Conditional Expectation on von Neumann algebras
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A linear map $phi$ from a von Neumann algebra M to the sub algebra N is called a conditional expectation when $phi$ has the following properties.
1)$phi(I)=I$, 2) $phi(x_1yx_2)=x_1phi(y)x_2$ whenever $x_1,x_2in M$ and $yin N.$
Can anyone explain me why this map is called conditional expectation? How it is related to the classical case? Thanks in advance.
functional-analysis conditional-expectation von-neumann-algebras
asked Sep 6 at 5:53
rkmath
828
add a comment |Â
up vote
4
down vote
favorite
A linear map $phi$ from a von Neumann algebra M to the sub algebra N is called a conditional expectation when $phi$ has the following properties.
1)$phi(I)=I$, 2) $phi(x_1yx_2)=x_1phi(y)x_2$ whenever $x_1,x_2in M$ and $yin N.$
Can anyone explain me why this map is called conditional expectation? How it is related to the classical case? Thanks in advance.
functional-analysis conditional-expectation von-neumann-algebras
asked Sep 6 at 5:53
rkmath
828
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
A linear map $phi$ from a von Neumann algebra M to the sub algebra N is called a conditional expectation when $phi$ has the following properties.
1)$phi(I)=I$, 2) $phi(x_1yx_2)=x_1phi(y)x_2$ whenever $x_1,x_2in M$ and $yin N.$
Can anyone explain me why this map is called conditional expectation? How it is related to the classical case? Thanks in advance.
functional-analysis conditional-expectation von-neumann-algebras
asked Sep 6 at 5:53
rkmath
828
A linear map $phi$ from a von Neumann algebra M to the sub algebra N is called a conditional expectation when $phi$ has the following properties.
1)$phi(I)=I$, 2) $phi(x_1yx_2)=x_1phi(y)x_2$ whenever $x_1,x_2in M$ and $yin N.$
Can anyone explain me why this map is called conditional expectation? How it is related to the classical case? Thanks in advance.
functional-analysis conditional-expectation von-neumann-algebras
functional-analysis conditional-expectation von-neumann-algebras
asked Sep 6 at 5:53
rkmath
828
asked Sep 6 at 5:53
rkmath
828
asked Sep 6 at 5:53
rkmath
828
asked Sep 6 at 5:53
rkmath
828
asked Sep 6 at 5:53
rkmath
828
828
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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I assume, by the "classical case", you mean the conditional expectations used in probability theory.
Let $(Omega, Sigma, mathbb P)$ be a probability space with $Sigma$ its $sigma$-algebra and let $Sigma_0 subset Sigma$ be a sub-$sigma$-algebra. You have a natural inclusion
$$ L^infty(Omega, Sigma_0) subset L^infty(Omega, Sigma)$$
Both are von Neumann algebras and the inclusion preserves the probability measure $mathbb P$. Classical probability theory gives you a conditional expectation $$mathbb E(cdot | Sigma_0): L^infty(Omega, Sigma) to L^infty(Omega, Sigma_0).$$ This is a (weak-$ast$ continuous) conditinal expectation in the sense of von Neumann algebras, i.e. it is unit-preserving and $$X , mathbbE( Y | Sigma_0) , Z = mathbbE(X , Y , Z | Sigma_0),$$ for every $Y, Z$ $Sigma_0$-measurable.
If you want a hint on how to obtain the conditional expectation $mathbbE( cdot | Sigma_0)$ just use that, since the inclusion above is $mathbb P$-preserving, it extends to an inclusion $j: L^1(Omega, Sigma_0;mathbb P) to L^1(Omega, Sigma; mathbb P)$. Dualizing that inclusion gives the conditional expectation. The same proof works in (finite) von Neumann algebras if the von Neumann sub-algebra is unital.
answered Sep 6 at 12:20
Adrián González-Pérez
66138
Thank you. Can you please elaborate me what you mean by 'dualizing the inclusions' in the last paragraph.? What goes wrong in the case of a general von Neumann algebra(not finite)?
