$\DeclareMathOperator\PSL{PSL}$Let $U\subset\mathbb C^2$ be an open set, $f:U\to \PSL(2,\mathbb C)$ a holomorphic map. If the image of $f$ is contained in $\operatorname{PSU}(2,\mathbb C)$, I guess that it can only be a constant map. Is it correct ?
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$\begingroup$ I think so because holomorphic map are open, and PSU(2,C) is closed inside PSL(2,C). No? $\endgroup$– Nicolas HemelsoetCommented Sep 2, 2021 at 13:26
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1$\begingroup$ @NicolasHemelsoet in higher dimension holomorphic maps need not be open, e.g., $(z,w)\mapsto (z,zw)$. $\endgroup$– YCorCommented Sep 2, 2021 at 13:35
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2$\begingroup$ At a generic point (where the rank of differential is maximal), the image is locally a complex submanifold. But, as we see after translation at identity and passing to the Lie algebra, locally the only complex submanifold contained in PSU(2) is 0-dimensional. So indeed $f$ is locally constant (of course if $U$ is not connected it can fail to be constant). $\endgroup$– YCorCommented Sep 2, 2021 at 13:37
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Condition $U\subseteq \mathbb{C}^2$ can be replaced by $U\subseteq \mathbb{C}^1$ (just restrict your map on a little disk in a complex line in $U$. Moreover, the restricted map lifts to a map into $SU(2)$. Then existence of a map to $SU(2)$ essentially means that you have two analytic functions satisfying $|f|^2+|g|^2=1$. Such functions must be constant. To see this write $f=u+iv,\; g=u_1+iv_1$, differentiate $u^2+v^2+u_1^2+v_1^2=1$ twice wrt $x$ and $y$, then add, and you you obtain that sum of squares of all first derivatives is $0$.
answered Sep 2, 2021 at 14:26
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$\begingroup$ Your reduction ("can be replaced") is correct if you replace "constant" with "locally constant" in the conclusion. $\endgroup$– YCorCommented Sep 2, 2021 at 15:19
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$\begingroup$ Of course, to avoid trivialities, I assume that $U$ is connected. $\endgroup$ Commented Sep 2, 2021 at 18:25