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comment How to show $[\omega]=0$ implies $[\omega^n]=0$?
@JasonDeVito Fixed it, don't know what I was thinking on Aug 11 '13
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revised How to show $[\omega]=0$ implies $[\omega^n]=0$?
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comment Example of a pair of non-cobordant manifolds
The cone on $\Bbb C P^2$ is not a topological manifold. In fact, the cone on any space $X$ that does not have the same integral homology as a sphere cannot possibly be a manifold, because for such spaces we have $\tilde{H}_\ast(X, X - \text{cone pt}; \Bbb Z) \not \cong \Bbb Z$.
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May
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comment How many differential forms on the complex plane?
@GiuseppeNegro I don't know if I would agree with that fact. $z$ and $\bar{z}$ are not vectors in $\Bbb C$. $\Bbb C$ has real basis $\{1, i\}$ and a real basis induces a complex basis in the complexification (although here it might look a little confusing because the basis elements $\{1, i\}$ are not the same as the scalars $1$ and $i$...). A general statement would be that if $\{b_1, \dots, b_n\}$ is a real basis for $V$, then $\{b_1 \otimes 1, \dots, b_n \otimes 1\}$ is a complex basis for $V \otimes_{\Bbb R} \Bbb C$.
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comment How many differential forms on the complex plane?
@GiuseppeNegro $V \otimes_{\Bbb R} \Bbb C$ has real dimension $2 \dim(V)$ and complex dimension $\dim(V)$. This matches your original question: $\Bbb C$ has real dimension $2$, so $\Bbb C \otimes_{\Bbb R} \Bbb C$ has complex dimension $2$ (and a complex basis is given by $\{dz, d\bar{z}\}$).
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comment How many differential forms on the complex plane?
@GiuseppeNegro They are different. We can complexify any real vector space $V$ my taking $V \otimes_{\Bbb R} \Bbb C$, and complexification makes no reference to a complex structure on $J$. For example, $J(dz - d\bar{z}) = i(dz + d\bar{z}) = 2\, dx \neq i(dz - d\bar{z}) = -2\, dy$, so we explicitly see that $J$ is not the same as multiplication by $i$.