nuclear $C^*$ algebra
If $(A_i)$ is a sequence of nuclear $C^*$ algebras,Is $oplus_{c_0}A_i$ ($c_0$ direct sum)and $prod A_i$($ell ^infty $ direct sum) also nuclear?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 19 '18 at 3:31
mathrookie
809512
add a comment |
If $(A_i)$ is a sequence of nuclear $C^*$ algebras,Is $oplus_{c_0}A_i$ ($c_0$ direct sum)and $prod A_i$($ell ^infty $ direct sum) also nuclear?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 19 '18 at 3:31
mathrookie
809512
add a comment |
If $(A_i)$ is a sequence of nuclear $C^*$ algebras,Is $oplus_{c_0}A_i$ ($c_0$ direct sum)and $prod A_i$($ell ^infty $ direct sum) also nuclear?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 19 '18 at 3:31
mathrookie
809512
If $(A_i)$ is a sequence of nuclear $C^*$ algebras,Is $oplus_{c_0}A_i$ ($c_0$ direct sum)and $prod A_i$($ell ^infty $ direct sum) also nuclear?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Nov 19 '18 at 3:31
mathrookie
809512
asked Nov 19 '18 at 3:31
mathrookie
809512
asked Nov 19 '18 at 3:31
mathrookie
809512
asked Nov 19 '18 at 3:31
mathrookie
809512
asked Nov 19 '18 at 3:31
mathrookie
809512
809512
add a comment |
add a comment |
1 Answer
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Fix $ain bigoplus_nA_n$ and $varepsilon>0$. For each $n$, there exist ucp maps $varphi_n:A_nto M_{k(n)}(mathbb C)$ and $psi_varepsilon:M_{k(n)}(mathbb C)to A_n$ such that $|psi_ncircvarphi_n(a_n)-a_n|<varepsilon$.
There is also $m$ such that $|a_n|<varepsilon$ for all $ngeq m$. Write $a_0$ for the truncation of $a$ to its first $m$ entries; then $|a-a_0|<varepsilon$. Then the maps
$$
varphi:bigoplus_nA_nto bigoplus_{n=1}^mM_{k(n)}(mathbb C), psi:bigoplus_{n=1}^mM_{k(n)}(mathbb C)to bigoplus_nA_n
$$
given by $$varphi(b)=bigoplus_{n=1}^mvarphi_n(b_n), psi(bigoplus_{n=1}^m c_n)=bigoplus_{n=1}^m psi_n(c_n)$$
satisfy
begin{align}
|psicircvarphi(a)-a|
&leq|psicircvarphi(a)-psicircvarphi(a_0)|+|psicircvarphi(a_0)-a_0|+|a_0-a|\ \
&leq 2|a_0-a|+|psicircvarphi(a_0)-a_0|<3varepsilon.
end{align}
By making this work over the finite sets $Fsubset bigoplus_nA_n$ we obtain net ${varphi_F}$ and ${psi_F}$ such that $|psi_Fcircvarphi_F(a)-a|to0$ for all $ain bigoplus_nA_n$.
The diret product, on the other hand is not nuclear; it's not even exact. For instance $M=prod_n M_n(mathbb C)$ is not nuclear. It is well-known that nuclear algebras are exact, and that exactness passes to subalgebras. The full C$^*$-algebra of $mathbb F_2$ is known to be non-exact, and to be residually finite; this means that there exists a faithful representation $pi:C^*(mathbb F_2)to M=prod_nM_n(mathbb C)$. So $M$ cannot be exact, and in particular it is not nuclear.
answered Nov 19 '18 at 4:19
Martin Argerami
124k1176174
Exsctly. $ $
– Martin Argerami
Nov 19 '18 at 10:16
Do you mean "the universal $C^ast$ algebra of $mathbb F_2$", right? The reduced $C^ast$-algebra is exact and simple, therefore can not be residually finite.
– Adrián González-Pérez
Nov 20 '18 at 12:49
@Adríán: indeed, thanks for noticing.
– Martin Argerami
Nov 20 '18 at 15:36
If $A$ is any $C^*$-subalgebra of $prod_n M_n(Bbb C)$ which properly contains $bigoplus_nM_n(Bbb C)$,can we use $C^*(F_2)$ to conclude that $A$ is not nuclear?
– mathrookie
Dec 15 '18 at 18:08
Nuclearity does not pass to subalgebras. In any case, I don't think that $C_r^*(mathbb F_2)$ embeds in $bigoplus_nM_n(mathbb C)$.
– Martin Argerami
Dec 15 '18 at 18:26
|
show 1 more comment
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1 Answer
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Fix $ain bigoplus_nA_n$ and $varepsilon>0$. For each $n$, there exist ucp maps $varphi_n:A_nto M_{k(n)}(mathbb C)$ and $psi_varepsilon:M_{k(n)}(mathbb C)to A_n$ such that $|psi_ncircvarphi_n(a_n)-a_n|<varepsilon$.
