Can distinct elements of a $C^*$-algebra be separated by a maximal left ideal?
Let $A$ be a $C^*$-algebra, and let $fneq gin A$. Does there exist a maximal ideal $Jtrianglelefteq A$ with $f+Jneq g+J$?
I'm particularly interested in the case of $A=B(mathcal H)$, and why things aren't obvious. In this case, $fneq g$ implies that there is some $hinmathcal H$ with $f(h)neq g(h)$, so the left ideal of the annihilator of $h$ certainly separates them. However, $text{Ann}_{B(mathcal H)}(h)$ is not maximal, and I don't immediately see how to upgrade this to a maximal ideal.
An alternative approach is to attempt to find an irreducible representation. We have a good family of representations from GNS: for each positive linear functional $rho$ on $A$, we have a representation $pi_rho$ on a Hilbert space, with the image of $1$ being cyclic for the representation. If $fneq g$, then at least one of these representations has $fcdot [1]neq gcdot [1]$, because the GNS representation is faithful. I know that the annihilator of a simple module is the intersection of maximal left ideals, so if I believe that if I can find a simple $A$-module on which $f$ and $g$ act differently, there is such an ideal.
c-star-algebras
asked Dec 10 '18 at 0:59
Ashwin Trisal
1,2141516
add a comment |
Let $A$ be a $C^*$-algebra, and let $fneq gin A$. Does there exist a maximal ideal $Jtrianglelefteq A$ with $f+Jneq g+J$?
I'm particularly interested in the case of $A=B(mathcal H)$, and why things aren't obvious. In this case, $fneq g$ implies that there is some $hinmathcal H$ with $f(h)neq g(h)$, so the left ideal of the annihilator of $h$ certainly separates them. However, $text{Ann}_{B(mathcal H)}(h)$ is not maximal, and I don't immediately see how to upgrade this to a maximal ideal.
An alternative approach is to attempt to find an irreducible representation. We have a good family of representations from GNS: for each positive linear functional $rho$ on $A$, we have a representation $pi_rho$ on a Hilbert space, with the image of $1$ being cyclic for the representation. If $fneq g$, then at least one of these representations has $fcdot [1]neq gcdot [1]$, because the GNS representation is faithful. I know that the annihilator of a simple module is the intersection of maximal left ideals, so if I believe that if I can find a simple $A$-module on which $f$ and $g$ act differently, there is such an ideal.
c-star-algebras
asked Dec 10 '18 at 0:59
Ashwin Trisal
1,2141516
add a comment |
Let $A$ be a $C^*$-algebra, and let $fneq gin A$. Does there exist a maximal ideal $Jtrianglelefteq A$ with $f+Jneq g+J$?
I'm particularly interested in the case of $A=B(mathcal H)$, and why things aren't obvious. In this case, $fneq g$ implies that there is some $hinmathcal H$ with $f(h)neq g(h)$, so the left ideal of the annihilator of $h$ certainly separates them. However, $text{Ann}_{B(mathcal H)}(h)$ is not maximal, and I don't immediately see how to upgrade this to a maximal ideal.
An alternative approach is to attempt to find an irreducible representation. We have a good family of representations from GNS: for each positive linear functional $rho$ on $A$, we have a representation $pi_rho$ on a Hilbert space, with the image of $1$ being cyclic for the representation. If $fneq g$, then at least one of these representations has $fcdot [1]neq gcdot [1]$, because the GNS representation is faithful. I know that the annihilator of a simple module is the intersection of maximal left ideals, so if I believe that if I can find a simple $A$-module on which $f$ and $g$ act differently, there is such an ideal.
c-star-algebras
asked Dec 10 '18 at 0:59
Ashwin Trisal
1,2141516
Let $A$ be a $C^*$-algebra, and let $fneq gin A$. Does there exist a maximal ideal $Jtrianglelefteq A$ with $f+Jneq g+J$?
I'm particularly interested in the case of $A=B(mathcal H)$, and why things aren't obvious. In this case, $fneq g$ implies that there is some $hinmathcal H$ with $f(h)neq g(h)$, so the left ideal of the annihilator of $h$ certainly separates them. However, $text{Ann}_{B(mathcal H)}(h)$ is not maximal, and I don't immediately see how to upgrade this to a maximal ideal.
