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

  • $\begingroup$ Please do double check your typing: the title is the very first thing people see of your question! $\endgroup$ Sep 21, 2011 at 21:04

3 Answers 3


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.


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).


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."

  • $\begingroup$ Christopher -- could you perhaps elaborate on how exactly you apply the maximum modulus principle to this? $\endgroup$
    – algori
    Sep 21, 2011 at 22:14
  • $\begingroup$ 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. $\endgroup$ Sep 21, 2011 at 22:22
  • $\begingroup$ Given that the domain is connected, of course. Otherwise, $f(z)$ can assume a different constant for each of its connected components. $\endgroup$ Sep 21, 2011 at 22:23
  • $\begingroup$ 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 $\mathbb{C}\to S^3$? $\endgroup$
    – algori
    Sep 21, 2011 at 22:34
  • $\begingroup$ 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. $\endgroup$ Sep 21, 2011 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.