– rkmath
Sep 6 at 15:05
1
If you have a bounded map $T:X to Y$ between Baach spaces you can defince $T^*:Y^ast to X^ast$ in the natural way by $T^*(lambda)(x) = lambda(T(x))$, that what i mean by "dualizing" (I am using that $(L^1)^ast = L^infty$)
– Adrián González-Pérez
Sep 6 at 17:52
1
In the case of non-finite algebras you risk that the restriction of the trace to the smaller algebra may not be semifinite. You can already observe that in the case of $mathbbR$ with the Lebesgue measure $m$. Take the sub-$sigma$-algebra $Sigma_0$ generated by $(infty,0]$ and $[0,infty)$. The measure $m_Sigma_0$ is identically $infty$ and so the inclusion of the $Sigma_0$-measurable functions do not extend to $L^1$.
– Adrián González-Pérez
Sep 6 at 17:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I assume, by the "classical case", you mean the conditional expectations used in probability theory.
Let $(Omega, Sigma, mathbb P)$ be a probability space with $Sigma$ its $sigma$-algebra and let $Sigma_0 subset Sigma$ be a sub-$sigma$-algebra. You have a natural inclusion
$$ L^infty(Omega, Sigma_0) subset L^infty(Omega, Sigma)$$
Both are von Neumann algebras and the inclusion preserves the probability measure $mathbb P$. Classical probability theory gives you a conditional expectation $$mathbb E(cdot | Sigma_0): L^infty(Omega, Sigma) to L^infty(Omega, Sigma_0).$$ This is a (weak-$ast$ continuous) conditinal expectation in the sense of von Neumann algebras, i.e. it is unit-preserving and $$X , mathbbE( Y | Sigma_0) , Z = mathbbE(X , Y , Z | Sigma_0),$$ for every $Y, Z$ $Sigma_0$-measurable.
If you want a hint on how to obtain the conditional expectation $mathbbE( cdot | Sigma_0)$ just use that, since the inclusion above is $mathbb P$-preserving, it extends to an inclusion $j: L^1(Omega, Sigma_0;mathbb P) to L^1(Omega, Sigma; mathbb P)$. Dualizing that inclusion gives the conditional expectation. The same proof works in (finite) von Neumann algebras if the von Neumann sub-algebra is unital.
answered Sep 6 at 12:20
Adrián González-Pérez
66138
Thank you. Can you please elaborate me what you mean by 'dualizing the inclusions' in the last paragraph.? What goes wrong in the case of a general von Neumann algebra(not finite)?
– rkmath
Sep 6 at 15:05
1
If you have a bounded map $T:X to Y$ between Baach spaces you can defince $T^*:Y^ast to X^ast$ in the natural way by $T^*(lambda)(x) = lambda(T(x))$, that what i mean by "dualizing" (I am using that $(L^1)^ast = L^infty$)
– Adrián González-Pérez
Sep 6 at 17:52
1
In the case of non-finite algebras you risk that the restriction of the trace to the smaller algebra may not be semifinite. You can already observe that in the case of $mathbbR$ with the Lebesgue measure $m$. Take the sub-$sigma$-algebra $Sigma_0$ generated by $(infty,0]$ and $[0,infty)$. The measure $m_Sigma_0$ is identically $infty$ and so the inclusion of the $Sigma_0$-measurable functions do not extend to $L^1$.
– Adrián González-Pérez
Sep 6 at 17:55
add a comment |Â
up vote
1
down vote
accepted
I assume, by the "classical case", you mean the conditional expectations used in probability theory.
Let $(Omega, Sigma, mathbb P)$ be a probability space with $Sigma$ its $sigma$-algebra and let $Sigma_0 subset Sigma$ be a sub-$sigma$-algebra. You have a natural inclusion
$$ L^infty(Omega, Sigma_0) subset L^infty(Omega, Sigma)$$
Both are von Neumann algebras and the inclusion preserves the probability measure $mathbb P$. Classical probability theory gives you a conditional expectation $$mathbb E(cdot | Sigma_0): L^infty(Omega, Sigma) to L^infty(Omega, Sigma_0).$$ This is a (weak-$ast$ continuous) conditinal expectation in the sense of von Neumann algebras, i.e. it is unit-preserving and $$X , mathbbE( Y | Sigma_0) , Z = mathbbE(X , Y , Z | Sigma_0),$$ for every $Y, Z$ $Sigma_0$-measurable.
