Does there exist a holomorphic map from a neighborhood of $\mathbb C$ to $S^3 \subseteq \mathbb C^2$?
3 Answers
Suppose such a map exists and is nonconstant. 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 $\mathbb{C}\to S^{2n1},n>2$ 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 nonintegrable).
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? $\endgroup$– algoriSep 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 1dimensional version of the maximum principle. I agree that the 2dimensional version also holds, but is it possible to prove it without first proving that there are no nonconstant functions $\mathbb{C}\to S^3$? $\endgroup$– algoriSep 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