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It is a basic and "intuition request" question. I have asked it on StackExchange yet it is probably to specialized for it since there were no answears.

Generalised complex structure is defined to be a field of endomorphisms $\mathcal{J}$ of the big tangent bundle $T^{big}M=TM \oplus T^*M$ such that $\mathcal{J}^2=-I$ and being orthogonal with respect to "natural paring" - neutral metric - $<X+ \xi, Y+\eta>=\xi(Y)+\eta(X)$. The first condition gives reduction of structure group of $T^{big}M$ to complex group, my question is why do we require the second condition?

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  • $\begingroup$ Are classical complex structures orthogonal with respect to some metric? $\endgroup$
    – Vít Tuček
    Commented Jan 23, 2017 at 17:16
  • $\begingroup$ This is not a metric on the manifold. It is the natural pairing between tangent and cotangent bundles, induced by the interior product: it's natural to require orthogonality with respect to it. In a sense, I would say, it's just to keep tangent and cotangent bundles "equal but different". $\endgroup$
    – daniele
    Commented Jan 24, 2017 at 11:05
  • $\begingroup$ @daniele I don't think anyone claimed it was a bonfide metric. $\endgroup$
    – Josh Lackman
    Commented Jan 24, 2017 at 21:17
  • $\begingroup$ mathoverflow.net/questions/139305/… $\endgroup$
    – user21574
    Commented Feb 23, 2017 at 6:01

1 Answer 1

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This condition implies that $J $ preserves the natural pairing. This is quite typical; if one puts a metric on a complex manifold one usually requires the analogous condition to hold, making it a Kahler manifold.

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