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Equivalence of projections in a von Neumann algebra
Clash Royale CLAN TAG#URR8PPP
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Let $Msubset B(H)$ be a von Neumann algebra. Also $p$ and $q$ are projections in $M$ such that $pMp$ is star isomorphic to $qMq$. Does this imply that $p$ and $q$ are equivalent in the Murray-von Neumann sense?
functional-analysis operator-algebras von-neumann-algebras
edited Jul 18 at 15:24
RayDansh
884215
asked Jul 18 at 15:21
rkmath
607
add a comment |Â
up vote
2
down vote
favorite
Let $Msubset B(H)$ be a von Neumann algebra. Also $p$ and $q$ are projections in $M$ such that $pMp$ is star isomorphic to $qMq$. Does this imply that $p$ and $q$ are equivalent in the Murray-von Neumann sense?
functional-analysis operator-algebras von-neumann-algebras
edited Jul 18 at 15:24
RayDansh
884215
asked Jul 18 at 15:21
rkmath
607
Do you require $M$ to be a factor? In any case there are counterexamples. For instance take $M = R$ the hyperfinite factor. Any $p R p$ gives another hyperfinite factor.
– Adrián González-Pérez
Jul 19 at 14:44
The notion you are after is probably that of "fundamental group"
– Adrián González-Pérez
Jul 19 at 14:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $Msubset B(H)$ be a von Neumann algebra. Also $p$ and $q$ are projections in $M$ such that $pMp$ is star isomorphic to $qMq$. Does this imply that $p$ and $q$ are equivalent in the Murray-von Neumann sense?
functional-analysis operator-algebras von-neumann-algebras
edited Jul 18 at 15:24
RayDansh
884215
asked Jul 18 at 15:21
rkmath
607
Let $Msubset B(H)$ be a von Neumann algebra. Also $p$ and $q$ are projections in $M$ such that $pMp$ is star isomorphic to $qMq$. Does this imply that $p$ and $q$ are equivalent in the Murray-von Neumann sense?
functional-analysis operator-algebras von-neumann-algebras
edited Jul 18 at 15:24
RayDansh
884215
asked Jul 18 at 15:21
rkmath
607
edited Jul 18 at 15:24
RayDansh
884215
edited Jul 18 at 15:24
RayDansh
884215
edited Jul 18 at 15:24
RayDansh
884215
884215
asked Jul 18 at 15:21
rkmath
607
asked Jul 18 at 15:21
rkmath
607
asked Jul 18 at 15:21
rkmath
607
607
Do you require $M$ to be a factor? In any case there are counterexamples. For instance take $M = R$ the hyperfinite factor. Any $p R p$ gives another hyperfinite factor.
– Adrián González-Pérez
Jul 19 at 14:44
The notion you are after is probably that of "fundamental group"
– Adrián González-Pérez
Jul 19 at 14:44
add a comment |Â
Do you require $M$ to be a factor? In any case there are counterexamples. For instance take $M = R$ the hyperfinite factor. Any $p R p$ gives another hyperfinite factor.
– Adrián González-Pérez
Jul 19 at 14:44
The notion you are after is probably that of "fundamental group"
– Adrián González-Pérez
Jul 19 at 14:44
Do you require $M$ to be a factor? In any case there are counterexamples. For instance take $M = R$ the hyperfinite factor. Any $p R p$ gives another hyperfinite factor.
– Adrián González-Pérez
Jul 19 at 14:44
Do you require $M$ to be a factor? In any case there are counterexamples. For instance take $M = R$ the hyperfinite factor. Any $p R p$ gives another hyperfinite factor.
– Adrián González-Pérez
Jul 19 at 14:44
The notion you are after is probably that of "fundamental group"
– Adrián González-Pérez
Jul 19 at 14:44
The notion you are after is probably that of "fundamental group"
– Adrián González-Pérez
Jul 19 at 14:44
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
No. Take $M=mathbb C^2subset M_2(mathbb C)$, $p=E_11$, $q=E_22$. Then
$$
pMp=mathbb Coplus 0simeq 0oplusmathbb C=qMq,
$$
but $p$ and $q$ are not equivalent (equivalence in a commutative algebra is equality).
answered Jul 18 at 15:23
Martin Argerami
116k1071164
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No. Take $M=mathbb C^2subset M_2(mathbb C)$, $p=E_11$, $q=E_22$. Then
$$
pMp=mathbb Coplus 0simeq 0oplusmathbb C=qMq,
$$
but $p$ and $q$ are not equivalent (equivalence in a commutative algebra is equality).
answered Jul 18 at 15:23
Martin Argerami
116k1071164
add a comment |Â
up vote
2
down vote
accepted
No. Take $M=mathbb C^2subset M_2(mathbb C)$, $p=E_11$, $q=E_22$. Then
$$
pMp=mathbb Coplus 0simeq 0oplusmathbb C=qMq,
$$
but $p$ and $q$ are not equivalent (equivalence in a commutative algebra is equality).
answered Jul 18 at 15:23
Martin Argerami
116k1071164
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No. Take $M=mathbb C^2subset M_2(mathbb C)$, $p=E_11$, $q=E_22$. Then
$$
pMp=mathbb Coplus 0simeq 0oplusmathbb C=qMq,
$$
but $p$ and $q$ are not equivalent (equivalence in a commutative algebra is equality).
answered Jul 18 at 15:23
Martin Argerami
116k1071164
No. Take $M=mathbb C^2subset M_2(mathbb C)$, $p=E_11$, $q=E_22$. Then
$$
pMp=mathbb Coplus 0simeq 0oplusmathbb C=qMq,
$$
but $p$ and $q$ are not equivalent (equivalence in a commutative algebra is equality).
answered Jul 18 at 15:23
Martin Argerami
116k1071164
answered Jul 18 at 15:23
Martin Argerami
116k1071164
answered Jul 18 at 15:23
Martin Argerami
116k1071164
answered Jul 18 at 15:23
Martin Argerami
116k1071164
116k1071164
add a comment |Â
add a comment |Â
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Do you require $M$ to be a factor? In any case there are counterexamples. For instance take $M = R$ the hyperfinite factor. Any $p R p$ gives another hyperfinite factor.
– Adrián González-Pérez
Jul 19 at 14:44
The notion you are after is probably that of "fundamental group"
– Adrián González-Pérez
Jul 19 at 14:44