Inclusion of multiplier algebras [closed]
0
$begingroup$
If $B$ is a $C^*$-subalgebra of $A$, i.e there exists an inclusion map $phicolon B rightarrow A$, can we conclude that there exists a $*$-homomorphism beween the multiplier algebras $bar{phi} colon M(B)rightarrow M(A)$ ?
operator-algebras c-star-algebras
edited Dec 15 '18 at 10:27
André S.
2,126314
asked Dec 15 '18 at 4:55
mathrookiemathrookie
832512
$endgroup$
closed as off-topic by Saad, Abcd, Shailesh, metamorphy, José Carlos Santos Dec 26 '18 at 19:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
If $B$ is a $C^*$-subalgebra of $A$, i.e there exists an inclusion map $phicolon B rightarrow A$, can we conclude that there exists a $*$-homomorphism beween the multiplier algebras $bar{phi} colon M(B)rightarrow M(A)$ ?
operator-algebras c-star-algebras
edited Dec 15 '18 at 10:27
André S.
2,126314
asked Dec 15 '18 at 4:55
mathrookiemathrookie
832512
$endgroup$
closed as off-topic by Saad, Abcd, Shailesh, metamorphy, José Carlos Santos Dec 26 '18 at 19:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
0
0
$begingroup$
If $B$ is a $C^*$-subalgebra of $A$, i.e there exists an inclusion map $phicolon B rightarrow A$, can we conclude that there exists a $*$-homomorphism beween the multiplier algebras $bar{phi} colon M(B)rightarrow M(A)$ ?
operator-algebras c-star-algebras
edited Dec 15 '18 at 10:27
André S.
2,126314
asked Dec 15 '18 at 4:55
mathrookiemathrookie
832512
$endgroup$
If $B$ is a $C^*$-subalgebra of $A$, i.e there exists an inclusion map $phicolon B rightarrow A$, can we conclude that there exists a $*$-homomorphism beween the multiplier algebras $bar{phi} colon M(B)rightarrow M(A)$ ?
operator-algebras c-star-algebras
operator-algebras c-star-algebras
edited Dec 15 '18 at 10:27
André S.
2,126314
asked Dec 15 '18 at 4:55
mathrookiemathrookie
832512
edited Dec 15 '18 at 10:27
André S.
2,126314
asked Dec 15 '18 at 4:55
mathrookiemathrookie
832512
edited Dec 15 '18 at 10:27
André S.
2,126314
edited Dec 15 '18 at 10:27
André S.
2,126314
edited Dec 15 '18 at 10:27
André S.
2,126314
2,126314
asked Dec 15 '18 at 4:55
mathrookiemathrookie
832512
asked Dec 15 '18 at 4:55
mathrookiemathrookie
832512
asked Dec 15 '18 at 4:55
mathrookiemathrookie
832512
832512
closed as off-topic by Saad, Abcd, Shailesh, metamorphy, José Carlos Santos Dec 26 '18 at 19:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Abcd, Shailesh, metamorphy, José Carlos Santos Dec 26 '18 at 19:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Yes, if $B$ sits inside $A$ non-degenerately. See [Prop. 2.1, Hilbert C*-Modules: A Toolkit for Operator Algebraists].
More precisely: If $(e_i)$ is an approximate unit for $B$ one may define
$$
bar phi(x)big (phi(e_i)a big) = phi(xe_i)a qquad ( x in M(B), a in A).
$$
Add: If $phi colon A to B$ is any surjective $*$-hom. you also have an extension to the multiplier algebras.
answered Dec 15 '18 at 9:03
André S.André S.
2,126314
$endgroup$
$begingroup$
I might be missing something, but I don't see what you are doing here. Are you taking limits over $i$? It seems like you are saying that an approximate unit of $B$ is also an approximate unit for $A$? That's not usually the case. And I don't see the relation with Lance's argument, where $B$ is an ideal and one obtains a map $Ato M(B)$.
$endgroup$
– Martin Argerami
Dec 15 '18 at 14:14
$begingroup$
Oh I forgot to mention that $(e_i)$ is an approximate unit for $B$ and by assumption $(phi(e_i))$ is an approximate unit for $A$. That means that the map $B hookrightarrow A hookrightarrow M(A)$ is non-degenerate. Then, if you want to apply the result in Lance's book, note that $B$ is an ideal in $M(B)$ and hence you can extend the above mentioned map to a $*$-hom. $M(B) to M(A)$.
$endgroup$
– André S.
Dec 15 '18 at 15:52
$begingroup$
I see. That's quite a requirement; not sure how often it happens.
$endgroup$
– Martin Argerami
Dec 15 '18 at 18:12
$begingroup$
I cannot really quantify it. However in classification, where I come from, it quite common that $*$-homomorphisms are lifts of Cu-morphisms, where you have control of the "rank" and the image of strictly positive elements.
$endgroup$
– André S.
Dec 15 '18 at 19:37
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Yes, if $B$ sits inside $A$ non-degenerately. See [Prop. 2.1, Hilbert C*-Modules: A Toolkit for Operator Algebraists].
