What is the purpose of making no reference to operators on a Hilbert space (Gelfand-Naimark)?












1












$begingroup$


I read from here pag. 209




The residue ring of an *-ring is an *-ring itself. Hence follows that if R is a closed *-subring of the ring of operators in Hilbert space, then any its residue ring can be imbedded into the ring of operators in Hilbert space




and from wiki




.. C*-algebras since it established the possibility of considering a C*-algebra as an abstract algebraic entity without reference to particular realizations as an operator algebra.




Ok, but if Von Neumann he needed to use ring of operator why we need to abstract its work to avoid to use an operator algebra?



I know that quantum mechanics formalism use Von Neumann approach in many aspects, not Gelfand-Naimark that I prefer instead.






gelfand-representation







share|cite|improve this question









$endgroup$












  • $begingroup$
    The advantage of abstracting is that often objects don't occur as a particular realization but only as an abstract thing and the particular realization can be non-unique or hard to construct. To be more specific in this contect: A C*-algebra can occure without an embedding into a ring of operators of some hilbert space. The construction of such a ring can be hard and non-unique. And actually, the construction to show that every C*-algebra can be seen as an operator algebra makes the hilbert space very large.
    $endgroup$
    – Paul K
    Jan 22 at 18:28












  • $begingroup$
    mm.. so Von Neumann work refer to a very small hilbert space ? Why not replace Von Neumann work with Gelfand-Naimark abstraction and use it in theoretical quantum physics ? because I see that Von Neumann work drive to close 'pure states' into projective space. Is not this a limitation?
    $endgroup$
    – Jack
    Jan 22 at 18:36










  • $begingroup$
    I don't know anything about theoretical physics and I only know a bit about C*-algebras and Von Neumann algebras. What I wanted to say is that the GNS-construction constructs for every state a pre-hilbert space which is then completed. Afterwards from all of these hilbert spaces you construct the hilbert space in whose operator ring your C*-algebra can be embedded. If some abstract C*-algebra occurs, you might not want to carry out this construction but just work with what you have.
    $endgroup$
    – Paul K
    Jan 22 at 18:51












  • $begingroup$
    Yes, but when you said the GNS-construction constructs for every state a pre-hilbert space which is then completed and in whose operator ring your C*-algebra can be embedded this is a very good news because I understand that Von Neumann make something more respect to GNS that I see that add more structure to 'limitate every each state. If you think a Von Neumann Algebra is -algebra of bounded operators on a Hilbert space that is *closed in the weak operator topology and contains the identity operator.
    $endgroup$
    – Jack
    Jan 22 at 19:05










  • $begingroup$
    In GNS construction we don't have more conditions that characterize each state (respect to Von Neumann work), is correct? But if is correct a Gelfand state is more weak of Von Neumann state. Is correct this ?
    $endgroup$
    – Jack
    Jan 22 at 19:06
















1












$begingroup$


I read from here pag. 209




The residue ring of an *-ring is an *-ring itself. Hence follows that if R is a closed *-subring of the ring of operators in Hilbert space, then any its residue ring can be imbedded into the ring of operators in Hilbert space




and from wiki




.. C*-algebras since it established the possibility of considering a C*-algebra as an abstract algebraic entity without reference to particular realizations as an operator algebra.




Ok, but if Von Neumann he needed to use ring of operator why we need to abstract its work to avoid to use an operator algebra?



I know that quantum mechanics formalism use Von Neumann approach in many aspects, not Gelfand-Naimark that I prefer instead.






gelfand-representation







share|cite|improve this question









$endgroup$












  • $begingroup$
    The advantage of abstracting is that often objects don't occur as a particular realization but only as an abstract thing and the particular realization can be non-unique or hard to construct. To be more specific in this contect: A C*-algebra can occure without an embedding into a ring of operators of some hilbert space. The construction of such a ring can be hard and non-unique. And actually, the construction to show that every C*-algebra can be seen as an operator algebra makes the hilbert space very large.
    $endgroup$
    – Paul K
    Jan 22 at 18:28












  • $begingroup$
    mm.. so Von Neumann work refer to a very small hilbert space ? Why not replace Von Neumann work with Gelfand-Naimark abstraction and use it in theoretical quantum physics ? because I see that Von Neumann work drive to close 'pure states' into projective space. Is not this a limitation?
    $endgroup$
    – Jack
    Jan 22 at 18:36










  • $begingroup$
    I don't know anything about theoretical physics and I only know a bit about C*-algebras and Von Neumann algebras. What I wanted to say is that the GNS-construction constructs for every state a pre-hilbert space which is then completed. Afterwards from all of these hilbert spaces you construct the hilbert space in whose operator ring your C*-algebra can be embedded. If some abstract C*-algebra occurs, you might not want to carry out this construction but just work with what you have.
    $endgroup$
    – Paul K
    Jan 22 at 18:51












