bio | website | math.washington.edu/~kovacs |
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location | Seattle, WA | |
age | ||
visits | member for | 4 years, 9 months |
seen | yesterday | |
stats | profile views | 7,390 |
I am an algebraic geometer.
Jul 1 |
revised |
The canonical bundle of an infinitesimal deformation
added 4 characters in body |
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?
added 1 character in body |
Jun 11 |
revised |
Projective dimension of zero module
deleted 3 characters in body |
Jun 2 |
awarded | Necromancer |
May 28 |
awarded | Nice Answer |
May 27 |
comment |
how does one understand GRR? (Grothendieck Riemann Roch)
@GregorBruns: You are absolutely correct, I left out a term (I had curves in my mind), but I now corrected it. Thanks for spotting the omission! |
May 27 |
revised |
how does one understand GRR? (Grothendieck Riemann Roch)
corrected the exponential Chern character of L and the consequences of this correction. |
May 26 |
comment |
On the number of irreducible components of an exceptional divisor
@JeskoHüttenhain:It's actually not surprising. Think about the case of a quadric cone and a ruling. If you blow up the line with the reduced scheme structure, you get the same as if you blew up the point, because it is not a Cartier divisor. But if you blow up the double line, then nothing happens, because you blew up a Cartier divisor. Similar things can happen if X is smooth, but Z is singular. |
May 26 |
answered | On the number of irreducible components of an exceptional divisor |
May 6 |
comment |
Could we extend any line bundle on the smooth part of a singular curve to a line bundle on the whole curve?
On a curve a non-empty subset is open if and only if its complement is finite. $X_0$ contains the open subset $U_0$ and hence the complement of $X_0$ is contained in the complement of $U_0$, which is finite. OK? |
May 4 |
answered | Could we extend any line bundle on the smooth part of a singular curve to a line bundle on the whole curve? |
Apr 28 |
comment |
Determine existence of irreducible variety in given homology class
Also, you would probably want to ask this differently. Say, not for a general homology class, but say a class in $\mathrm{Pic}\, X$. Otherwise, besides solving the Schottky problem you are also in danger of proving the Hodge Conjecture. |
Apr 28 |
comment |
Simple example of a ring which is normal but not CM
@VA: Nice reference! |
Apr 14 |
comment |
Cohen-Macaulayness of the direct image of the canonical sheaf
Just to extend on Karl's excellent answer: a slightly more general situation (that is, not just for $\omega_X$) is handled in Kollár's recent book (goo.gl/JE2CyM). Look at section 2.5.In particular, Theorem 2.74 gives exactly what Karl is saying with the substitution $\mathscr G=\omega_X$. |
Apr 12 |
answered | Question about a divisor and its image |