Mixing
Completely positive map is $*$-homomorphism
$begingroup$
Suppose $A$ is a $C^*$ algebra and we have a completely positive contractive map $f colon Arightarrow B(H)$ such that $sup_{a,b in A}lVert f(ab)-f(a)f(b)rVert =0$. Can we conclude that $f$ is a $*$-homomorphism?
operator-theory operator-algebras c-star-algebras
asked Jan 11 at 18:18
mathrookiemathrookie
899512
$endgroup$
add a comment |
$begingroup$
Suppose $A$ is a $C^*$ algebra and we have a completely positive contractive map $f colon Arightarrow B(H)$ such that $sup_{a,b in A}lVert f(ab)-f(a)f(b)rVert =0$. Can we conclude that $f$ is a $*$-homomorphism?
operator-theory operator-algebras c-star-algebras
asked Jan 11 at 18:18
mathrookiemathrookie
899512
$endgroup$
$begingroup$
As a small add-up. This will still be true is you change completely positive by completely bounded by a (nontrivial) theorem of Haagerup: Every completely bounded homomorphism is equivalent to a $ast$-homomorphism.
$endgroup$
– Adrián González-Pérez
Jan 14 at 11:33
$begingroup$
What is the definition of C.B.homomorphism?
$endgroup$
– mathrookie
Jan 14 at 19:46
$begingroup$
A multiplicative linear map such that all of its matrix amplifications $id_{M_k} otimes varphi$ are uniformly bounded.
$endgroup$
– Adrián González-Pérez
Jan 15 at 13:07
add a comment |
$begingroup$
Suppose $A$ is a $C^*$ algebra and we have a completely positive contractive map $f colon Arightarrow B(H)$ such that $sup_{a,b in A}lVert f(ab)-f(a)f(b)rVert =0$. Can we conclude that $f$ is a $*$-homomorphism?
operator-theory operator-algebras c-star-algebras
asked Jan 11 at 18:18
mathrookiemathrookie
899512
$endgroup$
Suppose $A$ is a $C^*$ algebra and we have a completely positive contractive map $f colon Arightarrow B(H)$ such that $sup_{a,b in A}lVert f(ab)-f(a)f(b)rVert =0$. Can we conclude that $f$ is a $*$-homomorphism?
operator-theory operator-algebras c-star-algebras
operator-theory operator-algebras c-star-algebras
asked Jan 11 at 18:18
mathrookiemathrookie
899512
asked Jan 11 at 18:18
mathrookiemathrookie
899512
edited Jan 12 at 13:08
user42761
asked Jan 11 at 18:18
mathrookiemathrookie
899512
asked Jan 11 at 18:18
mathrookiemathrookie
899512
asked Jan 11 at 18:18
mathrookiemathrookie
899512
899512
$begingroup$
As a small add-up. This will still be true is you change completely positive by completely bounded by a (nontrivial) theorem of Haagerup: Every completely bounded homomorphism is equivalent to a $ast$-homomorphism.
$endgroup$
– Adrián González-Pérez
Jan 14 at 11:33
$begingroup$
What is the definition of C.B.homomorphism?
$endgroup$
– mathrookie
Jan 14 at 19:46
$begingroup$
A multiplicative linear map such that all of its matrix amplifications $id_{M_k} otimes varphi$ are uniformly bounded.
$endgroup$
– Adrián González-Pérez
Jan 15 at 13:07
add a comment |
$begingroup$
As a small add-up. This will still be true is you change completely positive by completely bounded by a (nontrivial) theorem of Haagerup: Every completely bounded homomorphism is equivalent to a $ast$-homomorphism.
$endgroup$
– Adrián González-Pérez
Jan 14 at 11:33
$begingroup$
What is the definition of C.B.homomorphism?
$endgroup$
– mathrookie
Jan 14 at 19:46
$begingroup$
A multiplicative linear map such that all of its matrix amplifications $id_{M_k} otimes varphi$ are uniformly bounded.
