Compactness of unit ball in weak-operator topology.
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I'm reading Richard Kadison's book about operator algebras, and in the demonstration that the unit ball is compact in weak-operator topology, the author defines a function from the set of bounded operators on a Hilbert space $H$, to a product of disks:
beginalign*F:mathcalB(H)rightarrow&prod_x,yin HmathbbD_x,y\
Trightarrow & langle Tx,yrangle : x,yin H
endalign*
If we set the product topology on $prod_x,yin HmathbbD_x,y$, the function above is continuous because of the topology of $mathcalB(H)$ is induced by functions of that type, but I can't see why this function is a homeomorphism. Why the inverse is also continuous?
analysis
edited Dec 12 '15 at 15:56
Xiang Yu
1,669820
asked Nov 14 '12 at 2:35
Felipe Pérez
514313
add a comment |Â
up vote
2
down vote
favorite
I'm reading Richard Kadison's book about operator algebras, and in the demonstration that the unit ball is compact in weak-operator topology, the author defines a function from the set of bounded operators on a Hilbert space $H$, to a product of disks:
beginalign*F:mathcalB(H)rightarrow&prod_x,yin HmathbbD_x,y\
Trightarrow & langle Tx,yrangle : x,yin H
endalign*
If we set the product topology on $prod_x,yin HmathbbD_x,y$, the function above is continuous because of the topology of $mathcalB(H)$ is induced by functions of that type, but I can't see why this function is a homeomorphism. Why the inverse is also continuous?
analysis
edited Dec 12 '15 at 15:56
Xiang Yu
1,669820
asked Nov 14 '12 at 2:35
Felipe Pérez
514313
1
Actually $F$ should not be defined on all of $mathcalB(H)$, just on the unit ball, and $mathbb D_x,y$ should be the closed disk centred at $0$ with radius $|x| |y|$. If $T$ is not in the unit ball, $langle Tx, yrangle$ will be outside that disk for some $x,y$.
– Robert Israel
Nov 14 '12 at 5:06
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm reading Richard Kadison's book about operator algebras, and in the demonstration that the unit ball is compact in weak-operator topology, the author defines a function from the set of bounded operators on a Hilbert space $H$, to a product of disks:
beginalign*F:mathcalB(H)rightarrow&prod_x,yin HmathbbD_x,y\
Trightarrow & langle Tx,yrangle : x,yin H
endalign*
If we set the product topology on $prod_x,yin HmathbbD_x,y$, the function above is continuous because of the topology of $mathcalB(H)$ is induced by functions of that type, but I can't see why this function is a homeomorphism. Why the inverse is also continuous?
analysis
edited Dec 12 '15 at 15:56
Xiang Yu
1,669820
asked Nov 14 '12 at 2:35
Felipe Pérez
514313
I'm reading Richard Kadison's book about operator algebras, and in the demonstration that the unit ball is compact in weak-operator topology, the author defines a function from the set of bounded operators on a Hilbert space $H$, to a product of disks:
beginalign*F:mathcalB(H)rightarrow&prod_x,yin HmathbbD_x,y\
Trightarrow & langle Tx,yrangle : x,yin H
endalign*
If we set the product topology on $prod_x,yin HmathbbD_x,y$, the function above is continuous because of the topology of $mathcalB(H)$ is induced by functions of that type, but I can't see why this function is a homeomorphism. Why the inverse is also continuous?
analysis
edited Dec 12 '15 at 15:56
Xiang Yu
1,669820
asked Nov 14 '12 at 2:35
Felipe Pérez
514313
edited Dec 12 '15 at 15:56
Xiang Yu
1,669820
edited Dec 12 '15 at 15:56
Xiang Yu
1,669820
edited Dec 12 '15 at 15:56
Xiang Yu
1,669820
1,669820
asked Nov 14 '12 at 2:35
Felipe Pérez
514313
asked Nov 14 '12 at 2:35
Felipe Pérez
514313
asked Nov 14 '12 at 2:35
Felipe Pérez
514313
514313
1
Actually $F$ should not be defined on all of $mathcalB(H)$, just on the unit ball, and $mathbb D_x,y$ should be the closed disk centred at $0$ with radius $|x| |y|$. If $T$ is not in the unit ball, $langle Tx, yrangle$ will be outside that disk for some $x,y$.
