Mixing
minimal projections in finite dimensional von Neumann algebras
Clash Royale CLAN TAG#URR8PPP
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The algebras I'm working with are defined as follows
Let $mathcalH$ be a Hilbert space of finite dimension and denote by $mathcalB(mathcalH)$ the bounded operators on $mathcalH$. A matrix algebra $mathcalM$ is a *-subalgebra of $mathcalB(mathcalH)$.
Am I correct in saying that these are exactly the finite-dimensional von Neumann-algebras?
I need to prove the following:
The set of minimal projections in a commutative matrix algebra $mathcalM$ is a finite set $p_1,dots,p_k $ of pairwise orthogonal projections.
Note that I am working with the following definitions:
A projection $p$ is called minimal if there are no non-zero projections $q$ such that $q < p$, where $q < p$ means $q mathcalH subset p mathcalH$ (strictly) and $leq$ is defined similarly.
My proof is as follows. First I show that $mathcalM$ contains minimal projections. Since $mathcalM$ is finite-dimensional, we define $rank(p) = dim p mathcalH$. Because for any $p$ this is a finite number, we can either find $q in mathcalM$ such that $q < p$ or $p$ is minimal. Since $mathcalM$ contains a unit, we know that $mathcalM$ also has minimal projections. Now let $p$ and $q$ be two minimal projections. Because $mathcalM$ is commutative, $pq$ is also a projection and $pq leq p$ and $qp leq q$. Because $p$ and $q$ are minimal, we then have that either $pq = 0$ or $p=(pq=qp=)q$, so $p$ and $q$ are orthogonal. I still need to show that there are only finitely many and intuitively, this seems natural because of the orthogonality and the finite-dimensionality of $mathcalM$, but I can't seem to put my finger on the exact argument.
My question is this: The way I show that there are minimal projections in $mathcalM$, is that correct? Because I know that there are von Neumann-algebras without minimal projections, but then, I think, you should be able to costruct an infinite sequence $(p_1,dots)$ so that $p_i+1 leq p_i$, otherwise there would be a minimal projection. This is probably way beyond what I'm capable of, but is this true?
operator-algebras c-star-algebras von-neumann-algebras projection
edited Aug 2 at 16:42
Martin Argerami
115k1071164
asked Aug 2 at 12:50
user353840
1047
add a comment |Â
up vote
1
down vote
favorite
The algebras I'm working with are defined as follows
Let $mathcalH$ be a Hilbert space of finite dimension and denote by $mathcalB(mathcalH)$ the bounded operators on $mathcalH$. A matrix algebra $mathcalM$ is a *-subalgebra of $mathcalB(mathcalH)$.
Am I correct in saying that these are exactly the finite-dimensional von Neumann-algebras?
I need to prove the following:
The set of minimal projections in a commutative matrix algebra $mathcalM$ is a finite set $p_1,dots,p_k $ of pairwise orthogonal projections.
Note that I am working with the following definitions:
A projection $p$ is called minimal if there are no non-zero projections $q$ such that $q < p$, where $q < p$ means $q mathcalH subset p mathcalH$ (strictly) and $leq$ is defined similarly.
My proof is as follows. First I show that $mathcalM$ contains minimal projections. Since $mathcalM$ is finite-dimensional, we define $rank(p) = dim p mathcalH$. Because for any $p$ this is a finite number, we can either find $q in mathcalM$ such that $q < p$ or $p$ is minimal. Since $mathcalM$ contains a unit, we know that $mathcalM$ also has minimal projections. Now let $p$ and $q$ be two minimal projections. Because $mathcalM$ is commutative, $pq$ is also a projection and $pq leq p$ and $qp leq q$. Because $p$ and $q$ are minimal, we then have that either $pq = 0$ or $p=(pq=qp=)q$, so $p$ and $q$ are orthogonal. I still need to show that there are only finitely many and intuitively, this seems natural because of the orthogonality and the finite-dimensionality of $mathcalM$, but I can't seem to put my finger on the exact argument.
