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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!

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(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.

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