\input amstex
\def\YES{11}
\def\NO{10}
\def\C{\bold{C}}
\def\R{\bold{R}}
\def\Q{\bold{Q}}
\def\Z{\bold{Z}}
\def\U{\Cal U}
\def\I{\Cal I}
\def\eps{\varepsilon}
\def\phi{\varphi}
\def\empt{\varnothing}
\def\Re{\operatorname{Re}}
\def\Im{\operatorname{Im}}
\def\Res{\operatorname{Res}}
\def\Int{\operatorname{Int}}
\def\Area{\operatorname{Area}}
\def\d{\partial}
\def\bd{\bar\d }
\def\bc{\bar c }
\def\ba{\bar a }
\def\bb{\bar b }
\def\bA{\bar A }
\def\0{\bold{0}}
\def\<{\langle}
\def\>{\rangle}

\def\frac#1/#2{\leavevmode\kern.1em
    \raise.5ex\hbox{\the\scriptfont0 #1}\kern-.1em
    /\kern-.15em\lower.25ex\hbox{\the\scriptfont0 #2}}  %from TeXbook Ex.11.6


\documentstyle{amsppt}
\magnification 1200

\document
%\rightheadtext	{Potential at infinity of a polynomial image of disk}
\topmatter
%-----
\title          Potential at infinity of a polynomial image of the disk
\endtitle
%-----
\author 	S.Yu.Orevkov
\endauthor
%-----
\address        Steklov Math. Inst., Russ. Acad. Sci.,
                Gubkina 8, Moscow, Russia
\endaddress
%-----
\thanks         Partially supported by Grants RFFI-96-01-01218
		and DGICYT SAB95-0502
\endthanks
%-----
%\subjclass      Primary 31A25; Secondary 30C10, 30C55
%\endsubjclass
%-----
%\abstract
%                Let $D=q(\Delta)\subset\C$ where $\Delta$ is the unit disk
%                and
%                $q(t)=a_0 t + a_1 t^2 +...+a_n t^{n+1}$, $\,a_0=|a_0|>0$.
%                Let $p(z)=\pi^{-1}\int_D(z-\zeta)^{-1}
%                = c_0 z^{-1} + c_1 z^{-2} +...+ c_n z^{-(n+1)}$,
%                $\,c_0=|c_0|>0$ be the potential of $D$.
%                The correspondance $q\mapsto p$ can be regarded as a
%                mapping $\eta:\R_+\times\C^n\to\R_+\times\C^n$
%                %       such that $\eta(a_0,...,a_n)=(c_0,...,c_n)$.
%                It is shown that the jacobian of $\eta$ is equal to
%                %        {\d(\bar c_n,...,\bar c_1,c_0,c_1,...,c_n) \over
%                %        \d(a_n,...,a_1,a_0,\bar a_1,...,\bar a_n)}
%                $ \Res( q'(t), t^n\bar{q'}(t^{-1}))$.
%
%                As a corollary we obtain another proof of the known fact
%                that $\eta$ is a local diffeomorphism near $q$ if $q|_\Delta$
%                is an immersion.
%\endabstract
%-----
%\keywords
%  Potential theory, inverse problem, conformal mapping, moments.
%\endkeywords
\endtopmatter

Let $D$ be a domain in $\C$ which is the image $q(\Delta)$ of the unit disk 
$\Delta$ under the mapping $t\mapsto q(t)$ where
$$
        q(t)=a_0 t+a_1 t^2+...+a_n t^{n+1},\qquad a_0\in\R,\quad |a_0|>0
								\eqno(1)
$$ is a polynomial, univalent in $D$.

