I have a problem to understand the meaning of a complex measure; i.e., when someone writes ($i \equiv \sqrt{-1}$) $$ \int_{\mathbb{R}^2} d\mathrm{Re}z \, d\mathrm{Im}z \equiv \int_{\mathbb{C}} \frac{dz d\bar{z}}{2i} \quad (\ast) $$ The lefthand-side will yield a real number (after performing the integration over a real-valued function), while it is not obvious that the righthand-side yields a real number.
Furthermore, how can one obtain the equivalence relation? Is the factor $\frac{1}{2i}$ the Jacobian of some transformation like $$ z = \mathrm{Re} z + i \, \mathrm{Im} z ,\\ \bar{z} = \mathrm{Re} z - i \, \mathrm{Im} z ,\\ $$ So, the complex measure $dz d\bar{z}$ does not have the same meaning as a ‘simple’ complex integration which in complex calculus (integration over a path in the complex plane).

Please provide an explanation for the complex measure and the equivalence relation ($\ast$) above.

1. A similar question is asked here; yet no clear proof or justification is provided.
2. An example of the relation appears, for instance, in Altland, A. and B. D. Simons. “Condensed Matter Field Theory” (2nd ed., 2010), p. 102:

Altland, p102


You have to be aware that there are intagration of measures and integration of differiential forms. You cab see the difference when using the Change of variables formula. For example if $V,W$ are vector spaces of same dimension, $\lambda,\mu$ Lebesgue measures on $V$ and $W$ (or rather a Haar measure), and $f : V \rightarrow W$ smooth and compactely supported, then : $$\int_{x \in V} d\mu(f(x)) = \int_{x \in V} |\mathrm{Jac}(f(x))| ~ d\lambda(x).$$

In your context it is integration of differential form, and you need the following formula. If $V,W$ are oriented vector spaces not necessarily of the same dimension, $f : V \rightarrow W$ smooth, $v$ an invertible differential form on $V$ (i.e. volume form), $w$ a differential form on $W$ of degree $\dim V$ then $$\int_{x \in V} f^*v~(x) = \int_{x \in V} \mathrm{Jac}(f(x)) ~ v(x),$$ where $\mathrm{Jac}(f(x))$ is the unique number such that $f^*v~(x) = \mathrm{Jac}(f(x)) ~ v(x)$ (recall that $v$ is invertible, and it coincides with the determinant when $V$ and $W$ are of the same dimension).

In your situation $V=\mathbb{C}$, $W = \mathbb{C} \times \mathbb{C}$, $f : z=x+iy \mapsto (z,\bar{z})$, $v= dx \wedge dy$, and $w = dz_1 \wedge dz_2$, which yields : $$\int_{z \in \mathbb{C}} dz \wedge d\bar{z} = \int_{z \in \mathbb{C}} -2i ~ dx \wedge dy.$$

Note : $dz$, $d\bar{z}$, $dz.d\bar{z}$ are not measures nor complex measures. They are complex differential forms of degree 1 or 2.

  • $\begingroup$ Is there a way to perform an integration with the $dz$ and $d\bar{z}$ forms (i.e. in practice, with a specified function), without switching back to $dx$ and $dy$. $\endgroup$ – knzhou Feb 1 '17 at 18:02
  • $\begingroup$ @knzhou Yes. Choose a paramatrization $\phi : \mathbb{R}^2 \rightarrow \mathbb{C}$ (for example $\phi(x,y) = x+iy$). Then by definition :$\int_\mathbb{C} dz \wedge d\bar{z} = \int_{\mathbb{R}^2} w(\partial_1\phi(x,y),\partial_2\phi(x,y))~dxdy$, where $w(z_1,z_2) = z_1.\bar{z_2}-z_2.\bar{z_1}$ and the second integral is the Riemann or Lebesgue integral. $\endgroup$ – user10676 Feb 2 '17 at 15:07

Probably the easiest way to see this (up to a sign that has me confused --- are you sure the order of differentials is right in the formula you quoted) is in terms of differential forms. With $z=x+iy$ (where $x$ and $y$ are real variables), we have $\bar z=x-iy$ and so $$ dz\land d\bar z=(dx+i\,dy)\land(dx-i\,dy)=(dx\land dx)+i(dy\land dx) -i(dx\land dy)-(dy\land dy). $$ Since the wedge product is skew-symmetric, the first and last terms here vanish, and the other two terms combine to give $-2i\,dx\land dy$.

  • $\begingroup$ I am not sure about the order of the differentials; the books does not state that clearly (with mathematical rigour); lack of rigour is actually the root of this problem! $\endgroup$ – AlQuemist Dec 10 '15 at 17:09
  • $\begingroup$ What happens to the volume element $d z d \bar{z}$? Does it become a complex volume? If we use the absolute value of the Jacobian (as in a usual change of variables), one does not obtain the $-i$ factor. Does the current relation mean that the ‘orientation’ of the volume is not preserved upon the transformation to $z, \bar{z}$? $\endgroup$ – AlQuemist Dec 10 '15 at 20:32
  • $\begingroup$ @PhilosophiæNaturalis The product $dz\,d\bar z$ is a purely imaginary multiple of an area element. I see no reason to use the bsolute value of the Jacobian in this complex context. Nor do I see a reasonable meaning for "orientation" since that refers to the signs of real numbers. $\endgroup$ – Andreas Blass Dec 10 '15 at 21:11
  • $\begingroup$ Could you refer me to some pedagogical reference about this kind of complex measure? $\endgroup$ – AlQuemist Dec 10 '15 at 21:53

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