text:
\begin{document}

\title{Reilly-type inequalities for $p$-Laplacian on submanifolds in space forms}

\author{Hang Chen}
\address[Hang Chen]{Department of Applied Mathematics, Northwestern Polytechnical University, Xi' an 710129, P. R. China \\ email: chenhang86@nwpu.edu.cn}
\thanks{Chen supported by NSFC Grant No. 11601426}
\author{Guofang Wei}
\address[Guofang Wei]{Department of Mathematics, University of California, Santa Barbara CA 93106 \\ email: wei@math.ucsb.edu}
\thanks{GW partially supported by NSF DMS 1506393}

\begin{abstract}
Let $M$ be an $n$-dimensional closed orientable submanifold in an $N$-dimensional space form. When $1<p \le \frac n2 + 1$, we obtain an upper bound for the first nonzero eigenvalue of the $p$-Laplacian in terms of the mean curvature of $M$ and the curvature of the space form. This generalizes the Reilly inequality for the Laplacian \cite{Rei77, ESI92} to the $p$-Laplacian and extends the work of \cite{DM15} for the $p$-Laplacian.
\end{abstract}

\keywords {Reilly-type inequality, $p$-Laplacian, eigenvalue estimate, submanifolds.}

\subjclass[2000]{58C40, 53C42, 35P15}

\maketitle

\section{Introduction}
Let $M$ be an $n$-dimensional compact Riemannian manifold. The $p$-Laplacian ($p>1$) is a second order quasilinear elliptic operator on $M$ defined by
\begin{equation}
\pl u=\dive (|\nabla u|^{p-2}u).
\end{equation}
It is the usual Laplacian when $p=2$. Similar with Laplacian, one can consider the eigenvalue problem of $\pl$. A real number $\la$ is called a Dirichlet (or Neumann) eigenvalue if there exists a non-zero function $u$ satisfying the following equation with Dirichlet boundary condition  $u \equiv 0 $ on $\partial M$ (or  Neumann boundary condition $\nabla_{\nu}u \equiv 0$  on $\partial M$):
\begin{equation}
	\pl u=-\la |u|^{p-2} u\quad  \text{on $M$},
\end{equation}
where $\nu$ is the outward normal on $\partial M$.

So far, many estimates for the first eigenvalue of Laplacian have been generalized to $\pl$. For instance,
 Matei \cite{Mat00} extended Cheng's first Dirichlet eigenvalue comparison of balls \cite{Che75} to the $p$-Laplacian. For closed Riemannian manifolds with Ricci curvature bounded below by $(n-1)K$,  a sharp lower bound for the first nontrivial eigenvalue of $p$-Laplacian was obtained by Matei \cite{Mat00} for $K>0$, Valtora \cite{Val12} for $K=0$ and Naber-Valtora \cite{NV14} for general real number $K$ respectively, see also \cite{And15,CM17}. Recently, Seto-Wei \cite{SW17} gave various estimates of the first eigenvalue of the $p$-Laplacian on closed Riemannian manifolds with integral curvature condition.

Now we consider a Riemannian manifold $M$ without boundary. The first nonzero eigenvalue of $\pl$, denoted by $\la_{1,p}$,  has a Rayleigh type variational characterization (cf. \cite{Ver91}):
\begin{equation}
\la_{1,p}=\inf\left\{ \frac{\int_M |\nabla u|^p}{\int_M|u|^p} \Big| u\in W^{1,p} (M)\backslash \{0\}, \int_M|u|^{p-2} u =0 \right\}.
\end{equation}
When $M$ is a submanifold immersed in $\F$, where $\F$ is the $N$-dimensional simply connected space form of constant sectional curvature $c$ and represents the Euclidean space $\R$, the unit
sphere $\S$ and the hyperbolic space $\H$ for $c=0, 1$ and $-1$ respectively, there is a well-known estimate for the first  nonzero eigenvalue of Laplacian.
\begin{thm}[cf. \cite{Rei77,ESI92}]\label{thm_Rei}
	Let $M$ be an $n$-dimensional  closed orientable submanifold in an $N$-dimensional space form $\F$. Then the first non-zero eigenvalue $\lambda_{1}^{\Delta}$ of Laplacian satisfies
\begin{equation}\label{Rei}
\lambda_1^{\Delta}\leq\frac{n}{\vol(M)}\int_M\big(|\mathbf{H}|^2+c\big),
\end{equation}
where $\mathbf{H}$ is the mean curvature vector of $M$ in $\F$, and $\vol(M)$ is the volume of $M$. Moreover, the equality holds in \eqref{Rei} if and only if $M$ is minimally immersed in a geodesic sphere of radius $r_c$ of $\F$  with $r_0=\big(n\big/\lambda_1^{\Delta}\big)^{1/2}, r_1=\arcsin r_0$ and $r_{-1}=\arcsinh r_0$.
\end{thm}

 R. Reilly first proved \eqref{Rei} for $c=0$ in  \cite{Rei77}, hence  \eqref{Rei} is usually referred ``Reilly inequality". For the case $c=1$, by embedding the sphere $\S \rightarrow \mathbb R^{N+1}$, one can reduce it to the case $c=0$. In these case, one can move the center of $M$ to the origin and then the  coordinate functions can be used as test functions.  But this does not work for $c=-1$. The case $c=-1$   was proved by El Soufi and Ilias
in \cite{ESI92} by conformally mapping $\H$ to $\S$.

Recently, F. Du and J. Mao \cite{DM15} extended the eigenvalue estimate to  the $p$-Laplacian for $c=0,1$.

\begin{thm}[cf. Theorem 1.2 and Theorem 1.5 in \cite{DM15}]
		Let $M$ be an $n$-dimensional  closed orientable submanifold in an $N$-dimensional space form $\F$. For $p>1$, the first non-zero eigenvalue $\lambda_{1,p}$  of the $p$-Laplacian satisfies
\begin{numcases}
{\la_{1,p}\leq} N^{|2-p|/2}\frac{n^{p/2}}{\big(\vol(M)\big)^{p-1}}\Big(\int_M|\mathbf{H}|^{\frac{p}{(p-1)}}\Big)^{p-1}& for $c=0$, \label{Mao0}\\
(N+1)^{|2-p|/2}\frac{n^{p/2}}{\big(\vol(M)\big)^{p-1}}\Big(\int_M\big(1+|\mathbf{H}|^2\big)^{\frac{p}{2(p-1)}}\Big)^{p-1}& for $c=1$. \label{Mao1}
\end{numcases}
\end{thm}

 In this paper, we generalize the Reilly inequality \eqref{Rei} to the case of $p$-Laplacian for all $c$.

 Our main theorem is the following.
\begin{thm}\label{thm1.1}
	Let $M$ be an $n(\geq 2)$-dimensional  closed orientable submanifold in an $N$-dimensional space form $\F$. Then the first non-zero eigenvalue  $\lambda_{1,p}$ of the $p$-Laplacian satisfies
		\begin{numcases}
	{\la_{1,p}\leq}
	 (N+1)^{1-\frac{p}{2}}\frac{n^{p/2}}{\big(\vol(M)\big)^{p/2}}\Big(\int_M\big(c+|\mathbf{H}|^2\big)\Big)^{p/2} & for $1 < p\leq 2$,	\label{eq_thm1} \\
	 (N+1)^{\frac{p}{2}-1}
	 \frac{ n^{p/2}}{\vol(M)}\int_M(|c+|\mathbf{H}|^2|)^{p/2}& for  $2 < p \le \frac n2 +1$ \label{eq_thm2}.
	\end{numcases}
	Moreover, the equality holds if and only if $p=2$ and $M$ is minimally immersed in a geodesic sphere of radius $r_c$ of $\F$  with $r_0=\big(n\big/\lambda_1^{\Delta}\big)^{1/2}, r_1=\arcsin r_0$ and $r_{-1}=\arcsinh r_0$.
\end{thm}

\begin{rem}
When $p=2$ our estimate recovers \eqref{Rei}.

	When $1<p\leq 2$, we have $\frac{p}{2(p-1)}\geq 1$. Then by H\"{o}lder inequality, we have
	\begin{align*}
		\int_M\big(c+|\mathbf{H}|^2\big)&\leq \big(\vol(M)\big)^{1-\frac{2(p-1)}{p}}\Big(\int_M\big(c+|\mathbf{H}|^2\big)^{\frac{p}{2(p-1)}}\Big)^{\frac{2(p-1)}{p}}.
	\end{align*}
	Hence, the upper bound in \eqref{eq_thm1} is better than \eqref{Mao1} for $c=1$.

	When $2<p\leq \frac{n}{2}+1$, \eqref{eq_thm2} is weaker than \eqref{Mao0} and \eqref{Mao1} for $c=0$ and $c=1$. But \eqref{eq_thm2} is  a new estimate for $c=-1$.
\end{rem}

\begin{rem}
Matei \cite{Mat13} gave an upper bound for $\lambda_{1,p}$ in terms of conformal volume  when $1<p\leq \dim M$.
\end{rem}

This paper is organized as follows. In Section 2, we recall structure equations for a submanifold $M$ in $\F$. We also show how some geometric quantities change when the metric on $\F$ changes under the conformal transformation. In Section 3, we prove Theorem \ref{thm1.1}. To estimate the upper bound of $\lambda_{1,p}$, as the method in \cite{DM15} does not work for $c=-1$, we find suitable test functions by conformal transformation to a unit sphere.

\textbf{Acknowledgment}:
  This work is done while the first author is visiting UCSB. He would like to thank UCSB math department for the hospitality, and he also would like to acknowledge financial support from China Scholarship Council and Top International University Visiting Program for Outstanding Young scholars of Northwestern Polytechnical University.

\section{Preliminaries and notations}
In this section, we recall some well-known facts on geometry of submanifolds and conformal geometry by using the moving frame method.  We use the following convention on the ranges of indices except special declaration:
\begin{equation*}
1\leq i, j, k, \ldots \leq n;\quad
n+1\leq\alpha, \beta, \gamma, \ldots \leq N;\quad
1\leq A, B, C, \ldots \leq N.
\end{equation*}

\subsection{Structure equations for submanifolds}
Let $x$ be the immersion from $M^n$ to an  $N$-dimensional Riemannian manifold $(\bar{M},\bar{g})$. Then $M$ has an induced metric $g_M=x^*\bar{g}$.

We denote the Levi-Civita connections on $M$ and $\bar{M}$ by $\nabla$ and $\bar{\nabla}$ respectively. Choose an orthonormal frame $\{e_A\}_{A=1}^{N}$ on $\bar{M}$ such that $\{e_i\}_{i=1}^n$ are tangent to $M$ and $\{e_{\alpha}\}_{\alpha=n+1}^{N}$ are normal to $M$.  Let $\{\omega_A\}_{A=1}^{N}$ be the dual coframe of $\{e_A\}_{A=1}^{N}$. Then the structure equations of $\bar{M}$ are (cf. \cite{Chern1968}):
\begin{equation}
\left\{\begin{aligned}\label{eq21}
& d\omega_{A}=\sum_{B}\omega_{AB}\wedge\omega_{B},\quad \omega_{AB}+\omega_{BA}=0, \\
&d\omega_{AB}-\sum_C\omega_{AC}\wedge\omega_{CB}=-\frac{1}{2}\sum_{C,D}\bar{R}_{ABCD}\,\omega_C\wedge\omega_D,
\end{aligned}\right.
\end{equation}
where $\{\omega_{AB}\}$ are the connection forms on $\bar{M}$, and  $\bar{R}_{ABCD}$ are components of the curvature tensor of  $\bar{M}$.

Denote $x^{\ast}\omega_{A}=\theta_{A}, x^{\ast}\omega_{AB}=\theta_{AB}$, then restricted to $M$, we have (cf. \cite{Chern1968})
\begin{equation}\label{eq23}
\theta_{\alpha}=0, \quad\theta_{i\alpha}=\sum_j h_{ij}^{\alpha}\theta_{j}.
\end{equation}
and
\begin{equation}
\left\{\begin{aligned}\label{eq22}
&d\theta_{i}=\sum_{j}\theta_{ij}\wedge\theta_{j}, \quad\theta_{ij}+\theta_{ji}=0,\\
& d\theta_{ij}-\sum_k\theta_{ik}\wedge\theta_{kj}=-\frac{1}{2}\sum_{k,l}R_{ijkl}\,\theta_k\wedge\theta_l;
\end{aligned}
\right.
\end{equation}
where $R_{ijkl}$ are components of the curvature tensor of $M$, and $h_{ij}^{\alpha}$ are components of the second fundamental form of $M$ in $\bar{M}$.

We take $\bar{M}=\F$, then $\bar{R}_{ABCD}=c(\delta_{AC}\delta_{BD}-\delta_{AD}\delta_{BC})$. Pulling back \eqref{eq21} by $x$ and using \eqref{eq23} and \eqref{eq22}, we obtain
the Gauss equations
\begin{align}
\label{gauss}
R_{ijkl}&=(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})c+\sum_\alpha( h_{ik}^{\alpha}h_{jl}^{\alpha}-h_{il}^{\alpha}h_{jk}^{\alpha}),\\

R&=n(n-1)c+n^2H^2-S,\label{gauss3}
\end{align}
where $R$ is the scalar curvature of $M$, $S=\sum\limits_{\alpha,i,j}(h_{ij}^{\alpha})^2$ is the
norm square of the second fundamental form, $\mathbf{H}=\sum\limits_\alpha H^{\alpha}e_\alpha=\frac{1}{n}
\sum\limits_\alpha(\sum\limits_i h_{ii}^{\alpha})e_\alpha$ is the mean curvature  vector of $M$, and $H=|\mathbf{H}|$.

\subsection{Conformal relations} In this subsection, we focus on how curvature and the second fundamental form change under the conformal transformation. Although these relations are well-known (cf. \cite{Che73a,Che74}),  we give a brief proof for readers' convenience,  by using the moving frame method.

