Gene wrote: << Can someone give a geometrical or topological description of the shape of the complex projective space CP^n ? While CP^1 is the 2-sphere, some searching reveals a few results. CP^n is simply connected (in contrast to RP^n). CP^n for n>1 is not homeomorphic to the sphere S^2n. The quotient of CP^2 by complex conjugation is S^4. I'm writing a paper on the optics of polarized light, and I need to understand CP^2 and CP^3.

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In general, CP^n is the union of closed 2k-balls for 0 <= k <= n: CP^n = e_0 u e_2 u e_4 u ... e_2n. The e_2k are glued to one another in order with the glueing maps independent of which CP^n we're talking about. So to get CP^n from CP^(n-1), we take a CP^(n-1) and and e_2n, whose boundary is S^(2n-1). We then glue this S^(2n-1) to CP^(n-1) by thinking of S^(2n-1) as {(z_1,...,z_n) in C^n | |z_1|^2 + ... + |z_n|^2 = 1}. Then one definition of CP^(n-1) is to quotient out S^(n-1) by the action of the circle: (z_1,...,z_n) ~ (c z_1,...,c z_n) where c is an arbitrary complex number with |c| = 1. The equivalence classes constitute CP^(n-1). Then, to construct CP^n from CP^(n-1) you glue on the e_2n by this very same quotient map from S^(2n-1) = bd(e_2n) -> CP^(n-1). So to get CP^2 from CP^1 = S^2, you take a 4-ball e_4 and glue it to S^2 by mapping its boundary S^3 to S^2 via the Hopf fibration. If you start out from any point of the (real) 4-manifold CP^2 and march out in all directions (there is an S^3 of them), then you just get a larger and larger 4-ball, until suddenly at one stage you just get an S^2 (and that's the whole thing), with the sphere of directions at a certain radius being identified by that circle action, i.e., the Hopf fibration. CP^3, and all higher CP^n's, are described analogously. Despite this apparently asymmetric definition, it's clear from the quotient description that in fact each point of CP^n is like any other; there is an self-isometry carrying every point to any other. It's not isotropic as a real manifold, but as a complex manifold, all complex "directions" at any point look like every other. --Dan Even though kleptomaniacs can't help themselves, they do.