6
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ The image is pairs $([v] \in {\mathbb P}({\mathbb C}^3), [t] \in {\mathbb P}(Alt^2 {\mathbb C}^3)$ such that $v \wedge t = 0$. $\endgroup$
    – Allen Knutson
    Jun 26, 2010 at 2:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.