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Parameterizing and inverting analytic mappings with unit Jacobian | | Timur Sadykov
; | | Date: |
2 Jan 2022 | | Abstract: | Let $x=(x_1,ldots,x_n)in {
m f C}^n$ be a vector of complex variables,
denote by $A=(a_{jk})$ a square matrix of size $ngeq 2,$ and let
$varphiinmathcal{O}(Omega)$ be an analytic function defined in a nonempty
domain $Omegasubset {
m f C}.$ We investigate the family of mappings $$
f=(f_1,ldots,f_n):{
m f C}^n
ightarrow {
m f C}^n, quad
f[A,varphi](x):=x+varphi(Ax) $$ with the coordinates $$ f_j : x mapsto x_j +
varphileft(sumlimits_{k=1}^n a_{jk}x_k
ight), quad j=1,ldots,n $$ 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
$f[U,varphi](x)$ is a nonzero constant for any $x$ and moreover for any
analytic function $varphiinmathcal{O}(Omega).$ 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
$H(s), sin {
m f C}^n$ such that for any matrix $U$ as above the mapping
$x+left((Uodot H(s))x
ight)^d$ defined by the Hadamard product $Uodot H(s)$
has unit Jacobian. We prove any such mapping to be polynomially invertible and
provide an explicit recursive formula for its inverse. | | Source: | arXiv, 2201.00332 | | Services: | Forum | Review | PDF | Favorites | |
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