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Universität/Hochschule Universelles Geradenbündel
xiao_shi_tou_
Senior Letzter Besuch: in der letzten Woche
Dabei seit: 12.08.2014
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Hallo zusammen.
Bei der Definition des $\theta$-Divisors auf der Jakobischen einer Kurve $C/S$ begannen wir mit folgendem Satz:

"Sei $\lineb$ ein universelles Geradenbündel vom Grad $g-1$ (relativ zu $\pi$) auf $\Jac_{C/S}^{g-1}\tm S C\overset{\pi}{\to} \Jac_{C/S}^{g-1}$."

Was könnte das sein, ein universelles Geradenbündel?

Ich bin mir nicht sicher was $C/S$ hier ist, da es nicht gesagt wurde, aber ich nehme mal an das ist ein projektiver (eigentlicher) flacher Morphismus $C\to S$ sodass die geometrischen Fasern $C_{\cl{s}}=C\ot_S \sp{\cl{\kappa(s)}}$ reguläre zusammenhängende Kurven sind, denn so waren die Voraussetzungen an $C\to S$ als wir die Jakobische konstruiert haben.

$\viele$


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”己所不欲,勿施于人“(Konfuzius)
PS: Falls ich plötzlich aufhöre in einem Thread zu antworten, dann kann es sein, dass ich es vergessen habe. Ihr könnt mir in diesem Fall eine Private Nachricht schicken.
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Triceratops
Aktiv Letzter Besuch: in der letzten Woche
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Wie habt ihr denn $\mathrm{Jac}^g_{C/S}$ definiert? Und kannst du etwas dazu sagen, was $SC$ ist? (Oder meintest du $\times_S~ C$ eventuell?)

Die Jacobische (ohne Gradangabe) $J_{C/S}$ wird jedenfalls (zum Beispiel in Kapitel 9 von Néron Models von BLR) mit Hilfe einer universellen Eigenschaft (Identitätskomponente des relativen Picardfunktors) definiert, hier wäre also das universelle Geradenbündel in der Definition mit drin.



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kurtg
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Hi,

$\mathrm{Jac}^g_{C/S}$ ist vermutlich $\mathrm{Pic}^g_{C/S}$. Universell heißt, dass jedes Geradenbündel vom Grad $g$ Pullback des universellen ist. Das müsste in BLR erklärt sein. Das Picardschema zerfällt in die $\mathrm{Pic}^g$s, die alle $\mathrm{Pic}^0$-Torseure sind.

Der Theta-Divisor ist die $(g-1)$-fache Summe der Kurve vom Geschlecht $g$ unter der Abel-Jacobi-Abbildung und eine kanonische prinzipale Polarisierung der Jacobischen einer Kurve.



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xiao_shi_tou_
Senior Letzter Besuch: in der letzten Woche
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2019-10-11 15:21 - kurtg in Beitrag No. 2 schreibt:
Hi,

$\mathrm{Jac}^g_{C/S}$ ist vermutlich $\mathrm{Pic}^g_{C/S}$. Universell heißt, dass jedes Geradenbündel vom Grad $g$ Pullback des universellen ist. Das müsste in BLR erklärt sein. Das Picardschema zerfällt in die $\mathrm{Pic}^g$s, die alle $\mathrm{Pic}^0$-Torseure sind.

Der Theta-Divisor ist die $(g-1)$-fache Summe der Kurve vom Geschlecht $g$ unter der Abel-Jacobi-Abbildung und eine kanonische prinzipale Polarisierung der Jacobischen einer Kurve.

Hallo!
Vielen Dank euch beiden.

Wir haben $\Jac_{C/S}^g$ nie wirklich definiert, aber wir haben in einer anderen Vorlesung gezeigt, dass der relative Picard Funktor $\Pic_{C/S}^g$ der Geradenbündel vom faserweisen Grad $g$ welche "auf $S$ trivial" sind unter gewissen Voraussetzungen darstellbar ist und daher nehme ich an, dass $\Jac_{C/S}^g$ das darstellende Objekt sein soll, also das was man normalerweise mit $\Pic_{C/S}^g$ bezeichnet.

