Does a closed right ideal of a C$^*$-algebra have a C$^*$-algebra?
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$A$ is an infinite dimensional C$^*$-algebra and $Jsubset A$ is a closed right ideal. $A$ and $J$ are infinite dimensional(as a vector space). I want to find an infinite dimensional C$^*$-algebra subset of $J$. How can I find it?
I know an infinite dimensional C$^*$-algebra has an infinite dimensional commutative C$^*$-subalgebra. So if $A_1$ is infinite dimensional commutative C$^*$-subalgebra of $A$, Is the set $A_1cap J$ an infinite dimensional C$^*$-algebra? If no, so what can I do?
functional-analysis ideals c-star-algebras
asked Nov 15 at 16:27
Dadrahm
3278
add a comment |
up vote
1
down vote
favorite
$A$ is an infinite dimensional C$^*$-algebra and $Jsubset A$ is a closed right ideal. $A$ and $J$ are infinite dimensional(as a vector space). I want to find an infinite dimensional C$^*$-algebra subset of $J$. How can I find it?
I know an infinite dimensional C$^*$-algebra has an infinite dimensional commutative C$^*$-subalgebra. So if $A_1$ is infinite dimensional commutative C$^*$-subalgebra of $A$, Is the set $A_1cap J$ an infinite dimensional C$^*$-algebra? If no, so what can I do?
functional-analysis ideals c-star-algebras
asked Nov 15 at 16:27
Dadrahm
3278
What is $I$ in your first paragraph?
– Aweygan
Nov 15 at 17:20
Execuse me for my mistake
– Dadrahm
Nov 15 at 17:29
1
No worries. Well in general if $J$ is a one-sided ideal, then $Jcap J^*$ is a two-sided ideal, hence a $C^*$-subalgebra, although I am not sure if there is any satisfying criteria for determining whether or not this ideal is infinite-dimensional, or even non-trivial.
– Aweygan
Nov 15 at 17:35
But if $J={ fin B(ell^2): f^*((1,0,0,0,ldots ))=0 } subseteq B(ell^2)$ then $Jcap J^* $ isn't an two sided ideal.
– Dadrahm
Nov 15 at 18:31
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$A$ is an infinite dimensional C$^*$-algebra and $Jsubset A$ is a closed right ideal. $A$ and $J$ are infinite dimensional(as a vector space). I want to find an infinite dimensional C$^*$-algebra subset of $J$. How can I find it?
I know an infinite dimensional C$^*$-algebra has an infinite dimensional commutative C$^*$-subalgebra. So if $A_1$ is infinite dimensional commutative C$^*$-subalgebra of $A$, Is the set $A_1cap J$ an infinite dimensional C$^*$-algebra? If no, so what can I do?
functional-analysis ideals c-star-algebras
asked Nov 15 at 16:27
Dadrahm
3278
$A$ is an infinite dimensional C$^*$-algebra and $Jsubset A$ is a closed right ideal. $A$ and $J$ are infinite dimensional(as a vector space). I want to find an infinite dimensional C$^*$-algebra subset of $J$. How can I find it?
I know an infinite dimensional C$^*$-algebra has an infinite dimensional commutative C$^*$-subalgebra. So if $A_1$ is infinite dimensional commutative C$^*$-subalgebra of $A$, Is the set $A_1cap J$ an infinite dimensional C$^*$-algebra? If no, so what can I do?
functional-analysis ideals c-star-algebras
functional-analysis ideals c-star-algebras
asked Nov 15 at 16:27
Dadrahm
3278
asked Nov 15 at 16:27
Dadrahm
3278
edited Nov 16 at 6:21
asked Nov 15 at 16:27
Dadrahm
3278
asked Nov 15 at 16:27
Dadrahm
3278
asked Nov 15 at 16:27
Dadrahm
3278
3278
What is $I$ in your first paragraph?
– Aweygan
Nov 15 at 17:20
Execuse me for my mistake
– Dadrahm
Nov 15 at 17:29
1
No worries. Well in general if $J$ is a one-sided ideal, then $Jcap J^*$ is a two-sided ideal, hence a $C^*$-subalgebra, although I am not sure if there is any satisfying criteria for determining whether or not this ideal is infinite-dimensional, or even non-trivial.