– Robert Israel
Nov 14 '12 at 5:06
add a comment |Â
1
Actually $F$ should not be defined on all of $mathcalB(H)$, just on the unit ball, and $mathbb D_x,y$ should be the closed disk centred at $0$ with radius $|x| |y|$. If $T$ is not in the unit ball, $langle Tx, yrangle$ will be outside that disk for some $x,y$.
– Robert Israel
Nov 14 '12 at 5:06
1
1
Actually $F$ should not be defined on all of $mathcalB(H)$, just on the unit ball, and $mathbb D_x,y$ should be the closed disk centred at $0$ with radius $|x| |y|$. If $T$ is not in the unit ball, $langle Tx, yrangle$ will be outside that disk for some $x,y$.
– Robert Israel
Nov 14 '12 at 5:06
Actually $F$ should not be defined on all of $mathcalB(H)$, just on the unit ball, and $mathbb D_x,y$ should be the closed disk centred at $0$ with radius $|x| |y|$. If $T$ is not in the unit ball, $langle Tx, yrangle$ will be outside that disk for some $x,y$.
– Robert Israel
Nov 14 '12 at 5:06
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
For the same reason. If the net of numbers $langle T_jx,yrangle$ converges to $langle Tx,yrangle$ for qll $x,y $, then this means precisely that $T_jto T$ in the WOT.
answered Nov 14 '12 at 2:41
Martin Argerami
117k1071165
1
Careful: these are not first-countable spaces, so sequences don't tell you the whole story. But you can talk about nets...
– Robert Israel
Nov 14 '12 at 5:00
1
I never said sequences, did I? ;)
– Martin Argerami
Nov 14 '12 at 5:06
add a comment |Â
up vote
1
down vote
It is a homeomorphism onto its image (not onto the whole product) because it maps each basic weak-operator open set to a product-topology open set in the image.
answered Nov 14 '12 at 2:40
Robert Israel
306k22201443
If I take a basic open set on $mathcalB(H)$ containing $T_0$ of the form $TinmathcalB(H) : <(T-T_0)x_j,y_j> < epsilon (j=1,... , m) $, can I characterize its image under F ?
– Felipe Pérez
Nov 14 '12 at 3:16
1
As I commented, you want to restrict to the unit ball of $cal B(H)$. The image under $F$ is those $w in prod_x,y mathbb D_x,y$ such that $w_x_j, y_j - langle T_0 x_j, y_j rangle < epsilon$ for $j = 1 ldots m$ and $w_ax+by,z = a w_x,z + bw_y,z$ and $w_x,ay+bz = a w_x,y + bw_x,z$ for all scalars $a,b$ and all $x,y,z in H$.
– Robert Israel
Nov 14 '12 at 5:14
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
For the same reason. If the net of numbers $langle T_jx,yrangle$ converges to $langle Tx,yrangle$ for qll $x,y $, then this means precisely that $T_jto T$ in the WOT.
answered Nov 14 '12 at 2:41
Martin Argerami
117k1071165
1
Careful: these are not first-countable spaces, so sequences don't tell you the whole story. But you can talk about nets...
– Robert Israel
Nov 14 '12 at 5:00
1
I never said sequences, did I? ;)
– Martin Argerami
Nov 14 '12 at 5:06
add a comment |Â
up vote
3
down vote
For the same reason. If the net of numbers $langle T_jx,yrangle$ converges to $langle Tx,yrangle$ for qll $x,y $, then this means precisely that $T_jto T$ in the WOT.
answered Nov 14 '12 at 2:41
Martin Argerami
117k1071165
1
Careful: these are not first-countable spaces, so sequences don't tell you the whole story. But you can talk about nets...