Let $p(z)=\pi^{-1}\int_D(z-\zeta)^{-1}d\mu(\zeta)$ be the
potential of $D$
%(by $\mu$ here and below we denote the usual mesure on $\C$,
%i.e. $d\mu(z)=dx\,dy$ for $z=x+iy$).
($d\mu(x+iy)=dx\,dy$).
The inverse problem of potential theory is the problem of reconstructing
$D$ given the germ of $p$ at $\infty$ (see [1] and references therein).
For $|z|\gg1$ one has
$$
  p(z)=\sum_{k\ge0}{c_k\over z^{k+1}},\qquad\text{where}\quad
	c_k={1\over\pi}\int_D\zeta^k d\mu(\zeta)
	   ={1\over\pi}\int_\Delta q(t)^k q'(t) \overline{q'(t)}\,d\mu(t).
                                                                \eqno(2)
$$
($c_k$ are the moments of $D$).
Using the right hand side of (2) one can define $p(z)$ for any polynomial
$q(t)$ of the form (1), not necessarily univalent.
%It can be inerpreted as the potential of a domain with self-overlappings.
Since
$$
	{1\over\pi}\int_\Delta t^k \bar t^m\,d\mu(t)
	=\cases 1/(k+1),&\text{if $k=m$}\\ 0,&\text{if $k\ne m$,}
         \endcases                                              \eqno(3)
$$
(2) allows us to express $c_k$ as polynomials 
in $a_0,a_1,...,a_n,\bar a_1,...,\bar a_n$,
with rational coefficients.  
%The least exponent of $t$
%in $q(t)^k q'(t)\overline{q'(t)}$ is $k$ and the greatest exponent of
%$\bar t$ is $n$, hence,
It follows from (2), (3) that
$c_k=0$ for $k>n$.
%Since $c_0=\Area(D)/\pi$ we have
 Hence,
$$
	p(z)=c_0 z^{-1} + c_1 z^{-2} +...+c_n z^{-(n+1)},
	\qquad c_0\in\R,\quad |c_0|>0
								\eqno(4)
$$
Thus, we obtain a polynomial mapping $\eta:V^+\to W^+$ where
$V$ (resp. $W$)
is the vector space over $\R$, isomorphic to $\R\times\C^n$, with 
the coordinates $(a_0,...,a_n)$ (resp. $(c_0,...,c_n)$), 
$V^+$ (resp. $W^+$) is the half-spaces $a_0>0$, (resp. $c_0>0$) and
$\eta$ is defined by $\eta(q)=p$.
We identify points of $V^+$ (resp. $W^+$) with polynomials $q$ of the form (1)
(resp. of the form (4)).


Denote by $j(\eta)$
%$|J|=\det J$, $\,J=||\d(c,\bar c)/\d(a,\bar a)||$
the jacobian of
$\eta$ with respect to the volume forms
$da_n\wedge...\wedge da_1\wedge da_0\wedge d\ba_1\wedge...\wedge d\ba_n$ and
$dc_n\wedge...\wedge dc_1\wedge dc_0\wedge d\bc_1\wedge...\wedge d\bc_n$.
Given $q\in V$, let us define $t_1,...,t_n$ by 
$q'(t)=\prod_{i=1}^n(1-t_i t)$.
%The main result of this note is the following

\proclaim{ Theorem }
$j(\eta) =2a_0^{n^2+n+1}\Res_t(t^n\bar{q'}(t^{-1}),\,q'(t))
= 2a_0^{n^2+3n+1}\prod_{i=1}^n\prod_{j=1}^n(1-t_i\bar t_j)$.
\endproclaim

\remark{ Remark } (P.~Etingof) A similar formula appears in the Conformal
Field Theory as a formula for the cocycle defining the central extension 
of the complexification of the group Diff($S^1$).
\endremark

%Let $\U_n\subset\R_+\times\C^n$ be the subsets of 
%univalent in $K\}$ and
Let $\I_n=\{q\in V^+\;|\;q'$ has no root in $\Delta\}$.
%(we identify here a point $(a_0,...,a_n)\in\R_+\times\C^n$ with a
%polynomial $q$ defined by (1)). 
%In other words, $\I_n$ is the space of domains with self-overlappings
%(or, immersed domains) defined by polynomials of degree $\le n$.

