I'm reading the book Complex Algebraic Geometry and Hodge Theory, and the definitions of $\partial,\bar\partial$. Given a form $\alpha$ of type $(p,q)$, the differential $d\alpha$ splits into parts of type $(p+1,q)$ and $(p,q+1)$, which we define as $\partial\alpha$ and $\bar\partial\alpha$, respectively. The author then remarks that by definition we have the relation


I fail to see why this holds. For instance, if $f$ is a $(0,0)$-form, i.e. $f$ is a $C^{\infty}$ function $M\to\Bbb C$, then we can write

$$df=\sum_i\frac{\partial f}{\partial z_i}dz_i+\sum_i\frac{\partial f}{\partial\bar z_i}d\bar z_i,$$

and we have that $\partial f$ is the first summand, and $\bar\partial f$ is the second. But then, we have

$$\bar\partial\bar f=\sum_i\frac{\partial\bar f}{\partial\bar z_i}d\bar z_i.$$

However, I fail to see why taking the "complex conjugate" here gives us $\partial f$. I may be misinterpreting definitions, but it's not spelled out all that clearly in the book. Could somebody help me see what is going on here?


$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}\newcommand{\Cpx}{\mathbf{C}}$Write a complex-valued function as $f = u + iv$ with $u$ and $v$ real-valued, and use $$ \frac{\dd}{\dd z} = \frac{1}{2}\left(\frac{\dd}{\dd x} - i\, \frac{\dd}{\dd y}\right),\qquad \frac{\dd}{\dd \bar{z}} = \frac{1}{2}\left(\frac{\dd}{\dd x} + i\, \frac{\dd}{\dd y}\right) $$ to see that \begin{align*} \frac{\dd f}{\dd z} &= \frac{\dd (u + iv)}{\dd z} = \frac{1}{2}\left[\frac{\dd u}{\dd z} + i\, \frac{\dd v}{\dd z}\right] = \frac{1}{2}\left[\left(\frac{\dd u}{\dd x} - i\, \frac{\dd u}{\dd y}\right) + i \left(\frac{\dd v}{\dd x} - i\, \frac{\dd v}{\dd y}\right)\right], \\ \frac{\dd \bar{f}}{\dd \bar{z}} &= \frac{\dd (u - iv)}{\dd \bar{z}} = \frac{1}{2}\left[\frac{\dd u}{\dd \bar{z}} - i\, \frac{\dd v}{\dd \bar{z}}\right] = \frac{1}{2}\left[\left(\frac{\dd u}{\dd x} + i\, \frac{\dd u}{\dd y}\right) - i \left(\frac{\dd v}{\dd x} + i\, \frac{\dd v}{\dd y}\right)\right] = \overline{\,\frac{\dd f}{\dd z}\,}. \end{align*} The property extends to smooth forms because "a conjugate of a product is the product of the conjugates".

  • 1
    $\begingroup$ Perfect! Thanks $\endgroup$ Mar 28 '17 at 20:26

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