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Does there exist a holomorphic map from a neighborhood of $\mathbb C$ to $S^3 \subseteq \mathbb C^2$?

edited Sep 21 2011 at 21:18

asked Sep 21 2011 at 20:54

Sita
85●4

	Please do double check your typing: the title is the very first thing people see of your question! – Mariano Suárez-Alvarez Sep 21 2011 at 21:04
	Thanks for pointing out the typo. – Sita Sep 21 2011 at 21:19

algori · Accepted Answer · 2011-09-22 00:56:21Z UTC

Suppose such a map exists and is non-constant. Let $f,g$ be its components. By composing the map with a holomorphic function on the right and with an element of $U(2)$ on the left, we can assume that in a neighborhood of 0 we have $f(z)=z$ (i.e., we can take $f$ as a local coordinate) and $g(z)=1+az+\cdots$ with $a\neq 0$. Now let $b$ be a complex number such that $Re(ab)\neq 0$ and set $h(t)=bt, t\in\mathbb{R}$. Then on the curve given by $t\mapsto h(t)$ the equation $|f|^2+|g|^2=1$ becomes $$b^2t^2+1+2Re(ab)t+ o(t)=1,$$ which should hold for all $t$ sufficiently close to 0. This is impossible.

upd: this argument generalizes to maps $\mathbb{C}\to S^{2n-1},n>2$ and also implies the following "generalized maximum principle": for a holomorphic map $f:D\to\mathbb{C}^n,D\subset \mathbb{C}$ connected and contained in the closure of its interior, the maximum of $|f|$ can't be attained in an interior point, unless $f$ is constant. To see this one can notice that the moduli of the components of $f$ are subharmonic, so their sum, $|f|^2$, satisfies the maximum principle.

Dmitri · Answer 2 · 2011-09-21 21:33:51Z UTC

Yes, such a map exists, and any such map has to be a map to a point in $\mathbb S^3$, because has to sends any holomorphic curve in the neighbourhood of $\mathbb C$ to a holomorphic curve in $S^3$, but $S^3$ does not contain any holomorphic curve (since the distribution of complex directions in $S^3$ is non-integrable).

Christopher A. Wong · Answer 3 · 2011-09-22 00:48:38Z UTC

0

EDIT: The following is incorrect:

"Yes, but on each connected subset of the domain, the map will need to be constant; this follows from the maximum modulus principle."

edited Sep 22 2011 at 0:48

answered Sep 21 2011 at 21:48

Christopher A. Wong
358●1●9

	Christopher -- could you perhaps elaborate on how exactly you apply the maximum modulus principle to this? – algori Sep 21 2011 at 22:14
	If the map, say $f(z)$, sends to $S^3$, then the modulus of the map is constant. So then $\|f(z)\|$ is maximized on some interior point $z_0$; the maximum modulus principle states that a holomorphic map satisfying this must be constant. – Christopher A. Wong Sep 21 2011 at 22:22
	Given that the domain is connected, of course. Otherwise, $f(z)$ can assume a different constant for each of its connected components. – Christopher A. Wong Sep 21 2011 at 22:23
	Christopher -- I was just wondering whether you were applying the 1-dimensional version of the maximum principle. I agree that the 2-dimensional version also holds, but is it possible to prove it without first proving that there are no non-constant functions $\mathbb{C}\to S^3$? – algori Sep 21 2011 at 22:34
	The maximum modulus principle relies on the open mapping property for holomorphic functions, so were to have learned complex analysis all over again, I would probably encounter this result first before encountering the $\mathbb{C} \rightarrow S^3$ result. – Christopher A. Wong Sep 21 2011 at 22:54

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Holomorphic map from a neighborhood in $\mathbb C$ to S^3

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