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Holomorphic Quotient Maps on Projective Space

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex] \mathbb{CP}^n [/itex] be n-dimensional complex projective space, and let [itex] \pi: \mathbb C^{n+1}\setminus\{0\} \to \mathbb{CP}^n [/itex] be the quotient map taking [itex] \pi(z_1,\ldots,z_{n+1}) = [z_1,\ldots,z_{n+1}] [/itex] where the square brackets represent the equivalence class of lines through this point.

    Show that [itex] \pi [/itex] is smooth.

    3. The attempt at a solution

    There's no doubt in my mind I can do this, once I verify what the question is asking me to do. Here's my rationale, but I'm not sure if it's correct.

    Since we've claimed that [itex] \mathbb{CP}^n [/itex] is n-dimensional, we must be considering it with a complex structure rather than a real structure. This means that, as a smooth manifold, it is sufficient to show that its composition with any coordinate chart is smooth. That is, [itex] \phi_i \circ \pi : \mathbb{C}^{n+1}\setminus\{0\} \to \mathbb C^n [/itex] is smooth, where
    [tex] \phi_i([z_1,\ldots, z_{n+1}]) = \left( \frac{z_1}{z_i}, \ldots, \frac{z_{i-1}}{z_i}, \frac{z_{i+1}}{z_i}, \ldots, \frac{z_{n+1}}{z_i} \right) [/tex]
    with domain [itex] z_i > 0 [/itex].

    Okay, so assuming that's correct, we want to show that this function is smooth. But what is "smooth" in this context? I can't ever recall talking about "smooth" complex functions, only holomorphic/analytic complex function. Is it sufficient to show this is holomorphic, or do I need more?
     
  2. jcsd
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