What does $(1,1)$-form $\omega$ to be real mean?

Question

In Morrow & Kodaira's book Complex manifolds, p.84: Let $X$ be a complex manifold with a Hermitian metric $h$, to prove the associated $(1,1)$-form $\omega=\frac{\sqrt{-1}}{2}h_{i\bar j}dz^i\wedge d\bar z^j$ to be real, the authors give the proof: $$ \overline{\omega(\xi,\eta)}=\overline{\frac{\sqrt{-1}}{2}h_{i\bar j}(\xi^i\eta^{\bar j}-\eta^i\xi^{\bar j})}=-\frac{\sqrt{-1}}{2}h_{j\bar i}(\xi^{\bar i}\eta^j-\eta^{\bar i}\xi^j)=\omega(\xi,\eta). $$

It seems that the authors assume $\xi=\xi^i\frac{\partial}{\partial z^i}+\xi^{\bar i}\frac{\partial}{\partial \bar z^i}$ and $\eta=\eta^i\frac{\partial}{\partial z^i}+\eta^{\bar i}\frac{\partial}{\partial \bar z^i}$, and in their book, $\xi^{\bar i}=\overline{\xi^i}$, here my question is: why the the coefficients $\xi^{\bar i}$ before $\frac{\partial}{\partial \bar z^i}$ are chosen such that $\xi^{\bar i}=\overline{\xi^i}$? not just freely chosen? for example, $\xi=\frac{\partial}{\partial z}$?

Answer 1

There are a few reasons for why it is desirable to have a real $2$-form rather than a complex one - the first that comes to mind is that the tensor obtained from $\omega$ by $g(u,v):=\omega(u,Jv)$ is then a Riemannian metric, which is certainly a good object to have canonically defined on your manifold.

One way to check that $\omega$ is real, is to plug in real vectors and see if it spits out a real number. This is what the authors are doing in the book you are reading: notice that if you decompose a vector $\xi$ in its $(1,0)$ and $(0,1)$ components, $\xi=\xi^i\partial_{z^i}+\xi^{\bar{i}}\partial_{\bar{z}^i}$, the condition for the vector to be real is exactly $\overline{\xi^i}=\xi^{\bar{i}}$. You can certainly also evaluate $\omega$ on complex vectors, makes perfect sense, but this just won't tell you whether $\omega$ is a real 2-form or not.

It is also possible to prove that $\omega$ is real by directly computing its components in a local system of coordinates for the Hermitian manifold. Then $\omega=\frac{i}{2}h_{a\bar{b}}\operatorname{d}z^a\wedge\operatorname{d}\bar{z}^b$, and if you conjugate this you get $$\bar{\omega}=-\frac{i}{2}\overline{h_{a\bar{b}}}\operatorname{d}\bar{z}^a\wedge\operatorname{d}z^b=\frac{i}{2}\overline{h_{a\bar{b}}}\operatorname{d}z^b\wedge\operatorname{d}\bar{z}^a=\frac{i}{2}h_{b\bar{a}}\operatorname{d}z^b\wedge\operatorname{d}\bar{z}^a=\omega$$ where we used that $h$ is Hermitian (so that $\overline{h_{a\bar{b}}}=h_{b\bar{a}}$), and in the last passage you have to renominate the indices ($a\leftrightarrow b$). This second proof is essentially the same as the one proposed in the book, it just avoids explicitly evaluating $\omega$ on vectors.