\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 $j0$, 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 $\$. Then (3) means $\=\delta_{km}/(k+1)$. 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 $$ \xalignat1 \sum_j A_{k,j}\<\d_m(q^j q'),q'\> &=\big\<(\d_m q)\cdot{d\over dt}\sum A_{k,j}q^j, q'\big\> +\big\<(\d_m q')\cdot\sum A_{k,j}q^j, q'\big\> \\ &=\< t^{m+1}\cdot kt^{k-1}, q'\>+ \< (m+1)t^m\cdot t^k, q'\> \\ &=(m+k+1)\=(m+k+1)\ba_{m+k}, \\ \sum_j A_{k,j}\&=\ =(m-k+1)a_{m-k} \endxalignat $$ 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}\=(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'\>+\)= (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