# $(1,0)$-forms on a complex manifold

I'm reading in a paper: "Let $\theta$ be a $(1,0)$-form with respect to $I$ ($I \theta = -i \theta$)". Here $I$ is the almost complex structure.

Any idea why it says $I \theta = -i \theta$ and not $I \theta = i \theta$?

NB: It is not a typo, the author repeats this several times.

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What paper? Some more context would probably be helpful. – Micah Aug 12 '14 at 19:44
I don't think it will help, but here it is: see top of p.16 in malkoun.org/malkoun_thesis.pdf – Seub Aug 12 '14 at 19:51

It is a matter of convention, and some authors do not follow it. If $Z$ is a $(1,0)$ vector field, then $IZ = iZ$. If $\alpha$ is a $(1,0)$-form, then one has $I^* \alpha = i \alpha$, where $I^*(\alpha)(-) = \alpha(I-)$ is the (pointwise) pullback of $\alpha$ by $I$. Ok, so one can, say, define $I$ as $I^*$. But then you have a rather annoying sign showing up. Namely, denote by $g$ the map $T\otimes \mathbb{C} \to T^* \otimes \mathbb{C}$ obtained by complexifying the map $T \to T^*$ induced by the metric (also denoted by $g$). Then you have $g \circ I = -I^* \circ g$. In order to avoid the appearance of a minus sign, some authors instead define $I = -I^*$ instead of $+I^*$. This issue is briefly discussed in Chern's complex manifolds without potential theory (along similar lines).
Another advantage of the $-I^*$ convention in hyperkähler geometry is the following. Consider $K = IJ$. This implies that $K^* = J^* I^*$ (since the $*$ operation is contravariant). This can be annoying, if one uses the $+I^*$,$+J^*$,..., convention. However, using the $-I^*$,..., convention, one recovers the familiar $K = IJ$, also for $1$-forms.