There is also $m$ such that $|a_n|<varepsilon$ for all $ngeq m$. Write $a_0$ for the truncation of $a$ to its first $m$ entries; then $|a-a_0|<varepsilon$. Then the maps
$$
varphi:bigoplus_nA_nto bigoplus_{n=1}^mM_{k(n)}(mathbb C), psi:bigoplus_{n=1}^mM_{k(n)}(mathbb C)to bigoplus_nA_n
$$
given by $$varphi(b)=bigoplus_{n=1}^mvarphi_n(b_n), psi(bigoplus_{n=1}^m c_n)=bigoplus_{n=1}^m psi_n(c_n)$$
satisfy
begin{align}
|psicircvarphi(a)-a|
&leq|psicircvarphi(a)-psicircvarphi(a_0)|+|psicircvarphi(a_0)-a_0|+|a_0-a|\ \
&leq 2|a_0-a|+|psicircvarphi(a_0)-a_0|<3varepsilon.
end{align}
By making this work over the finite sets $Fsubset bigoplus_nA_n$ we obtain net ${varphi_F}$ and ${psi_F}$ such that $|psi_Fcircvarphi_F(a)-a|to0$ for all $ain bigoplus_nA_n$.
The diret product, on the other hand is not nuclear; it's not even exact. For instance $M=prod_n M_n(mathbb C)$ is not nuclear. It is well-known that nuclear algebras are exact, and that exactness passes to subalgebras. The full C$^*$-algebra of $mathbb F_2$ is known to be non-exact, and to be residually finite; this means that there exists a faithful representation $pi:C^*(mathbb F_2)to M=prod_nM_n(mathbb C)$. So $M$ cannot be exact, and in particular it is not nuclear.
answered Nov 19 '18 at 4:19
Martin Argerami
124k1176174
Exsctly. $ $
– Martin Argerami
Nov 19 '18 at 10:16
Do you mean "the universal $C^ast$ algebra of $mathbb F_2$", right? The reduced $C^ast$-algebra is exact and simple, therefore can not be residually finite.
– Adrián González-Pérez
Nov 20 '18 at 12:49
@Adríán: indeed, thanks for noticing.
– Martin Argerami
Nov 20 '18 at 15:36
If $A$ is any $C^*$-subalgebra of $prod_n M_n(Bbb C)$ which properly contains $bigoplus_nM_n(Bbb C)$,can we use $C^*(F_2)$ to conclude that $A$ is not nuclear?
– mathrookie
Dec 15 '18 at 18:08
Nuclearity does not pass to subalgebras. In any case, I don't think that $C_r^*(mathbb F_2)$ embeds in $bigoplus_nM_n(mathbb C)$.
– Martin Argerami
Dec 15 '18 at 18:26
|
show 1 more comment
Fix $ain bigoplus_nA_n$ and $varepsilon>0$. For each $n$, there exist ucp maps $varphi_n:A_nto M_{k(n)}(mathbb C)$ and $psi_varepsilon:M_{k(n)}(mathbb C)to A_n$ such that $|psi_ncircvarphi_n(a_n)-a_n|<varepsilon$.
There is also $m$ such that $|a_n|<varepsilon$ for all $ngeq m$. Write $a_0$ for the truncation of $a$ to its first $m$ entries; then $|a-a_0|<varepsilon$. Then the maps
$$
varphi:bigoplus_nA_nto bigoplus_{n=1}^mM_{k(n)}(mathbb C), psi:bigoplus_{n=1}^mM_{k(n)}(mathbb C)to bigoplus_nA_n
$$
given by $$varphi(b)=bigoplus_{n=1}^mvarphi_n(b_n), psi(bigoplus_{n=1}^m c_n)=bigoplus_{n=1}^m psi_n(c_n)$$
satisfy
begin{align}
|psicircvarphi(a)-a|
&leq|psicircvarphi(a)-psicircvarphi(a_0)|+|psicircvarphi(a_0)-a_0|+|a_0-a|\ \
&leq 2|a_0-a|+|psicircvarphi(a_0)-a_0|<3varepsilon.
end{align}
By making this work over the finite sets $Fsubset bigoplus_nA_n$ we obtain net ${varphi_F}$ and ${psi_F}$ such that $|psi_Fcircvarphi_F(a)-a|to0$ for all $ain bigoplus_nA_n$.
