Timeline for Polarization induces alternating pairing on homology
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 2, 2022 at 20:29 | comment | added | 57Jimmy | @DanPetersen But in this pairing there is no multiplication involved, right? So where does the Koszul sign rule come into play? | |
| Jul 2, 2022 at 20:14 | comment | added | 57Jimmy | @ali If $\phi:A \to B$ is a map of abelian varieties, the dual map is $\phi^*:B^* \to A^*$. If $B=A^*$, then $B^*=A^{**} \cong A$ and $\phi^*$ is again in $\mathrm{Hom}(A,A^*)$. | |
| Jul 2, 2022 at 9:28 | comment | added | Dan Petersen | Are you familiar with the Koszul sign rule ? In brief, an alternating pairing on $H_1$ is in fact what you would expect from a symmetric construction. Compare with the cup product in the cohomology of a space $X$: it is induced by the diagonal $X \to X \times X$, so it should clearly be commutative as the diagonal is $S_2$-invariant, but in fact it satisfies $\alpha \beta = (-1)^{\vert \alpha \vert \vert \beta \vert} \beta\alpha$ and in particular classes in odd degree anticommute. This type of commutativity twisted by the Koszul rule is in fact the natural one! | |
| Jul 1, 2022 at 12:44 | comment | added | ali | what do you mean $\phi^*=\phi$? the domain of one morphism is $A$ and the other is $A^*$! Any way the definition is like this: a polarisation gives you an ample line bundle $\lambda$ on $A$, this gives you an element of $H^2(A,\mathbb{Q})$ and $H^2(A,\mathbb{Q})$ is the second exterior power of $H^1(A,\mathbb{Q})$ e.g the space of antisymmetric forms on H^1. | |
| Jul 1, 2022 at 12:08 | comment | added | abx | The polarization corresponds to a hermitian form on $H^1(A,\Bbb{C})$, whose imaginary part is an integer-valued skew-symmetric form. You'll find this in any book on Abelian varieties, for instance Mumford's. | |
| Jul 1, 2022 at 11:52 | history | asked | 57Jimmy | CC BY-SA 4.0 |