Mixing
Isomorphism between von Neumann algebras
$begingroup$
Let $R_{1}$ and $R_{2}$ be two von Neumann a algebras with wot dense sub algebras $U_{1}$ and $U_{2}$. Suppose $varphi$ is a * isomorphism from $U_{1}$ onto $U_{2}$. Is there always an isomorphism $phi$ between $R_{1}$ and $R_{2}$ which is an extension of $varphi$?
general-topology operator-algebras von-neumann-algebras
edited Jan 20 at 17:02
ervx
10.3k31338
asked Jun 28 '18 at 5:33
rkmathrkmath
1018
$endgroup$
add a comment |
$begingroup$
Let $R_{1}$ and $R_{2}$ be two von Neumann a algebras with wot dense sub algebras $U_{1}$ and $U_{2}$. Suppose $varphi$ is a * isomorphism from $U_{1}$ onto $U_{2}$. Is there always an isomorphism $phi$ between $R_{1}$ and $R_{2}$ which is an extension of $varphi$?
general-topology operator-algebras von-neumann-algebras
edited Jan 20 at 17:02
ervx
10.3k31338
asked Jun 28 '18 at 5:33
rkmathrkmath
1018
$endgroup$
$begingroup$
What if $U_2=R_2$ but $U_1$ is a proper subalgebra of $R_1$? Then you obviously cannot extend $varphi$ to an injective function. Or am I missing something?
$endgroup$
– freakish
Jun 28 '18 at 10:11
add a comment |
$begingroup$
Let $R_{1}$ and $R_{2}$ be two von Neumann a algebras with wot dense sub algebras $U_{1}$ and $U_{2}$. Suppose $varphi$ is a * isomorphism from $U_{1}$ onto $U_{2}$. Is there always an isomorphism $phi$ between $R_{1}$ and $R_{2}$ which is an extension of $varphi$?
general-topology operator-algebras von-neumann-algebras
edited Jan 20 at 17:02
ervx
10.3k31338
asked Jun 28 '18 at 5:33
rkmathrkmath
1018
$endgroup$
Let $R_{1}$ and $R_{2}$ be two von Neumann a algebras with wot dense sub algebras $U_{1}$ and $U_{2}$. Suppose $varphi$ is a * isomorphism from $U_{1}$ onto $U_{2}$. Is there always an isomorphism $phi$ between $R_{1}$ and $R_{2}$ which is an extension of $varphi$?
general-topology operator-algebras von-neumann-algebras
general-topology operator-algebras von-neumann-algebras
edited Jan 20 at 17:02
ervx
10.3k31338
asked Jun 28 '18 at 5:33
rkmathrkmath
1018
edited Jan 20 at 17:02
ervx
10.3k31338
asked Jun 28 '18 at 5:33
rkmathrkmath
1018
edited Jan 20 at 17:02
ervx
10.3k31338
edited Jan 20 at 17:02
ervx
10.3k31338
edited Jan 20 at 17:02
ervx
10.3k31338
10.3k31338
asked Jun 28 '18 at 5:33
rkmathrkmath
1018
asked Jun 28 '18 at 5:33
rkmathrkmath
1018
asked Jun 28 '18 at 5:33
rkmathrkmath
1018
1018
$begingroup$
What if $U_2=R_2$ but $U_1$ is a proper subalgebra of $R_1$? Then you obviously cannot extend $varphi$ to an injective function. Or am I missing something?
$endgroup$
– freakish
Jun 28 '18 at 10:11
add a comment |
$begingroup$
What if $U_2=R_2$ but $U_1$ is a proper subalgebra of $R_1$? Then you obviously cannot extend $varphi$ to an injective function. Or am I missing something?
$endgroup$
– freakish
Jun 28 '18 at 10:11
$begingroup$
What if $U_2=R_2$ but $U_1$ is a proper subalgebra of $R_1$? Then you obviously cannot extend $varphi$ to an injective function. Or am I missing something?
$endgroup$
– freakish
Jun 28 '18 at 10:11
$begingroup$
What if $U_2=R_2$ but $U_1$ is a proper subalgebra of $R_1$? Then you obviously cannot extend $varphi$ to an injective function. Or am I missing something?
$endgroup$
– freakish
Jun 28 '18 at 10:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is no.
Let $F_2$ be the free group on two generators, then the reduced group $C^ast$-algebra $C^ast_r(F_2)$ is wot dense in the group von Neumann algebra $L(F_2) subseteq mathbb B(L^2(F_2))$. As $C^ast_r(F_2)$ is separable, simple, and not type I, there is a faithful irreducible representation $pi colon C^ast_r(F_2) to mathbb B(L^2(F_2))$, so the image $A:=pi(C^ast_r(F_2))$ is $ast$-isomorphic to $C^ast_r(F_2)$, and $A'' = mathbb B(L^2(F_2)) not cong L(F_2)$.
answered Jun 28 '18 at 11:01
Jamie GabeJamie Gabe
862
$endgroup$
add a comment |
$begingroup$
This is not even true when $U_1=U_2$. For instance if you take $U_1=U_2=UHF(2^infty)$, then by taking the von Neumann algebra obtained via GNS with the trace, you get $mathcal R_1$ the hyperfinite II$_1$-factor. While by using an appropriate state $psi_lambda$, you can get $mathcal R_2$ to be one Power's III$_lambda$ factors.
answered Jun 28 '18 at 14:25
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The answer is no.
