Kähler form convention

Question

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?

Answer 1

"I believe there is one natural way of identifying elements of an exterior product ..."

I completely sympathize with you. There are so many situations in life where I see one natural way of doing things and other people resist. The fools.

The problem here, as one of my old teachers used to say, is that when there are two ways of doing something there will exist two people for whom one method is completely natural and the other heretic, and they will not agree on which method is the natural one. The mathematical way out of this problem is to define precisely what we mean by "natural", which we've settled on meaning "satisfies a universal property". I don't know enough category theory to prove what I'm going to say now, but I think there's no universal property that does away with the source of your frustration.

It actually seems this is exactly the situation we find ourselves in now. To you, it is natural to embed the antisymmetric algebra in the tensor algebra in one way, which results in a $2^n$ fudge factor when calculating volume forms. To differential geometers, it is natural to choose another embedding that ensures the volume forms of a Kahler form and a Riemannian metric agree. Since all of Kahler geometry rests on applying results of Riemannian geometry to complex situations, I'm quite happy with the second convention since it takes away all those $2^n$ factors that I just know I'd get wrong every single time.

In the end this is a bad (= non-canonical) situation where we have to make an arbitrary choice, and that choice will depend a lot on what you intend to do with your exterior algebra. If you want to do differential geometry there's a choice that makes your life easier, and if you want to do something else then maybe another choice is good for you.