How to see if a (1,1)-form stems from a hermitian metric In Griffiths & Harris Principles of Algebraic Geometry on page 28 it is explained how you derive a real differential (1,1)-form $\omega$ from a hermitian metric $ds^2$ on a complex monifold $M$. Namely, once you find a coframe $(\varphi_1,\dots,\varphi_n)$ of (1,0) forms such that
$$ds^2 = \sum \varphi_i \otimes \overline{\varphi_i}$$
$\omega$ is given by
$$\omega= -\frac{1}{2} \Im(ds^2) = \frac{i}{2} \sum \varphi_i \wedge \overline{\varphi_i}.$$
On page 29 G&H try to explain if once a real differential (1,1)-form $\omega$ is given this stems from a hermitian metric $ds^2$. They say this is the case if and only if
$$i \langle \omega(z), v \wedge \overline v \rangle >0$$
for every $z \in M$ and every $v \in {T_z}'(M)$ (I guess they mean $v \ne 0$).
My problems/questions:


*

*I'm not sure (but I have an assumption) what is meant by $\langle \omega(z), v \wedge \overline v \rangle$. Can you explain to me what that means and why it is well-defined (if it means what I assume well-definedness is not clear to me).

*Why is this the condition which ensures $\omega$ arises from a hermitian metric?

 A: Actually, this is off by a sign. What they are giving is the condition for the bilinear form on the complexified tangent space to be positive definite. A $2$-form (in particular, this $(1,1)$-form) needs to be evaluated on an ordered pair of tangent vectors — writing $\omega(v,\bar v)$ is the same as contracting the $2$-form with the $2$-vector $v\wedge\bar v$. 
To check the sign, note for example that the Kähler form for $\Bbb C^n$ is
$\omega = \frac i2 \sum dz^j\wedge d\bar z^j$, and if we take $v=\partial/\partial z^k$, we get
$$\omega(v\wedge\bar v) = \omega(v,\bar v) = \tfrac i2 \sum dz^j\big(\frac{\partial}{\partial z^k}\big)d\bar z^j\big(\frac{\partial}{\partial \bar z^k}\big) - dz^j\big(\frac{\partial}{\partial\bar z^k}\big)d\bar z^j\big(\frac{\partial}{\partial z^k}\big) = \tfrac i2.$$
This tells us that we need $-i\omega(v,\bar v)>0$ for positive-definiteness. You can check the general case. Indeed, $\omega = \frac i2\sum h_{j\bar k}dz^j\wedge d\bar z^k$ is positive-definite if and only if the matrix $[h_{j\bar k}]$ is positive-definite hermitian.