  • $begingroup$
    Yes, but when you said the GNS-construction constructs for every state a pre-hilbert space which is then completed and in whose operator ring your C*-algebra can be embedded this is a very good news because I understand that Von Neumann make something more respect to GNS that I see that add more structure to 'limitate every each state. If you think a Von Neumann Algebra is -algebra of bounded operators on a Hilbert space that is *closed in the weak operator topology and contains the identity operator.
    $endgroup$
    – Jack
    Jan 22 at 19:05










  • $begingroup$
    In GNS construction we don't have more conditions that characterize each state (respect to Von Neumann work), is correct? But if is correct a Gelfand state is more weak of Von Neumann state. Is correct this ?
    $endgroup$
    – Jack
    Jan 22 at 19:06














1












1








1





$begingroup$


I read from here pag. 209




The residue ring of an *-ring is an *-ring itself. Hence follows that if R is a closed *-subring of the ring of operators in Hilbert space, then any its residue ring can be imbedded into the ring of operators in Hilbert space




and from wiki




.. C*-algebras since it established the possibility of considering a C*-algebra as an abstract algebraic entity without reference to particular realizations as an operator algebra.




Ok, but if Von Neumann he needed to use ring of operator why we need to abstract its work to avoid to use an operator algebra?



I know that quantum mechanics formalism use Von Neumann approach in many aspects, not Gelfand-Naimark that I prefer instead.






gelfand-representation







share|cite|improve this question









$endgroup$




I read from here pag. 209




The residue ring of an *-ring is an *-ring itself. Hence follows that if R is a closed *-subring of the ring of operators in Hilbert space, then any its residue ring can be imbedded into the ring of operators in Hilbert space




and from wiki




.. C*-algebras since it established the possibility of considering a C*-algebra as an abstract algebraic entity without reference to particular realizations as an operator algebra.




Ok, but if Von Neumann he needed to use ring of operator why we need to abstract its work to avoid to use an operator algebra?



I know that quantum mechanics formalism use Von Neumann approach in many aspects, not Gelfand-Naimark that I prefer instead.






gelfand-representation




gelfand-representation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 22 at 18:21









JackJack

215




215












  • $begingroup$
    The advantage of abstracting is that often objects don't occur as a particular realization but only as an abstract thing and the particular realization can be non-unique or hard to construct. To be more specific in this contect: A C*-algebra can occure without an embedding into a ring of operators of some hilbert space. The construction of such a ring can be hard and non-unique. And actually, the construction to show that every C*-algebra can be seen as an operator algebra makes the hilbert space very large.
    $endgroup$
    – Paul K
    Jan 22 at 18:28












  • $begingroup$
    mm.. so Von Neumann work refer to a very small hilbert space ? Why not replace Von Neumann work with Gelfand-Naimark abstraction and use it in theoretical quantum physics ? because I see that Von Neumann work drive to close 'pure states' into projective space. Is not this a limitation?
    $endgroup$
    – Jack
    Jan 22 at 18:36










  • $begingroup$
    I don't know anything about theoretical physics and I only know a bit about C*-algebras and Von Neumann algebras. What I wanted to say is that the GNS-construction constructs for every state a pre-hilbert space which is then completed. Afterwards from all of these hilbert spaces you construct the hilbert space in whose operator ring your C*-algebra can be embedded. If some abstract C*-algebra occurs, you might not want to carry out this construction but just work with what you have.
    $endgroup$
    – Paul K
    Jan 22 at 18:51












  • $begingroup$
    Yes, but when you said the GNS-construction constructs for every state a pre-hilbert space which is then completed and in whose operator ring your C*-algebra can be embedded this is a very good news because I understand that Von Neumann make something more respect to GNS that I see that add more structure to 'limitate every each state. If you think a Von Neumann Algebra is -algebra of bounded operators on a Hilbert space that is *closed in the weak operator topology and contains the identity operator.
    $endgroup$
    – Jack
    Jan 22 at 19:05










  • $begingroup$
    In GNS construction we don't have more conditions that characterize each state (respect to Von Neumann work), is correct? But if is correct a Gelfand state is more weak of Von Neumann state. Is correct this ?
    $endgroup$
    – Jack
    Jan 22 at 19:06


















  • $begingroup$
    The advantage of abstracting is that often objects don't occur as a particular realization but only as an abstract thing and the particular realization can be non-unique or hard to construct. To be more specific in this contect: A C*-algebra can occure without an embedding into a ring of operators of some hilbert space. The construction of such a ring can be hard and non-unique. And actually, the construction to show that every C*-algebra can be seen as an operator algebra makes the hilbert space very large.
    $endgroup$
    – Paul K
    Jan 22 at 18:28












  • $begingroup$
    mm.. so Von Neumann work refer to a very small hilbert space ? Why not replace Von Neumann work with Gelfand-Naimark abstraction and use it in theoretical quantum physics ? because I see that Von Neumann work drive to close 'pure states' into projective space. Is not this a limitation?
    $endgroup$
    – Jack
    Jan 22 at 18:36