$endgroup$
– Adrián González-Pérez
Jan 15 at 13:07
$begingroup$
As a small add-up. This will still be true is you change completely positive by completely bounded by a (nontrivial) theorem of Haagerup: Every completely bounded homomorphism is equivalent to a $ast$-homomorphism.
$endgroup$
– Adrián González-Pérez
Jan 14 at 11:33
$begingroup$
As a small add-up. This will still be true is you change completely positive by completely bounded by a (nontrivial) theorem of Haagerup: Every completely bounded homomorphism is equivalent to a $ast$-homomorphism.
$endgroup$
– Adrián González-Pérez
Jan 14 at 11:33
$begingroup$
What is the definition of C.B.homomorphism?
$endgroup$
– mathrookie
Jan 14 at 19:46
$begingroup$
What is the definition of C.B.homomorphism?
$endgroup$
– mathrookie
Jan 14 at 19:46
$begingroup$
A multiplicative linear map such that all of its matrix amplifications $id_{M_k} otimes varphi$ are uniformly bounded.
$endgroup$
– Adrián González-Pérez
Jan 15 at 13:07
$begingroup$
A multiplicative linear map such that all of its matrix amplifications $id_{M_k} otimes varphi$ are uniformly bounded.
$endgroup$
– Adrián González-Pérez
Jan 15 at 13:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The conditions immediately imply that $f$ is an algebra homomorphism. To show that it preserves adjoints, let $ain A$ be self-adjoint. Then $a=a_+-a_-$, where $a_+,a_-$ are positive elements of $A$. Then $f(a)=f(a_+)-f(a_-)$ is a difference of positive elements (as $f$ is a positive map), hence self-adjoint. If now $bin A$ is arbitrary, write $b=b_1+ib_2$, where $b_1,b_2$ are self-adjoint. Then
$$f(b)^*=(f(b_1)+if(b_2))^*=f(b_1)-if(b_2)=f(b_1-ib_2)=f(b^*),$$
and therefore $f$ is a $*$-homomorphism.
Indeed, the above work proves the following proposition:
Let $A,B$ be $C^*$-algebras, and let $f:Ato B$ be a positive linear map. Then $f$ is $*$-preserving, i.e., $f(a)^*=f(a^*)$ for all $ain A$.
answered Jan 12 at 2:38
AweyganAweygan
14.2k21441
$endgroup$
add a comment |
$begingroup$
Assuming that the sup is take over all $a,bin A$, then the condition $sup_{a,b}|f(ab)-f(a)f(b)|=0$ is exactly the same as $f(ab)=f(a)f(b)$ for all $a,b$. Neither contractive not completely positive are relevant in that case. All you need is $f$ to be linear and preserve adjoints.
answered Jan 12 at 2:23
Martin ArgeramiMartin Argerami
127k1182181
$endgroup$
$begingroup$
I believe the point of the question was to ask whether or not the conditions imply that $f$ is $*$-preserving.
$endgroup$
– Aweygan
Jan 12 at 2:41
2
$begingroup$
A (completely) positive map is $*$-preserving.
$endgroup$
– Martin Argerami
Jan 12 at 2:41
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%2f3070174%2fcompletely-positive-map-is-homomorphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The conditions immediately imply that $f$ is an algebra homomorphism. To show that it preserves adjoints, let $ain A$ be self-adjoint. Then $a=a_+-a_-$, where $a_+,a_-$ are positive elements of $A$. Then $f(a)=f(a_+)-f(a_-)$ is a difference of positive elements (as $f$ is a positive map), hence self-adjoint. If now $bin A$ is arbitrary, write $b=b_1+ib_2$, where $b_1,b_2$ are self-adjoint. Then
$$f(b)^*=(f(b_1)+if(b_2))^*=f(b_1)-if(b_2)=f(b_1-ib_2)=f(b^*),$$
and therefore $f$ is a $*$-homomorphism.
