Definition of the joint spectrum of Hilbert space operators
$begingroup$
I see in this paper the following definition:
How ${bf A''}$ is defined? Is there a relation between $sigma({bf A})$ and $sigma_H({bf A})$? Note that $sigma_H({bf A})$ is defined as:
Definition: $(lambda_1,lambda_2,cdots,lambda_n)notin sigma_H({bf A})$ if there exist operators $U_1,cdots,U_n,V_1,cdots,V_n in mathcal{B}(mathcal{H})$ such that
$$sum_{1leq k leq n}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq n}(A_k-lambda_k I)V_k =I.$$
functional-analysis operator-theory definition operator-algebras
asked Dec 25 '18 at 8:06
StudentStudent
2,4902524
$endgroup$
add a comment |
$begingroup$
I see in this paper the following definition:
How ${bf A''}$ is defined? Is there a relation between $sigma({bf A})$ and $sigma_H({bf A})$? Note that $sigma_H({bf A})$ is defined as:
Definition: $(lambda_1,lambda_2,cdots,lambda_n)notin sigma_H({bf A})$ if there exist operators $U_1,cdots,U_n,V_1,cdots,V_n in mathcal{B}(mathcal{H})$ such that
$$sum_{1leq k leq n}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq n}(A_k-lambda_k I)V_k =I.$$
functional-analysis operator-theory definition operator-algebras
asked Dec 25 '18 at 8:06
StudentStudent
2,4902524
$endgroup$
add a comment |
$begingroup$
I see in this paper the following definition:
How ${bf A''}$ is defined? Is there a relation between $sigma({bf A})$ and $sigma_H({bf A})$? Note that $sigma_H({bf A})$ is defined as:
Definition: $(lambda_1,lambda_2,cdots,lambda_n)notin sigma_H({bf A})$ if there exist operators $U_1,cdots,U_n,V_1,cdots,V_n in mathcal{B}(mathcal{H})$ such that
$$sum_{1leq k leq n}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq n}(A_k-lambda_k I)V_k =I.$$
functional-analysis operator-theory definition operator-algebras
asked Dec 25 '18 at 8:06
StudentStudent
2,4902524
$endgroup$
I see in this paper the following definition:
How ${bf A''}$ is defined? Is there a relation between $sigma({bf A})$ and $sigma_H({bf A})$? Note that $sigma_H({bf A})$ is defined as:
Definition: $(lambda_1,lambda_2,cdots,lambda_n)notin sigma_H({bf A})$ if there exist operators $U_1,cdots,U_n,V_1,cdots,V_n in mathcal{B}(mathcal{H})$ such that
$$sum_{1leq k leq n}U_k(A_k-lambda_k I)=I;hbox{and};;sum_{1leq k leq n}(A_k-lambda_k I)V_k =I.$$
functional-analysis operator-theory definition operator-algebras
functional-analysis operator-theory definition operator-algebras
asked Dec 25 '18 at 8:06
StudentStudent
2,4902524
asked Dec 25 '18 at 8:06
StudentStudent
2,4902524
edited Dec 25 '18 at 17:50
Student
asked Dec 25 '18 at 8:06
StudentStudent
2,4902524
asked Dec 25 '18 at 8:06
StudentStudent
2,4902524
asked Dec 25 '18 at 8:06
StudentStudent
2,4902524
2,4902524
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The double commutant $A''$ is defined as $(A')'$, where
$$
A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
$$
and
$$
A''={Sin B(H): ST=TS forall Tin A'}.
$$
The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.
The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
$$
M''=overline{M}^{rm sot}.
$$
answered Dec 26 '18 at 6:02
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
$begingroup$
Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
$endgroup$
– Student
Dec 26 '18 at 7:53
$begingroup$
Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
$endgroup$
– Student
Dec 26 '18 at 7:54
$begingroup$
Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
$endgroup$
– Martin Argerami
Dec 26 '18 at 13:10
$begingroup$
Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
$endgroup$
– Student
Dec 29 '18 at 11:11
$begingroup$
What I said about the inclusion seems to be wrong. I have deleted that paragraph.
$endgroup$
– Martin Argerami
Dec 29 '18 at 18:15
|
show 2 more comments
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
The double commutant $A''$ is defined as $(A')'$, where
$$
A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
$$
and
$$
A''={Sin B(H): ST=TS forall Tin A'}.
$$
The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.
The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
$$
M''=overline{M}^{rm sot}.
$$
answered Dec 26 '18 at 6:02
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
$begingroup$
Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
$endgroup$
– Student
Dec 26 '18 at 7:53
$begingroup$
Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
$endgroup$
– Student
Dec 26 '18 at 7:54
$begingroup$
Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
$endgroup$
– Martin Argerami
Dec 26 '18 at 13:10
$begingroup$
Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
$endgroup$
– Student
Dec 29 '18 at 11:11
$begingroup$
What I said about the inclusion seems to be wrong. I have deleted that paragraph.
