Timeline for Is there a covering of Prym variety?
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6 events
| when toggle format | what | by | license | comment | |
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| Jan 19, 2016 at 17:09 | comment | added | roy smith | Explicitly, if D' = j(D) denotes the involution applied to a divisor D, and q is a chosen point of C, the map C^(d)-->P takes a divisor D to (the class of ) D-D'-d(q-q'). If you are willing to map to the other component of the kernel, just send D to the class of D-D'. | |
| Jan 19, 2016 at 2:13 | comment | added | roy smith | If j is the involution of J(C) induced by the sheet interchange involution on the curve C, then (1-j):J(C)-->P is a surjection whose composition with the abel map gives an abel-prym map C-->P. Adding the images of any d points of C, gives a map C^(d)-->P which, if my computation is right, maps the fundamental class of C^(d) to 2^d times the fundamental class of P, hence the map is surjective and finite. This may avoid the need to understand the finite subgroups. | |
| Jan 17, 2016 at 15:43 | comment | added | Dimitri Koshelev | Jason, thank you for answering my question. How can I explicitely construct the scheme $\Gamma$ and the morphism $\pi$? | |
| Jan 16, 2016 at 21:31 | comment | added | roy smith | nice answer. Over the complex numbers at least, there is a topological approach as well generalizing the topological proof of jacobi inversion. I.e. the g fold pontrjagin product of the homology class of the abel curve C in J(C) is, as I recall, g! times the fundamental class of J(C), hence the abel map from C^(g)-->J(C) is surjective of degree one. Since the abel prym curve C in the prym variety has twice this minimal class, the same computation apparently shows that d fold sums of the abel prym curve give the prym variety via a map of degre 2^(d) from the d fold symmetric procduct. | |
| S Jan 16, 2016 at 19:55 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S Jan 16, 2016 at 19:55 | history | made wiki | Post Made Community Wiki by Jason Starr |