Holomorphic map from a neighborhood in $\mathbb C$ to S^3 - MathOverflow most recent 30 from http://mathoverflow.net2013-05-18T23:09:15Zhttp://mathoverflow.net/feeds/question/76082http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/76082/holomorphic-map-from-a-neighborhood-in-mathbb-c-to-s3Holomorphic map from a neighborhood in $\mathbb C$ to S^3Sita2011-09-21T20:54:48Z2011-09-22T00:56:21Z
<p>Does there exist a holomorphic map from a neighborhood of $\mathbb C$ to $S^3 \subseteq \mathbb C^2$?</p>
http://mathoverflow.net/questions/76082/holomorphic-map-from-a-neighborhood-in-mathbb-c-to-s3/76084#76084Answer by Dmitri for Holomorphic map from a neighborhood in $\mathbb C$ to S^3Dmitri2011-09-21T21:33:51Z2011-09-21T21:33:51Z<p>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).</p>
http://mathoverflow.net/questions/76082/holomorphic-map-from-a-neighborhood-in-mathbb-c-to-s3/76088#76088Answer by Christopher A. Wong for Holomorphic map from a neighborhood in $\mathbb C$ to S^3Christopher A. Wong2011-09-21T21:48:17Z2011-09-22T00:48:38Z<p>EDIT: The following is <em>incorrect</em>:</p>
<p>"Yes, but on each connected subset of the domain, the map will need to be constant; this follows from the maximum modulus principle."</p>
http://mathoverflow.net/questions/76082/holomorphic-map-from-a-neighborhood-in-mathbb-c-to-s3/76090#76090Answer by algori for Holomorphic map from a neighborhood in $\mathbb C$ to S^3algori2011-09-21T22:01:30Z2011-09-22T00:56:21Z<p>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.</p>
<p>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.</p>