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Could any one tell me how to show :

$\mathbb{C}P^n ≅ SU ( n + 1)/ U ( n)$?

$\mathbb{C}P^n ≅ S^{2 n +1}/ U (1)$ ?

what is transitive group of $\mathbb{C}P^n$?

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1 Answer 1

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This really depends on how you define $\mathbb{C} P^n$. For example if we define $\mathbb{C} P^n$ as $$\mathbb{C} P^n = S^{2n-1}/((z_0,\ldots,z_n) \sim \lambda(z_0,\ldots,z_n))$$ where $\lambda \in \mathbb{C}$ and $|\lambda| = 1$ then your second equation follows almost immediately (once you identify $U(1) \simeq S^1$.

The first one, at least for me, is a little bit harder. There is an action of $SU(n+1)$ on $\mathbb{C}^{n+1}$, which induces a transitive action on $\mathbb{C} P^n$. The stabilizer (isotropy group) of this action turns out to be the image of the embedding $U(n) \to SU(n+1)$, taking the matrix $A$ to the matrix $ \begin{pmatrix} A & 0\\ 0 & \text{det}A^{-1} \end{pmatrix}$

Recall that a set $X$ is a homogenous space if there is a transitive action $G \times X \to X$. Let $G_x$ by the stabilizer. Then it is a theorem that there is a bijection $O(X) \simeq G/G_x$, where $O(x)$ is the orbit of $x$. In our case this gives $\mathbb{C}P^n \simeq SU(n+1)/U(n)$

You should check out 'Representation of Compact Lie Groups' by Brocker, tom Dieck and 'Topology of Lie Groups I and II' by Mimura and Toda.

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