Now assume that $\bar{M}$ is equipped with a new metric $\tbg=e^{2\rho}\bar{g}$ which is conformal to $\bar{g}$, where $\rho\in C^{\infty}(\bar{M})$. Then $\{\et_A=e^{-\rho}e_A\}$ is an orthonormal frame of $(\bar{M},\tbg)$, and $\{\tom_A=e^{\rho}\omega_A\}$ is the dual coframe of  $\{\et_A\}$. The structure equations of $(\bar{M},\tbg)$ are given by

\begin{equation}
\left\{\begin{aligned}\label{eq25}
&d\tom_{A}=\sum\limits_{B}\tom_{AB}\wedge\tom_{B},\\
&\tom_{AB}+\tom_{BA}=0,
\end{aligned}\right.
\end{equation}
where $\{\tom_{AB}\}$ are the connection forms on $(\bar{M},\tbg)$.
 Denoting $\tg_M=x^{\ast}\tbg, \ x^{\ast}\tom_{A}=\tt_{A}, \ x^{\ast}\tom_{AB}=\tt_{AB}$, then restricted to $(M,\tg_M)$, we have
\begin{equation}\label{eq29}\tt_{\alpha}=0, \quad\tt_{i\alpha}=\sum_j \th_{ij}^{\alpha}\tt_{j},
\end{equation}
and
\begin{equation}
\left\{
\begin{aligned}\label{eq28}
&\displaystyle d\tt_{i}=\sum_{j}\tt_{ij}\wedge\tt_{j}, \quad\tt_{ij}+\tt_{ji}=0, \\
\displaystyle &d\tt_{ij}-\sum_k\tt_{ik}\wedge\tt_{kj}=-\frac{1}{2}\sum_{k,l}\tR_{ijkl}\,\tt_k\wedge\tt_l;
\end{aligned}
\right.\end{equation}
where $\tR_{ijkl}$ are components of the curvature tensor of $(M, \tg_M)$ and $\th_{ij}^{\alpha}$ are components of the second fundamental form of $(M, \tg_M)$ in $(\bar{M}, \tbg)$.

From (\ref{eq21}) and (\ref{eq25}), we can solve
\begin{equation}
\begin{aligned}\label{eq26}
\tom_{AB}=\omega_{AB}+\rho_A\omega_B-\rho_B\omega_A,
\end{aligned}
\end{equation}
where $\rho_A$ is the covariant derivative of $\rho$ with respect to $e_A$, i.e. $d\rho=\sum_A\rho_Ae_A$.

We derive from \eqref{eq22}, \eqref{eq26} and \eqref{eq28}
\begin{align}
e^{2\rho}\tR_{ijkl}=&R_{ijkl}-(\rho_{ik}\delta_{jl}+\rho_{jl}\delta_{ik}-\rho_{il}\delta_{jk}-\rho_{jk}\delta_{il})\nonumber\\
&+(\rho_i\rho_k\delta_{jl}+\rho_j\rho_l\delta_{ik}-\rho_j\rho_k\delta_{il}-\rho_i\rho_l\delta_{jk})\nonumber\\
&-|\nabla\rho|^2(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}).\label{eq209}

\end{align}

By pulling back (\ref{eq26}) to $M$ by $x$ and using (\ref{eq23}) and (\ref{eq29}), we have
\begin{equation}\label{eq208}
\th_{ij}^{\alpha}=e^{-\rho}(h_{ij}^{\alpha}-\rho_{\alpha}\delta_{ij}), \quad\tH^{\alpha}=e^{-\rho}(H^{\alpha}-\rho_{\alpha}),
\end{equation}
from this, it is easy to show  the well-known relation
\begin{equation}\label{eq_um}
e^{2\rho}(\tilde{S}-n\tH^2)=S-nH^2.
\end{equation}

\section{Proof of Theorem \ref{thm1.1}}
In this section, we prove our main result, Theorem \ref{thm1.1}. At first, we recall some  lemmas from \cite{Mat13} to our setting.  For convenience of the reader, we also give the proof here as we need the proof to analyse the equality case.
\begin{lem}[cf. Lemma 2.6 in \cite{Mat13}]\label{lem3.1}
Let $x: M\to \F$ be the immersion from an $n$-dimensional closed orientable submanifold to an $N$-dimensional space form $\F$. Then for $p>1$, there exists a regular conformal map $\Gamma: \F\to \S\subset\mathbb{R}^{N+1}$ such that  the immersion $\Phi=\Gamma\circ x=(\Phi^1, \cdots, \Phi^{N+1})$ satisfies that
\begin{equation}\label{eq_lem31}
\int_M |\Phi^A|^{p-2}\Phi^A\,dv_M=0,~A=1,\ldots,N+1.
\end{equation}
\end{lem}

\begin{proof}
	The main idea of Lemma \ref{lem3.1} is inspired by the case $p=2$ (cf. \cite{LY82,ESI00}).

	First observe that there is the standard conformal map $\Pi_c$ from $\F$ to $\S$. Here $\Pi_c$ is identity when $c=1$, and $\Pi_c$ can be given by the stereographic projection when $c=0$ or $c=-1$.

	For any $a\in \S$, consider the flow  $\gamma_t^a$ generated by the  vector field $V_a(x)=a-\l x,a\r x$ on $\S$. In fact, $\gamma_t^a=\pi_a^{-1}(e^t\pi_a(x)), x\in \S$, where $\pi_a$ is the stereographic projection of pole $a$. It is easy to see that  $\gamma_0^a$ is identity map on $\S$, and $\gamma_t^a(x)\to a$ for any $x\in\S$ as $t\to +\infty$.

	We claim there is a $\gamma_t^a$ such that $\Gamma=\gamma_t^a\circ \Pi_c$ satisfies the required property \eqref{eq_lem31}. If not, we can define a map $F(t,a): [0,+\infty) \times \S\to \S$ as follows:

	\begin{equation}
		F(t,a)=\frac{(\int_M |\Phi^1|^{p-2}\Phi^1,\cdots,\int_M |\Phi^{N+1}|^{p-2}\Phi^{N+1})}{\|(\int_M |\Phi^1|^{p-2}\Phi^1,\cdots,\int_M |\Phi^{N+1}|^{p-2}\Phi^{N+1})\|}.
	\end{equation}

	Now $F(0,\cdot)$ maps any  $a\in \S$ to a fixed point in $\S$. And $F(+\infty, \cdot)$ maps $a=(a^1,\cdots, a^{N+1})\in\S$ to $\frac{(|a^1|^{p-2}a^1,\cdots,  |a^{N+1}|^{p-2} a^{N+1})}{\|(|a^1|^{p-2}a^1,\cdots,  |a^{N+1}|^{p-2} a^{N+1})\|}$, which is bijective. So $\deg F(0,\cdot)=0$ and $\deg F(+\infty,\cdot)$ is odd. But $F(\cdot,\cdot)$ gives a homotopy between  $F(0,\cdot)$ and $F(+\infty,\cdot)$, which is a contradiction. So we complete the proof.
\end{proof}

 	Using the test function constructed in the above lemma, we can get an upper bound for $\lambda_{1,p}$  in terms of the conformal function, compare Lemma 2.7 in \cite{Mat13}.
\begin{lem}\label{lem3.2}
 	Let $M$ be an $n(\geq 2)$-dimensional  closed orientable submanifold in an $N$-dimensional space form $\F$.  Denote by $h_c$  the standard metric on $\F$  and assume $\Gamma^{\ast}h_1=e^{2\rho}h_c$, where $\Gamma$ is the conformal map in Lemma \ref{lem3.1}. Then we have, for all $p>1$,
 	\begin{equation}\label{eq3.3}
 		\lambda_{1,p}\vol(M)\leq (N+1)^{|1-p/2|}n^{p/2}\int_M(e^{2\rho})^{p/2}.
 	\end{equation}
 \end{lem}
 \begin{proof}
  	By Lemma \ref{lem3.1}, we can choose $\Phi^A$ as the test function, so
 	\begin{equation}\label{eq3.7}
 	\la_{1,p}\int_M|\Phi^A|^p\leq|\nabla\Phi^A|^p, \quad 1\leq A\leq N+1.
 	\end{equation}
 	Note that $\sum_{A=1}^{N+1}|\Phi^A|^2=1$, then $|\Phi^A|\leq 1$. We also have
 	\begin{equation}\label{eq32}
 	\sum_{A=1}^{N+1}|\nabla\Phi^A|^2=\sum_{i=1}^{n}|\nabla_{e_i}\Phi|^2=ne^{2\rho}.
 	\end{equation}

 	When $1<p\leq 2$, we have
 	\begin{equation}\label{eq3.8}
 	|\Phi^A|^2\leq |\Phi^A|^p,
 	\end{equation}
 	Then by using \eqref{eq3.7}, \eqref{eq32}, \eqref{eq3.8} and the H\"{o}lder inequality, we have
 	\begin{align*}
 	\la_{1,p}\vol(M)&=\la_{1,p}\sum_{A=1}^{N+1}\int_M|\Phi^A|^2\\
 	&\leq\la_{1,p}\sum_{A=1}^{N+1}\int_M|\Phi^A|^p\leq\int_M\sum_{A=1}^{N+1}|\nabla\Phi^A|^p\\
 	&\leq (N+1)^{1-p/2}\int_M\Big(\sum_{A=1}^{N+1}|\nabla\Phi^A|^2\Big)^{p/2}\\
 	&=(N+1)^{1-p/2}\int_M(ne^{2\rho})^{p/2}.
 	\end{align*}
 This is \eqref{eq3.3}.

 When $p\geq 2$, the H\"{o}lder inequality gives
 \begin{equation}\label{eq3-6}
 	1=\sum_{A=1}^{N+1}|\Phi^A|^2\leq (N+1)^{1-\frac 2p}(\sum_{A=1}^{N+1}|\Phi^A|^p)^{2/p},
 \end{equation}
 from which we have
 \begin{equation}\label{eq3-7}
 	\lambda_{1,p}\vol(M)\leq (N+1)^{\frac p2-1}\Big(\sum_{A=1}^{N+1}\lambda_{1,p}\int_M|\Phi^A|^p\Big).
 \end{equation}
On the other hand, we have
\begin{equation}\label{eq3-8}
	\sum_{A=1}^{N+1}|\nabla\Phi^A|^p\leq(\sum_{A=1}^{N+1}|\nabla\Phi^A|^2)^{p/2}=(ne^{2\rho})^{p/2}.
\end{equation}
Hence \eqref{eq3.3} follows from \eqref{eq3.7}, \eqref{eq3-7} and \eqref{eq3-8}.
 \end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]

When $1<p \le 2$,  then $p/2\leq 1$. Using Lemma \ref{lem3.2}  and  the  H\"{o}lder inequality,  we have
\begin{align*}
\la_{1,p}\vol(M)& \le (N+1)^{1-p/2}n^{p/2}\int_M(e^{2\rho})^{p/2}\\
&\leq (N+1)^{1-p/2}n^{p/2}\big(\vol(M)\big)^{1-p/2}\Big(\int_Me^{2\rho}\Big)^{p/2}.
\end{align*}

Note that we can compute $e^{2\rho}$ using the conformal relations and Gauss equations as follows. 	We take $\bar{M}=\F,\ \tg=h_c, \ \tbg=\Gamma^*h_1$ in Subsection 2.2.
From \eqref{gauss3}, the Gauss equations for the immersion $x$ and the immersion $\Phi=\Gamma\circ x$ are respectively:
\begin{align}
R=&n(n-1)c+n(n-1)H^2+(nH^2-S),\label{G1}\\
\tR=&n(n-1)+n(n-1)\tH^2+(n\tH^2-\tilde{S}).\label{G2}
\end{align}
Contracting \eqref{eq209} we have
\begin{equation}\label{eq_scal}
e^{2\rho}\tR=R-(n-2)(n-1)|\nabla \rho|^2-2(n-1)\Delta\rho.
\end{equation}

Now from \eqref{G1}, \eqref{G2},  \eqref{eq_scal},  \eqref{eq208} and  \eqref{eq_um}, we derive
\begin{align*}
&n(n-1)(e^{2\rho}-c)+n(n-1)\sum_{\alpha}(H^\alpha-\rho_{\alpha})^2-n(n-1)H^2\\
=&-(n-2)(n-1)|\nabla \rho|^2-2(n-1)\Delta\rho,
\end{align*}
divided by $n(n-1)$ on both sides, we obtain 	\begin{equation}\label{eq_relation}
	e^{2\rho}=(c+|\mathbf{H}|^2)-\frac{2}{n}\Delta \rho-\frac{n-2}{n}|\nabla\rho|^2-|(\bar{\nabla}\rho)^\bot-\mathbf{H}|^2.
\end{equation}
Hence,
\begin{align*}
\la_{1,p}\vol(M)
&\leq (N+1)^{1-p/2}n^{p/2}\big(\vol(M)\big)^{1-p/2}\Big(\int_Me^{2\rho}\Big)^{p/2}\\
&\leq (N+1)^{1-p/2}\frac{n^{p/2}}{\big(\vol(M)\big)^{p/2-1}}\Big(\int_M\big(c+|\mathbf{H}|^2\big)\Big)^{p/2},
\end{align*}
which is equivalent to \eqref{eq_thm1}.

	When $p>2$, we cannot use $\int_M(e^{2\rho})$ to control $\int_M(e^{2\rho})^{p/2}$ by applying H\"{o}lder inequality directly.
	Instead multiplying $e^{(p-2)\rho}$ on both sides of \eqref{eq_relation}, and then integrating on $M$ (cf. \cite{CL11}), we obtain
\begin{equation}\label{eq4.1}
\int_Me^{p\rho}\leq\int_M(c+|\mathbf{H}|^2)e^{(p-2)\rho}-\int_M\frac{n-2-2(p-2)}{n}e^{(p-2)}|\nabla\rho|^2\leq\int_M(c+|\mathbf{H}|^2)e^{(p-2)\rho}.
\end{equation}
where we used the assumption $n\geq 2p-2$.