– Aweygan
Nov 15 at 17:35
But if $J={ fin B(ell^2): f^*((1,0,0,0,ldots ))=0 } subseteq B(ell^2)$ then $Jcap J^* $ isn't an two sided ideal.
– Dadrahm
Nov 15 at 18:31
add a comment |
What is $I$ in your first paragraph?
– Aweygan
Nov 15 at 17:20
Execuse me for my mistake
– Dadrahm
Nov 15 at 17:29
1
No worries. Well in general if $J$ is a one-sided ideal, then $Jcap J^*$ is a two-sided ideal, hence a $C^*$-subalgebra, although I am not sure if there is any satisfying criteria for determining whether or not this ideal is infinite-dimensional, or even non-trivial.
– Aweygan
Nov 15 at 17:35
But if $J={ fin B(ell^2): f^*((1,0,0,0,ldots ))=0 } subseteq B(ell^2)$ then $Jcap J^* $ isn't an two sided ideal.
– Dadrahm
Nov 15 at 18:31
What is $I$ in your first paragraph?
– Aweygan
Nov 15 at 17:20
What is $I$ in your first paragraph?
– Aweygan
Nov 15 at 17:20
Execuse me for my mistake
– Dadrahm
Nov 15 at 17:29
Execuse me for my mistake
– Dadrahm
Nov 15 at 17:29
1
1
No worries. Well in general if $J$ is a one-sided ideal, then $Jcap J^*$ is a two-sided ideal, hence a $C^*$-subalgebra, although I am not sure if there is any satisfying criteria for determining whether or not this ideal is infinite-dimensional, or even non-trivial.
– Aweygan
Nov 15 at 17:35
No worries. Well in general if $J$ is a one-sided ideal, then $Jcap J^*$ is a two-sided ideal, hence a $C^*$-subalgebra, although I am not sure if there is any satisfying criteria for determining whether or not this ideal is infinite-dimensional, or even non-trivial.
– Aweygan
Nov 15 at 17:35
But if $J={ fin B(ell^2): f^*((1,0,0,0,ldots ))=0 } subseteq B(ell^2)$ then $Jcap J^* $ isn't an two sided ideal.
– Dadrahm
Nov 15 at 18:31
But if $J={ fin B(ell^2): f^*((1,0,0,0,ldots ))=0 } subseteq B(ell^2)$ then $Jcap J^* $ isn't an two sided ideal.
– Dadrahm
Nov 15 at 18:31
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
You can't do that in general. Many C$^*$-algebras are simple. In such a case if $Bsubset J$ is a C$^*$-algebra, then
$$
B=Bcap B^*subset Jcap J^*={0},
$$
since $Jcap J^*$ is a closed ideal.
answered Nov 15 at 20:53
Martin Argerami
121k1073172
Why $Jcap J^*=0$ ? $Jcap J^*$ is a two sided ideal of $J$ and not a two sided ideal of $A$, So why $Jcap J^*=0$
– Dadrahm
Nov 16 at 6:10
But if J is simple then $Jcap J^*=0$. Is there an infinite dimensional simple closed right ideal for ecah infinite dimensional $C^*$-algebra?
– Dadrahm
Nov 16 at 6:16
I don't really follow what you are saying; a right ideal is not an algebra, so what does "$J $ is simple" mean, and what would "ideal of $J $ mean? And, again, if $J $ is a right ideal of $A $, then $ J^*cap J $ is a bilateral ideal of $A $.
– Martin Argerami
Nov 16 at 12:49
Why $Jcap J^*$ is a two sided ideal of A?
