Full flag $Fl_{\mathbb C}(3)$

Question

How we can see that the full complex flag when $n=3$ is equivalent to one of these spaces:

• $\{(u,v)\in \mathbb CP^2\times \mathbb CP^2 ; u\perp v\}$ and what is dimension over $\mathbb C$ here?
• $\{(l,W); l\in Gr(1,3),W\in Gr(2,3),l\subset W\}$?

I know that the full flag $Fl_{\mathbb C}(3)= U(3)/(\mathbb S^1)^{\times 3}$ but I can't deduce that is the same as above spaces (or at least one of them)!

Answer 1

The right space is the second!

For completeness, let $\mathbb{V}$ be a complex vector space of dimension $n$; I recall that a flag of $\mathbb{V}$ is a strictly increasing sequence of vector subspaces of $\mathbb{V}$ $$ \{\underline{0}\}=\mathbb{V}_0<\mathbb{V}_1<\dots<\mathbb{V}_r\leq\mathbb{V}\,\text{where:}\,\forall i\in\{1,\dots,r\},\dim\mathbb{V}_i=m_i,\,dim\mathbb{V}_{i\displaystyle/\mathbb{V}_{i-1}}=d_i, $$ the $r$-tuple $(m_1,\dots,m_r)$ is called signature of the flag; and if the signature is $(1,\dots,n)$ then the flag is called complete.

Let $\mathcal{F}(n;m_1,\dots,m_r)$ be the set of the flags of $\mathbb{V}$ with signature $(m_1,\dots,m_r)$, let $F\in\mathcal{F}(n;m_1,\dots,m_r)$; by definition: $$ F\equiv\mathbb{V}_0<\mathbb{V}_1<\mathbb{V}_2<\dots<\mathbb{V}_r\leq\mathbb{V} $$ and let $\{e_1\dots,e_n\}$ be a basis of $\mathbb{V}$ such that: $$ \forall i\in\{1,\dots,r\},\,\mathbb{V}_i=\langle e_1,\dots,e_{m_i}\rangle; $$ then $\forall M\in\mathrm{GL}(n,\,\mathbb{C}),i\in\{1,\dots,r\},\,M\cdot\mathbb{V}_i=\langle M\cdot e_1,\dots,M\cdot e_{m_i}\rangle$ are well defined subspaces of $\mathbb{V}$; in particular, the flag $$ \mathcal{F}(n;m_1,\dots,m_r)\ni M\cdot F\equiv\mathbb{V}_0< M\cdot\mathbb{V}_1<\dots<M\cdot\mathbb{V}_r\leq\mathbb{V} $$ is well defined.

After all this, one can define the action $$ \alpha:(M,F)\in\mathrm{GL}(n,\mathbb{C})\times\mathcal{F}(n;m_1,\dots,m_r)\to M\cdot F\in\mathcal{F}(n;m_1,\dots,m_r); $$ one (easily) prove that $\alpha$ is a transitive action and therefore $\mathcal{F}(n;m_1,\dots,m_r)$ is in bijection with $$ \mathrm{GL}(n,\mathbb{C})_{\displaystyle/P(d_1,\dots,d_r)}, $$ where $P(d_1,\dots,d_r)$ is the (closed) subgroup of $\mathrm{GL}(n,\mathbb{C})$ generated by the matrices of the following type $$ \forall i\leq j\in\{1,\dots,r\},\,A_i^j\in\mathbb{C}_{d_i}^{d_j},\,\begin{pmatrix} A_1^1 & A_1^2 & \dots & A_1^r\\ \underline{0}_{d_2}^{d_1} & A_2^2 & \dots & A_2^r\\ \vdots & \ddots & \ddots & \vdots\\ \underline{0}_{d_r}^{d^1} & \underline{0}_{d_r}^{d_2} & \dots & A_r^r \end{pmatrix}. $$ In particular, one can restrict the focus on the isometries of $\mathbb{V}$; one can define the action $$ \widetilde{\alpha}:(M,F)\in\mathrm{U}(n)\times\mathcal{F}(n;m_1,\dots,m_r)\to M\cdot F\in\mathcal{F}(n;m_1,\dots,m_r) $$ and as for $\alpha$, one prove that $\widetilde{\alpha}$ is a transitive action and therefore $\mathcal{F}(n;m_1,\dots,m_r)$ is in bijection with $ \mathrm{U}(n)_{\displaystyle/\mathrm{U}(d_1)\times\dots\mathrm{U}(d_r)}$.

Is it all clear?

Furthermore: the first space is the space of (complete) orthogonal flag $\mathcal{F}^{\perp}(3;1,2,3)$ of $\mathbb{C}^3$.