My question is this: The way I show that there are minimal projections in $mathcalM$, is that correct? Because I know that there are von Neumann-algebras without minimal projections, but then, I think, you should be able to costruct an infinite sequence $(p_1,dots)$ so that $p_i+1 leq p_i$, otherwise there would be a minimal projection. This is probably way beyond what I'm capable of, but is this true?
operator-algebras c-star-algebras von-neumann-algebras projection
edited Aug 2 at 16:42
Martin Argerami
115k1071164
asked Aug 2 at 12:50
user353840
1047
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The algebras I'm working with are defined as follows
Let $mathcalH$ be a Hilbert space of finite dimension and denote by $mathcalB(mathcalH)$ the bounded operators on $mathcalH$. A matrix algebra $mathcalM$ is a *-subalgebra of $mathcalB(mathcalH)$.
Am I correct in saying that these are exactly the finite-dimensional von Neumann-algebras?
I need to prove the following:
The set of minimal projections in a commutative matrix algebra $mathcalM$ is a finite set $p_1,dots,p_k $ of pairwise orthogonal projections.
Note that I am working with the following definitions:
A projection $p$ is called minimal if there are no non-zero projections $q$ such that $q < p$, where $q < p$ means $q mathcalH subset p mathcalH$ (strictly) and $leq$ is defined similarly.
My proof is as follows. First I show that $mathcalM$ contains minimal projections. Since $mathcalM$ is finite-dimensional, we define $rank(p) = dim p mathcalH$. Because for any $p$ this is a finite number, we can either find $q in mathcalM$ such that $q < p$ or $p$ is minimal. Since $mathcalM$ contains a unit, we know that $mathcalM$ also has minimal projections. Now let $p$ and $q$ be two minimal projections. Because $mathcalM$ is commutative, $pq$ is also a projection and $pq leq p$ and $qp leq q$. Because $p$ and $q$ are minimal, we then have that either $pq = 0$ or $p=(pq=qp=)q$, so $p$ and $q$ are orthogonal. I still need to show that there are only finitely many and intuitively, this seems natural because of the orthogonality and the finite-dimensionality of $mathcalM$, but I can't seem to put my finger on the exact argument.
My question is this: The way I show that there are minimal projections in $mathcalM$, is that correct? Because I know that there are von Neumann-algebras without minimal projections, but then, I think, you should be able to costruct an infinite sequence $(p_1,dots)$ so that $p_i+1 leq p_i$, otherwise there would be a minimal projection. This is probably way beyond what I'm capable of, but is this true?
operator-algebras c-star-algebras von-neumann-algebras projection
edited Aug 2 at 16:42
Martin Argerami
115k1071164
asked Aug 2 at 12:50
user353840
1047
The algebras I'm working with are defined as follows
Let $mathcalH$ be a Hilbert space of finite dimension and denote by $mathcalB(mathcalH)$ the bounded operators on $mathcalH$. A matrix algebra $mathcalM$ is a *-subalgebra of $mathcalB(mathcalH)$.
Am I correct in saying that these are exactly the finite-dimensional von Neumann-algebras?
I need to prove the following:
The set of minimal projections in a commutative matrix algebra $mathcalM$ is a finite set $p_1,dots,p_k $ of pairwise orthogonal projections.
Note that I am working with the following definitions:
A projection $p$ is called minimal if there are no non-zero projections $q$ such that $q < p$, where $q < p$ means $q mathcalH subset p mathcalH$ (strictly) and $leq$ is defined similarly.
My proof is as follows. First I show that $mathcalM$ contains minimal projections. Since $mathcalM$ is finite-dimensional, we define $rank(p) = dim p mathcalH$. Because for any $p$ this is a finite number, we can either find $q in mathcalM$ such that $q < p$ or $p$ is minimal. Since $mathcalM$ contains a unit, we know that $mathcalM$ also has minimal projections. Now let $p$ and $q$ be two minimal projections. Because $mathcalM$ is commutative, $pq$ is also a projection and $pq leq p$ and $qp leq q$. Because $p$ and $q$ are minimal, we then have that either $pq = 0$ or $p=(pq=qp=)q$, so $p$ and $q$ are orthogonal. I still need to show that there are only finitely many and intuitively, this seems natural because of the orthogonality and the finite-dimensionality of $mathcalM$, but I can't seem to put my finger on the exact argument.
