If the pushforward of $\mu$ under the group action is invariant, then $\mu$ must be the Haar measure.

Question

Let $G=\text{PU}(d)$ the projective unitary group, acting on the complex projective space $X=\mathbb{P}(\mathbb{C}^d)$, in the usual way.

Let $\mu:\mathcal{B}(G)\to[0,\infty]$ be a Radon measure, where $\mathcal{B}(G)$ denotes the Borel sets of $G$. For every $x\in X$, denote by $\mu_x:\mathcal{B}(X)\to[0,\infty]$ the pushforward measure of $\mu$ under the group action $g\mapsto gx$, which is given by $\mu_x(A)=\mu(\{g\in G\,|\, gx\in A\})$.

Consider the following property:

$(\star)$ For all $x\in X$, $\mu_x$ is $G$-invariant, meaning that $$\forall x\in X, \forall h\in G, \forall A\in\mathcal{B}(X): \mu_x(hA)=\mu_x(A).$$

Note that if $\mu$ is the Haar measure of $G$, then $(\star)$ holds. Is the converse true? That is, if $(\star)$ holds, must $\mu$ be the Haar measure?

I believe the answer is yes, but can't seem find an argument. Is there a way to write $\mu$ in terms of the measures $\{\mu_x\,|\,x\in X\}$ that would allow to conclude that $\mu$ must be itself invariant using the invariance of the pushforward measures $\mu_x$?

Answer 1

Let G be a compact Lie group, and H a closed subgroup. Let X=G/H. Suppose there exists a representation V of G with V^H=0. I claim that under these conditions, the answer to your question is no.

Such a V exists in all cases with d>2 in your setup with G=PU(d) and H=PU(d-1). (e.g. using Weyl branching law).

Let $\pi(g)$ be a matrix coefficient of V. Define a measure on G by $$\mu = (1+\epsilon \Re(\pi(g)))\,dg$$ where $\epsilon>0$ is chosen to be sufficiently small so that the ratio of $\mu$ with the Haar measure $dg$ is positive.

Then the measure $\mu_x$ constructed in the question from this $\mu$ agrees with the one obtained from the Haar measure. This boils down to the vanishing of the integral $$ \int_H \pi(ahb)\,dh$$ for any $a,b\in G$, which in turn is derived from the fact that $\int_H hv\,dh=0$ for all $v\in V$, which is where the condition V^H=0 comes into play.

Answer 2

Alternatively one can simply use disintegrations. Indeed, $G=PU(d)$ acts transitively on $X=\mathbb{P}(\mathbb{C}^d)$ with each stabilizer isomorphic to $H=PU(d-1)$. The idea is roughly that we may think $G\cong X\times H$ (locally or measure theoretically), and a measure on $G$ may be invariant in the first coordinate but possibly not in the second coordinate.

Let $\nu$ be any Borel probability measure on $H$ and let $\eta_{G/H}$ be the unique probability measure on $G/H$ invariant under translations by elements in $G$. Then we can define a Borel probability measure $\mu$ on $G$ by declaring for $\phi:G\to\mathbb{R}$ an anonymous bounded measurable function

$$\int_G \phi\, d\mu = \int_{G/H} \int_H \phi(xh)\, d\nu(h) \, d\eta_{G/H}(xH).$$

(Recall that any such disintegration defines a unique measure.)

Denote by $\pi:G\to G/H$ the canonical projection. Then for $\psi:G/H\to \mathbb{R}$ an anonymous bounded measurable function, we have

\begin{align*} \int_{G/H} \psi(xH) \,d \pi_\ast(\mu)(xH) &=\int_G \psi\circ \pi(x) \, d\mu(x)\\ &= \int_{G/H} \int_H \psi\circ \pi(xh)\, d\nu(h) \, d\eta_{G/H}(xH)\\ &= \int_{G/H} \int_H \psi\circ \pi(x)\, d\nu(h) \, d\eta_{G/H}(xH)\\ &= \int_{G/H} \psi(xH) \, d\eta_{G/H}(xH), \end{align*}

so that $\pi_\ast(\mu)=\eta_{G/H}$. On the other hand, for $h_0\in H$ we have (since the fibers of $\pi:G\to G/H$ are invariant under right translations $R_\bullet$ by elements in $H$):

\begin{align*} \int_G \phi(xh_0)\, d\mu(x) &= \int_{G/H} \int_H \phi(xhh_0)\, d\nu(h) \, d\eta_{G/H}(xH)\\ &= \int_{G/H} \int_H \phi(x\hbar)\, d(R_{h_0})_\ast(\nu)(\hbar) \, d\eta_{G/H}(xH), \end{align*}

so that if $\mu$ were invariant under translations by elements in $G$ (and hence in $H$), then $\nu$ would have to be invariant under translations by elements in $H$. But there are probability measures on $H$ that are not Haar (if $H\neq \{1\}$, that is, if $d\in\mathbb{Z}_{\geq3}$).