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 $\mu_x$?
Best Answer
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.