UHF-algebra tracial state is faithful?












2












$begingroup$


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.










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$endgroup$

















    2












    $begingroup$


    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.










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      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.










      share|cite|improve this question











      $endgroup$




      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






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      share|cite|improve this question













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      edited Jan 2 at 17:28









      Martin Argerami

      126k1182180




      126k1182180










      asked Jan 2 at 17:15









      k76u4vkweek547v7k76u4vkweek547v7

      463317




      463317






















          1 Answer
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          $begingroup$

          The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











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          1 Answer
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          active

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          1












          $begingroup$

          The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.






          share|cite|improve this answer









          $endgroup$



          The universal UHF algebra has $K_0$-group $mathbb Q$ and the unique tracial state is faithful because the algebra is simple.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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