functional calculus under the $*$ homomorphism
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If $A,B$ are two $C^*$ algebras,$psi:A rightarrow B$ is a non $*$ homomorphism.Suppose $b$ is a nonzero normal element in $B$,we have a $*$ isometric isomorphism $phi:C(sigma_B(b))to C^*(b,b^*),; fmapsto f(b)$
My question is:does there exist a $fin C(sigma_B(b)),ain A$ such that $psi(a) =f(b)$ is nonzero?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked yesterday
mathrookie
691512
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up vote
0
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If $A,B$ are two $C^*$ algebras,$psi:A rightarrow B$ is a non $*$ homomorphism.Suppose $b$ is a nonzero normal element in $B$,we have a $*$ isometric isomorphism $phi:C(sigma_B(b))to C^*(b,b^*),; fmapsto f(b)$
My question is:does there exist a $fin C(sigma_B(b)),ain A$ such that $psi(a) =f(b)$ is nonzero?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked yesterday
mathrookie
691512
This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
– André S.
yesterday
I have reedited the question
– mathrookie
22 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $A,B$ are two $C^*$ algebras,$psi:A rightarrow B$ is a non $*$ homomorphism.Suppose $b$ is a nonzero normal element in $B$,we have a $*$ isometric isomorphism $phi:C(sigma_B(b))to C^*(b,b^*),; fmapsto f(b)$
My question is:does there exist a $fin C(sigma_B(b)),ain A$ such that $psi(a) =f(b)$ is nonzero?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked yesterday
mathrookie
691512
If $A,B$ are two $C^*$ algebras,$psi:A rightarrow B$ is a non $*$ homomorphism.Suppose $b$ is a nonzero normal element in $B$,we have a $*$ isometric isomorphism $phi:C(sigma_B(b))to C^*(b,b^*),; fmapsto f(b)$
My question is:does there exist a $fin C(sigma_B(b)),ain A$ such that $psi(a) =f(b)$ is nonzero?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked yesterday
mathrookie
691512
asked yesterday
mathrookie
691512
edited 22 hours ago
asked yesterday
mathrookie
691512
asked yesterday
mathrookie
691512
asked yesterday
mathrookie
691512
691512
This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
– André S.
yesterday
I have reedited the question
– mathrookie
22 hours ago
add a comment |
This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
– André S.
yesterday
I have reedited the question
– mathrookie
22 hours ago
This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
– André S.
yesterday
This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
– André S.
yesterday
I have reedited the question
– mathrookie
22 hours ago
I have reedited the question
– mathrookie
22 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
$$
psi(x,y)=(x,0), b=(0,1).
$$
Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.
answered 22 hours ago
Martin Argerami
121k1073172
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
$$
psi(x,y)=(x,0), b=(0,1).
$$
Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.
answered 22 hours ago
Martin Argerami
121k1073172
add a comment |
up vote
1
down vote
accepted
This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
$$
psi(x,y)=(x,0), b=(0,1).
$$
Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.
answered 22 hours ago
Martin Argerami
121k1073172
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
$$
psi(x,y)=(x,0), b=(0,1).
$$
Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.
answered 22 hours ago
Martin Argerami
121k1073172
This can fail even if $psi$ is a $*$-homomorphism. Take $A=B=mathbb Coplusmathbb C$, and
$$
psi(x,y)=(x,0), b=(0,1).
$$
Then $psi(A)=mathbb Coplus 0$, while $C^*(b)=0oplusmathbb C$; so your equality can only occur when $psi(a)=0=f(b)$.
answered 22 hours ago
Martin Argerami
121k1073172
answered 22 hours ago
Martin Argerami
121k1073172
answered 22 hours ago
Martin Argerami
121k1073172
answered 22 hours ago
Martin Argerami
121k1073172
121k1073172
add a comment |
add a comment |
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This is surely the case, if you assume that $f(a) in A$, which you suggest by your notation. Probably you want to know something else.
– André S.
yesterday
I have reedited the question
– mathrookie
22 hours ago