Question

The volume in the orthogonal group is measured by the Haar measure, which is the up to scaling unique measure that is invariant under the group operation. I consider the usual metric that is induced by the spectral norm |M| = max |Mx| where x ranges over all vectors of length 1 and the vector norm is the Euclidean one. A \delta-ball is the set of all orthogonal matrices that have distance less or equal \delta to a fixed matrix M. Because of the invariance of the Haar measure, for a fixed \delta, all \delta-balls have the same volume.

Answer 1

The volume of the delta-ball of the special orthogonal group can be computed exactly by applying the Weyl integration formula: (Without loss of generality, we assume that the delta-ball is around the unit group element).

a. One notices (Again due to the invariance under the Haar measure) that the characteristic function of the delta ball is a class function. Thus upon the application of the Weyl integration formula we are left only with the radial part on the eigenvalues which is a $\lfloor N/2\rfloor$-dimensional integral for $\mathrm{SO}(N)$. Here, the radial integral is described explicitely.

b. The eigenvalues of an orthogonal matrix of dimension $N=2m+1$ are $1$ and $m$ pairs $\exp(i \phi_ m)$ and $\exp(-i \phi_ m)$, $0\leq\phi_ 1 \leq\ldots\leq\phi_m \leq\pi$. In the case of even dimensions, the unit eigenvalue is absent.

c. The delta-ball condition on the eigenvalues becomes:

$$ |\exp(i\phi_k)-1|\leq\delta , $$ which implies: $$\phi_k\leq 2 \arcsin\sqrt{\delta/2}.$$

d. Applying the Weyl integration formula, we obtain for the odd case $\mathrm{SO}(2m+1)$:

$$ \mathrm{Vol}(\delta\mathrm{-ball}) = \frac{2^{m^2}}{\pi^m m!} \int_{\phi_1\leq\ldots\leq\phi_m \leq 2 \arcsin\sqrt{\delta/2}} \prod_{1\leq j < k \leq m} (\cos\phi_k-\cos\phi_j)^2 \prod_l \sin^2(\phi_l) d\phi_1 \cdots d\phi_k. $$

e. For the even dimensional case, the only changes are $2^{m^2}$ is replaced $2^{(m-1)^2}$ and the sine terms are absent.