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This question seems like it would be very hard to do directly. I wouldn't know where to begin. I was wondering if anyone had a very slick proof of this. The only thing I think is easy is that its connected. The rest seems like the only way I know how to do it would be way to hard to construct easily. I would appreciate it if someone had an easy solution as I am studying for a qual. Thank you.

Recall that the complex projective space $\mathbb{C}P^d$ is the quotient space of $\mathbb{C}^{d+1} \backslash \{0\}$ under the equivalence relation $x \sim y$ if and only if there is a $\lambda \in \mathbb{C}$ with $v = \lambda $w. Prove that $\mathbb{C}^ d$ is a compact, connected, orientable manifold of dimension $2^d$.

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You might profit from a general perspective: $v\mapsto \lambda w$ is a free proper $\mathbb{C}^*$ action, hence its quotient is a manifold and in fact the map is a principal fiber bundle. Check out section 1.4 of math.toronto.edu/mein/teaching/action.pdf –  Neal Apr 5 '13 at 3:46
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1 Answer

Hints:

Compactness: You can alternatively realize $\mathbb{P}^n(\mathbb{C})$ as the coset space $S^{2n+1}/U(1)$.

Connectedness: Same hint.

Orientable: Complex projective space has the structure of a complex manifold--this about why this allows you to create a chart whose overlaps have positive Jacobians.

Dimension: same hint as for compactness and connectedness.

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So Complex Manifolds are always orientable? –  Susan Apr 5 '13 at 3:12
    
Also, what is $U(1)$? –  Susan Apr 5 '13 at 3:13
    
@Susan Yes, and $U(1)$ is the 1-dimensional unitary group. –  Alex Youcis Apr 5 '13 at 3:13
    
And the qoutient of a compact set by a closed set is compact? –  Susan Apr 5 '13 at 3:18
    
@Susan The quotient of a compact set by ANYTHING is closed. Compactness is preserved under continuous maps. –  Alex Youcis Apr 5 '13 at 3:21
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