I need someone to explain for me the unitary groups $U(1)$, $U(2)$ and $U(3)$ and their actions: Specifically:
- $U(3)/U(2)$
- $U(3)/U(2)\times U(1)$
- $U(3)/U(1)\times U(1) \times U(1)$
I have seen somewhere that $U(3)/U(2)\times U(1)=\mathbb CP^2$ but I don't know the proof! If it is so, and if $\mathbb CP^1= \mathbb C^1 \sqcup \{\infty\}$, then can we write the set $U(3)/U(1)\times U(1) \times U(1)$ in this way?
The description of all three homogeneous spaces can be derived from the standard linear action of $U(3)$ on $\mathbb C^3$. Since unitary maps preserve the lenght of vectors, this restricts to a transitive action on the Unit sphere in $\mathbb C^3$ which is $S^5$. The stabilizer of a point in $S^5$ is easily seen to be isomorphic to $U(2)$, whence $U(3)/U(2)\cong S^5$.
The linear action also induces a transitive actions of $U(3)$ on the space $\mathbb CP^2$ of 1-dimensional complex subspaces in $\mathbb C^3$, and the stabilizer of such a subspace is isomorphic to $U(2)\times U(1)$. (The factor $U(2)$ is the stabilizer of a unit vector in that line as above and the $U(1)$-factor comex from the fact that you can form complex multiples of that vector). This gives the isomorphism $U(3)/(U(2)\times U(1))$ with $\mathbb CP^2$ that you mention. In the same way, you can identify $U(3)/(U(2)\times U(1))$ with the space of two dimensional complex subspaces of $\mathbb C^3$.
Finally, $U(3)$ also acts transitively on the space of all flags consisting of a 1-dimensional complex subspace cotained in a 2-dimensional complex subspace of $\mathbb C^3$. The stabilizer of such a flag is isomorphic to $U(1)\times U(1)\times U(1)$, whence the last homogeneous space can be identified with the space of all such flags.