Do all *-isomorphisms between von Neumann algebras preserve strong operator topology?
Do all $*$-isomorphisms between von Neumann algebras preserve the strong operator topology?
Seems clearly true for $*$-isomorphisms with a unitary implementation, but I don't see the answer for other cases ... perhaps there is an easy argument from the fact that von Neumann algebras are closed in this topology, but I've spent a while looking for one and don't see it.
operator-algebras von-neumann-algebras
asked Nov 28 '18 at 18:02
Doug McLellanDoug McLellan
38418
add a comment |
Do all $*$-isomorphisms between von Neumann algebras preserve the strong operator topology?
Seems clearly true for $*$-isomorphisms with a unitary implementation, but I don't see the answer for other cases ... perhaps there is an easy argument from the fact that von Neumann algebras are closed in this topology, but I've spent a while looking for one and don't see it.
operator-algebras von-neumann-algebras
asked Nov 28 '18 at 18:02
Doug McLellanDoug McLellan
38418
add a comment |
Do all $*$-isomorphisms between von Neumann algebras preserve the strong operator topology?
Seems clearly true for $*$-isomorphisms with a unitary implementation, but I don't see the answer for other cases ... perhaps there is an easy argument from the fact that von Neumann algebras are closed in this topology, but I've spent a while looking for one and don't see it.
operator-algebras von-neumann-algebras
asked Nov 28 '18 at 18:02
Doug McLellanDoug McLellan
38418
Do all $*$-isomorphisms between von Neumann algebras preserve the strong operator topology?
Seems clearly true for $*$-isomorphisms with a unitary implementation, but I don't see the answer for other cases ... perhaps there is an easy argument from the fact that von Neumann algebras are closed in this topology, but I've spent a while looking for one and don't see it.
operator-algebras von-neumann-algebras
operator-algebras von-neumann-algebras
asked Nov 28 '18 at 18:02
Doug McLellanDoug McLellan
38418
asked Nov 28 '18 at 18:02
Doug McLellanDoug McLellan
38418
asked Nov 28 '18 at 18:02
Doug McLellanDoug McLellan
38418
asked Nov 28 '18 at 18:02
Doug McLellanDoug McLellan
38418
asked Nov 28 '18 at 18:02
Doug McLellanDoug McLellan
38418
38418
add a comment |
add a comment |
1 Answer
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No. Take $M$ to be any II$_1$-factor. Let $pi:Mto B(H)$ be an irreducible representation (it exists because you can do GNS of a pure state). As $M$ is simple (as a C$^*$-algebra!), $pi$ is injective. And $pi(M)$ is dense in $B(H)$, but it cannot be everything.
So $pi:Mtopi(M)$ is a $*$-isomorphism that does not preserve the sot/wot/ultrasot/ultrawot topologies.
As mentioned in the comments, if $Msubset B(H)$ and $Nsubset B(K)$ are von Neumann algebras (in the usual "double commutant" sense) then a $*$-isomorphism $pi:Mto N$ is sot-continuous by passing through normality.
answered Nov 28 '18 at 18:32
Martin ArgeramiMartin Argerami
124k1176175
But in this case $pi$ is not a $ast$-isomorphism since it is not surjective.
– Adrián González-Pérez
Nov 29 '18 at 12:17
I am also not convinced since $ast$-homomorphism must preserve the suprema of ascending families of projections and that will imply normality for $pi$. Perhaps I am confused about that.
– Adrián González-Pérez
Nov 29 '18 at 12:18
1
@Adrián: of course, I can take $pi(M)$ to be the codomain. And you seem to have a misunderstanding of the equivalence normal $iff$ sot-continuous (and you need selfadjoints, not just projections). The equivalence is true, if you are doing it in a von Neumann algebra. Unless you claim that every monotone-complete C$^*$-algebra is a von Neumann algebra.
– Martin Argerami
Nov 29 '18 at 15:09
Thanks! I didn't realize that.
– Adrián González-Pérez
Nov 29 '18 at 15:25
1
No problem. I have mixed feelings about the fact that von Neumann algebras are always considered represented. On the one hand, it makes a lot of sense because it is how you would usually use them. But, on the other hand, because this is usually not really considered in textbooks, together with AW$^*$-algebras becoming unfashionable a few decades ago, there is little knowledge about all this (including me).
– Martin Argerami
Nov 29 '18 at 15:33
|
show 2 more comments
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No. Take $M$ to be any II$_1$-factor. Let $pi:Mto B(H)$ be an irreducible representation (it exists because you can do GNS of a pure state). As $M$ is simple (as a C$^*$-algebra!), $pi$ is injective. And $pi(M)$ is dense in $B(H)$, but it cannot be everything.
So $pi:Mtopi(M)$ is a $*$-isomorphism that does not preserve the sot/wot/ultrasot/ultrawot topologies.
As mentioned in the comments, if $Msubset B(H)$ and $Nsubset B(K)$ are von Neumann algebras (in the usual "double commutant" sense) then a $*$-isomorphism $pi:Mto N$ is sot-continuous by passing through normality.
answered Nov 28 '18 at 18:32
Martin ArgeramiMartin Argerami
124k1176175
But in this case $pi$ is not a $ast$-isomorphism since it is not surjective.
