# Confusion about complex differential forms

I follow Kobayashi "Differential Geometry of Complex Vector Bundles", pages 11-12, prop. 4.9. Given a rank-$$r$$ Hermitian holomorphic vector bundle $$(E,h)$$ over a complex manifold $$M$$, there exists a unique $$h$$-connection $$D$$ such that $$D^{0,1}=\bar\partial$$.

The proof works in a local holomorphic frame $$s=(s_1,...,s_r)$$ and uses uniqueness to construct as follows: Consider the connection form $$\omega^j_i$$: $$0=\bar\partial s_i=D^{0,1}(s_i)=s_j{\omega^{0,1}}^j_i$$ so $$\omega^j_i$$ is of type $$(1,0)$$.

Since $$D$$ is an $$h$$-connection, $$dh_{ij}=h(Ds_i,s_j)+h(s_i,Ds_j)=\omega_i^ah_{a\bar j}+h_{ib}\bar \omega^b_{\bar j}.$$

Now the claim is that since $$\omega$$ is $$(1,0)$$ then $$d'h_{ij}=\omega^a_ih_{a\bar j}$$.

Now here is my confusion: This is (as good as I understand) supposed to mean that (in local coordinates on M) $$\bar \omega(\partial_{z^l})$$=0. But this conjugation of $$\omega$$ is on the target, i.e. conjugate the values of $$\omega$$ and not the domain, i.e. $$\partial_{z^l}$$.

In this case, it is unlikely that $$\bar \omega(\partial_{z^l})=0$$ since this would mean that $$\omega(\partial_{z^l})=0$$ so $$\omega$$ is both type $$(1,0)$$ and $$(0,1)$$, that is: zero.

The question comes down to this: There are two ways to conjugate a form - conjugate the input, and conjugate the output. In this calculation, I understand that we use the latter kind, but the conclusion that the type is exchanged assumes the former kind.

• These $s^i$ are not holomorphic coordinates; they are holomorphic local sections defined in a common open set. – Ben McKay Jul 11 '19 at 10:28
• Correct, I will edit. – Or Kedar Jul 11 '19 at 11:28

Forget about vector fields $$\partial_{z^{\mu}}$$ and $$\partial_{z^{\bar\mu}}$$. Just think about 1-forms: $$\omega^j_i = \Gamma^j_{i\mu}dz^{\mu}$$, with $$C^{\infty}$$ functions $$\Gamma^j_{i\mu}(z)$$. Then it is clear why $$\partial h$$ is the $$(1,0)$$-part of $$dh$$, and so if we write out $$h^{-1}dh=\omega+\bar\omega$$ as $$(1,0)$$ and $$(0,1)$$-parts. It is always easiest to forget about vector fields and work directly with forms.
To work with vector fields, you have to start with $$\partial_{x^{\mu}},\partial_{y^{\mu}}$$ spanning the real tangent bundle. We work on a complex or almost complex manifold, but think of it as almost complex. Every complex valued 1-form splits uniquely into $$J$$-linear and $$J$$-conjugate linear parts. When you complexify the real tangent bundle, you face the old $$J$$ and also a new $$i$$. You now have $$J$$ extended to be $$i$$-complex linear, with $$i$$ the imaginary unit you complexified with. You define the holomorphic and conjugate holomorphic tangent bundles to lie inside the complexified real tangent bundle, and check that $$(1,0)$$-forms vanish on the conjugate holomorphic tangent bundle, but only by declaring that $$(0,1)$$-forms are $$i$$-linear and $$J$$-conjugate linear, while $$(1,0)$$-forms are $$i$$-linear and $$J$$-linear too. This is messy, so just work with differential forms.