spectrum of $C^*$ algebras












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When $A=bigoplus B(Bbb C^n)$ ($c_0$ direct sum),how to compute the spectrum of $A$ ?What about the conclusion If we replace the $ell ^infty $ direct sum with $c_0$ direct sum?










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    $begingroup$


    When $A=bigoplus B(Bbb C^n)$ ($c_0$ direct sum),how to compute the spectrum of $A$ ?What about the conclusion If we replace the $ell ^infty $ direct sum with $c_0$ direct sum?










    share|cite|improve this question









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      0





      $begingroup$


      When $A=bigoplus B(Bbb C^n)$ ($c_0$ direct sum),how to compute the spectrum of $A$ ?What about the conclusion If we replace the $ell ^infty $ direct sum with $c_0$ direct sum?










      share|cite|improve this question









      $endgroup$




      When $A=bigoplus B(Bbb C^n)$ ($c_0$ direct sum),how to compute the spectrum of $A$ ?What about the conclusion If we replace the $ell ^infty $ direct sum with $c_0$ direct sum?







      operator-theory operator-algebras c-star-algebras von-neumann-algebras






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      asked Dec 10 '18 at 14:29









      mathrookiemathrookie

      889512




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          $begingroup$

          Let $tau$ be a character of $A$. The composition $taucircpi_n$ where $pi_n$ is the projection onto the $n^{rm th}$ coordinate, is a character of $M_n(mathbb C)$; which has none unless $n=1$.



          So the only character of $A$ is $alongmapsto a_1$, assuming that the first coordinate is one-dimensional.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Pro Argerami,why the character of $M_n(Bbb C)$ has none?
            $endgroup$
            – mathrookie
            Dec 10 '18 at 21:53






          • 1




            $begingroup$
            Because the kernel is an ideal in $M_n(mathbb C)$.
            $endgroup$
            – Martin Argerami
            Dec 10 '18 at 22:25











          Your Answer





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          1 Answer
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          active

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          2












          $begingroup$

          Let $tau$ be a character of $A$. The composition $taucircpi_n$ where $pi_n$ is the projection onto the $n^{rm th}$ coordinate, is a character of $M_n(mathbb C)$; which has none unless $n=1$.



          So the only character of $A$ is $alongmapsto a_1$, assuming that the first coordinate is one-dimensional.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Pro Argerami,why the character of $M_n(Bbb C)$ has none?
            $endgroup$
            – mathrookie
            Dec 10 '18 at 21:53






          • 1




            $begingroup$
            Because the kernel is an ideal in $M_n(mathbb C)$.
            $endgroup$
            – Martin Argerami
            Dec 10 '18 at 22:25
















          2












          $begingroup$

          Let $tau$ be a character of $A$. The composition $taucircpi_n$ where $pi_n$ is the projection onto the $n^{rm th}$ coordinate, is a character of $M_n(mathbb C)$; which has none unless $n=1$.



          So the only character of $A$ is $alongmapsto a_1$, assuming that the first coordinate is one-dimensional.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Pro Argerami,why the character of $M_n(Bbb C)$ has none?
            $endgroup$
            – mathrookie
            Dec 10 '18 at 21:53






          • 1




            $begingroup$
            Because the kernel is an ideal in $M_n(mathbb C)$.
            $endgroup$
            – Martin Argerami
            Dec 10 '18 at 22:25














          2












          2








          2





          $begingroup$

          Let $tau$ be a character of $A$. The composition $taucircpi_n$ where $pi_n$ is the projection onto the $n^{rm th}$ coordinate, is a character of $M_n(mathbb C)$; which has none unless $n=1$.



          So the only character of $A$ is $alongmapsto a_1$, assuming that the first coordinate is one-dimensional.






          share|cite|improve this answer









          $endgroup$



          Let $tau$ be a character of $A$. The composition $taucircpi_n$ where $pi_n$ is the projection onto the $n^{rm th}$ coordinate, is a character of $M_n(mathbb C)$; which has none unless $n=1$.



          So the only character of $A$ is $alongmapsto a_1$, assuming that the first coordinate is one-dimensional.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 21:40









          Martin ArgeramiMartin Argerami

          126k1182181




          126k1182181












          • $begingroup$
            Pro Argerami,why the character of $M_n(Bbb C)$ has none?
            $endgroup$
            – mathrookie
            Dec 10 '18 at 21:53






          • 1




            $begingroup$
            Because the kernel is an ideal in $M_n(mathbb C)$.
            $endgroup$
            – Martin Argerami
            Dec 10 '18 at 22:25


















          • $begingroup$
            Pro Argerami,why the character of $M_n(Bbb C)$ has none?
            $endgroup$
            – mathrookie
            Dec 10 '18 at 21:53






          • 1




            $begingroup$
            Because the kernel is an ideal in $M_n(mathbb C)$.
            $endgroup$
            – Martin Argerami
            Dec 10 '18 at 22:25
















          $begingroup$
          Pro Argerami,why the character of $M_n(Bbb C)$ has none?
          $endgroup$
          – mathrookie
          Dec 10 '18 at 21:53




          $begingroup$
          Pro Argerami,why the character of $M_n(Bbb C)$ has none?
          $endgroup$
          – mathrookie
          Dec 10 '18 at 21:53




          1




          1




          $begingroup$
          Because the kernel is an ideal in $M_n(mathbb C)$.
          $endgroup$
          – Martin Argerami
          Dec 10 '18 at 22:25




          $begingroup$
          Because the kernel is an ideal in $M_n(mathbb C)$.
          $endgroup$
          – Martin Argerami
          Dec 10 '18 at 22:25


















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