GNS representation of a nuclear $C^*$-algebra The 2019 Stack Overflow Developer Survey Results...
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GNS representation of a nuclear $C^*$-algebra
The 2019 Stack Overflow Developer Survey Results Are InDoes an irreducible operator generate a nuclear $C^{*}$-algebra?Non-existence Tracial statesA proposition about residually finite dimensional C*-algebraGNS construction: Finding a representationGNS construction and separability.GNS representation doubttracial state on a unital infinite dimensional simple $C^*$ algebrasimple nuclear $C^*$ algebraconnection between unital $C^*$ algebra and finite von neumann algebraGNS representation
$begingroup$
Suppose $A$ is a nuclear $C^*$-algebra with a tracial state $psi$, $(pi_{psi},H_{psi})$ is the GNS reprsentation with respect to $psi$.
My question:
Does there exist $A$ which satisfy the above condition and $A/ker(pi_{psi})=K(H)$, where $H$ is separable and infinite dimensional.
Can anyone show me an example? Thanks!
functional-analysis operator-theory operator-algebras c-star-algebras
edited Mar 21 at 17:03
Andrews
1,2812423
asked Mar 21 at 15:46
mathrookiemathrookie
936512
$endgroup$
add a comment |
$begingroup$
Suppose $A$ is a nuclear $C^*$-algebra with a tracial state $psi$, $(pi_{psi},H_{psi})$ is the GNS reprsentation with respect to $psi$.
My question:
Does there exist $A$ which satisfy the above condition and $A/ker(pi_{psi})=K(H)$, where $H$ is separable and infinite dimensional.
Can anyone show me an example? Thanks!
functional-analysis operator-theory operator-algebras c-star-algebras
edited Mar 21 at 17:03
Andrews
1,2812423
asked Mar 21 at 15:46
mathrookiemathrookie
936512
$endgroup$
add a comment |
$begingroup$
Suppose $A$ is a nuclear $C^*$-algebra with a tracial state $psi$, $(pi_{psi},H_{psi})$ is the GNS reprsentation with respect to $psi$.
My question:
Does there exist $A$ which satisfy the above condition and $A/ker(pi_{psi})=K(H)$, where $H$ is separable and infinite dimensional.
Can anyone show me an example? Thanks!
functional-analysis operator-theory operator-algebras c-star-algebras
edited Mar 21 at 17:03
Andrews
1,2812423
asked Mar 21 at 15:46
mathrookiemathrookie
936512
$endgroup$
Suppose $A$ is a nuclear $C^*$-algebra with a tracial state $psi$, $(pi_{psi},H_{psi})$ is the GNS reprsentation with respect to $psi$.
My question:
Does there exist $A$ which satisfy the above condition and $A/ker(pi_{psi})=K(H)$, where $H$ is separable and infinite dimensional.
Can anyone show me an example? Thanks!
functional-analysis operator-theory operator-algebras c-star-algebras
functional-analysis operator-theory operator-algebras c-star-algebras
edited Mar 21 at 17:03
Andrews
1,2812423
asked Mar 21 at 15:46
mathrookiemathrookie
936512
edited Mar 21 at 17:03
Andrews
1,2812423
asked Mar 21 at 15:46
mathrookiemathrookie
936512
edited Mar 21 at 17:03
Andrews
1,2812423
edited Mar 21 at 17:03
Andrews
1,2812423
edited Mar 21 at 17:03
Andrews
1,2812423
1,2812423
asked Mar 21 at 15:46
mathrookiemathrookie
936512
asked Mar 21 at 15:46
mathrookiemathrookie
936512
asked Mar 21 at 15:46
mathrookiemathrookie
936512
936512
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, if $psine 0$. You can define a state on $A/kerpi_psi$ by
$$
tildepsi(a+kerpi_psi)=psi(a).
$$
This is well-defined since $psi=0$ on $kerpi_psi$. So $psi$ is a tracial state on $A/kerpi_psi$. And $K(H)$ does not admit a tracial state.
answered Mar 21 at 18:29
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
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
$begingroup$
No, if $psine 0$. You can define a state on $A/kerpi_psi$ by
$$
tildepsi(a+kerpi_psi)=psi(a).
$$
This is well-defined since $psi=0$ on $kerpi_psi$. So $psi$ is a tracial state on $A/kerpi_psi$. And $K(H)$ does not admit a tracial state.
answered Mar 21 at 18:29
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
No, if $psine 0$. You can define a state on $A/kerpi_psi$ by
$$
tildepsi(a+kerpi_psi)=psi(a).
$$
This is well-defined since $psi=0$ on $kerpi_psi$. So $psi$ is a tracial state on $A/kerpi_psi$. And $K(H)$ does not admit a tracial state.
answered Mar 21 at 18:29
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
No, if $psine 0$. You can define a state on $A/kerpi_psi$ by
$$
tildepsi(a+kerpi_psi)=psi(a).
$$
This is well-defined since $psi=0$ on $kerpi_psi$. So $psi$ is a tracial state on $A/kerpi_psi$. And $K(H)$ does not admit a tracial state.
answered Mar 21 at 18:29
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
No, if $psine 0$. You can define a state on $A/kerpi_psi$ by
$$
tildepsi(a+kerpi_psi)=psi(a).
$$
This is well-defined since $psi=0$ on $kerpi_psi$. So $psi$ is a tracial state on $A/kerpi_psi$. And $K(H)$ does not admit a tracial state.
answered Mar 21 at 18:29
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 21 at 18:29
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 21 at 18:29
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 21 at 18:29
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
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