# Volume form on a complex manifold vs. volume form on the underlying real manifold

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?

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
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 $X$.
• 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 $\Omega\wedge\overline{\Omega}$. Aug 26, 2016 at 16:39