On the other hand, by Young's inequality, we have
\begin{equation}\label{eq4.2}
\int_M(c+|\mathbf{H}|^2)e^{(p-2)\rho}\leq\frac{1}{p/2}\int_M(|c+|\mathbf{H}|^2|)^{p/2}+\frac{1}{p/(p-2)}\int_Me^{p\rho}.
\end{equation}
Hence we obtain
\begin{equation}\label{eq4.3}
	\int_Me^{p\rho}\leq\int_M(|c+|\mathbf{H}|^2|)^{p/2}.
\end{equation}
from \eqref{eq4.1} and \eqref{eq4.2}. Putting \eqref{eq4.3} into \eqref{eq3.3}, we obtain \eqref{eq_thm2}.

Now we check the equality case. When the equality holds in \eqref{eq_thm1},  by checking \eqref{eq3.7} and \eqref{eq3.8},  we must have $|\Phi^A|^2=|\Phi^A|^p$
and $\pl \Phi^A=-\lambda_{1,p} |\Phi^A|^{p-2}\Phi^A$
for each $A=1,\cdots, N+1$. If $1<p<2$, then $|\Phi^A|=0$ or $1$. But $\sum_{A=1}^{N+1}|\Phi^A|^2=1$,  so there is exactly one $A$ such that $|\Phi^A|=1$ and then $\lambda_{1,p}=0$, which is a contradiction.  Hence $p=2$ and it reduces to the Laplacian case. We can complete the proof by using Theorem \ref{thm_Rei}.

When equality holds in \eqref{eq_thm2}, suppose $p>2$,  then \eqref{eq3-7} and \eqref{eq3-8} must become the equalities, which means
\begin{equation*}
	|\Phi^1|^p=\cdots=|\Phi^{N+1}|^p,
\end{equation*}
and there exists some $A$ such that $|\nabla\Phi^A|=0$. So $\Phi^A$ is constant and then $\lambda_{1,p}=0$, which is a contradiction.

\end{proof}

\end{document}

config: arxiv


TITLE: Why do we multiply to cancel out?
QUESTION [0 upvotes]: Say we have $\frac{g}{4} = 3.2$ why do we multiply each side by 4 to get g by itself? I don’t get why it doesn’t stay $\frac{g}{4}$ to be honest

REPLY [0 votes]: If $g$ is the price of a pizza, then your equation could describe the cost of a quarter of that pizza (a slice).
That slice costs $3$ dollars and $20$ cents.
How much does a whole pizza cost? (It's an expensive 100% Italian bufalina)


TITLE: Is the polarization of light changed by gravity?
QUESTION [2 upvotes]: The Gravitational_redshift shows, that the wavelength of light gets altered in a gravitational field. But what about polarization of light? I imagine that e.g. by tidal forces circular polarized light could get deformed into elliptic polarized light. Is this possible? Or is polarization of light an invariant quantity in a gravitational field?

REPLY [4 votes]: Fermi transport will be the equations that describe how each polarisation component changes direction, as the ray passes through the gravitational field (or rather, curved spacetime). So for circular polarisation to become eliptical, you need to check whether othogonal components cease to be orthogonal (when Fermi-Walker transported along an arbitrary trajectory). It turns out that inner products are preserved (so the angle of polarisation may be altered depending on the trajectory through space-time, but the ellipticity shouldn't change in vacuum). However, this assumes the short-wavelength (ray-optics) limit, so you might get a more complex answer by considering extremely low frequencies.


TITLE: Factorials in Pascals Triangle
QUESTION [36 upvotes]: Hi,
I asked this question of Keith Conrad, and he suggested that I try posting here.  One of my students observed that the only instances of factorials in the interior of Pascal's triangle are $\binom{4}{2}=3!$ and $\binom{10}{3}=\binom{16}{2}=5!$.  I checked the first 500 rows, and he's right up to that point.
This is a special case of the apparently unsolved problem of finding non-trivial solutions to n!=a!b!c!...  The special feature here is that I need (a+b)!=a!b!c!, and that seems like a special enough case to have been treated by someone.  Unfortunately, literature searches have been fruitless, because every paper about Pascal's triangle contains the word "factorial" somewhere.
My best idea (which I can't make work) is to show that the powers of 7 in the equation (a+b)!=a!b!c! can't be made to match up unless neither side is a multiple of 7.  Then exhaustive searching can show that the above are the only non-trivial solutions.
Thanks greatly for any ideas that anyone might have.

REPLY [8 votes]: The latest issue of the Journal of Number Theory (February 2016 issue) contains a paper by Nair and Shorey which gives a conditional resolution of this problem.  They consider the problem of finding solutions to
$$
n! = a_1! a_2! \cdots a_t!,
$$
with $n>a_1 \ge a_2 \ge \ldots a_t>1$ and $t>1$.   Obviously we can assume that $a_1 \le (n-2)$, for if $a_1 =n-1$ we are just looking at $n=a_2!\cdots a_t!$.
Then Theorem 4 of the cited paper shows that assuming Baker's explicit $abc$-conjecture, the only solutions to this equation are
$$
7!3!3!2!=9!; 7!6!=10!; 14!5!2!=16!.
$$
Here the explicit $abc$ conjecture of Baker states that if $a$, $b$ and $c$ are coprime positive integers with $a+b=c$ and $\omega$ denotes the number of distinct prime factors of $abc$ then
$$
c< \frac{6}{5} N \frac{(\log N)^{\omega}}{\omega!},
$$
with
$$
N = \prod_{p|abc} p
$$
denoting the radical of $abc$.


TITLE: Measure given by an integral
QUESTION [0 upvotes]: Find $\mu( \mathbb Q \cap [0,1] )$ and $\mu( \mathbb Q \cup [0,1] )$, where $\mathcal L(\mathbb R)$ is the Lebesgue sigma algebra and the function $\mu: \mathcal L(\mathbb{R})\to [0,\infty]$ maps $A \mapsto \int_{A} x^2  \ d \lambda(x)$.
Just how can I plug the sets in the integral? Thanks

REPLY [1 votes]: Here are two useful facts:

If $A$ is a set of measure zero and $f$ is any measurable function, then
$ \int_A f \ d\lambda = 0.$
If $A = [a,b]$ is an interval and $f$ is Riemann integrable, then $\int_A f \ d\lambda$ is equal to the Riemann integral of $f$ over $[a,b]$.

Using the first fact, it should be possible to solve the case $A = \mathbb Q \cap [0,1]$ immediately.
For $A = \mathbb Q \cup [0,1]$, we could split $A$ into two disjoint subsets:
$$ A = A_1 \cup A_2,$$ $$ {\rm with \ \ } A_1 = [0,1] {\rm \ \ \ \ and \ \ \ \ } A_2 = \mathbb Q \cap \left( (-\infty, 0) \cup (1, \infty)\right).$$
Now observe that $A_1$ is an interval and $A_2$ is of measure zero...


TITLE: Eigenvalues for discretization matrix in Poisson equation with finite difference
QUESTION [0 upvotes]: I am trying to find the eigenvalues for the discretization matrix in the Poisson equation using the Chebyshev polynomials, i.e.
$$ -u''(x) = f(x), x \in [0,1],\;\; u(0)=u(1)=0 $$
Discretize the spatial domain uniformly: $x_i = ih$ with $h=1/m, i \in \{0,1,\ldots,m\}$. Let
$$A = \begin{bmatrix} 2&-1&0&\cdots&0\\
                      -1&2&-1&\cdots&0\\
\cdots&\cdots&-1&2&-1\\
0&0&0&2&-1\\
\end{bmatrix}$$
Then the solution is approximated in finite difference by solving
$$Au = f$$
My question is how to find the eigenvealues of A using the Chebyshev polynomials. I can show the recurrence relation for $u_{k+1}$ is a special case of the Chebyshev polynomials, i.e
$$A[u^p_1,u^p_2, \cdots, u^p_{m-1}]^T \Rightarrow (A u^p)_j = -\frac{1}{h^2} \left(u_{j-1}^p -2u_j^p + u_{j+1}^p  \right) = \lambda_p u_j^p$$
$$\Rightarrow u_{j+1}^p = (2-h^2 \lambda_p)u_j - u_{j-1}$$
is a special case of
$$U_{j+1}(x) = 2xU_j(x)-U_{j-1}(z)$$
by setting $x = 1 - \frac{h^2 \lambda_p}{2}$. So with representation of the Chebyshev polynomial, I can have
$$ u_j= U_j(1 - \frac{h^2 \lambda_p}{2}) =  \frac{\sin\left[(j+1)\cos^{-1}\left(1 - \frac{h^2\lambda_p}{2}\right)\right]}{\sqrt{1 - \left(1 - \frac{h^2\lambda_p}{2}\right)^2}}$$
But how can we find the the eigenvalue from the above equation? We know that for Chebyshev polynomials $U_0(x)=1$, but I don't think $u_0 = 0 \neq U_0(1 - \frac{h^2 \lambda_p}{2})$.
Any help/hints will be greatly appreciated.
Thanks.

REPLY [0 votes]: I am not sure that your recurrence relation can be deal with by Chebychev polynomials. Indeed the term in front of the $u_j$ should be dependent on $j$ for that.
Rather, from the solutions of the continuous eigenvalue problem you could look for an eigenvector/value by plugging $v_p=e^{iqp}$ in the expression $-v_{p+1}+2v_p-v_{p-1}$ and get
$$-v_{p+1}+2v_p-v_{p-1}=(2-e^{iq}-e^{-iq})v_p=2(1-\cos q)v_p=4\sin^2(q/2)v_p$$
To fulfil boundary conditions, set $w_p=\sin(qp)=\frac{1}{2i}(v_p-v_{-p})$ which verifies $w_0=w_{m}=0$ if $q=\frac{k\pi}{m}$ with $k=1,\ldots,m-1$.
Then $w_p$ verifies $w_0=w_{m}=0$ and
$$-w_{p+1}+2w_p-w_{p-1}=4\sin^2(\frac{k\pi}{2m})w_p,\qquad k=1,\ldots,m-1$$
which is exactly $Aw=4\sin^2(\frac{k\pi}{2m})w$ (your matrix $A$ is of size $m-1$).
Thus you got your eigenvectors and values.


TITLE: Evaluating the indefinite integral $\int \frac{2x+3} {{x^2-2x-3}}\mathrm{d}x$
QUESTION [0 upvotes]: Evaluating an indefinite integral it is simple.
Hello it is simple but I can not. Can you help me? evaluate the following integral:
$$\int \frac{2x+3} {{x^2-2x-3}}\mathrm{d}x$$
Give me show step by step solutions please.
Thank you very much.

REPLY [0 votes]: Hint : computational purpose only

$ \int \frac {px+q}{ax^2+bx+c}dx=\frac{p}{2a}log|{ax^2+bx+c}|+(q-\frac{pb}{2a})log|\frac{2ax+b-\sqrt{\Delta}}{2ax+b+\sqrt{\Delta}}|$

note: use this result only for $b^2-4ac>0$

In your problem

$a=1,b=-2,c=-3,p=2,q=3,\Delta=16$
$\int \frac {2x+3}{x^2-2x-3}dx$
$=\frac{2}{2(1)}log|x^2-2x-3|+(3-\frac{(2)(-2)}{2(1)})\frac{1}{\sqrt{16}}log|\frac{2x-2-\sqrt{16}}{2x-2+\sqrt{16}}|$
$=1log|x^2-2x-3|+\frac{5}{4}log|\frac{x-3}{x+1}|+c$
Where "c" is integral constant


TITLE: If $\varphi(a) = \varphi(b)$ and $K=\ker \varphi$ why does $aK = bK$?
QUESTION [0 upvotes]: If $\varphi(a) = \varphi(b)$ and $K = \ker \varphi$
then $a^{-1}b \in K$ and $b^{-1}a \in K$
where $aK$ is left coset such that $aK = \{ ak | k \in K \}$
then $b \in aK$ and $a \in bK$
I follow that so far. Then, my textbook says
$aK=bK$
Why is this so? Can someone do a simple proof?

REPLY [0 votes]: Hint: For any subgroup $H$ and $h\in H$, $H=\{hh': h'\in H\}$.  So, $K=\{a^{-1}bk: k\in K\}$.


TITLE: Prove that the locus of the incenter of the $\Delta PSS'$ is an ellipse of eccentricity $\sqrt{\frac{2e}{1+e}}$
QUESTION [1 upvotes]: Let $S$ and $S'$ be the foci of an ellipse whose eccentricity is $e$.$P$ is a variable point on the ellipse.Prove that the locus of the incenter of the $\Delta PSS'$ is an ellipse of eccentricity $\sqrt{\frac{2e}{1+e}}$.
Let $P$ be $(a\cos \theta,b \sin \theta)$.  Let the incenter of the triangle $PSS'$ be $(h,k)$.  The formula for the incenter of a triangle whose side lengths are $a,b,c$ and whose vertices have coordinates $\left(x_1,y_1\right)$, $\left(x_2,y_2\right)$, and $\left(x_3,y_3\right)$ is $$\left(\frac{ax_1+bx_2+cx_3}{a+b+c},\frac{ay_1+by_2+cy_3}{a+b+c}\right)\,.$$
Then,
$$h=\frac{2c\cdot a\cos \theta+PS'\cdot c-PS\cdot c}{2c+PS'+PS}\text{ and }k=\frac{2c\cdot a\sin \theta}{2c+PS'+PS}\,,$$
but I could not find the relationship between $h$ and $k$, whence I could not find the eccentricity of this ellipse.

