My aim is to study some special representations of $SO(3)$ using characters, and to do this I need an explicit expression of the Haar measure on $SO(3)$. I've found some "versions" of the Haar measure on $SO(3)$, and the most common one seem to be the acquired by parameterizing $SO(3)$ using Euler angles (can be found here).

Since the characters of $SO(3)$ only depend on the angle from the axis-angle representation, it seems as if it would be suitable to find a Haar measure based on this parameterization. However, I don't really know where to start, and I've searched the web like a maniac.

Could someone provide some help?

Thanks in advance!


The Haar measure can be explicitly calculated as follows in terms of a given parametrization $R(\phi,\theta,\psi)$ of $SO(3)$.

The first step is to define a metric thereon, which can be done by means of the Killing form: denoting by $X$, $Y$ and $Z$ the generators of the $so(3)$ algebra, $$ [X,Y]=Z\,,\qquad [Y,Z]=X\,,\qquad [Z,X]=Y\,, $$ the Killing form on $so(3)$ is defined as the symmetric, bilinear form $(\cdot\,,\cdot)$ on $so(3)$ satisfying $$ (X,X)=(Y,Y)=(Z,Z)=1\,,\qquad (X,Y)=(Y,Z)=(Z,X)=0\,. $$ In fact, one can check that this Killing form is indeed invariant under the Lie product, $([A,B],C)+(B,[A,C])=0$.

Now that we have a metric on the tangent space to the identity, we can define a metric at any point as follows: let $\dot R=\frac{d}{dt}R(t)$ be some vector, defined at the point $R=R(t)$; we may transport it back to the tangent space to the identity by means of the map induced by left multiplication. In this case, this simply means taking $\Omega=R^T\dot R$, which is indeed an element of the Lie algebra. Then, we may define a metric $ds^2$ by means of $ds^2=(\Omega, \Omega)\,dt^2$, the so-called Killing metric.

More explicitly, in the fundamental representation, we have $$ (A,B)=\frac{1}{2}\mathrm{tr}(A^TB)\, $$ and $$ ds^2=\frac{1}{2}\mathrm{tr}(\Omega^T\Omega)\,dt^2\,. $$ For instance, employing the Euler parametrization $$ R(\phi,\theta,\psi)=R_z(\phi)R_y(\theta)R_z(\psi) $$ with $\phi,\psi\in[0,2\pi)$ and $\theta\in[0,\pi)$, $$ R_z(\phi)=\left( \begin{matrix} \cos\phi & -\sin\phi & 0\\ \sin\phi & \cos\phi &0\\ 0 & 0 & 1 \end{matrix} \right) \,,\qquad R_y(\theta)=\left( \begin{matrix} \cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta &0 & \cos\theta \\ \end{matrix} \right) $$ you should obtain (after a somewhat lengthy calculation) $$ ds^2= d\phi^2 + 2\cos\theta\, d\phi\, d\psi + d\theta^2 + d\psi^2\,. $$ The determinant of this metric, $(\sin\theta)^2$, provides the Haar measure/volume form by means of $$ \omega = \sqrt{\mathrm{det}g}\,d\phi\wedge d\theta\wedge d\psi= \sin\theta \,d\phi \wedge d\theta\wedge d\psi. $$ Note that this measure is automatically right- and left-invariant, since so is the Killing metric.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.