Dependence of operator topologies in a $C^*$ algebra on the representation
Let $A$ be a $C^*$ algebra. Given a faithful representation $pi:Ato mathcal{B}(H)$, we can define the weak operator topology with respect to $pi$ as initial with respect to the maps $amapsto langle pi(a) x,yrangle$ for each $x,yinmathcal{H}$. However, this definition depends on our representation $pi$.
Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?
functional-analysis operator-algebras c-star-algebras
asked Sep 25 '18 at 22:56
Ashwin TrisalAshwin Trisal
1,2441516
add a comment |
Let $A$ be a $C^*$ algebra. Given a faithful representation $pi:Ato mathcal{B}(H)$, we can define the weak operator topology with respect to $pi$ as initial with respect to the maps $amapsto langle pi(a) x,yrangle$ for each $x,yinmathcal{H}$. However, this definition depends on our representation $pi$.
Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?
functional-analysis operator-algebras c-star-algebras
asked Sep 25 '18 at 22:56
Ashwin TrisalAshwin Trisal
1,2441516
add a comment |
Let $A$ be a $C^*$ algebra. Given a faithful representation $pi:Ato mathcal{B}(H)$, we can define the weak operator topology with respect to $pi$ as initial with respect to the maps $amapsto langle pi(a) x,yrangle$ for each $x,yinmathcal{H}$. However, this definition depends on our representation $pi$.
Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?
functional-analysis operator-algebras c-star-algebras
asked Sep 25 '18 at 22:56
Ashwin TrisalAshwin Trisal
1,2441516
Let $A$ be a $C^*$ algebra. Given a faithful representation $pi:Ato mathcal{B}(H)$, we can define the weak operator topology with respect to $pi$ as initial with respect to the maps $amapsto langle pi(a) x,yrangle$ for each $x,yinmathcal{H}$. However, this definition depends on our representation $pi$.
Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?
functional-analysis operator-algebras c-star-algebras
functional-analysis operator-algebras c-star-algebras
asked Sep 25 '18 at 22:56
Ashwin TrisalAshwin Trisal
1,2441516
asked Sep 25 '18 at 22:56
Ashwin TrisalAshwin Trisal
1,2441516
asked Sep 25 '18 at 22:56
Ashwin TrisalAshwin Trisal
1,2441516
asked Sep 25 '18 at 22:56
Ashwin TrisalAshwin Trisal
1,2441516
asked Sep 25 '18 at 22:56
Ashwin TrisalAshwin Trisal
1,2441516
1,2441516
add a comment |
add a comment |
1 Answer
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No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.
For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).
Then we have
$$
overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
$$
a II$_1$-factor. While
$$
overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
$$
a type I factor.
The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.
answered Sep 26 '18 at 15:04
Martin ArgeramiMartin Argerami
124k1176175
It would be really interesting to know what the downvoter things is wrong with this answer.
– Martin Argerami
Sep 30 '18 at 18:57
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.
For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).
Then we have
$$
overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
$$
a II$_1$-factor. While
$$
overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
$$
a type I factor.
The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.
answered Sep 26 '18 at 15:04
Martin ArgeramiMartin Argerami
124k1176175
It would be really interesting to know what the downvoter things is wrong with this answer.
– Martin Argerami
Sep 30 '18 at 18:57
add a comment |
No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.
For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).
Then we have
$$
overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
$$
a II$_1$-factor. While
$$
overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
$$
a type I factor.
The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.
answered Sep 26 '18 at 15:04
Martin ArgeramiMartin Argerami
124k1176175
It would be really interesting to know what the downvoter things is wrong with this answer.
– Martin Argerami
Sep 30 '18 at 18:57
add a comment |
No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.
For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).
Then we have
$$
overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
$$
a II$_1$-factor. While
$$
overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
$$
a type I factor.
The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.
answered Sep 26 '18 at 15:04
Martin ArgeramiMartin Argerami
124k1176175
No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.
For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).
Then we have
$$
overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
$$
a II$_1$-factor. While
$$
overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
$$
a type I factor.
The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.
answered Sep 26 '18 at 15:04
Martin ArgeramiMartin Argerami
124k1176175
edited Dec 12 '18 at 21:43
answered Sep 26 '18 at 15:04
Martin ArgeramiMartin Argerami
124k1176175
answered Sep 26 '18 at 15:04
Martin ArgeramiMartin Argerami
124k1176175
answered Sep 26 '18 at 15:04
Martin ArgeramiMartin Argerami
124k1176175
124k1176175
It would be really interesting to know what the downvoter things is wrong with this answer.
– Martin Argerami
Sep 30 '18 at 18:57
add a comment |
It would be really interesting to know what the downvoter things is wrong with this answer.
– Martin Argerami
Sep 30 '18 at 18:57
It would be really interesting to know what the downvoter things is wrong with this answer.
– Martin Argerami
Sep 30 '18 at 18:57
It would be really interesting to know what the downvoter things is wrong with this answer.
– Martin Argerami
Sep 30 '18 at 18:57
add a comment |
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