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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jul 6, 2013 at 6:58 vote accept Michael Albanese
Jul 5, 2013 at 15:22 comment added diverietti Exactly. Moreover, the coefficients $c_{jk\lambda\mu}$ come from $\Theta(E)$. I mean that $\Theta(E)=\sum c_{jk\lambda\mu}dz_j\wedge d\bar z_k\otimes e_\lambda^*\otimes e_\mu$, and these $c_{jk\lambda\mu}$ are the ones which satisfy the hermitian relations above.
Jul 4, 2013 at 4:22 comment added Michael Albanese So does that mean in your previous answer you want $\theta_E(v\otimes\sigma, v\otimes\sigma) = h(\Theta(E)(v, \bar{v})\cdot\sigma, \sigma)$? Furthermore, just to be clear, the coefficients $c_{jk\lambda\mu}$ come from $i\Theta(E)$ not $\Theta(E)$, correct?
S Jun 27, 2013 at 6:08 history suggested Fred Daniel Kline CC BY-SA 3.0
fixed syntax of array
Jun 27, 2013 at 4:06 review Suggested edits
S Jun 27, 2013 at 6:08
Jun 23, 2013 at 22:25 history edited diverietti CC BY-SA 3.0
added 475 characters in body
Jun 23, 2013 at 22:11 history edited diverietti CC BY-SA 3.0
added 127 characters in body; added 84 characters in body
Jun 23, 2013 at 22:00 history answered diverietti CC BY-SA 3.0