An alternative approach is to attempt to find an irreducible representation. We have a good family of representations from GNS: for each positive linear functional $rho$ on $A$, we have a representation $pi_rho$ on a Hilbert space, with the image of $1$ being cyclic for the representation. If $fneq g$, then at least one of these representations has $fcdot [1]neq gcdot [1]$, because the GNS representation is faithful. I know that the annihilator of a simple module is the intersection of maximal left ideals, so if I believe that if I can find a simple $A$-module on which $f$ and $g$ act differently, there is such an ideal.
c-star-algebras
c-star-algebras
asked Dec 10 '18 at 0:59
Ashwin Trisal
1,2141516
asked Dec 10 '18 at 0:59
Ashwin Trisal
1,2141516
asked Dec 10 '18 at 0:59
Ashwin Trisal
1,2141516
asked Dec 10 '18 at 0:59
Ashwin Trisal
1,2141516
asked Dec 10 '18 at 0:59
Ashwin Trisal
1,2141516
1,2141516
add a comment |
add a comment |
2 Answers
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$forall aneq 0$, there is a pure state $tau$ on $A$ such that $tau(a^*a)=|a^*a|neq 0$(Theorem 5.1.11, [1]). Let $N_tau={a|tau(a^*a)=0},$ then $N_tau$ is a maximal left ideal(Theorem 5.3.5, [1]), and $anotin N_tau$.
[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.
answered Dec 10 '18 at 4:37
C.Ding
1,3931321
add a comment |
Your ideal is maximal when $A=B(H)$. Let $J={T: Th=0}$. Let $Sin B(H)setminus J$. Then $Shne0$. Choose $k$ with $langle Sh,krangle=1$, and let $Wx=langle x,krangle,h$. Put $R=I-WS$. Then
$$ Rh=h-WSh=h-h=0, $$ so $Rin J$. Then $$ I=R+WSin J+mathbb C,WS,$$ and $J$ is maximal (its codimension is 1).
answered Dec 10 '18 at 3:40
Martin Argerami
124k1176174
But $J$ is not an ideal.
– C.Ding
Dec 10 '18 at 3:59
I might be missing something. In what sense is $J $ not a left ideal?
– Martin Argerami
Dec 10 '18 at 4:14
The OP needs an ideal, not just an left ideal.
– C.Ding
Dec 10 '18 at 4:19
Separating points by maximal ideals in $B (H) $? And, did you read the title of the question?
– Martin Argerami
Dec 10 '18 at 4:21
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$forall aneq 0$, there is a pure state $tau$ on $A$ such that $tau(a^*a)=|a^*a|neq 0$(Theorem 5.1.11, [1]). Let $N_tau={a|tau(a^*a)=0},$ then $N_tau$ is a maximal left ideal(Theorem 5.3.5, [1]), and $anotin N_tau$.
[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.
answered Dec 10 '18 at 4:37
C.Ding
1,3931321
add a comment |
$forall aneq 0$, there is a pure state $tau$ on $A$ such that $tau(a^*a)=|a^*a|neq 0$(Theorem 5.1.11, [1]). Let $N_tau={a|tau(a^*a)=0},$ then $N_tau$ is a maximal left ideal(Theorem 5.3.5, [1]), and $anotin N_tau$.
[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.
answered Dec 10 '18 at 4:37
C.Ding
1,3931321
add a comment |
$forall aneq 0$, there is a pure state $tau$ on $A$ such that $tau(a^*a)=|a^*a|neq 0$(Theorem 5.1.11, [1]). Let $N_tau={a|tau(a^*a)=0},$ then $N_tau$ is a maximal left ideal(Theorem 5.3.5, [1]), and $anotin N_tau$.
[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.
answered Dec 10 '18 at 4:37
C.Ding
1,3931321
$forall aneq 0$, there is a pure state $tau$ on $A$ such that $tau(a^*a)=|a^*a|neq 0$(Theorem 5.1.11, [1]). Let $N_tau={a|tau(a^*a)=0},$ then $N_tau$ is a maximal left ideal(Theorem 5.3.5, [1]), and $anotin N_tau$.
[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.
answered Dec 10 '18 at 4:37
C.Ding
1,3931321
answered Dec 10 '18 at 4:37
C.Ding
1,3931321
answered Dec 10 '18 at 4:37
C.Ding
1,3931321
answered Dec 10 '18 at 4:37
C.Ding
1,3931321
1,3931321
add a comment |
add a comment |
Your ideal is maximal when $A=B(H)$. Let $J={T: Th=0}$. Let $Sin B(H)setminus J$. Then $Shne0$. Choose $k$ with $langle Sh,krangle=1$, and let $Wx=langle x,krangle,h$. Put $R=I-WS$. Then
$$ Rh=h-WSh=h-h=0, $$ so $Rin J$. Then $$ I=R+WSin J+mathbb C,WS,$$ and $J$ is maximal (its codimension is 1).
answered Dec 10 '18 at 3:40
Martin Argerami
124k1176174
But $J$ is not an ideal.
