bio | website | math.washington.edu/~kovacs |
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location | Seattle, WA | |
age | ||
visits | member for | 4 years, 6 months |
seen | yesterday | |
stats | profile views | 7,146 |
I am an algebraic geometer.
Mar 16 |
answered | Vanishing of the top Chern class of a vector bundle |
Mar 3 |
revised |
Global section of very ample line bundles and its value on stalks
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Mar 3 |
comment |
Global section of very ample line bundles and its value on stalks
Karl, you are right. I was only thinking about 0 or not zero. |
Mar 3 |
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Global section of very ample line bundles and its value on stalks
To get a surjective map to a direct sum you need to get a single element that maps to the various choices. If the same point is repeated, then on the right hand side you can choose an element which has a component that is zero at that point and another component that is non-zero. Then to get surjectivity you would have to find a single section on the left hand side that maps to this element, but it could only map onto one of those components. |
Mar 3 |
answered | Global section of very ample line bundles and its value on stalks |
Mar 2 |
comment |
Reference for Arakelov's theorem: $K^2_f=0$ iff $f$ is locally trivial
(con't-ed) This may seem a trivial point to you now, but it can lead to disaster if you don't keep your notions in order. As an exercise compare the push forwards $f_*K_f$ and $f_*\omega_{X/B}$... |
Mar 2 |
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Reference for Arakelov's theorem: $K^2_f=0$ iff $f$ is locally trivial
@Vesselin: of course you did. I didn't think you meant the tensor power, but no one can know what you would have written if you meant something else than what you have actually written! And just to be a bit nitpicking, there is no such thing as the self-intersection of a line bundle. That's my point. Cycles have self-intersections and line bundles have associated divisors which are cycles. (OK, you could define self-intersection of a line bundle as the self-intersection of the associated divisor, but that's besides the point)... |
Mar 2 |
answered | Reference for Arakelov's theorem: $K^2_f=0$ iff $f$ is locally trivial |
Mar 2 |
comment |
Reference for Arakelov's theorem: $K^2_f=0$ iff $f$ is locally trivial
Please don't write $K^2=\omega^2$!! One of them is a divisor, the other is a sheaf. There is a reason these notions are distinguished. $\omega^2$ is the line bundle corresponding to the divisor $2K$ and not the number $K^2$. Of course, what you have in mind is different, but then why not write down what you have in mind? It's not that hard. You could have written for instance that $K^2=c_1(\omega)^2$... |
Feb 20 |
comment |
Extending holomorphic forms
For $p=\dim X$, the extendibility is almost the same as being lc. That suggests that lc is a natural class. Having said that I am not claiming that for $1$-forms it cannot hold in more generality. After all our paper with Graf is saying that for $1$-forms there is a stronger extension theorem than for forms with $p>1$. Then again, I would expect it to be relatively easy to find examples that are worse than log-canonical where it fails even for $1$-forms. The difficulty in going beyond the lc case is that I don't see a natural class of singularities where it would hold. |
Feb 19 |
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$R^{\dim X-\dim Y}f_{\ast}\omega_X \simeq \omega_Y$ in positive characteristic?
BTW, you exchanged $X$ and $Y$ everywhere where there is an $R^d\pi_*$... |
Feb 19 |
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$R^{\dim X-\dim Y}f_{\ast}\omega_X \simeq \omega_Y$ in positive characteristic?
Karl, I meant to post an answer with essentially the same conclusion (without appealing to your test submodule). I don't see how that spectral sequence would help with the injectivity. That is equivalent to saying that $R^d\pi_*\omega_X$ is torsion-free (which is also shown by Kollár in char=0). The OP claims that it is locally free... I don't see why. |
Feb 18 |
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$R^{\dim X-\dim Y}f_{\ast}\omega_X \simeq \omega_Y$ in positive characteristic?
... and the case of $d=0$ of this question is straightforward. |
Feb 18 |
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$R^{\dim X-\dim Y}f_{\ast}\omega_X \simeq \omega_Y$ in positive characteristic?
@MatthieuRomagny: No. They handle the case when $d=0$ and even pose this as an open problem. |
Feb 18 |
answered | Extending holomorphic forms |
Feb 17 |
comment |
$R^{\dim X-\dim Y}f_{\ast}\omega_X \simeq \omega_Y$ in positive characteristic?
Where did you get the $[-d]$ in the first displayed equation? Or alternatively you should drop the $^\bullet$ from the dualizing complexes. If you do that, then you end up with either $\mathscr Ext^0$ or $^{-d}$, but without the $^\bullet$'s. (cont'd) |
Feb 10 |
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Definition and sigularity of Ramified covers
I can confirm that Karl is completely right: 1) we meant the codimension $1$ points of $X$ that lie in $\mathrm{Sing}X$ and 2) our main concern was to allow covers of non-normal schemes. An interesting point that shows why one needs to exclude codimension $1$ points is that for instance the Hurwitz formula (see page 65) fails for the normalization map. More generally, the normalization map works sort of the opposite way than ramified covers (see what I mean in (5.7) on page 191) and then one runs into difficulty comparing discrepancies.... |
Feb 1 |
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Can a rational map be extended without using resolution of indeterminacies?
Limits don't work, because they're not unique. Think of a blow-up. A point might have to be mapped to a higher dimensional object, that's why you need a resolution. If you do a push-out, that will contract the locus where the indeterminacy should map, so for instance in the example above, you will get the identity map $C'\to C'$. |
Feb 1 |
answered | Can a rational map be extended without using resolution of indeterminacies? |
Jan 31 |
revised |
Blowing-up and direct image sheaf.
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