= Solution
Set $d_0=0$ and assume $0<d_1<\cdots<d_m<n$. Choose the first member of the <flag manifold> as a $d_1$-plane, contributing $d_1(n-d_1)$ complex dimensions. After choosing $L_{i-1}$, choose $L_i/L_{i-1}$ as a $(d_i-d_{i-1})$-plane in $V/L_{i-1}$. Its <Grassmannian> has dimension $(d_i-d_{i-1})(n-d_i)$. Adding these fiber dimensions gives
$$
\boxed{\dim_{\mathbb C}F_{d_1,\ldots,d_m}(V)=\sum_{i=1}^m(d_i-d_{i-1})(n-d_i).}
$$
For $V=\mathbb C^4$ the results are
$$
\boxed{\dim_{\mathbb C}F_1=3,\qquad\dim_{\mathbb C}F_2=4,\qquad\dim_{\mathbb C}F_{12}=5.}
$$
Here $F_1=\mathbb{CP}^3$ is <projective twistor space>, $F_2=\operatorname{Gr}(2,4)$ is <complexified compactified Minkowski space>, and $F_{12}$ is the <twistor correspondence> space of incident pairs consisting of a spacetime point and a twistor on its <twistor line>. Over $F_2$ its fiber is $\mathbb{CP}^1$; over $F_1$ its fiber is $\mathbb{CP}^2$, consistent with $4+1=3+2=5$. These are complex dimensions; the underlying real dimensions are $6,8,10$, while a real spacetime slice of $F_2$ has real dimension four.
Back to article page
