Suppose you want to work with complete flags $\mathbb{F}_3$ on $\mathbb{C}^3$. Given a flag

$$ \{0\}\leq V_1\leq V_2 \leq \mathbb{C}^3$$

you can think of $V_1$ as the span of a vector $\vec{u}$, and then you can choose a vector $\vec{v}$ that is Hermitian orthogonal to $\vec{u}$ so that $V_2=<\vec{u},\vec{v}>$. Finally you can choose $\vec{w}$ so that it is Hermitian orthogonal to $<\vec{u},\vec{v}>$. This gives an embedding

$$\mathbb{F}_3\rightarrow \mathbb{C}P(2)^3 .$$

Since the Hermitian inner product involves complex conjugates, this embedding cannot possibly be holomorphic. For instance if the first line in homogeneous coordinates is $[a,b,c]$ and the second is $[d,e,f]$ then they satisfy $a\overline{d}=b\overline{e}+c\overline{f}=0$ where the overline indicates complex conjugate. Is there some way of playing around with complex structures to fix this? Is there a similar map, that is better behaved?


1 Answer 1


I think you can use wedge products. Choose $v \in V_1$, then $u \in V_2$, which is linearly independent. Map the flag to $([v], [v \wedge u]) \in (CP^{2})^2$. This should be well defined and holomorphic.

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    $\begingroup$ The image is pairs $([v] \in {\mathbb P}({\mathbb C}^3), [t] \in {\mathbb P}(Alt^2 {\mathbb C}^3)$ such that $v \wedge t = 0$. $\endgroup$ Jun 26, 2010 at 2:21

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