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Finite dimensional von Neumann algebra [on hold]
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How to prove that finite dimensional von Neumann algebra is direct sum of matrix algebras?
operator-algebras c-star-algebras von-neumann-algebras
edited Aug 2 at 19:31
Martin Argerami
115k1071164
asked Aug 2 at 18:20
mathlover
10518
put on hold as off-topic by José Carlos Santos, John Ma, Mostafa Ayaz, Stefan4024, rschwieb Aug 2 at 23:03
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." РJos̩ Carlos Santos, John Ma, Mostafa Ayaz, Stefan4024, rschwieb
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How to prove that finite dimensional von Neumann algebra is direct sum of matrix algebras?
operator-algebras c-star-algebras von-neumann-algebras
edited Aug 2 at 19:31
Martin Argerami
115k1071164
asked Aug 2 at 18:20
mathlover
10518
put on hold as off-topic by José Carlos Santos, John Ma, Mostafa Ayaz, Stefan4024, rschwieb Aug 2 at 23:03
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." РJos̩ Carlos Santos, John Ma, Mostafa Ayaz, Stefan4024, rschwieb
All right ideals are finitely generated, hence summands. QED. But next time, please don’t dump your questions without some context. For example, maybe you expect to reprice the entire artinian wedderburn theorem: I don’t know, because you didn’t say.
– rschwieb
Aug 2 at 23:05
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up vote
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How to prove that finite dimensional von Neumann algebra is direct sum of matrix algebras?
operator-algebras c-star-algebras von-neumann-algebras
edited Aug 2 at 19:31
Martin Argerami
115k1071164
asked Aug 2 at 18:20
mathlover
10518
How to prove that finite dimensional von Neumann algebra is direct sum of matrix algebras?
operator-algebras c-star-algebras von-neumann-algebras
edited Aug 2 at 19:31
Martin Argerami
115k1071164
asked Aug 2 at 18:20
mathlover
10518
edited Aug 2 at 19:31
Martin Argerami
115k1071164
edited Aug 2 at 19:31
Martin Argerami
115k1071164
edited Aug 2 at 19:31
Martin Argerami
115k1071164
115k1071164
asked Aug 2 at 18:20
mathlover
10518
asked Aug 2 at 18:20
mathlover
10518
asked Aug 2 at 18:20
mathlover
10518
10518
put on hold as off-topic by José Carlos Santos, John Ma, Mostafa Ayaz, Stefan4024, rschwieb Aug 2 at 23:03
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." РJos̩ Carlos Santos, John Ma, Mostafa Ayaz, Stefan4024, rschwieb
put on hold as off-topic by José Carlos Santos, John Ma, Mostafa Ayaz, Stefan4024, rschwieb Aug 2 at 23:03
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." РJos̩ Carlos Santos, John Ma, Mostafa Ayaz, Stefan4024, rschwieb
All right ideals are finitely generated, hence summands. QED. But next time, please don’t dump your questions without some context. For example, maybe you expect to reprice the entire artinian wedderburn theorem: I don’t know, because you didn’t say.
– rschwieb
Aug 2 at 23:05
add a comment |Â
All right ideals are finitely generated, hence summands. QED. But next time, please don’t dump your questions without some context. For example, maybe you expect to reprice the entire artinian wedderburn theorem: I don’t know, because you didn’t say.
– rschwieb
Aug 2 at 23:05
All right ideals are finitely generated, hence summands. QED. But next time, please don’t dump your questions without some context. For example, maybe you expect to reprice the entire artinian wedderburn theorem: I don’t know, because you didn’t say.
– rschwieb
Aug 2 at 23:05
All right ideals are finitely generated, hence summands. QED. But next time, please don’t dump your questions without some context. For example, maybe you expect to reprice the entire artinian wedderburn theorem: I don’t know, because you didn’t say.
– rschwieb
Aug 2 at 23:05
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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accepted
You first show that there is a finite (unique) set $p_1,ldots,p_m$ of minimal central projections.
Then you prove that $p_jMp_j=p_jM$ is a factor. Since $M=bigoplus_j p_jM$, all that remains is to show that a finite-dimensional factor is isomorphic to $M_n(mathbb C)$ for some $n$. You start with a minimal projection $e_11$, and construct pairwise orthogonal minimal projections $e_11,ldots,e_nn$. Since we are now in a factor, minimal projections are equivalent. So there exist partial isometries $e_12,ldots,e_1n$ such that
$$
e_11=e_1j^*e_1j, e_jj=e_1je_1j^*.
$$
With these partial isometries we define
$$
e_kj=e_1k^*e_1j.
$$
And then one can check that these behave exactly as the matrix units in $M_n(mathbb C)$, namely
$$
e_kje_st=delta_js,e_kt,
$$
and $e_11+cdots+e_nn=I_n$.
answered Aug 2 at 19:43
Martin Argerami
115k1071164
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You first show that there is a finite (unique) set $p_1,ldots,p_m$ of minimal central projections.