My question is this: The way I show that there are minimal projections in $mathcalM$, is that correct? Because I know that there are von Neumann-algebras without minimal projections, but then, I think, you should be able to costruct an infinite sequence $(p_1,dots)$ so that $p_i+1 leq p_i$, otherwise there would be a minimal projection. This is probably way beyond what I'm capable of, but is this true?
operator-algebras c-star-algebras von-neumann-algebras projection
edited Aug 2 at 16:42
Martin Argerami
115k1071164
asked Aug 2 at 12:50
user353840
1047
edited Aug 2 at 16:42
Martin Argerami
115k1071164
edited Aug 2 at 16:42
Martin Argerami
115k1071164
edited Aug 2 at 16:42
Martin Argerami
115k1071164
115k1071164
asked Aug 2 at 12:50
user353840
1047
asked Aug 2 at 12:50
user353840
1047
asked Aug 2 at 12:50
user353840
1047
1047
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Yes, the finite-dimensional von Neumann algebras, the finite-dimensional C$^*$-algebras, and the finite-dimensional $*$-algebras are exactly the same.
Your argument is correct, but there is no need to use the inclusion $mathcal Msubset B(mathcal H)$. Simply if $mathcal M$ has no minimal projections, you would have a strict sequence $p_1geq p_2geqcdots$ Now take $p_1-p_2, p_2-p_3, p_3-p_4,ldots $ and show that they are linearly independent.
Then you need to show that there is a maximal family of pairwise orthogonal minimal projections. Because $mathcal M$ is finite, this is trivial: keep adding new orthogonal minimal projections until you run out of space.
And then you show that the sum of a maximal family of minimal projections is the identity: if such sum is $p$ and $I-pne0$, then there is a minimal projection $qleq I-p$, contradicting the maximality.
answered Aug 2 at 16:42
Martin Argerami
115k1071164
Do you maybe have a hint how to show the linear independence? I can't seem to find it.
– user353840
Aug 2 at 20:39
It's a sequence of pairwise orthogonal projections, so what you need to show is that such a sequence is linearly independent. If that's still not obvious to you, try to show that pairwise orthogonal projections in, say, $mathbb C^3$ are linearly independent.
– Martin Argerami
Aug 2 at 21:16
But I find that $(p_1-p_2)^2 = p_1 + p_2 - 2p_1p_2$, so even if I assume they commute, they are not even projections. What am I missing?
– user353840
Aug 2 at 21:44
That $p_2leq p_1$.
– Martin Argerami
Aug 2 at 22:54
So, if I use the definition above, i.e. that $p_2 leq p_1$ means $p_2mathcalH subseteq p_1 mathcalH$, I find that $p_2p_1 = p_2$ which gets me further, but then I use that $mathcalM subseteq mathcalB(mathcalH)$. Is there another way to show this?
– user353840
Aug 3 at 13:55
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes, the finite-dimensional von Neumann algebras, the finite-dimensional C$^*$-algebras, and the finite-dimensional $*$-algebras are exactly the same.
Your argument is correct, but there is no need to use the inclusion $mathcal Msubset B(mathcal H)$. Simply if $mathcal M$ has no minimal projections, you would have a strict sequence $p_1geq p_2geqcdots$ Now take $p_1-p_2, p_2-p_3, p_3-p_4,ldots $ and show that they are linearly independent.
Then you need to show that there is a maximal family of pairwise orthogonal minimal projections. Because $mathcal M$ is finite, this is trivial: keep adding new orthogonal minimal projections until you run out of space.
And then you show that the sum of a maximal family of minimal projections is the identity: if such sum is $p$ and $I-pne0$, then there is a minimal projection $qleq I-p$, contradicting the maximality.
answered Aug 2 at 16:42
Martin Argerami
115k1071164
Do you maybe have a hint how to show the linear independence? I can't seem to find it.
– user353840
Aug 2 at 20:39
It's a sequence of pairwise orthogonal projections, so what you need to show is that such a sequence is linearly independent. If that's still not obvious to you, try to show that pairwise orthogonal projections in, say, $mathbb C^3$ are linearly independent.
– Martin Argerami
Aug 2 at 21:16
But I find that $(p_1-p_2)^2 = p_1 + p_2 - 2p_1p_2$, so even if I assume they commute, they are not even projections. What am I missing?
– user353840
Aug 2 at 21:44
That $p_2leq p_1$.
