Confusion about Positive Curvature in Holomorphic Bundles.

Solution 1:

First of all, let me tell you that you are far from "un idiota." Signs and factors of $\sqrt{-1}$ in complex geometry are inconsistent in books and papers and are often the bane of the existence of those of us who work(ed) in the field. Second of all, let me say that I adore Griffiths and Harris (and have taught courses out of parts of it), but it is known to have errors, mostly small but--as you've discovered--annoying.

OK, so, of course you're right about the meaning of skew-hermitian. But, in fact, it's really the upper right of the matrix that should be labeled $A$ and consists of $(1,0)$ forms.† The various transformation rules ($g^{-1}$ on the right, for instance) indicate that we're really working with row vectors, not column vectors. So the correct formulas should in fact be $$\Theta_S = \Theta_E\big|_S - A\wedge A^* \quad\text{and}\quad \Theta_Q = \Theta_E\big|_Q - A^*\wedge A.$$

Next: Here's the reason for the factor $\sqrt{-1}$. The notion of positivity of curvature, as you'll see, will tie in with real cohomology. If we go back to basics, remember that in $\Bbb C$, we have $dz\wedge d\bar z = -2\sqrt{-1}dx\wedge dy$ and we all agree that $dx\wedge dy$ gives the (positive) area form for $\Bbb C$. And, indeed, $-\sqrt{-1}\big(\sqrt{-1}dz\wedge d\bar z\big)\big(\frac{\partial}{\partial z},\frac{\partial}{\partial\bar z}\big)>0$. In general, the $(1,1)$-form $\sqrt{-1}\sum h_{\alpha\bar \beta} dz_\alpha\wedge dz^{\bar\beta}$ will be positive iff $(h_{\alpha\bar\beta})$ is a positive-definite hermitian matrix. So, if you unwind this, Griffiths and Harris's sign check on curvature is in fact correct.

Last, checking positivity is a pointwise matter. Given any holomorphic tangent vector $v$ at $x$, you can certainly choose local holomorphic coordinates near $x$ so that $v=\partial/\partial z_\alpha$ for some $\alpha$. So we're OK on that, too.

†EDIT: In case you want the argument, we're doing the connection matrix with respect to a unitary frame field. If you instead think of an adapted holomorphic frame field $Z_i$, then we'll have (for $i,j\le\text{rank}\,S<q\le n$) $e_i = \sum g_i^j Z_j$ for some smooth functions $g_i^j$ and $$A_{i\bar q} = \langle\nabla e_i,e_q\rangle = \langle \nabla\big(\sum g_i^j Z_j\big),e_q\rangle = \sum dg_i^j \langle Z_j,e_q\rangle + \sum g_i^j\langle \nabla Z_j,e_q\rangle;$$ the first term vanishes and the second term consists of $(1,0)$-forms because the connection is compatible with the holomorphic structure and the $Z_j$ are holomorphic sections. (Note that this argument works for sections of the subbundle but not for sections of the quotient bundle.)