28,134 reputation
26199
bio website math.washington.edu/~kovacs
location Seattle, WA
age
visits member for 4 years, 11 months
seen 14 hours ago
I am an algebraic geometer.

Aug
22
comment Varieties with an ample vector bundle mapping to their tangent bundle
No, it isn't. That morphism is only on $\mathbb P^1$ and it maps to the restriction of $T_V$ to $\mathbb P^1$. So, you really just get a morpohism $T_{\mathbb{P}^1}\rightarrow T_V\left|_{\mathbb P^1}\right.$. It also contradicts the Andreatta and Wiśniewski result mentioned, so if this were correct, that theorem would be toast.
Aug
22
answered Vanishing of sheaf cohomology with compact support
Aug
15
revised Vanishing for ideal sheaves on spaces with only rational singularities
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Aug
15
comment Vanishing for ideal sheaves on spaces with only rational singularities
Karl, you are right. When I started I meant to say this. Then I got interrupted and forgot what I had planned. I guess I'm getting old... I'll edit the answer accordingly.
Aug
15
answered Vanishing for ideal sheaves on spaces with only rational singularities
Aug
1
awarded  birational-geometry
Jul
28
awarded  Nice Answer
Jul
25
awarded  Good Answer
Jul
24
revised Intuition behind the Kodaira Vanishing Theorem?
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Jul
23
answered Intuition behind the Kodaira Vanishing Theorem?
Jul
23
comment Intuition behind the Kodaira Vanishing Theorem?
I'm not sure what you are asking. There are many proofs available. I think the easiest way to see that a theorem is true is to read a proof. I personally like Kollár's proof: goo.gl/bzE5Ho
Jul
22
answered pull back of an ample line bundle under a blow up
Jul
1
revised The canonical bundle of an infinitesimal deformation
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Jul
1
comment The canonical bundle of an infinitesimal deformation
Right, of course. I guess I got stock in the proper world by the wording of the question... :)
Jun
30
answered The canonical bundle of an infinitesimal deformation
Jun
26
comment Morphisms contracting a family of curves
Hi Roy, this sounds good. It's a nice proof assuming that $Y$ is projective. The statement is still true assuming that $g$ is proper and $f$ is arbitrary (plus the connectivity assumptions). (Which is what I had in mind all along so I thought that there cannot be such a simple proof. But there is one with $Y$ projective). Cheers! :)
Jun
25
comment Morphisms contracting a family of curves
Roy: I am not sure how that works if you only know that a special fiber is contracted. In that case the general hyperplane section will miss the point where that special fiber maps and thus does not give any information. No?
Jun
25
answered Morphisms contracting a family of curves
Jun
12
revised is there a pattern here showing up or it's simply a coincidence?
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Jun
11
revised Projective dimension of zero module
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