Let $\mathbb{P^nC}$ the complex projective space and $O(1)$ the dual bundle to the tautological bundle $O(-1)= \{([x], z) \in \mathbb{P^nC} \times \mathbb{C}^n \ \vert \ z \in \mathbb{C}^* \cdot x \}$. I have a lot of problems to solve following exercise from my complex geometry class:
Show that $c_1(O(1))(-,-)$ is positive (that means to show that $c_1(-, J(-))$ is Riemannian metric, the $J$ from almost complex structure of $\mathbb{P^nC}$).
recall the endomorphism $J$ on $T\mathbb{P^nC}$ is here literally the multiplication by $i$ since every complex manifold $M$ has natural structure of almost complex manifold $(M,J)$ where endomorphism $J \in End(TM)$ has property $J^2= -Id$.
The suggested strategy includes two hints:
The positivity of $c_1(O(1))(-,-)$ can be calculated in one fixed point of $\mathbb{P^nC} $ and then one can apply the $SU(n+1)$-action on $O(1)$.
Find a local holomorphic section of $O(-1)$ and consider the induced riemannian metric on it.
I don't know from which side I should approach the problem. The 1. hint I think exploits that $O(1)$ is invariant by $SU(n+1)$-action: namely for all $M \in SU(n+1): M^*O(1) \cong O(1)$. This allows to choose arbitrary point/chart where the claim can be locally checked.
The reason why it's true I think is a consequence from $SU(n+1)$-invariance of tautological bundle $O(-1)$ which can be explicitly computed and $\mathbb{P^nC} \times \mathbb{C} \cong O(1) \otimes O(-1)$.
From here the actually problems start: first of all $c_1(O(1))(-,-)$ is a priori a homology class living in $H^2(\mathbb{P^nC})$. This class can be represented by arbitrary curvature $\Omega^{O(1)}= \nabla^2$ for arbitrary connection $\nabla$ on $O(1)$. That is as homology class we have
$$[\frac{\Omega^{O(1)}}{2 \pi i}] =[\frac{\nabla^2}{2 \pi i}]=[c_1(O(1))] \in H^2(\mathbb{P^nC})$$
But how a class can be considered as a well defined metric/ $2$-form? Why is $c_1(O(1))(-,J(-))$ well defined metric? I conjecture that this could make sense if we show that in our special case the choice of a connection $\nabla^{O(1)}$ is unique. Is this true if we know that $O(1)$ is holomorphic bundle?
Next, the idea 2. I not understand. If we choose for example an affine chart $U \cong \mathbb{C}^n$ around eg $p=[1:0:...:0]$, choose a holomorphic section on $O(-1)$ and consider the induced riemannian metric on it, how is it related to $c_1(-, J(-))$ at long last?
I think the suggested canonical choice of a section $s: U \cong \mathbb{C}^n \to O(-1)$ is inteded to be nothing but
$$v=(v_1,...,v_n) \mapsto ([1:v_1:...v_n], (1,v_1,...,v_n)) \in O(-1)$$ The metric $h$ on $O(-1)$ is locally induced by standard hermitian $<-,->$ on $\mathbb{C}^n$ by $h((1, v), (1, w))= 1+ \langle v,w\rangle $ for $v, w \in \mathbb{C}^n$. laborious calculation might show it's well defined.
Now how this induced metric $h$ is related to $c_1(-, J(-))$? I not know how to utilize the 2. hint in order to verify $c_1(-, J(-))$ is Riemannian.
Concerning your two hints:
For (1), the hint only make sense if the curvature form $\Theta$ is $SU(n+1)$ invariant: $L_A ^* \Theta = \Theta$ for all $A\in SU(n+1)$, where $L_A [v] := [Av]$ for all $[v]\in \mathbb P^n$. So you might need to construct a connection so that the curvature form is invariant.
