bio | website | math.washington.edu/~kovacs |
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location | Seattle, WA | |
age | ||
visits | member for | 3 years, 9 months |
seen | Jul 8 at 15:31 | |
stats | profile views | 6,497 |
I am an algebraic geometer.
Jul 3 |
awarded | Popular Question |
Jul 2 |
awarded | Curious |
Jun 28 |
comment |
Vanishing theorems for pluri-canonical bundle
And if you look at the proof in the cited paper, the author first comments that this is a special case of Kollár's vanishing and immediately goes to saying that he will only prove it for the case $\dim X=\dim Y$. I think he called it GR vanishing mistakenly. It's not a big deal to me, but Kollár's vanishing is a lot harder than GR vanishing, so he should get the credit for it. |
Jun 27 |
answered | Vanishing theorems for pluri-canonical bundle |
May 12 |
awarded | Nice Answer |
May 10 |
comment |
Is the Kähler cone of a toric variety always simplicial?
Thanks!!!!!!!!!! |
May 10 |
comment |
Is the Kähler cone of a toric variety always simplicial?
Could you tell me the definition of a simplicial cone? Thanks |
Apr 17 |
comment |
Degree and quasi projective family
Why can't you take the closure $\bar V$ of V in $\mathbb P^n\times \mathbb P^m$ and apply your argument in the projective case? It seems to me that $\deg V_p\leq \deg (\bar V)_p$, so this should be OK. |
Apr 10 |
answered | A covering lemma of Kawamata |
Apr 3 |
revised |
An affine singular surface
added 428 characters in body |
Apr 3 |
comment |
An affine singular surface
Indeed, I hesitated writing that about being "more familiar" as it surely depends on one's point of view. Also, at the end it is the same thing. From the fact that the exceptional curve is a smooth rational curve with self-intersection $-n$ it follows easily that the resolution of the projectivized cone is actually $\mathbb F_n$. |
Apr 3 |
revised |
An affine singular surface
added 95 characters in body |
Apr 3 |
answered | An affine singular surface |
Mar 28 |
answered | Recognizing a Mukai flop |
Mar 16 |
comment |
Kawamata-Log-Terminal pairs
@ggelli: If $\tilde\Delta\subset X$ is a smooth Cartier divisor, then $(X,\varepsilon\tilde\Delta)$ is klt for any $1>\varepsilon>0$ essentially by the argument you're providing. On the other hand, if you want to change or refine your question, you should do it by editing the question and not by posting a comment to an answer. |
Mar 9 |
revised |
Making Hironaka's theorem explicit for hypersurfaces
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Mar 8 |
answered | Making Hironaka's theorem explicit for hypersurfaces |
Feb 20 |
comment |
Bertini type theorem
No,of course not. I was just answering to what you wrote. Sure, you need to resolve the indeterminacies. So it is a projective closure somewhere else. I don't see the problem. I should edit what I wrote, because it maybe ambiguous. I'll do that soon. |
Feb 20 |
comment |
Bertini type theorem
Smoothness is not needed, so there is no need for a resolution. Just take a projective closure. |
Feb 19 |
comment |
Ample divisors on $\mathbb{P}^3$ blow-up along single point
Uh, did you mean that the blow up is "Fano" instead of "ample"? |