REPLY [0 votes]: Just separate cos and sin from your incentre and square it then add it you get your locus and eccentricity


TITLE: Multiplying TRS Affine Transformations
QUESTION [0 upvotes]: To represent 2D affine transformations I'm using translation, rotation and scale. When I transform a vector I compose the TRS matrix, that is I scale then rotate and finally translate the vector.
That is, given $T, R, S$
$$
T = \begin{bmatrix} t_1 \\ t_2 \end{bmatrix} \\
R = \alpha \\
S = \begin{bmatrix} s_1 \\ s_2 \end{bmatrix}
$$
where $s_1 \neq 0$ and $s_2 \neq 0$, I compose the matrix
$$
M(T,R,S) =
\begin{bmatrix}
s_1 \cos \alpha & -s_2 \sin \alpha & t_1 \\
s_1 \sin \alpha & s_2 \cos \alpha & t_2 \\
0 & 0 & 1
\end{bmatrix}
$$
and finally use matrix multiplication to transform the vector.
The problem I'm having is when decomposing several $T, R, S$ transformations. With the case the scale is uniform ($s_1 = s_2$) I can decompose two transformations as
$$
M(T,R,S) = M(T',R',S') \cdot M(T'',R'',S'') \\
T =
\begin{bmatrix}
s'(t''_1\cos \alpha'- t''_2\sin \alpha') + t'_1 \\
s'(t''_1\sin \alpha'+ t''_2\cos \alpha') + t'_2
\end{bmatrix} \\
R = \alpha' + \alpha'' \\
S = \begin{bmatrix} s's'' \\ s's'' \end{bmatrix}
$$
however this only works when scale is uniform.
Is it possible to decompose the above $M(T,R,S)$ matrix to $T,R,S$ form when scale is non-uniform?
When I try it I only end up trying to simplify $a \cos \alpha \cos \beta - b \sin \alpha \sin \beta, b \neq a$ to the form $a' \cos \alpha'$ with no success.

REPLY [1 votes]: Seems it's possible that scaling, rotating and repeating can add shear so José was right, with non-uniform scale it's not possible.


TITLE: $xy=zw$ substitution
QUESTION [4 upvotes]: In Olympiad math context (mainly AoPS), I often see substitution $(x,\ y,\ z,\ w)=(ab,\ cd,\ ad,\ bc)$ under $xy=zw$.
The following is a construction of $(a,b,c,d)$ which I came up with.

For a prime $p$ and nonzero integer $n$, we let $\nu_p(n)$ denote the largest integer $e$ with $p^e$ dividing $n$.
For an arbitrary prime $p$, let $$(s,\ t,\ u,\ v)=(\nu_p(x),\ \nu_p(y),\ \nu_p(z),\ \nu_p(w))$$ We have $s+t=u+v$.
Without loss of generality we may assume $s=\min\{s,t,u,v\}$.
We let $(a,b,c,d)$ satisfy
\begin{eqnarray}
\nu_p(a)&=&s \\
\nu_p(b)&=&0 \\
\nu_p(c)&=&u \\
\nu_p(d)&=&v-s
\end{eqnarray}
Now we have $\nu_p(a)+\nu_p(b)=\nu_p(x)$, $\nu_p(c)+\nu_p(d)=\nu_p(y)$, etc.
We repeat this operation and the construction is done.

The question is:
Does this substitution have a particular name that one can easily refer to?

REPLY [1 votes]: Your property is related to equidivisibility, which is defined in a more general context.
A monoid $M$ is said to be equidivisible if for each quadruple $(x, y, z, t) \in M^4$ satisfying $xy = zw$, there exists $h \in M$ such that either $x = zh$ and $hy = t$ or $xh = z$ and $y = ht$.
Thus either
$$
(x, y, z, t) = (ab, cd, ad, bc),
$$
with $a = z$, $b = h$, $c = y$ and $d = 1$, or
$$
(x, y, z, t) = (ab, dc, ad, bc),
$$
with $a = x$, $b = 1$, $c = t$ and $d = h$. Note the difference between the terms $cd$ and $dc$, which of course disappears in a commutative setting.
Levi's lemma shows that any free monoid is equidivisible.
[1] McKnight, J. D., Jr. and Storey, A. J., Equidivisible semigroups, J. Algebra 12 (1969), 24-48.


text:
\begin{document}

\title[Hele-Shaw flow]
{Hele-Shaw flow on weakly hyperbolic surfaces}

\author[Hedenmalm]
{H\aa{}kan Hedenmalm}

\address{Hedenmalm: Department of Mathematics\\
The Royal Institute of Technology\\
S -- 100 44 Stockholm\\
Sweden}

\email{haakanh@math.kth.se}

\thanks{Both authors wish to thank the G\"oran Gustafsson
Foundation for generous support.}

\author[Olofsson]{Anders Olofsson}
\address{Olofsson: Department of Mathematics\\
The Royal Institute of Technology\\
S -- 100 44 Stockholm\\
Sweden}
\email{ao@math.kth.se}

\subjclass{Primary 35R35, 35Q35; Secondary 31A05, 31C12, 53B20, 76D27}

\keywords{Hele-Shaw flow, biharmonic Green function, mean value property}

\begin{abstract}
We consider the Hele-Shaw flow that arises from injection of two-dimensional
fluid into a point of a curved surface. The resulting fluid domains have
and are more or less determined implicitly by a mean value property for
harmonic functions. We improve on the results of Hedenmalm and Shimorin
\cite{HS} and obtain essentially the same conclusions while imposing a
weaker curvature condition on the surface. Incidentally, the curvature
condition is the same as the one that appears in Hedenmalm and Perdomo's
paper \cite{HP}, where the problem of finding smooth area minimizing surfaces
for a given curvature form under a natural normalizing condition was
considered. Probably there are deep reasons behind this coincidence.
\end{abstract}

\maketitle

\section{Introduction}

Let ${\mathbf\Omega}$ be a simply connected Riemann surface with a
$C^\infty$-smooth metric $\diff{\mathbf s}$.
We consider the case when the surface ${\mathbf\Omega}$ is conformally
equivalent to a simply connected planar domain $\Omega$, which we pick to be
the unit disk $\Omega=\D$, and the metric $\diff{\mathbf s}$ is given by
\begin{equation}\label{metric}
\diff{\mathbf s}(z)^2=\omega(z)\,\vert \diff z\vert^2,\qquad z\in\Omega,
\end{equation}
for some positive $C^\infty$-smooth weight function $\omega$ in $\Omega$.
This means that we use {\sl isothermal coordinates}, which is basically
always possible;
the only real restriction imposed is that we do not allow ${\mathbf\Omega}$ to
be conformally  equivalent to the sphere or the plane. The {\sl area form}
of the surface ${\mathbf\Omega}$ is (with suitable normalization)
$$\diff {\mathbf\Sigma}(z)=\omega(z)\,\diff\Sigma(z),\qquad z\in\Omega=\D,$$
where
$$\diff\Sigma(z)=\frac{\diff x\diff y}{\pi},\qquad z=x+\imag y,$$
is the normalized area element in the plane.

The Gaussian curvature function associated to the metric $\diff{\bf s}$
given by (\ref{metric}) is the function ${\boldsymbol\kappa}$ defined by
$$
{\boldsymbol\kappa}(z)=
-\frac{2}{\omega(z)}\,\Delta(\log\omega)(z),\qquad z\in\D,
$$
where
$$\Delta=\Delta_z=\frac{\partial^2}{\partial z\partial\bar z}
=\frac14\,\bigg(\frac{\partial^2}{\partial x^2}+
\frac{\partial^2}{\partial y^2}\bigg),\qquad z=x+\imag y,$$
is the (normalized) Laplacian in the plane. The curvature form associated
to the metric $\diff{\bf s}$ given by (\ref{metric}) is the $2$-form
${\mathbf K}$ defined by
$$
{\mathbf K}(z)={\boldsymbol\kappa}(z)\,\diff{\mathbf\Sigma}(z)=
-2\,\Delta(\log\omega)(z)\,\diff\Sigma(z),\qquad z\in\D.
$$

In the recent paper \cite{HP}, Hedenmalm and Perdomo studied the following
problem: Suppose the curvature form $\mathbf K$ is given and
$C^\infty$-smooth on the closed disk $\bar\D$, while the metric
$\diff{\mathbf s}$ is to be found. If we ask of this metric which we are
looking for that it is $C^\infty$-smooth on $\bar\D$ and has the
{\sl mean value property}
\begin{equation}
h(0)=\int_\D h(z)\,\diff {\mathbf \Sigma}(z)=\int_\D h(z)\,\omega(z)\,
\diff {\Sigma}(z)
\label{eq-meanvalue}
\end{equation}
for all bounded harmonic functions $h$ in $\D$, can we then find such a
metric, and is it unique? It turns out that this is indeed so if the condition
\begin{equation}\label{curvaturecond}
{\mathbf K}(z)+\frac{1}{2}\,{\mathbf K}_\mathbb{H}(z)\leq0,\qquad z\in\D,
\end{equation}
is met, where ${\mathbf K}_\mathbb{H}$ denotes the curvature form for
the surface $\D$ when equipped with the (Poincar\'e) metric
of constant curvature $-1$:
$${\mathbf K}_\mathbb{H}(z)=-\frac{4\,\diff\Sigma(z)}{(1-\vert z\vert^2)^2},
\qquad z\in\D.$$
Moreover, the result is fairly sharp, in the following sense: If the number
$\frac12$ is replaced by the slightly larger number $0.52$ in condition
(\ref{curvaturecond}), the resulting condition is too weak to force the
existence of a $C^\infty$-smooth bordered surface $\mathbf\Omega=\langle
\D,\diff{\mathbf s}\rangle=\langle\D,\omega\rangle$ with the mean value
property for harmonic functions and the given curvature form
${\mathbf K}$. It was conjectured in \cite{HP} that the numerically
obtained number $0.52$ could be replaced by $\frac12+\varepsilon$, for any
positive $\varepsilon$.
We shall at times refer to (\ref{curvaturecond}) as the condition of
{\sl weak hyperbolicity}.

It is a natural question to ask whether there are subdomains $D(t)$, depending
on the parameter $t$, $0<t<1$, with the modified mean value property
\begin{equation}
t\,h(0)=\int_{D(t)} h(z)\,d{\mathbf \Sigma}(z)=\int_{D(t)} h(z)\,\omega(z)\,
d{\Sigma}(z)
\label{eq-meanvalue-t}
\end{equation}
for all bounded harmonic functions $h$ in $D(t)$. We make the {\sl a priori}
assumption that the domains $D(t)$ are simply connected with $C^\infty$-smooth
boundaries. Then, if we look at the curvature condition (\ref{curvaturecond})
and use the fact that the curvature form for the Poincar\'e metric decreases
as the domain gets smaller, we see that the analogous curvature condition
for ${\mathbf\Omega}(t)=\langle D(t),\omega\rangle$ is fulfilled, so by the
Hedenmalm-Perdomo theorem, the surfaces ${\mathbf\Omega}(t)$ exist
as abstract surfaces, being determined uniquely by the given curvature form
and the mean value property (\ref{eq-meanvalue-t}).
However, this does not tell us that the abstract surface
${\mathbf\Omega}(t)$ forms a subregion of $\mathbf\Omega$. It would be
desirable to know that this is so.
If this works, then perhaps it will eventually be possible to understand the
Hedenmalm-Perdomo theorem in terms of the following growth process:

\noindent (1) for $t$ close to $0$, we form the surface ${\mathbf\Omega}(t)$,
by fairly elementary means;

\noindent (2) as $t$ increases, we iteratively add abstractly infinitesimally
thin boundary layers to the surface ${\mathbf\Omega}(t)$, while checking
that the curvature condition (\ref{curvaturecond}) permits us to do so.

Here, we shall obtain the existence of the smooth simply connected subdomains
$D(t)$ of $\D$ with (\ref{eq-meanvalue-t}) under the weak hyperbolicity
condition (\ref{curvaturecond})
in the category of {\sl real-analytic surfaces}. Moreover, we shall prove
that the boundary curves $\partial D(t)$ are real-analytically smooth, and
that we get an extra degree of analytic smoothness at the origin for
$t\to0$. The growth of the domains $D(t)$ is a certain kind of {\sl Hele-Shaw
flow}, modelling the growth of a patch of two-dimensional fluid as fluid
is injected at the origin. We extend the Hele-Shaw exponential mapping theorem
obtained by Hedenmalm and Shimorin in \cite{HS} to weakly hyperbolic surfaces.
From the point of view of differential geometry, it is perhaps surprising that
this is possible, because it does not work globally for the ordinary
exponential mapping. After all, it is rather easy to construct a
real-analytically smooth surface ${\mathbf\Omega}=\langle\D,\omega\rangle$
whose curvature form $\mathbf K$ satisfies (\ref{curvaturecond}) while two
geodesics emanating from the origin meet at some other point; we supply an
example of such a situation in Section~\ref{exwhs}. Probably this discrepancy
can be explained by the viscosity of the fluid as compared with the lack
thereof in the photonic gas, which is the physical manifestation of the metric
flow.

The techniques used in the paper are basically similar to what was used
by Hedenmalm-Shimorin in \cite{HS}. The main difference is that the Green
function $\Gamma_1$ for the weighted biharmonic operator $\Delta(1-|z|^2)^{-1}
\Delta$ is used in place of the biharmonic Green function $\Gamma$. We
suspect that the main results of this paper are false if the constant
$\frac12$ in the condition (\ref{curvaturecond}) is replaced by
$\frac12+\varepsilon$, for any fixed positive $\varepsilon$. In another vein,
we suspect that the results obtained here for real-analytic surfaces will
remain valid for $C^\infty$-smooth surfaces as well.

\section{Preliminaries on Hele-Shaw domains}\label{prelHSdomains}

Let $\omega$ be a $C^\infty$-smooth weight function on $\bar\D$ which is
strictly positive (that is, it is positive at all points).
We denote by $G$ the Green function for the Laplacian $\Delta$ on $\D$:
\begin{equation*}
G(z,\zeta)=\log\left\vert\frac{z-\zeta}{1-\bar\zeta z}\right\vert^2,
\qquad(z,\zeta)\in\bar\D\times\D.
\label{Greenfn}
\end{equation*}
For $0<t<+\infty$, we consider the function
$$
V_t(z)=t\,G(z,0)-
\int_\D G(z,\zeta)\,\omega(\zeta)\,\diff\Sigma(\zeta),
\qquad z\in\D.
$$
This function $V_t$ is smooth in $\D\setminus\{0\}$, vanishes on
the unit circle $\T=\partial\D$, and has a logarithmic singularity at the
origin.
We denote by $\widehat{V}_t$ the smallest superharmonic majorant of $V_t$
in $\D$. The set $D(t)$ is now defined by
\begin{equation*}
D(t)=D(t;\omega)=
\Big\{z\in\D:\,\, V_t(z)<\widehat{V}_t(z)\Big\}.
\end{equation*}
The sets $D(t)$ are known to have a number of basic
properties (see~\cite{HS}).
For instance, the sets $D(t)$ are open, connected and increasing in the
positive parameter  $t$.

