Proving that the Flag Variety $Fl(n;m_1,m_2)$ is connected.

Question

I wish to prove that the flag variety $Fl(n;m_1,m_2) = \{ W_1 \subset W_2 \subset V | dimW_i = m_i \}$, for $0 \le m_1 \le m_2 \le n$ where V is an n-dimensional vector space over $\mathbb{C}$ and $W_1, W_2$ are vector subspaces of V, is connected. I'm following a similar method we have been using in lectures for other Lie groups.

I have already shown that $U(n)$ acts transitively on $Fl(n;m_1,m_2)$ and that $U(n)$ is connected. I'm wondering if what follows is proof that $Fl(n;m_1,m_2)$ is connected.

$Fl(n;m_1,m_2)=U(n)/U(n)_{W_1 \subset W_2}$,since $U(n)$ acts transitively on $Fl(n;m_1,m_2)$. Thus, since $U(n)$ is connected, if we can show that $U(n)_{W_1 \subset W_2}$ is connected then it follows that $Fl(n;m_1,m_2)$ is connected.

Consider $U(n)_{W_1 \subset W_2}=\{F\in U(n) |F(W_1)=W_1, F(W_2)=W_2 \}$. It consists of maps with matrices of the form:

A=$ \left( \begin{array}{ccc} A_{m_1,m_1} & 0 & 0 \\ 0 & B_{m_2-m_1,m_2-m_1} & 0 \\ 0 & 0 & C_{n-m_2,n-m_2} \end{array} \right)$

Where the index of the block denotes its size. This is because $F(W_2)$ will be orthogonal to $W_1$ and will be linear multiples of $e_{m_1+1},...,e_{m_2}$, and $F(W_1)$ will be orthogonal to $W_2$ and will be linear multiples of $e_{1},...,e_{m_1}$, where $e_1,...,e_n$ is the orthonormal basis of $Fl(n;m_1,m_2)$.

Now I think that I have to show that the blocks in the matrix are themselves contained in a connected component of a Lie group, which would mean that I could connect each of them to the identity matrix of the respective size and thus connect any matrix in the stabiliser subgroup I'm considering to the identity, but I'm not quite sure how to proceed. Could anyone please give me any hints as to the next step?

Many thanks.

Answer 1

I think you were already done after the introductory remark:

If the (continuous) action of a connected group $G$ on a space $X$ is transitive, the $X$ is connected.

In fact, assume $X=U_1\cup U_2$ with $U_1,U_2$ open and $U_1\cap U_2=\emptyset$. Select $x\in X$. Then the map $f\colon G\to X$, $g\mapsto gx$ is continuous, hence $f^{-1}(U_1), f^{-1}(U_2)$ are open and are disjoint and $f^{-1}(U_1)\cup f^{-1}(U_2)=G$. By connectedness of $G$, one of $f^{-1}(U_1), f^{-1}(U_2)$ is empty. But by transitivity, $f$ is onto, hence $f^{-1}(U_i)=\emptyset$ implies $U_i=\emptyset$ and we conclude that $X$ is connected.