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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.

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2 Answers 2

Let $\Gamma$ be a sofic group. Gabor Elek and Endre Szabo showed here, that for any field $k$ and $a,b \in k[\Gamma]$ with $ab=1$ one has $ba=1$. Hence, coming up with a cleverly chosen group where this fails would provide a counterexample. Note that $k=\mathbb C$ is not a good start since Kaplansky showed long ago that the implication holds for fields of characteristic zero. However, for $k= GF(2)$ one might be lucky.

Let us consider $k=GF(2)$ for now. One strategy could be to start with $a = \sum_{g \in F} g$ and $b = \sum_{h \in K} h$ for some finite sets $F,K \subset G$. If $ab=1$, then a number of relations must hold: We quickly convince ourselves that $F$ and $K$ must have an odd number of elements and there exists a self-matching of the set $F \times K \setminus (f,k)$ such that matched pairs $(f',k') \sim (f'',k'')$ satisfy $f'k' = f''k''$ and $f=k^{-1}$ for the special unmatched pair. You can now turn everything around and start with an abstract group with generators $F \cup K$ and relations as above coming from an arbitrarily chosen self-matching. In the group ring of this abstract group, we will have $ab=1$, but why do we have $ba=1$? I was working on this for a while but could not come up with a counterexample. On the other hand, if $F$ and $K$ are large, I cannot believe that $ba=1$ will always hold.

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This is similar to the zero divisors conjecture, and looks as hopeless: too many relatively short relations. –  Mark Sapir Oct 27 '10 at 18:11
    
I still think this idea is tantalizing, though. Perhaps if we aimed at it with a computer in order to "stumble" on a counterexample! –  Jon Bannon Oct 27 '10 at 20:20
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@Jon: Computer won't help because you won't be able to show using a computer that certain relations do not hold. One can try to use Gromov random groups method and assign the matching randomly. In the case of zero divisors, though the "Gromov density" was 1/2, which means we can conclude nothing. I suspect that this is the case here too. –  Mark Sapir Oct 27 '10 at 21:07
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If I remember correctly, Gromov told me that somebody (in Orsay?) made a computer experiment with random groups, trying to show that property (T) occurs with probability 1 for density between $1/3$ and $1/2$ (which is proved by Gromov). The results were disappointing because the probability converges very slowly, and one needs huge number of relations (and very large relations) to get probability anywhere close to 1. –  Mark Sapir Oct 28 '10 at 0:07
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@ Andreas: Kate Juschenko showed me your answer after I mentioned exactly the same approach. I tried it few years ago and here is what I found. I wrote a simple program in GAP which randomly choses the way to match pairs of FxK, produces the corresponding group presentation with generators F,K, and simplifies it via Tietze transformations. I then analized about 10^5 random matchings for various cardinalities of F and K. In all cases the resulting groups happened to be obviously sofic (even residually finite), so ba=1 follows. For |F|,|K|-> infinity, the group was Z with probability -> 1. –  Denis Osin Feb 23 '11 at 6:49

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.

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Indeed, for free Burnside groups the word problem is solvable, so such Tarski monsters exist. (Just a comment to supplement.) –  Jon Bannon Oct 30 '10 at 23:33
    
The paper of Ad-Nov is not online, but the statement is here: eom.springer.de/B/b130300.htm –  Jon Bannon Oct 30 '10 at 23:39

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