# “Barred” Tensor Indices in Complex Manifolds

I'm having an embarrassingly hard time straightening out how to work with the "barred" indices that show up in tensors on complex manifolds. For example, the Kahler form $\omega = \frac{i}{2}g_{i \bar{j}}dz^{i} \wedge d\bar{z}^{\bar{j}}$. Is it correct to say that these barred indices only serve to denote an index that is contracted with either a $d\bar{z}$ or $\partial/\partial \bar{z}$ in some general tensor? In other words, we could I think equivalently write the above Kahler form as $\omega = \frac{i}{2}g_{i j}dz^{i} \wedge d\bar{z}^{j}$, correct? I suppose this barred notation is simply convenient when writing things out in coordinates, so we can read off the holomorphic and anti-holomorphic components.

The precise problem in which this tripped me up, is arguing that the $g_{i \bar{j}}$ coming from the above Kahler form is actually a Hermitian matrix in local coordinates. Clearly, without the barred notation, we would say a matrix is Hermitian if its entries satisfy $g_{ij}=(g_{ji})^{*}$. Can someone perhaps help me reason through how the barred indices are affected by this complex conjugation? I know the Kahler form is real, so it should equal its complex conjugate, but even for a non-real form there is a way to conjugate components, and the Hermitian condition I think is something extra. I'd very much appreciate a little nudging here!

(First, try to avoid using $i=\sqrt{-1}$ when you're also using $i$ as an index. First pitfall with complex geometry. :) ) The custom is to write $$\sum g_{i\bar j} dz^i\wedge dz^{\bar j}$$ so that the summation convention is consistent for both unbarred and barred indices. The hermitian condition is then consistent as well: To say that $\left[g_{i\bar j}\right]$ is a hermitian matrix is to say that $g_{j\bar i} = \overline{g_{i\bar j}}$; note that $i$ gets barred and $\bar j$ gets unbarred on the right.
Now, there is one more issue to sort out (and I don't want to write a whole lecture here). Is $d\bar z$ being fed a holomorphic tangent vector $v$ or an antiholomorphic tangent vector $\bar v$? In the former case, we interpret $d\bar z(v)$ as $\overline{dz(v)}$; in the latter, we interpret $d\bar z(\bar v)$ as the same thing. So you have to wrestle with how you're dealing with the complexified tangent bundle.