Extension of a von Neumann algebra by a von Neumann algebra
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Let $A,B,C$ be $3$ unital $C^*$ algebras. Assume that we have the following short exact sequence of $C^*$-algebras:
$$0to Ato Cto Bto 0$$
Assume that $A,B$ are generated by their projections. Is $C$ necessarily generated by its projections, too?
Assume that $A,B$ are von Neumann algebras, is $C$ necessarily a von Neumann algebra, too?
Does the last question has an obvious answer when $A,B$ (hence $C$) are commutative algebras?
measure-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 18 at 6:22
Ali Taghavi
186329
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reputation from Ali Taghavi ending tomorrow.
This question has not received enough attention.
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Let $A,B,C$ be $3$ unital $C^*$ algebras. Assume that we have the following short exact sequence of $C^*$-algebras:
$$0to Ato Cto Bto 0$$
Assume that $A,B$ are generated by their projections. Is $C$ necessarily generated by its projections, too?
Assume that $A,B$ are von Neumann algebras, is $C$ necessarily a von Neumann algebra, too?
Does the last question has an obvious answer when $A,B$ (hence $C$) are commutative algebras?
measure-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 18 at 6:22
Ali Taghavi
186329
This question has an open bounty worth +50
reputation from Ali Taghavi ending tomorrow.
This question has not received enough attention.
I don't know what you mean by a short exact sequence of unital algebras. If $C to B$ is a nontrivial surjection of unital algebras, its kernel is a nontrivial ideal of $C$, which is at best a nonunital algebra in general, and if it has a unit the map $A to C$ won't preserve units.
– Qiaochu Yuan
Nov 18 at 21:45
@Qiaochu In the question we do not assume that $Ato C$ preserve unit.
– Ali Taghavi
Nov 18 at 21:52
1
Your first question has surely a positive answer if you can lift projections. Otherwise I have the strong feeling that this will be false. This could also be compared to real rank zero algebras.
– André S.
Nov 20 at 11:43
1
This is true. However, the kernel will be $C_0(0,1)$ in this case, which is not generated by its priojections, hence not a counter example. If you have a counterexample to the real rank zero case, i.e. $A$ and $B$ are real rank zero but $C$ is not, it should be possible to even choose $C$ so that it is not generated by its projections.
– André S.
Nov 22 at 9:16
1
You would also have a counter example, if you can find a $sigma$-unital simple C*-algebra $A$ of real rank zero such that $M(A)$ is not generated by its projections. Indeed, in that case $M(A)/A$, the corona algebra, is generated by its projections.
– André S.
Nov 22 at 9:22
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $A,B,C$ be $3$ unital $C^*$ algebras. Assume that we have the following short exact sequence of $C^*$-algebras:
$$0to Ato Cto Bto 0$$
Assume that $A,B$ are generated by their projections. Is $C$ necessarily generated by its projections, too?
Assume that $A,B$ are von Neumann algebras, is $C$ necessarily a von Neumann algebra, too?
Does the last question has an obvious answer when $A,B$ (hence $C$) are commutative algebras?
measure-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 18 at 6:22
Ali Taghavi
186329
Let $A,B,C$ be $3$ unital $C^*$ algebras. Assume that we have the following short exact sequence of $C^*$-algebras:
$$0to Ato Cto Bto 0$$
Assume that $A,B$ are generated by their projections. Is $C$ necessarily generated by its projections, too?
Assume that $A,B$ are von Neumann algebras, is $C$ necessarily a von Neumann algebra, too?
Does the last question has an obvious answer when $A,B$ (hence $C$) are commutative algebras?
measure-theory operator-algebras c-star-algebras von-neumann-algebras
measure-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 18 at 6:22
Ali Taghavi
186329
asked Nov 18 at 6:22
Ali Taghavi
186329
edited Nov 19 at 8:43
asked Nov 18 at 6:22
Ali Taghavi
186329
asked Nov 18 at 6:22
Ali Taghavi
186329
asked Nov 18 at 6:22
Ali Taghavi
186329
186329
This question has an open bounty worth +50
reputation from Ali Taghavi ending tomorrow.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from Ali Taghavi ending tomorrow.
This question has not received enough attention.
I don't know what you mean by a short exact sequence of unital algebras. If $C to B$ is a nontrivial surjection of unital algebras, its kernel is a nontrivial ideal of $C$, which is at best a nonunital algebra in general, and if it has a unit the map $A to C$ won't preserve units.
– Qiaochu Yuan
Nov 18 at 21:45
@Qiaochu In the question we do not assume that $Ato C$ preserve unit.
– Ali Taghavi
Nov 18 at 21:52
1
Your first question has surely a positive answer if you can lift projections. Otherwise I have the strong feeling that this will be false. This could also be compared to real rank zero algebras.
– André S.
Nov 20 at 11:43
1
This is true. However, the kernel will be $C_0(0,1)$ in this case, which is not generated by its priojections, hence not a counter example. If you have a counterexample to the real rank zero case, i.e. $A$ and $B$ are real rank zero but $C$ is not, it should be possible to even choose $C$ so that it is not generated by its projections.
– André S.