– Dadrahm
Nov 16 at 12:51
a right ideal is not an algebra?! So I think I made some mistakes. Excuse me
– Dadrahm
Nov 16 at 12:53
|
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You can't do that in general. Many C$^*$-algebras are simple. In such a case if $Bsubset J$ is a C$^*$-algebra, then
$$
B=Bcap B^*subset Jcap J^*={0},
$$
since $Jcap J^*$ is a closed ideal.
answered Nov 15 at 20:53
Martin Argerami
121k1073172
Why $Jcap J^*=0$ ? $Jcap J^*$ is a two sided ideal of $J$ and not a two sided ideal of $A$, So why $Jcap J^*=0$
– Dadrahm
Nov 16 at 6:10
But if J is simple then $Jcap J^*=0$. Is there an infinite dimensional simple closed right ideal for ecah infinite dimensional $C^*$-algebra?
– Dadrahm
Nov 16 at 6:16
I don't really follow what you are saying; a right ideal is not an algebra, so what does "$J $ is simple" mean, and what would "ideal of $J $ mean? And, again, if $J $ is a right ideal of $A $, then $ J^*cap J $ is a bilateral ideal of $A $.
– Martin Argerami
Nov 16 at 12:49
Why $Jcap J^*$ is a two sided ideal of A?
– Dadrahm
Nov 16 at 12:51
a right ideal is not an algebra?! So I think I made some mistakes. Excuse me
– Dadrahm
Nov 16 at 12:53
|
show 6 more comments
up vote
1
down vote
You can't do that in general. Many C$^*$-algebras are simple. In such a case if $Bsubset J$ is a C$^*$-algebra, then
$$
B=Bcap B^*subset Jcap J^*={0},
$$
since $Jcap J^*$ is a closed ideal.
answered Nov 15 at 20:53
Martin Argerami
121k1073172
Why $Jcap J^*=0$ ? $Jcap J^*$ is a two sided ideal of $J$ and not a two sided ideal of $A$, So why $Jcap J^*=0$
– Dadrahm
Nov 16 at 6:10
But if J is simple then $Jcap J^*=0$. Is there an infinite dimensional simple closed right ideal for ecah infinite dimensional $C^*$-algebra?
– Dadrahm
Nov 16 at 6:16
I don't really follow what you are saying; a right ideal is not an algebra, so what does "$J $ is simple" mean, and what would "ideal of $J $ mean? And, again, if $J $ is a right ideal of $A $, then $ J^*cap J $ is a bilateral ideal of $A $.
– Martin Argerami
Nov 16 at 12:49
Why $Jcap J^*$ is a two sided ideal of A?
– Dadrahm
Nov 16 at 12:51
a right ideal is not an algebra?! So I think I made some mistakes. Excuse me
– Dadrahm
Nov 16 at 12:53
|
show 6 more comments
up vote
1
down vote
up vote
1
down vote
You can't do that in general. Many C$^*$-algebras are simple. In such a case if $Bsubset J$ is a C$^*$-algebra, then
$$
B=Bcap B^*subset Jcap J^*={0},
$$
since $Jcap J^*$ is a closed ideal.
answered Nov 15 at 20:53
Martin Argerami
121k1073172
You can't do that in general. Many C$^*$-algebras are simple. In such a case if $Bsubset J$ is a C$^*$-algebra, then
$$
B=Bcap B^*subset Jcap J^*={0},
$$
since $Jcap J^*$ is a closed ideal.
answered Nov 15 at 20:53
Martin Argerami
121k1073172
edited Nov 15 at 21:48
answered Nov 15 at 20:53
Martin Argerami
121k1073172
answered Nov 15 at 20:53
Martin Argerami
121k1073172
answered Nov 15 at 20:53
Martin Argerami
121k1073172
121k1073172
Why $Jcap J^*=0$ ? $Jcap J^*$ is a two sided ideal of $J$ and not a two sided ideal of $A$, So why $Jcap J^*=0$
– Dadrahm
Nov 16 at 6:10
But if J is simple then $Jcap J^*=0$. Is there an infinite dimensional simple closed right ideal for ecah infinite dimensional $C^*$-algebra?
– Dadrahm
Nov 16 at 6:16
I don't really follow what you are saying; a right ideal is not an algebra, so what does "$J $ is simple" mean, and what would "ideal of $J $ mean? And, again, if $J $ is a right ideal of $A $, then $ J^*cap J $ is a bilateral ideal of $A $.
