Why are ultraweak *-homomorphisms the `right' morphisms for von Neumann algebras (and say, not ultrastrong)? The Next CEO of Stack OverflowNon-commutative geometry from von Neumann algebras?Reference for the Gelfand-Neumark theorem for commutative von Neumann algebrasCan we recover a von Neumann algebra from its predual?Mono- and epi-morphisms for C*-algebrasWhich complete Boolean algebras arise as the algebras of projections of commutative von Neumann algebras?Morphism of von Neumann AlgebrasContinuity of a weight on its definition domain in a von Neumann algebraTrivializing unitary cocycles in abelian von Neumann algebras that are uniformly close to the trivial oneExamples of isomorphic W* algebra with non-homeomorphic weak topologyUltraweak topology in abelian von Neumann algebras
Why are ultraweak *-homomorphisms the `right' morphisms for von Neumann algebras (and say, not ultrastrong)?
The Next CEO of Stack OverflowNon-commutative geometry from von Neumann algebras?Reference for the Gelfand-Neumark theorem for commutative von Neumann algebrasCan we recover a von Neumann algebra from its predual?Mono- and epi-morphisms for C*-algebrasWhich complete Boolean algebras arise as the algebras of projections of commutative von Neumann algebras?Morphism of von Neumann AlgebrasContinuity of a weight on its definition domain in a von Neumann algebraTrivializing unitary cocycles in abelian von Neumann algebras that are uniformly close to the trivial oneExamples of isomorphic W* algebra with non-homeomorphic weak topologyUltraweak topology in abelian von Neumann algebras
$begingroup$
Both the ultraweak and ultrastrong topologies are intrinsic topologies in the sense that the image of a continuous (unital) $*$-homomorphism between von Neumann algebras (in either topology) is a von Neumann subalgebra of the target von Neumann algebra, unlike say just norm-continuous $*$-homomorphisms. One may then speak of the category of von Neumann algebras with morphisms as ultrastrong $*$-homomorphisms.
Why do we predominantly think of the ultraweak topology as the intrinsic one when presumably there could be many more topologies that are intrinsic in the above sense? I understand that the ultraweak topology is the weak-topology coming from the pre-dual and hence quite natural to study. But is there a guiding logical or category-theoretic principle that tells us to make this choice?
Thank you.
oa.operator-algebras von-neumann-algebras
edited Mar 21 at 14:38
YCor
28.5k484139
asked Mar 21 at 11:46
condexpcondexp
673
$endgroup$
add a comment |
$begingroup$
Both the ultraweak and ultrastrong topologies are intrinsic topologies in the sense that the image of a continuous (unital) $*$-homomorphism between von Neumann algebras (in either topology) is a von Neumann subalgebra of the target von Neumann algebra, unlike say just norm-continuous $*$-homomorphisms. One may then speak of the category of von Neumann algebras with morphisms as ultrastrong $*$-homomorphisms.
Why do we predominantly think of the ultraweak topology as the intrinsic one when presumably there could be many more topologies that are intrinsic in the above sense? I understand that the ultraweak topology is the weak-topology coming from the pre-dual and hence quite natural to study. But is there a guiding logical or category-theoretic principle that tells us to make this choice?
Thank you.
oa.operator-algebras von-neumann-algebras
edited Mar 21 at 14:38
YCor
28.5k484139
asked Mar 21 at 11:46
condexpcondexp
673
$endgroup$
1
$begingroup$
Ultrastrongly continuous $ast$-homomorphisms are normal, i.e. preserve suprema of bounded, increasing nets, hence they are ultraweakly continuous. So the class of ultraweakly continuous $ast$-homomorphisms is potentially broader; might be the same, I'm not sure.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:18
$begingroup$
The ultraweak topology is the coarsest topology such that the normal linear functionals are continuous. But perhaps there are coarser topologies ($mathcalT$) in which, under a *-homomorphism that is continuous in this topology the image of a von Neumann algebra is a von Neumann algebra. In other words, an ultraweak continuous $*$-homomorphism is automatically $mathcalT - mathcalT$-continuous. (Of course one could use trivial topologies. But my question is about a study of the collection of such topologies and identifying other potentially interesting topologies.)
$endgroup$
– condexp
Mar 21 at 12:30
1
$begingroup$
Well, weak operator topology also has this property and it is coarser than the ultraweak topology, but it is hardly intrinsic, because it depends on the particular representation as operators on a Hilbert space.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:57
add a comment |
$begingroup$
Both the ultraweak and ultrastrong topologies are intrinsic topologies in the sense that the image of a continuous (unital) $*$-homomorphism between von Neumann algebras (in either topology) is a von Neumann subalgebra of the target von Neumann algebra, unlike say just norm-continuous $*$-homomorphisms. One may then speak of the category of von Neumann algebras with morphisms as ultrastrong $*$-homomorphisms.
