Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer
    
Thanks! Thats exactly what I needed. –  Charlie Frohman Jun 25 '10 at 13:36
2  
The image is pairs $([v] \in {\mathbb P}({\mathbb C}^3), [t] \in {\mathbb P}(Alt^2 {\mathbb C}^3)$ such that $v \wedge t = 0$. –  Allen Knutson Jun 26 '10 at 2:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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