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Leaky Nun
2:00 AM
ok so $x/y$ is a uniformizer at $\infty$
$y^{-1} = (x/y)^3 + y^{-3}$
$(-y^{-2} - 3y^{-4}) \ \mathrm dy = 3(x/y)^2 \ \mathrm d(x/y)$
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Nov '19
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random cohomology for quantum nerds
covariance is unacceptable
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