finite dimensional $C^*$ subalgebra
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Given any $C^*$-algebra $A$ (not necessarily unital),can we construct a nonzero finite dimensional $C^*$-subalgebra of $A$?
operator-theory operator-algebras c-star-algebras
asked Dec 31 '18 at 18:38
mathrookiemathrookie
918512
$endgroup$
add a comment |
$begingroup$
Given any $C^*$-algebra $A$ (not necessarily unital),can we construct a nonzero finite dimensional $C^*$-subalgebra of $A$?
operator-theory operator-algebras c-star-algebras
asked Dec 31 '18 at 18:38
mathrookiemathrookie
918512
$endgroup$
add a comment |
$begingroup$
Given any $C^*$-algebra $A$ (not necessarily unital),can we construct a nonzero finite dimensional $C^*$-subalgebra of $A$?
operator-theory operator-algebras c-star-algebras
asked Dec 31 '18 at 18:38
mathrookiemathrookie
918512
$endgroup$
Given any $C^*$-algebra $A$ (not necessarily unital),can we construct a nonzero finite dimensional $C^*$-subalgebra of $A$?
operator-theory operator-algebras c-star-algebras
operator-theory operator-algebras c-star-algebras
asked Dec 31 '18 at 18:38
mathrookiemathrookie
918512
asked Dec 31 '18 at 18:38
mathrookiemathrookie
918512
asked Dec 31 '18 at 18:38
mathrookiemathrookie
918512
asked Dec 31 '18 at 18:38
mathrookiemathrookie
918512
asked Dec 31 '18 at 18:38
mathrookiemathrookie
918512
918512
add a comment |
add a comment |
1 Answer
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No. Every finite dimensional $C^*$-algebra is a von Neumann algebra, because it is a Banach dual space. $W^*$ algebras contain many projections -- in particular the range projections of their elements -- but there are $C^*$-algebras with no projections.
answered Dec 31 '18 at 21:02
Ashwin TrisalAshwin Trisal
1,2891516
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1 Answer
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1 Answer
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active
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$begingroup$
No. Every finite dimensional $C^*$-algebra is a von Neumann algebra, because it is a Banach dual space. $W^*$ algebras contain many projections -- in particular the range projections of their elements -- but there are $C^*$-algebras with no projections.
answered Dec 31 '18 at 21:02
Ashwin TrisalAshwin Trisal
1,2891516
$endgroup$
add a comment |
$begingroup$
No. Every finite dimensional $C^*$-algebra is a von Neumann algebra, because it is a Banach dual space. $W^*$ algebras contain many projections -- in particular the range projections of their elements -- but there are $C^*$-algebras with no projections.
answered Dec 31 '18 at 21:02
Ashwin TrisalAshwin Trisal
1,2891516
$endgroup$
add a comment |
$begingroup$
No. Every finite dimensional $C^*$-algebra is a von Neumann algebra, because it is a Banach dual space. $W^*$ algebras contain many projections -- in particular the range projections of their elements -- but there are $C^*$-algebras with no projections.
answered Dec 31 '18 at 21:02
Ashwin TrisalAshwin Trisal
1,2891516
$endgroup$
No. Every finite dimensional $C^*$-algebra is a von Neumann algebra, because it is a Banach dual space. $W^*$ algebras contain many projections -- in particular the range projections of their elements -- but there are $C^*$-algebras with no projections.
answered Dec 31 '18 at 21:02
Ashwin TrisalAshwin Trisal
1,2891516
answered Dec 31 '18 at 21:02
Ashwin TrisalAshwin Trisal
1,2891516
answered Dec 31 '18 at 21:02
Ashwin TrisalAshwin Trisal
1,2891516
answered Dec 31 '18 at 21:02
Ashwin TrisalAshwin Trisal
1,2891516
1,2891516
add a comment |
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