It is a basic fact that every type $II_{1}$ factor posesses a maximal abelian $*$-subalgebra (MASA). My question concerns the concrete realization of such subalgebras. For example, in the free group factor $L\mathbb{F}_{2}$ such a realization is easy, since each of the standard generators generates a MASA.
The Free Burnside group $G=B(2,665)=\langle a,b|g^{665} \rangle$ is infinite, by the work of Adyan and Novikov. Furthermore, the centralizer of any nonidentity element in $G$ is finite cyclic, and so the group is an i.c.c. group and the associated left group von Neumann algebra $LG$ is a type $II_{1}$ factor.
In $LG$, if we want to construct a MASA, it is obvious that the considering the von Neumann subalgebra generated by a given standard generator (and the identity) will not do.
In the free group factor $L\mathbb{F}_{2}$, the self-adjoint element that is the sum of all the standard unitary generators and their inverses generates a maximal abelian *-subalgebra known as the Laplacian MASA. (That this is indeed maximal abelian was originally proved in: T. Pytlik. Radial functions on free groups and a decomposition of the regular representation into irreducible components. J. Reine Angew. Math., 326:124–135, 1981.)
My specific question is this:
If one takes the sum in $LG$ of the standard unitary generators and their inverses, does the resulting element generate a maximal abelian *-subalgebra?
