Suppose $u \in \mathbb{R}^{n}$ is uniformly distributed on the unit sphere denoted by $S^{n-1}$. Let $\pi$ be the uniform distribution on this sphere. We are about to evaluate $$ L=\int_{S^{n-1}}\int_{S^{n-1}} e^{ u^{\prime} v} \pi(d u) \pi(d v). $$
The author of a paper says that, by spherical symmetry, we can fix $v=e_{1}=(1,0, \ldots, 0)$ to obtain $$ L=\int e^{u_{1}} \pi(d u), $$ with $u=\left(u_{1}, \ldots, u_{n}\right)$ uniform on $S^{n-1}$.
I do not understand how we can get $ L=\int e^{u_{1}} \pi(d u), $ by spherical symmetry.
The spherical symmetry comes in when evaluating the $u$ integral. Because of the symmetry, the $u$ integral does not depend on $v$. To see this, consider two values $v_1$ and $v_2$. Then there exists a rotated coordinate system for $u$, call it $u'$, such that $v_1' = v_2$. Thus $$ \int_{S^{n-1}} e^{v_1\cdot u}d\pi(u) = \int_{S^{n-1}}e^{v_1'\cdot u'}d\pi(u') = \int_{S^{n-1}}e^{v_2\cdot u'}d\pi(u') = \int_{S^{n-1}} e^{v_2\cdot u}d\pi(u). $$ Sowe can pick any particular value of $v$ to evaluate the integral at and get the same answer. The paper chooses $v = \hat{e}_1 = (1,0,0,0,...)$ for convenience, giving $$ \int_{S^{n-1}}\int_{S^{n-1}}e^{v\cdot u}d\pi(u)d\pi(v) = \int_{S^{n-1}}\int_{S^{n-1}}e^{\hat{e}_1\cdot u}d\pi(u)d\pi(v) = \int_{S^{n-1}}e^{u_1}d\pi(u). $$