Indeed, the above work proves the following proposition:
Let $A,B$ be $C^*$-algebras, and let $f:Ato B$ be a positive linear map. Then $f$ is $*$-preserving, i.e., $f(a)^*=f(a^*)$ for all $ain A$.
answered Jan 12 at 2:38
AweyganAweygan
14.2k21441
$endgroup$
add a comment |
$begingroup$
The conditions immediately imply that $f$ is an algebra homomorphism. To show that it preserves adjoints, let $ain A$ be self-adjoint. Then $a=a_+-a_-$, where $a_+,a_-$ are positive elements of $A$. Then $f(a)=f(a_+)-f(a_-)$ is a difference of positive elements (as $f$ is a positive map), hence self-adjoint. If now $bin A$ is arbitrary, write $b=b_1+ib_2$, where $b_1,b_2$ are self-adjoint. Then
$$f(b)^*=(f(b_1)+if(b_2))^*=f(b_1)-if(b_2)=f(b_1-ib_2)=f(b^*),$$
and therefore $f$ is a $*$-homomorphism.
Indeed, the above work proves the following proposition:
Let $A,B$ be $C^*$-algebras, and let $f:Ato B$ be a positive linear map. Then $f$ is $*$-preserving, i.e., $f(a)^*=f(a^*)$ for all $ain A$.
answered Jan 12 at 2:38
AweyganAweygan
14.2k21441
$endgroup$
add a comment |
$begingroup$
The conditions immediately imply that $f$ is an algebra homomorphism. To show that it preserves adjoints, let $ain A$ be self-adjoint. Then $a=a_+-a_-$, where $a_+,a_-$ are positive elements of $A$. Then $f(a)=f(a_+)-f(a_-)$ is a difference of positive elements (as $f$ is a positive map), hence self-adjoint. If now $bin A$ is arbitrary, write $b=b_1+ib_2$, where $b_1,b_2$ are self-adjoint. Then
$$f(b)^*=(f(b_1)+if(b_2))^*=f(b_1)-if(b_2)=f(b_1-ib_2)=f(b^*),$$
and therefore $f$ is a $*$-homomorphism.
Indeed, the above work proves the following proposition:
Let $A,B$ be $C^*$-algebras, and let $f:Ato B$ be a positive linear map. Then $f$ is $*$-preserving, i.e., $f(a)^*=f(a^*)$ for all $ain A$.
answered Jan 12 at 2:38
AweyganAweygan
14.2k21441
$endgroup$
The conditions immediately imply that $f$ is an algebra homomorphism. To show that it preserves adjoints, let $ain A$ be self-adjoint. Then $a=a_+-a_-$, where $a_+,a_-$ are positive elements of $A$. Then $f(a)=f(a_+)-f(a_-)$ is a difference of positive elements (as $f$ is a positive map), hence self-adjoint. If now $bin A$ is arbitrary, write $b=b_1+ib_2$, where $b_1,b_2$ are self-adjoint. Then
$$f(b)^*=(f(b_1)+if(b_2))^*=f(b_1)-if(b_2)=f(b_1-ib_2)=f(b^*),$$
and therefore $f$ is a $*$-homomorphism.
Indeed, the above work proves the following proposition:
Let $A,B$ be $C^*$-algebras, and let $f:Ato B$ be a positive linear map. Then $f$ is $*$-preserving, i.e., $f(a)^*=f(a^*)$ for all $ain A$.
answered Jan 12 at 2:38
AweyganAweygan
14.2k21441
answered Jan 12 at 2:38
AweyganAweygan
14.2k21441
answered Jan 12 at 2:38
AweyganAweygan
14.2k21441
answered Jan 12 at 2:38
AweyganAweygan
14.2k21441
14.2k21441
add a comment |
add a comment |
$begingroup$
Assuming that the sup is take over all $a,bin A$, then the condition $sup_{a,b}|f(ab)-f(a)f(b)|=0$ is exactly the same as $f(ab)=f(a)f(b)$ for all $a,b$. Neither contractive not completely positive are relevant in that case. All you need is $f$ to be linear and preserve adjoints.
answered Jan 12 at 2:23
Martin ArgeramiMartin Argerami
127k1182181
$endgroup$
$begingroup$
I believe the point of the question was to ask whether or not the conditions imply that $f$ is $*$-preserving.
