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Jan 29, 2016 at 5:04 history edited Honglu CC BY-SA 3.0
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Jan 28, 2016 at 19:31 history edited Honglu CC BY-SA 3.0
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Jan 28, 2016 at 16:56 answer added Will Sawin timeline score: 2
Jan 28, 2016 at 16:29 history edited Honglu CC BY-SA 3.0
correct mistakes in the motivation
Jan 28, 2016 at 16:27 comment added Daniel Litt This will be pretty rare--the morphism in question is a map of Hodge structures, so it is surjective (hence an isomorphism by Lefschetz) iff it is an isomorphism on every Hodge component. The obstruction to surjectivity on $H^n(X, \mathcal{O}_X)\to H^n(Y, \mathcal{O}_Y)$ is the kernel of the map $H^{n+1}(X, \mathcal{O}_X(-Y))\to H^{n+1}(X, \mathcal{O}_X)$. But for $Y$ sufficiently ample, this kernel will be huge.
Jan 28, 2016 at 16:27 history edited Honglu CC BY-SA 3.0
Add my motivation and stuff like that.
Jan 28, 2016 at 16:04 history edited Honglu CC BY-SA 3.0
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Jan 28, 2016 at 16:01 comment added Honglu @JasonStarr Right. But I only need the restriction to be surjective. I think in your example it's still ok. Probably it's better for me to change boundary divisors into torus invariant divisors.
Jan 28, 2016 at 10:25 comment added Jason Starr What do you mean by "boundary divisors"? For the smooth projective toric variety $X=\mathbb{P}^1\times \mathbb{P}^2$ with the standard action of $\mathbb{G}_m\times (\mathbb{G}_m)^2$, one of the irreducible components of the complement of the open orbit is $Y=\{0\}\times \mathbb{P}^2$. The restriction map $H^2(X;\mathbb{Q})\to H^2(Y;\mathbb{Q})$ is not injective.
Jan 28, 2016 at 1:24 history asked Honglu CC BY-SA 3.0