Associated to the domains $D(t)$, we consider the number
$$
T=T(\omega)=\sup\Big\{t\in]0,+\infty[:\  D(t)\Subset\D\Big\},
$$
the {\sl termination time} for the Hele-Shaw flow. It is known that
$0<T\leq+\infty$ (see~\cite[Proposition~2.8]{HS}).
For $0<t<T$, the domain $D(t)$ is called a {\sl Hele-Shaw domain}.
For $t\geq T$, the domain $D(t)$ is sometimes referred to as a
{\sl generalized Hele-Shaw domain}.

\section{A weighted biharmonic Green function}\label{wbhGfcn}

In this section we shall need some properties of the Green function
for the weighted biharmonic operator
$\Delta(1-\vert z\vert^2)^{-1}\Delta$.
We review the results needed and refer to~\cite{HP} for details.

The Green function for the weighted biharmonic operator
$\Delta(1-\vert z\vert^2)^{-1}\Delta$ is the
function $\Gamma_1$ on $\bar\D\times\D$
solving (in an appropriate sense) for fixed $\zeta\in\D$
the boundary value problem
\begin{equation}
\left\{
\begin{array}{rclr}
\Delta_z(1-\vert z\vert^2)^{-1}\Delta_z\Gamma_1(z,\zeta)&=&
\delta_\zeta(z)&\quad z\in\D,\\
\Gamma_1(z,\zeta)&=&0 & z\in\T,\\
\nabla_z\Gamma_1(z,\zeta)&=&0 & z\in\T;
\end{array}
\right.
\end{equation}
here $\delta_\zeta$ is the (unit) Dirac mass at $\zeta$ and
$\T=\partial\D$ is the unit circle.
The Green function $\Gamma_1$ has the explicit expression
\begin{align}\notag
\Gamma_1(z,\zeta)&=
\bigg\{\vert z-\zeta\vert^2-
\frac{1}{4}\,\big\vert z^2-\zeta^2\big\vert^2\bigg\}G(z,\zeta)+
\frac{1}{8}\big(1-\vert z\vert^2\big)\big(1-\vert\zeta\vert^2\big)\\
&\times\bigg\{7-\vert z\vert^2-\vert\zeta\vert^2-\vert z\zeta\vert^2-
4\re(z\bar\zeta)-2\big(1-\vert z\vert^2\big)\big(1-\vert\zeta\vert^2\big)
\frac{1-\vert z\zeta\vert^2}{\vert1-z\bar\zeta\vert^2}\bigg\},
\label{G1dfn}
\end{align}
where
$$
G(z,\zeta)=\log\left\vert\frac{z-\zeta}{1-\bar\zeta z}\right\vert^2,
\qquad(z,\zeta)\in\bar\D\times\D,
$$
is the Green function for the Laplacian in $\D$.
We mention that the above
explicit expression for $\Gamma_1$ first appeared in~\cite{Hprobl}.
A derivation of formula (\ref{G1dfn}) can be found in~\cite{W2}.

It is known that the Green function $\Gamma_1$ is positive in the bidisk:
$$
\Gamma_1(z,\zeta)>0,\qquad(z,\zeta)\in\D\times\D;
$$
see \cite[Proposition~3.4]{Hprobl} or \cite[Lemma~2.2]{HP}.

The function $H_1$ defined by
$$
H_1(z,\zeta)=\big(1-\vert\zeta\vert^2\big)
\bigg\{\frac{1}{2}\big(3-\vert\zeta\vert^2\big)
\frac{1-\vert z\zeta\vert^2}{\vert1-z\bar\zeta\vert^2}+
\big(1-\vert\zeta\vert^2\big)\re\bigg[\frac{z\bar\zeta}
{(1-z\bar\zeta)^2}\bigg]\bigg\},
$$
for $(z,\zeta)\in\bar\D\times\bar\D$ with $z\neq\zeta$, has been coined the
{\sl harmonic compensator}; it is positive on $\bar\D\times\bar\D$.
Indeed, by harmonicity in $z$, it is enough to consider
$(z,\zeta)\in\T\times\D$, in which case we have that
\begin{align*}
H_1(z,\zeta)&\geq\big(1-\vert\zeta\vert^2\big)
\bigg\{\frac{1}{2}\big(3-\vert\zeta\vert^2\big)
\frac{1-\vert\zeta\vert^2}{\vert 1-z\bar\zeta\vert^2}-
\big(1-\vert\zeta\vert^2\big)
\frac{\vert\zeta\vert}{\vert 1-z\bar\zeta\vert^2}\bigg\}=\\
&=\frac{1}{2}\,\frac{(1-\vert\zeta\vert^2)^2}{\vert1-z\bar\zeta\vert^2}
\big(1-\vert\zeta\vert\big)\big(3+\vert\zeta\vert\big)>0.
\end{align*}
The functions $\Gamma_1$ and $H_1$ are related by the formula
\begin{equation}\label{lplG1}
\Delta_z\Gamma_1(z,\zeta)=
\big(1-\vert z\vert^2\big)\big(G(z,\zeta)+H_1(z,\zeta)\big),
\qquad(z,\zeta)\in\bar\D\times\D,
\end{equation}
which can be verified by straightforward computation (see~\cite{HP}).

The following lemma establishes an integral representation
in terms of the functions $\Gamma_1$ and $H_1$.

\begin{lem}\label{reprformula}
Let $u$ be a smooth function on $\bar\D$ such that $u=0$ on $\T$.
Then $u$ admits the representation
$$
u(\zeta)=\int_\D\Gamma_1(z,\zeta)\,\Delta(1-\vert z\vert^2)^{-1}
\Delta u(z)\,\diff\Sigma(z)+\frac{1}{2}\int_\T H_1(z,\zeta)\,
\partial_n u(z)\,\diff\sigma(z),
\quad\zeta\in\D,
$$
where $\partial_n$ denotes differentiation in the
inward normal direction and $\diff\sigma$ is normalized arc length measure on
$\T$.
\end{lem}

A more general representation formula is planned to appear elsewhere.
Therefore we omit the proof of Lemma~\ref{reprformula}.
The lemma can also be obtained by repeated applications
of Green's formula.
We recall that with our normalizations Green's formula
(Green's second identity) takes the form
\begin{equation}\label{Greenformula}
\int_\Omega\big(u\Delta v-v\Delta u\big)\,\diff\Sigma=
\frac{1}{2}\int_{\partial\Omega}\big(v\partial_n u-
u\partial_nv\big)\,\diff\sigma,
\end{equation}
where $\diff\sigma=\vert\diff z\vert/2\pi$.

The following lemma is a simple consequence of the fact that $\Gamma_1$ is
positive. The technique is analogous to that which was developed for the
biharmonic Green function $\Gamma$ in \cite{DKSS}; see also \cite{H91,H92,HP}.

\begin{lem}\label{nucontremb}
Let $\nu$ be a smooth function on $\bar\D$ such that
$$z\mapsto\frac{\nu(z)}{1-\vert z\vert^2}$$
is subharmonic in $\D$.
Assume also that $\nu$ is reproducing at the origin in the sense that
$$
h(0)=\int_\D h(z)\,\nu(z)\,\diff\Sigma(z)
$$
holds for every harmonic polynomial $h$.
Then the inequality
$$
\int_\D u(z)\,2\big(1-\vert z\vert^2\big)\,\diff\Sigma(z)\leq
\int_\D u(z)\,\nu(z)\,\diff\Sigma(z)
$$
holds for every $u\in C^\infty(\bar\D)$ which is subharmonic in $\D$.
\end{lem}

\begin{proof}
We consider the function $\Phi$ defined by
$$
\Phi(z)=
\int_\D G(z,\zeta)\,\big[\nu(\zeta)-2\big(1-\vert\zeta\vert^2\big)\big]\,
\diff\Sigma(\zeta),
\qquad z\in\D,
$$
where $G$ is the Green function for the Laplacian in $\D$.
By construction $\Phi\vert_\T=0$.
Since $\Delta\Phi=\nu-2(1-\vert z\vert^2)$ annihilates harmonic polynomials,
an application of Green's formula (\ref{Greenformula})
shows that also $\partial_n\Phi\vert_\T=0$.
By Lemma~\ref{reprformula}, the function $\Phi$ has the representation
\begin{multline*}
\Phi(z)=\int_\D\Gamma_1(z,\zeta)\,
\Delta_\zeta(1-\vert\zeta\vert^2)^{-1}
\Delta_\zeta\Phi(\zeta)\,\diff\Sigma(\zeta)\\
=\int_\D\Gamma_1(z,\zeta)\,
\Delta_\zeta\bigg[\frac{\nu(\zeta)}{1-\vert\zeta\vert^2}\bigg]\,
\diff\Sigma(\zeta)\geq0.
\end{multline*}
Another application of Green's formula now shows that
\begin{multline*}
\int_\D u(z)\big[\nu(z)-2\big(1-\vert z\vert^2\big)\big]\,\diff\Sigma(z)=
\int_\D u(z)\,\Delta\Phi(z)\,\diff\Sigma(z)\\=
\int_\D\Delta u(z)\,\Phi(z)\,\diff\Sigma(z)\geq0,
\end{multline*}
which concludes the proof.
\end{proof}

In the proof of the next lemma, we shall use some easy properties
of Bergman kernel functions. Let us denote by $A^2_\alpha(\D)$, for
$-1<\alpha<\infty$, the weighted Bergman space of all analytic functions in
$\D$ that are square integrable with respect to the measure
$$\diff \Sigma_\alpha(z)=\omega_\alpha(z)\,\diff \Sigma(z),$$
where the weight $\omega_\alpha$ is as follows:
$$
\omega_\alpha(z)=(1+\alpha)\big(1-\vert z\vert^2\big)^\alpha,\qquad z\in\D.
$$
The Bergman kernel function for the space $A^2_\alpha(\D)$ is the function
$$
K_\alpha(z,\zeta)=\frac{1}{\big(1-z\bar\zeta\big)^{2+\alpha}},
\qquad(z,\zeta)\in\D\times\D;
$$
see~\cite[Proposition~1.1.4]{HKZ}. The reproducing property of the kernel
function $K_\alpha$ asserts that
$$
f(z)=\int_\D K_\alpha(z,\zeta)\,f(\zeta)\,\omega_\alpha(z)\,
\diff\Sigma(\zeta),\qquad z\in\D,
$$
for $f\in A^2_\alpha(\D)$.

The following lemma is inspired by~\cite[Proposition~3.2]{HP}.

\begin{lem}\label{nuonT}
Let $\nu$ be a nonnegative smooth weight function satisfying
the assumptions of Lemma~\ref{nucontremb}.
Let $e^{\imag\theta}\in\T$.
Then either $\nu(e^{\imag\theta})>0$, or
$\nu(e^{\imag\theta})=0$ and $\partial_n\nu(e^{\imag\theta})>0$.
\end{lem}

\begin{proof}
Without loss of generality, we may assume that $e^{\imag\theta}=1$. To
reach a contradiction, we assume that the conclusion of the lemma does not
hold. Then there is a positive constant $C$ such that
$$\nu(z)\leq C\vert z-1\vert^2,\qquad z\in\D.$$

Fix $r$ in the interval $0<r<1$. By Lemma~\ref{nucontremb} applied to the
function
$$u(z)=\big\vert K_1(z,r)\big\vert^2=\frac{1}{|1-rz|^6},$$
we have that
\begin{equation}\label{absurdineq}
\frac1{(1-r^2)^3}=\int_\D\big\vert K_1(z,r)\big\vert^2\,2(1-\vert z\vert^2)
\,\diff\Sigma(z)\leq
C\int_\D\big\vert K_1(z,r)\big\vert^2\,\vert z-1\vert^2\, \diff\Sigma(z).
\end{equation}
The leftmost equality in (\ref{absurdineq}) holds because of the
reproducing property of $K_1$. We turn to estimating the right hand side
in (\ref{absurdineq}). By the elementary inequality
$$r^2\vert z-1\vert^2\leq\vert 1-rz\vert^2,\qquad z\in\D,$$
we have that
$$
\int_\D\big\vert K_1(z,r)\big\vert^2\,\vert z-1\vert^2\, \diff\Sigma(z)\leq
\frac{1}{r^2}\int_\D\frac{1}{\vert 1-rz\vert^4}\,
\diff\Sigma(z)=\frac{1}{r^2(1-r^2)^2},
$$
where the last equality follows from the reproducing property of $K_0$.
Thus, by (\ref{absurdineq}), we have that
$$
\frac{1}{(1-r^2)^3}\leq\frac{C}{r^2(1-r^2)^2},
$$
which is clearly impossible for $r$ close to $1$.
\end{proof}

The next lemma is analogous to \cite[Lemma~5.2]{HS}.

\begin{lem}\label{wbhgestimate}
Let $u$ be a smooth real-valued function in $\bar\D\setminus\{0\}$
with a logarithmic singularity at the origin such that
$$
\Delta(1-\vert z\vert^2)^{-1}\Delta u=\Delta\delta_0-\mu\qquad\text{in}\ \D,
$$
where $0\leq\mu\in C^\infty(\D)$.
Suppose that $u\vert_\T=0$ and $\partial_n u\leq0$ on $\T$.
Then
$$
u(z)\leq\log\vert z\vert^2+\frac{3}{2}-2\vert z\vert^2+
\frac{1}{2}\vert z\vert^4<0,\qquad z\in\D.
$$
\end{lem}

\begin{proof}
Recall that the kernels $\Gamma_1$ and $H_1$  are both positive.
By the representation formula in Lemma~\ref{reprformula} we have that
$$
u(\zeta)\leq\int_\D\Gamma_1(z,\zeta)\,\Delta\delta_0(z)\,\diff\Sigma(z)=
\Delta_z\Gamma_1(0,\zeta)=\log\vert\zeta\vert^2+\frac{3}{2}-
2\vert\zeta\vert^2+\frac{1}{2}\vert\zeta\vert^4,
$$
where the integration is to be interpreted in the usual sense of
distribution theory. The proof is complete.
\end{proof}