\proclaim{ Corollary 1 }{(\rm see [1])} The restriction of $\eta$
onto $\Int{\I_n}$ is an immersion.
\endproclaim

\proclaim{ Corollary 2 }
The closure of the fold of $\eta$ is the (real) hypersurface 
$F=\{q\in V^+\;|\;\exists t_i,\,q'(t_i)=0$ and $|t_i|=1\}$.
%Clearly, that $\Int\I_n$ is a connected component of $V^+-F$.
\endproclaim

\proclaim{ Corollary 3 } 
The closure of the branching set of $\eta$ is the codimension 2 subvariety 
$B=\{q\in V^+\;|\;\exists t_i\ne t_j,\, t_i\bar t_j=1\}$
\endproclaim

\demo{ Proof of Theorem } 
Denote $\d/\d a_m$ by $\d_m$ and $\d/\d\bar a_m$ by $\bar\d_m$, $\,m=0,...,n$.
($\d_0=\bd_0$). Put
$$
  J=
  \left(\matrix
    \bd_n  c_n\;\dots\;\bd_1  c_n\;\;\d_0  c_n\;\;\d_1  c_n\;\dots\;\d_n  c_n\\
    \cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad
	\cdot\quad\cdot\\
    \bd_n  c_1\;\dots\;\bd_1  c_1\;\;\d_0  c_1\;\;\d_1  c_1\;\dots\;\d_n  c_1\\
    \bd_n  c_0\;\dots\;\bd_1  c_0\;\;\d_0  c_0\;\;\d_1  c_0\;\dots\;\d_n  c_0\\
    \bd_n\bc_1\;\dots\;\bd_1\bc_1\;\;\d_0\bc_1\;\;\d_1\bc_1\;\dots\;\d_n\bc_1\\
    \cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad\cdot\quad
	\cdot\quad\cdot\\
    \bd_n\bc_n\;\dots\;\bd_1\bc_n\;\;\d_0\bc_n\;\;\d_1\bc_n\;\dots\;\d_n\bc_n
  \endmatrix\right)
  \,,\;\;
$$
\def\nn{\leavevmode\kern.0em				%%% fraction
    \raise.0ex\hbox{\the\scriptfont0 (n+1)}\kern-.0em
}  %from TeXbook Ex.11.6
$$
A=\pmatrix
\format            \c&\;   \c&\;  \c&\;\;\c&\;\;\c&\;\c&\;\c\\
	     A_{n,n} &       &\0      & 0   \\
	\vdots       &\ddots &        &\vdots&    &\0       \\
	     A_{1,n} &\hdots &A_{1,1} & 0 \\
	0&\hdots&0& 1    &0&\hdots &0\\
                     &       & & 0    &\bA_{1,1}&\hdots&\bA_{1,n} \\
                     &\0     & &\vdots&       &\ddots&\vdots  \\
                     &       & & 0    & \0    &      &\bA_{n,n}
  \endpmatrix
\,,\;
M=
	{1\over2a_0}
		\pmatrix
	                    -\nn\ba_n                 \\
	     2a_0E  \qquad\quad   \vdots  \qquad\quad    \0         \\
	                    -2\ba_1                   	\\
	    0\;\;\hdots\;\;0\quad\; a_0 \quad\;      0\;\;\hdots\;\;0	\\
	                    2a_1                       	\\
	        \0  \qquad\quad   \vdots  \qquad\quad        2a_0E     \\
	                    \nn a_n                   
  \endpmatrix
$$
where $A_{k,j}$, $\,j\ge k>1$ are polynomials 
in $a_0^{-1},a_n,...,a_1,a_0,\bar a_1,...,\bar a_n$,
such that 
$
        \sum_{j\ge k} A_{k,j} q^j = t^k.        % \eqno(4)
$
Let us calculate $AJ$.
Put $A_{k,j}=0$ for $j<k$, $A_{0,0}=1$, $A_{0,j}=0$ for $j>0$, 
and $a_m=0$ unless $0\le m\le n$.
Denote ${1\over\pi}\int_\Delta f(t)\overline{g(t)}\,d\mu(t)$ by $\<f,g\>$.