– Martin Argerami
Aug 2 at 22:54
So, if I use the definition above, i.e. that $p_2 leq p_1$ means $p_2mathcalH subseteq p_1 mathcalH$, I find that $p_2p_1 = p_2$ which gets me further, but then I use that $mathcalM subseteq mathcalB(mathcalH)$. Is there another way to show this?
– user353840
Aug 3 at 13:55
 |Â
show 5 more comments
up vote
0
down vote
Yes, the finite-dimensional von Neumann algebras, the finite-dimensional C$^*$-algebras, and the finite-dimensional $*$-algebras are exactly the same.
Your argument is correct, but there is no need to use the inclusion $mathcal Msubset B(mathcal H)$. Simply if $mathcal M$ has no minimal projections, you would have a strict sequence $p_1geq p_2geqcdots$ Now take $p_1-p_2, p_2-p_3, p_3-p_4,ldots $ and show that they are linearly independent.
Then you need to show that there is a maximal family of pairwise orthogonal minimal projections. Because $mathcal M$ is finite, this is trivial: keep adding new orthogonal minimal projections until you run out of space.
And then you show that the sum of a maximal family of minimal projections is the identity: if such sum is $p$ and $I-pne0$, then there is a minimal projection $qleq I-p$, contradicting the maximality.
answered Aug 2 at 16:42
Martin Argerami
115k1071164
Do you maybe have a hint how to show the linear independence? I can't seem to find it.
– user353840
Aug 2 at 20:39
It's a sequence of pairwise orthogonal projections, so what you need to show is that such a sequence is linearly independent. If that's still not obvious to you, try to show that pairwise orthogonal projections in, say, $mathbb C^3$ are linearly independent.
– Martin Argerami
Aug 2 at 21:16
But I find that $(p_1-p_2)^2 = p_1 + p_2 - 2p_1p_2$, so even if I assume they commute, they are not even projections. What am I missing?
– user353840
Aug 2 at 21:44
That $p_2leq p_1$.
– Martin Argerami
Aug 2 at 22:54
So, if I use the definition above, i.e. that $p_2 leq p_1$ means $p_2mathcalH subseteq p_1 mathcalH$, I find that $p_2p_1 = p_2$ which gets me further, but then I use that $mathcalM subseteq mathcalB(mathcalH)$. Is there another way to show this?
– user353840
Aug 3 at 13:55
 |Â
show 5 more comments
up vote
0
down vote
up vote
0
down vote
Yes, the finite-dimensional von Neumann algebras, the finite-dimensional C$^*$-algebras, and the finite-dimensional $*$-algebras are exactly the same.
Your argument is correct, but there is no need to use the inclusion $mathcal Msubset B(mathcal H)$. Simply if $mathcal M$ has no minimal projections, you would have a strict sequence $p_1geq p_2geqcdots$ Now take $p_1-p_2, p_2-p_3, p_3-p_4,ldots $ and show that they are linearly independent.
Then you need to show that there is a maximal family of pairwise orthogonal minimal projections. Because $mathcal M$ is finite, this is trivial: keep adding new orthogonal minimal projections until you run out of space.
And then you show that the sum of a maximal family of minimal projections is the identity: if such sum is $p$ and $I-pne0$, then there is a minimal projection $qleq I-p$, contradicting the maximality.
answered Aug 2 at 16:42
Martin Argerami
115k1071164
Yes, the finite-dimensional von Neumann algebras, the finite-dimensional C$^*$-algebras, and the finite-dimensional $*$-algebras are exactly the same.
Your argument is correct, but there is no need to use the inclusion $mathcal Msubset B(mathcal H)$. Simply if $mathcal M$ has no minimal projections, you would have a strict sequence $p_1geq p_2geqcdots$ Now take $p_1-p_2, p_2-p_3, p_3-p_4,ldots $ and show that they are linearly independent.
Then you need to show that there is a maximal family of pairwise orthogonal minimal projections. Because $mathcal M$ is finite, this is trivial: keep adding new orthogonal minimal projections until you run out of space.
And then you show that the sum of a maximal family of minimal projections is the identity: if such sum is $p$ and $I-pne0$, then there is a minimal projection $qleq I-p$, contradicting the maximality.
answered Aug 2 at 16:42
Martin Argerami
115k1071164
answered Aug 2 at 16:42
Martin Argerami
115k1071164
answered Aug 2 at 16:42
Martin Argerami
115k1071164
answered Aug 2 at 16:42
Martin Argerami
115k1071164
115k1071164
Do you maybe have a hint how to show the linear independence? I can't seem to find it.
