Mixing
Can $C^*$ algebra $A$ be decomposed?
Clash Royale CLAN TAG#URR8PPP
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If $A$ is a $C^*$ algebra ,$B$ is a finite dimensional $C^*$ subalgebra of $A$.Does there exists a $*$ subalgebra of $C$ such that $A=B oplus C$?
operator-theory operator-algebras c-star-algebras
asked Jul 22 at 8:35
mathrookie
437211
 |Â
show 1 more comment
up vote
-1
down vote
favorite
If $A$ is a $C^*$ algebra ,$B$ is a finite dimensional $C^*$ subalgebra of $A$.Does there exists a $*$ subalgebra of $C$ such that $A=B oplus C$?
operator-theory operator-algebras c-star-algebras
asked Jul 22 at 8:35
mathrookie
437211
What am I, a goldfish?
– Kenny Lau
Jul 22 at 8:37
1
The joke is not funny at all!
– mathrookie
Jul 22 at 9:00
It's not a joke, but rather an accusation that you posted the same question twice and hoped that nobody would notice.
– Kenny Lau
Jul 22 at 9:01
It's very impolite to call someone a goldfish
– mathrookie
Jul 22 at 9:20
I was calling myself.
– Kenny Lau
Jul 22 at 9:27
 |Â
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
If $A$ is a $C^*$ algebra ,$B$ is a finite dimensional $C^*$ subalgebra of $A$.Does there exists a $*$ subalgebra of $C$ such that $A=B oplus C$?
operator-theory operator-algebras c-star-algebras
asked Jul 22 at 8:35
mathrookie
437211
If $A$ is a $C^*$ algebra ,$B$ is a finite dimensional $C^*$ subalgebra of $A$.Does there exists a $*$ subalgebra of $C$ such that $A=B oplus C$?
operator-theory operator-algebras c-star-algebras
asked Jul 22 at 8:35
mathrookie
437211
edited Jul 22 at 8:47
asked Jul 22 at 8:35
mathrookie
437211
asked Jul 22 at 8:35
mathrookie
437211
asked Jul 22 at 8:35
mathrookie
437211
437211
What am I, a goldfish?
– Kenny Lau
Jul 22 at 8:37
1
The joke is not funny at all!
– mathrookie
Jul 22 at 9:00
It's not a joke, but rather an accusation that you posted the same question twice and hoped that nobody would notice.
– Kenny Lau
Jul 22 at 9:01
It's very impolite to call someone a goldfish
– mathrookie
Jul 22 at 9:20
I was calling myself.
– Kenny Lau
Jul 22 at 9:27
 |Â
show 1 more comment
What am I, a goldfish?
– Kenny Lau
Jul 22 at 8:37
1
The joke is not funny at all!
– mathrookie
Jul 22 at 9:00
It's not a joke, but rather an accusation that you posted the same question twice and hoped that nobody would notice.
– Kenny Lau
Jul 22 at 9:01
It's very impolite to call someone a goldfish
– mathrookie
Jul 22 at 9:20
I was calling myself.
– Kenny Lau
Jul 22 at 9:27
What am I, a goldfish?
– Kenny Lau
Jul 22 at 8:37
What am I, a goldfish?
– Kenny Lau
Jul 22 at 8:37
1
1
The joke is not funny at all!
– mathrookie
Jul 22 at 9:00
The joke is not funny at all!
– mathrookie
Jul 22 at 9:00
It's not a joke, but rather an accusation that you posted the same question twice and hoped that nobody would notice.
– Kenny Lau
Jul 22 at 9:01
It's not a joke, but rather an accusation that you posted the same question twice and hoped that nobody would notice.
– Kenny Lau
Jul 22 at 9:01
It's very impolite to call someone a goldfish
– mathrookie
Jul 22 at 9:20
It's very impolite to call someone a goldfish
– mathrookie
Jul 22 at 9:20
I was calling myself.
– Kenny Lau
Jul 22 at 9:27
I was calling myself.
– Kenny Lau
Jul 22 at 9:27
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
No. In such an algebra $A$, both $B$ and $C$ would be nontrivial ideals. So if you take any simple C$^*$-algebra that has finite-dimensional subalgebras, your decomposition does not exist.
Such an example could be $A=$UHF$(2^infty)$, which has lots of projections. Take a nontrivial projection $pin A$, and put $B=mathbb C, p$. Then $B$ is one-dimensional but not an ideal, and thus the decomposition does not exit.
