UHF-algebra tracial state is faithful?
$begingroup$
It is known that a unital UHF-algebra has a unique tracial state, it is true that it is true that this trace is normal and faithful? I am particularly interested in the universal UHF-algebra, i.e. the one with $K_0$ group isomorphic to $mathbb{Q} cap [0,1]$.
I been looking for this in the literature without any success. On the other hand it is clear that if we restrict the trace to any finite factor it is faithful there.
operator-algebras c-star-algebras trace
edited Jan 2 at 17:28
Martin Argerami
126k1182180
asked Jan 2 at 17:15
k76u4vkweek547v7k76u4vkweek547v7
463317
$endgroup$
add a comment |
$begingroup$
It is known that a unital UHF-algebra has a unique tracial state, it is true that it is true that this trace is normal and faithful? I am particularly interested in the universal UHF-algebra, i.e. the one with $K_0$ group isomorphic to $mathbb{Q} cap [0,1]$.
I been looking for this in the literature without any success. On the other hand it is clear that if we restrict the trace to any finite factor it is faithful there.
operator-algebras c-star-algebras trace
edited Jan 2 at 17:28
Martin Argerami
126k1182180
asked Jan 2 at 17:15
k76u4vkweek547v7k76u4vkweek547v7
463317
$endgroup$
add a comment |
$begingroup$
It is known that a unital UHF-algebra has a unique tracial state, it is true that it is true that this trace is normal and faithful? I am particularly interested in the universal UHF-algebra, i.e. the one with $K_0$ group isomorphic to $mathbb{Q} cap [0,1]$.
I been looking for this in the literature without any success. On the other hand it is clear that if we restrict the trace to any finite factor it is faithful there.
operator-algebras c-star-algebras trace
edited Jan 2 at 17:28
Martin Argerami
126k1182180
asked Jan 2 at 17:15
k76u4vkweek547v7k76u4vkweek547v7
463317
$endgroup$
It is known that a unital UHF-algebra has a unique tracial state, it is true that it is true that this trace is normal and faithful? I am particularly interested in the universal UHF-algebra, i.e. the one with $K_0$ group isomorphic to $mathbb{Q} cap [0,1]$.
I been looking for this in the literature without any success. On the other hand it is clear that if we restrict the trace to any finite factor it is faithful there.
operator-algebras c-star-algebras trace
operator-algebras c-star-algebras trace
edited Jan 2 at 17:28
Martin Argerami
126k1182180
asked Jan 2 at 17:15
k76u4vkweek547v7k76u4vkweek547v7
463317
edited Jan 2 at 17:28
Martin Argerami
126k1182180
asked Jan 2 at 17:15
k76u4vkweek547v7k76u4vkweek547v7
463317
edited Jan 2 at 17:28
Martin Argerami
126k1182180
edited Jan 2 at 17:28
Martin Argerami
126k1182180
edited Jan 2 at 17:28
Martin Argerami
126k1182180
126k1182180
asked Jan 2 at 17:15
k76u4vkweek547v7k76u4vkweek547v7
463317
asked Jan 2 at 17:15
k76u4vkweek547v7k76u4vkweek547v7
463317
asked Jan 2 at 17:15
k76u4vkweek547v7k76u4vkweek547v7
463317
463317
add a comment |
add a comment |
1 Answer
1
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oldest
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The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.
answered Jan 2 at 21:18
André S.André S.
2,126314
$endgroup$
$begingroup$
I just realized that for a trace the left ideal of elements with zero norm trace is also a right ideal. Thanks!
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:21
1
$begingroup$
Also, "normal" is not really defined for the tracial state on the UHF-algebra, but of course the extension to the double dual will be normal.
$endgroup$
– André S.
Jan 2 at 21:22
$begingroup$
When you say double dual, this is equivalent to taking the double commutant of the GNS representation (which in this case is an irreducible faithful representation)?
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:25
1
$begingroup$
Yes exactly, the GNS coming from the unique trace.
$endgroup$
– André S.
Jan 2 at 21:46
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.
answered Jan 2 at 21:18
André S.André S.
2,126314
$endgroup$
$begingroup$
I just realized that for a trace the left ideal of elements with zero norm trace is also a right ideal. Thanks!