Why do we predominantly think of the ultraweak topology as the intrinsic one when presumably there could be many more topologies that are intrinsic in the above sense? I understand that the ultraweak topology is the weak-topology coming from the pre-dual and hence quite natural to study. But is there a guiding logical or category-theoretic principle that tells us to make this choice?
Thank you.
oa.operator-algebras von-neumann-algebras
edited Mar 21 at 14:38
YCor
28.5k484139
asked Mar 21 at 11:46
condexpcondexp
673
$endgroup$
Both the ultraweak and ultrastrong topologies are intrinsic topologies in the sense that the image of a continuous (unital) $*$-homomorphism between von Neumann algebras (in either topology) is a von Neumann subalgebra of the target von Neumann algebra, unlike say just norm-continuous $*$-homomorphisms. One may then speak of the category of von Neumann algebras with morphisms as ultrastrong $*$-homomorphisms.
Why do we predominantly think of the ultraweak topology as the intrinsic one when presumably there could be many more topologies that are intrinsic in the above sense? I understand that the ultraweak topology is the weak-topology coming from the pre-dual and hence quite natural to study. But is there a guiding logical or category-theoretic principle that tells us to make this choice?
Thank you.
oa.operator-algebras von-neumann-algebras
oa.operator-algebras von-neumann-algebras
edited Mar 21 at 14:38
YCor
28.5k484139
asked Mar 21 at 11:46
condexpcondexp
673
edited Mar 21 at 14:38
YCor
28.5k484139
asked Mar 21 at 11:46
condexpcondexp
673
edited Mar 21 at 14:38
YCor
28.5k484139
edited Mar 21 at 14:38
YCor
28.5k484139
edited Mar 21 at 14:38
YCor
28.5k484139
28.5k484139
asked Mar 21 at 11:46
condexpcondexp
673
asked Mar 21 at 11:46
condexpcondexp
673
asked Mar 21 at 11:46
condexpcondexp
673
673
1
$begingroup$
Ultrastrongly continuous $ast$-homomorphisms are normal, i.e. preserve suprema of bounded, increasing nets, hence they are ultraweakly continuous. So the class of ultraweakly continuous $ast$-homomorphisms is potentially broader; might be the same, I'm not sure.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:18
$begingroup$
The ultraweak topology is the coarsest topology such that the normal linear functionals are continuous. But perhaps there are coarser topologies ($mathcalT$) in which, under a *-homomorphism that is continuous in this topology the image of a von Neumann algebra is a von Neumann algebra. In other words, an ultraweak continuous $*$-homomorphism is automatically $mathcalT - mathcalT$-continuous. (Of course one could use trivial topologies. But my question is about a study of the collection of such topologies and identifying other potentially interesting topologies.)
$endgroup$
– condexp
Mar 21 at 12:30
1
$begingroup$
Well, weak operator topology also has this property and it is coarser than the ultraweak topology, but it is hardly intrinsic, because it depends on the particular representation as operators on a Hilbert space.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:57
add a comment |
1
$begingroup$
Ultrastrongly continuous $ast$-homomorphisms are normal, i.e. preserve suprema of bounded, increasing nets, hence they are ultraweakly continuous. So the class of ultraweakly continuous $ast$-homomorphisms is potentially broader; might be the same, I'm not sure.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:18
$begingroup$
The ultraweak topology is the coarsest topology such that the normal linear functionals are continuous. But perhaps there are coarser topologies ($mathcalT$) in which, under a *-homomorphism that is continuous in this topology the image of a von Neumann algebra is a von Neumann algebra. In other words, an ultraweak continuous $*$-homomorphism is automatically $mathcalT - mathcalT$-continuous. (Of course one could use trivial topologies. But my question is about a study of the collection of such topologies and identifying other potentially interesting topologies.)
$endgroup$
– condexp
Mar 21 at 12:30
1
$begingroup$
Well, weak operator topology also has this property and it is coarser than the ultraweak topology, but it is hardly intrinsic, because it depends on the particular representation as operators on a Hilbert space.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:57
1
1
$begingroup$
Ultrastrongly continuous $ast$-homomorphisms are normal, i.e. preserve suprema of bounded, increasing nets, hence they are ultraweakly continuous. So the class of ultraweakly continuous $ast$-homomorphisms is potentially broader; might be the same, I'm not sure.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:18
$begingroup$
Ultrastrongly continuous $ast$-homomorphisms are normal, i.e. preserve suprema of bounded, increasing nets, hence they are ultraweakly continuous. So the class of ultraweakly continuous $ast$-homomorphisms is potentially broader; might be the same, I'm not sure.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:18
$begingroup$
The ultraweak topology is the coarsest topology such that the normal linear functionals are continuous. But perhaps there are coarser topologies ($mathcalT$) in which, under a *-homomorphism that is continuous in this topology the image of a von Neumann algebra is a von Neumann algebra. In other words, an ultraweak continuous $*$-homomorphism is automatically $mathcalT - mathcalT$-continuous. (Of course one could use trivial topologies. But my question is about a study of the collection of such topologies and identifying other potentially interesting topologies.)
