Timeline for Hilbert's Syzygy Theorem in the bigraded case
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15 events
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| Dec 17, 2022 at 21:35 | answer | added | Jason Starr | timeline score: 0 | |
| Dec 16, 2022 at 10:03 | comment | added | Carnby | @JasonStarr ... of the question I posted? It's just because in the graded case it can be proved in a rather elementary way, and I'd like to explore first this possibility before getting into all this sheaf-theoretic machinery you exposed. Again, thank you very much for time, your comments are really appreciated. | |
| Dec 16, 2022 at 9:59 | comment | added | Carnby | @JasonStarr Thank you very much for the explanation and the references. I'll take a careful look at all this concepts because there are some of them I'm totally unfamiliar with. The question I posted arose when I was studying the concept of regularity of sheaves and the possibility of using it to define the Hilbert scheme for products of projective spaces from a 'bigraded point of view', in a similar way we do for the Hilbert scheme in the usual projective space. Anyway, I'd like to introduce all this stuff in a more elementary way, so, do you know if there is a purely algebraic proof... | |
| Dec 16, 2022 at 3:17 | comment | added | Jason Starr | Olsson and I considered writing this up for the Hilbert polynomials section of our article on Quot schemes for stacks, but we decided against it since we could not see a low-tech way to extend this to the standard generators of the K-group of the stack $B\textbf{GL}_m$. | |
| Dec 16, 2022 at 3:15 | comment | added | Jason Starr | . . . Thus, in the K-group of $X$, there exist finitely many elements $\mathcal{B}_{e_1,\dots,e_r}=\bigotimes_{i=1}^n(\mathcal{A}\otimes \mathcal{L}_i)^{\otimes(-e_i)}$ with $0\leq e_i\leq r_i$ such that every element $\bigotimes_{i=1}^n \mathcal{L}_i^{\otimes d_i}$ is a linear combination of the elements $\mathcal{B}_{e_1,\dots,e_r}$ with coefficients that are polynomials in $(d_1,\dots,d_r)$. This gives the existence of a "multigraded Hilbert polynomial" on $X$. | |
| Dec 16, 2022 at 3:12 | comment | added | Jason Starr | Actually, you need much less than Hirzebruch-Riemann-Roch. Snapper's method works just fine: for every finite set of invertible sheaves, $(\mathcal{L}_1,\dots,\mathcal{L}_n)$, on a projective scheme $X$, there exists a very ample invertible sheaf $\mathcal{A}$ such that each invertible sheaf $\mathcal{A}\otimes \mathcal{L}_i$ is also very ample. In the K-group of $\mathbb{P}^r$, each $[\mathcal{O}(d)]$ is a linear combination $\sum_{e=0}^r f_e(d)[\mathcal{O}(-e)]$ where $f_e(d)$ is a polynomial of degree $r$ . . . | |
| Dec 16, 2022 at 0:55 | comment | added | Jason Starr | The existence of such a Hilbert polynomial in two variables follows from Hirzebruch-Riemann-Roch. | |
| S Dec 15, 2022 at 2:51 | review | First questions | |||
| Dec 15, 2022 at 7:28 | |||||
| S Dec 15, 2022 at 2:51 | history | asked | Carnby | CC BY-SA 4.0 |