– Robert Israel
Nov 14 '12 at 5:00
1
I never said sequences, did I? ;)
– Martin Argerami
Nov 14 '12 at 5:06
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For the same reason. If the net of numbers $langle T_jx,yrangle$ converges to $langle Tx,yrangle$ for qll $x,y $, then this means precisely that $T_jto T$ in the WOT.
answered Nov 14 '12 at 2:41
Martin Argerami
117k1071165
For the same reason. If the net of numbers $langle T_jx,yrangle$ converges to $langle Tx,yrangle$ for qll $x,y $, then this means precisely that $T_jto T$ in the WOT.
answered Nov 14 '12 at 2:41
Martin Argerami
117k1071165
edited Aug 27 at 11:40
answered Nov 14 '12 at 2:41
Martin Argerami
117k1071165
answered Nov 14 '12 at 2:41
Martin Argerami
117k1071165
answered Nov 14 '12 at 2:41
Martin Argerami
117k1071165
117k1071165
1
Careful: these are not first-countable spaces, so sequences don't tell you the whole story. But you can talk about nets...
– Robert Israel
Nov 14 '12 at 5:00
1
I never said sequences, did I? ;)
– Martin Argerami
Nov 14 '12 at 5:06
add a comment |Â
1
Careful: these are not first-countable spaces, so sequences don't tell you the whole story. But you can talk about nets...
– Robert Israel
Nov 14 '12 at 5:00
1
I never said sequences, did I? ;)
– Martin Argerami
Nov 14 '12 at 5:06
1
1
Careful: these are not first-countable spaces, so sequences don't tell you the whole story. But you can talk about nets...
– Robert Israel
Nov 14 '12 at 5:00
Careful: these are not first-countable spaces, so sequences don't tell you the whole story. But you can talk about nets...
– Robert Israel
Nov 14 '12 at 5:00
1
1
I never said sequences, did I? ;)
– Martin Argerami
Nov 14 '12 at 5:06
I never said sequences, did I? ;)
– Martin Argerami
Nov 14 '12 at 5:06
add a comment |Â
up vote
1
down vote
It is a homeomorphism onto its image (not onto the whole product) because it maps each basic weak-operator open set to a product-topology open set in the image.
answered Nov 14 '12 at 2:40
Robert Israel
306k22201443
If I take a basic open set on $mathcalB(H)$ containing $T_0$ of the form $TinmathcalB(H) : <(T-T_0)x_j,y_j> < epsilon (j=1,... , m) $, can I characterize its image under F ?
– Felipe Pérez
Nov 14 '12 at 3:16
1
As I commented, you want to restrict to the unit ball of $cal B(H)$. The image under $F$ is those $w in prod_x,y mathbb D_x,y$ such that $w_x_j, y_j - langle T_0 x_j, y_j rangle < epsilon$ for $j = 1 ldots m$ and $w_ax+by,z = a w_x,z + bw_y,z$ and $w_x,ay+bz = a w_x,y + bw_x,z$ for all scalars $a,b$ and all $x,y,z in H$.
– Robert Israel
Nov 14 '12 at 5:14
add a comment |Â
up vote
1
down vote
It is a homeomorphism onto its image (not onto the whole product) because it maps each basic weak-operator open set to a product-topology open set in the image.
answered Nov 14 '12 at 2:40
Robert Israel
306k22201443
If I take a basic open set on $mathcalB(H)$ containing $T_0$ of the form $TinmathcalB(H) : <(T-T_0)x_j,y_j> < epsilon (j=1,... , m) $, can I characterize its image under F ?
– Felipe Pérez
Nov 14 '12 at 3:16
1
As I commented, you want to restrict to the unit ball of $cal B(H)$. The image under $F$ is those $w in prod_x,y mathbb D_x,y$ such that $w_x_j, y_j - langle T_0 x_j, y_j rangle < epsilon$ for $j = 1 ldots m$ and $w_ax+by,z = a w_x,z + bw_y,z$ and $w_x,ay+bz = a w_x,y + bw_x,z$ for all scalars $a,b$ and all $x,y,z in H$.