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:21
1
$begingroup$
Also, "normal" is not really defined for the tracial state on the UHF-algebra, but of course the extension to the double dual will be normal.
$endgroup$
– André S.
Jan 2 at 21:22
$begingroup$
When you say double dual, this is equivalent to taking the double commutant of the GNS representation (which in this case is an irreducible faithful representation)?
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:25
1
$begingroup$
Yes exactly, the GNS coming from the unique trace.
$endgroup$
– André S.
Jan 2 at 21:46
add a comment |
$begingroup$
The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.
answered Jan 2 at 21:18
André S.André S.
2,126314
$endgroup$
$begingroup$
I just realized that for a trace the left ideal of elements with zero norm trace is also a right ideal. Thanks!
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:21
1
$begingroup$
Also, "normal" is not really defined for the tracial state on the UHF-algebra, but of course the extension to the double dual will be normal.
$endgroup$
– André S.
Jan 2 at 21:22
$begingroup$
When you say double dual, this is equivalent to taking the double commutant of the GNS representation (which in this case is an irreducible faithful representation)?
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:25
1
$begingroup$
Yes exactly, the GNS coming from the unique trace.
$endgroup$
– André S.
Jan 2 at 21:46
add a comment |
$begingroup$
The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.
answered Jan 2 at 21:18
André S.André S.
2,126314
$endgroup$
The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.
answered Jan 2 at 21:18
André S.André S.
2,126314
answered Jan 2 at 21:18
André S.André S.
2,126314
answered Jan 2 at 21:18
André S.André S.
2,126314
answered Jan 2 at 21:18
André S.André S.
2,126314
2,126314
$begingroup$
I just realized that for a trace the left ideal of elements with zero norm trace is also a right ideal. Thanks!
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:21
1
$begingroup$
Also, "normal" is not really defined for the tracial state on the UHF-algebra, but of course the extension to the double dual will be normal.
$endgroup$
– André S.
Jan 2 at 21:22
$begingroup$
When you say double dual, this is equivalent to taking the double commutant of the GNS representation (which in this case is an irreducible faithful representation)?
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:25
1
$begingroup$
Yes exactly, the GNS coming from the unique trace.
$endgroup$
– André S.
Jan 2 at 21:46
add a comment |
$begingroup$
I just realized that for a trace the left ideal of elements with zero norm trace is also a right ideal. Thanks!
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:21
1
$begingroup$
Also, "normal" is not really defined for the tracial state on the UHF-algebra, but of course the extension to the double dual will be normal.
$endgroup$
– André S.
Jan 2 at 21:22
$begingroup$
When you say double dual, this is equivalent to taking the double commutant of the GNS representation (which in this case is an irreducible faithful representation)?
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:25
1
$begingroup$
Yes exactly, the GNS coming from the unique trace.
$endgroup$
– André S.
Jan 2 at 21:46
$begingroup$
I just realized that for a trace the left ideal of elements with zero norm trace is also a right ideal. Thanks!
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:21
$begingroup$
I just realized that for a trace the left ideal of elements with zero norm trace is also a right ideal. Thanks!
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:21
1
1
$begingroup$
Also, "normal" is not really defined for the tracial state on the UHF-algebra, but of course the extension to the double dual will be normal.
$endgroup$
– André S.
Jan 2 at 21:22
$begingroup$
Also, "normal" is not really defined for the tracial state on the UHF-algebra, but of course the extension to the double dual will be normal.
$endgroup$
– André S.
Jan 2 at 21:22
$begingroup$
When you say double dual, this is equivalent to taking the double commutant of the GNS representation (which in this case is an irreducible faithful representation)?
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:25
$begingroup$
When you say double dual, this is equivalent to taking the double commutant of the GNS representation (which in this case is an irreducible faithful representation)?
$endgroup$
– k76u4vkweek547v7
Jan 2 at 21:25
1
1
$begingroup$
Yes exactly, the GNS coming from the unique trace.
$endgroup$
– André S.
Jan 2 at 21:46
$begingroup$
Yes exactly, the GNS coming from the unique trace.
$endgroup$
– André S.
Jan 2 at 21:46
add a comment |
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