The diret product, on the other hand is not nuclear; it's not even exact. For instance $M=prod_n M_n(mathbb C)$ is not nuclear. It is well-known that nuclear algebras are exact, and that exactness passes to subalgebras. The full C$^*$-algebra of $mathbb F_2$ is known to be non-exact, and to be residually finite; this means that there exists a faithful representation $pi:C^*(mathbb F_2)to M=prod_nM_n(mathbb C)$. So $M$ cannot be exact, and in particular it is not nuclear.
answered Nov 19 '18 at 4:19
Martin Argerami
124k1176174
Exsctly. $ $
– Martin Argerami
Nov 19 '18 at 10:16
Do you mean "the universal $C^ast$ algebra of $mathbb F_2$", right? The reduced $C^ast$-algebra is exact and simple, therefore can not be residually finite.
– Adrián González-Pérez
Nov 20 '18 at 12:49
@Adríán: indeed, thanks for noticing.
– Martin Argerami
Nov 20 '18 at 15:36
If $A$ is any $C^*$-subalgebra of $prod_n M_n(Bbb C)$ which properly contains $bigoplus_nM_n(Bbb C)$,can we use $C^*(F_2)$ to conclude that $A$ is not nuclear?
– mathrookie
Dec 15 '18 at 18:08
Nuclearity does not pass to subalgebras. In any case, I don't think that $C_r^*(mathbb F_2)$ embeds in $bigoplus_nM_n(mathbb C)$.
– Martin Argerami
Dec 15 '18 at 18:26
|
show 1 more comment
Fix $ain bigoplus_nA_n$ and $varepsilon>0$. For each $n$, there exist ucp maps $varphi_n:A_nto M_{k(n)}(mathbb C)$ and $psi_varepsilon:M_{k(n)}(mathbb C)to A_n$ such that $|psi_ncircvarphi_n(a_n)-a_n|<varepsilon$.
There is also $m$ such that $|a_n|<varepsilon$ for all $ngeq m$. Write $a_0$ for the truncation of $a$ to its first $m$ entries; then $|a-a_0|<varepsilon$. Then the maps
$$
varphi:bigoplus_nA_nto bigoplus_{n=1}^mM_{k(n)}(mathbb C), psi:bigoplus_{n=1}^mM_{k(n)}(mathbb C)to bigoplus_nA_n
$$
given by $$varphi(b)=bigoplus_{n=1}^mvarphi_n(b_n), psi(bigoplus_{n=1}^m c_n)=bigoplus_{n=1}^m psi_n(c_n)$$
satisfy
begin{align}
|psicircvarphi(a)-a|
&leq|psicircvarphi(a)-psicircvarphi(a_0)|+|psicircvarphi(a_0)-a_0|+|a_0-a|\ \
&leq 2|a_0-a|+|psicircvarphi(a_0)-a_0|<3varepsilon.
end{align}
By making this work over the finite sets $Fsubset bigoplus_nA_n$ we obtain net ${varphi_F}$ and ${psi_F}$ such that $|psi_Fcircvarphi_F(a)-a|to0$ for all $ain bigoplus_nA_n$.
The diret product, on the other hand is not nuclear; it's not even exact. For instance $M=prod_n M_n(mathbb C)$ is not nuclear. It is well-known that nuclear algebras are exact, and that exactness passes to subalgebras. The full C$^*$-algebra of $mathbb F_2$ is known to be non-exact, and to be residually finite; this means that there exists a faithful representation $pi:C^*(mathbb F_2)to M=prod_nM_n(mathbb C)$. So $M$ cannot be exact, and in particular it is not nuclear.
answered Nov 19 '18 at 4:19
Martin Argerami
124k1176174
Fix $ain bigoplus_nA_n$ and $varepsilon>0$. For each $n$, there exist ucp maps $varphi_n:A_nto M_{k(n)}(mathbb C)$ and $psi_varepsilon:M_{k(n)}(mathbb C)to A_n$ such that $|psi_ncircvarphi_n(a_n)-a_n|<varepsilon$.
There is also $m$ such that $|a_n|<varepsilon$ for all $ngeq m$. Write $a_0$ for the truncation of $a$ to its first $m$ entries; then $|a-a_0|<varepsilon$. Then the maps
$$
varphi:bigoplus_nA_nto bigoplus_{n=1}^mM_{k(n)}(mathbb C), psi:bigoplus_{n=1}^mM_{k(n)}(mathbb C)to bigoplus_nA_n
$$
given by $$varphi(b)=bigoplus_{n=1}^mvarphi_n(b_n), psi(bigoplus_{n=1}^m c_n)=bigoplus_{n=1}^m psi_n(c_n)$$
satisfy
begin{align}
|psicircvarphi(a)-a|
&leq|psicircvarphi(a)-psicircvarphi(a_0)|+|psicircvarphi(a_0)-a_0|+|a_0-a|\ \
&leq 2|a_0-a|+|psicircvarphi(a_0)-a_0|<3varepsilon.
end{align}
By making this work over the finite sets $Fsubset bigoplus_nA_n$ we obtain net ${varphi_F}$ and ${psi_F}$ such that $|psi_Fcircvarphi_F(a)-a|to0$ for all $ain bigoplus_nA_n$.