More precisely: If $(e_i)$ is an approximate unit for $B$ one may define
$$
bar phi(x)big (phi(e_i)a big) = phi(xe_i)a qquad ( x in M(B), a in A).
$$
Add: If $phi colon A to B$ is any surjective $*$-hom. you also have an extension to the multiplier algebras.
answered Dec 15 '18 at 9:03
André S.André S.
2,126314
$endgroup$
$begingroup$
I might be missing something, but I don't see what you are doing here. Are you taking limits over $i$? It seems like you are saying that an approximate unit of $B$ is also an approximate unit for $A$? That's not usually the case. And I don't see the relation with Lance's argument, where $B$ is an ideal and one obtains a map $Ato M(B)$.
$endgroup$
– Martin Argerami
Dec 15 '18 at 14:14
$begingroup$
Oh I forgot to mention that $(e_i)$ is an approximate unit for $B$ and by assumption $(phi(e_i))$ is an approximate unit for $A$. That means that the map $B hookrightarrow A hookrightarrow M(A)$ is non-degenerate. Then, if you want to apply the result in Lance's book, note that $B$ is an ideal in $M(B)$ and hence you can extend the above mentioned map to a $*$-hom. $M(B) to M(A)$.
$endgroup$
– André S.
Dec 15 '18 at 15:52
$begingroup$
I see. That's quite a requirement; not sure how often it happens.
$endgroup$
– Martin Argerami
Dec 15 '18 at 18:12
$begingroup$
I cannot really quantify it. However in classification, where I come from, it quite common that $*$-homomorphisms are lifts of Cu-morphisms, where you have control of the "rank" and the image of strictly positive elements.
$endgroup$
– André S.
Dec 15 '18 at 19:37
add a comment |
2
$begingroup$
Yes, if $B$ sits inside $A$ non-degenerately. See [Prop. 2.1, Hilbert C*-Modules: A Toolkit for Operator Algebraists].
More precisely: If $(e_i)$ is an approximate unit for $B$ one may define
$$
bar phi(x)big (phi(e_i)a big) = phi(xe_i)a qquad ( x in M(B), a in A).
$$
Add: If $phi colon A to B$ is any surjective $*$-hom. you also have an extension to the multiplier algebras.
answered Dec 15 '18 at 9:03
André S.André S.
2,126314
$endgroup$
$begingroup$
I might be missing something, but I don't see what you are doing here. Are you taking limits over $i$? It seems like you are saying that an approximate unit of $B$ is also an approximate unit for $A$? That's not usually the case. And I don't see the relation with Lance's argument, where $B$ is an ideal and one obtains a map $Ato M(B)$.
$endgroup$
– Martin Argerami
Dec 15 '18 at 14:14
$begingroup$
Oh I forgot to mention that $(e_i)$ is an approximate unit for $B$ and by assumption $(phi(e_i))$ is an approximate unit for $A$. That means that the map $B hookrightarrow A hookrightarrow M(A)$ is non-degenerate. Then, if you want to apply the result in Lance's book, note that $B$ is an ideal in $M(B)$ and hence you can extend the above mentioned map to a $*$-hom. $M(B) to M(A)$.
$endgroup$
– André S.
Dec 15 '18 at 15:52
$begingroup$
I see. That's quite a requirement; not sure how often it happens.
$endgroup$
– Martin Argerami
Dec 15 '18 at 18:12
$begingroup$
I cannot really quantify it. However in classification, where I come from, it quite common that $*$-homomorphisms are lifts of Cu-morphisms, where you have control of the "rank" and the image of strictly positive elements.
$endgroup$
– André S.
Dec 15 '18 at 19:37
add a comment |
2
2
2
$begingroup$
Yes, if $B$ sits inside $A$ non-degenerately. See [Prop. 2.1, Hilbert C*-Modules: A Toolkit for Operator Algebraists].
More precisely: If $(e_i)$ is an approximate unit for $B$ one may define
$$
bar phi(x)big (phi(e_i)a big) = phi(xe_i)a qquad ( x in M(B), a in A).
$$
Add: If $phi colon A to B$ is any surjective $*$-hom. you also have an extension to the multiplier algebras.
answered Dec 15 '18 at 9:03
André S.André S.
2,126314
$endgroup$
Yes, if $B$ sits inside $A$ non-degenerately. See [Prop. 2.1, Hilbert C*-Modules: A Toolkit for Operator Algebraists].
More precisely: If $(e_i)$ is an approximate unit for $B$ one may define
$$
bar phi(x)big (phi(e_i)a big) = phi(xe_i)a qquad ( x in M(B), a in A).
$$
Add: If $phi colon A to B$ is any surjective $*$-hom. you also have an extension to the multiplier algebras.
answered Dec 15 '18 at 9:03
André S.André S.
2,126314
edited Dec 15 '18 at 19:41
answered Dec 15 '18 at 9:03
André S.André S.
2,126314
answered Dec 15 '18 at 9:03
André S.André S.
2,126314
answered Dec 15 '18 at 9:03
André S.André S.
2,126314
2,126314
$begingroup$
I might be missing something, but I don't see what you are doing here. Are you taking limits over $i$? It seems like you are saying that an approximate unit of $B$ is also an approximate unit for $A$? That's not usually the case. And I don't see the relation with Lance's argument, where $B$ is an ideal and one obtains a map $Ato M(B)$.
