spectrum of $C^*$ algebras
$begingroup$
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?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 10 '18 at 14:29
mathrookiemathrookie
911512
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add a comment |
$begingroup$
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?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 10 '18 at 14:29
mathrookiemathrookie
911512
$endgroup$
add a comment |
$begingroup$
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?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 10 '18 at 14:29
mathrookiemathrookie
911512
$endgroup$
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
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 10 '18 at 14:29
mathrookiemathrookie
911512
asked Dec 10 '18 at 14:29
mathrookiemathrookie
911512
asked Dec 10 '18 at 14:29
mathrookiemathrookie
911512
asked Dec 10 '18 at 14:29
mathrookiemathrookie
911512
asked Dec 10 '18 at 14:29
mathrookiemathrookie
911512
911512
add a comment |
add a comment |
1 Answer
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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.
answered Dec 10 '18 at 21:40
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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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.
answered Dec 10 '18 at 21:40
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 10 '18 at 21:40
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 10 '18 at 21:40
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
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.
answered Dec 10 '18 at 21:40
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 10 '18 at 21:40
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 10 '18 at 21:40
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 10 '18 at 21:40
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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