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Parikh and Parnes showed that one can define a isometry-invariant finitely-additive conditional probability for all pairs of subsets of the interval. Can one extend this result to bounded subsets of $\mathbb R^2$? Of course, due to the non-amenability of SO($n$), $n>2$, one can't do it in $\mathbb R^n$ for $n>2$.

One way to put the question more precisely is to ask whether there is a Popper function $P$ on $[0,1]^2$ such that (a) every non-empty set is normal and (b) we have the following weak invariance condition: $P(A|B)=P(\rho A|\rho B)$ for every isometry $\rho$ such that $A,B,\rho A,\rho B$ are all subsets of $[0,1]^2$.

I've tried to leverage Winfried Just's construction of a bounded paradoxical subset of $\mathbb R^2$ to provide a counterexample, but with only the weak invariance condition, I can't do it. (A stronger invariance condition would be $P(A|B)=P(\rho A|\sigma B)$ for every pair of isometries $\rho,\sigma$ such that $A,B,\sigma B$ are subsets of $[0,1]^2$ and $A\subseteq B$ and $\rho A\subseteq \sigma B$. With that invariance condition, the bounded paradoxical subset of $\mathbb R^2$ shows we can't have such a $P$.)

I don't think the Parikh and Parnes construction using a Lebesgue sample generalizes to $\mathbb R^2$. That construction will, I think, fail when the isometry group has exponential growth.

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Actually, there is no way the Parikh and Parnes construction could generalize to $\mathbb R^2$, since it requires constructing a finitely additive hyperreal-valued measure that is almost invariant in the sense that $(\mu(\rho A)-\mu(A))/\mu(A)$ is infinitesimal for non-empty $A$ and isometry $\rho$. But such a measure would rule out bounded paradoxical subsets. –  Alexander Pruss Jul 16 '13 at 3:09

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The answer seems to be negative. Here's a quick and rough sketch.

Say that $C:2^\Omega \times (2^\Omega \backslash \{\varnothing\})\to [0,\infty]$ is a comparative probability on $\Omega$ iff $C(-,Y)$ is a finitely additive measure with $C(Y,Y)=1$, and $C(X,Y)C(Y,Z)=C(X,Z)$ whenever the product on the lhs is defined (i.e., isn't zero times infinity). Comparative probabilities and Popper functions making all non-empty subsets normal are interdefinable: given such a Popper function $P$, let $C(X,Y)=P(X,X\cup Y)/P(Y,X\cup Y)$ (where $x/0=\infty$ if $x>0$). Analogous invariance conditions hold for comparative probabilities and Popper functions. So I'll just show that there is no weakly invariant comparative probability on a large square $\Omega$ (and hence on any square), where weak invariance is the condition $C(\rho A,\rho B)=C(A,B)$ assuming all four sets fit in $\Omega$, and strong invariance is the condition $C(\rho A,B)=C(A,\rho B)=C(A,B)$, where $\rho$ is an isometry.

Fact: We can bootstrap from the weaker invariance condition $C(\rho A,\rho B)=C(A,B)$ to the stronger invariance condition $C(\rho A,B)=C(A,\rho B)=C(A,B)$ for $A,B\subseteq [-1,1]^2$, as long as $\Omega=[-L,L]^2$ for a sufficiently large $L$, size dependent on $\rho$.

Once we have this bootstrapping, we simply let $E$ be Just's bounded paradoxical subset of the unit disc and choose $L$ sufficiently large to guarantee that the stronger invariance condition holds for all the isometries $\rho$ used in Just's paradoxical decomposition.

So we just need to prove the bootstrapping fact. Throughout, I will assume that $L$ is sufficiently large to make the arguments go through, i.e., to make all the transformations fit inside $[-L,L]^2$.

All we need to prove is that $C(A,\rho B)=C(A,B)$ for any isometry $\rho$ (assuming $L$ is big enough relative to facts about $\rho$), since we then have $C(\rho A,B)=C(\rho A,\rho B)=C(A,B)$.

First, we prove $C(A,\rho B)=C(A,B)$ for $\rho$ a reflection.

We have $C(A,\rho A)=C(\rho A,\rho\rho A)=C(\rho A,A)$ by the weaker invariance condition. Moreover, $1=C(A,A)=C(A,\rho A)C(\rho A,A)$ if the rhs is defined, which it is since both factors are equal. Thus $C(A,\rho A)=C(\rho A,A)=1$. Thus, $C(B,A)=C(B,A)C(A,\rho A)=C(B,\rho A)$ for any $B$. (For this argument to work, $L$ must be large enough that $B\cup \rho B$ fit in $[-L,L]$.)

Now suppose $\rho$ is a rotation about some point $z$. Let $\sigma$ be a reflection about a line that goes through $z$. Observe that $\sigma\rho = \rho^{-1}\sigma$. And we've already proved our strong invariance for $\sigma$. Then $C(A,\rho A)=C(A,\sigma\rho A)=C(A,\rho^{-1}\sigma A)=C(\rho A,\rho\rho^{-1}\sigma A)=C(\rho A,\sigma A)=C(\rho A,A)$.

But $1=C(A,A)=C(A,\rho A)C(\rho A,A)$. Thus, $C(A,\rho A)=C(\rho A,A)=1$. Thus, $C(B,A)=C(B,\rho A)C(\rho A,A)=C(B,\rho A)$. (For this argument to work, $L$ must be large enough that $A\cup \rho A\cup \sigma\rho A\cup \sigma A$ fit in $[-L,L]$.)

But all isometries can be written in terms of rotations and reflections.

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