The diret product, on the other hand is not nuclear; it's not even exact. For instance $M=prod_n M_n(mathbb C)$ is not nuclear. It is well-known that nuclear algebras are exact, and that exactness passes to subalgebras. The full C$^*$-algebra of $mathbb F_2$ is known to be non-exact, and to be residually finite; this means that there exists a faithful representation $pi:C^*(mathbb F_2)to M=prod_nM_n(mathbb C)$. So $M$ cannot be exact, and in particular it is not nuclear.
answered Nov 19 '18 at 4:19
Martin Argerami
124k1176174
edited Nov 20 '18 at 15:37
answered Nov 19 '18 at 4:19
Martin Argerami
124k1176174
answered Nov 19 '18 at 4:19
Martin Argerami
124k1176174
answered Nov 19 '18 at 4:19
Martin Argerami
124k1176174
124k1176174
Exsctly. $ $
– Martin Argerami
Nov 19 '18 at 10:16
Do you mean "the universal $C^ast$ algebra of $mathbb F_2$", right? The reduced $C^ast$-algebra is exact and simple, therefore can not be residually finite.
– Adrián González-Pérez
Nov 20 '18 at 12:49
@Adríán: indeed, thanks for noticing.
– Martin Argerami
Nov 20 '18 at 15:36
If $A$ is any $C^*$-subalgebra of $prod_n M_n(Bbb C)$ which properly contains $bigoplus_nM_n(Bbb C)$,can we use $C^*(F_2)$ to conclude that $A$ is not nuclear?
– mathrookie
Dec 15 '18 at 18:08
Nuclearity does not pass to subalgebras. In any case, I don't think that $C_r^*(mathbb F_2)$ embeds in $bigoplus_nM_n(mathbb C)$.
– Martin Argerami
Dec 15 '18 at 18:26
|
show 1 more comment
Exsctly. $ $
– Martin Argerami
Nov 19 '18 at 10:16
Do you mean "the universal $C^ast$ algebra of $mathbb F_2$", right? The reduced $C^ast$-algebra is exact and simple, therefore can not be residually finite.
– Adrián González-Pérez
Nov 20 '18 at 12:49
@Adríán: indeed, thanks for noticing.
– Martin Argerami
Nov 20 '18 at 15:36
If $A$ is any $C^*$-subalgebra of $prod_n M_n(Bbb C)$ which properly contains $bigoplus_nM_n(Bbb C)$,can we use $C^*(F_2)$ to conclude that $A$ is not nuclear?
– mathrookie
Dec 15 '18 at 18:08
Nuclearity does not pass to subalgebras. In any case, I don't think that $C_r^*(mathbb F_2)$ embeds in $bigoplus_nM_n(mathbb C)$.
– Martin Argerami
Dec 15 '18 at 18:26
Exsctly. $ $
– Martin Argerami
Nov 19 '18 at 10:16
Exsctly. $ $
– Martin Argerami
Nov 19 '18 at 10:16
Do you mean "the universal $C^ast$ algebra of $mathbb F_2$", right? The reduced $C^ast$-algebra is exact and simple, therefore can not be residually finite.
– Adrián González-Pérez
Nov 20 '18 at 12:49
Do you mean "the universal $C^ast$ algebra of $mathbb F_2$", right? The reduced $C^ast$-algebra is exact and simple, therefore can not be residually finite.
– Adrián González-Pérez
Nov 20 '18 at 12:49
@Adríán: indeed, thanks for noticing.
– Martin Argerami
Nov 20 '18 at 15:36
@Adríán: indeed, thanks for noticing.
– Martin Argerami
Nov 20 '18 at 15:36
If $A$ is any $C^*$-subalgebra of $prod_n M_n(Bbb C)$ which properly contains $bigoplus_nM_n(Bbb C)$,can we use $C^*(F_2)$ to conclude that $A$ is not nuclear?
– mathrookie
Dec 15 '18 at 18:08
If $A$ is any $C^*$-subalgebra of $prod_n M_n(Bbb C)$ which properly contains $bigoplus_nM_n(Bbb C)$,can we use $C^*(F_2)$ to conclude that $A$ is not nuclear?
– mathrookie
Dec 15 '18 at 18:08
Nuclearity does not pass to subalgebras. In any case, I don't think that $C_r^*(mathbb F_2)$ embeds in $bigoplus_nM_n(mathbb C)$.
– Martin Argerami
Dec 15 '18 at 18:26
Nuclearity does not pass to subalgebras. In any case, I don't think that $C_r^*(mathbb F_2)$ embeds in $bigoplus_nM_n(mathbb C)$.
– Martin Argerami
Dec 15 '18 at 18:26
|
show 1 more comment
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