Then (3) means $\<t^k,t^m\>=\delta_{km}/(k+1)$.
Note that $\d_m q=t^{m+1}$, $\d_m q'=(m+1)t^m$ and
$$
	(\d_m q^j)\cdot q' = {dq^j\over dq}\cdot\d_m q\cdot q'
			   = {dq^j\over dq}\cdot{dq\over dt}\cdot\d_m q
			   = {dq^j\over dt}\cdot\d_m q.
$$
Hence, for any $m\ge 0$ we have
$$
\xalignat1
	\sum_j A_{k,j}\<\d_m(q^j q'),q'\>
		&=\big\<(\d_m q)\cdot{d\over dt}\sum A_{k,j}q^j, q'\big\>
		 +\big\<(\d_m q')\cdot\sum A_{k,j}q^j, q'\big\>		\\
		&=\< t^{m+1}\cdot kt^{k-1}, q'\>+
		  \< (m+1)t^m\cdot t^k, q'\>			\\
                &=(m+k+1)\<t^{m+k},q'\>=(m+k+1)\ba_{m+k},       \\
        \sum_j A_{k,j}\<q^j q',\d_m q'\>&=\<q't^k,(m+1)t^m\>
        =(m-k+1)a_{m-k}
\endxalignat
$$
Therefore, the entries of the right upper quadrant of $AJ$
(resp. the left upper quadrant, the upper part of the central column)
are equal resp. to
$$
\xalignat1
\sum A_{k,j}\d_m c_j&=\sum A_{k,j}\<\d_m(q^j q'),q'\>=(m+k+1)\ba_{m+k},
\qquad m>0,\\
\sum A_{k,j}\bd_m c_j&=\sum A_{k,j}\<q^j q',\d_m q'\>=(m-k+1) a_{m-k},
\qquad\;\; m>0,\\
\sum A_{k,j}\d_0 c_j&=\sum A_{k,j}(\<\d_0(q^j q'),q'\>+\<q^j q',\d_0 q'\>)=
(k+1)\ba_k-(k-1)a_{-k}.
\endxalignat
$$
Each entry of the lower part of $AJ$ is conjugated to the centrally symmetric
entry of the upper one. Thus,
$$
AJ=
\pmatrix
\format\c&\c&\;\;\c&\;\;\c&\c&\c&\;\;\c&\;\;\c&\c\\
  a_0    &          &\0    &    &\nn\ba_n  &          &      &  \0       &\\
2a_1     &a_0       &      &     &n\ba_{n-1}&\nn\ba_n  &      &          &\\
\vdots   &2a_1      &\ddots&      &\vdots   &n\ba_{n-1}&\ddots&         &\\
n a_{n-1}&\vdots    &\ddots&a_0    &2\ba_1  &\vdots&\ddots &\nn\ba_n  &\\
\nn a_n  &n a_{n-1}&\hdots&2\ba_1  &2a_0 &2\ba_1 &\hdots&n\ba_{n-1}&\nn\ba_n\\
          &\nn a_n &\ddots&\vdots   &2a_1   &a_0&\ddots &\vdots  &n\ba_{n-1}\\
          &        &\ddots&n\ba_{n-1}&\vdots &  &\ddots   &2\ba_1&\vdots\\
          &       &      &\nn\ba_n  &n a_{n-1}&  &         &a_0  &2a_1\\
          &   \0 &      &          &\nn a_n   &  &\0       &    &a_0
\endpmatrix
$$
Multiplying this matrix by $M$ from the right, we replace the central
column with $(0,...,0,a_0,2a_1,...,(n+1)a_n)^t$ and obtain 
the matrix whose upper row is $(a_0,0,...,0)$ and the complementary minor 
of the $a_0$ is the transposed Sylvester matrix for the resultant of
$a_0+2a_1t+...+(n+1)a_n t^n$ and 
$(n+1)\ba_n + n\ba_{n-1}t+... +\ba_0 t^n$.
Thus, 
$$
\det A\det J\det M = a_0 \Res_t(q',t^n\bar{q'}(t^{-1})).
$$
Clearly that $\det M=\frac1/2$ and $A_{k,k}=a_0^{-k}$
which implies $\det A=\prod A_{k,k}^2=a_0^{-n(n+1)}$.
It remains to note that the reversing the order of $\d a_m$'s in $J$
changes the sign the same
way as the swapping the arguments of the resultant.
\qed
\enddemo


