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Jon Bannon
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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 by Pytlikin: T. Pytlik. http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdfRadial 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?

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 by Pytlik: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdf)

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?

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?

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Jon Bannon
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  • 112

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 generatorsgenerator (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 by Pytlik: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdf)

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?

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 considering the standard generators 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 by Pytlik: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdf)

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?

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 by Pytlik: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdf)

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?

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Jon Bannon
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InIt 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 von Neumann algebrafactor $L\mathbb{F}_{2}$, the self-adjoint element that such a realization is the sumeasy, since each 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 by Pytlik: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdf)

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.

If one takes the sum inIn $LG$, if we want to construct a MASA, it is obvious that considering the standard generators 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, does the resulting element generate generates a maximal abelian *-subalgebra? known as the Laplacian MASA. (That this is indeed maximal abelian was originally proved by Pytlik: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdf)

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?

In the group von Neumann algebra $L\mathbb{F}_{2}$, 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 by Pytlik: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdf)

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.

If one takes the sum in $LG$ of the standard unitary generators and their inverses, does the resulting element generate a maximal abelian *-subalgebra?

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 considering the standard generators 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 by Pytlik: http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10035.pdf)

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?

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Jon Bannon
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Jon Bannon
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