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.


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.

  • $\begingroup$ You really needn't waste your reputation points on a (retroactive) bounty. I don't know if you can get them back, but if you can, please do. :) $\endgroup$ Oct 1, 2016 at 2:14
  • $\begingroup$ No worries. I got plenty of upvotes for this question. =D $\endgroup$ Oct 2, 2016 at 14:00
  • $\begingroup$ Thank you for bringing this question up. I met exactly the same question with you, and this post really helps! $\endgroup$ Aug 19, 2020 at 0:21

1 Answer 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.)

  • $\begingroup$ I see!! The conventions i'm thinking of would give $\Theta=d\theta+\theta\wedge \theta$ as in the book by Wells. But here the formula has a minus sign. So for G-H the top right matrix is $(1,0)$, thanks for pointing it out! Also, I definitely see why I would want a form like $idz\wedge d\bar z$ to be positive. Ditto the argument for the base change of forms and holomorphic tangent vectors. Thank you for your answer! I will accept it tomorrow. $\endgroup$ Sep 27, 2016 at 22:47
  • 1
    $\begingroup$ You're most welcome. Keep learning! $\endgroup$ Sep 27, 2016 at 22:52

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