Let $F_2$ be the free group on two generators, then the reduced group $C^ast$-algebra $C^ast_r(F_2)$ is wot dense in the group von Neumann algebra $L(F_2) subseteq mathbb B(L^2(F_2))$. As $C^ast_r(F_2)$ is separable, simple, and not type I, there is a faithful irreducible representation $pi colon C^ast_r(F_2) to mathbb B(L^2(F_2))$, so the image $A:=pi(C^ast_r(F_2))$ is $ast$-isomorphic to $C^ast_r(F_2)$, and $A'' = mathbb B(L^2(F_2)) not cong L(F_2)$.
answered Jun 28 '18 at 11:01
Jamie GabeJamie Gabe
862
$endgroup$
add a comment |
$begingroup$
The answer is no.
Let $F_2$ be the free group on two generators, then the reduced group $C^ast$-algebra $C^ast_r(F_2)$ is wot dense in the group von Neumann algebra $L(F_2) subseteq mathbb B(L^2(F_2))$. As $C^ast_r(F_2)$ is separable, simple, and not type I, there is a faithful irreducible representation $pi colon C^ast_r(F_2) to mathbb B(L^2(F_2))$, so the image $A:=pi(C^ast_r(F_2))$ is $ast$-isomorphic to $C^ast_r(F_2)$, and $A'' = mathbb B(L^2(F_2)) not cong L(F_2)$.
answered Jun 28 '18 at 11:01
Jamie GabeJamie Gabe
862
$endgroup$
add a comment |
$begingroup$
The answer is no.
Let $F_2$ be the free group on two generators, then the reduced group $C^ast$-algebra $C^ast_r(F_2)$ is wot dense in the group von Neumann algebra $L(F_2) subseteq mathbb B(L^2(F_2))$. As $C^ast_r(F_2)$ is separable, simple, and not type I, there is a faithful irreducible representation $pi colon C^ast_r(F_2) to mathbb B(L^2(F_2))$, so the image $A:=pi(C^ast_r(F_2))$ is $ast$-isomorphic to $C^ast_r(F_2)$, and $A'' = mathbb B(L^2(F_2)) not cong L(F_2)$.
answered Jun 28 '18 at 11:01
Jamie GabeJamie Gabe
862
$endgroup$
The answer is no.
Let $F_2$ be the free group on two generators, then the reduced group $C^ast$-algebra $C^ast_r(F_2)$ is wot dense in the group von Neumann algebra $L(F_2) subseteq mathbb B(L^2(F_2))$. As $C^ast_r(F_2)$ is separable, simple, and not type I, there is a faithful irreducible representation $pi colon C^ast_r(F_2) to mathbb B(L^2(F_2))$, so the image $A:=pi(C^ast_r(F_2))$ is $ast$-isomorphic to $C^ast_r(F_2)$, and $A'' = mathbb B(L^2(F_2)) not cong L(F_2)$.
answered Jun 28 '18 at 11:01
Jamie GabeJamie Gabe
862
answered Jun 28 '18 at 11:01
Jamie GabeJamie Gabe
862
answered Jun 28 '18 at 11:01
Jamie GabeJamie Gabe
862
answered Jun 28 '18 at 11:01
Jamie GabeJamie Gabe
862
862
add a comment |
add a comment |
$begingroup$
This is not even true when $U_1=U_2$. For instance if you take $U_1=U_2=UHF(2^infty)$, then by taking the von Neumann algebra obtained via GNS with the trace, you get $mathcal R_1$ the hyperfinite II$_1$-factor. While by using an appropriate state $psi_lambda$, you can get $mathcal R_2$ to be one Power's III$_lambda$ factors.
answered Jun 28 '18 at 14:25
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
add a comment |
$begingroup$
This is not even true when $U_1=U_2$. For instance if you take $U_1=U_2=UHF(2^infty)$, then by taking the von Neumann algebra obtained via GNS with the trace, you get $mathcal R_1$ the hyperfinite II$_1$-factor. While by using an appropriate state $psi_lambda$, you can get $mathcal R_2$ to be one Power's III$_lambda$ factors.
answered Jun 28 '18 at 14:25
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
add a comment |
$begingroup$
This is not even true when $U_1=U_2$. For instance if you take $U_1=U_2=UHF(2^infty)$, then by taking the von Neumann algebra obtained via GNS with the trace, you get $mathcal R_1$ the hyperfinite II$_1$-factor. While by using an appropriate state $psi_lambda$, you can get $mathcal R_2$ to be one Power's III$_lambda$ factors.
answered Jun 28 '18 at 14:25
Martin ArgeramiMartin Argerami
128k1182183
$endgroup$
This is not even true when $U_1=U_2$. For instance if you take $U_1=U_2=UHF(2^infty)$, then by taking the von Neumann algebra obtained via GNS with the trace, you get $mathcal R_1$ the hyperfinite II$_1$-factor. While by using an appropriate state $psi_lambda$, you can get $mathcal R_2$ to be one Power's III$_lambda$ factors.
answered Jun 28 '18 at 14:25
Martin ArgeramiMartin Argerami
128k1182183
answered Jun 28 '18 at 14:25
Martin ArgeramiMartin Argerami
128k1182183
answered Jun 28 '18 at 14:25
Martin ArgeramiMartin Argerami
128k1182183
answered Jun 28 '18 at 14:25
Martin ArgeramiMartin Argerami
128k1182183
128k1182183
add a comment |
add a comment |
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$begingroup$
What if $U_2=R_2$ but $U_1$ is a proper subalgebra of $R_1$? Then you obviously cannot extend $varphi$ to an injective function. Or am I missing something?
$endgroup$
– freakish
Jun 28 '18 at 10:11