identify GNS construction as asubalgebra of $R^{omega}$












1












$begingroup$


enter image description hereenter image description here



I have two questions In lemma6.5.5.



1.why can we identify the GNS representation of $prod M_{k(n)}(Bbb C)/bigoplus M_{k(n)}(Bbb C)$ with respect to $tau _{omega}$ with a subalgebra of $R^{omega}$.
2.why is the Gns construction unique?






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    In the second question, do you mean, why $tau_omega circ sigma pi$ is the canonical group trace? The fact that the GNS constructions under equal traces seems quite straightforward.
    $endgroup$
    – Adrián González-Pérez
    Dec 7 '18 at 11:51
















1












$begingroup$


enter image description hereenter image description here



I have two questions In lemma6.5.5.



1.why can we identify the GNS representation of $prod M_{k(n)}(Bbb C)/bigoplus M_{k(n)}(Bbb C)$ with respect to $tau _{omega}$ with a subalgebra of $R^{omega}$.
2.why is the Gns construction unique?






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    In the second question, do you mean, why $tau_omega circ sigma pi$ is the canonical group trace? The fact that the GNS constructions under equal traces seems quite straightforward.
    $endgroup$
    – Adrián González-Pérez
    Dec 7 '18 at 11:51














1












1








1





$begingroup$


enter image description hereenter image description here



I have two questions In lemma6.5.5.



1.why can we identify the GNS representation of $prod M_{k(n)}(Bbb C)/bigoplus M_{k(n)}(Bbb C)$ with respect to $tau _{omega}$ with a subalgebra of $R^{omega}$.
2.why is the Gns construction unique?






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







share|cite|improve this question









$endgroup$




enter image description hereenter image description here



I have two questions In lemma6.5.5.



1.why can we identify the GNS representation of $prod M_{k(n)}(Bbb C)/bigoplus M_{k(n)}(Bbb C)$ with respect to $tau _{omega}$ with a subalgebra of $R^{omega}$.
2.why is the Gns construction unique?






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




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 5:23









mathrookiemathrookie

845512




845512












  • $begingroup$
    In the second question, do you mean, why $tau_omega circ sigma pi$ is the canonical group trace? The fact that the GNS constructions under equal traces seems quite straightforward.
    $endgroup$
    – Adrián González-Pérez
    Dec 7 '18 at 11:51


















  • $begingroup$
    In the second question, do you mean, why $tau_omega circ sigma pi$ is the canonical group trace? The fact that the GNS constructions under equal traces seems quite straightforward.
    $endgroup$
    – Adrián González-Pérez
    Dec 7 '18 at 11:51
















$begingroup$
In the second question, do you mean, why $tau_omega circ sigma pi$ is the canonical group trace? The fact that the GNS constructions under equal traces seems quite straightforward.
$endgroup$
– Adrián González-Pérez
Dec 7 '18 at 11:51




$begingroup$
In the second question, do you mean, why $tau_omega circ sigma pi$ is the canonical group trace? The fact that the GNS constructions under equal traces seems quite straightforward.
$endgroup$
– Adrián González-Pérez
Dec 7 '18 at 11:51










1 Answer
1






active

oldest

votes


















0












$begingroup$

For the first point:



Just assume $k(n) > 1$ for infinitely many $n$ (otherwise the claim is trivial). Note that
$$
R = overline{bigotimes}_n M_{k(n)}
$$
Any sequence $(x_n)_n in prod M_{k(n)}$ gives a sequence in $prod R$ by $x_n mapsto 1^{otimes (n - 1)} otimes x_n$. This gives a map $prod M_k(n) to prod R$. Compose that with the quotient map to get $prod M_k(n) to R^omega$. Recall that $R^omega = prod R / J_omega$. Now check that the $bigoplus M_{k(n)}$ lays inside the tracial ideal $J_omega$ and that the traces agree.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it injective ?
    $endgroup$
    – André S.
    Dec 7 '18 at 13:02











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

For the first point:



Just assume $k(n) > 1$ for infinitely many $n$ (otherwise the claim is trivial). Note that
$$
R = overline{bigotimes}_n M_{k(n)}
$$
Any sequence $(x_n)_n in prod M_{k(n)}$ gives a sequence in $prod R$ by $x_n mapsto 1^{otimes (n - 1)} otimes x_n$. This gives a map $prod M_k(n) to prod R$. Compose that with the quotient map to get $prod M_k(n) to R^omega$. Recall that $R^omega = prod R / J_omega$. Now check that the $bigoplus M_{k(n)}$ lays inside the tracial ideal $J_omega$ and that the traces agree.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it injective ?
    $endgroup$
    – André S.
    Dec 7 '18 at 13:02
















0












$begingroup$

For the first point:



Just assume $k(n) > 1$ for infinitely many $n$ (otherwise the claim is trivial). Note that
$$
R = overline{bigotimes}_n M_{k(n)}
$$
Any sequence $(x_n)_n in prod M_{k(n)}$ gives a sequence in $prod R$ by $x_n mapsto 1^{otimes (n - 1)} otimes x_n$. This gives a map $prod M_k(n) to prod R$. Compose that with the quotient map to get $prod M_k(n) to R^omega$. Recall that $R^omega = prod R / J_omega$. Now check that the $bigoplus M_{k(n)}$ lays inside the tracial ideal $J_omega$ and that the traces agree.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it injective ?
    $endgroup$
    – André S.
    Dec 7 '18 at 13:02














0












0








0





$begingroup$

For the first point:



Just assume $k(n) > 1$ for infinitely many $n$ (otherwise the claim is trivial). Note that
$$
R = overline{bigotimes}_n M_{k(n)}
$$
Any sequence $(x_n)_n in prod M_{k(n)}$ gives a sequence in $prod R$ by $x_n mapsto 1^{otimes (n - 1)} otimes x_n$. This gives a map $prod M_k(n) to prod R$. Compose that with the quotient map to get $prod M_k(n) to R^omega$. Recall that $R^omega = prod R / J_omega$. Now check that the $bigoplus M_{k(n)}$ lays inside the tracial ideal $J_omega$ and that the traces agree.






share|cite|improve this answer









$endgroup$



For the first point:



Just assume $k(n) > 1$ for infinitely many $n$ (otherwise the claim is trivial). Note that
$$
R = overline{bigotimes}_n M_{k(n)}
$$
Any sequence $(x_n)_n in prod M_{k(n)}$ gives a sequence in $prod R$ by $x_n mapsto 1^{otimes (n - 1)} otimes x_n$. This gives a map $prod M_k(n) to prod R$. Compose that with the quotient map to get $prod M_k(n) to R^omega$. Recall that $R^omega = prod R / J_omega$. Now check that the $bigoplus M_{k(n)}$ lays inside the tracial ideal $J_omega$ and that the traces agree.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 11:59









Adrián González-PérezAdrián González-Pérez

1,086139




1,086139












  • $begingroup$
    Is it injective ?
    $endgroup$
    – André S.
    Dec 7 '18 at 13:02


















  • $begingroup$
    Is it injective ?
    $endgroup$
    – André S.
    Dec 7 '18 at 13:02
















$begingroup$
Is it injective ?
$endgroup$
– André S.
Dec 7 '18 at 13:02




$begingroup$
Is it injective ?
$endgroup$
– André S.
Dec 7 '18 at 13:02


















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