Any other simple real rank zero example would do.
answered Jul 22 at 15:15
Martin Argerami
116k1071164
Is there a splitting lemma for $C^*$ algebra?If $I$ is finite dimensional ideal generated by projections,Can we conclude that $A=I oplus A/I$?
– mathrookie
Jul 22 at 18:27
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
No. In such an algebra $A$, both $B$ and $C$ would be nontrivial ideals. So if you take any simple C$^*$-algebra that has finite-dimensional subalgebras, your decomposition does not exist.
Such an example could be $A=$UHF$(2^infty)$, which has lots of projections. Take a nontrivial projection $pin A$, and put $B=mathbb C, p$. Then $B$ is one-dimensional but not an ideal, and thus the decomposition does not exit.
Any other simple real rank zero example would do.
answered Jul 22 at 15:15
Martin Argerami
116k1071164
Is there a splitting lemma for $C^*$ algebra?If $I$ is finite dimensional ideal generated by projections,Can we conclude that $A=I oplus A/I$?
– mathrookie
Jul 22 at 18:27
add a comment |Â
up vote
1
down vote
accepted
No. In such an algebra $A$, both $B$ and $C$ would be nontrivial ideals. So if you take any simple C$^*$-algebra that has finite-dimensional subalgebras, your decomposition does not exist.
Such an example could be $A=$UHF$(2^infty)$, which has lots of projections. Take a nontrivial projection $pin A$, and put $B=mathbb C, p$. Then $B$ is one-dimensional but not an ideal, and thus the decomposition does not exit.
Any other simple real rank zero example would do.
answered Jul 22 at 15:15
Martin Argerami
116k1071164
Is there a splitting lemma for $C^*$ algebra?If $I$ is finite dimensional ideal generated by projections,Can we conclude that $A=I oplus A/I$?
– mathrookie
Jul 22 at 18:27
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
No. In such an algebra $A$, both $B$ and $C$ would be nontrivial ideals. So if you take any simple C$^*$-algebra that has finite-dimensional subalgebras, your decomposition does not exist.
Such an example could be $A=$UHF$(2^infty)$, which has lots of projections. Take a nontrivial projection $pin A$, and put $B=mathbb C, p$. Then $B$ is one-dimensional but not an ideal, and thus the decomposition does not exit.
Any other simple real rank zero example would do.
answered Jul 22 at 15:15
Martin Argerami
116k1071164
No. In such an algebra $A$, both $B$ and $C$ would be nontrivial ideals. So if you take any simple C$^*$-algebra that has finite-dimensional subalgebras, your decomposition does not exist.
Such an example could be $A=$UHF$(2^infty)$, which has lots of projections. Take a nontrivial projection $pin A$, and put $B=mathbb C, p$. Then $B$ is one-dimensional but not an ideal, and thus the decomposition does not exit.
Any other simple real rank zero example would do.
answered Jul 22 at 15:15
Martin Argerami
116k1071164
answered Jul 22 at 15:15
Martin Argerami
116k1071164
answered Jul 22 at 15:15
Martin Argerami
116k1071164
answered Jul 22 at 15:15
Martin Argerami
116k1071164
116k1071164
Is there a splitting lemma for $C^*$ algebra?If $I$ is finite dimensional ideal generated by projections,Can we conclude that $A=I oplus A/I$?
– mathrookie
Jul 22 at 18:27
add a comment |Â
Is there a splitting lemma for $C^*$ algebra?If $I$ is finite dimensional ideal generated by projections,Can we conclude that $A=I oplus A/I$?
– mathrookie
Jul 22 at 18:27
Is there a splitting lemma for $C^*$ algebra?If $I$ is finite dimensional ideal generated by projections,Can we conclude that $A=I oplus A/I$?
– mathrookie
Jul 22 at 18:27
Is there a splitting lemma for $C^*$ algebra?If $I$ is finite dimensional ideal generated by projections,Can we conclude that $A=I oplus A/I$?
– mathrookie
Jul 22 at 18:27
add a comment |Â
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What am I, a goldfish?
– Kenny Lau
Jul 22 at 8:37
1
The joke is not funny at all!
– mathrookie
Jul 22 at 9:00
It's not a joke, but rather an accusation that you posted the same question twice and hoped that nobody would notice.
– Kenny Lau
Jul 22 at 9:01
It's very impolite to call someone a goldfish
– mathrookie
Jul 22 at 9:20
I was calling myself.
– Kenny Lau
Jul 22 at 9:27