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Consider a closed complex manifold $X=(M,J)$ of complex dimension $n$, where $M$ is a real smooth manifold of (real) dimension $2n$ and $J$ an integrable almost complex structure on it.

The holomorphic tangent bundle of $X$ is $T_X\cong T_X^{1,0}$, so a volume form for $X$ will be a nowhere vanishing section of $$\bigwedge^n T_X^*\cong\bigwedge^n(T_X^{1,0})^*, $$ i.e., a global section of the canonical bundle $\Omega^{n,0}_X=K_X$.

Up to this point everything seems to make sense, unless I'm making a mistake somewhere.

My confusion arises when we consider the underlying smooth manifold $M$. For $M$, the volume form is a section of $\Omega^{2n}_M$, which by Hodge decomposition is itself isomorphic to $\Omega^{n,n}_X$. Therefore, I conclude that a volume form of $M$ is a section of this vector bundle, which is clearly different from the other volume forms we got before.

I'm pretty sure the first approach I mention is the correct one, yet can't find the mistake in the second argument. Could anyone help me understand where am I going astray?

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1 Answer 1

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A nowhere zero section $\Omega$ of $K_X$ is called a holomorphic volume form, but it is not a volume form on the underlying smooth manifold $X$ precisely for the reason you point out. However, in any coordinate chart $(U, (z^1, \dots, z^n))$, $\Omega = fdz^1\wedge\dots\wedge dz^n$ for some nowhere-zero holomorphic function $f$ on $U$, so

\begin{align*} \Omega\wedge\overline{\Omega} &= fdz^1\wedge\dots\wedge dz^n\wedge\overline{(fdz^1\wedge\dots\wedge dz^n)}\\ &= fdz^1\wedge\dots\wedge dz^n\wedge(\bar{f}d\bar{z}^1\wedge\dots\wedge d\bar{z}^n)\\ &= |f|^2dz^1\wedge\dots\wedge dz^n\wedge d\bar{z}^1\wedge\dots\wedge d\bar{z}^n\\ &= (-1)^{n(n-1)/2}|f|^2dz^1\wedge d\bar{z}^1\wedge\dots\wedge dz^n\wedge d\bar{z}^n\\ &= (-1)^{n(n-1)/2}|f|^2(-2idx^1\wedge dy^1)\wedge\dots\wedge(-2idx^n\wedge dy^n)\\ &= (-1)^{n(n-1)/2}(-2i)^n|f|^2dx^1\wedge dy^1\wedge\dots\wedge dx^n\wedge dy^n. \end{align*}

As $f$ is nowhere-vanishing, $\Omega\wedge\overline{\Omega}$ is a nowhere vanishing $2n$-form and hence a volume form on the smooth $2n$-dimensional manifold $X$.

Note, another distinction is that an orientable smooth manifold always admits a volume form, but a complex manifold need not admit a holomorphic volume form. A complex manifold which admits a holomorphic volume is called a Calabi-Yau manifold.

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  • $\begingroup$ Awesome, so just yo be completely sure: both my arguments are correct, but strictly speaking $\Omega$ is not a volume form, yet it gives rise to a legitimate volume form $\Omega\wedge\overline{\Omega}$. $\endgroup$
    – user347489
    Aug 26, 2016 at 16:39
  • $\begingroup$ Absolutely correct. $\endgroup$ Aug 26, 2016 at 16:41
  • $\begingroup$ Great! Thanks, Michael. $\endgroup$
    – user347489
    Aug 26, 2016 at 16:42

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