Proof that $\frac{dx\ dy}{x^2+y^2}$ is a Haar measure on the multiplicative group $\mathbb C\setminus\{0\}$

Question

How can it be proven that for every Borel subset of $\mathbb{C}\setminus\{0\}$ as A we have $\mu(cA)=\mu(A)$? $$ ∬_{cA} \frac{dx\ dy}{x^2+y^2}=\iint_{A} \frac{dx\ dy}{x^2+y^2} $$ I'm confused...

Answer 1

Here is a nice general result (see Folland's Real Analysis, Chapter 11, Exercises 3 and 4): if you have two elements $g,h\in G$ (where $G$ is locally compact of course) and the action of $g$ on $h$ (by left multiplication) can be written as $A_gh+b_{g}$, where $A_g$ is a linear transformation and $b_{g}$ is an affine part (meaning $g$ action on $h$ is an affine mapping), then the Haar measure is given by $\frac{1}{|\det(A_g)|}$. This tacitly assumes that $G$ is homeomorphic to $\mathbb{R}^n$ in some way and comes from a change of variable (hence the occurrence of something resembling a Jacobian). I'll illustrate this with your example so it is clearer.

In your case if $z_1,z_2\in\mathbb{C}\setminus\{0\}$, then write $z_1 = x_1+iy_1$ and $z_2 = x_2+iy_2$. What you have then is that $z_1z_2 = x_1x_2-y_1y_2 + i(x_1y_2+y_1x_2)$. We can view this as a matrix product (associating the real part to the first coordinate and imaginary part to the second coordinate):

$$\left(\begin{array}{cc} x_1 & -y_1 \\ y_1 & x_1\end{array}\right)\left(\begin{array}{c} x_2 \\ y_2\end{array}\right).$$

Then $A_{z_1} = \left(\begin{array}{cc} x_1 & -y_1 \\ y_1 & x_1\end{array}\right)$ and $b_{z_1} = 0$. It's easy to see that $\det(A_{z_1}) = x_1^2+y_1^2$ which immediately gives you your result.

Alternatively: show that

$$\int_{\mathbb{C}\setminus\{0\}}\frac{f(x,y)}{x^2+y^2}\,dxdy$$

is left-invariant. By the construction of the Haar measure we know that left-invariant measures correspond to left-invariant integrals and vice versa so this is an easy check. The above technique is used for constructing the Haar measure for simple examples but it is not necessary to do that if all you need to do is check something is a Haar measure.

Answer 2

Note $\mathrm{d}x \, \mathrm{d}y = \frac{1}{2i}\mathrm{d}z \, \mathrm{d}\bar{z}$

(edit: I think I have the sign wrong: $\mathrm{d}x \, \mathrm{d}y = -\frac{1}{2i}\mathrm{d}z \, \mathrm{d}\bar{z}$. This error doesn't affect the argument below)

I will instead prove the following: if I make the substitution $z = cw$, then

$$\begin{align} \frac{1}{2i}\mathrm{d}z \, \mathrm{d}\bar{z} &= \frac{1}{2i}\mathrm{d}(cw) \, \mathrm{d}(\overline{cw}) \\&=\frac{1}{2i} c \mathrm{d}w \, \bar{c} d\overline{w} \\&= \frac{\|c\|^2}{2 i} \mathrm{d}w \, \mathrm{d}\bar{w} \end{align}$$

Consequently,

$$\frac{\mathrm{d}z \, \mathrm{d}\bar{z}}{2i \|z\|^2} = \frac{\mathrm{d}w \, \mathrm{d}\bar{w}}{2i \|w\|^2} $$

and thus this differential form is invariant under the transformation $z \mapsto cz$ for any nonzero complex $c$.