$endgroup$
– Martin Argerami
Dec 15 '18 at 14:14
$begingroup$
Oh I forgot to mention that $(e_i)$ is an approximate unit for $B$ and by assumption $(phi(e_i))$ is an approximate unit for $A$. That means that the map $B hookrightarrow A hookrightarrow M(A)$ is non-degenerate. Then, if you want to apply the result in Lance's book, note that $B$ is an ideal in $M(B)$ and hence you can extend the above mentioned map to a $*$-hom. $M(B) to M(A)$.
$endgroup$
– André S.
Dec 15 '18 at 15:52
$begingroup$
I see. That's quite a requirement; not sure how often it happens.
$endgroup$
– Martin Argerami
Dec 15 '18 at 18:12
$begingroup$
I cannot really quantify it. However in classification, where I come from, it quite common that $*$-homomorphisms are lifts of Cu-morphisms, where you have control of the "rank" and the image of strictly positive elements.
$endgroup$
– André S.
Dec 15 '18 at 19:37
add a comment |
$begingroup$
I might be missing something, but I don't see what you are doing here. Are you taking limits over $i$? It seems like you are saying that an approximate unit of $B$ is also an approximate unit for $A$? That's not usually the case. And I don't see the relation with Lance's argument, where $B$ is an ideal and one obtains a map $Ato M(B)$.
$endgroup$
– Martin Argerami
Dec 15 '18 at 14:14
$begingroup$
Oh I forgot to mention that $(e_i)$ is an approximate unit for $B$ and by assumption $(phi(e_i))$ is an approximate unit for $A$. That means that the map $B hookrightarrow A hookrightarrow M(A)$ is non-degenerate. Then, if you want to apply the result in Lance's book, note that $B$ is an ideal in $M(B)$ and hence you can extend the above mentioned map to a $*$-hom. $M(B) to M(A)$.
$endgroup$
– André S.
Dec 15 '18 at 15:52
$begingroup$
I see. That's quite a requirement; not sure how often it happens.
$endgroup$
– Martin Argerami
Dec 15 '18 at 18:12
$begingroup$
I cannot really quantify it. However in classification, where I come from, it quite common that $*$-homomorphisms are lifts of Cu-morphisms, where you have control of the "rank" and the image of strictly positive elements.
$endgroup$
– André S.
Dec 15 '18 at 19:37
$begingroup$
I might be missing something, but I don't see what you are doing here. Are you taking limits over $i$? It seems like you are saying that an approximate unit of $B$ is also an approximate unit for $A$? That's not usually the case. And I don't see the relation with Lance's argument, where $B$ is an ideal and one obtains a map $Ato M(B)$.
$endgroup$
– Martin Argerami
Dec 15 '18 at 14:14
$begingroup$
I might be missing something, but I don't see what you are doing here. Are you taking limits over $i$? It seems like you are saying that an approximate unit of $B$ is also an approximate unit for $A$? That's not usually the case. And I don't see the relation with Lance's argument, where $B$ is an ideal and one obtains a map $Ato M(B)$.
$endgroup$
– Martin Argerami
Dec 15 '18 at 14:14
$begingroup$
Oh I forgot to mention that $(e_i)$ is an approximate unit for $B$ and by assumption $(phi(e_i))$ is an approximate unit for $A$. That means that the map $B hookrightarrow A hookrightarrow M(A)$ is non-degenerate. Then, if you want to apply the result in Lance's book, note that $B$ is an ideal in $M(B)$ and hence you can extend the above mentioned map to a $*$-hom. $M(B) to M(A)$.
$endgroup$
– André S.
Dec 15 '18 at 15:52
$begingroup$
Oh I forgot to mention that $(e_i)$ is an approximate unit for $B$ and by assumption $(phi(e_i))$ is an approximate unit for $A$. That means that the map $B hookrightarrow A hookrightarrow M(A)$ is non-degenerate. Then, if you want to apply the result in Lance's book, note that $B$ is an ideal in $M(B)$ and hence you can extend the above mentioned map to a $*$-hom. $M(B) to M(A)$.
$endgroup$
– André S.
Dec 15 '18 at 15:52
$begingroup$
I see. That's quite a requirement; not sure how often it happens.
$endgroup$
– Martin Argerami
Dec 15 '18 at 18:12
$begingroup$
I see. That's quite a requirement; not sure how often it happens.
$endgroup$
– Martin Argerami
Dec 15 '18 at 18:12
$begingroup$
I cannot really quantify it. However in classification, where I come from, it quite common that $*$-homomorphisms are lifts of Cu-morphisms, where you have control of the "rank" and the image of strictly positive elements.
$endgroup$
– André S.
Dec 15 '18 at 19:37
$begingroup$
I cannot really quantify it. However in classification, where I come from, it quite common that $*$-homomorphisms are lifts of Cu-morphisms, where you have control of the "rank" and the image of strictly positive elements.
$endgroup$
– André S.
Dec 15 '18 at 19:37
add a comment |
