The trace as an integral over the projective space

Question

Let $(E,h)$ be a Hermitian vector space of dimension $n$ and $u\in End(E)$. We have an expression of the trace of $u$ as the integral $$Tr(u)=\frac{n}{A}\int_{S}\langle v,uv\rangle d\mu$$ where $S$ is the unit sphere in $E$ and $A$ its volume (see Integral around unit sphere of inner product).

Is there a similar expression but using the projective space instead of the sphere? That is some equality that looks like $$Tr(u)=C\int_{\mathbb{P}E}\frac{\langle v,uv\rangle}{||v||^2} d\mu_{FS}$$ for some constant $C$ and $d\mu_{FS}$ being the Fubini-Study volume form.

Answer 1

Yes, this is true as a consequence of the fact that the unitary group induced by $h$ acts by isometries on the projective space endowed with the Fubini-Study metric.

Let $a\in\mathrm{End}(E)$. By the polar decomposition, there exists a unitary $u$ such that $a^\ast=|a|u$, where $|a|=(a^\ast a)^{1/2}$. Then $$ \int\frac{\langle \xi,aa^\ast\xi\rangle}{\|\xi\|^2}\,d\mu_{\mathrm{FS}}(\xi)=\int\frac{\|\,|a|u\xi\,\|^2}{\|\xi\|^2}\,d\mu_{\mathrm{FS}}(\xi)=\int\frac{\|\,|a|\xi\,\|^2}{\|\xi\|^2}\,d\mu_{\mathrm{FS}}(\xi)=\int\frac{\langle \xi,a^\ast a\xi\rangle}{\|\xi\|^2}\,d\mu_{\mathrm{FS}}(\xi). $$ Thus the functional $$ \phi\colon\mathrm{End}(E)\to\mathbb C,\,a\mapsto \int\frac{\langle\xi,a\xi\rangle}{\|\xi\|^2}\,d\mu_{\mathrm{FS}}(\xi) $$ satisfies $\phi(aa^\ast)=\phi(a^\ast a)$ for all $a\in \mathrm{End}(E)$. But it is known that the trace is (up to scaling) the only linear functional on $\mathrm{End}(E)$ with this property.