strict topology on multiplier algebras
1
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Suppose $A$ is a $C^*$ algebra,$M(A)$ is the multiplier algebra.If $S$ is a subset of $M(A)$ which is compact for the strict topology on $M(A)$,is $S$ also a subset of $M(M(A))$ which is compact for the strict topology on $M(M(A))$?
operator-theory operator-algebras c-star-algebras
asked Jan 10 at 15:44
mathrookiemathrookie
916512
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add a comment |
1
$begingroup$
Suppose $A$ is a $C^*$ algebra,$M(A)$ is the multiplier algebra.If $S$ is a subset of $M(A)$ which is compact for the strict topology on $M(A)$,is $S$ also a subset of $M(M(A))$ which is compact for the strict topology on $M(M(A))$?
operator-theory operator-algebras c-star-algebras
asked Jan 10 at 15:44
mathrookiemathrookie
916512
$endgroup$
add a comment |
1
1
1
$begingroup$
Suppose $A$ is a $C^*$ algebra,$M(A)$ is the multiplier algebra.If $S$ is a subset of $M(A)$ which is compact for the strict topology on $M(A)$,is $S$ also a subset of $M(M(A))$ which is compact for the strict topology on $M(M(A))$?
operator-theory operator-algebras c-star-algebras
asked Jan 10 at 15:44
mathrookiemathrookie
916512
$endgroup$
Suppose $A$ is a $C^*$ algebra,$M(A)$ is the multiplier algebra.If $S$ is a subset of $M(A)$ which is compact for the strict topology on $M(A)$,is $S$ also a subset of $M(M(A))$ which is compact for the strict topology on $M(M(A))$?
operator-theory operator-algebras c-star-algebras
operator-theory operator-algebras c-star-algebras
asked Jan 10 at 15:44
mathrookiemathrookie
916512
asked Jan 10 at 15:44
mathrookiemathrookie
916512
asked Jan 10 at 15:44
mathrookiemathrookie
916512
asked Jan 10 at 15:44
mathrookiemathrookie
916512
asked Jan 10 at 15:44
mathrookiemathrookie
916512
916512
add a comment |
add a comment |
1 Answer
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Note that $M(A)$ is unital, so $M(M(A))=M(A)$.
answered Jan 10 at 22:10
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
$begingroup$
Like I said, with the usual definitions the multiplier algebra of $M(A)$ is $M(A)$, since $M(A)$ is unital. This is simply the fact that the multiplier algebra of a unital $A$ is $A$: if $xin M(A)$, then $x=x,1in A$.
$endgroup$
– Martin Argerami
Jan 11 at 1:52
$begingroup$
I guess. It's confusing, because $M(M(A))$ doesn't really make sense to be mentioned as it's just $M(A)$.
$endgroup$
– Martin Argerami
Jan 11 at 2:01
$begingroup$
I still don't know what you are talking about. You have $M(M(A))=M(A)$, so I don't really know what you mean when you talk about different things in two algebras that are the same.
$endgroup$
– Martin Argerami
Jan 11 at 2:21
add a comment |
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1 Answer
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$begingroup$
Note that $M(A)$ is unital, so $M(M(A))=M(A)$.
answered Jan 10 at 22:10
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
$begingroup$
Like I said, with the usual definitions the multiplier algebra of $M(A)$ is $M(A)$, since $M(A)$ is unital. This is simply the fact that the multiplier algebra of a unital $A$ is $A$: if $xin M(A)$, then $x=x,1in A$.
$endgroup$
– Martin Argerami
Jan 11 at 1:52
$begingroup$
I guess. It's confusing, because $M(M(A))$ doesn't really make sense to be mentioned as it's just $M(A)$.
$endgroup$
– Martin Argerami
Jan 11 at 2:01
$begingroup$
I still don't know what you are talking about. You have $M(M(A))=M(A)$, so I don't really know what you mean when you talk about different things in two algebras that are the same.
$endgroup$
– Martin Argerami
Jan 11 at 2:21
add a comment |
1
$begingroup$
Note that $M(A)$ is unital, so $M(M(A))=M(A)$.
answered Jan 10 at 22:10
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
$begingroup$
Like I said, with the usual definitions the multiplier algebra of $M(A)$ is $M(A)$, since $M(A)$ is unital. This is simply the fact that the multiplier algebra of a unital $A$ is $A$: if $xin M(A)$, then $x=x,1in A$.