– Robert Israel
Nov 14 '12 at 5:14
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is a homeomorphism onto its image (not onto the whole product) because it maps each basic weak-operator open set to a product-topology open set in the image.
answered Nov 14 '12 at 2:40
Robert Israel
306k22201443
It is a homeomorphism onto its image (not onto the whole product) because it maps each basic weak-operator open set to a product-topology open set in the image.
answered Nov 14 '12 at 2:40
Robert Israel
306k22201443
answered Nov 14 '12 at 2:40
Robert Israel
306k22201443
answered Nov 14 '12 at 2:40
Robert Israel
306k22201443
answered Nov 14 '12 at 2:40
Robert Israel
306k22201443
306k22201443
If I take a basic open set on $mathcalB(H)$ containing $T_0$ of the form $TinmathcalB(H) : <(T-T_0)x_j,y_j> < epsilon (j=1,... , m) $, can I characterize its image under F ?
– Felipe Pérez
Nov 14 '12 at 3:16
1
As I commented, you want to restrict to the unit ball of $cal B(H)$. The image under $F$ is those $w in prod_x,y mathbb D_x,y$ such that $w_x_j, y_j - langle T_0 x_j, y_j rangle < epsilon$ for $j = 1 ldots m$ and $w_ax+by,z = a w_x,z + bw_y,z$ and $w_x,ay+bz = a w_x,y + bw_x,z$ for all scalars $a,b$ and all $x,y,z in H$.
– Robert Israel
Nov 14 '12 at 5:14
add a comment |Â
If I take a basic open set on $mathcalB(H)$ containing $T_0$ of the form $TinmathcalB(H) : <(T-T_0)x_j,y_j> < epsilon (j=1,... , m) $, can I characterize its image under F ?
– Felipe Pérez
Nov 14 '12 at 3:16
1
As I commented, you want to restrict to the unit ball of $cal B(H)$. The image under $F$ is those $w in prod_x,y mathbb D_x,y$ such that $w_x_j, y_j - langle T_0 x_j, y_j rangle < epsilon$ for $j = 1 ldots m$ and $w_ax+by,z = a w_x,z + bw_y,z$ and $w_x,ay+bz = a w_x,y + bw_x,z$ for all scalars $a,b$ and all $x,y,z in H$.
– Robert Israel
Nov 14 '12 at 5:14
If I take a basic open set on $mathcalB(H)$ containing $T_0$ of the form $TinmathcalB(H) : <(T-T_0)x_j,y_j> < epsilon (j=1,... , m) $, can I characterize its image under F ?
– Felipe Pérez
Nov 14 '12 at 3:16
If I take a basic open set on $mathcalB(H)$ containing $T_0$ of the form $TinmathcalB(H) : <(T-T_0)x_j,y_j> < epsilon (j=1,... , m) $, can I characterize its image under F ?
– Felipe Pérez
Nov 14 '12 at 3:16
1
1
As I commented, you want to restrict to the unit ball of $cal B(H)$. The image under $F$ is those $w in prod_x,y mathbb D_x,y$ such that $w_x_j, y_j - langle T_0 x_j, y_j rangle < epsilon$ for $j = 1 ldots m$ and $w_ax+by,z = a w_x,z + bw_y,z$ and $w_x,ay+bz = a w_x,y + bw_x,z$ for all scalars $a,b$ and all $x,y,z in H$.
– Robert Israel
Nov 14 '12 at 5:14
As I commented, you want to restrict to the unit ball of $cal B(H)$. The image under $F$ is those $w in prod_x,y mathbb D_x,y$ such that $w_x_j, y_j - langle T_0 x_j, y_j rangle < epsilon$ for $j = 1 ldots m$ and $w_ax+by,z = a w_x,z + bw_y,z$ and $w_x,ay+bz = a w_x,y + bw_x,z$ for all scalars $a,b$ and all $x,y,z in H$.
– Robert Israel
Nov 14 '12 at 5:14
add a comment |Â
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1
Actually $F$ should not be defined on all of $mathcalB(H)$, just on the unit ball, and $mathbb D_x,y$ should be the closed disk centred at $0$ with radius $|x| |y|$. If $T$ is not in the unit ball, $langle Tx, yrangle$ will be outside that disk for some $x,y$.
– Robert Israel
Nov 14 '12 at 5:06