0

My question is simple. I had read about the existence of Chern connection in the Huybrechts book Complex Geometry p. 177 but I don't remember meaning of some notation :

enter image description here

Here the $A$ is the connection matrix with respect to the $\{e_i\}$

Q. What is the $\partial $? I know the $\bar{\partial}$ as follows (his book p.109):

enter image description here

; i.e., $\bar{\partial}_E : \mathcal{A}^{0}(E) = \mathcal{A}^{0,0}(E) \to \mathcal{A}^{0,1}(E)$

But what $\partial$ does mean? defined similarly as in the proof of the Lemma 2.6.23 (in the above image)?

Plantation
  • 2,417

1 Answers1

1

Note that $H$ is a matrix of smooth complex-valued functions on $X$. The expression $\partial H$ is notation for the matrix of $(1,0)$-forms $[\partial h_{ij}]$, i.e. $\partial$ acts on the matrix $H$ by acting on each entry.

The Dolbeault operator $\bar{\partial}_E$ is defined for any holomorphic vector bundle (this is Lemma $2.6.23$). As you can see from the proof, the construction relies on the transition functions of $E$ being holomorphic, i.e. in the kernel of $\bar{\partial}$. For an analogous global operator $\partial_E$ to exist, one would need the transition functions to be in the kernel of $\partial$ which, for a holomorphic trivialisation, would imply they are locally constant.

Michael Albanese
  • 99,526
  • Yes. $\partial H$ maybe means that $[\partial_X h_{ij}] $, where

    $ \partial_X := \prod ^{1,0}{X} \circ d_X : \mathcal{A}^{0,0}_X(X) \to \mathcal{A}^{1,0}{X}(X) $ as in the definition 2.6.9 in his book p.106

    Now I'am considering when $\partial_E$ (if defined) is same as $\partial$

    Anyway, thank you for answer!

     – Plantation Mar 21 '22 at 13:32
  • 1
    If $\partial_E$ is defined, it acts on sections of $E$ (and can be extended to act on $(p,q)$-forms with values in $E$), while $\partial$ acts on $(p, q)$-forms. – Michael Albanese Mar 21 '22 at 13:35
  • Yes. Thank you. I will check it!. Uhm, if we let $E$ be the trivial bundle, then perhaps $\partial_E$ is same as $\partial$ ? – Plantation Mar 21 '22 at 13:47
  • If $E$ is the trivial line bundle, then they do indeed coincide. – Michael Albanese Mar 21 '22 at 13:52
  • Yes. O.K. ~ I see. – Plantation Mar 21 '22 at 13:53