State space of finite dimensional, abelian C*-algebra is a simplex.

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I am looking for a proof that the state space of a finite dimensional C*-algebra is a simplex and, vice versa, if the state space is a simplex, the C*-algebra is abelian. I've found one proof, but it involves representations and the GNS construction, which is way beyond the scope of what I need.

I was hoping that because I only need the result in finite dimensions, there would be a simple proof. Does anyone have an idea?

functional-analysis operator-algebras c-star-algebras

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asked Jul 31 at 15:53

user353840

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up vote
3
down vote

favorite

1

I am looking for a proof that the state space of a finite dimensional C*-algebra is a simplex and, vice versa, if the state space is a simplex, the C*-algebra is abelian. I've found one proof, but it involves representations and the GNS construction, which is way beyond the scope of what I need.

I was hoping that because I only need the result in finite dimensions, there would be a simple proof. Does anyone have an idea?

functional-analysis operator-algebras c-star-algebras

share|cite|improve this question

asked Jul 31 at 15:53

user353840

1097

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

I am looking for a proof that the state space of a finite dimensional C*-algebra is a simplex and, vice versa, if the state space is a simplex, the C*-algebra is abelian. I've found one proof, but it involves representations and the GNS construction, which is way beyond the scope of what I need.

I was hoping that because I only need the result in finite dimensions, there would be a simple proof. Does anyone have an idea?

functional-analysis operator-algebras c-star-algebras

share|cite|improve this question

asked Jul 31 at 15:53

user353840

1097

I am looking for a proof that the state space of a finite dimensional C*-algebra is a simplex and, vice versa, if the state space is a simplex, the C*-algebra is abelian. I've found one proof, but it involves representations and the GNS construction, which is way beyond the scope of what I need.

I was hoping that because I only need the result in finite dimensions, there would be a simple proof. Does anyone have an idea?

functional-analysis operator-algebras c-star-algebras

share|cite|improve this question

asked Jul 31 at 15:53

user353840

1097

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 15:53

user353840

1097

asked Jul 31 at 15:53

user353840

1097

asked Jul 31 at 15:53

user353840

1097

1097

add a comment | 

add a comment | 

1 Answer
1

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oldest

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up vote
2
down vote



accepted

For the finite-dimensional case:

You have $mathcal A=bigoplus_k=1^m M_n_k(mathbb C)subset M_n(mathbb C)$, where $n=n_1+cdots+n_m$.

• Any linear functional on $mathcal A$ is of the form $f(X)=sum_k,jbeta_k,jX_k,j$. This is precisely $$f(X)=operatornameTr(BX),$$ where $B=[beta_k,j]$. We may assume that $Bin mathcal A$.




• If $f$ is a state, it follows that $Bgeq0$, $operatornameTr(B)=1$.




• Within this picture, one can show that the pure states (the extreme points of the state space) are given by the rank-one projections. If $n_kgeq 2$ for at least one index, then there are uncountably many of them.

When $mathcal A$ is abelian, $n_1=cdots=n_m=1$, and there are finitely many rank-one projections. Namely, $mathcal A=mathbb C^n$, and the pure states are given by the canonical basis (so, for instance, $mathbb C^2$ has the pure state $(1,0)$ and $(0,1)$).

If the state space is a simplex (finitely many pure states), then $mathcal A$ is (finite-dimensional and) abelian. First because the pure states separate points (because the states do); if $mathcal A$ is infinite-dimensional, there exists an infinite linearly independent set; and so one needs enough pure states to distinguish infinitely many distinct linear combinations. Once we know that $mathcal A$ is finite-dimensional, by the bullets above the only way to have finitely many pure states is to have $n_1=cdots=n_m=1$, so $mathcal A$ is abelian.

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answered Jul 31 at 19:31

Martin Argerami

115k1071164

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

For the finite-dimensional case:

You have $mathcal A=bigoplus_k=1^m M_n_k(mathbb C)subset M_n(mathbb C)$, where $n=n_1+cdots+n_m$.

• Any linear functional on $mathcal A$ is of the form $f(X)=sum_k,jbeta_k,jX_k,j$. This is precisely $$f(X)=operatornameTr(BX),$$ where $B=[beta_k,j]$. We may assume that $Bin mathcal A$.




• If $f$ is a state, it follows that $Bgeq0$, $operatornameTr(B)=1$.




• Within this picture, one can show that the pure states (the extreme points of the state space) are given by the rank-one projections. If $n_kgeq 2$ for at least one index, then there are uncountably many of them.

When $mathcal A$ is abelian, $n_1=cdots=n_m=1$, and there are finitely many rank-one projections. Namely, $mathcal A=mathbb C^n$, and the pure states are given by the canonical basis (so, for instance, $mathbb C^2$ has the pure state $(1,0)$ and $(0,1)$).

If the state space is a simplex (finitely many pure states), then $mathcal A$ is (finite-dimensional and) abelian. First because the pure states separate points (because the states do); if $mathcal A$ is infinite-dimensional, there exists an infinite linearly independent set; and so one needs enough pure states to distinguish infinitely many distinct linear combinations. Once we know that $mathcal A$ is finite-dimensional, by the bullets above the only way to have finitely many pure states is to have $n_1=cdots=n_m=1$, so $mathcal A$ is abelian.

share|cite|improve this answer

answered Jul 31 at 19:31

Martin Argerami

115k1071164

add a comment | 

up vote
2
down vote



accepted

For the finite-dimensional case:

You have $mathcal A=bigoplus_k=1^m M_n_k(mathbb C)subset M_n(mathbb C)$, where $n=n_1+cdots+n_m$.

