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One can describe a $\mathbb{CP}^{N-1}$ manifold with a Fubini-Study metric $g^{FS}$, and there is a connection one form $v$ on it. A is connection one form(gauge field) of a line bundle($\mathcal{O}(1)$) on $\mathbb{CP}^{N-1}$ whose first Chern class generates the integral cohomology group $H^2(\mathbb{CP}^{N-1},Z)$. I have problem here:

1.Why $v$ is a pull back of $A$?

2.Why A generates cohomology group $H^2(\mathbb{CP}^{N-1},Z)$?

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A does not generate the cohomology of $\mathbb C P^{N-1}$ ($A$ is not even globally defined on $\mathbb C P^{N-1}$). The curvature of the connection is the cohomology class of the Fubini-Study Kahler form and that is what generates the cohomology ring. –  Eric O. Korman Mar 16 '14 at 15:05
thanks for your answer.but why curvature generates the $H^2$ –  Xin Wang Mar 16 '14 at 15:49

1 Answer 1

up vote 4 down vote accepted

Here's an outline that should work.

  1. Show (or accept as given) that $H^2(\mathbb CP^{n-1}; \mathbb Z) \simeq \mathbb Z$. The easiest way I know how to do this is to use cellular cohomology. I think Hatcher does this but a google search will bring up lots of hits.
  2. Working in a chart you can compute that the curvature, $F$, of $A$ is the Fubini-Study metric (or maybe its negative). By Chern-Weil theory, $\frac{i}{2\pi} F$ represents the first Chern class of $\mathcal O(1)$ and so lies in $H^2(\mathbb C P^{n-1}; \mathbb Z)$. Thus it is equal to $nx$ where $x$ is a generator of $H^2(\mathbb CP^{n-1};\mathbb Z)$ and $n$ is an integer. Now you just need to show that $n = \pm 1$. It is sufficient to show that you get $\pm 1$ when you evaluate $\frac{i}{2\pi} F$ on some closed embedded 2-manifold. I'd suggest evaluating this on some embedding of $\mathbb C P^1$ (i.e. choose a natural inclusion $\mathbb C P^1 \to \mathbb C P^{n-1}$, pullback $\frac{i}{2\pi} F$ by it, and integrate over $\mathbb C P^1$ and check that you get $\pm 1$).
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