Timeline for Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2012 at 16:46 | comment | added | Mina | ok I will post it as a question :). | |
| Oct 24, 2012 at 12:26 | comment | added | Mina | Yes but does it follow then that $\omega_{J}^{n} = c_{n} \Omega \wedge \Omega$, where $c_{n}$ is a constant depending only on $n$, where $n = dim_{\mathbb{C}}M$? | |
| Oct 23, 2012 at 18:19 | comment | added | Spiro Karigiannis | You should not ask a new question as an answer to an existing question. You can either create a new question or post a comment. But it's not at all clear what you mean. A hyperKahler manifold is Calabi-Yau in an $S^2$ worth of ways. This is clear, because the triple $\omega_I$, $\omega_J$, and $\omega_K$ are all parallel with respect to the Calabi-Yau metric, so the $\Omega$ that Peter defines is parallel, thus the pair $(\omega_J, \Omega = (\omega_K + i \omega_I)^{\dim_{\mathbb C} M}$ is a Calabi-Yau structure. | |
| Oct 23, 2012 at 14:12 | history | answered | Mina | CC BY-SA 3.0 |