TITLE: Properties of a non-sofic group
QUESTION [23 upvotes]: This question is, essentially, a comment of Mark Sapir. I think it deserves to be a question.
A countable, discrete group $\Gamma$ is sofic if for every $\epsilon>0$ and finite subset $F$ of $\Gamma$ there exists an $(\epsilon,F)$-almost action of $\Gamma$. See, for example Theorem 3.5 of the nice survey of Pestov http://arxiv.org/PS_cache/arxiv/pdf/0804/0804.3968v8.pdf.
Gromov asked whether all countable discrete groups are sofic. It is now widely believed that there should be a counterexample to this.
Since most groups are sofic, it would be useful to have a collection of properties that would imply that a group is not sofic...so one can then construct a beast having such properties.

What are some abstract properties of $\Gamma$ that would imply $\Gamma$ is not sofic?

An open question of Nate Brown asks whether all one-relator groups are sofic. I'd be interested to know what properties of a one-relator group $\Gamma$ would imply that $\Gamma$ is not sofic.

REPLY [5 votes]: Sofic groups fulfill determinant conjecture. 
This implies in particular that there exists a natural constant $c$ such that given a matrix $M$ over the integral  group ring of a given sofic group $G$, we have that
$$
|tr_{vN} \exp(-cM) - \dim_{vN}\ker M| < \frac{1}{3}.
$$
This can be used to show that some problems about the group are decidable. Suppose a group $G$ is torsion-free, has decidable word problem, fulfills Atiyah conjecture, and is sofic. Then there is an algortihm which decides whether a given matrix $M$ over the integral group ring has non-trivial kernel, as an operator on $[l^2(G)]^{\dim M}$. 
Indeed, given $M$ it's easy to bound its $l^2$ norm and based on this to decide how many terms in $tr_{vN}\exp(-cM)$ have to be computed in order to be less than $\frac{1}{6}$ from the actual value of $tr_{vN} \exp(-cM)$. Call this approximation $a$ (it can be computed since the word problem is decidable). Now, because $G$ is torsion free and fulfills Atiyah conjecture, we know that $\dim_{vN}\ker M$ is an integer, and it's equal to $0$ iff $M$ has trivial kernel. So $M$ kas trivial kernel if and only if $a<\frac{1}{2}$
Similar algorithm works if a group has bounded torsion, since $\frac{1}{3}$ in the first equation can be exchanged with any postivie real number. I seem to have read that there exist Tarski monsters with decidable word problem. That means that in principle :-) one could try to show that there's no such algorithm for these Tarski monsters and arrive at the conclusion that either these monsters are non-sofic or they don't fulfill Atiyah conjecture.