Mixing
The Hahn-Hellinger Theorem
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People can tell the question is not up to the mark, but I felt without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras for myself. So can any body help me out clearing the concept of Hahn-Hellinger Theorem, what is the game playing inside the theorem, I tried many books and I am not able to get with clarity. More specifically how cyclic vectors are connected with multiplicity of self-adjoint operator, this theorem say any self-adjoint operator in $mathcalH$ is equivalent to a multiplication operator $M_z$ on $oplus L^2(X,mu_i)$, how it looks the case at least for finite dimensional case, how direct sum take cares multiplicity of the eigenvalue that I did not get, Also it is not clear all the measure joined to single measure. Further why cyclic decomposition remembers the spectral multiplicity. Please help. Thanks in advance
operator-algebras von-neumann-algebras
asked Jul 20 at 16:10
mathlover
10518
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up vote
0
down vote
favorite
People can tell the question is not up to the mark, but I felt without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras for myself. So can any body help me out clearing the concept of Hahn-Hellinger Theorem, what is the game playing inside the theorem, I tried many books and I am not able to get with clarity. More specifically how cyclic vectors are connected with multiplicity of self-adjoint operator, this theorem say any self-adjoint operator in $mathcalH$ is equivalent to a multiplication operator $M_z$ on $oplus L^2(X,mu_i)$, how it looks the case at least for finite dimensional case, how direct sum take cares multiplicity of the eigenvalue that I did not get, Also it is not clear all the measure joined to single measure. Further why cyclic decomposition remembers the spectral multiplicity. Please help. Thanks in advance
operator-algebras von-neumann-algebras
asked Jul 20 at 16:10
mathlover
10518
2
"without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras": a dubious claim, at best. Arguably, it is much easier to approach the spectral theorem via von Neumann algebras, not vice versa. A normal operator N on a Hilbert space H generates a commutative von Neumann algebra A, whose Gelfand spectrum S is a measurable space. Furthermore, H is an A-module, and the category of A-modules is equivalent to the category of Hilbert bundles over S, yielding a Hilbert bundle B over S. The dimensions of fibers of B are precisely spectral multiplicities.
– Dmitri Pavlov
Jul 21 at 16:19
@Dmitri Pavlov can you please ellaborate how you connecting modules and hilbert bundles. dimension of fibers to spectral theorem, I will be helpful if you give the litle more clarification
– mathlover
Jul 21 at 18:38
A few more details can be found in my notes dmitripavlov.org/notes/2018s-6325.pdf, see, in particular, 9.11, 9.14, 9.20. Many things in the text are hyperlinked and will lead you to the relevant definitions.
– Dmitri Pavlov
Jul 21 at 22:50
Thanks for the link.
– mathlover
Jul 22 at 4:15
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
People can tell the question is not up to the mark, but I felt without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras for myself. So can any body help me out clearing the concept of Hahn-Hellinger Theorem, what is the game playing inside the theorem, I tried many books and I am not able to get with clarity. More specifically how cyclic vectors are connected with multiplicity of self-adjoint operator, this theorem say any self-adjoint operator in $mathcalH$ is equivalent to a multiplication operator $M_z$ on $oplus L^2(X,mu_i)$, how it looks the case at least for finite dimensional case, how direct sum take cares multiplicity of the eigenvalue that I did not get, Also it is not clear all the measure joined to single measure. Further why cyclic decomposition remembers the spectral multiplicity. Please help. Thanks in advance
operator-algebras von-neumann-algebras
asked Jul 20 at 16:10
mathlover
10518
People can tell the question is not up to the mark, but I felt without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras for myself. So can any body help me out clearing the concept of Hahn-Hellinger Theorem, what is the game playing inside the theorem, I tried many books and I am not able to get with clarity. More specifically how cyclic vectors are connected with multiplicity of self-adjoint operator, this theorem say any self-adjoint operator in $mathcalH$ is equivalent to a multiplication operator $M_z$ on $oplus L^2(X,mu_i)$, how it looks the case at least for finite dimensional case, how direct sum take cares multiplicity of the eigenvalue that I did not get, Also it is not clear all the measure joined to single measure. Further why cyclic decomposition remembers the spectral multiplicity. Please help. Thanks in advance
operator-algebras von-neumann-algebras
asked Jul 20 at 16:10
mathlover
10518
edited Jul 20 at 16:41
asked Jul 20 at 16:10
mathlover
10518
asked Jul 20 at 16:10
mathlover
10518
asked Jul 20 at 16:10
mathlover
10518
10518
2
"without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras": a dubious claim, at best. Arguably, it is much easier to approach the spectral theorem via von Neumann algebras, not vice versa. A normal operator N on a Hilbert space H generates a commutative von Neumann algebra A, whose Gelfand spectrum S is a measurable space. Furthermore, H is an A-module, and the category of A-modules is equivalent to the category of Hilbert bundles over S, yielding a Hilbert bundle B over S. The dimensions of fibers of B are precisely spectral multiplicities.
