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Let $\Gamma$ be a discrete group acting on an infinite-dimensional Banach space $X$ by linear isometries.

Is there a probability measure (non-atomic, not supported on a finite dimensional subspace) on Borel subsets of $B_X$, the unit ball of $X$, that is invariant under this action? If not in general then under what assumptions?

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    $\begingroup$ The Dirac measure at $0$? $\endgroup$ Mar 22, 2016 at 9:30
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    $\begingroup$ oh, come on.... ;) $\endgroup$
    – user89292
    Mar 22, 2016 at 9:57
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    $\begingroup$ A sufficient condition is that $\Gamma$ preserves a finite-dimensional subspace. If this is too trivial for your taste, consider replacing "non-atomic" by "not supported on a finite-dimensional subspace". $\endgroup$ Mar 22, 2016 at 11:06
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    $\begingroup$ 1°) What does strongly continuous mean for a discrete group action ? 2°) Would assuming $\Gamma=\mathbb Z$ be useful to you? $\endgroup$ Mar 22, 2016 at 16:24
  • $\begingroup$ consider $B(x,\epsilon) \subset B_X$ the ball of radius $\epsilon$ around some $x \in B_X$. then $\mu(B(x,\epsilon)) = \mu(\gamma B(x,\epsilon)) = \mu( B(\gamma x,\epsilon))$ for any $\gamma \in \Gamma$. hence if $\Gamma$ is countably infinite, no, not even finitely supported. if $\Gamma$ is finite, there are the discrete distributions $H_{x_0}(x) = \sum_{\gamma \in \Gamma} \delta( x - \gamma x_0)$ where $x_0 \in B_X$ is fixed, from which (by filtering $\delta$ in some different directions on $B_X$) you can get some continuous distributions but only on a finite dimensional subset of $B_X$ $\endgroup$
    – reuns
    Mar 23, 2016 at 1:30

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The answer is: yes iff $X$ contains a finite dimensional invariant subspace of arbitrary large dimension (as you see, it has really nothing to do with the Banach structure of $X$ only with its quality as a $\Gamma$-representation). Equivalently, the answer is: yes iff $X_0$ (defined below) is of infinite dimension.

First example (which you will not like): take a trivial representation of $\Gamma$ on $X$ and any probability measure supported on $B_X$ (but not on a finite dimensional subspace).

You want to disregard the first example as non-ergodic.

Second example (very explicit): Take $\Gamma= \mathbb Z$ and let it act by an irrational rotation (by $\sqrt{2}$, as I said I'll be explicit) on $S^1$. Consider $X=L^2(S^1)$. In $B_X$ take the subset $Y$ consisting of all characteristic functions of half circles in $S^1$ (normalized). $Y$ is an $S^1$ equivariant copy of $S^1$ in $B_X$ and we can endow it with the Haar measure of $S^1$. This measure satisfies everything you wished for.

The second example is typical in the sense that the $\Gamma$-invariant measure is actually invariant under a compact group enveloping $\Gamma$.

I will explain below that if $X$ has no finite dimensional invariant subspaces (other than $\{0\}$) then the only invariant probability measure on $X$ is the Dirac measure at $0$ (for $X$ Hilbert this is Lemma 5.6 in "Amenable Invariant Random Subgroups" by Bader*-Duchesne-Lecureux). More generally, let $X_0$ be the closure in $X$ of the vector space generated by all invariant finite dimensional subspaces. Then every invariant probability measure on $X$ is supported on $X_0$.

It follows that if $X_0$ is finite dimensional then every invariant measure is indeed supported on finite dimensional invariant subspace. Note that if $X_0$ is infinite dimensional then it contains a sequence of finite dimensional invariant subspaces of arbitrary large dimensions and summing up the Lebesgue measures of the corresponding spheres (with coefficients so that the total mass is 1) gives an invariant probability measure which is not supported on a finite dimensional invariant subspace.

In what follows I will discuss $\Gamma$-invariant measures on the Banach space $X$ (not necessarily on the unit ball). I will assume that $X$ is separable, which probably is enough for you (e.g if $\Gamma$ is countable, or more generally, if $\Gamma$ is locally compact second countable and it acts strongly continuously on $X$). Fix a probability measure $\nu$ on the orbit space $X_0/G$, where $G$ is the Bohr compactification of $\Gamma$ (recall that the $\Gamma$ action on $X_0$ extends to $G$) and set $$\mu=\int_{X_0/G} (\text{The normalized Haar measure on the orbit})~ d\nu(\text{orbit}).$$ Then $\mu$ is clearly a $\Gamma$-invariant measure and it is supported on $X_0$. Conversely, every $\Gamma$-invariant measure is given in this form. In particular, every ergodic $\Gamma$-measure is supported (on $X_0$ and) on a unique $G$-orbit, and it is the Haar measure on it. Note that by ergodic decomposition it is enough to prove the last statement.

Claim: let $\Gamma$ be a group acting isometrically on a complete, separable metric space $Y$ preserving a fully supported, ergodic probability measure $\mu$. Then there is a $\Gamma$-equivariant homeomorphism $Y\simeq G/L$, where $G$ is the Bohr compactification of $\Gamma$ and $L<G$ a closed subgroup, and under this identification $\mu$ corresponds to the Haar measure.

The essential part of the claim is the compactness of $Y$, which follows by completeness from total-boundedness, which is easy (generalize the standard claim that $Y$ is bounded, which is proven by taking a compact set of more than half the weight). Then the isometry group of $Y$ is compact by Arzela-Ascoli and the $\Gamma$ action extends to $G$. By ergodicity (and the fact that $Y/G$ is nice) we get that $\mu$ is supported on a unique orbit and everything else follows easily.

Now, for the statement we had regarding ergodic measures on $X$, take $Y$ to be the support of an ergodic measure on $X$, which I assume to be total. Observe that the action of $G$ on $Y$ extends to a linear isometric action on $X$. Call Peter-Weyl to see that $X=X_0$.

$*$ Hi.

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  • $\begingroup$ by the way @user89334 do you have any idea if it is possible to find a probability measure whose class is preserved instead? $\endgroup$
    – user89292
    Mar 24, 2016 at 8:56
  • $\begingroup$ Finding a probability measure whose class is invariant (even ergodic) is just too easy. For example you can find such a measure on $\Gamma$ itself (a positive summable to 1 function), fix a point $x\in X$ and push your measure via the orbit map. No assumption on the representation here. I suppose that there is no reasonable classification of invariant classes on $X$, even under strong assumptions on $X$ and the $\Gamma$-representation. $\endgroup$
    – Uri Bader
    Mar 24, 2016 at 11:01
  • $\begingroup$ On the other hand, if you put assumptions on your measure classes you might get some non-trivial restrictions. For example, if your class is stationary (wrt a generating probability measure on $\Gamma$) then it already must be $\Gamma$-invariant (this follows from the metric-ergodicity of the Furstenberg-Poisson boundary). This puts you back in the setting of your question (and its answer). $\endgroup$
    – Uri Bader
    Mar 24, 2016 at 11:02
  • $\begingroup$ That is very helpful, thanks again. $\endgroup$
    – user89292
    Mar 25, 2016 at 8:55

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