$endgroup$
– condexp
Mar 21 at 12:30
$begingroup$
The ultraweak topology is the coarsest topology such that the normal linear functionals are continuous. But perhaps there are coarser topologies ($mathcalT$) in which, under a *-homomorphism that is continuous in this topology the image of a von Neumann algebra is a von Neumann algebra. In other words, an ultraweak continuous $*$-homomorphism is automatically $mathcalT - mathcalT$-continuous. (Of course one could use trivial topologies. But my question is about a study of the collection of such topologies and identifying other potentially interesting topologies.)
$endgroup$
– condexp
Mar 21 at 12:30
1
1
$begingroup$
Well, weak operator topology also has this property and it is coarser than the ultraweak topology, but it is hardly intrinsic, because it depends on the particular representation as operators on a Hilbert space.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:57
$begingroup$
Well, weak operator topology also has this property and it is coarser than the ultraweak topology, but it is hardly intrinsic, because it depends on the particular representation as operators on a Hilbert space.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A $*$-homomorphism between two von Neumann algebras is weak* to weak* continuous if and only if it is ultrastrong to ultrastrong continuous. See Proposition III.2.2.2 of Blackadar's book (which, basically, answers all questions of this type that you might have).
answered Mar 21 at 13:40
Nik WeaverNik Weaver
22k150131
$endgroup$
add a comment |
$begingroup$
A von Neumann algebra is a $C^*$-algebra $A$ that admits a predual,
i.e., a Banach space $A_*$ such that there is an isomorphism $Ato(A_*)^*$.
A morphism of von Neumann algebras is a morphism of $C^*$-algebras $Ato B$ that admits a predual,
i.e., a morphism of Banach spaces $B_*to A_*$ such that
$(A_*)^*to (B_*)^*$ is isomorphic to $Ato B$.
The weak topology induced by the predual on $A$ is precisely the ultraweak topology,
and so ultraweakly continuous morphisms are precisely those morphisms that admit a predual.
edited Mar 21 at 15:29
LSpice
2,87822627
answered Mar 21 at 14:32
Dmitri PavlovDmitri Pavlov
13.7k43486
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
A $*$-homomorphism between two von Neumann algebras is weak* to weak* continuous if and only if it is ultrastrong to ultrastrong continuous. See Proposition III.2.2.2 of Blackadar's book (which, basically, answers all questions of this type that you might have).
answered Mar 21 at 13:40
Nik WeaverNik Weaver
22k150131
$endgroup$
add a comment |
$begingroup$
A $*$-homomorphism between two von Neumann algebras is weak* to weak* continuous if and only if it is ultrastrong to ultrastrong continuous. See Proposition III.2.2.2 of Blackadar's book (which, basically, answers all questions of this type that you might have).
answered Mar 21 at 13:40
Nik WeaverNik Weaver
22k150131
$endgroup$
add a comment |
$begingroup$
A $*$-homomorphism between two von Neumann algebras is weak* to weak* continuous if and only if it is ultrastrong to ultrastrong continuous. See Proposition III.2.2.2 of Blackadar's book (which, basically, answers all questions of this type that you might have).
answered Mar 21 at 13:40
Nik WeaverNik Weaver
22k150131
$endgroup$
A $*$-homomorphism between two von Neumann algebras is weak* to weak* continuous if and only if it is ultrastrong to ultrastrong continuous. See Proposition III.2.2.2 of Blackadar's book (which, basically, answers all questions of this type that you might have).
answered Mar 21 at 13:40
Nik WeaverNik Weaver
22k150131
answered Mar 21 at 13:40
Nik WeaverNik Weaver
22k150131
answered Mar 21 at 13:40
Nik WeaverNik Weaver
22k150131
answered Mar 21 at 13:40
Nik WeaverNik Weaver
22k150131
22k150131
add a comment |
add a comment |
$begingroup$
A von Neumann algebra is a $C^*$-algebra $A$ that admits a predual,
i.e., a Banach space $A_*$ such that there is an isomorphism $Ato(A_*)^*$.
