I've been wondering about this for a while and I have my ideas about the answer, but I would like to make sure once and for all that I'm not missing something.

Let's look at this from a purely linear algebra perspective: let $h$ be a Hermitian inner product on a complex vector space. Should the Kähler form of $h$, say $\omega$, be defined by $h = g + i\omega$, $h = g -i\omega$, $h = g - 2i\omega$, etc?

Here are my thoughts: let's consider the simplest example $h = dz \otimes d\overline{z}$ on $\mathbb{C}$.

Contrary to what the accepted answer says here, I believe there is **one** natural way of identifying elements of an exterior product $\Lambda^2 V$ (which is a quotient of $V \otimes V$) and antisymmetric tensors in $V \otimes V$, and that is
$$x \wedge y \stackrel{\sim}{\to} {x \otimes y - y \otimes x \over 2}$$
This is because it is natural (in my opinion) to ask that this arrow should be a section of the projection map $V \otimes V \to \Lambda^2V$. The same observation goes for the symmetric product.

Coming back to our example, we thus get the expression $$ h = dz \otimes d\overline{z} \approx dz d\overline{z} + dz \wedge d\overline{z} = dx^2 + dy^2 - 2i \,dx \wedge dy~~.$$

So in this example if we want $\omega$ to be the area form $dx \wedge dy$ for the Riemannian metric $g = dx^2 + dy^2$ associated to $h$, I guess we should define $\omega$ by $$h = g - 2i \omega~$$ (in other words $$ \omega = -{1 \over 2} \mathrm{Im} \,h = {i \over 2} dz \wedge d\overline{z}~.)$$

However, it seems to me (maybe) that most authors who seem to care about that convention matter choose instead $h = g -i\omega$.

I guess it's not the end of the world if $\omega = 2 d\mathrm{vol}_g$ instead of $d\mathrm{vol}_g$ (for complex curves), and more importantly that $\omega(x,y) = g(Ix, y)$ is more pleasant than $\omega(x,y) = {1 \over 2} g(Ix, y)$. But I'd like to make sure this is intentional.

Any thoughts on the question?