Reference request: extension of $*$-homomorphism to multiplier algebra

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

Let $A$ be a $C^*$-algebra and $f:ArightarrowmathbbC$ a $^*$-homomorphism. Does $f$ always extend to a $^*$-homomorphism $tildef:M(A)rightarrowmathbbC$, where $M(A)$ is the multiplier algebra of $A$?

I suspect the answer is no, but I'm not sure how to show this. In case the answer is yes, a proof or reference would be greatly appreciated.

Thanks!

functional-analysis operator-theory operator-algebras c-star-algebras

share|cite|improve this question

edited Jul 19 at 4:41

Martin Argerami

116k1071164

asked Jul 19 at 3:24

ougoah

1,065610

add a comment | 

up vote
2
down vote

favorite

Let $A$ be a $C^*$-algebra and $f:ArightarrowmathbbC$ a $^*$-homomorphism. Does $f$ always extend to a $^*$-homomorphism $tildef:M(A)rightarrowmathbbC$, where $M(A)$ is the multiplier algebra of $A$?

I suspect the answer is no, but I'm not sure how to show this. In case the answer is yes, a proof or reference would be greatly appreciated.

Thanks!

functional-analysis operator-theory operator-algebras c-star-algebras

share|cite|improve this question

edited Jul 19 at 4:41

Martin Argerami

116k1071164

asked Jul 19 at 3:24

ougoah

1,065610

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $A$ be a $C^*$-algebra and $f:ArightarrowmathbbC$ a $^*$-homomorphism. Does $f$ always extend to a $^*$-homomorphism $tildef:M(A)rightarrowmathbbC$, where $M(A)$ is the multiplier algebra of $A$?

I suspect the answer is no, but I'm not sure how to show this. In case the answer is yes, a proof or reference would be greatly appreciated.

Thanks!

functional-analysis operator-theory operator-algebras c-star-algebras

share|cite|improve this question

edited Jul 19 at 4:41

Martin Argerami

116k1071164

asked Jul 19 at 3:24

ougoah

1,065610

Let $A$ be a $C^*$-algebra and $f:ArightarrowmathbbC$ a $^*$-homomorphism. Does $f$ always extend to a $^*$-homomorphism $tildef:M(A)rightarrowmathbbC$, where $M(A)$ is the multiplier algebra of $A$?

I suspect the answer is no, but I'm not sure how to show this. In case the answer is yes, a proof or reference would be greatly appreciated.

Thanks!

functional-analysis operator-theory operator-algebras c-star-algebras

share|cite|improve this question

edited Jul 19 at 4:41

Martin Argerami

116k1071164

asked Jul 19 at 3:24

ougoah

1,065610

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 4:41

Martin Argerami

116k1071164

edited Jul 19 at 4:41

Martin Argerami

116k1071164

edited Jul 19 at 4:41

Martin Argerami

116k1071164

116k1071164

asked Jul 19 at 3:24

ougoah

1,065610

asked Jul 19 at 3:24

ougoah

1,065610

asked Jul 19 at 3:24

ougoah

1,065610

1,065610

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

The answer is yes. This depends on two facts:

• $A$ is an ideal in $M(A)$.




• if $Jsubset B$ is an ideal, any non-degenerate representation $Jto B(H)$ can be extended to a representation $Bto B(H)$. This is for instance Lemma I.9.14 in Davidson's C$^*$-Algebras By Example. Here we have $J=A$, $B=M(A)$, $H=mathbb C$.

The argument works for any $*$-homomorphism $Ato B(H)$.

share|cite|improve this answer

edited Jul 19 at 4:32

answered Jul 19 at 4:25

Martin Argerami

116k1071164

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856225%2freference-request-extension-of-homomorphism-to-multiplier-algebra%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The answer is yes. This depends on two facts:

• $A$ is an ideal in $M(A)$.




• if $Jsubset B$ is an ideal, any non-degenerate representation $Jto B(H)$ can be extended to a representation $Bto B(H)$. This is for instance Lemma I.9.14 in Davidson's C$^*$-Algebras By Example. Here we have $J=A$, $B=M(A)$, $H=mathbb C$.

The argument works for any $*$-homomorphism $Ato B(H)$.

share|cite|improve this answer

edited Jul 19 at 4:32

answered Jul 19 at 4:25

Martin Argerami

116k1071164

add a comment | 

up vote
2
down vote



accepted

The answer is yes. This depends on two facts:

• $A$ is an ideal in $M(A)$.




• if $Jsubset B$ is an ideal, any non-degenerate representation $Jto B(H)$ can be extended to a representation $Bto B(H)$. This is for instance Lemma I.9.14 in Davidson's C$^*$-Algebras By Example. Here we have $J=A$, $B=M(A)$, $H=mathbb C$.

The argument works for any $*$-homomorphism $Ato B(H)$.

share|cite|improve this answer

edited Jul 19 at 4:32

answered Jul 19 at 4:25

Martin Argerami

116k1071164

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The answer is yes. This depends on two facts:

• $A$ is an ideal in $M(A)$.




• if $Jsubset B$ is an ideal, any non-degenerate representation $Jto B(H)$ can be extended to a representation $Bto B(H)$. This is for instance Lemma I.9.14 in Davidson's C$^*$-Algebras By Example. Here we have $J=A$, $B=M(A)$, $H=mathbb C$.

The argument works for any $*$-homomorphism $Ato B(H)$.

share|cite|improve this answer

edited Jul 19 at 4:32

answered Jul 19 at 4:25

Martin Argerami

116k1071164

The answer is yes. This depends on two facts:

• $A$ is an ideal in $M(A)$.




• if $Jsubset B$ is an ideal, any non-degenerate representation $Jto B(H)$ can be extended to a representation $Bto B(H)$. This is for instance Lemma I.9.14 in Davidson's C$^*$-Algebras By Example. Here we have $J=A$, $B=M(A)$, $H=mathbb C$.

The argument works for any $*$-homomorphism $Ato B(H)$.

share|cite|improve this answer

edited Jul 19 at 4:32

answered Jul 19 at 4:25

Martin Argerami

116k1071164

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 4:32

edited Jul 19 at 4:32

edited Jul 19 at 4:32

answered Jul 19 at 4:25

Martin Argerami

116k1071164

answered Jul 19 at 4:25

Martin Argerami

116k1071164

answered Jul 19 at 4:25

Martin Argerami

116k1071164

116k1071164

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856225%2freference-request-extension-of-homomorphism-to-multiplier-algebra%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email