Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and let $X$ be a compact connected $C^\infty$-smooth manifold. Let $F \colon \Omega \times X \to X$ and $\bar{F} \colon \Omega \times X \to X$ be measurable functions such that for each $\omega$, $F(\omega,\,\cdot\,) \colon X \to X$ is a $C^\infty$-diffeomorphism with inverse $\bar{F}(\omega,\,\cdot\,) \colon X \to X$.

Suppose we have a probability measure $\rho$ on $X$, with strictly positive $C^\infty$-smooth density everywhere [as interpreted via charts], such that $$ \rho(A) \ = \ \mathbb{P} \otimes \rho(F^{-1}(A)) \ = \ \mathbb{P} \otimes \rho(\bar{F}^{-1}(A)) $$ for all $A \in \mathcal{B}(X)$. Does it follow that $\mathbb{P}$-almost every $\omega \in \Omega$ has the property that for all $A \in \mathcal{B}(X)$,

$\hspace{50mm} \rho(A) \ = \ \rho(\{x \in X : F(\omega,x) \in A\}) \ $?

Motivation. Leaving aside the compactness assumption, take $X=(0,1)$ and $\Omega=\{1,2\}$ with $\mathbb{P}(\{1\})=\mathbb{P}(\{2\})=\frac{1}{2}$. Writing $f=F(1,\cdot)$ and $g=F(2,\cdot)$, I think it should be easy to show that for $\rho$ being the Lebesgue measure,

  • $\rho(\cdot)=\mathbb{P} \otimes \rho(\bar{F}^{-1}(\cdot))$ if and only if $g(x)=2x-f(x)$;
  • $\rho(\cdot)=\mathbb{P} \otimes \rho(F^{-1}(\cdot))$ if and only if $g$ is the inverse of the map $x \mapsto 2x - f^{-1}(x)$;

and therefore, my very strong intuition (although I don't see an immediate proof) is that we cannot have $\rho(\cdot) = \mathbb{P} \otimes \rho(F^{-1}(\cdot)) = \mathbb{P} \otimes \rho(\bar{F}^{-1}(\cdot))$ unless $f=g=\mathrm{id}_X$: we would require that $x \mapsto 2x-f(x)$ is the inverse of $x \mapsto 2x-f^{-1}(x)$, which is the same as saying that the vertical reflection in $\{y=x\}$ of the diagonal reflection in $\{y=x\}$ of $\mathrm{graph}(f)$ coincides with the diagonal reflection in $\{y=x\}$ of the vertical reflection in $\{y=x\}$ of $\mathrm{graph}(f)$ - and my intuition is that this implies that $f$ is the identity function.

My question above addresses whether, in some sense, this reasoning generalises. I first wondered whether my conjectured result holds in the topological setting, with $X$ being any compact metric space and $\rho$ any probability measure on $X$; but as shown in the answer to this question, the answer is no. The counterexample given is very nice, and at the intuitive level (at least under my potentially nonsense intuition) the total disconnectedness of the given counterexample seems important for how my intuitive reasoning about the pair of maps on $(0,1)$ does not extend to this counterexample. So now I am wondering whether having a smooth structure rectifies the situation. [Of course, I might still be wrong even about the case of a pair of maps on $(0,1)$, although this would very much surprise me!]


1 Answer 1


Discontinuity is not crucial at all in the example given in the answer to this question, and the same phenomenon is present in the smooth setup as well. Namely, the free group can be replaced with $SL(2,\mathbb R)$ (or any non-compact semi-simple Lie group) and the space of infinite words with the boundary circle of the hyperbolic plane (or the "flag space" of the associated Riemannian symmetric space).

More precisely, let $G$ be the group $SL(2,\mathbb R)$. It acts in the standard way by linear fractional transformations on the augmented complex plane and preserves the augmented real line $S^1$ ($\equiv$ the boundary circle of the hyperbolic plane in the upper half-plane model). Let $K=SO(2)\subset G$ be the group of rotations, and let $m$ be a bi-$K$-invariant probability measure on $G$ (in particular, $m$ is preserved by the group inversion). Now one can put $(\Omega,\mathbb P)=(G,m)$ and $X=S^1$ with the map $\Omega\times X\to X$ being the aformentioned boundary action. The unique $K$-invariant measure $\rho$ on $S^1$ (the "Haar measure") is then $m$-stationary, but not $G$-invariant.

PS Stationarity alone is not sufficient to imply invariance (as these examples illustrate). It is reversibility that is essentially equivalent to invariance.

EDIT (too long for a comment) In the above example I really wanted the measure $m$ to be symmetric (i.e., to be preserved by the group inversion). The reason is that you are talking about measures which are stationary simultaneously with respect to $m$ and the reflected measure $\check m$ (the image of $m$ under group inversion). I do not recall any works where measures which are simultaneously $m$- and $\check m$-stationary would be considered (and it might be possible that in some situations all such measures must indeed be invariant - this seems to be an interesting and open question). On the other hand, if $m$ is symmetric, then we are talking about just one stationarity condition, and this situation is quite well understood.

  • $\begingroup$ Thank you! When you said "$m$ is preserved by the group inversion", is this really what you meant? Isn't the point rather that since $m$ is bi-$K$-invariant, it follows that the image-measure $\bar{m}$ of $m$ under group inversion is also bi-$K$-invariant, and therefore $\rho$ is both $m$-stationary and $\bar{m}$-stationary? Also, presumably you can weaken bi-$K$-invariant to just invariant under conjugation by elements of $K$? $\endgroup$ Oct 22, 2019 at 15:14
  • $\begingroup$ So I think a simple concrete example would be: Expressing $\mathbb{S}^1=\mathbb{R}/(2\pi\mathbb{Z})$, take $\Omega=(-\pi,\pi)$ with $\mathbb{P}=\frac{1}{2\pi}.\mathrm{Lebesgue}$, and let $F(\omega,[x])=[x+\varepsilon\sin(x-\omega)]$ for some $\varepsilon \in (0,1)$. (What's funny is that I've studied random maps of this kind before, and it didn't occurred to me that this is precisely a counterexample to what I had been thinking.) $\endgroup$ Oct 22, 2019 at 15:18
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    $\begingroup$ I have added an explanation why I wanted the measure $m$ to be symmetric (too long for a comment) $\endgroup$
    – R W
    Oct 23, 2019 at 16:00

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