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For any complex manifold $M$ we have the usual decomposition of the cotangent bundle into the direct sum $\Omega^{(1,0)} \oplus \Omega^{(0,1)}$. I know these two sub-modules as the holomorphic and anti-holomorphic forms respectively. However, I have been told that this is the terminology usually used by physicists. How to mathematicians refer to $\Omega^{(1,0)}$ and $\Omega^{(0,1)}$?

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 Holomorphic and anti-holomorphic forms :-) – Francesco Polizzi Dec 16 2010 at 23:07
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 The issue is that you might want the phrase "holomorphic form" on a complex manifold to refer to something that is $\overline{\partial}$-closed as well as $(p,0)$. I prefer this, as the notion of holomorphic form lives entirely within the world of holomorphic geometry - i.e. you can define the sheaf $\Omega ^p _{hol}$ of holomorphic p-forms without reference to an underlying real manifold structure. I find calling them just (1,0) or (0,1)-forms avoids potential confusion. – Sam Gunningham Dec 17 2010 at 2:19
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 Actually, in Wells's "Differential Analysis on Complex Manifolds", $\Omega^{(p,q)}$ are called differential forms of type $(p,q)$, even when either $p$ or $q$ is zero. He reserves the terminology holomorphic for forms which are in the kernel of the $\overline\partial$ operator. See, for instance, Example 2.12 in the third edition of that book. (This used to be an answer, but I think that it is best left as a comment.) – José Figueroa-O'Farrill Dec 17 2010 at 3:00
The motivation for only calling $\bar \partial$ closed $(p,0)$-forms holomorphic comes from that when you write a $(p,0)$-form in local coordinates, its coefficient functions will be holomorphic exactly when the form is $\bar \partial$ closed. – Gunnar Magnusson Dec 17 2010 at 8:06
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 Perhaps this is unnecessary, but to drive home the point, $\overline{z}dz$ has no right to be called holomorphic. It's better to call it type (1,0). – Donu Arapura Dec 17 2010 at 13:58

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