa} &
\Gamma^2_{12} = -\frac{\pd_1 \kappa}{2\kappa} & 
\Gamma^2_{22} = -\frac{\pd_2 \kappa}{2\kappa}
\end{aligned}
\end{equation*}
Therefore we arrive at the following equations on the vector field
$\lambda_i$:
\begin{equation}
\label{eq:13}
\begin{aligned}
&\pd_1 \lambda_2 + \pd_2 \lambda_1 = 
- \frac{\pd_2 \kappa}{\kappa}\lambda_1    - \frac{\pd_1
  \kappa}{\kappa}\lambda_2
\\
&\pd_1 \lambda_1 + \pd_2 \lambda_2 = - 2\kappa   
\end{aligned}
\end{equation}
\comment{I think this system is worth thinking about. For special
  cases of $\kappa$ we can easily find solutions. }
\comment{Another thing we can use is Norden's fundamental differential
  equation for vector field $\nabla_i\lambda_j = a_i \lambda_j + A_i
  \widetilde\lambda_j$ which he introduced in the Theory of
  Surfaces. You have at least an electronic version of this book among
  those books on the disks. This gives some invariants of vector field. \textit{JR: Excuse me Misha. You can put it in any site, with password if you want and I can download it. Only put the Norden book of "Theory of Surfaces".} }

\section{Partial Solutions to the equation \fref{eq:3}}
\label{sect:part_sol}


The equation \fref{eq:3} was studied much (see,
e.g. \cite{Carstea_Vicinescu}). 
\comment{Possibly we should include solutions given in this
  paper. Also it would be interesting to look through [3] from this paper. \textit{JR. Misha, what do you want.}}


In \fref{eq:3} we set $f(t,u^1,u^2,t)=e^{-4\varphi(t,u^1,u^2,t)}$,
then we get the equation 
\begin{equation*}
\frac{\pd}{\pd t}f = f^2 \Delta_E \log f  
\end{equation*}
from which we obtain 
\begin{equation}
  \label{eq:14}
\frac{\pd}{\pd t}f = f \Delta_E f - (\grad_E f)^2, 
\end{equation}
where the lower index $E$ means that the differential operators are
taken with respect to the Euclidean metric. In what follows we will
drop $E$.

On the vector space of smooth functions $C^\infty(\R^2)$ let us take
the tensor 
\begin{equation}
  \label{eq:15}
q(f,g) = \frac{1}{2}(f \Delta g + g \Delta f) - \grad f \cdot \grad g.  
\end{equation}
which has type $(2,1)$.
Then \fref{eq:14} is rewritten as follows
\begin{equation}
  \label{eq:16}
\frac{\pd}{\pd t} f = q(f,f)  
\end{equation}
that is turns to an ``infinite-dimensional ordinary differential equation system''. 
If now we choose a subalgebra $V \subset C^\infty(\R^2)$ 
spanned by functions $\gamma_i(u^a)$ and closed with respect to the differentiation, then we
get (in general situation) an infinite  differential equation system. 
Then $f(t,u^a) = \varphi^i(t) \gamma_i(u^a)$ is a solution to
\fref{eq:14} if and only if $\varphi^i(t)$ is a solution of the
ordinary differential equation system:
\begin{equation}
  \label{eq:17}
\frac{d}{dt} \varphi^k = q^k_{ij} \varphi^i \varphi^j,
\end{equation}
where $q^k_{ij} \gamma_k = q(\gamma_i,\gamma_j)$.

This idea makes it possible to construct various examples of solutions.

\subsection{Example of solution}

Let us consider the one-dimensional case, that is $f=f(t,u^1)$
(geometrically this corresponds to the case of surfaces of
revolution). In what follows we set $u^1=x$.
Let $\gamma_k = e^{kx}$. Then one can easily find that  
\begin{equation*}
q(\gamma_i,\gamma_j) = \frac{1}{2}(i-j)^2 \gamma_{i+j},  
\end{equation*}
therefore the system \fref{eq:17} turns into the system
\begin{equation*}
  \frac{d}{dt}\varphi_k = \frac{1}{2} \sum\limits_{i+j=k} (i-j)^2
  \varphi_i \varphi_j.
\end{equation*}
Rewrite this system as follows.
For $k=0,1$, we have 
\begin{equation*}
\frac{d}{dt}\varphi_0=0, \quad \frac{d}{dt}\varphi_1=\varphi_0\varphi_1 
\end{equation*}
hence $\varphi_0(t)=a_0$ is constant, 
and $\varphi_1=a_1 \exp(a_0 t)$.
For $k \ge 2$, we have
\begin{equation}
\label{eq:18}
  \frac{d}{dt}\varphi_k = k^2 \varphi_0 \varphi_k +  \eta_k,
\end{equation}
where
\begin{equation}
\label{eq:19}
\eta_k =  \frac{1}{2} \sum\limits_{j=1}^{k-1} (k-2j)^2 \varphi_{k-j} \varphi_j.
\end{equation}
Now, we arrive at the following recurrent procedure. Since $\eta_k$ in
\fref{eq:18} depend on $\varphi_0, \ldots, \varphi_{k-1}$ and does not
depend on $\varphi_k$ (see \fref{eq:19}),  and $\varphi_0 = a_0$ is
constant, we solve the ordinary differential equation \fref{eq:18}
with respect to $\varphi_k$ and get 
\begin{equation}
\label{eq:20}
\varphi_k(t) = \{a_k + \int_0^{t} \eta_k(t) \exp(-a_0 k^2 u) du \}
\exp(a_0 k^2 t).
\end{equation}
Thus we can recurrently find functions $\varphi_k(t)$.  Now, \emph{if
the series $\sum_k^\infty \varphi_k(t) e^{kx}$ converges, 
the function $f(t,u) = \sum_k^\infty \varphi_k(t) e^{kx}$ is the
solution of \emph{\fref{eq:14}} such that $f(0,x)=\sum_{k=0} a_k e^{kx}$}.


Certainly, the convergence of the series depends on the coefficients
$a_i$. For example, take the initial condition $f(0,x) = 1 + e^{2x}$, 
then $a_0=1$, $a_2=1$ and all other $a_k$ are zero.
We have $\varphi_0(t)=1$, $\varphi_1(t)=0$, $\eta_2(t)=0$, then from
\fref{eq:20} we get $\varphi_2(t)=e^{4t}$. Then $\eta_3(t)=0$, hence
$\varphi_3(t)=0$. By induction one can prove that $\eta_k(t)=0$ and
$\varphi_k(t)=0$ for $k \ge 4$, and, finally, we get the solution 
$f(t,u)=1+e^{4t+2x}$.

\comment{Surely, we need more examples here.}

\begin{thebibliography}{99}

\bibitem{Kagan} Kagan V.F. {Foundations of Surface Theory} V. I, 
OGIZ, Gostehizdat, 1947

\bibitem{DubrovinNovikovFomenko}
Dubrovin B.A., Novikov S.P., Fomenko A.T. \emph{Modern
Geometry}, Nauka, Moscow  (1979)

\bibitem{Norden}
Norden A.P. \emph{Theory of Surfaces}, Moscow, 1956.

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