Complex integration over sphere

Question

Consider complex plane and unit sphere $\mathbb{S}^1 = \lbrace z \in \mathbb{C} : |z|^2 = 1 \rbrace$. It is quite simple that for example by a parameterization, we have

$$\int_{\mathbb{S}^1} z\, d z = \int_{0}^{2\pi} e^{i \theta} \, d\theta = 0.$$

Same goes for $z^k$ where $k \in \mathbb{N}$. I'm wondering, how does it work in higher dimensions. So consider $\mathbb{C}^2$ and a unit sphere $\mathbb{S}_{\mathbb{C}}^2 = \lbrace z \in \mathbb{C}^2 : |z_1| ^2 +|z_2|^2 = 1 \rbrace$ (which I think works like $\mathbb{S}^3 \subset \mathbb{R}^4$, right?) equipped with an $SO(4)$-invariant measure d$m$. How can I calculate for example

$$\int_{\mathbb{S}_{\mathbb{C}}^2} z_1{z_2}^2\, dm?$$

Can I find the right parameterization there, too? Or should I translate this to real numbers and use spherical coordinates?

Answer 1

First of all, the complex integral is not an integral with respect to the invariant (or any other) measure. Indeed, for a path $\gamma$ from $z_0$ to $z_1$, $\int_{\gamma} 1 dz= z_1-z_0$, but $\int_\gamma 1 dm$ is a real non-negative number.

This is reflected in the fact that if $z=e^{i\theta}$ then $dz=i e^{i\theta} d\theta$ (note that this way $\int_{S^1} z^{-1} dz=\int_0^{2\pi} i d\theta=2\pi i$, as it should be).

So you have to decide which integral -- over $dz$ or over $dm$ -- you want. In either case, one can say that the integral of $z^k$ over the circle is zero because it is a complex number invariant under rotations (except of course $\int_{S^1} z^{-1} dz=2\pi i$ and $\int_{S^1} 1 dm=2\pi$). Namely, if we fix a complex number $\alpha$ with $|\alpha|=1$, the variable substitution $w=\alpha z$ gives $\int_{S^1} z^k dm=\int_{S^1} w^k dm=\int_{S^1} \alpha^k z^k dm=\alpha^k \int_{S^1} z^k dm$. If $k\neq 0$ this is only possible if $\int_{S^1} z^k dm=0$.

This is applicable to any complex monomial integrated over $SO(2n)$-invariant measure on the unit sphere -- pick any variable $z_j$ not in zeroth power and rotate the space by $z_j \to \alpha z_j$.

Answer 2

Let $H_1^+=\{z\in \mathbb{C} :|z|^2=1, \Re(z)>0 \}$ and $H_1^-=\{z\in \mathbb{C} :|z|^2=1, \Re(z)<0 \}$ denote the (one dimensional) hemispheres of the complex unit circle. Quite clearly, the map $\phi:H_1^+\to H_1^-$ taking $z\mapsto \phi(z)=-z$ can be seen as a change of coordinates, and we see that $$\int_{H_1^+}z\,dz=-\int_{H_1^-}z\,dz\implies \int_{\mathbb{S}_1} z\,dz=0,$$ since the equator has measure $0$. This idea generalizes nicely to higher dimensions. For example, if we take $H_2^+=\{z\in \mathbb{C}^2:|z_1|^2+|z_2|^2=1, \Re(z_1)>0\}$ and $H_2^-=\{z\in \mathbb{C}^2:|z_1|^2+|z_2|^2=1, \Re(z_1)<0\}$ as before (and implicitly consider the change of coordinates from $H_2^+$ to $H_2^-$ taking $(z_1,z_2)\mapsto(-z_1,z_2)$), we see that $$\int_{H_2^+}z_1{z_2}^2\,dm=-\int_{H_2^-}z_1{z_2}^2\,dm\implies \int_{\mathbb{S}_2} z_1{z_2}^2\,dm=0.$$

The downside of this approach is that, if your integrand gets any more complicated, this won't get you very far. But in cases such as the one you consider, this should work.