– user353840
Aug 2 at 20:39
It's a sequence of pairwise orthogonal projections, so what you need to show is that such a sequence is linearly independent. If that's still not obvious to you, try to show that pairwise orthogonal projections in, say, $mathbb C^3$ are linearly independent.
– Martin Argerami
Aug 2 at 21:16
But I find that $(p_1-p_2)^2 = p_1 + p_2 - 2p_1p_2$, so even if I assume they commute, they are not even projections. What am I missing?
– user353840
Aug 2 at 21:44
That $p_2leq p_1$.
– Martin Argerami
Aug 2 at 22:54
So, if I use the definition above, i.e. that $p_2 leq p_1$ means $p_2mathcalH subseteq p_1 mathcalH$, I find that $p_2p_1 = p_2$ which gets me further, but then I use that $mathcalM subseteq mathcalB(mathcalH)$. Is there another way to show this?
– user353840
Aug 3 at 13:55
 |Â
show 5 more comments
Do you maybe have a hint how to show the linear independence? I can't seem to find it.
– user353840
Aug 2 at 20:39
It's a sequence of pairwise orthogonal projections, so what you need to show is that such a sequence is linearly independent. If that's still not obvious to you, try to show that pairwise orthogonal projections in, say, $mathbb C^3$ are linearly independent.
– Martin Argerami
Aug 2 at 21:16
But I find that $(p_1-p_2)^2 = p_1 + p_2 - 2p_1p_2$, so even if I assume they commute, they are not even projections. What am I missing?
– user353840
Aug 2 at 21:44
That $p_2leq p_1$.
– Martin Argerami
Aug 2 at 22:54
So, if I use the definition above, i.e. that $p_2 leq p_1$ means $p_2mathcalH subseteq p_1 mathcalH$, I find that $p_2p_1 = p_2$ which gets me further, but then I use that $mathcalM subseteq mathcalB(mathcalH)$. Is there another way to show this?
– user353840
Aug 3 at 13:55
Do you maybe have a hint how to show the linear independence? I can't seem to find it.
– user353840
Aug 2 at 20:39
Do you maybe have a hint how to show the linear independence? I can't seem to find it.
– user353840
Aug 2 at 20:39
It's a sequence of pairwise orthogonal projections, so what you need to show is that such a sequence is linearly independent. If that's still not obvious to you, try to show that pairwise orthogonal projections in, say, $mathbb C^3$ are linearly independent.
– Martin Argerami
Aug 2 at 21:16
It's a sequence of pairwise orthogonal projections, so what you need to show is that such a sequence is linearly independent. If that's still not obvious to you, try to show that pairwise orthogonal projections in, say, $mathbb C^3$ are linearly independent.
– Martin Argerami
Aug 2 at 21:16
But I find that $(p_1-p_2)^2 = p_1 + p_2 - 2p_1p_2$, so even if I assume they commute, they are not even projections. What am I missing?
– user353840
Aug 2 at 21:44
But I find that $(p_1-p_2)^2 = p_1 + p_2 - 2p_1p_2$, so even if I assume they commute, they are not even projections. What am I missing?
– user353840
Aug 2 at 21:44
That $p_2leq p_1$.
– Martin Argerami
Aug 2 at 22:54
That $p_2leq p_1$.
– Martin Argerami
Aug 2 at 22:54
So, if I use the definition above, i.e. that $p_2 leq p_1$ means $p_2mathcalH subseteq p_1 mathcalH$, I find that $p_2p_1 = p_2$ which gets me further, but then I use that $mathcalM subseteq mathcalB(mathcalH)$. Is there another way to show this?
– user353840
Aug 3 at 13:55
So, if I use the definition above, i.e. that $p_2 leq p_1$ means $p_2mathcalH subseteq p_1 mathcalH$, I find that $p_2p_1 = p_2$ which gets me further, but then I use that $mathcalM subseteq mathcalB(mathcalH)$. Is there another way to show this?
– user353840
Aug 3 at 13:55
 |Â
show 5 more comments
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