– C.Ding
Dec 10 '18 at 3:59
I might be missing something. In what sense is $J $ not a left ideal?
– Martin Argerami
Dec 10 '18 at 4:14
The OP needs an ideal, not just an left ideal.
– C.Ding
Dec 10 '18 at 4:19
Separating points by maximal ideals in $B (H) $? And, did you read the title of the question?
– Martin Argerami
Dec 10 '18 at 4:21
add a comment |
Your ideal is maximal when $A=B(H)$. Let $J={T: Th=0}$. Let $Sin B(H)setminus J$. Then $Shne0$. Choose $k$ with $langle Sh,krangle=1$, and let $Wx=langle x,krangle,h$. Put $R=I-WS$. Then
$$ Rh=h-WSh=h-h=0, $$ so $Rin J$. Then $$ I=R+WSin J+mathbb C,WS,$$ and $J$ is maximal (its codimension is 1).
answered Dec 10 '18 at 3:40
Martin Argerami
124k1176174
But $J$ is not an ideal.
– C.Ding
Dec 10 '18 at 3:59
I might be missing something. In what sense is $J $ not a left ideal?
– Martin Argerami
Dec 10 '18 at 4:14
The OP needs an ideal, not just an left ideal.
– C.Ding
Dec 10 '18 at 4:19
Separating points by maximal ideals in $B (H) $? And, did you read the title of the question?
– Martin Argerami
Dec 10 '18 at 4:21
add a comment |
Your ideal is maximal when $A=B(H)$. Let $J={T: Th=0}$. Let $Sin B(H)setminus J$. Then $Shne0$. Choose $k$ with $langle Sh,krangle=1$, and let $Wx=langle x,krangle,h$. Put $R=I-WS$. Then
$$ Rh=h-WSh=h-h=0, $$ so $Rin J$. Then $$ I=R+WSin J+mathbb C,WS,$$ and $J$ is maximal (its codimension is 1).
answered Dec 10 '18 at 3:40
Martin Argerami
124k1176174
Your ideal is maximal when $A=B(H)$. Let $J={T: Th=0}$. Let $Sin B(H)setminus J$. Then $Shne0$. Choose $k$ with $langle Sh,krangle=1$, and let $Wx=langle x,krangle,h$. Put $R=I-WS$. Then
$$ Rh=h-WSh=h-h=0, $$ so $Rin J$. Then $$ I=R+WSin J+mathbb C,WS,$$ and $J$ is maximal (its codimension is 1).
answered Dec 10 '18 at 3:40
Martin Argerami
124k1176174
answered Dec 10 '18 at 3:40
Martin Argerami
124k1176174
answered Dec 10 '18 at 3:40
Martin Argerami
124k1176174
answered Dec 10 '18 at 3:40
Martin Argerami
124k1176174
124k1176174
But $J$ is not an ideal.
– C.Ding
Dec 10 '18 at 3:59
I might be missing something. In what sense is $J $ not a left ideal?
– Martin Argerami
Dec 10 '18 at 4:14
The OP needs an ideal, not just an left ideal.
– C.Ding
Dec 10 '18 at 4:19
Separating points by maximal ideals in $B (H) $? And, did you read the title of the question?
– Martin Argerami
Dec 10 '18 at 4:21
add a comment |
But $J$ is not an ideal.
– C.Ding
Dec 10 '18 at 3:59
I might be missing something. In what sense is $J $ not a left ideal?
– Martin Argerami
Dec 10 '18 at 4:14
The OP needs an ideal, not just an left ideal.
– C.Ding
Dec 10 '18 at 4:19
Separating points by maximal ideals in $B (H) $? And, did you read the title of the question?
– Martin Argerami
Dec 10 '18 at 4:21
But $J$ is not an ideal.
– C.Ding
Dec 10 '18 at 3:59
But $J$ is not an ideal.
– C.Ding
Dec 10 '18 at 3:59
I might be missing something. In what sense is $J $ not a left ideal?
– Martin Argerami
Dec 10 '18 at 4:14
I might be missing something. In what sense is $J $ not a left ideal?
– Martin Argerami
Dec 10 '18 at 4:14
The OP needs an ideal, not just an left ideal.
– C.Ding
Dec 10 '18 at 4:19
The OP needs an ideal, not just an left ideal.
– C.Ding
Dec 10 '18 at 4:19
Separating points by maximal ideals in $B (H) $? And, did you read the title of the question?
– Martin Argerami
Dec 10 '18 at 4:21
Separating points by maximal ideals in $B (H) $? And, did you read the title of the question?
– Martin Argerami
Dec 10 '18 at 4:21
add a comment |
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