When $\mu$ is concentrated at a single point $u_0$ on the sphere, we have $\mathbb{E}[uu^T] = u_0u_0^T$ and $\mathscr{M}(\mu) = \|I_n - u_0u_0^T\| = 1$, achieving the upper bound of $\mathscr{M}(\mu)$.
For a completely uniform distribution over the sphere, we have $\mathbb{E}[uu^T] = \frac{1}{n}I_n$. This result relies on the isotropic nature of the uniform distribution on the sphere. It can be derived from the fact that the trace of $\mathbb{E}[uu^T]$ must equal 1 (since each $u$ is a unit vector) and by symmetry, each diagonal entry of $\mathbb{E}[uu^T]$ is the same and all off-diagonal entries are zero. In this case, $\mathscr{M}(\mu) = \|I_n - \frac{1}{n}I_n\| = \|(1 - \frac{1}{n})I_n\| = 1 - \frac{1}{n}$.
For a discrete mixture distribution $\mu = \sum_{m=1}^N p_m\delta(u_m)$, where $\sum_m p_m = 1$, $p_m > 0$, and $u_m \in S^{n-1}$, we have:
$$\mathbb{E}[uu^T] = \sum_{m=1}^N p_mu_mu_m^T$$
The computation of $\mathscr{M}(\mu)$ in this case depends on how the vectors $u_m$ are positioned on the sphere and their corresponding probabilities $p_m$.
$\mathscr{M}(\mu)$ serves as a measure of non-uniformity, quantifying the maximum deviation of $\mu$ from the uniform distribution in any direction. Values close to 1 indicate high concentration, while values close to 0 indicate a more uniform spread. For any fixed distribution type that is not perfectly uniform, $\mathscr{M}(\mu)$ tends to increase with $n$, approaching 1 as $n$ approaches infinity: $\lim_{n \to \infty} \mathscr{M}(\mu) = 1$. Even small deviations from uniformity in any one dimension can lead to large overall deviations in high-dimensional space. Conversely, for the uniform distribution, $\mathscr{M}(\mu) = 1 - \frac{1}{n}$ approaches 0 as $n$ increases, reflecting less deviation from uniformity in higher dimensions.
Also $\mathscr{M}(\mu)$ is invariant under rotations of the sphere. For any orthogonal matrix $Q$, we have $\mathscr{M}(Q\mu) = \mathscr{M}(\mu)$, where $Q\mu$ denotes the pushed-forward measure.