Then you prove that $p_jMp_j=p_jM$ is a factor. Since $M=bigoplus_j p_jM$, all that remains is to show that a finite-dimensional factor is isomorphic to $M_n(mathbb C)$ for some $n$. You start with a minimal projection $e_11$, and construct pairwise orthogonal minimal projections $e_11,ldots,e_nn$. Since we are now in a factor, minimal projections are equivalent. So there exist partial isometries $e_12,ldots,e_1n$ such that
$$
e_11=e_1j^*e_1j, e_jj=e_1je_1j^*.
$$
With these partial isometries we define
$$
e_kj=e_1k^*e_1j.
$$
And then one can check that these behave exactly as the matrix units in $M_n(mathbb C)$, namely
$$
e_kje_st=delta_js,e_kt,
$$
and $e_11+cdots+e_nn=I_n$.
answered Aug 2 at 19:43
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
You first show that there is a finite (unique) set $p_1,ldots,p_m$ of minimal central projections.
Then you prove that $p_jMp_j=p_jM$ is a factor. Since $M=bigoplus_j p_jM$, all that remains is to show that a finite-dimensional factor is isomorphic to $M_n(mathbb C)$ for some $n$. You start with a minimal projection $e_11$, and construct pairwise orthogonal minimal projections $e_11,ldots,e_nn$. Since we are now in a factor, minimal projections are equivalent. So there exist partial isometries $e_12,ldots,e_1n$ such that
$$
e_11=e_1j^*e_1j, e_jj=e_1je_1j^*.
$$
With these partial isometries we define
$$
e_kj=e_1k^*e_1j.
$$
And then one can check that these behave exactly as the matrix units in $M_n(mathbb C)$, namely
$$
e_kje_st=delta_js,e_kt,
$$
and $e_11+cdots+e_nn=I_n$.
answered Aug 2 at 19:43
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You first show that there is a finite (unique) set $p_1,ldots,p_m$ of minimal central projections.
Then you prove that $p_jMp_j=p_jM$ is a factor. Since $M=bigoplus_j p_jM$, all that remains is to show that a finite-dimensional factor is isomorphic to $M_n(mathbb C)$ for some $n$. You start with a minimal projection $e_11$, and construct pairwise orthogonal minimal projections $e_11,ldots,e_nn$. Since we are now in a factor, minimal projections are equivalent. So there exist partial isometries $e_12,ldots,e_1n$ such that
$$
e_11=e_1j^*e_1j, e_jj=e_1je_1j^*.
$$
With these partial isometries we define
$$
e_kj=e_1k^*e_1j.
$$
And then one can check that these behave exactly as the matrix units in $M_n(mathbb C)$, namely
$$
e_kje_st=delta_js,e_kt,
$$
and $e_11+cdots+e_nn=I_n$.
answered Aug 2 at 19:43
Martin Argerami
115k1071164
You first show that there is a finite (unique) set $p_1,ldots,p_m$ of minimal central projections.
Then you prove that $p_jMp_j=p_jM$ is a factor. Since $M=bigoplus_j p_jM$, all that remains is to show that a finite-dimensional factor is isomorphic to $M_n(mathbb C)$ for some $n$. You start with a minimal projection $e_11$, and construct pairwise orthogonal minimal projections $e_11,ldots,e_nn$. Since we are now in a factor, minimal projections are equivalent. So there exist partial isometries $e_12,ldots,e_1n$ such that
$$
e_11=e_1j^*e_1j, e_jj=e_1je_1j^*.
$$
With these partial isometries we define
$$
e_kj=e_1k^*e_1j.
$$
And then one can check that these behave exactly as the matrix units in $M_n(mathbb C)$, namely
$$
e_kje_st=delta_js,e_kt,
$$
and $e_11+cdots+e_nn=I_n$.
answered Aug 2 at 19:43
Martin Argerami
115k1071164
answered Aug 2 at 19:43
Martin Argerami
115k1071164
answered Aug 2 at 19:43
Martin Argerami
115k1071164
answered Aug 2 at 19:43
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
All right ideals are finitely generated, hence summands. QED. But next time, please don’t dump your questions without some context. For example, maybe you expect to reprice the entire artinian wedderburn theorem: I don’t know, because you didn’t say.
– rschwieb
Aug 2 at 23:05