1
$\begingroup$

In $\mathbb{C}^n,\ n\geq 2$, there is no bijection between unit disk $B^n(0,1)$ and unit polydisk $P^n(0,1)$. But if we wish to find injective holomorphic mapping from unit disk to polydisk(whose image contains origin), inclusion is the obvious mapping or suitable automorphisms of unit disk(which is sort of inclusion, after applying automorphism). But can we go beyond unit disk in polydisk, by means of injective holomorphic mapping. Are there results in this direction available?

$\endgroup$
6
  • 2
    $\begingroup$ I'm not exactly sure what your question is - to find an injective map from the unit ball to the polydisk with the image not in the unit ball? You can map $B^n(0,1)$ to some small in the "corner" of the polydisk. $\endgroup$
    – Wojowu
    Commented Jun 3, 2020 at 15:59
  • $\begingroup$ I want to see, if we can find a map, whose image contains origin, but also consists points out of $B^n(0,1)$. Edited my question a bit. $\endgroup$
    – Jean
    Commented Jun 3, 2020 at 16:08
  • $\begingroup$ How about $z\mapsto 2z$? $\endgroup$
    – Henri
    Commented Jun 4, 2020 at 6:24
  • $\begingroup$ @Henri Its image not be within polydisk $P^n(0,1)$ $\endgroup$
    – Jean
    Commented Jun 4, 2020 at 9:02
  • $\begingroup$ Ok, I hadn't seen that you wanted the image to be contained in the polydisk. $\endgroup$
    – Henri
    Commented Jun 4, 2020 at 12:23

1 Answer 1

2
$\begingroup$

Take $f(z_1,z_2):=(z_1,sz_2+(1-s)z_1)$, $0<s<1$. For example, $s=1/2$. Then $f$ is an injective holomorphic mapping from $B(0,1)$ to $P(0,1)$, ($n=2$), sending $0$ to $0$. Moreover, $\|f(s,0)\|>1$ for $s<1$, near $1$, i.e. $f(s,0)\notin B(0,1)$..

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .