On proof of weak operator closed ness of kernel of the representation
Let $pi:L^{infty}(X,mu)mapsto B(mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $mu$ is a probability measure and $mathcal{H}$ is a seperable Hilbert space. Prove that $text{ ker }pi$ is weak operator closed.
operator-algebras von-neumann-algebras
asked Nov 25 '18 at 11:31
mathlover
15119
add a comment |
Let $pi:L^{infty}(X,mu)mapsto B(mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $mu$ is a probability measure and $mathcal{H}$ is a seperable Hilbert space. Prove that $text{ ker }pi$ is weak operator closed.
operator-algebras von-neumann-algebras
asked Nov 25 '18 at 11:31
mathlover
15119
What would "weak operator topology" mean on $L^infty$?
– Martin Argerami
Nov 25 '18 at 20:06
Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
– mathlover
Nov 27 '18 at 12:35
Ok. And what would "representation" mean here? In terms of continuity?
– Martin Argerami
Nov 27 '18 at 15:20
Thinking the algebra as multiplication operator in $L^2(X,mu)$
– mathlover
Nov 28 '18 at 5:05
add a comment |
Let $pi:L^{infty}(X,mu)mapsto B(mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $mu$ is a probability measure and $mathcal{H}$ is a seperable Hilbert space. Prove that $text{ ker }pi$ is weak operator closed.
operator-algebras von-neumann-algebras
asked Nov 25 '18 at 11:31
mathlover
15119
Let $pi:L^{infty}(X,mu)mapsto B(mathcal{H})$ be a representation of an abelian von Neumann algebra. Where $mu$ is a probability measure and $mathcal{H}$ is a seperable Hilbert space. Prove that $text{ ker }pi$ is weak operator closed.
operator-algebras von-neumann-algebras
operator-algebras von-neumann-algebras
asked Nov 25 '18 at 11:31
mathlover
15119
asked Nov 25 '18 at 11:31
mathlover
15119
asked Nov 25 '18 at 11:31
mathlover
15119
asked Nov 25 '18 at 11:31
mathlover
15119
asked Nov 25 '18 at 11:31
mathlover
15119
15119
What would "weak operator topology" mean on $L^infty$?
– Martin Argerami
Nov 25 '18 at 20:06
Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
– mathlover
Nov 27 '18 at 12:35
Ok. And what would "representation" mean here? In terms of continuity?
– Martin Argerami
Nov 27 '18 at 15:20
Thinking the algebra as multiplication operator in $L^2(X,mu)$
– mathlover
Nov 28 '18 at 5:05
add a comment |
What would "weak operator topology" mean on $L^infty$?
– Martin Argerami
Nov 25 '18 at 20:06
Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
– mathlover
Nov 27 '18 at 12:35
Ok. And what would "representation" mean here? In terms of continuity?
– Martin Argerami
Nov 27 '18 at 15:20
Thinking the algebra as multiplication operator in $L^2(X,mu)$
– mathlover
Nov 28 '18 at 5:05
What would "weak operator topology" mean on $L^infty$?
– Martin Argerami
Nov 25 '18 at 20:06
What would "weak operator topology" mean on $L^infty$?
– Martin Argerami
Nov 25 '18 at 20:06
Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
– mathlover
Nov 27 '18 at 12:35
Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
– mathlover
Nov 27 '18 at 12:35
Ok. And what would "representation" mean here? In terms of continuity?
– Martin Argerami
Nov 27 '18 at 15:20
Ok. And what would "representation" mean here? In terms of continuity?
– Martin Argerami
Nov 27 '18 at 15:20
Thinking the algebra as multiplication operator in $L^2(X,mu)$
– mathlover
Nov 28 '18 at 5:05
Thinking the algebra as multiplication operator in $L^2(X,mu)$
– mathlover
Nov 28 '18 at 5:05
add a comment |
1 Answer
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It's not clear what you are asking. There are two possibilities:
You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.
You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.
answered Nov 28 '18 at 17:18
Martin Argerami
124k1176174
add a comment |
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1 Answer
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1 Answer
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It's not clear what you are asking. There are two possibilities:
You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.
You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.
answered Nov 28 '18 at 17:18
Martin Argerami
124k1176174
add a comment |
It's not clear what you are asking. There are two possibilities:
You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.
You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.
answered Nov 28 '18 at 17:18
Martin Argerami
124k1176174
add a comment |
It's not clear what you are asking. There are two possibilities:
You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.
You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.
answered Nov 28 '18 at 17:18
Martin Argerami
124k1176174
It's not clear what you are asking. There are two possibilities:
You are requiring $pi$ to be wot continuous. In that case, $kerpi=pi^{-1}({0})$ is closed.
You are not requiring $pi$ to be wot continuous. In that case the assertion is not true. Take $X=mathbb N$ with the counting measure. Fix a free ultrafilter $omega$, and let $pi:ell^infty(mathbb N)tomathbb C$ be $pi(x)=lim_{ntoomega}x_n$. Then $pi$ is a representation, and $kerpisupset c_0$. But $c_0$ is wot dense in $ell^infty(mathbb N)$.
answered Nov 28 '18 at 17:18
Martin Argerami
124k1176174
edited Nov 28 '18 at 18:28
answered Nov 28 '18 at 17:18
Martin Argerami
124k1176174
answered Nov 28 '18 at 17:18
Martin Argerami
124k1176174
answered Nov 28 '18 at 17:18
Martin Argerami
124k1176174
124k1176174
add a comment |
add a comment |
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What would "weak operator topology" mean on $L^infty$?
– Martin Argerami
Nov 25 '18 at 20:06
Means thinking $L^{infty}(X,mu)$ sitting inside $B(L^2(X,mu))$
– mathlover
Nov 27 '18 at 12:35
Ok. And what would "representation" mean here? In terms of continuity?
– Martin Argerami
Nov 27 '18 at 15:20
Thinking the algebra as multiplication operator in $L^2(X,mu)$
– mathlover
Nov 28 '18 at 5:05