pairwise orthogonal projections in an inseparable $C^* $ algebra
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If $A$ is a separable $C^*$ algebra,then there are at most countable pairwise orthogonal projections.If $A$ is inseparable,how many pairwise orthogonal projections in $A$? If it has, is it uncountable?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Nov 13 at 14:07
mathrookie
697512
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If $A$ is a separable $C^*$ algebra,then there are at most countable pairwise orthogonal projections.If $A$ is inseparable,how many pairwise orthogonal projections in $A$? If it has, is it uncountable?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Nov 13 at 14:07
mathrookie
697512
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $A$ is a separable $C^*$ algebra,then there are at most countable pairwise orthogonal projections.If $A$ is inseparable,how many pairwise orthogonal projections in $A$? If it has, is it uncountable?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Nov 13 at 14:07
mathrookie
697512
If $A$ is a separable $C^*$ algebra,then there are at most countable pairwise orthogonal projections.If $A$ is inseparable,how many pairwise orthogonal projections in $A$? If it has, is it uncountable?
functional-analysis operator-theory operator-algebras c-star-algebras
functional-analysis operator-theory operator-algebras c-star-algebras
asked Nov 13 at 14:07
mathrookie
697512
asked Nov 13 at 14:07
mathrookie
697512
edited Nov 13 at 14:14
asked Nov 13 at 14:07
mathrookie
697512
asked Nov 13 at 14:07
mathrookie
697512
asked Nov 13 at 14:07
mathrookie
697512
697512
add a comment |
add a comment |
1 Answer
1
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2
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accepted
It may have zero. For instance, take
$$
A=prod_{tin[0,1]} C_0(mathbb R).
$$
With the norm $|a|=sup{|a_t|: tin[0,1]}$, this is a non-separable C$^*$-algebra. And it has no projections other that $0$, since any projection has to be a product of projections and $C_0(mathbb R)$ has no nonzero projections.
As s.harp mentions, any family of pairwise orthogonal projections will have cardinality less than or equal that of $A$.
answered Nov 13 at 17:52
Martin Argerami
121k1073172
Also: the upper bound should be given by the cardinality of $A$.
– s.harp
Nov 13 at 18:14
Indeed. I edited that in.
– Martin Argerami
Nov 13 at 18:16
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It may have zero. For instance, take
$$
A=prod_{tin[0,1]} C_0(mathbb R).
$$
With the norm $|a|=sup{|a_t|: tin[0,1]}$, this is a non-separable C$^*$-algebra. And it has no projections other that $0$, since any projection has to be a product of projections and $C_0(mathbb R)$ has no nonzero projections.
As s.harp mentions, any family of pairwise orthogonal projections will have cardinality less than or equal that of $A$.
answered Nov 13 at 17:52
Martin Argerami
121k1073172
Also: the upper bound should be given by the cardinality of $A$.
– s.harp
Nov 13 at 18:14
Indeed. I edited that in.
– Martin Argerami
Nov 13 at 18:16
add a comment |
up vote
2
down vote
accepted
It may have zero. For instance, take
$$
A=prod_{tin[0,1]} C_0(mathbb R).
$$
With the norm $|a|=sup{|a_t|: tin[0,1]}$, this is a non-separable C$^*$-algebra. And it has no projections other that $0$, since any projection has to be a product of projections and $C_0(mathbb R)$ has no nonzero projections.
As s.harp mentions, any family of pairwise orthogonal projections will have cardinality less than or equal that of $A$.
answered Nov 13 at 17:52
Martin Argerami
121k1073172
Also: the upper bound should be given by the cardinality of $A$.
– s.harp
Nov 13 at 18:14
Indeed. I edited that in.
– Martin Argerami
Nov 13 at 18:16
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It may have zero. For instance, take
$$
A=prod_{tin[0,1]} C_0(mathbb R).
$$
With the norm $|a|=sup{|a_t|: tin[0,1]}$, this is a non-separable C$^*$-algebra. And it has no projections other that $0$, since any projection has to be a product of projections and $C_0(mathbb R)$ has no nonzero projections.
As s.harp mentions, any family of pairwise orthogonal projections will have cardinality less than or equal that of $A$.
answered Nov 13 at 17:52
Martin Argerami
121k1073172
It may have zero. For instance, take
$$
A=prod_{tin[0,1]} C_0(mathbb R).
$$
With the norm $|a|=sup{|a_t|: tin[0,1]}$, this is a non-separable C$^*$-algebra. And it has no projections other that $0$, since any projection has to be a product of projections and $C_0(mathbb R)$ has no nonzero projections.
As s.harp mentions, any family of pairwise orthogonal projections will have cardinality less than or equal that of $A$.
answered Nov 13 at 17:52
Martin Argerami
121k1073172
edited Nov 13 at 18:16
answered Nov 13 at 17:52
Martin Argerami
121k1073172
answered Nov 13 at 17:52
Martin Argerami
121k1073172
answered Nov 13 at 17:52
Martin Argerami
121k1073172
121k1073172
Also: the upper bound should be given by the cardinality of $A$.
– s.harp
Nov 13 at 18:14
Indeed. I edited that in.
– Martin Argerami
Nov 13 at 18:16
add a comment |
Also: the upper bound should be given by the cardinality of $A$.
– s.harp
Nov 13 at 18:14
Indeed. I edited that in.
– Martin Argerami
Nov 13 at 18:16
Also: the upper bound should be given by the cardinality of $A$.
– s.harp
Nov 13 at 18:14
Also: the upper bound should be given by the cardinality of $A$.
– s.harp
Nov 13 at 18:14
Indeed. I edited that in.
– Martin Argerami
Nov 13 at 18:16
Indeed. I edited that in.
– Martin Argerami
Nov 13 at 18:16
add a comment |
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