If you want a hint on how to obtain the conditional expectation $mathbbE( cdot | Sigma_0)$ just use that, since the inclusion above is $mathbb P$-preserving, it extends to an inclusion $j: L^1(Omega, Sigma_0;mathbb P) to L^1(Omega, Sigma; mathbb P)$. Dualizing that inclusion gives the conditional expectation. The same proof works in (finite) von Neumann algebras if the von Neumann sub-algebra is unital.
answered Sep 6 at 12:20
Adrián González-Pérez
66138
Thank you. Can you please elaborate me what you mean by 'dualizing the inclusions' in the last paragraph.? What goes wrong in the case of a general von Neumann algebra(not finite)?
– rkmath
Sep 6 at 15:05
1
If you have a bounded map $T:X to Y$ between Baach spaces you can defince $T^*:Y^ast to X^ast$ in the natural way by $T^*(lambda)(x) = lambda(T(x))$, that what i mean by "dualizing" (I am using that $(L^1)^ast = L^infty$)
– Adrián González-Pérez
Sep 6 at 17:52
1
In the case of non-finite algebras you risk that the restriction of the trace to the smaller algebra may not be semifinite. You can already observe that in the case of $mathbbR$ with the Lebesgue measure $m$. Take the sub-$sigma$-algebra $Sigma_0$ generated by $(infty,0]$ and $[0,infty)$. The measure $m_Sigma_0$ is identically $infty$ and so the inclusion of the $Sigma_0$-measurable functions do not extend to $L^1$.
– Adrián González-Pérez
Sep 6 at 17:55
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I assume, by the "classical case", you mean the conditional expectations used in probability theory.
Let $(Omega, Sigma, mathbb P)$ be a probability space with $Sigma$ its $sigma$-algebra and let $Sigma_0 subset Sigma$ be a sub-$sigma$-algebra. You have a natural inclusion
$$ L^infty(Omega, Sigma_0) subset L^infty(Omega, Sigma)$$
Both are von Neumann algebras and the inclusion preserves the probability measure $mathbb P$. Classical probability theory gives you a conditional expectation $$mathbb E(cdot | Sigma_0): L^infty(Omega, Sigma) to L^infty(Omega, Sigma_0).$$ This is a (weak-$ast$ continuous) conditinal expectation in the sense of von Neumann algebras, i.e. it is unit-preserving and $$X , mathbbE( Y | Sigma_0) , Z = mathbbE(X , Y , Z | Sigma_0),$$ for every $Y, Z$ $Sigma_0$-measurable.
If you want a hint on how to obtain the conditional expectation $mathbbE( cdot | Sigma_0)$ just use that, since the inclusion above is $mathbb P$-preserving, it extends to an inclusion $j: L^1(Omega, Sigma_0;mathbb P) to L^1(Omega, Sigma; mathbb P)$. Dualizing that inclusion gives the conditional expectation. The same proof works in (finite) von Neumann algebras if the von Neumann sub-algebra is unital.
answered Sep 6 at 12:20
Adrián González-Pérez
66138
I assume, by the "classical case", you mean the conditional expectations used in probability theory.
Let $(Omega, Sigma, mathbb P)$ be a probability space with $Sigma$ its $sigma$-algebra and let $Sigma_0 subset Sigma$ be a sub-$sigma$-algebra. You have a natural inclusion
$$ L^infty(Omega, Sigma_0) subset L^infty(Omega, Sigma)$$
Both are von Neumann algebras and the inclusion preserves the probability measure $mathbb P$. Classical probability theory gives you a conditional expectation $$mathbb E(cdot | Sigma_0): L^infty(Omega, Sigma) to L^infty(Omega, Sigma_0).$$ This is a (weak-$ast$ continuous) conditinal expectation in the sense of von Neumann algebras, i.e. it is unit-preserving and $$X , mathbbE( Y | Sigma_0) , Z = mathbbE(X , Y , Z | Sigma_0),$$ for every $Y, Z$ $Sigma_0$-measurable.
If you want a hint on how to obtain the conditional expectation $mathbbE( cdot | Sigma_0)$ just use that, since the inclusion above is $mathbb P$-preserving, it extends to an inclusion $j: L^1(Omega, Sigma_0;mathbb P) to L^1(Omega, Sigma; mathbb P)$. Dualizing that inclusion gives the conditional expectation. The same proof works in (finite) von Neumann algebras if the von Neumann sub-algebra is unital.
answered Sep 6 at 12:20
Adrián González-Pérez
66138
answered Sep 6 at 12:20
Adrián González-Pérez
66138
answered Sep 6 at 12:20
Adrián González-Pérez
66138
answered Sep 6 at 12:20
Adrián González-Pérez
66138
66138
Thank you. Can you please elaborate me what you mean by 'dualizing the inclusions' in the last paragraph.? What goes wrong in the case of a general von Neumann algebra(not finite)?
