Timeline for Kaehler differentials on a nodal curve
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 8, 2013 at 14:06 | vote | accept | mkemeny | ||
| Jul 8, 2013 at 12:24 | history | edited | user36560 | CC BY-SA 3.0 |
missed $y$ in the differential form.
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| Jul 8, 2013 at 11:48 | history | edited | user36560 | CC BY-SA 3.0 |
added 170 characters in body
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| Jul 7, 2013 at 14:51 | comment | added | mkemeny | Thank you! I did not realize that completion was flat so I did not know how to use the analytic computation rigorously. So in conclusion the correct formula is $\Omega_C / T \simeq \omega_c \otimes I_p$, where $T$ is the torsion subsheaf. | |
| Jul 7, 2013 at 13:36 | history | edited | user36560 | CC BY-SA 3.0 |
added 52 characters in body
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| Jul 7, 2013 at 13:31 | comment | added | user36560 | @mkemeny: you are right. I will edit my answer. | |
| Jul 7, 2013 at 12:10 | comment | added | Alexander Chervov | is it isomorphic as a module to functions on normalizations ? | |
| Jul 7, 2013 at 10:59 | comment | added | mkemeny | Locally anaytically about $p$, $T_p$ is generated by the torsion differential $tds$. Since we have $st=0$ this cannot be a multiple of $ds/s$, so $\Omega_p \neq m_p\omega_p$, at least in the analytic case. | |
| Jul 7, 2013 at 10:40 | comment | added | user36560 | @AlexanderChervov: it is the $m_p$ in my answer above. The stalk $\omega_p$ is generated by $ds/s$, and $\Omega$ is generated by $ds=s.(ds/s)$ and $dt=-t.(ds/s)$, so $\Omega_p=(s,t)\omega_p=m_p \omega_p$. | |
| Jul 7, 2013 at 9:53 | comment | added | Alexander Chervov | Can you comment what the authors of paper mean by ideal sheaves of node ? | |
| Jul 7, 2013 at 9:29 | review | First posts | |||
| Jul 7, 2013 at 9:45 | |||||
| Jul 7, 2013 at 9:10 | history | answered | user36560 | CC BY-SA 3.0 |