Definition of $dz_i\otimes d\bar{z_j}(\frac{\partial}{\partial z_m},c\frac{\partial}{\partial z_n})$

Is $dz_i\otimes d\bar{z_j}(\frac{\partial}{\partial z_m},c\frac{\partial}{\partial z_n}):=dz_i(\frac{\partial}{ \partial z_m})d\bar{z_j}(\bar{c}\frac{\partial}{\partial \bar{z_n}})$?

It seems that by the word above, it follows that $dz_i\otimes d\bar{z_j}(\frac{\partial}{\partial z_i},c\frac{\partial}{\partial z_j}):=dz_i(\frac{\partial}{ \partial z_i})d\bar{z_j}(\bar{c}\frac{\partial}{\partial \bar{z_j}})=\bar{c}$.

But I don't know if it's a convention to take the conjugate of the coordinate corresponding to the $\bar{z_j}$ or is there a reason? Because in my view (maybe not right), $dz_i\otimes d\bar{z_j}(\frac{\partial}{\partial z_i},c\frac{\partial}{\partial z_j})$ should be $dz_i(\frac{\partial}{\partial z_i})d\bar{z_j}(c\frac{\partial}{\partial z_j})=1\cdot 0=0$.

Reference: Griffith and Harris p.27

Solution 1:

There are indeed various typos and issues in Griffiths and Harris. Signs in complex geometry are always a bugaboo, too; even Chern had the sign wrong in defining $c_1$ in the first edition of his Complex Manifolds without Potential Theory.

G/H are inconsistent here. A hermitian metric should be given on the complex vector space $T'_z(M)$, so, as such we feed in two (holomorphic) tangent vectors. Let's think about $\Bbb C$: To get sesquilinearity, we need to take $dz\otimes d\bar z$, with the understanding that $d\bar z(v) = \overline{dz(v)}$. Then $\|v\|^2 = dz\otimes d\bar z(v,v)$. On the other hand, if you think of the hermitian metric as giving a bilinear map on $T'_z(M)\times T''_z(M)$, then we take $\|v\|^2 = dz\otimes d\bar z(v,\bar v)$. Because we're going to work primarily with differential forms we really prefer to work with the latter interpretation, thinking of the hermitian metric as a $(2,0)$ tensor on the complexified tangent space.

By the way, there's a sign error on the top of p. 29. There should be a negative sign in the first displayed equation giving the definition of a positive $(1,1)$-form.