$endgroup$
– Martin Argerami
Jan 11 at 1:52
$begingroup$
I guess. It's confusing, because $M(M(A))$ doesn't really make sense to be mentioned as it's just $M(A)$.
$endgroup$
– Martin Argerami
Jan 11 at 2:01
$begingroup$
I still don't know what you are talking about. You have $M(M(A))=M(A)$, so I don't really know what you mean when you talk about different things in two algebras that are the same.
$endgroup$
– Martin Argerami
Jan 11 at 2:21
add a comment |
1
1
1
$begingroup$
Note that $M(A)$ is unital, so $M(M(A))=M(A)$.
answered Jan 10 at 22:10
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
Note that $M(A)$ is unital, so $M(M(A))=M(A)$.
answered Jan 10 at 22:10
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 22:10
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 22:10
Martin ArgeramiMartin Argerami
128k1182183
answered Jan 10 at 22:10
Martin ArgeramiMartin Argerami
128k1182183
128k1182183
$begingroup$
Like I said, with the usual definitions the multiplier algebra of $M(A)$ is $M(A)$, since $M(A)$ is unital. This is simply the fact that the multiplier algebra of a unital $A$ is $A$: if $xin M(A)$, then $x=x,1in A$.
$endgroup$
– Martin Argerami
Jan 11 at 1:52
$begingroup$
I guess. It's confusing, because $M(M(A))$ doesn't really make sense to be mentioned as it's just $M(A)$.
$endgroup$
– Martin Argerami
Jan 11 at 2:01
$begingroup$
I still don't know what you are talking about. You have $M(M(A))=M(A)$, so I don't really know what you mean when you talk about different things in two algebras that are the same.
$endgroup$
– Martin Argerami
Jan 11 at 2:21
add a comment |
$begingroup$
Like I said, with the usual definitions the multiplier algebra of $M(A)$ is $M(A)$, since $M(A)$ is unital. This is simply the fact that the multiplier algebra of a unital $A$ is $A$: if $xin M(A)$, then $x=x,1in A$.
$endgroup$
– Martin Argerami
Jan 11 at 1:52
$begingroup$
I guess. It's confusing, because $M(M(A))$ doesn't really make sense to be mentioned as it's just $M(A)$.
$endgroup$
– Martin Argerami
Jan 11 at 2:01
$begingroup$
I still don't know what you are talking about. You have $M(M(A))=M(A)$, so I don't really know what you mean when you talk about different things in two algebras that are the same.
$endgroup$
– Martin Argerami
Jan 11 at 2:21
$begingroup$
Like I said, with the usual definitions the multiplier algebra of $M(A)$ is $M(A)$, since $M(A)$ is unital. This is simply the fact that the multiplier algebra of a unital $A$ is $A$: if $xin M(A)$, then $x=x,1in A$.
$endgroup$
– Martin Argerami
Jan 11 at 1:52
$begingroup$
Like I said, with the usual definitions the multiplier algebra of $M(A)$ is $M(A)$, since $M(A)$ is unital. This is simply the fact that the multiplier algebra of a unital $A$ is $A$: if $xin M(A)$, then $x=x,1in A$.
$endgroup$
– Martin Argerami
Jan 11 at 1:52
$begingroup$
I guess. It's confusing, because $M(M(A))$ doesn't really make sense to be mentioned as it's just $M(A)$.
$endgroup$
– Martin Argerami
Jan 11 at 2:01
$begingroup$
I guess. It's confusing, because $M(M(A))$ doesn't really make sense to be mentioned as it's just $M(A)$.
$endgroup$
– Martin Argerami
Jan 11 at 2:01
$begingroup$
I still don't know what you are talking about. You have $M(M(A))=M(A)$, so I don't really know what you mean when you talk about different things in two algebras that are the same.
$endgroup$
– Martin Argerami
Jan 11 at 2:21
$begingroup$
I still don't know what you are talking about. You have $M(M(A))=M(A)$, so I don't really know what you mean when you talk about different things in two algebras that are the same.
$endgroup$
– Martin Argerami
Jan 11 at 2:21
add a comment |
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