Given is the following claim (proof of Lemma 4.2. in http://annals.math.princeton.edu/wp-content/uploads/annals-v170-n3-p09-p.pdf) $$ \int_{S^{n-1}} (v,v \otimes v - \frac{I_{n}}{n}) \: d\mu = 0$$ where $S^{n-1}$ is the unit sphere in $\mathbb{R}^n$ and $\mu$ is the Haar measure. Could someone explain why this holds, please? :)
I read that $S^{n-1}$ can be understood as the homogeneous space $SO(n)/SO(n-1)$ but I still don't understand why this claim holds.
Assuming the expression $(v,v\otimes v-I_n/n)$ is a tuple, the first component of which is the vector $v$ and the second the matrix $v\otimes v-I_n/n$, the answer drops out of the rotational invariance of the Haar measure $\mu$. Whatever $a=\int_{S^{n-1}}v\,d\mu$ is, it must obey $Oa=a$ for all $O\in\operatorname{SO}(n)$. If $a\ne0$ there exists an $O$ so that $Oa=-a$; hence $a=0$. Similarly, $b=\int_{S^{n-1}}v\otimes v-I_n/n\,d\mu$ is a matrix such that $ObO'=b$ for all orthogonal $O$. This implies $b$ is a multiple of the identity matrix. It is easy to check that for each $v\in S^{n-1}$ the trace of $v\otimes v$ equals $1$, so finally $b=0$.