$endgroup$
– Aweygan
Jan 12 at 2:41
2
$begingroup$
A (completely) positive map is $*$-preserving.
$endgroup$
– Martin Argerami
Jan 12 at 2:41
add a comment |
$begingroup$
Assuming that the sup is take over all $a,bin A$, then the condition $sup_{a,b}|f(ab)-f(a)f(b)|=0$ is exactly the same as $f(ab)=f(a)f(b)$ for all $a,b$. Neither contractive not completely positive are relevant in that case. All you need is $f$ to be linear and preserve adjoints.
answered Jan 12 at 2:23
Martin ArgeramiMartin Argerami
127k1182181
$endgroup$
$begingroup$
I believe the point of the question was to ask whether or not the conditions imply that $f$ is $*$-preserving.
$endgroup$
– Aweygan
Jan 12 at 2:41
2
$begingroup$
A (completely) positive map is $*$-preserving.
$endgroup$
– Martin Argerami
Jan 12 at 2:41
add a comment |
$begingroup$
Assuming that the sup is take over all $a,bin A$, then the condition $sup_{a,b}|f(ab)-f(a)f(b)|=0$ is exactly the same as $f(ab)=f(a)f(b)$ for all $a,b$. Neither contractive not completely positive are relevant in that case. All you need is $f$ to be linear and preserve adjoints.
answered Jan 12 at 2:23
Martin ArgeramiMartin Argerami
127k1182181
$endgroup$
Assuming that the sup is take over all $a,bin A$, then the condition $sup_{a,b}|f(ab)-f(a)f(b)|=0$ is exactly the same as $f(ab)=f(a)f(b)$ for all $a,b$. Neither contractive not completely positive are relevant in that case. All you need is $f$ to be linear and preserve adjoints.
answered Jan 12 at 2:23
Martin ArgeramiMartin Argerami
127k1182181
answered Jan 12 at 2:23
Martin ArgeramiMartin Argerami
127k1182181
answered Jan 12 at 2:23
Martin ArgeramiMartin Argerami
127k1182181
answered Jan 12 at 2:23
Martin ArgeramiMartin Argerami
127k1182181
127k1182181
$begingroup$
I believe the point of the question was to ask whether or not the conditions imply that $f$ is $*$-preserving.
$endgroup$
– Aweygan
Jan 12 at 2:41
2
$begingroup$
A (completely) positive map is $*$-preserving.
$endgroup$
– Martin Argerami
Jan 12 at 2:41
add a comment |
$begingroup$
I believe the point of the question was to ask whether or not the conditions imply that $f$ is $*$-preserving.
$endgroup$
– Aweygan
Jan 12 at 2:41
2
$begingroup$
A (completely) positive map is $*$-preserving.
$endgroup$
– Martin Argerami
Jan 12 at 2:41
$begingroup$
I believe the point of the question was to ask whether or not the conditions imply that $f$ is $*$-preserving.
$endgroup$
– Aweygan
Jan 12 at 2:41
$begingroup$
I believe the point of the question was to ask whether or not the conditions imply that $f$ is $*$-preserving.
$endgroup$
– Aweygan
Jan 12 at 2:41
2
2
$begingroup$
A (completely) positive map is $*$-preserving.
$endgroup$
– Martin Argerami
Jan 12 at 2:41
$begingroup$
A (completely) positive map is $*$-preserving.
$endgroup$
– Martin Argerami
Jan 12 at 2:41
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.
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%2f3070174%2fcompletely-positive-map-is-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
$begingroup$
As a small add-up. This will still be true is you change completely positive by completely bounded by a (nontrivial) theorem of Haagerup: Every completely bounded homomorphism is equivalent to a $ast$-homomorphism.
$endgroup$
– Adrián González-Pérez
Jan 14 at 11:33
$begingroup$
What is the definition of C.B.homomorphism?
$endgroup$
– mathrookie
Jan 14 at 19:46
$begingroup$
A multiplicative linear map such that all of its matrix amplifications $id_{M_k} otimes varphi$ are uniformly bounded.
$endgroup$
– Adrián González-Pérez
Jan 15 at 13:07