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accepted Intuition for Kähler manifolds?
Dec
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revised Algebraic methods to compute the cohomology ring of the complex topology of a variety?
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comment Intuition for Kähler manifolds?
@GunnarÞórMagnússon I think that is conceptually the clearest explanation. If you post it as an answer, I would be happy to accept it!
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comment Is there a proof of Bézout's theorem via residue theory?
@TakumiMurayama Actually I think I'm being really stupid. Consider for example $f_1 = x, f_2 = y$. in $\mathbb P^2$. We have the right residue at $0$, and then I think we pick up the extra stuff when we switch charts and consider the line at infinity, so it all works out. The sum is zero, but one of the terms is exactly what we want, so we can just rearrange and get Bezout.
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comment Is there a proof of Bézout's theorem via residue theory?
@TakumiMurayama So the global residue theorem is on p. 656, I think. But now I'm confused about something simple: the residue theorem says a certain sum is zero, but we want a sum that equals $\Pi d_i$. In particular, if we take the form to be $\frac{df_1}{f_1}\wedge\dots\wedge \frac{df_i}{f_i}$, doesn't the global residue theorem tells us the sum of the residues is $0$ and not the product of the degrees, as we desire?
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comment Other ways to compute this integral?
@Unit You did the straightforward thing. I don't see anything wrong with that aesthetically.
Oct
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comment Unconventional mathematics books
Mumford wrote approvingly of this book in his Notices article "Calculus Reform--For the Millions."
Oct
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comment Other ways to compute this integral?
Perhaps use residues? This method already seems fairly painless...
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