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YCor
YCor
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This is a comment on the definition of being "finitely generated".

There is a difference between algebraically finitely generated and topological finitely generated Hilbert $C^*$-modules. Consider $C_0((0,1])$ - the $C^*$-algebra of all continuous functions on the left-open unit interval with the usual metric topology. It is a Hilbert $C([0,1])$-module as a maximal ideal of that $C^*$-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in $C([0,1])$, in a modular sense the generating set $\{x(t)=t\}$ suffices to generate $C_0((0,1])$ as a $C([0,1])$-module topologically. However, $C_0((0,1])$ is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert $C^*$-modules among countably generated ones.

There is a difference between algebraically finitely generated and topological finitely generated Hilbert $C^*$-modules. Consider $C_0((0,1])$ - the $C^*$-algebra of all continuous functions on the left-open unit interval with the usual metric topology. It is a Hilbert $C([0,1])$-module as a maximal ideal of that $C^*$-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in $C([0,1])$, in a modular sense the generating set $\{x(t)=t\}$ suffices to generate $C_0((0,1])$ as a $C([0,1])$-module topologically. However, $C_0((0,1])$ is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert $C^*$-modules among countably generated ones.

This is a comment on the definition of being "finitely generated".

There is a difference between algebraically finitely generated and topological finitely generated Hilbert $C^*$-modules. Consider $C_0((0,1])$ - the $C^*$-algebra of all continuous functions on the left-open unit interval with the usual metric topology. It is a Hilbert $C([0,1])$-module as a maximal ideal of that $C^*$-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in $C([0,1])$, in a modular sense the generating set $\{x(t)=t\}$ suffices to generate $C_0((0,1])$ as a $C([0,1])$-module topologically. However, $C_0((0,1])$ is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert $C^*$-modules among countably generated ones.

Post Undeleted by Stefan Kohl♦
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David Roberts ♦
David Roberts ♦
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There is a difference between algebraically finitely generated and topological finitely generated Hilbert C-modules. Consider C_0((0,1]) - the C$C^*$-modules. Consider $C_0((0,1])$ - the $C^*$-algebra of all continuous functions on the left-open unit interval with the usual metric topology. It is a Hilbert C([0,1])$C([0,1])$-module as a maximal ideal of that C-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in C([0,1]), in a modular sense the generating set {x(t)=t} suffices to generated C_0((0,1]) as a C([0,1])-module topologically. However, C_0((0,1]) is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert C$C^*$-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in $C([0,1])$, in a modular sense the generating set $\{x(t)=t\}$ suffices to generate $C_0((0,1])$ as a $C([0,1])$-module topologically. However, $C_0((0,1])$ is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert $C^*$-modules among countably generated ones.

There is a difference between algebraically finitely generated and topological finitely generated Hilbert C-modules. Consider C_0((0,1]) - the C-algebra of all continuous functions on the left-open unit interval with the usual metric topology. It is a Hilbert C([0,1])-module as a maximal ideal of that C-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in C([0,1]), in a modular sense the generating set {x(t)=t} suffices to generated C_0((0,1]) as a C([0,1])-module topologically. However, C_0((0,1]) is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert C-modules among countably generated ones.

There is a difference between algebraically finitely generated and topological finitely generated Hilbert $C^*$-modules. Consider $C_0((0,1])$ - the $C^*$-algebra of all continuous functions on the left-open unit interval with the usual metric topology. It is a Hilbert $C([0,1])$-module as a maximal ideal of that $C^*$-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in $C([0,1])$, in a modular sense the generating set $\{x(t)=t\}$ suffices to generate $C_0((0,1])$ as a $C([0,1])$-module topologically. However, $C_0((0,1])$ is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert $C^*$-modules among countably generated ones.

Post Deleted by Stefan Kohl♦
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Michael Frank
Michael Frank
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There is a difference between algebraically finitely generated and topological finitely generated Hilbert C-modules. Consider C_0((0,1]) - the C-algebra of all continuous functions on the left-open unit interval with the usual metric topology. It is a Hilbert C([0,1])-module as a maximal ideal of that C-algebra. By the Theorem of Weierstrass the finite polynomials in one variable are norm-dense in C([0,1]), in a modular sense the generating set {x(t)=t} suffices to generated C_0((0,1]) as a C([0,1])-module topologically. However, C_0((0,1]) is not projective and not algebraically finitely generated. Kasparov subsummized this kind of Hilbert C-modules among countably generated ones.