Existence of Non-Sofic Groups

Introduce the central open question: can every countable group be approximated by permutations?

• A countable group is sofic if finite tables can be approximated by permutations of finite sets.
• Weiss asked whether a non-sofic group exists.
• The soficity conjecture is the claim that every countable group is sofic.

Make the definition of soficity precise using the Hamming metric on symmetric groups.

• For p,q in Sym(Y), d_H(p,q) is the fraction of points moved differently.
• Maps p_n : H -> Sym(Y_n) must almost satisfy multiplication.
• Every nonidentity element must move almost every point.
• Taking disjoint copies ensures |Y_n| tends to infinity.

Check that the learner can recognize the components of a sofic approximation.

• Identify the multiplier condition.
• Recognize the nonidentity faithfulness condition.
• Apply the definition to a simple example.

Introduce the three ingredients used to turn a sofic approximation into a contradiction.

• Property (T) is a uniform spectral gap for unitary representations.
• An expander family has bounded degree and a uniform edge-expansion constant.
• LEF means every finite multiplication table embeds exactly into a finite group.
• LEF is stronger than soficity; groups like Thompson's V are not LEF.

Explain what property (T) does to the geometry of a sofic approximation.

• Sofic approximations of property-(T) groups split into disjoint bounded-degree expanders.
• The expansion constant is uniform along the approximation.
• The number of components may grow without bound.
• Property (T) constrains approximations, but does not rule them out.

State the single-expander criterion for a commuting group to be LEF.

• If H has property (T) and one sofic approximation of H×J has a single expanding H-generator graph, then J is LEF.
• The H-generator graph must expand on the whole model set.
• Kun–Thom upgrades expansion to exact finite embeddings of J.

Let the learner predict whether a union of many expanders already forces the commuting group to be LEF.

• Predict whether many expanding components suffice.
• Reflect on how commuting generators can move between components.
• Compare with the single-expander Kun–Thom obstruction.

Introduce Proposition 2.3: from many expanding components, recover one component for a commuting direct product.

• The direct product Λ×B shows a union alone is insufficient.
• Nested conjugation is needed: t_i Γ t_i^{-1} ≤ Γ and t_1 J t_1^{-1} ≤ Γ.
• A median-size function f forces transported components to inject into new components.
• Restricting to one component gives a sofic approximation of Γ×J on a single expander.

Realize the expander-matching criterion inside the unit group of the binary Leavitt algebra.

• R = LF2(1,2) with generators s_i,t_i satisfying t_i s_j = δ_ij and s_0 t_0 + s_1 t_1 = 1.
• G = EL_D(R) ≅ EL_9(R) is a property-(T) subgroup of R×.
• There are Γ, u, v, and J ≤ G with Γ×J ≤ G, uJu^{-1} ≤ Γ, and G = 〈Γ,u,v〉.
• The relation t_0s_0 = 1 ≠ s_0t_0 uses non-units, so it is not itself a non-soficity witness.

Check that the Leavitt example satisfies the hypotheses of Proposition 2.3.

• Match each hypothesis of Proposition 2.3 to the Leavitt setup.
• Choose the commuting subgroup and the nested conjugation data.
• Connect J ≅ Thompson's group V to the non-LEF contradiction.

Close the contradiction: if G were sofic, Proposition 2.3 would force J ≅ V to be LEF, but V is not.

• Thompson's group V is finitely presented, infinite, and simple.
• Infinite simple finitely presented groups are not LEF.
• Proposition 2.3 would force J ≅ V to be LEF if G were sofic.
• Contradiction: G is non-sofic, hence R× is non-sofic.

State the final result and separate it from nearby open problems.

• The soficity conjecture is false.
• Hyperlinearity and Connes' embedding conjecture for group von Neumann algebras remain open.
• Surjunctivity of R× is unknown; either answer would be interesting.
• The method combines expander matching with a Leavitt-algebra construction.