we-solve (beta)

Math-1832 · mathematics · Group theory

Some Questions — Question 41

Status Partially solved · unsolvedmath.comLLM status Open unverified

Statement

Definitions and mathematical statement

Let a countable discrete group $\Gamma$ act by measure-preserving Borel transformations on a standard atomless probability space $(X,\mu)$. The action is essentially free if almost every point has trivial stabilizer. Its orbit equivalence relation is $R=\{(x,\gamma x):x\in X,\gamma\in\Gamma\}$, considered modulo null sets.

A graphing of $R$ is a countable family of measure-preserving partial Borel bijections $\phi_i:A_i\to B_i$ whose graphs lie in $R$ and whose generated equivalence relation equals $R$ almost everywhere. Its cost is $\sum_i\mu(A_i)$, possibly infinite. Define \[ \operatorname{cost}(\Gamma\curvearrowright X) =\inf_{\Phi\text{ graphing }R}\sum_i\mu(A_i). \] The domain measures count each chosen partial bijection once, rather than separately counting its inverse.

Does every countable group have fixed price? That is, for any two essentially free probability-measure-preserving actions of the same group, must their costs agree? The common value is allowed to be infinite and may depend on the group.

The two actions can occur on different probability spaces, and neither is assumed ergodic or equivalent to the other. The condition of essential freeness is part of the question: arbitrary actions with stabilizers need not have the same orbit-relation cost. The infimum defining each cost ranges over all graphings of that orbit relation, not merely the transformations arising from one preferred finite generating set of the group.

Short English statement

Do all essentially free probability-measure-preserving actions of a countable group have the same cost?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-1832-A1 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Math-1832

The witness is one group with finite generators carrying two essentially free actions preserving probability. Their costs lie in disjoint ranges, so the cost cannot be independent of the action as requested. This is presented as a claimed negative answer.

Some Questions — Question 41

OpenAI's claim

OpenAI presents a counterexample in the selected manuscript. A finitely generated amalgam is claimed to have free probability-preserving actions of different costs, refuting fixed price.

Mathematical statement of the claimed result

Result 1.
Let $A=F(a,b_1,\ldots,b_{99})$, $w=ab_1ab_2\cdots ab_{99}a$, $J=\langle b_1,\ldots,b_{99},w\rangle$, and \[ \Gamma=A*_J(J\times\langle t\rangle),\qquad\chi(a)=1,\quad\chi(b_i)=\chi(t)=0. \] Let $X$ be the free conull part of the Bernoulli shift $[0,1]^\Gamma$, and $Y_M=X\times\mathbb Z/M\mathbb Z$ with action $(x,r)g=(xg,r+\chi(g))$. Both carry their natural probability measures. Set $K=e^{96}96^{99}/95^{95}$, choose $0<\alpha<1/200$ with $K\alpha^3<1/2$, and put $\eta=\alpha/100$. The asserted orbit-equivalence-relation costs satisfy \[ \operatorname{Cost}(\Gamma\curvearrowright X)\ge1+\eta,\qquad \operatorname{Cost}(\Gamma\curvearrowright Y_M)\le1+99/M\quad(M>100). \] An integer $M>\max\{100,99/\eta\}$ therefore separates the costs of two essentially free probability-preserving actions of the same group.

Outline of the argument

Finite-height extensions of the amalgam yield graphings with costs approaching one. A relative graphing-compression inequality and Bernoulli finite-permutation models give a separate cost lower bound, supported by a planar word lemma; the two action constructions therefore have unequal costs.

Formal verification

Family 259 has no Lean scope page in this release. Its selected result is not presented as formalized. This site has not executed the Lean code. The scope descriptions do not certify the remaining manuscript arguments.

Status caveat

OpenAI attributes these manuscripts to work by an internal model. Its release cautions that arguments without formalization can contain mistakes. This catalogue has not independently checked the proofs; the labels report the mathematical claims attributed to OpenAI.

Sources

Go to contribution →