$endgroup$
– Martin Argerami
Dec 29 '18 at 18:15
|
show 2 more comments
$begingroup$
The double commutant $A''$ is defined as $(A')'$, where
$$
A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
$$
and
$$
A''={Sin B(H): ST=TS forall Tin A'}.
$$
The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.
The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
$$
M''=overline{M}^{rm sot}.
$$
answered Dec 26 '18 at 6:02
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
$begingroup$
Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
$endgroup$
– Student
Dec 26 '18 at 7:53
$begingroup$
Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
$endgroup$
– Student
Dec 26 '18 at 7:54
$begingroup$
Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
$endgroup$
– Martin Argerami
Dec 26 '18 at 13:10
$begingroup$
Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
$endgroup$
– Student
Dec 29 '18 at 11:11
$begingroup$
What I said about the inclusion seems to be wrong. I have deleted that paragraph.
$endgroup$
– Martin Argerami
Dec 29 '18 at 18:15
|
show 2 more comments
$begingroup$
The double commutant $A''$ is defined as $(A')'$, where
$$
A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
$$
and
$$
A''={Sin B(H): ST=TS forall Tin A'}.
$$
The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.
The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
$$
M''=overline{M}^{rm sot}.
$$
answered Dec 26 '18 at 6:02
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
The double commutant $A''$ is defined as $(A')'$, where
$$
A'={Tin B(H): TA_j=A_jT, j=1,ldots,n}
$$
and
$$
A''={Sin B(H): ST=TS forall Tin A'}.
$$
The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.
The Double Commutant Theorem says that, if $Msubset B(H)$ is a $*$-algebra, then
$$
M''=overline{M}^{rm sot}.
$$
answered Dec 26 '18 at 6:02
Martin ArgeramiMartin Argerami
130k1184185
edited Dec 29 '18 at 18:15
answered Dec 26 '18 at 6:02
Martin ArgeramiMartin Argerami
130k1184185
answered Dec 26 '18 at 6:02
Martin ArgeramiMartin Argerami
130k1184185
answered Dec 26 '18 at 6:02
Martin ArgeramiMartin Argerami
130k1184185
130k1184185
$begingroup$
Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
$endgroup$
– Student
Dec 26 '18 at 7:53
$begingroup$
Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
$endgroup$
– Student
Dec 26 '18 at 7:54
$begingroup$
Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
$endgroup$
– Martin Argerami
Dec 26 '18 at 13:10
$begingroup$
Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
$endgroup$
– Student
Dec 29 '18 at 11:11
$begingroup$
What I said about the inclusion seems to be wrong. I have deleted that paragraph.
$endgroup$
– Martin Argerami
Dec 29 '18 at 18:15
|
show 2 more comments
$begingroup$
Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
$endgroup$
– Student
Dec 26 '18 at 7:53
$begingroup$
Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
$endgroup$
– Student
Dec 26 '18 at 7:54
$begingroup$
Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
$endgroup$
– Martin Argerami
Dec 26 '18 at 13:10
$begingroup$
Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
$endgroup$
– Student
Dec 29 '18 at 11:11
$begingroup$
What I said about the inclusion seems to be wrong. I have deleted that paragraph.
$endgroup$
– Martin Argerami
Dec 29 '18 at 18:15
$begingroup$
Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
$endgroup$
– Student
Dec 26 '18 at 7:53
$begingroup$
Thank you for your answer. I think it used it in the paper because this joint spectrum is equal to the joint approximate point spectrum of commuting normal operators.
$endgroup$
– Student
Dec 26 '18 at 7:53
$begingroup$
Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
$endgroup$
– Student
Dec 26 '18 at 7:54
$begingroup$
Also I find today that this spectrum is egal to the Harte spectrum in the case of normal operators: cambridge.org/core/journals/…
$endgroup$
– Student
Dec 26 '18 at 7:54
$begingroup$
Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
$endgroup$
– Martin Argerami
Dec 26 '18 at 13:10
$begingroup$
Yes. If the $A_j $ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used.
$endgroup$
– Martin Argerami
Dec 26 '18 at 13:10
$begingroup$
Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
$endgroup$
– Student
Dec 29 '18 at 11:11
$begingroup$
Dear Professor. Please see the following paper page 3 projecteuclid.org/euclid.pjm/1102867217 (because perhaps there is a small problem in the inclusion given in your answer.
$endgroup$
– Student
Dec 29 '18 at 11:11
$begingroup$
What I said about the inclusion seems to be wrong. I have deleted that paragraph.
$endgroup$
– Martin Argerami
Dec 29 '18 at 18:15
$begingroup$
What I said about the inclusion seems to be wrong. I have deleted that paragraph.
$endgroup$
– Martin Argerami
Dec 29 '18 at 18:15
|
show 2 more comments
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