Papers
Topics
Authors
Recent
Search
2000 character limit reached

Parameterizing and inverting analytic mappings with unit Jacobian

Published 2 Jan 2022 in math.CV | (2201.00332v1)

Abstract: Let $x=(x_1,\ldots,x_n)\in {\rm \bf C}n$ be a vector of complex variables, denote by $A=(a_{jk})$ a square matrix of size $n\geq 2,$ and let $\varphi\in\mathcal{O}(\Omega)$ be an analytic function defined in a nonempty domain $\Omega\subset {\rm \bf C}.$ We investigate the family of mappings $$ f=(f_1,\ldots,f_n):{\rm \bf C}n\rightarrow {\rm \bf C}n, \quad fA,\varphi:=x+\varphi(Ax) $$ with the coordinates $$ f_j : x \mapsto x_j + \varphi\left(\sum\limits_{k=1}n a_{jk}x_k\right), \quad j=1,\ldots,n $$ whose Jacobian is identically equal to a nonzero constant for any $x$ such that all of $f_j$ are well-defined. Let $U$ be a square matrix such that the Jacobian of the mapping $fU,\varphi$ is a nonzero constant for any $x$ and moreover for any analytic function $\varphi\in\mathcal{O}(\Omega).$ We show that any such matrix $U$ is uniquely defined, up to a suitable permutation similarity of matrices, by a partition of the dimension $n$ into a sum of $m$ positive integers together with a permutation on $m$ elements. For any $d=2,3,\ldots$ we construct $n$-parametric family of square matrices $H(s), s\in {\rm \bf C}n$ such that for any matrix $U$ as above the mapping $x+\left((U\odot H(s))x\right)d$ defined by the Hadamard product $U\odot H(s)$ has unit Jacobian. We prove any such mapping to be polynomially invertible and provide an explicit recursive formula for its inverse.

Summary

Paper to Video (Beta)

Whiteboard

No one has generated a whiteboard explanation for this paper yet.

Open Problems

We haven't generated a list of open problems mentioned in this paper yet.

Continue Learning

We haven't generated follow-up questions for this paper yet.

Authors (1)

Collections

Sign up for free to add this paper to one or more collections.