Faithful normal states on C-Star algebras [closed]

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Let $A$ be a c-star algebra acting on a non separable Hilbert space. Can one always define a faithful normal state on it?

functional-analysis c-star-algebras

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asked Jul 16 at 15:42

rkmath

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closed as off-topic by Alex Francisco, Strants, Mostafa Ayaz, Taroccoesbrocco, José Carlos Santos Jul 16 at 22:04

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Let $A$ be a c-star algebra acting on a non separable Hilbert space. Can one always define a faithful normal state on it?

functional-analysis c-star-algebras

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asked Jul 16 at 15:42

rkmath

607

closed as off-topic by Alex Francisco, Strants, Mostafa Ayaz, Taroccoesbrocco, José Carlos Santos Jul 16 at 22:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Strants, Mostafa Ayaz, Taroccoesbrocco, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

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up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $A$ be a c-star algebra acting on a non separable Hilbert space. Can one always define a faithful normal state on it?

functional-analysis c-star-algebras

share|cite|improve this question

asked Jul 16 at 15:42

rkmath

607

Let $A$ be a c-star algebra acting on a non separable Hilbert space. Can one always define a faithful normal state on it?

functional-analysis c-star-algebras

share|cite|improve this question

asked Jul 16 at 15:42

rkmath

607

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 15:42

rkmath

607

asked Jul 16 at 15:42

rkmath

607

asked Jul 16 at 15:42

rkmath

607

607

closed as off-topic by Alex Francisco, Strants, Mostafa Ayaz, Taroccoesbrocco, José Carlos Santos Jul 16 at 22:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Strants, Mostafa Ayaz, Taroccoesbrocco, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Alex Francisco, Strants, Mostafa Ayaz, Taroccoesbrocco, José Carlos Santos Jul 16 at 22:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Strants, Mostafa Ayaz, Taroccoesbrocco, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

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No. That's the reason one considers weights.

For an easy example consider the von Neumann algebra $ell^infty(mathbb R)$. Then, if $e_t$ denotes the canonical elements (that is, $e_t(r)=delta_r,t$) you have the net of projections
$$
p_t=sum_sleq te_t.
$$
This net converges strongly to the identity. If you had a faithful normal state $f$, we would have $f(p_t)to f(I)=1$. This would imply that $f(e_t)=0$ for all $t$, a contradiction (you can check this by getting $f(p_t+varepsilon-p_t)$ arbitrarily small, and putting a sequence in between).

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answered Jul 16 at 21:11

Martin Argerami

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1 Answer
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1 Answer
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active

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up vote
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No. That's the reason one considers weights.

For an easy example consider the von Neumann algebra $ell^infty(mathbb R)$. Then, if $e_t$ denotes the canonical elements (that is, $e_t(r)=delta_r,t$) you have the net of projections
$$
p_t=sum_sleq te_t.
$$
This net converges strongly to the identity. If you had a faithful normal state $f$, we would have $f(p_t)to f(I)=1$. This would imply that $f(e_t)=0$ for all $t$, a contradiction (you can check this by getting $f(p_t+varepsilon-p_t)$ arbitrarily small, and putting a sequence in between).

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answered Jul 16 at 21:11

Martin Argerami

116k1071164

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up vote
1
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No. That's the reason one considers weights.

For an easy example consider the von Neumann algebra $ell^infty(mathbb R)$. Then, if $e_t$ denotes the canonical elements (that is, $e_t(r)=delta_r,t$) you have the net of projections
$$
p_t=sum_sleq te_t.
$$
This net converges strongly to the identity. If you had a faithful normal state $f$, we would have $f(p_t)to f(I)=1$. This would imply that $f(e_t)=0$ for all $t$, a contradiction (you can check this by getting $f(p_t+varepsilon-p_t)$ arbitrarily small, and putting a sequence in between).

share|cite|improve this answer

answered Jul 16 at 21:11

Martin Argerami

116k1071164

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up vote
1
down vote

up vote
1
down vote

No. That's the reason one considers weights.

For an easy example consider the von Neumann algebra $ell^infty(mathbb R)$. Then, if $e_t$ denotes the canonical elements (that is, $e_t(r)=delta_r,t$) you have the net of projections
$$
p_t=sum_sleq te_t.
$$
This net converges strongly to the identity. If you had a faithful normal state $f$, we would have $f(p_t)to f(I)=1$. This would imply that $f(e_t)=0$ for all $t$, a contradiction (you can check this by getting $f(p_t+varepsilon-p_t)$ arbitrarily small, and putting a sequence in between).

share|cite|improve this answer

answered Jul 16 at 21:11

Martin Argerami

116k1071164

No. That's the reason one considers weights.

For an easy example consider the von Neumann algebra $ell^infty(mathbb R)$. Then, if $e_t$ denotes the canonical elements (that is, $e_t(r)=delta_r,t$) you have the net of projections
$$
p_t=sum_sleq te_t.
$$
This net converges strongly to the identity. If you had a faithful normal state $f$, we would have $f(p_t)to f(I)=1$. This would imply that $f(e_t)=0$ for all $t$, a contradiction (you can check this by getting $f(p_t+varepsilon-p_t)$ arbitrarily small, and putting a sequence in between).

share|cite|improve this answer

answered Jul 16 at 21:11

Martin Argerami

116k1071164

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 21:11

Martin Argerami

116k1071164

answered Jul 16 at 21:11

Martin Argerami

116k1071164

answered Jul 16 at 21:11

Martin Argerami

116k1071164

116k1071164

add a comment | 

add a comment |