– rkmath
Sep 6 at 15:05
1
If you have a bounded map $T:X to Y$ between Baach spaces you can defince $T^*:Y^ast to X^ast$ in the natural way by $T^*(lambda)(x) = lambda(T(x))$, that what i mean by "dualizing" (I am using that $(L^1)^ast = L^infty$)
– Adrián González-Pérez
Sep 6 at 17:52
1
In the case of non-finite algebras you risk that the restriction of the trace to the smaller algebra may not be semifinite. You can already observe that in the case of $mathbbR$ with the Lebesgue measure $m$. Take the sub-$sigma$-algebra $Sigma_0$ generated by $(infty,0]$ and $[0,infty)$. The measure $m_Sigma_0$ is identically $infty$ and so the inclusion of the $Sigma_0$-measurable functions do not extend to $L^1$.
– Adrián González-Pérez
Sep 6 at 17:55
add a comment |Â
Thank you. Can you please elaborate me what you mean by 'dualizing the inclusions' in the last paragraph.? What goes wrong in the case of a general von Neumann algebra(not finite)?
– rkmath
Sep 6 at 15:05
1
If you have a bounded map $T:X to Y$ between Baach spaces you can defince $T^*:Y^ast to X^ast$ in the natural way by $T^*(lambda)(x) = lambda(T(x))$, that what i mean by "dualizing" (I am using that $(L^1)^ast = L^infty$)
– Adrián González-Pérez
Sep 6 at 17:52
1
In the case of non-finite algebras you risk that the restriction of the trace to the smaller algebra may not be semifinite. You can already observe that in the case of $mathbbR$ with the Lebesgue measure $m$. Take the sub-$sigma$-algebra $Sigma_0$ generated by $(infty,0]$ and $[0,infty)$. The measure $m_Sigma_0$ is identically $infty$ and so the inclusion of the $Sigma_0$-measurable functions do not extend to $L^1$.
– Adrián González-Pérez
Sep 6 at 17:55
Thank you. Can you please elaborate me what you mean by 'dualizing the inclusions' in the last paragraph.? What goes wrong in the case of a general von Neumann algebra(not finite)?
– rkmath
Sep 6 at 15:05
Thank you. Can you please elaborate me what you mean by 'dualizing the inclusions' in the last paragraph.? What goes wrong in the case of a general von Neumann algebra(not finite)?
– rkmath
Sep 6 at 15:05
1
1
If you have a bounded map $T:X to Y$ between Baach spaces you can defince $T^*:Y^ast to X^ast$ in the natural way by $T^*(lambda)(x) = lambda(T(x))$, that what i mean by "dualizing" (I am using that $(L^1)^ast = L^infty$)
– Adrián González-Pérez
Sep 6 at 17:52
If you have a bounded map $T:X to Y$ between Baach spaces you can defince $T^*:Y^ast to X^ast$ in the natural way by $T^*(lambda)(x) = lambda(T(x))$, that what i mean by "dualizing" (I am using that $(L^1)^ast = L^infty$)
– Adrián González-Pérez
Sep 6 at 17:52
1
1
In the case of non-finite algebras you risk that the restriction of the trace to the smaller algebra may not be semifinite. You can already observe that in the case of $mathbbR$ with the Lebesgue measure $m$. Take the sub-$sigma$-algebra $Sigma_0$ generated by $(infty,0]$ and $[0,infty)$. The measure $m_Sigma_0$ is identically $infty$ and so the inclusion of the $Sigma_0$-measurable functions do not extend to $L^1$.
– Adrián González-Pérez
Sep 6 at 17:55
In the case of non-finite algebras you risk that the restriction of the trace to the smaller algebra may not be semifinite. You can already observe that in the case of $mathbbR$ with the Lebesgue measure $m$. Take the sub-$sigma$-algebra $Sigma_0$ generated by $(infty,0]$ and $[0,infty)$. The measure $m_Sigma_0$ is identically $infty$ and so the inclusion of the $Sigma_0$-measurable functions do not extend to $L^1$.
– Adrián González-Pérez
Sep 6 at 17:55
add a comment |Â
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