4
$\begingroup$

I'm working through "Lecture on Kahler Geometry" by Andrei Moroianu, and am stuck on Lemma 11.7 (p. 85).

The lemma says:

For every section $Y$ of the complex vector bundle $(TM, J)$ the $\overline{\partial}$-operator, as a $TM$-valued $(0,1)$-form is given by

$$\overline{\partial}^\nabla Y (X) := \frac{1}{2} (\nabla_X Y + J \nabla_{JX}Y + J(\nabla_YJ)X)$$ where $\nabla$ denotes the Levi-Civita connection of any Hermitian metric $h$ on $M$.

The proof starts by proving the Leibniz rule, recalling that $(\overline{\partial} f) (X) = \frac{1}{2} \partial_{(X+iJX)} f$, then

$$(\overline{\partial}^\nabla f Y)(X) = f \frac{1}{2} (\nabla_X Y + J \nabla_{JX}Y + J(\nabla_YJ)X) + \frac{1}{2} ((\partial_Xf)Y + (\partial_{JX}f)JY) = f \overline{\partial}^\nabla Y(X) + \overline{\partial}f (X)Y$$

My question is how $\overline{\partial}f (X)Y = (\frac{1}{2} ((\partial_Xf)Y + (\partial_{JX}f)JY))$ when we're given $(\overline{\partial} f) (X) = \frac{1}{2} \partial_{(X+iJX)} f$?

Any help would be greatly apperiated.

$\endgroup$
2
  • $\begingroup$ Are you missing some equal signs? $\endgroup$ Commented Apr 1, 2021 at 0:55
  • $\begingroup$ Yes, I just fixed it. Sorry about that. $\endgroup$
    – Andy Stump
    Commented Apr 1, 2021 at 1:15

1 Answer 1

2
$\begingroup$

The Dolbeault operator $\overline{\partial}$ can be thought of as the $(0,1)$--part of the flat connection $d$, i.e., the exterior derivative. An import property of a connection $\nabla$ is that $$\nabla_{X+Y} Z = \nabla_X Z + \nabla_Y Z,$$ for tangent vectors $X,Y,Z$, and for smooth functions $f$, $$\nabla_{f X} Y = f \nabla_X Y.$$

Now, if $(\overline{\partial}f)(X) : = \frac{1}{2}\partial_{X + \sqrt{-1} J X} (f)$, then $$(\overline{\partial} f)(X)Y = \frac{1}{2}\partial_{X+\sqrt{-1} JX}(f)Y = \frac{1}{2}\partial_X (f)Y + \frac{\sqrt{-1}}{2}\partial_{JX}(f)Y = \frac{1}{2}\partial_X(f) + \frac{1}{2}\partial_{JX}(f)JY.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .