Moscow Mathematical Journal
Volume 23, Issue 3, July–September 2023 pp. 369–400.
Parameterizing and Inverting Analytic Mappings with Unit Jacobian
Let $x=(x_1,\dots,x_n)\in\mathbb{C}^n$ 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 $\Omega\subset\mathbb{C}$.
We investigate the family of mappings
$$
f=(f_1,\dots,f_n)\colon \mathbb{C}^n\rightarrow\mathbb{C}^n, \quad f[A,\varphi](x):=x+\varphi(Ax)
$$
with the coordinates
$$
f_j \colon x \mapsto x_j + \varphi\biggl(\sum\limits_{k=1}^n a_{jk}x_k\biggr), \quad j=1,\dots,n,
$$
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 $f[U,\varphi](x)$ 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 $d=2,3,\dots$ we construct $n$-parametric family of square matrices $H(s)$, $s\in\mathbb{C}^n$,
such that for any matrix $U$ as above the mapping $x+((U\odot H(s))x)^d$ 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. 2020 Math. Subj. Class. 14R15, 32H50.
Authors:
Timur Sadykov
Author institution:Plekhanov Russian University of Economics, 115054, Moscow, Russia
Summary:
Keywords: Jacobian conjecture, polynomial invertibility, Hadamard product, permutation similarity.
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