(2): If I understand correctly, the hint suggests that you consider $O(-1)$ as a subbundle of $\mathbb P^n \times \mathbb C^{n+1}$ and use the metric induced from this trivial bundle. You figured out that already: just take the standard hermitian metric $h(u, v) = u\cdot \bar v$ on $\mathbb C^{n+1}$.
Now, to answer some of your questions in the comment.
What you know is that the curvature form is given by $\Theta = \nabla ^2$. This is just definition. But now that we have a metric, that gives us a canonical choice of connection: the Chern connection: In particular, locally if $s_\alpha$ is a local holomorphic basis of the line bundle on $U_\alpha$ and $h_\alpha = h(s_\alpha, s_\alpha)$ ($= |s_\alpha|_h^2$), then the Chern connection is given by $$ \nabla = d + \partial \log h_\alpha$$ and locally the curvature form is given by $$\Theta = d(\partial log h_\alpha) = \bar\partial\partial \log h_\alpha.$$
Now we see how we use the induced metric from the trivial bundle to calculate $\Theta$: in the coordinate $U_0= \{[z]\in \mathbb P^n : z_0 \neq 0\}$ we have the following chart $$ (z_1, \cdots , z_n ) \mapsto [1, z_1, \cdots, z_n]$$ and the local holomorphic basis is given by $$ s_0(z_1, \cdots, z_n) = (1, z_1, \cdots, z_n).$$ Then \begin{align*} h_0 &= 1+ |z_1|^2 + \cdots + |z_n|^2 = 1+ |z|^2,\\ \partial \log h_0 &= \frac{1}{1+|z|^2} (\bar z_1 d z _1 + \cdots + \bar z_n d z_n)\\ \Theta = \bar\partial\partial \log h_0&= -\frac{1}{(1+|z|^2)^2}( z_1 d \bar z _1 + \cdots + z_n d \bar z_n)\wedge (\bar z_1 d z _1 + \cdots + \bar z_n d z_n) \\ & \ \ + \frac{1}{1+|z|^2} (d\bar z_1 \wedge dz_1 + \cdots + d\bar z_n \wedge dz_n). \end{align*}
Write $\omega = \frac{i}{2\pi} \Theta$ (Caution: this is different from $\frac{1}{2\pi i}\Theta$ as you suggested. But it should be $\frac{i}{2\pi}$, which you can check in Griffiths-Harris, p.141). Then at the origin $0 = (0,\cdots, 0)$ we have $$\omega (0)= \frac{i}{2\pi}(dz_1 \wedge d \bar z_1 + \cdots + dz_n \wedge d\bar z_n).$$
Thus, at $0$, $$ \omega \left(\frac{\partial}{\partial z_i}, \frac{\partial}{ \partial \bar z_j}\right) = -\frac{i}{2\pi} \delta_{ij}. $$
Lastly, writing $g(X, Y) = \omega (X, JY)$,
\begin{align*} g\left(\frac{\partial}{\partial z_i}, \frac{\partial}{ \partial \bar z_j}\right)&=\omega \left(\frac{\partial}{\partial z_i}, J\frac{\partial}{ \partial \bar z_j}\right)\\ &=\omega \left(\frac{\partial}{\partial z_i}, -i\frac{\partial}{ \partial \bar z_j}\right) \\ &= -\frac{1}{2\pi} \delta_{ij} \end{align*} Thus $g$ is negative.
We recover the curvature form of $O(1)$, we use that $O(1)$ is the dual line bundle of $O(-1)$, and for any connection $\nabla $ defined on $O(-1)$ one can define the dual connection by the product rule:
$$ d(s^* (e)) = (\nabla^* s^*) (e) + s^* (\nabla e)$$
for any local section $s^*$, $e$ in $O(1), O(-1)$ respectively. Then the connection one form of $\nabla^*$ is negative of that of $\nabla$, and thus the curvature two forms are negative to each other. In particular, we have shown that the first Chern class of $O(1)$ can be represented by $-\omega$, which is a positive 2 form.