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Let $G=\langle a,b | R \rangle$ be a one-relator group. When can the left group von Neumann algebra $LG$ be isomorphic to a free group factor? Jesse and Andreas have "trapped the lion" pretty well with their comments below.

A bit more modest related question: if $L_{a}$ and $L_{b}$ are the unitary elements in $LG$ corresponding to $a$ and $b$, respectively, can the free entropy of ($L_{a}+L_{a}^{*},L_{b}+L_{b}^{*}$) be finite? For the definition of free entropy, see Voiculescu's survey: http://arxiv.org/PS_cache/math/pdf/0103/0103168v1.pdf.

(It is possible for a set of generators of a type $II_{1}$-factor that is not a free group factor to have finite free entropy. Nate Brown establishes this in http://arxiv.org/abs/math/0403294.)

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    $\begingroup$ Following the $\ell^2$-Betti numerology, this should never happen if $G$ is torsionfree (unless $G$ is abelian). Linnell and Dicks showed that the first $\ell^2$-Betti number of a torsionfree 2-generator 1-relator group vanishes. If it where isomorphic to an interpolated free group factor $L\mathbb F_t$, then one would expect that $t=1$ (being equal to the first $\ell^2$-Betti number plus $1$. $\endgroup$
    – Andreas Thom
    Commented Dec 7, 2010 at 18:10
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    $\begingroup$ To go along with what Andreas wrote, if you allow torsion then I believe it was Dykema and Radulescu who showed that the groups $\langle a, b \ | \ b^k \rangle$ for $2 \leq k < \infty$ always give interpolated free group factors $L\mathbb F_t$, with $1 < t < 2$. $\endgroup$
    – Jesse Peterson
    Commented Dec 7, 2010 at 21:35
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    $\begingroup$ Jesse has answered this question as asked. Thanks! (I should have looked more carefully for such results before asking the question!!!) $\endgroup$
    – Jon Bannon
    Commented Dec 7, 2010 at 22:31
  • $\begingroup$ I've changed the question a bit to allow for more thoughts. $\endgroup$
    – Jon Bannon
    Commented Dec 7, 2010 at 22:37
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    $\begingroup$ Also, do you mean free entropy when you say free entropy dimension above? The free entropy dimension is always finite when the factor embeds into $R^\omega$. $\endgroup$
    – Jesse Peterson
    Commented Dec 7, 2010 at 23:15

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I post this for future reference... I've just come across this nice result:

http://www.math.jussieu.fr/~pfima/Documents/Baumslag-Solitar-Groups.pdf

It turns out that the group factors associated to certain non-residually finite Baumslag-Solitar groups are prime, have no Cartan, and are not solid...whereas free group factors are solid. The result here also proves (Appealing to Ozawa's solid von Neumann algebra paper) that the group von Neumann algebra of such a Baumslag -Solitar group cannot be isomorphic to the group von Neumann algebra of an I.C.C. hyperbolic group.

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