I have some very basics question on the Haar measure on $SU(2)$.

What I understood from definition of Haar Measure is that it is a measure that ensure me to have the property :

$$ \int f(gh) d \mu(g) = \int f(h) d \mu(h) $$

Where $g$ and $h$ are element of a Lie group. And $f$ a function from the Lie group to the complex numbers (for example).

In practice, if we parametrize the $SU(2)$ matrices by the angles $\theta, \phi, \chi$, I will have :

$$ \int_{SU(2)} f(x) d \mu(x) = \int_{0}^{\pi} d \theta \int_{0}^{\pi} d \phi \int_{0}^{2\pi} d \psi f(\theta,\psi,\phi) sin^2(\theta) sin(\psi) $$

Where an element of $SU(2)$ can be written as :

$$ g=\left[ \matrix { x1+ix2 &x3+ix4 \\ -x3+ix4&x1-ix2&\\ } \right]$$

With : $x1=cos(\theta)$, $x2=sin(\theta)cos(\phi)$, $x3=sin(\theta)sin(\phi)cos(\psi)$, $x4=sin(\theta)sin(\phi)sin(\psi)$.

My question :

When we have $g_1$ and $g_2$ in $SU(2)$, the angles $\theta$,$\phi$,$\psi$ of their product will be the sum of the angles of the first and the second ? I could compute it by hand but it can be long so I would like to know if it is true ?

And so if it is true, if I compute the integral of $f(\theta_1+\theta_2,\psi_1+\psi_2,\phi_1+\phi_2)$ with $d\theta_1$,$d \phi_1$, $d \psi_1$, I would end up with the same result as if I have integrated $f(\theta_1,\psi_1,\phi_1)$ right ?


1 Answer 1


The answer to your first question is no. It would mean that the group SU(2) were commutative, which it is not.


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