Die Einbettung einer Kurve in seine Jacobische, die Konstruktion und Definition des Theta Divisors und das durch den Theta Divisor gegebene Skalarprodukt haben wir aus Zeitmangel nur sehr skizzenhaft (bzw. das erste Thema gar nicht) behandelt. Ich habe aber in Milnes Notizen schon etwas nachgelesen und schaue mir auch die anderen Quellen noch genauer an.

Diese Themen haben wir für den Beweis von Mumfords Ungleichung und den Beweis von Faltings-Mordell verwendet.

Mit $\tm SC$ meinte ich $\tm_S C$, ein Tippfehler.

Ich kann jetzt erstmal die Quellen zu diesen Standardthemen befragen, aber erstmal versuche ich noch den Beweis von Mordell-Weil zu verstehen, wie er in Mumfords Buch steht, da der Beweis aus der Vorlesung mich etwas überfordert hatte.

Vielen Dank für das Engagement.
XST

\(\endgroup\)


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xiao_shi_tou_
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2019-10-11 15:21 - kurtg in Beitrag No. 2 schreibt:
Hi,
Universell heißt, dass jedes Geradenbündel vom Grad $g$ Pullback des universellen ist.

Hallo!
Verwirrend finde ich nur, dass wir erst ein Universelles G.B. genommen haben und damit den $\theta$-Divisor konstruiert haben.
Dann haben wir erst die Eigenschaft bewiesen, dass jedes G.B. vom Grad $g$ der Pullback von $\c{O}_{\Jac_{C/S}^{g-1}}(\theta)$ ist.

Ich kann das auch gerne ausführen wie wir das gemacht haben.

\(\endgroup\)


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xiao_shi_tou_
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2019-10-10 20:59 - Triceratops in Beitrag No. 1 schreibt:
Wie habt ihr denn $\mathrm{Jac}^g_{C/S}$ definiert? Und kannst du etwas dazu sagen, was $SC$ ist? (Oder meintest du $\times_S~ C$ eventuell?)

Die Jacobische (ohne Gradangabe) $J_{C/S}$ wird jedenfalls (zum Beispiel in Kapitel 9 von Néron Models von BLR) mit Hilfe einer universellen Eigenschaft (Identitätskomponente des relativen Picardfunktors) definiert, hier wäre also das universelle Geradenbündel in der Definition mit drin.

Hi.
Ich glaube, dass es so gemeint ist.
Das macht auf den ersten Blick auch Sinn.
$\viele$
\(\endgroup\)


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Beitrag 4 sieht aber so aus, dass ihr die universelle Eigenschaft nicht als Definition genommen habt, sondern eine explizite Konstruktion, wovon dann erst die universelle Eigenschaft nachgewiesen worden ist. Das (universelle) Geradenbündel muss dann allerdings auch explizit konstruiert worden sein. Mehr kann man dazu jetzt ohne eure Konstruktion vermutlich nicht sagen.



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2019-10-11 22:16 - Triceratops in Beitrag No. 6 schreibt:
Beitrag 4 sieht aber so aus, dass ihr die universelle Eigenschaft nicht als Definition genommen habt, sondern eine explizite Konstruktion, wovon dann erst die universelle Eigenschaft nachgewiesen worden ist. Das (universelle) Geradenbündel muss dann allerdings auch explizit konstruiert worden sein. Mehr kann man dazu jetzt ohne eure Konstruktion vermutlich nicht sagen.

Hallo Triceratops,
die Konstruktion die wir in der Vorlesung hatten findest du hier.
Diese Vorlesung ist noch nicht komplett, aber die Konstruktion des Theta-Divisors steht dort schon.

Ich habe Fragen mit dem Symbol $\qst$ versehen.
Vielleicht weißt du ja etwas dazu.
Vielen Dank für Deine Hilfe.
Viele Grüße
XST
\(\endgroup\)


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