– Adrián González-Pérez
Nov 29 '18 at 12:17
I am also not convinced since $ast$-homomorphism must preserve the suprema of ascending families of projections and that will imply normality for $pi$. Perhaps I am confused about that.
– Adrián González-Pérez
Nov 29 '18 at 12:18
1
@Adrián: of course, I can take $pi(M)$ to be the codomain. And you seem to have a misunderstanding of the equivalence normal $iff$ sot-continuous (and you need selfadjoints, not just projections). The equivalence is true, if you are doing it in a von Neumann algebra. Unless you claim that every monotone-complete C$^*$-algebra is a von Neumann algebra.
– Martin Argerami
Nov 29 '18 at 15:09
Thanks! I didn't realize that.
– Adrián González-Pérez
Nov 29 '18 at 15:25
1
No problem. I have mixed feelings about the fact that von Neumann algebras are always considered represented. On the one hand, it makes a lot of sense because it is how you would usually use them. But, on the other hand, because this is usually not really considered in textbooks, together with AW$^*$-algebras becoming unfashionable a few decades ago, there is little knowledge about all this (including me).
– Martin Argerami
Nov 29 '18 at 15:33
|
show 2 more comments
No. Take $M$ to be any II$_1$-factor. Let $pi:Mto B(H)$ be an irreducible representation (it exists because you can do GNS of a pure state). As $M$ is simple (as a C$^*$-algebra!), $pi$ is injective. And $pi(M)$ is dense in $B(H)$, but it cannot be everything.
So $pi:Mtopi(M)$ is a $*$-isomorphism that does not preserve the sot/wot/ultrasot/ultrawot topologies.
As mentioned in the comments, if $Msubset B(H)$ and $Nsubset B(K)$ are von Neumann algebras (in the usual "double commutant" sense) then a $*$-isomorphism $pi:Mto N$ is sot-continuous by passing through normality.
answered Nov 28 '18 at 18:32
Martin ArgeramiMartin Argerami
124k1176175
But in this case $pi$ is not a $ast$-isomorphism since it is not surjective.
– Adrián González-Pérez
Nov 29 '18 at 12:17
I am also not convinced since $ast$-homomorphism must preserve the suprema of ascending families of projections and that will imply normality for $pi$. Perhaps I am confused about that.
– Adrián González-Pérez
Nov 29 '18 at 12:18
1
@Adrián: of course, I can take $pi(M)$ to be the codomain. And you seem to have a misunderstanding of the equivalence normal $iff$ sot-continuous (and you need selfadjoints, not just projections). The equivalence is true, if you are doing it in a von Neumann algebra. Unless you claim that every monotone-complete C$^*$-algebra is a von Neumann algebra.
– Martin Argerami
Nov 29 '18 at 15:09
Thanks! I didn't realize that.
– Adrián González-Pérez
Nov 29 '18 at 15:25
1
No problem. I have mixed feelings about the fact that von Neumann algebras are always considered represented. On the one hand, it makes a lot of sense because it is how you would usually use them. But, on the other hand, because this is usually not really considered in textbooks, together with AW$^*$-algebras becoming unfashionable a few decades ago, there is little knowledge about all this (including me).
– Martin Argerami
Nov 29 '18 at 15:33
|
show 2 more comments
No. Take $M$ to be any II$_1$-factor. Let $pi:Mto B(H)$ be an irreducible representation (it exists because you can do GNS of a pure state). As $M$ is simple (as a C$^*$-algebra!), $pi$ is injective. And $pi(M)$ is dense in $B(H)$, but it cannot be everything.
So $pi:Mtopi(M)$ is a $*$-isomorphism that does not preserve the sot/wot/ultrasot/ultrawot topologies.
As mentioned in the comments, if $Msubset B(H)$ and $Nsubset B(K)$ are von Neumann algebras (in the usual "double commutant" sense) then a $*$-isomorphism $pi:Mto N$ is sot-continuous by passing through normality.
answered Nov 28 '18 at 18:32
Martin ArgeramiMartin Argerami
124k1176175
No. Take $M$ to be any II$_1$-factor. Let $pi:Mto B(H)$ be an irreducible representation (it exists because you can do GNS of a pure state). As $M$ is simple (as a C$^*$-algebra!), $pi$ is injective. And $pi(M)$ is dense in $B(H)$, but it cannot be everything.
So $pi:Mtopi(M)$ is a $*$-isomorphism that does not preserve the sot/wot/ultrasot/ultrawot topologies.
As mentioned in the comments, if $Msubset B(H)$ and $Nsubset B(K)$ are von Neumann algebras (in the usual "double commutant" sense) then a $*$-isomorphism $pi:Mto N$ is sot-continuous by passing through normality.
answered Nov 28 '18 at 18:32
Martin ArgeramiMartin Argerami
124k1176175
edited Nov 29 '18 at 21:16
answered Nov 28 '18 at 18:32
Martin ArgeramiMartin Argerami
124k1176175
answered Nov 28 '18 at 18:32
Martin ArgeramiMartin Argerami
124k1176175
answered Nov 28 '18 at 18:32
Martin ArgeramiMartin Argerami
124k1176175
124k1176175
But in this case $pi$ is not a $ast$-isomorphism since it is not surjective.
