Mixing
Projections correspond to double duals of $C(X)$-algebras fibers
Let $A$ be a $C(X)$-algebra ($X$ compact).
For $xneq y$ in $X$, we have the Glimm ideals in $A$:
$I=C_0(Xsetminus {x})A$ and $J=C_0(Xsetminus {y})A$.
The fibers are denoted by $A_x=A/I$ and $A_y=A/J$.
Choose central projections $p,qin A^{**}$ satisfying:
$I^{**}=pA^{**}$ and $J^{**}=qA^{**}$.
It follows that $A_x^{**}=(1-p)A^{**}$ and $A_y^{**}= (1-q)A^{**}$.
I want to show that $1-p$ and $1-q$ are orthogonal projections.
Probably one has to look on representation of $A$ that vanish on the distinct Glimm ideals $I$ and $J$ and to extend them to representations on the double duals.
I would be happy for any help!
Thanks.
functional-analysis operator-algebras c-star-algebras von-neumann-algebras
asked Nov 21 '18 at 9:48
Shirly Geffen
1,326614
add a comment |
Let $A$ be a $C(X)$-algebra ($X$ compact).
For $xneq y$ in $X$, we have the Glimm ideals in $A$:
$I=C_0(Xsetminus {x})A$ and $J=C_0(Xsetminus {y})A$.
The fibers are denoted by $A_x=A/I$ and $A_y=A/J$.
Choose central projections $p,qin A^{**}$ satisfying:
$I^{**}=pA^{**}$ and $J^{**}=qA^{**}$.
It follows that $A_x^{**}=(1-p)A^{**}$ and $A_y^{**}= (1-q)A^{**}$.
I want to show that $1-p$ and $1-q$ are orthogonal projections.
Probably one has to look on representation of $A$ that vanish on the distinct Glimm ideals $I$ and $J$ and to extend them to representations on the double duals.
I would be happy for any help!
Thanks.
functional-analysis operator-algebras c-star-algebras von-neumann-algebras
asked Nov 21 '18 at 9:48
Shirly Geffen
1,326614
What do you mean by a $C(X)$-algebra? A Banach algebra that is also $C(X)$-bimodular and for which the multiplication is $C(X)$-bimodular?
– Adrián González-Pérez
Nov 22 '18 at 14:16
2
@Adrián: yes and no. There is a "twist". A $C(X)$-algebra is a C$^*$-algebra $A$ such that there is a unital $*$-homomorphism $rho:C(X)to Z(M(A))$. It is a standard notion in the world of C$^*$-algebras.
– Martin Argerami
Nov 22 '18 at 18:02
add a comment |
Let $A$ be a $C(X)$-algebra ($X$ compact).
For $xneq y$ in $X$, we have the Glimm ideals in $A$:
$I=C_0(Xsetminus {x})A$ and $J=C_0(Xsetminus {y})A$.
The fibers are denoted by $A_x=A/I$ and $A_y=A/J$.
Choose central projections $p,qin A^{**}$ satisfying:
$I^{**}=pA^{**}$ and $J^{**}=qA^{**}$.
It follows that $A_x^{**}=(1-p)A^{**}$ and $A_y^{**}= (1-q)A^{**}$.
I want to show that $1-p$ and $1-q$ are orthogonal projections.
Probably one has to look on representation of $A$ that vanish on the distinct Glimm ideals $I$ and $J$ and to extend them to representations on the double duals.
I would be happy for any help!
Thanks.
functional-analysis operator-algebras c-star-algebras von-neumann-algebras
asked Nov 21 '18 at 9:48
Shirly Geffen
1,326614
Let $A$ be a $C(X)$-algebra ($X$ compact).
For $xneq y$ in $X$, we have the Glimm ideals in $A$:
$I=C_0(Xsetminus {x})A$ and $J=C_0(Xsetminus {y})A$.
The fibers are denoted by $A_x=A/I$ and $A_y=A/J$.
Choose central projections $p,qin A^{**}$ satisfying:
$I^{**}=pA^{**}$ and $J^{**}=qA^{**}$.
It follows that $A_x^{**}=(1-p)A^{**}$ and $A_y^{**}= (1-q)A^{**}$.
I want to show that $1-p$ and $1-q$ are orthogonal projections.
Probably one has to look on representation of $A$ that vanish on the distinct Glimm ideals $I$ and $J$ and to extend them to representations on the double duals.
I would be happy for any help!
Thanks.
functional-analysis operator-algebras c-star-algebras von-neumann-algebras
functional-analysis operator-algebras c-star-algebras von-neumann-algebras
asked Nov 21 '18 at 9:48
Shirly Geffen
1,326614
asked Nov 21 '18 at 9:48
Shirly Geffen
1,326614
asked Nov 21 '18 at 9:48
Shirly Geffen
1,326614
asked Nov 21 '18 at 9:48
Shirly Geffen
1,326614
asked Nov 21 '18 at 9:48
Shirly Geffen
1,326614
1,326614
What do you mean by a $C(X)$-algebra? A Banach algebra that is also $C(X)$-bimodular and for which the multiplication is $C(X)$-bimodular?
– Adrián González-Pérez
Nov 22 '18 at 14:16
2
@Adrián: yes and no. There is a "twist". A $C(X)$-algebra is a C$^*$-algebra $A$ such that there is a unital $*$-homomorphism $rho:C(X)to Z(M(A))$. It is a standard notion in the world of C$^*$-algebras.
– Martin Argerami
Nov 22 '18 at 18:02
add a comment |
What do you mean by a $C(X)$-algebra? A Banach algebra that is also $C(X)$-bimodular and for which the multiplication is $C(X)$-bimodular?
– Adrián González-Pérez
Nov 22 '18 at 14:16
2
@Adrián: yes and no. There is a "twist". A $C(X)$-algebra is a C$^*$-algebra $A$ such that there is a unital $*$-homomorphism $rho:C(X)to Z(M(A))$. It is a standard notion in the world of C$^*$-algebras.
– Martin Argerami
Nov 22 '18 at 18:02
What do you mean by a $C(X)$-algebra? A Banach algebra that is also $C(X)$-bimodular and for which the multiplication is $C(X)$-bimodular?
– Adrián González-Pérez
Nov 22 '18 at 14:16
What do you mean by a $C(X)$-algebra? A Banach algebra that is also $C(X)$-bimodular and for which the multiplication is $C(X)$-bimodular?
– Adrián González-Pérez
Nov 22 '18 at 14:16
2
2
@Adrián: yes and no. There is a "twist". A $C(X)$-algebra is a C$^*$-algebra $A$ such that there is a unital $*$-homomorphism $rho:C(X)to Z(M(A))$. It is a standard notion in the world of C$^*$-algebras.
– Martin Argerami
Nov 22 '18 at 18:02
@Adrián: yes and no. There is a "twist". A $C(X)$-algebra is a C$^*$-algebra $A$ such that there is a unital $*$-homomorphism $rho:C(X)to Z(M(A))$. It is a standard notion in the world of C$^*$-algebras.
– Martin Argerami
Nov 22 '18 at 18:02
add a comment |
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What do you mean by a $C(X)$-algebra? A Banach algebra that is also $C(X)$-bimodular and for which the multiplication is $C(X)$-bimodular?
– Adrián González-Pérez
Nov 22 '18 at 14:16
2
@Adrián: yes and no. There is a "twist". A $C(X)$-algebra is a C$^*$-algebra $A$ such that there is a unital $*$-homomorphism $rho:C(X)to Z(M(A))$. It is a standard notion in the world of C$^*$-algebras.
– Martin Argerami
Nov 22 '18 at 18:02