– Martin Argerami
Nov 16 at 12:49
Why $Jcap J^*$ is a two sided ideal of A?
– Dadrahm
Nov 16 at 12:51
a right ideal is not an algebra?! So I think I made some mistakes. Excuse me
– Dadrahm
Nov 16 at 12:53
|
show 6 more comments
Why $Jcap J^*=0$ ? $Jcap J^*$ is a two sided ideal of $J$ and not a two sided ideal of $A$, So why $Jcap J^*=0$
– Dadrahm
Nov 16 at 6:10
But if J is simple then $Jcap J^*=0$. Is there an infinite dimensional simple closed right ideal for ecah infinite dimensional $C^*$-algebra?
– Dadrahm
Nov 16 at 6:16
I don't really follow what you are saying; a right ideal is not an algebra, so what does "$J $ is simple" mean, and what would "ideal of $J $ mean? And, again, if $J $ is a right ideal of $A $, then $ J^*cap J $ is a bilateral ideal of $A $.
– Martin Argerami
Nov 16 at 12:49
Why $Jcap J^*$ is a two sided ideal of A?
– Dadrahm
Nov 16 at 12:51
a right ideal is not an algebra?! So I think I made some mistakes. Excuse me
– Dadrahm
Nov 16 at 12:53
Why $Jcap J^*=0$ ? $Jcap J^*$ is a two sided ideal of $J$ and not a two sided ideal of $A$, So why $Jcap J^*=0$
– Dadrahm
Nov 16 at 6:10
Why $Jcap J^*=0$ ? $Jcap J^*$ is a two sided ideal of $J$ and not a two sided ideal of $A$, So why $Jcap J^*=0$
– Dadrahm
Nov 16 at 6:10
But if J is simple then $Jcap J^*=0$. Is there an infinite dimensional simple closed right ideal for ecah infinite dimensional $C^*$-algebra?
– Dadrahm
Nov 16 at 6:16
But if J is simple then $Jcap J^*=0$. Is there an infinite dimensional simple closed right ideal for ecah infinite dimensional $C^*$-algebra?
– Dadrahm
Nov 16 at 6:16
I don't really follow what you are saying; a right ideal is not an algebra, so what does "$J $ is simple" mean, and what would "ideal of $J $ mean? And, again, if $J $ is a right ideal of $A $, then $ J^*cap J $ is a bilateral ideal of $A $.
– Martin Argerami
Nov 16 at 12:49
I don't really follow what you are saying; a right ideal is not an algebra, so what does "$J $ is simple" mean, and what would "ideal of $J $ mean? And, again, if $J $ is a right ideal of $A $, then $ J^*cap J $ is a bilateral ideal of $A $.
– Martin Argerami
Nov 16 at 12:49
Why $Jcap J^*$ is a two sided ideal of A?
– Dadrahm
Nov 16 at 12:51
Why $Jcap J^*$ is a two sided ideal of A?
– Dadrahm
Nov 16 at 12:51
a right ideal is not an algebra?! So I think I made some mistakes. Excuse me
– Dadrahm
Nov 16 at 12:53
a right ideal is not an algebra?! So I think I made some mistakes. Excuse me
– Dadrahm
Nov 16 at 12:53
|
show 6 more comments
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What is $I$ in your first paragraph?
– Aweygan
Nov 15 at 17:20
Execuse me for my mistake
– Dadrahm
Nov 15 at 17:29
1
No worries. Well in general if $J$ is a one-sided ideal, then $Jcap J^*$ is a two-sided ideal, hence a $C^*$-subalgebra, although I am not sure if there is any satisfying criteria for determining whether or not this ideal is infinite-dimensional, or even non-trivial.
– Aweygan
Nov 15 at 17:35
But if $J={ fin B(ell^2): f^*((1,0,0,0,ldots ))=0 } subseteq B(ell^2)$ then $Jcap J^* $ isn't an two sided ideal.
– Dadrahm
Nov 15 at 18:31