– Adrián González-Pérez
Nov 29 '18 at 12:17
I am also not convinced since $ast$-homomorphism must preserve the suprema of ascending families of projections and that will imply normality for $pi$. Perhaps I am confused about that.
– Adrián González-Pérez
Nov 29 '18 at 12:18
1
@Adrián: of course, I can take $pi(M)$ to be the codomain. And you seem to have a misunderstanding of the equivalence normal $iff$ sot-continuous (and you need selfadjoints, not just projections). The equivalence is true, if you are doing it in a von Neumann algebra. Unless you claim that every monotone-complete C$^*$-algebra is a von Neumann algebra.
– Martin Argerami
Nov 29 '18 at 15:09
Thanks! I didn't realize that.
– Adrián González-Pérez
Nov 29 '18 at 15:25
1
No problem. I have mixed feelings about the fact that von Neumann algebras are always considered represented. On the one hand, it makes a lot of sense because it is how you would usually use them. But, on the other hand, because this is usually not really considered in textbooks, together with AW$^*$-algebras becoming unfashionable a few decades ago, there is little knowledge about all this (including me).
– Martin Argerami
Nov 29 '18 at 15:33
|
show 2 more comments
But in this case $pi$ is not a $ast$-isomorphism since it is not surjective.
– Adrián González-Pérez
Nov 29 '18 at 12:17
I am also not convinced since $ast$-homomorphism must preserve the suprema of ascending families of projections and that will imply normality for $pi$. Perhaps I am confused about that.
– Adrián González-Pérez
Nov 29 '18 at 12:18
1
@Adrián: of course, I can take $pi(M)$ to be the codomain. And you seem to have a misunderstanding of the equivalence normal $iff$ sot-continuous (and you need selfadjoints, not just projections). The equivalence is true, if you are doing it in a von Neumann algebra. Unless you claim that every monotone-complete C$^*$-algebra is a von Neumann algebra.
– Martin Argerami
Nov 29 '18 at 15:09
Thanks! I didn't realize that.
– Adrián González-Pérez
Nov 29 '18 at 15:25
1
No problem. I have mixed feelings about the fact that von Neumann algebras are always considered represented. On the one hand, it makes a lot of sense because it is how you would usually use them. But, on the other hand, because this is usually not really considered in textbooks, together with AW$^*$-algebras becoming unfashionable a few decades ago, there is little knowledge about all this (including me).
– Martin Argerami
Nov 29 '18 at 15:33
But in this case $pi$ is not a $ast$-isomorphism since it is not surjective.
– Adrián González-Pérez
Nov 29 '18 at 12:17
But in this case $pi$ is not a $ast$-isomorphism since it is not surjective.
– Adrián González-Pérez
Nov 29 '18 at 12:17
I am also not convinced since $ast$-homomorphism must preserve the suprema of ascending families of projections and that will imply normality for $pi$. Perhaps I am confused about that.
– Adrián González-Pérez
Nov 29 '18 at 12:18
I am also not convinced since $ast$-homomorphism must preserve the suprema of ascending families of projections and that will imply normality for $pi$. Perhaps I am confused about that.
– Adrián González-Pérez
Nov 29 '18 at 12:18
1
1
@Adrián: of course, I can take $pi(M)$ to be the codomain. And you seem to have a misunderstanding of the equivalence normal $iff$ sot-continuous (and you need selfadjoints, not just projections). The equivalence is true, if you are doing it in a von Neumann algebra. Unless you claim that every monotone-complete C$^*$-algebra is a von Neumann algebra.
– Martin Argerami
Nov 29 '18 at 15:09
@Adrián: of course, I can take $pi(M)$ to be the codomain. And you seem to have a misunderstanding of the equivalence normal $iff$ sot-continuous (and you need selfadjoints, not just projections). The equivalence is true, if you are doing it in a von Neumann algebra. Unless you claim that every monotone-complete C$^*$-algebra is a von Neumann algebra.
– Martin Argerami
Nov 29 '18 at 15:09
Thanks! I didn't realize that.
– Adrián González-Pérez
Nov 29 '18 at 15:25
Thanks! I didn't realize that.
– Adrián González-Pérez
Nov 29 '18 at 15:25
1
1
No problem. I have mixed feelings about the fact that von Neumann algebras are always considered represented. On the one hand, it makes a lot of sense because it is how you would usually use them. But, on the other hand, because this is usually not really considered in textbooks, together with AW$^*$-algebras becoming unfashionable a few decades ago, there is little knowledge about all this (including me).
– Martin Argerami
Nov 29 '18 at 15:33
No problem. I have mixed feelings about the fact that von Neumann algebras are always considered represented. On the one hand, it makes a lot of sense because it is how you would usually use them. But, on the other hand, because this is usually not really considered in textbooks, together with AW$^*$-algebras becoming unfashionable a few decades ago, there is little knowledge about all this (including me).
– Martin Argerami
Nov 29 '18 at 15:33
|
show 2 more comments
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