Timeline for determinant of integrals of forms
Current License: CC BY-SA 3.0
15 events
when toggle format | what | by | license | comment | |
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May 28, 2014 at 12:34 | answer | added | Jason Starr | timeline score: 1 | |
S May 27, 2014 at 15:20 | history | suggested | user21574 | CC BY-SA 3.0 |
typo edited
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May 27, 2014 at 15:14 | review | Suggested edits | |||
S May 27, 2014 at 15:20 | |||||
May 27, 2014 at 14:39 | comment | added | detted92 | I edited the question | |
May 27, 2014 at 14:38 | history | edited | detted92 | CC BY-SA 3.0 |
added 538 characters in body
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May 27, 2014 at 14:31 | comment | added | detted92 | Umm, good point. What is then the operation dual to cup-product? | |
May 27, 2014 at 14:26 | comment | added | David Loeffler | You can't take the cup-product of homology classes! Cup-product is an operation on cohomology, so the cycle you're integrating over is nonsensical as soon as $n > 1$. | |
May 27, 2014 at 14:09 | comment | added | detted92 | Thanks Jason. Ok, let me ask this question: you take an abelian variety of dimension $d$, a basis of $H^0(A, \Omega^1_A)$, you pick some cycles $\gamma_1, \ldots, \gamma_d$ and you write the corresponding matrix. Is it possible to express its determinant as the integral of $\omega_1 \wedge \cdots \wedge \omega_d$ (now an element of $H^0(A, \Omega^d)$) over a suitable cycle in $H_d(A(\mathbb{C}), \mathbb{Q})$? | |
May 27, 2014 at 14:07 | review | First posts | |||
May 27, 2014 at 14:49 | |||||
May 27, 2014 at 14:06 | comment | added | detted92 | The cup product | |
May 27, 2014 at 14:04 | comment | added | abx | What is $\gamma _1\cup\ldots \cup \gamma _n$? | |
May 27, 2014 at 14:04 | comment | added | Jason Starr | "In that case both sides are zero, no?" Why? Certainly the integral you write is zero, but that does not imply that the determinant is zero. That is why I raised my objection. | |
May 27, 2014 at 14:01 | comment | added | detted92 | In that case both sides are zero, no? But you are right that the case I have in mind is $X$ an abelian variety of dimension $n$ and $\omega_i$ a basis of $H^0(X, \Omega^1_X)$ | |
May 27, 2014 at 13:59 | comment | added | Jason Starr | What if the dimension of $X$ is less than $n$? Do you want to assume that $X$ is a complex torus and $\omega_1,\dots,\omega_n$ form a basis? | |
May 27, 2014 at 13:51 | history | asked | detted92 | CC BY-SA 3.0 |