\section{Simply connectedness of Hele-Shaw domains}\label{simplyconnHSdomain}

In this section we shall show that each Hele-Shaw domain $D(t)$, $0<t<T$,
is simply connected and has real-analytic Jordan boundary.
First we need a lemma.

\begin{lem}\label{Westimate}
Let $\nu$ be a strictly positive smooth function on $\bar\D$ such that
the function
$$z\mapsto\frac{\nu(z)}{1-\vert z\vert^2}$$
is subharmonic in $\D$. Let
$$
W(z)=\log\vert z\vert^2-\int_\D G(z,\zeta)\,\nu(\zeta)\,
\diff\Sigma(\zeta),
\qquad z\in\D,
$$
and denote by $\widehat{W}$ the smallest superharmonic majorant
of $W$ in $\D$.
Assume that the coincidence set
$$\Big\{z\in\D:\ W(z)=\widehat{W}(z)\Big\}$$
is a compact subset of $\D$. Then
$$W(z)\le\log\vert z\vert^2+\frac{3}{2}-2\vert z\vert^2+
\frac{1}{2}\vert z\vert^4<0,\qquad z\in\D;$$
in particular, the coincidence set is empty.
\end{lem}

\begin{proof}
We first introduce some preliminary notation.
We consider the functions
$$
W_r(z)=r\log\vert z\vert^2-\int_\D G(z,\zeta)\,\nu(\zeta)\,\diff\Sigma(\zeta),
\qquad z\in\D,
$$
for $1\leq r<\infty$, and note that $W_1=W$.
By construction, $W_r$ vanishes on $\T$.
Denote by $\widehat{W}_r$ the smallest superharmonic majorant of
$W_r$ in $\D$, and write
$$B(r)=\Big\{z\in\D:\ W_r(z)<\widehat{W}_r(z)\Big\}.$$
The set $B(r)$ gets bigger as $r$ increases; see~\cite[Proposition~2.7]{HS}.
If $W_r\leq0$ in $\D$, then $\partial_n W_r\leq0$ on $\T$.
Conversely, if  $\partial_n W_r\leq0$ on $\T$,
then Lemma~\ref{wbhgestimate} applied to the function $r^{-1}W_r$
shows that
\begin{equation}\label{Wrestimate}
W_r(z)\leq r\Big(\log\vert z\vert^2+\frac{3}{2}-2\vert z\vert^2+
\frac{1}{2}\vert z\vert^4\Big)<0,\qquad z\in\D.
\end{equation}

If $\partial_n W_1\leq0$ on $\T$, we are done.
Assume, to reach a contradiction, that
$\max_\T\partial_n W_1$ is strictly positive.
In view of the definition of $W_r$, we have that
$$
\partial_n W_r(z)=-2(r-1)+\partial_n W_1(z),\qquad z\in\T.
$$
This formula makes evident that
$$\max_\T\partial_n W_r>0\quad \text{for}\quad
1\leq r<r_1=1+\max_\T\partial_n W_1/2$$
while $\max_\T\partial_n W_r\leq0$ for $r\geq r_1$.

For $1\leq r<r_1$, the function $W_r$ attains a positive maximum at some
point $z(r)\in\D$. Clearly, $z(r)\in\D\setminus B(r)\subset \D\setminus B(1)$.
Let $\{r_j\}_{j=2}^\infty$ be a sequence with $1\leq r_j<r_1$ and
$r_j\to r_1$ as $j\to+\infty$. In view of the assumed compactness of
$\D\setminus B(1)$, we may by passing to a subsequence assume that
$z(r_j)$ converges to a point $z_1\in\D\setminus B(1)$ as $j\to+\infty$.
Since $W_{r_j}(z(r_j))>0$, we obtain in the limit that $W_{r_1}(z_1)\geq0$,
which contradicts (\ref{Wrestimate}) for $r=r_1$. It follows that $B(1)=\D$,
and that the estimate (\ref{Wrestimate}) holds for all $r$, $1\le r<+\infty$.
\end{proof}

We may now derive the asserted properties of $D(t)$. We recall the definition
of the termination time $T=T(\omega)$ for the Hele-Shaw flow.

\begin{thm}\label{simplyconnected}
Let ${\mathbf\Omega}=\langle\D,\omega\rangle$ be a
simply connected Riemann surface with a metric
$\diff{\mathbf s}^2=\omega\vert\diff z\vert^2$.

Assume that the weight function $\omega$ is real-analytic and
strictly positive in $\D$ and that the curvature condition
\begin{equation}\label{curvaturecondition}
{\mathbf K}(z)+\frac{1}{2}\,{\mathbf K}_\mathbb{H}(z)\leq0,
\qquad z\in\D,
\end{equation}
is satisfied.
Then, for $0<t<T$,
the Hele-Shaw domain $D(t)=D(t;\omega)$ is simply connected.
\end{thm}

\begin{proof}
Without loss of generality, we can assume that $\omega$ is smooth up to the
boundary. It is known that $\partial D(t)$ has a local Schwarz function
at every point and that this implies that $D(t)$
can have at most finitely many holes
(see~\cite[Section~4]{HS}; the argument uses the work of Sakai~\cite{Sakai}).
Let $D_{\bullet}(t)$ be the simply connected domain obtained from
$D(t)$ by adding all the interior holes.
Let $\varphi:\D\to D_\bullet(t)$ be a Riemann map with $\varphi(0)=0$.
In view of the regularity of $\partial D(t)$ implied by the existence of a
local Schwarz function, the map $\varphi$ extends analytically to a
neighborhood of the closed disk $\bar\D$. Let $B=\varphi^{-1}(D(t))$,
and note that $\D\setminus B$ is a compact subset of $\D$.
Introduce the function
\begin{equation}\label{dfnnu}
\nu(z)=\frac{1}{t}\,
(\omega\circ\varphi)(z)\,\vert\varphi'(z)\vert^2,\qquad z\in\D.
\end{equation}
By the curvature assumption (\ref{curvaturecondition}),
the function
$$z\mapsto\frac{\nu(z)}{1-\vert z\vert^2}$$
is logarithmically subharmonic in $\D$ and, hence, subharmonic there.

The set $B$ can be interpreted as the non-coincidence set of
an obstacle problem. A computation shows that
$$
\Delta\big[V_t\circ\varphi(z)\big]=t\,\delta_0(z)-
\omega\circ\varphi(z)\,\,\vert\varphi'(z)\vert^2,\qquad z\in \D,
$$
where the potential function $V_t$ is as in Section~\ref{prelHSdomains}.
Let $W$ be defined in terms of $\nu$ as in Lemma~\ref{Westimate}.
It is easy to see that
$$
W(z)=\frac{1}{t}
\Big(V_t\circ\varphi(z)-P\big[V_t\circ\varphi\vert_\T\big](z)\Big),
\qquad z\in\D.
$$
where $P[\cdot]$ denotes the usual Poisson integral in $\D$.
By~\cite[Proposition~2.9(a)]{HS}, the function $\widehat{V}_t$
is also the smallest superharmonic majorant for $V_t$ in $D_\bullet(t)$.
By conformal invariance, the function $\widehat{V}_t\circ\varphi$
is the smallest superharmonic majorant for $V_t\circ\varphi$ in $\D$,
and we have that
$$
\widehat W(z)=\frac{1}{t}
\Big(\widehat V_t\circ\varphi(z)-P\big[V_t\circ\varphi\vert_\T\big](z)\Big),
\qquad z\in\D.
$$
It is now clear that
$$B=\Big\{z\in\D:\ W(z)<\widehat W(z)\Big\}.$$
Lemma~\ref{Westimate} shows that $B=\D$, which means that
$D(t)=D_\bullet(t)$ is simply connected. The proof is complete.
\end{proof}

\begin{thm}
Let ${\mathbf\Omega}=\langle\D,\omega\rangle$ be as in
Theorem~\ref{simplyconnected}.
Then the boundary $\partial D(t)$ of the Hele-Shaw domain is a real-analytic
Jordan curve for $0<t<T$.
\label{realanalytic}
\end{thm}

\begin{proof}
In view of the apriori regularity of $\partial D(t)$
(see~\cite[Section~4]{HS}) which follows from Sakai's theorem \cite{Sakai},
as well as from Theorem~\ref{simplyconnected}, the boundary $\partial D(t)$
is real-analytic with the exception of at most finitely many cusp
or contact points (see \cite{HS} for the terminology).
Let $\varphi:\D\to D(t)$ be a Riemann map with $\varphi(0)=0$.
The regularity of $\partial D(t)$ afforded by the existence of a local
Schwarz function shows that the map $\varphi$ extends analytically to a
neighborhood of $\bar\D$. The cusp points of $\partial D(t)$ correspond to
points $z\in\T$ such that $\varphi'(z)=0$.
Arguing as in~\cite[Subsection~5.3]{HS}
we see that the mean value identity
$$
h(0)=\int_\D h(z)\,\nu(z)\,\diff\Sigma(z)
$$
holds for harmonic polynomials $h$, where $\nu$ is given by (\ref{dfnnu}).
The non-vanishing of $\varphi'$ on $\T$ now follows by Lemma~\ref{nuonT}.

If $D(t)$ has a contact point, then, for $t'=t+\delta$,
with $\delta>0$ sufficiently small, the domain $D(t')$ has a hole,
which is impossible by Theorem~\ref{simplyconnected}. For details of this
argument, see~\cite[Subsection~5.3]{HS}.
\end{proof}

\section{Modification of a lemma of Korenblum}\label{Korenblum}

The following lemma is inspired by~\cite[Lemma~8.3]{HS}
which is due to Boris Korenblum.

\begin{lem}\label{Klemma}
Fix $\alpha$, $0<\alpha<+\infty$. Let $\omega$ be a nonnegative function
such that the function
\begin{equation}\label{omegaash}
z\mapsto\log\frac{\omega(z)}{(1-\vert z\vert^2)^{2\alpha}}
\end{equation}
is subharmonic in $\D$.
Assume also that $\omega$ is area-integrable in $\D$ and
reproducing at the origin in the sense that
$$
h(0)=\int_\D h(z)\,\omega(z)\,\diff\Sigma(z)
$$
holds for all harmonic polynomials $h$.
Then, for a specific positive constant $c_{p,\alpha}$ depending only on
$p,\alpha$, we have
\begin{equation}\label{omegalengthestimate}
\int_0^1\omega(r)^p\, \diff r\leq c_{p,\alpha}<+\infty
\end{equation}
for $$0<p<\frac{\pi}{\pi+2\alpha(4-\pi)}.$$
\end{lem}

\begin{rem} The constant $c_{p,\alpha}$ is given by equation (\ref{cpa})
below.
\end{rem}

\begin{proof}
The proof depends on the formula
\begin{equation}\label{Pkformula}
\int_0^1\frac{1}{(1-r)^2}\,\chi_{\D(r,1-r)}(z)\,\diff r=
\frac{1-\vert z\vert^2}{\vert 1-z\vert^2},\qquad z\in\D,
\end{equation}
where $\chi_D$ denotes the characteristic function for the set $D$ and
$\D(z_0,r)$ is the open disk with center $z_0$ and radius $r$.
The formula (\ref{Pkformula}) is obtained by straightforward computation.

By the reproducing property of $\omega$, we have
$$\int_\D\frac{1-\vert rz\vert^2}{\vert 1-rz\vert^2}
\,\omega(z)\,\diff\Sigma(z)=1,
\qquad 0<r<1,$$
and an application of Fatou's lemma shows that
\begin{equation}\label{omegabalayage}
\int_\D
\frac{1-\vert z\vert^2}{\vert 1-z\vert^2}\,
\omega(z)\,\diff\Sigma(z)\leq1.
\end{equation}

We now turn to the proof of (\ref{omegalengthestimate}).
Fix an $r$ with $0<r<1$. By the subharmonicity assumption (\ref{omegaash}),
we have that
\begin{multline*}
\log\frac{\omega(r)}{(1-r^2)^{2\alpha}}\leq
\frac{1}{(1-r)^2}\int_{\D(r,1-r)}\log\omega(z)\,\diff\Sigma(z)\\
+\frac{2\alpha}{(1-r)^2}\int_{\D(r,1-r)}\log\frac{1}{1-\vert z\vert^2}\,
\diff\Sigma(z).
\end{multline*}
By an application of the geometric-arithmetic mean inequality,
we see that
\begin{equation}\label{omegaestimate}
\frac{\omega(r)}{(1-r^2)^{2\alpha}}\leq

F(r)^{2\alpha}
\frac{1}{(1-r)^2}\int_{\D(r,1-r)}\omega(z)\,\diff\Sigma(z),
\end{equation}
where we have written $F(r)$ for the quantity
\begin{equation}\label{Fdfn}
F(r)=\exp\bigg\{
\frac{1}{(1-r)^2}\int_{\D(r,1-r)}\log\frac{1}{1-\vert z\vert^2}\,
\diff\Sigma(z)\bigg\},
\qquad 0<r<1.
\end{equation}
It is clear that $F(r)$ is bounded from above away from the right end-point
$r=1$. Below, we will show that $F(r)$ admits the estimate
\begin{equation}\label{Festimate}
F(r)\leq \frac{1}{(1-r)^{4/\pi+\varepsilon}},\qquad r_\varepsilon<r<1,
\end{equation}
for every positive $\varepsilon$, where $0<r_\varepsilon<1$.
In view of (\ref{Pkformula}), (\ref{omegabalayage}), and
(\ref{omegaestimate}), we have that
$$
\int_0^1\frac{\omega(r)}{(1-r^2)^{2\alpha}F(r)^{2\alpha}}\,\diff r\leq
\int_0^1\frac{1}{(1-r)^2}\int_{\D(r,1-r)}\omega(z)\,\diff\Sigma(z)
\,\diff r\leq1.
$$
An application of H\"older's inequality now gives that
\begin{equation}
\int_0^1\omega(r)^p\, \diff r\leq
\bigg\{
\int_0^1\big[(1-r^2)F(r)\big]^{2\alpha p/(1-p)}\,\diff r
\bigg\}^{1-p}
=c_{p,\alpha},
\label{cpa}
\end{equation}
where the rightmost equality is used to define the constant $c_{p,\alpha}$.
The estimate (\ref{omegalengthestimate}) now follows.

