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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!

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3 Answers 3

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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.

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I will add an extension to the answer by Brightsun, since the reply posted above does not explicitly answer the original question.

I understand "angle-axis" representation as a parametrisation of the group manifold of $SO(3)$ in terms of an angle of rotation $\theta$ and an axis of rotation $\hat{n}$ of unit length. Explicitly, the group manifold of $SO(3)$ in terms of $\theta$ and $\hat{n}(\beta_1,\beta_2)$ is parametrised by coordinates $(\theta,\beta_1,\beta_2)$. In these coordinates, a general group element $R(\theta,\beta_1,\beta_2) \in SO(3)$ is given by $$R(\theta,\beta_1,\beta_2)=\mathbb{I}+\sin (\theta)K(\beta_1,\beta_2)+(1-\cos (\theta))K^2,$$ where $\mathbb{I}$ is the unit matrix in three dimenions. The matrix $K(\beta_1,\beta_2)$ is given in terms of the axis of rotation $\hat{n}=(\hat{n}_1,\hat{n}_2,\hat{n}_3)$ of unit length, $$K(\beta_1,\beta_2)=\left( \begin{matrix} 0&-\hat{n}_3&\hat{n}_2\\ \hat{n}_3&0&-\hat{n}_1\\ -\hat{n}_2&\hat{n}_1&0 \end{matrix} \right).$$ The axis of rotation $\hat{n}(\beta_1,\beta_2)$ is explicitly given by $$\hat{n}(\beta_1,\beta_2)=\left( \begin{matrix} \sin(\beta_1)\cos(\beta_2)\\ \sin(\beta_1)\sin(\beta_2)\\ \cos(\beta_1) \end{matrix} \right).$$ The rotation matrix $R(\theta,\beta_1,\beta_2)$ corresponds to a rotation of an angle $\theta$ around the axis $\hat{n}(\beta_1,\beta_2)$.

With the procedure described by Brightsun, using $R(t) \equiv R(\theta(t),\beta_1(t),\beta_2(t))$, the Haar measure $\omega$ of $SO(3)$ in the axis-angle representation is $$\omega=4\sin(\frac{\theta}{2})^2\sin(\beta_1) d\theta \ \wedge d\beta_1\wedge d\beta_2.$$ Note that the determinant of the Killing metric can be read of from $\omega$.

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    $\begingroup$ Can you clarify how the Haar measure is normalized here? Your final expression does not seem to be unit-normalized, integrating to $8\pi^2$. What's the role of the $4$ there? Can you point to reliable references showing this Haar measure and its normalization? $\endgroup$
    – E.P.
    Jun 21, 2021 at 18:24
  • $\begingroup$ (All of which are, of course, details. The core of the answer is good.) $\endgroup$
    – E.P.
    Jun 21, 2021 at 18:24
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I found this question searching for an explicit form of Haar measure in Cartesian components of axis-angle representation. Maybe someone will find worked out answer useful.

Quickest way to the correct result for me was to notice first that picking a rotation matrix at random wrt Haar measure and acting on a fixed, unit $\mathbb{R}^3$ vector should give vectors that are uniformly distributed on a sphere. From this one can conclude that angle of rotation should have density proportional to $1 - \cos(\alpha)$.

Recalling that surface area of $S^2$ is $4\pi r^2$ we arrive at

$$ \omega = \frac{1 - \cos(\sqrt{x^2+y^2+z^2})}{4 \pi^2 (x^2 +y^2 + z^2)} dx\wedge dy \wedge dz $$

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