Timeline for Holomorphic map from a neighborhood in $\mathbb C$ to S^3
Current License: CC BY-SA 3.0
15 events
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| Sep 22, 2011 at 1:10 | comment | added | algori | Christopher -- the argument in Dmitri's answer or mine shows that it holds for any $B=\mathbb{C}^n$; I've edited my answer to include that. In the paper you cite there are references to proofs for general $B$. | |
| Sep 22, 2011 at 0:53 | comment | added | Christopher A. Wong | At least for the case $B = \mathbb{C}^2$, the result appears to hold, according to the following: jstor.org/stable/2035432 | |
| Sep 22, 2011 at 0:48 | history | edited | Christopher A. Wong | CC BY-SA 3.0 |
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| Sep 22, 2011 at 0:47 | comment | added | Christopher A. Wong | Yes, you are right. I just realized that only a much weaker result holds. I retract my response | |
| Sep 22, 2011 at 0:43 | comment | added | algori | Christopher -- ok but none of such maps is open unless $dim B=1$. | |
| Sep 22, 2011 at 0:40 | comment | added | Christopher A. Wong | This should hold for any holomorphic map $\mathbb{C} \rightarrow B$, where $B$ is a Banach space over $\mathbb{C}$. | |
| Sep 22, 2011 at 0:37 | comment | added | algori | Christopher -- are you in fact talking about the maximum principle for maps from $\mathbb{C}$ to $\mathbb{C}$? | |
| Sep 22, 2011 at 0:32 | comment | added | Christopher A. Wong | Let $U$ be an open neighborhood. Suppose that $f(z)$ is holomorphic on $U$ and $|f(z)|$ is maximized at $p \in U$. If $f$ is non-constant, then $f$ is an open map, hence $f(U)$ is open, hence there is an open neighborhood of $f(p)$ contained in $f(U)$, hence there are points in $f(U)$ such that $|f(z)| > |f(p)|$, a contradiction. Therefore $f$ is constant. | |
| Sep 21, 2011 at 23:03 | comment | added | algori | Christopher -- how exactly do you deduce the maximum modulus principle for maps $\mathbb{C}\to\mathbb{C}^2$ from the open mapping property? | |
| Sep 21, 2011 at 22:54 | comment | added | Christopher A. Wong | 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. | |
| Sep 21, 2011 at 22:34 | comment | added | algori | 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 $\mathbb{C}\to S^3$? | |
| Sep 21, 2011 at 22:23 | comment | added | Christopher A. Wong | Given that the domain is connected, of course. Otherwise, $f(z)$ can assume a different constant for each of its connected components. | |
| Sep 21, 2011 at 22:22 | comment | added | Christopher A. Wong | 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. | |
| Sep 21, 2011 at 22:14 | comment | added | algori | Christopher -- could you perhaps elaborate on how exactly you apply the maximum modulus principle to this? | |
| Sep 21, 2011 at 21:48 | history | answered | Christopher A. Wong | CC BY-SA 3.0 |