The finiteness of $c_{p,\alpha}$
for the asserted values of $p$

follows by (\ref{Festimate}).

Finally, we turn to the proof of (\ref{Festimate}).
By a change to polar coordinates in (\ref{Fdfn}), we arrive at the formula
$$
\log F(r)=
\frac{2}{\pi(1-r)^2}\int_{2r-1}^1 t\log\bigg[\frac{1}{1-t^2}\bigg]
\arcsin\bigg[\frac{\sqrt{(1-t^2)(t^2-(2r-1)^2)}}{2rt}\bigg]\,\diff t
$$
which is valid for $\frac12<r<1$.
We note that by the geometric-arithmetic mean inequality, we have that
$$
\frac{\sqrt{(1-t^2)[t^2-(2r-1)^2]}}{2rt}\leq\frac{1-r}{t}.
$$
From an elementary estimate of the arcsine, we find that
$$
\log F(r)\leq\frac{2}{\pi}\,\frac{1+\varepsilon}{1-r}
\int_{2r-1}^1\log\bigg[\frac{1}{1-t^2}\bigg]\,\diff t
$$
for $r$, $\frac12<r<1$, sufficiently close to $1$.
A straightforward computation shows that
$$
\int_{2r-1}^1\log\bigg[\frac{1}{1-t^2}\bigg]\,\diff t=
2(1-r)\log\frac{1}{1-r}+4(1-\log2)(1-r)+2r\log r.
$$
Using the elementary inequality $\log r\leq r-1$, we see that
$$
\int_{2r-1}^1\log\bigg[\frac{1}{1-t^2}\bigg]\,\diff t\leq
2(1-r)\log\frac{1}{1-r}
$$
for $2(1-\log2)\leq r<1$.
As we return to the function $F$, we see that
$$
\log F(r)\leq\frac{4}{\pi}(1+\varepsilon)\log\frac{1}{1-r}
$$
for $r$ close to $1$, which yields (\ref{Festimate}).
\end{proof}

\begin{rem}
Assume that $\omega$ is logarithmically subharmonic in $\D$ and
reproducing is the sense of Lemma~\ref{Klemma}.
Letting $\alpha\to0$ in (\ref{omegalengthestimate}), we see that
$$\int_0^1\omega(r)^p\diff r\leq1,\qquad 0<p<1.$$
By monotone convergence, we conclude that
$$\int_0^1\omega(r)\diff r\leq1.$$
In~\cite[Lemma~8.3]{HS}, this last inequality was shown to hold true
under the slightly weaker assumption that
$\omega$ is subharmonic and reproducing in $\D$.
\end{rem}

In terms of the curvature form ${\mathbf K}$ of
${\mathbf\Omega}=\langle\D,\omega\rangle$, the assumption (\ref{omegaash})
means that
\begin{equation}\label{curvaturealpha}
{\mathbf K}(z)+\alpha\,{\mathbf K}_\mathbb{H}(z)\leq0,\qquad z\in\D,
\end{equation}
as is easily verified by straightforward computation.
In this context, we recall that the Poincar\'e metric of constant curvature
$-1$ on $\D$ corresponds to the weight function
$$
\omega_\mathbb{H}(z)=\frac{4}{(1-\vert z\vert^2)^2},\qquad z\in\D.
$$

We shall only need Lemma~\ref{Klemma} for the parameter value $p=\frac12$.
In geometric language, the estimate in the lemma then says that
the length of the radial segment $[0,1[$ in the metric
(\ref{metric}) is less than or equal to $c_\alpha=c_{1/2,\alpha}$.
By a conformal mapping argument,
we arrive at the proposition below. The notation ${\mathbf B}(z_0,r)$
stands for a metric disk of radius $r$ about the point $z_0\in\D$; also,
as before, the parameter $T=T(\omega)$ is the termination time for the
Hele-Shaw flow emanating from the origin.

\begin{prop}\label{HSdestimate}
Let ${\mathbf\Omega}=\langle\D,\omega\rangle$ be a
simply connected Riemann surface with a metric
$\diff{\mathbf s}^2=\omega\vert\diff z\vert^2$.
Assume that the weight function $\omega$ is $C^2$-smooth and
strictly positive in $\D$ and that the curvature condition
$$
{\mathbf K}(z)+\alpha\,{\mathbf K}_\mathbb{H}(z)\leq0,\qquad z\in\D,
$$
is satisfied for some $\alpha$ with $0\leq\alpha<\pi/(8-2\pi)$.
Let $c_\alpha=c_{1/2,\alpha}$ be as in Lemma~\ref{Klemma}.

If for some $t$, $0<t<T$, the Hele-Shaw domain $D(t)$ is simply connected and
$\partial D(t)$ is a Jordan curve, then
$D(t)\subset {\mathbf B}(0,c_\alpha\sqrt{t})$.
\end{prop}

\begin{proof}

Let $z_0\in D(t)$, and let $\varphi:\D\to D(t)$ be the conformal map
such that $\varphi(0)=0$ and $\varphi^{-1}(z_0)\in[0,1[$.
Let $\nu$ be defined by (\ref{dfnnu}).
We shall apply Lemma~\ref{Klemma} to the weight $\nu$.
It is straightforward to see that $\nu$ satisfies the
subharmonicity assumption in Lemma~\ref{Klemma}.
We proceed to check the reproducing property of $\nu$.

A classical result of Torsten Carleman~\cite[Section~1]{Carleman}
asserts that the space of
analytic polynomials is dense in the Bergman space $A^p(D)$ ($0<p<+\infty$)
of all $p$-th power area integrable analytic functions in $D$ if
the domain $D$ is simply connected with Jordan boundary. Let $h$ be a
harmonic polynomial.
In view of Carleman's theorem, we can find a sequence $\{h_j\}$ of functions
harmonic in a neighbourhood of $\bar D(t)$ such that
$h_j\to h\circ\varphi^{-1}$ in $L^1(D(t))$.
By the mean value property of $D(t)$ (see~\cite[Theorem~2.3]{HS}), we have
$$
t\,h_j(0)=\int_{D(t)}h_j(z)\,\omega(z)\,\diff\Sigma(z),
$$
and a passage to the limit yields that
$$
t\,h(0)=\int_{D(t)}h\circ\varphi^{-1}(z)\,\omega(z)\,\diff\Sigma(z).
$$
An obvious change of variables now gives the reproducing property of $\nu$.

An application of Lemma~\ref{Klemma} gives that
$$\int_0^1\sqrt{\nu(r)}\,\diff r\leq c_\alpha.$$
This last inequality shows that the geodesic distance in the
metric (\ref{metric}) from the origin to the point $z_0$ is
less than $c_\alpha\sqrt t$.
We conclude that $D(t)\subset{\mathbf B}(0,c_\alpha\sqrt{t})$.
\end{proof}

We remark that for $\alpha=0$ the above proposition
recovers~\cite[Proposition~8.2]{HS}.
For a complete surface, we have the following corollary.

\begin{cor}
Let ${\mathbf\Omega}=\langle\D,\omega\rangle$
be as in Theorem~\ref{simplyconnected} and
assume in addition that ${\mathbf\Omega}$ is complete.
Then $T=+\infty$. This in turn implies that the Hele-Shaw flow covers
all of ${\mathbf\Omega}$, that is, $\D=\bigcup_{t>0}D(t)$.
\end{cor}

\begin{proof}
Since ${\mathbf\Omega}$ is complete, a famous theorem of Hopf, Rinow
and de Rham asserts that the metric disks ${\mathbf B}(0,r)$, $0<r<+\infty$,
are all precompact in ${\mathbf\Omega}=\langle\D,\omega\rangle$
(see~\cite[Theorem~I.10.3]{Helgason}). In view of Theorems
\ref{simplyconnected} and \ref{realanalytic}, the Hele-Shaw domains $D(t)$
are simply connected with real-analytic Jordan curve boundaries for as long
as $0<t<T$, that is, $D(t)\Subset\D$. Moreover, Proposition~\ref{HSdestimate}
guarantees that the growth of the Hele-Shaw domains is essentially no faster
than that of the metric disks. The conclusion is that $D(t)$ remains
precompact in $\D$ for all $t$, $0<t<+\infty$, so that $T=+\infty$.
In addition, an obstacle problem argument shows that the Hele-Shaw flow
covers all of ${\mathbf\Omega}$ (see~\cite[pp. 219-220]{HS}).
\end{proof}

\section{The Hele-Shaw exponential mapping}\label{HSexpsection}

Just as in~\cite{HS}, we have the following existence result regarding
the so called {\sl Hele-Shaw exponential mapping}.

\begin{thm}\label{HSexpmapping}
Let ${\mathbf\Omega}=\langle\D,\omega\rangle$ be as in
Theorem~\ref{simplyconnected}.
Let $z_0\in{\mathbf\Omega}$.
Then there exists a real-analytic mapping
$\Phi=\HSexp_{z_0}:\D(0,\sqrt{T})\to{\mathbf\Omega}$ such that
\begin{itemize}
\item $\Phi(0)=z_0$,
\item each ray $\{z\in\D(0,\sqrt{T})\setminus\{0\}:\ \arg(z)=\theta\}$ is
mapped mapped
onto a curve in ${\mathbf\Omega}$ which points in the same direction
as the ray at $z_0$,
\item the map $\Phi$ maps each pair consisting of a concentric circle
about the origin and a straight line passing through the origin
onto a pair of orthogonal curves, and
\item for each $0<r<\sqrt{T}$, the domain $\Phi(\D(0,r))$
equals the Hele-Shaw domain ${\bf D}(z_0,r^2)$.
\end{itemize}
The map $\Phi$ is uniquely determined in the class of $C^1$-mappings
satisfying the above four properties.
Furthermore, the map $\Phi$ has the asymptotics
$$
\Phi(z)=z_0+\omega(0)^{-1/2}z+O(\vert z\vert^2)\qquad\text{as}\
\vert z\vert\to0.
$$
If the surface ${\mathbf\Omega}$ is complete,
then $T=+\infty$, which in turn implies that the Hele-Shaw flow covers
all of ${\mathbf\Omega}$, that is, ${\mathbf\Omega}$ is the union of all
the domains ${\bf D}(z_0,t)$ over all positive $t$.
\end{thm}

\begin{proof}
Using properties of the weighted biharmonic Green function $\Gamma_1$
we have shown that the Hele-Shaw flow domain $D(t)$, $0<t<T$,
is simply connected and has real-analytic Jordan boundary
(see Section~\ref{wbhGfcn} and Section~\ref{simplyconnHSdomain}).
Also, in Section~\ref{Korenblum} we showed that $T=+\infty$ if
${\mathbf\Omega}$ is complete. Knowing this, the proof of
Theorem~\ref{HSexpmapping} proceeds exactly as in~\cite{HS}. We omit
the details.
\end{proof}

\section{An example of a weakly hyperbolic surface}\label{exwhs}

In this section, we give an example of a complete real-analytic
metric (\ref{metric}) satisfying (\ref{curvaturealpha})
for a given positive $\alpha$, which, for some $r$, $0<r<1$, has the circle
$r\T$ as a geodesic curve. By an obvious pull-back with a conformal self-map
of the unit disk, we obtain a metric of the kind referred to in the
introduction.

Let $I$ be a bounded interval on the real line $\R$. The differential
equation for a geodesic curve $\gamma:I\to\D$
in the metric (\ref{metric}) can be written
\begin{equation}\label{geodesiceq}
\gamma''(t)+\frac{1}{\omega(\gamma(t))}\,
\frac{\partial\omega}{\partial z}(\gamma(t))\,[\gamma'(t)]^2=0,
\qquad t\in I,
\end{equation}
where the curve $\gamma$ is to be parameterized proportionally to arc length,
that is, the quantity
$$\|\gamma'(t)\|_{\gamma(t)}^2=\omega(\gamma(t))\,\vert\gamma'(t)\vert^2$$
is to be constant.

We now specialize to a radial weight:
$$\omega(z)=\omega_0\big(\vert z\vert^2\big),\qquad z\in\D.$$
Let $\gamma_r$ be the circle $r\T$ parameterized by
$\gamma_r(t)=re^{\imag t}$ for real $t$, $-\pi<t\le\pi$. A straightforward
computation shows that $\gamma_r$ satisfies (\ref{geodesiceq}) if and only if
\begin{equation}\label{omega0eq}
\omega_0(r^2)+r^2\omega_0'(r^2)=0.
\end{equation}

Fix $\alpha$, $0<\alpha<+\infty$. We introduce a positive real parameter
$c$, and consider weights of the form
$$
\omega(z)=\frac{c}{(1-\vert z\vert^2)^2}+(1-\vert z\vert^2)^{2\alpha},
\qquad z\in\D.
$$
By comparison with the Poincar\'e metric, it is clear that the metric
(\ref{metric}) so obtained is complete. Since the function
$$z\mapsto\frac{\omega(z)}{(1-\vert z\vert^2)^{2\alpha}}$$
is the sum of two logarithmically subharmonic functions, it is itself
logarithmically subharmonic (see~\cite[Corollary~1.6.8]{HorSCV}).
This shows that (\ref{curvaturealpha}) holds.