  • $begingroup$
    I don't know anything about theoretical physics and I only know a bit about C*-algebras and Von Neumann algebras. What I wanted to say is that the GNS-construction constructs for every state a pre-hilbert space which is then completed. Afterwards from all of these hilbert spaces you construct the hilbert space in whose operator ring your C*-algebra can be embedded. If some abstract C*-algebra occurs, you might not want to carry out this construction but just work with what you have.
    $endgroup$
    – Paul K
    Jan 22 at 18:51












  • $begingroup$
    Yes, but when you said the GNS-construction constructs for every state a pre-hilbert space which is then completed and in whose operator ring your C*-algebra can be embedded this is a very good news because I understand that Von Neumann make something more respect to GNS that I see that add more structure to 'limitate every each state. If you think a Von Neumann Algebra is -algebra of bounded operators on a Hilbert space that is *closed in the weak operator topology and contains the identity operator.
    $endgroup$
    – Jack
    Jan 22 at 19:05










  • $begingroup$
    In GNS construction we don't have more conditions that characterize each state (respect to Von Neumann work), is correct? But if is correct a Gelfand state is more weak of Von Neumann state. Is correct this ?
    $endgroup$
    – Jack
    Jan 22 at 19:06
















$begingroup$
The advantage of abstracting is that often objects don't occur as a particular realization but only as an abstract thing and the particular realization can be non-unique or hard to construct. To be more specific in this contect: A C*-algebra can occure without an embedding into a ring of operators of some hilbert space. The construction of such a ring can be hard and non-unique. And actually, the construction to show that every C*-algebra can be seen as an operator algebra makes the hilbert space very large.
$endgroup$
– Paul K
Jan 22 at 18:28






$begingroup$
The advantage of abstracting is that often objects don't occur as a particular realization but only as an abstract thing and the particular realization can be non-unique or hard to construct. To be more specific in this contect: A C*-algebra can occure without an embedding into a ring of operators of some hilbert space. The construction of such a ring can be hard and non-unique. And actually, the construction to show that every C*-algebra can be seen as an operator algebra makes the hilbert space very large.
$endgroup$
– Paul K
Jan 22 at 18:28














$begingroup$
mm.. so Von Neumann work refer to a very small hilbert space ? Why not replace Von Neumann work with Gelfand-Naimark abstraction and use it in theoretical quantum physics ? because I see that Von Neumann work drive to close 'pure states' into projective space. Is not this a limitation?
$endgroup$
– Jack
Jan 22 at 18:36




$begingroup$
mm.. so Von Neumann work refer to a very small hilbert space ? Why not replace Von Neumann work with Gelfand-Naimark abstraction and use it in theoretical quantum physics ? because I see that Von Neumann work drive to close 'pure states' into projective space. Is not this a limitation?
$endgroup$
– Jack
Jan 22 at 18:36












$begingroup$
I don't know anything about theoretical physics and I only know a bit about C*-algebras and Von Neumann algebras. What I wanted to say is that the GNS-construction constructs for every state a pre-hilbert space which is then completed. Afterwards from all of these hilbert spaces you construct the hilbert space in whose operator ring your C*-algebra can be embedded. If some abstract C*-algebra occurs, you might not want to carry out this construction but just work with what you have.
$endgroup$
– Paul K
Jan 22 at 18:51






$begingroup$
I don't know anything about theoretical physics and I only know a bit about C*-algebras and Von Neumann algebras. What I wanted to say is that the GNS-construction constructs for every state a pre-hilbert space which is then completed. Afterwards from all of these hilbert spaces you construct the hilbert space in whose operator ring your C*-algebra can be embedded. If some abstract C*-algebra occurs, you might not want to carry out this construction but just work with what you have.
$endgroup$
– Paul K
Jan 22 at 18:51














$begingroup$
Yes, but when you said the GNS-construction constructs for every state a pre-hilbert space which is then completed and in whose operator ring your C*-algebra can be embedded this is a very good news because I understand that Von Neumann make something more respect to GNS that I see that add more structure to 'limitate every each state. If you think a Von Neumann Algebra is -algebra of bounded operators on a Hilbert space that is *closed in the weak operator topology and contains the identity operator.
$endgroup$
– Jack
Jan 22 at 19:05




$begingroup$
Yes, but when you said the GNS-construction constructs for every state a pre-hilbert space which is then completed and in whose operator ring your C*-algebra can be embedded this is a very good news because I understand that Von Neumann make something more respect to GNS that I see that add more structure to 'limitate every each state. If you think a Von Neumann Algebra is -algebra of bounded operators on a Hilbert space that is *closed in the weak operator topology and contains the identity operator.
$endgroup$
– Jack
Jan 22 at 19:05












$begingroup$
In GNS construction we don't have more conditions that characterize each state (respect to Von Neumann work), is correct? But if is correct a Gelfand state is more weak of Von Neumann state. Is correct this ?
$endgroup$
– Jack
Jan 22 at 19:06




$begingroup$
In GNS construction we don't have more conditions that characterize each state (respect to Von Neumann work), is correct? But if is correct a Gelfand state is more weak of Von Neumann state. Is correct this ?
$endgroup$
– Jack
Jan 22 at 19:06










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