A morphism of von Neumann algebras is a morphism of $C^*$-algebras $Ato B$ that admits a predual,
i.e., a morphism of Banach spaces $B_*to A_*$ such that
$(A_*)^*to (B_*)^*$ is isomorphic to $Ato B$.
The weak topology induced by the predual on $A$ is precisely the ultraweak topology,
and so ultraweakly continuous morphisms are precisely those morphisms that admit a predual.
edited Mar 21 at 15:29
LSpice
2,87822627
answered Mar 21 at 14:32
Dmitri PavlovDmitri Pavlov
13.7k43486
$endgroup$
add a comment |
$begingroup$
A von Neumann algebra is a $C^*$-algebra $A$ that admits a predual,
i.e., a Banach space $A_*$ such that there is an isomorphism $Ato(A_*)^*$.
A morphism of von Neumann algebras is a morphism of $C^*$-algebras $Ato B$ that admits a predual,
i.e., a morphism of Banach spaces $B_*to A_*$ such that
$(A_*)^*to (B_*)^*$ is isomorphic to $Ato B$.
The weak topology induced by the predual on $A$ is precisely the ultraweak topology,
and so ultraweakly continuous morphisms are precisely those morphisms that admit a predual.
edited Mar 21 at 15:29
LSpice
2,87822627
answered Mar 21 at 14:32
Dmitri PavlovDmitri Pavlov
13.7k43486
$endgroup$
add a comment |
$begingroup$
A von Neumann algebra is a $C^*$-algebra $A$ that admits a predual,
i.e., a Banach space $A_*$ such that there is an isomorphism $Ato(A_*)^*$.
A morphism of von Neumann algebras is a morphism of $C^*$-algebras $Ato B$ that admits a predual,
i.e., a morphism of Banach spaces $B_*to A_*$ such that
$(A_*)^*to (B_*)^*$ is isomorphic to $Ato B$.
The weak topology induced by the predual on $A$ is precisely the ultraweak topology,
and so ultraweakly continuous morphisms are precisely those morphisms that admit a predual.
edited Mar 21 at 15:29
LSpice
2,87822627
answered Mar 21 at 14:32
Dmitri PavlovDmitri Pavlov
13.7k43486
$endgroup$
A von Neumann algebra is a $C^*$-algebra $A$ that admits a predual,
i.e., a Banach space $A_*$ such that there is an isomorphism $Ato(A_*)^*$.
A morphism of von Neumann algebras is a morphism of $C^*$-algebras $Ato B$ that admits a predual,
i.e., a morphism of Banach spaces $B_*to A_*$ such that
$(A_*)^*to (B_*)^*$ is isomorphic to $Ato B$.
The weak topology induced by the predual on $A$ is precisely the ultraweak topology,
and so ultraweakly continuous morphisms are precisely those morphisms that admit a predual.
edited Mar 21 at 15:29
LSpice
2,87822627
answered Mar 21 at 14:32
Dmitri PavlovDmitri Pavlov
13.7k43486
edited Mar 21 at 15:29
LSpice
2,87822627
edited Mar 21 at 15:29
LSpice
2,87822627
edited Mar 21 at 15:29
LSpice
2,87822627
2,87822627
answered Mar 21 at 14:32
Dmitri PavlovDmitri Pavlov
13.7k43486
answered Mar 21 at 14:32
Dmitri PavlovDmitri Pavlov
13.7k43486
answered Mar 21 at 14:32
Dmitri PavlovDmitri Pavlov
13.7k43486
13.7k43486
add a comment |
add a comment |
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$begingroup$
Ultrastrongly continuous $ast$-homomorphisms are normal, i.e. preserve suprema of bounded, increasing nets, hence they are ultraweakly continuous. So the class of ultraweakly continuous $ast$-homomorphisms is potentially broader; might be the same, I'm not sure.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:18
$begingroup$
The ultraweak topology is the coarsest topology such that the normal linear functionals are continuous. But perhaps there are coarser topologies ($mathcalT$) in which, under a *-homomorphism that is continuous in this topology the image of a von Neumann algebra is a von Neumann algebra. In other words, an ultraweak continuous $*$-homomorphism is automatically $mathcalT - mathcalT$-continuous. (Of course one could use trivial topologies. But my question is about a study of the collection of such topologies and identifying other potentially interesting topologies.)
$endgroup$
– condexp
Mar 21 at 12:30
1
$begingroup$
Well, weak operator topology also has this property and it is coarser than the ultraweak topology, but it is hardly intrinsic, because it depends on the particular representation as operators on a Hilbert space.
$endgroup$
– Mateusz Wasilewski
Mar 21 at 12:57