• Any linear functional on $mathcal A$ is of the form $f(X)=sum_k,jbeta_k,jX_k,j$. This is precisely $$f(X)=operatornameTr(BX),$$ where $B=[beta_k,j]$. We may assume that $Bin mathcal A$.




• If $f$ is a state, it follows that $Bgeq0$, $operatornameTr(B)=1$.




• Within this picture, one can show that the pure states (the extreme points of the state space) are given by the rank-one projections. If $n_kgeq 2$ for at least one index, then there are uncountably many of them.

When $mathcal A$ is abelian, $n_1=cdots=n_m=1$, and there are finitely many rank-one projections. Namely, $mathcal A=mathbb C^n$, and the pure states are given by the canonical basis (so, for instance, $mathbb C^2$ has the pure state $(1,0)$ and $(0,1)$).

If the state space is a simplex (finitely many pure states), then $mathcal A$ is (finite-dimensional and) abelian. First because the pure states separate points (because the states do); if $mathcal A$ is infinite-dimensional, there exists an infinite linearly independent set; and so one needs enough pure states to distinguish infinitely many distinct linear combinations. Once we know that $mathcal A$ is finite-dimensional, by the bullets above the only way to have finitely many pure states is to have $n_1=cdots=n_m=1$, so $mathcal A$ is abelian.

share|cite|improve this answer

answered Jul 31 at 19:31

Martin Argerami

115k1071164

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

For the finite-dimensional case:

You have $mathcal A=bigoplus_k=1^m M_n_k(mathbb C)subset M_n(mathbb C)$, where $n=n_1+cdots+n_m$.

• Any linear functional on $mathcal A$ is of the form $f(X)=sum_k,jbeta_k,jX_k,j$. This is precisely $$f(X)=operatornameTr(BX),$$ where $B=[beta_k,j]$. We may assume that $Bin mathcal A$.




• If $f$ is a state, it follows that $Bgeq0$, $operatornameTr(B)=1$.




• Within this picture, one can show that the pure states (the extreme points of the state space) are given by the rank-one projections. If $n_kgeq 2$ for at least one index, then there are uncountably many of them.

When $mathcal A$ is abelian, $n_1=cdots=n_m=1$, and there are finitely many rank-one projections. Namely, $mathcal A=mathbb C^n$, and the pure states are given by the canonical basis (so, for instance, $mathbb C^2$ has the pure state $(1,0)$ and $(0,1)$).

If the state space is a simplex (finitely many pure states), then $mathcal A$ is (finite-dimensional and) abelian. First because the pure states separate points (because the states do); if $mathcal A$ is infinite-dimensional, there exists an infinite linearly independent set; and so one needs enough pure states to distinguish infinitely many distinct linear combinations. Once we know that $mathcal A$ is finite-dimensional, by the bullets above the only way to have finitely many pure states is to have $n_1=cdots=n_m=1$, so $mathcal A$ is abelian.

share|cite|improve this answer

answered Jul 31 at 19:31

Martin Argerami

115k1071164

For the finite-dimensional case:

You have $mathcal A=bigoplus_k=1^m M_n_k(mathbb C)subset M_n(mathbb C)$, where $n=n_1+cdots+n_m$.

• Any linear functional on $mathcal A$ is of the form $f(X)=sum_k,jbeta_k,jX_k,j$. This is precisely $$f(X)=operatornameTr(BX),$$ where $B=[beta_k,j]$. We may assume that $Bin mathcal A$.




• If $f$ is a state, it follows that $Bgeq0$, $operatornameTr(B)=1$.




• Within this picture, one can show that the pure states (the extreme points of the state space) are given by the rank-one projections. If $n_kgeq 2$ for at least one index, then there are uncountably many of them.

When $mathcal A$ is abelian, $n_1=cdots=n_m=1$, and there are finitely many rank-one projections. Namely, $mathcal A=mathbb C^n$, and the pure states are given by the canonical basis (so, for instance, $mathbb C^2$ has the pure state $(1,0)$ and $(0,1)$).

If the state space is a simplex (finitely many pure states), then $mathcal A$ is (finite-dimensional and) abelian. First because the pure states separate points (because the states do); if $mathcal A$ is infinite-dimensional, there exists an infinite linearly independent set; and so one needs enough pure states to distinguish infinitely many distinct linear combinations. Once we know that $mathcal A$ is finite-dimensional, by the bullets above the only way to have finitely many pure states is to have $n_1=cdots=n_m=1$, so $mathcal A$ is abelian.

share|cite|improve this answer

answered Jul 31 at 19:31

Martin Argerami

115k1071164

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 19:31

Martin Argerami

115k1071164

answered Jul 31 at 19:31

Martin Argerami

115k1071164

answered Jul 31 at 19:31

Martin Argerami

115k1071164

115k1071164

add a comment | 

add a comment | 

 

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