I am grateful to P.~Etingof and N.~Kruzhilin for usefull discussions.


\Refs
\ref\no1\by A.I.~Prilepko
\paper Potential theory, inverse problems in
\inbook in: Mathematical Encyclopedia
\vol 4 \pages 266--269
\endref

%\ref\no2\by B.~Gustafsson
%\paper On quadrature domains and an inverse problem in potential theory
%\jour J. d'Anal. Math. \vol 55 \yr 1990 \pages 172--216
%\endref
\endRefs

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%
%	The formula with the Curvature
%
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\enddocument


Let $q(t)$ be a polynomial of the form (1).
For $t\in\d\Delta$ (i.e. $|t|=1$) put
$$
K(q,t)=\Re{t q''(t)\over q'(t)}+1,\quad q'(t)\ne0;\qquad
	K(q,t)=\lim_{\theta\to 0} K(q,t e^{i\theta}),\quad q'(t)=0.
								\eqno(6)
$$

Let $S_1\subset V^+$ be the hypersurface consisting 
of polynomials $q$ such that $q'$ has a
root on the unit circle $\d\Delta$. It can be evidently parametrized as
follows. Let $\tilde V_1=\R\times\C^n$ be the space with coordinates
$(b_0,...,b_{n-1},t_0)$ whose half-space $V_1^+=\{b_0>0\}$ is identified with
pairs $(q_0,t_0)$ where
$$
	q_0=b_0 t + b_1 t^2 +...+ b_{n-1} t^n,\qquad b_0\in\R,\quad b_0>0
								\eqno(7)
$$
and $t_0\in\C$. Let $\tilde S_1\subset\tilde V_1^+$ be defined by 
$|t_0|=1$, and let $\iota_1:\tilde V_1\to V$ be the mapping
$(q_0,t_0) \mapsto q$, where $q$ is the polynomial of
the form (1) such that $q'(t)=(1-t_0 t)q_0'(t)$

Let $pr_0:V\to\C^n$ be the projection $(c_0,c_1,...,c_n)\mapsto(c_1,...,c_n)$.
Now we are going to compute the jacobian of the mapping 
$pr_0\circ\eta\circ\iota_1|_{\tilde S_1}$
in the coordinates $(b_0,...,b_{n-1},\phi)$ where $t_0=r e^{i\phi}$.

\proclaim{Lemma 2}
$$
{\d(c_n,c_{n-1},...,c_1,\bc_1,...,\bc_{n-1},\bc_n) \over
 \d(b_{n-1},...,b_1,b_0,\bb_1,...,\bb_{n-1},\phi)} =
K(q,t_0)\cdot\lim_{|t_0|\to1}
{\Res_t\big( t^n\bar{q'}(t^{-1}),\,q'(t)\big) \over
1-|t_0|^2}
$$ $$
= \big(K(q_0,t_0)+\frac1/2\big)\cdot|q_0'(t_0)|^2\cdot
  \Res_t\big( t^{n-1}\bar{q_0'}(t^{-1}),\,q_0'(t)\big).
$$
\endproclaim


\enddocument