Nov 22 at 9:16
1
You would also have a counter example, if you can find a $sigma$-unital simple C*-algebra $A$ of real rank zero such that $M(A)$ is not generated by its projections. Indeed, in that case $M(A)/A$, the corona algebra, is generated by its projections.
– André S.
Nov 22 at 9:22
|
show 2 more comments
I don't know what you mean by a short exact sequence of unital algebras. If $C to B$ is a nontrivial surjection of unital algebras, its kernel is a nontrivial ideal of $C$, which is at best a nonunital algebra in general, and if it has a unit the map $A to C$ won't preserve units.
– Qiaochu Yuan
Nov 18 at 21:45
@Qiaochu In the question we do not assume that $Ato C$ preserve unit.
– Ali Taghavi
Nov 18 at 21:52
1
Your first question has surely a positive answer if you can lift projections. Otherwise I have the strong feeling that this will be false. This could also be compared to real rank zero algebras.
– André S.
Nov 20 at 11:43
1
This is true. However, the kernel will be $C_0(0,1)$ in this case, which is not generated by its priojections, hence not a counter example. If you have a counterexample to the real rank zero case, i.e. $A$ and $B$ are real rank zero but $C$ is not, it should be possible to even choose $C$ so that it is not generated by its projections.
– André S.
Nov 22 at 9:16
1
You would also have a counter example, if you can find a $sigma$-unital simple C*-algebra $A$ of real rank zero such that $M(A)$ is not generated by its projections. Indeed, in that case $M(A)/A$, the corona algebra, is generated by its projections.
– André S.
Nov 22 at 9:22
I don't know what you mean by a short exact sequence of unital algebras. If $C to B$ is a nontrivial surjection of unital algebras, its kernel is a nontrivial ideal of $C$, which is at best a nonunital algebra in general, and if it has a unit the map $A to C$ won't preserve units.
– Qiaochu Yuan
Nov 18 at 21:45
I don't know what you mean by a short exact sequence of unital algebras. If $C to B$ is a nontrivial surjection of unital algebras, its kernel is a nontrivial ideal of $C$, which is at best a nonunital algebra in general, and if it has a unit the map $A to C$ won't preserve units.
– Qiaochu Yuan
Nov 18 at 21:45
@Qiaochu In the question we do not assume that $Ato C$ preserve unit.
– Ali Taghavi
Nov 18 at 21:52
@Qiaochu In the question we do not assume that $Ato C$ preserve unit.
– Ali Taghavi
Nov 18 at 21:52
1
1
Your first question has surely a positive answer if you can lift projections. Otherwise I have the strong feeling that this will be false. This could also be compared to real rank zero algebras.
– André S.
Nov 20 at 11:43
Your first question has surely a positive answer if you can lift projections. Otherwise I have the strong feeling that this will be false. This could also be compared to real rank zero algebras.
– André S.
Nov 20 at 11:43
1
1
This is true. However, the kernel will be $C_0(0,1)$ in this case, which is not generated by its priojections, hence not a counter example. If you have a counterexample to the real rank zero case, i.e. $A$ and $B$ are real rank zero but $C$ is not, it should be possible to even choose $C$ so that it is not generated by its projections.
– André S.
Nov 22 at 9:16
This is true. However, the kernel will be $C_0(0,1)$ in this case, which is not generated by its priojections, hence not a counter example. If you have a counterexample to the real rank zero case, i.e. $A$ and $B$ are real rank zero but $C$ is not, it should be possible to even choose $C$ so that it is not generated by its projections.
– André S.
Nov 22 at 9:16
1
1
You would also have a counter example, if you can find a $sigma$-unital simple C*-algebra $A$ of real rank zero such that $M(A)$ is not generated by its projections. Indeed, in that case $M(A)/A$, the corona algebra, is generated by its projections.
– André S.
Nov 22 at 9:22
You would also have a counter example, if you can find a $sigma$-unital simple C*-algebra $A$ of real rank zero such that $M(A)$ is not generated by its projections. Indeed, in that case $M(A)/A$, the corona algebra, is generated by its projections.
– André S.
Nov 22 at 9:22
|
show 2 more comments
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I don't know what you mean by a short exact sequence of unital algebras. If $C to B$ is a nontrivial surjection of unital algebras, its kernel is a nontrivial ideal of $C$, which is at best a nonunital algebra in general, and if it has a unit the map $A to C$ won't preserve units.
– Qiaochu Yuan
Nov 18 at 21:45
@Qiaochu In the question we do not assume that $Ato C$ preserve unit.
– Ali Taghavi
Nov 18 at 21:52
1
Your first question has surely a positive answer if you can lift projections. Otherwise I have the strong feeling that this will be false. This could also be compared to real rank zero algebras.
– André S.
Nov 20 at 11:43
1
This is true. However, the kernel will be $C_0(0,1)$ in this case, which is not generated by its priojections, hence not a counter example. If you have a counterexample to the real rank zero case, i.e. $A$ and $B$ are real rank zero but $C$ is not, it should be possible to even choose $C$ so that it is not generated by its projections.
– André S.
Nov 22 at 9:16
1
You would also have a counter example, if you can find a $sigma$-unital simple C*-algebra $A$ of real rank zero such that $M(A)$ is not generated by its projections. Indeed, in that case $M(A)/A$, the corona algebra, is generated by its projections.
– André S.
Nov 22 at 9:22