I'm trying to understand the principle that curvature decreases in holomorphic subbundles and increases in quotient bundles as shown in G-H (Griffiths Harris) page 78-79.
Setup:
Let $E\rightarrow M$ be a Hermitian holomorphic bundle, $S\subset E$ be a holomorphic subbundle and $Q=E/S$ be the quotient bundle. Q can be seen in unitary frames to be diffeomorphic to $S^{\perp}$. Hence $S$ and $Q$ are both Hermitian.
Let $D_E$, $D_S$ and $D_Q$ be the connections compatible with the complex structures (Compatibility with the metric and vanishing on holomorphic sections). By uniqueness of such connections, $D_S=\pi_S\circ (D_E|_{A^0(S)})$ where $A^0(S)$ are sections of $S$.
Now $A:=D_E|_{A^0(S)} -D_S$ is $C^\infty (M)$ linear and maps to sections in $S^{\perp}=Q$ with coefficients in $A^{(1,0)}(M)$ (Because of the definition of a connection and the compatibility conditions on $D_E$ and $D_S$). In short, $A\in A^{(1,0)}(Hom(S,Q))$.
Now, as in G-H, choosing a unitary frame ${e_i}$ for E (the first few vectors of which form a frame for $S$) we see that the matrix of forms for $D_E$ is $\theta_E = \begin{bmatrix} \theta_S & A^* \\ A & \theta_Q \end{bmatrix}$.
Remark 1: The above formula is incorrect since in a unitary frame $\theta_E = -\theta_E^*$. So we should have a $-$ sign along with the $A^*$.
Question 1: Is this correct?
Assuming this, the correct formulas are:
$\theta_E = \begin{bmatrix} \theta_S & -A^* \\ A & \theta_Q \end{bmatrix}$,
$\Theta_S=\Theta_E|_{S}-A^*\wedge A$,
$\Theta_Q=\Theta_E|_Q-A\wedge A^*$. These are derived from $\Theta = d\theta-\theta\wedge\theta$.
Definition: G-H defines $\Theta$ to be positive if for all $v$ - holomorphic tangent vectors, the matrix $-i(\Theta(x);v,\bar v)$ is positive definite, in particular hermitian. Here, I assume we are fixing a local unitary frame, evaluating the coefficients of (1,1)-forms of $\Theta(x)$ on $(v,\bar v)$ and then looking at the resulting matrix in $Hom(E_x,E_x)$.
Claim 1: is that $\Theta_Q \geq \Theta_E|_Q$, i.e. curvature increases in quotient bundles. Using the corrected formula, we want to show $-A\wedge A^*$ is positive.
Note, a computation shows (by definition?) $-A\wedge A^*$ is in $A^{(1,1)}(Hom(Q_x,Q_x))$ and the $(p,q)$ entry is $-\sum_{\alpha,\beta} \sum_k a_{pk}^\alpha \bar{a}_{qk}^\beta dz_\alpha\wedge d\bar{z}_\beta$. And following G-H, contracting with $(\partial{z_\alpha},\partial{\bar z_\alpha})$ gives the matrix $-A^\alpha {A^*}^\alpha$ in $Hom(Q_x,Q_x)$ and they conclude that this proves positivity.
Question 2: Since I am Un Idiota and the most I can do is follow instructions, I am assuming they are checking (as required by the definition) that $-i(-A^\alpha {A^*}^\alpha)$ is hermitian positive definite. But clearly the $i$ is spoiling things. Can I drop it from the definition? What am I missing? Positivity in other sources (e.g. https://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf defn 6.3 on page 338) have neither the $i$ nor the $-$ sign. But I need the $-$ sign so I can't drop both! Is there some standard viewpoint on the definition of positivity (Hopefully one that would fit consistently into this argument)?
Question 3: Why is it sufficient to check the positivity condition on vector pairs of the form $(\partial z_k,\partial \bar z_k)$? If I try to contract with $(\sum_{l} c_l \partial z_l, \sum_k \bar c_k \partial \bar z_k)$, I get a matrix in $Hom(Q_x,Q_x)$ I get something far uglier which doesn't seem to be of the form $MM^*$
Thank you for reading. I have 80 Rep now! Will put Bounty when it's eligible.