– Dmitri Pavlov
Jul 21 at 16:19
@Dmitri Pavlov can you please ellaborate how you connecting modules and hilbert bundles. dimension of fibers to spectral theorem, I will be helpful if you give the litle more clarification
– mathlover
Jul 21 at 18:38
A few more details can be found in my notes dmitripavlov.org/notes/2018s-6325.pdf, see, in particular, 9.11, 9.14, 9.20. Many things in the text are hyperlinked and will lead you to the relevant definitions.
– Dmitri Pavlov
Jul 21 at 22:50
Thanks for the link.
– mathlover
Jul 22 at 4:15
add a comment |Â
2
"without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras": a dubious claim, at best. Arguably, it is much easier to approach the spectral theorem via von Neumann algebras, not vice versa. A normal operator N on a Hilbert space H generates a commutative von Neumann algebra A, whose Gelfand spectrum S is a measurable space. Furthermore, H is an A-module, and the category of A-modules is equivalent to the category of Hilbert bundles over S, yielding a Hilbert bundle B over S. The dimensions of fibers of B are precisely spectral multiplicities.
– Dmitri Pavlov
Jul 21 at 16:19
@Dmitri Pavlov can you please ellaborate how you connecting modules and hilbert bundles. dimension of fibers to spectral theorem, I will be helpful if you give the litle more clarification
– mathlover
Jul 21 at 18:38
A few more details can be found in my notes dmitripavlov.org/notes/2018s-6325.pdf, see, in particular, 9.11, 9.14, 9.20. Many things in the text are hyperlinked and will lead you to the relevant definitions.
– Dmitri Pavlov
Jul 21 at 22:50
Thanks for the link.
– mathlover
Jul 22 at 4:15
2
2
"without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras": a dubious claim, at best. Arguably, it is much easier to approach the spectral theorem via von Neumann algebras, not vice versa. A normal operator N on a Hilbert space H generates a commutative von Neumann algebra A, whose Gelfand spectrum S is a measurable space. Furthermore, H is an A-module, and the category of A-modules is equivalent to the category of Hilbert bundles over S, yielding a Hilbert bundle B over S. The dimensions of fibers of B are precisely spectral multiplicities.
– Dmitri Pavlov
Jul 21 at 16:19
"without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras": a dubious claim, at best. Arguably, it is much easier to approach the spectral theorem via von Neumann algebras, not vice versa. A normal operator N on a Hilbert space H generates a commutative von Neumann algebra A, whose Gelfand spectrum S is a measurable space. Furthermore, H is an A-module, and the category of A-modules is equivalent to the category of Hilbert bundles over S, yielding a Hilbert bundle B over S. The dimensions of fibers of B are precisely spectral multiplicities.
– Dmitri Pavlov
Jul 21 at 16:19
@Dmitri Pavlov can you please ellaborate how you connecting modules and hilbert bundles. dimension of fibers to spectral theorem, I will be helpful if you give the litle more clarification
– mathlover
Jul 21 at 18:38
@Dmitri Pavlov can you please ellaborate how you connecting modules and hilbert bundles. dimension of fibers to spectral theorem, I will be helpful if you give the litle more clarification
– mathlover
Jul 21 at 18:38
A few more details can be found in my notes dmitripavlov.org/notes/2018s-6325.pdf, see, in particular, 9.11, 9.14, 9.20. Many things in the text are hyperlinked and will lead you to the relevant definitions.
– Dmitri Pavlov
Jul 21 at 22:50
A few more details can be found in my notes dmitripavlov.org/notes/2018s-6325.pdf, see, in particular, 9.11, 9.14, 9.20. Many things in the text are hyperlinked and will lead you to the relevant definitions.
– Dmitri Pavlov
Jul 21 at 22:50
Thanks for the link.
– mathlover
Jul 22 at 4:15
Thanks for the link.
– mathlover
Jul 22 at 4:15
add a comment |Â
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2
"without understanding the Hahn-Hellinger Theorem properly there is no point talking von Neumann algebras": a dubious claim, at best. Arguably, it is much easier to approach the spectral theorem via von Neumann algebras, not vice versa. A normal operator N on a Hilbert space H generates a commutative von Neumann algebra A, whose Gelfand spectrum S is a measurable space. Furthermore, H is an A-module, and the category of A-modules is equivalent to the category of Hilbert bundles over S, yielding a Hilbert bundle B over S. The dimensions of fibers of B are precisely spectral multiplicities.
– Dmitri Pavlov
Jul 21 at 16:19
@Dmitri Pavlov can you please ellaborate how you connecting modules and hilbert bundles. dimension of fibers to spectral theorem, I will be helpful if you give the litle more clarification
– mathlover
Jul 21 at 18:38
A few more details can be found in my notes dmitripavlov.org/notes/2018s-6325.pdf, see, in particular, 9.11, 9.14, 9.20. Many things in the text are hyperlinked and will lead you to the relevant definitions.
– Dmitri Pavlov
Jul 21 at 22:50
Thanks for the link.
– mathlover
Jul 22 at 4:15