functional calculus under the $*$ homomorphism
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 Nov 18 '18 at 20:13
mathrookie
809512
add a comment |
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 Nov 18 '18 at 20:13
mathrookie
809512
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.
Nov 18 '18 at 21:42
I have reedited the question
– mathrookie
Nov 19 '18 at 3:45
add a comment |
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 Nov 18 '18 at 20:13
mathrookie
809512
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 Nov 18 '18 at 20:13
mathrookie
809512
asked Nov 18 '18 at 20:13
mathrookie
809512
edited Nov 19 '18 at 3:45
asked Nov 18 '18 at 20:13
mathrookie
809512
asked Nov 18 '18 at 20:13
mathrookie
809512
asked Nov 18 '18 at 20:13
mathrookie
809512
809512
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.
Nov 18 '18 at 21:42
I have reedited the question
– mathrookie
Nov 19 '18 at 3:45
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.
Nov 18 '18 at 21:42
I have reedited the question
– mathrookie
Nov 19 '18 at 3:45
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.
Nov 18 '18 at 21:42
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.
Nov 18 '18 at 21:42
I have reedited the question
– mathrookie
Nov 19 '18 at 3:45
I have reedited the question
– mathrookie
Nov 19 '18 at 3:45
add a comment |
1 Answer
1
active
oldest
votes
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 Nov 19 '18 at 4:25
Martin Argerami
124k1176174
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004059%2ffunctional-calculus-under-the-homomorphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Nov 19 '18 at 4:25
Martin Argerami
124k1176174
add a comment |
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 Nov 19 '18 at 4:25
Martin Argerami
124k1176174
add a comment |
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 Nov 19 '18 at 4:25
Martin Argerami
124k1176174
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 Nov 19 '18 at 4:25
Martin Argerami
124k1176174
answered Nov 19 '18 at 4:25
Martin Argerami
124k1176174
answered Nov 19 '18 at 4:25
Martin Argerami
124k1176174
answered Nov 19 '18 at 4:25
Martin Argerami
124k1176174
124k1176174
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004059%2ffunctional-calculus-under-the-homomorphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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.
Nov 18 '18 at 21:42
I have reedited the question
– mathrookie
Nov 19 '18 at 3:45