We now show that the equation (\ref{omega0eq}) has a solution $r$, $0<r<1$.
A computation shows that
$$
\omega_0(r)+r\omega_0'(r)=c\,\frac{3-r}{(1-r)^3}+
(1-r)^{2\alpha-1}\big(1-(1+2\alpha)r\big).
$$
Note that the second term on the right hand side is
negative for $r>1/(1+2\alpha)$.
Thus, for $c$ small and positive, the function
$$r\mapsto\omega_0(r)+r\omega_0'(r)$$
attains negative values in $]0,1[$.
Taking the endpoints $r=0,1$ also into account, we see that
equation (\ref{omega0eq}) has at least two solutions $r$ in the interval
$]0,1[$ for small positive values of $c$.


TITLE: How to prove that two $p$-adic lattices are isomorphic?
QUESTION [9 upvotes]: Let $\mathbb{Z}_{p}$ be the ring of p-adic integers.
A pair $(L,<>)$ is called lattice if $L$ be a free $\mathbb{Z}_{p}$ module of finite rank and $<>:L×L \to \mathbb{Z}_{p}$be a nondegenerate symmetric bilinear
form on $\mathbb{Z}_{p}$.
Two lattices $L_1,L_2$ is called isomorphic if there exist  isomorphism of $\mathbb{Z}_{p}$ module $L_{1} \to L_{2}$ preserving $<>$.
Let $X_1,X_2$ be 2-adic lattices of rank 2 determined by matrices
$\begin{pmatrix}0&2^k&\\2^k&0&\end{pmatrix},\begin{pmatrix}2^{k+1}&2^k&\\2^k&2^{k+1}&\end{pmatrix}$.
How to prove $X_{1}\oplus X_{1} \cong X_{2}\oplus X_{2}$ and write this isomorphism explicitly ?

REPLY [6 votes]: I fiddled a little with your $\mathbb{Z}_2$-lattices, and believe that a lattice isomorphic of $X_1 \oplus X_1$ with $X_2 \oplus X_2$ is not hard to make. Clearly we may assume $k = 0$. The lattice $X_2$ may be identified with the additive group of the ring $A = \mathbb{Z}[\zeta_3]$, which is a quadratic ring extension of $\mathbb{Z}_2$---defined by$$\zeta_3^2 + \zeta_3 + 1 = 0$$---the inner product being given by$$\langle a, b\rangle = a\overline{b} + b\overline{a},$$where the overline is the automorphism of $A$ sending $\zeta_3$ to its inverse and each element of $\mathbb{Z}_2$ to itself. Hence $X_2 \oplus X_2$ can similarly be identified with $A \oplus A$. Now it is easy to see that $A$ has an element $u$ with $u\overline{u} = -1$. Then the subgroup$$H = \{(x, u.x) \mid x \in A\}$$of $A$ is totally isotropic, and so is$$I = \{(x, u.x.\zeta_3) \mid x \in A\},$$while $A\, \oplus$ is the direct sum of $H$ and $I$. The inner product is unimodular, so it identifies $H$ with the $\mathbb{Z}_2$-dual of $I$; hence one can now conclude that $A \oplus A$ is as a lattice isomorphic to $X_1 \oplus X_1$.


TITLE: solve equation $(2x + 1)e^{-2x} = 0.3$
QUESTION [1 upvotes]: How can I solve the equation
$$
(2x + 1)e^{-2x} = 0.3
$$
for $x$?
I don't know how to solve this kind of equations, even don't know what kind of knowledge should I use to solve this equation.

REPLY [0 votes]: As a hint
Using logarithms, we say:
$\ln[(2x + 1)e^{-2x}] = \ln(0.3)$
$\ln(e^{-2x}) + \ln(2x + 1) = \ln(0.3)$
$-2x + \ln(2x + 1) = \ln(0.3)$
$2x = \ln(2x + 1) - \ln(0.3)$
$2x = \ln\left(\dfrac{2x + 1}{0.3}\right)$
$e^{2x} = \dfrac{2x + 1}{0.3} = \dfrac{20}{3}x + \dfrac{10}{3}$
This is relatively easy to sketch without a calculator.
Having completed the sketch, you are then in a good position to apply a numerical method, as suggested by T_M in the comments and by P. Michaels in their answer.


TITLE: Seeing cosmic activity now, really means it happens millions/billions of years ago?
QUESTION [10 upvotes]: A Recent report about a cosmic burst 3.8 billion light years away. It is written as though it is happening now. However, my question is, if the event is 3.8 billion light years away, doesn't that mean we are continuously looking at history, or is it possible to detect activity in "realtime" despite the distance?

REPLY [17 votes]: You are absolutely correct.  The event happened 3.8 billion years ago, but it gets tedious and confusing to write in the past tense.  For example, which one of the following statements do you find easier to understand?

The high-energy radiation continued to brighten and fade for at least a week after the burst 3.8 billion years ago

or

More than a week later, high-energy radiation continues to brighten and fade from its location

As an example, see this article about writing in astronomy, under Miscellaneous Sticky Points:

Remember that statements about astronomical objects should always be in present tense ("the galaxy has a strong color gradient") unless you are specifically talking about a past event or an object that no longer exists ("the supernova progenitor star was a type O supergiant").

Hope that helps.

REPLY [4 votes]: From Wikipedia:

The finite speed of light is important in astronomy. Due to the vast distances involved, it can take a very long time for light to travel from its source to Earth. For example, it has taken 13 billion (13×109) years for light to travel to Earth from the faraway galaxies viewed in the Hubble Ultra Deep Field images. Those photographs, taken today, capture images of the galaxies as they appeared 13 billion years ago, when the universe was less than a billion years old. The fact that more distant objects appear to be younger, due to the finite speed of light, allows astronomers to infer the evolution of stars, of galaxies, and of the universe itself.

So this means that you actually see what happened a long time ago if it is very far away.

REPLY [2 votes]: This is true, but you should not forget that comparison of times depends on the frame of reference by relativity theory. In particular, from the point of view of the travelling light, no time has passed.
You might want to look at the cartoon at
http://xkcd.com/811/
Don't forget to read the mouse-over text.


TITLE: Flabby representable sheaves
QUESTION [2 upvotes]: Let $S$ be a scheme. Consider some representable moduli functor $\mathcal{M}:(Sch/S)^{op}\rightarrow Set$ represented by some scheme $M$. Then for each $V\in (Sch/S)^{op}$, let define
$$\mathcal{M}^{glob}(V):=\text{im}\left(\mathcal{M}(S)\rightarrow \mathcal{M}(V)\right).$$
This defines a subfunctor $\mathcal{M}^{glob}\subset \mathcal{M}$. I'm interested if this functor is again representable. I tried showing that this is a open or closed subfunctor, but with no succes.
In a bit more generality, consdier a sheaf representable $\mathcal{F}$ on some site $\mathcal{C}$ with an initial object $X$. Then we can define a maximal flabby sub-pre-sheaf $\mathcal{F}^{glob}$ by defining
$$\mathcal{F}^{glob}(V):=\text{im}\left(\mathcal{F}(X)\rightarrow \mathcal{F}(V)\right).$$
Is this a sheaf again, and is it representable?

REPLY [1 votes]: Let $\mathcal{C}$ be a small site (or, at least, cofinally small) and let $\textbf{Psh} (\mathcal{C})$ be the category of presheaves on $\mathcal{C}$. There is a functor $\Gamma : \textbf{Psh} (\mathcal{C}) \to \textbf{Set}$ represented by the terminal presheaf (which may or may not be representable in $\mathcal{C}$, at this level of generality), and it has a left adjoint $\Delta : \textbf{Set} \to \textbf{Psh} (\mathcal{C})$ that sends every set $A$ to the "constant" presheaf defined by $(\Delta A) (U) = A$. We have a counit morphism $\epsilon_F : \Delta \Gamma F \to F$ for every presheaf $F$, and your construction is precisely the image of this morphism. Expressed this way, the failure of $\operatorname{Im} \epsilon_F \subseteq F$ to be a sheaf becomes unsurprising: usually we have to sheafify the presheaf image to obtain a sheaf.
If your goal is to construct a flabby (pre)sheaf, then it would be inappropriate to sheafify $\operatorname{Im} \epsilon_F$: sheafification can destroy flabbiness. On the other hand, if we work with presheaves then representability is a rather strong condition: indeed, representable presheaves are projective, so the epimorphism $\Delta \Gamma F \to \operatorname{Im} \epsilon_F$ would be split. But that would make $\operatorname{Im} \epsilon_F$ a retract of a constant presheaf, hence also constant – not very interesting, I think.
Finally, let me remark that the notion of flabby (pre)sheaf does not seem to be appropriate for non-localic sites. The point of flabby sheaves of modules on a topological space or locale is that they are acyclic with respect to the global sections functor, but I don't think this is true for a general site.


TITLE: Markov Jump Process - Transition Probabilities
QUESTION [1 upvotes]: For the transition probability of a Markov jump process from state $x$ to $y$ in a small time interval $\Delta t$, what is the meaning of $o(\Delta t)$?
$\alpha$ is a constant $\in[0,1)$.

REPLY [0 votes]: That is expressing the fact that the probability to have two or more jumps in the time $\Delta T$ is negligible with respect to $\alpha \Delta t$, and this fact is expressed in the little o notation.
It could also be expressed in "Big O notation" as $$
\alpha \,\Delta t + O\left( {\Delta t^2 } \right)
$$


TITLE: unique solution of pde using method of characteristics
QUESTION [2 upvotes]: I'm solving the pde: $\begin{cases}
xu_x + yu_y = 1 + y^2 \\
u(x, 1) = x + 1
\end{cases}$
Make sure to include all pictures of characteristics and justify whether or not you have found the unique solution.
I've found a solution using the method of characteristics: $u = \ln(y)+\frac{y^2}{2}+\frac{x}{y}+\frac{1}{2}$. How can I argue using the picture of projections on $(x,y)$-plane that the solution is not uniquely determine $u(x, y)$ for $y < 0$?

REPLY [1 votes]: The projections of the characteristics are lines through the origin $x/y=c_1$ (btw, this is the picture you need). From the boundary conditions, at first sight it seems that every characteristic curve has a initial value assigned as we are given the value for $u$ along a line cutting every characteristic curve (the line $y=1$ or $(x,1)$ ) e. g. To pick a value for $c_1$ unambiguously determines (apparently) the value for $u$ along the line $x=c_1y$, so is,  $u_0=u(c_1,1)=1+c_1$ then $u(c_1y,y)=\ln |y|+y^2/2+c_1+1/2$
But I said "apparently" because in fact the projections of the characteristics are rays from the origin, not lines through the origin. The formula for the general solution, $u = \ln|y|+\frac{y^2}{2}+f\left(\frac{x}{y}\right)$ shows that in any case is defined the value for $y=0$. So is, the characteristics are cut in two for each value of $c_1$ We have $u(c_1y,y)=\ln |y|+y^2/2+c_1+1/2$ for $y>0$, but for $y<0$ we can have the perfectly compatible equation $u(c_1y,y)=\ln |y|+y^2/2+g(c_1)$, being $g$ any function, because the given boundary conditions don't cut the rays laying into the half plane with $y<0$: is not given any value for $u$ at some point of these rays.


TITLE: Prove that the determinant of polynomials is zero
QUESTION [2 upvotes]: Prove that this determinant is zero (this matrix is $n\times n$):
$$\begin{vmatrix}
f_1(a_1) & f_1(a_2) & \cdots & f_1(a_n) \\
f_2(a_1) & f_2(a_2) & \cdots & f_2(a_n) \\
\vdots   & \vdots   & \ddots & \vdots_\strut  \\
f_n(a_1) & f_n(a_2) & \cdots & f_n(a_n)
\end{vmatrix}$$
Every $f_i$ is a polynomial of degree $n-2$ maximum.
I had an idea that, because there are $n$ polynomials and the degree of each one is at most $n-2$, the lines of the matrix must be linearly dependent and therefore the determinant is $0$? I don't know how to write a formal and clear solution to this, I think my idea is pretty rough and need to be explained clearly.. can anyone help me please?

REPLY [2 votes]: Consider the matrix $A$ where the entry $a_{ij}$ is the coefficient of $x^{j-1}$ in the polynomial $f_i$.
What is the maximal rank of this $n \times (n-1)$ matrix $A$?
What does this tell you about the linear independence of the rows?
What does this tell you about the linear independence of the polynomials?


TITLE: "Minimal" definition fo Wiener process
QUESTION [1 upvotes]: Almost everywhere (for example on Wiki) the Wiener process is defined as a process $W(t)$ such that:

$W(0)=0 $;
It has independent increments;
The increments in an interval $\Delta t$ are gaussian with expected value 0 and variance $D\Delta t$;
The process is sample continuous;

Is this definition minimal? Or could the 4) be dropped?
My idea is trying to prove the fact that the following  holds $\forall \epsilon>0$:
$$lim_{\Delta t \rightarrow0}P(|\Delta W|> \epsilon)=0$$
where $\Delta W$ is the increment of the process in $\Delta t$. This should be a sufficient condition for sample continuity (sometimes it is called Lindeberg condition) that is valid for Markov processes according to Feller "Handbook of stochastic methods". Of course the definition of the Wiener process entails that the process is Markov because it has independent increments.

REPLY [1 votes]: The first $3$ conditions are not quite enough to show that $W$ has continuous sample paths, but they will ensure that there exists a modification that does have sample paths by Kolmogorov's continuity theorem.  For example, let $W$ be a standard Brownian motion and define $X(t) = W(t)1_{W(t) \ne 1}$.  For any fixed $t$ we have that $X(t) = W(t)$ a.s. so the first $3$ conditions will hold for $X$ ($X(0)=0$, $X(t)-X(s)$ is independent of $X(u)-X(v)$ for any $v < u < s < t$, and $X(t)-X(s) \stackrel{d}{=}W(t)-W(s) \sim N(0,t-s)$), but $X(t)$ does not have continuous sample paths.