network profile
| bio | website | |
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| location | ||
| age | ||
| visits | member for | 11 months |
| seen | Dec 27 '12 at 14:04 | |
| stats | profile views | 15 |
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Nov 9 |
comment |
Global sections of the line bundle $\mathcal{O}(D)$ Errata. Everything that I wrote in the last comments is wrong! For example, $1/x_0$ as a rational function on $\mathbb{P}^1$ of course has degree zero and divisor $-\{0\}+\infty$ which is different from $D=\{0\}$. |
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Nov 9 |
comment |
Global sections of the line bundle $\mathcal{O}(D)$ Errata. I think as $\eta_\alpha\in\mathcal{K}(X)$ we must take one of the $\eta_\alpha$'s such that $(\eta_\alpha)\neq 0$. |
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Nov 9 |
comment |
Global sections of the line bundle $\mathcal{O}(D)$ And if you picked a different $\alpha$, say $\alpha'$, $s_D$ would be multiplied by the global invertible function $\psi_{\alpha\alpha'}$. |
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Nov 9 |
comment |
Global sections of the line bundle $\mathcal{O}(D)$ Conjecture: $s_D$ is in fact determined only up to global invertible functions $\lambda\in\mathcal{O}_X^{\;*}(X)$, and, in terms of the $\eta_\alpha$'s, $s_D$ is obtained as follows: fix an $\alpha$ and consider $\eta_\alpha$ as a global rational function on $X$, then, for any $\beta$, $s_D|_{U_\beta}=\eta_\alpha^{-1}|_{U_\beta}$. |
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Nov 9 |
awarded | Commentator |
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Nov 9 |
comment |
Global sections of the line bundle $\mathcal{O}(D)$ Well, in this case we can take $\eta_0=x_0$ and $\eta_1=1$ (the constant function $1$ on $U_1$). So $\psi_{01}=x_0$ on $U_{01}=\mathbb{C}\setminus\{0\}$. So I guess it's $s_D=1/x_0$, thought of as a global rational function on $\mathbb{P}^1$. As I remarked in the question, $s_D|_{U_\alpha}$ cannot be equal to $\eta_\alpha^{-1}$ for every $\alpha$: indeed $s_D|_{U_1}=1/x_0\neq 1=\eta_1^{-1}$. Is that all right? |
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Nov 9 |
comment |
Global sections of the line bundle $\mathcal{O}(D)$ I realize maybe there can't be an explicit description of $s_D$ in terms of $\eta_\alpha$, because in fact the $\eta_alpha$'s are determined up to a coboundary for $\mathcal{O}_X^{\;*}$. For $X=\mathbb{P}^1$ and $D=\{0\}$, how can we describe $s_D$ in terms of homogeneus coordinates $[x_0:x_1]$ on $\mathbb{P}^1$? |
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Nov 5 |
asked | Global sections of the line bundle $\mathcal{O}(D)$ |
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Nov 1 |
revised |
(Weil divisors : Cartier divisors) = (p-Cycles : ? ) added 70 characters in body |
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Nov 1 |
revised |
(Weil divisors : Cartier divisors) = (p-Cycles : ? ) added 70 characters in body |
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Nov 1 |
asked | (Weil divisors : Cartier divisors) = (p-Cycles : ? ) |
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Oct 29 |
comment |
Generically non degenerate quadratic forms on a scheme Wait. I was talking about the map of bundles not the corresponding map of sheaves. Take for example an inclusion $\mathcal{O}(-1)\to\mathcal{O}$ on $\mathbb{P}^1$: it's an inclusion of sheaves (it's injective on stalks), but it is not an injective map of line bundles (it's not injective on fibers). |
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Oct 27 |
comment |
Generically non degenerate quadratic forms on a scheme (when I say the map is zero at a point $x$, I mean on the fiber at $x$ not at the stalk at $x$) |
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Oct 27 |
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Generically non degenerate quadratic forms on a scheme I think non-degenerate means $\det(V)\to\det(V^*)\otimes\ell^{n}$ is never zero on $X$, and generically non-degenerate means it can be zero at most along a divisor. Right? (assume $X$ irreducible) |
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Oct 27 |
accepted | Generically non degenerate quadratic forms on a scheme |
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Oct 26 |
awarded | Supporter |
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Oct 26 |
comment |
Sum of inverse squares of denominators Even if this answer is ok, I think I should accept coffeemath's answer instead, because it's more complete. |
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Oct 26 |
accepted | Sum of inverse squares of denominators |
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Oct 26 |
comment |
Sum of inverse squares of denominators Great. Thank you! |
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Oct 26 |
revised |
Sum of inverse squares of denominators deleted 5 characters in body |
