Mixing
Condition expectation on maximal group C*-algebra
$begingroup$
I am trying to understand some basic $C^*$-algebraic terminology in a context specific to group $C^*$-algebras, and would appreciate some help from experts to whom the meaning of these things is clear.
Suppose $G$ is a locally compact topological group, and let $C^*(G)$ denote the maximal group $C^*$-algebra. Consider the map
$$C_c(G)rightarrowmathbb{C},qquad fmapsto f(e).$$
I am wondering:
- Does this map extend to what is called a "conditional expectation" on $C^*(G)$?
- If so, is this conditional expectation "faithful"?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Jan 26 at 7:43
ougoahougoah
1,256710
$endgroup$
add a comment |
$begingroup$
I am trying to understand some basic $C^*$-algebraic terminology in a context specific to group $C^*$-algebras, and would appreciate some help from experts to whom the meaning of these things is clear.
Suppose $G$ is a locally compact topological group, and let $C^*(G)$ denote the maximal group $C^*$-algebra. Consider the map
$$C_c(G)rightarrowmathbb{C},qquad fmapsto f(e).$$
I am wondering:
- Does this map extend to what is called a "conditional expectation" on $C^*(G)$?
- If so, is this conditional expectation "faithful"?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Jan 26 at 7:43
ougoahougoah
1,256710
$endgroup$
add a comment |
$begingroup$
I am trying to understand some basic $C^*$-algebraic terminology in a context specific to group $C^*$-algebras, and would appreciate some help from experts to whom the meaning of these things is clear.
Suppose $G$ is a locally compact topological group, and let $C^*(G)$ denote the maximal group $C^*$-algebra. Consider the map
$$C_c(G)rightarrowmathbb{C},qquad fmapsto f(e).$$
I am wondering:
- Does this map extend to what is called a "conditional expectation" on $C^*(G)$?
- If so, is this conditional expectation "faithful"?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Jan 26 at 7:43
ougoahougoah
1,256710
$endgroup$
I am trying to understand some basic $C^*$-algebraic terminology in a context specific to group $C^*$-algebras, and would appreciate some help from experts to whom the meaning of these things is clear.
Suppose $G$ is a locally compact topological group, and let $C^*(G)$ denote the maximal group $C^*$-algebra. Consider the map
$$C_c(G)rightarrowmathbb{C},qquad fmapsto f(e).$$
I am wondering:
- Does this map extend to what is called a "conditional expectation" on $C^*(G)$?
- If so, is this conditional expectation "faithful"?
functional-analysis operator-theory operator-algebras c-star-algebras
functional-analysis operator-theory operator-algebras c-star-algebras
asked Jan 26 at 7:43
ougoahougoah
1,256710
asked Jan 26 at 7:43
ougoahougoah
1,256710
asked Jan 26 at 7:43
ougoahougoah
1,256710
asked Jan 26 at 7:43
ougoahougoah
1,256710
asked Jan 26 at 7:43
ougoahougoah
1,256710
1,256710
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer to the second is negative. In fact, it is a well known characterization of amenability.
Indeed, a group $G$ is amenable iff the trace $tau:C^ast(G) to mathbb C$, that you define as $$f mapsto f(e)$$ is faithful. To see that, just use that if $tau$ is faithful so shall its associated GNS representation be. But the image of the associated GNS is $C_lambda^ast(G)$ and $C_lambda^ast(G)$ and $C^ast(G)$ are isomorphic iff $G$ is amenable.
The first is true since $tau$ is given by composing $q:C^ast(G) to C^ast_lambda(G)$ with the trace of $C_lambda^ast(G)$. But you can express that trace as a vector state $tau(x) = langle delta_e, x , delta_e rangle$.
answered Jan 26 at 23:59
Adrián González-PérezAdrián González-Pérez
1,206139
$endgroup$
2
$begingroup$
Is the answer also for general locally compact groups? or just discrete?
$endgroup$
– Shirly Geffen
Jan 27 at 15:07
1
$begingroup$
Perhaps I should add a clarification to the answer. The characterization of amenability works in the case of locally compact groups but with a modification. In that case $tau$ is not a trace but a tracial weight (in the case of unimodular groups) or a non-tracial weight in the case of general non-unimodular groups. In either case it still holds that the unbounded functional $tau: C_lambda^ast(G) _+ to [0,infty]$ (now is only well-defined in the positive cone of the algebra) is a faithful weight iff $G$ is amenable.
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:13
1
$begingroup$
For the first question: It also holds that you can extend $tau$ to a weight -not a conditional expectation- (although some people think of weights as "unbounded conditional expectations"). Of course, it would not be given by a vector state but by a sum of vector states. In the case in which the group is second countable you can take that sum to be countable. The construction of this is a little bit more technical. It is described in detail in Pedersen's book "$C^ast$-algebras and their automorphism groups".
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:17
1
$begingroup$
For the first question, if you want $tau$ to be a conditional expectation you need to impose that $G$ is discrete. Otherwise $tau(mathbb{1})$ would be infinity since, intuitively, $lambda_e$ is given by $f$ being a Dirac delta over the unit $e in G$. And evaluation in $e$ would give $infty$. This argument can be made rigorous by taking an increasing approximate unit on $C_lambda^ast(G)$
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:22
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The answer to the second is negative. In fact, it is a well known characterization of amenability.
Indeed, a group $G$ is amenable iff the trace $tau:C^ast(G) to mathbb C$, that you define as $$f mapsto f(e)$$ is faithful. To see that, just use that if $tau$ is faithful so shall its associated GNS representation be. But the image of the associated GNS is $C_lambda^ast(G)$ and $C_lambda^ast(G)$ and $C^ast(G)$ are isomorphic iff $G$ is amenable.
The first is true since $tau$ is given by composing $q:C^ast(G) to C^ast_lambda(G)$ with the trace of $C_lambda^ast(G)$. But you can express that trace as a vector state $tau(x) = langle delta_e, x , delta_e rangle$.
answered Jan 26 at 23:59
Adrián González-PérezAdrián González-Pérez
1,206139
$endgroup$
2
$begingroup$
Is the answer also for general locally compact groups? or just discrete?
$endgroup$
– Shirly Geffen
Jan 27 at 15:07
1
$begingroup$
Perhaps I should add a clarification to the answer. The characterization of amenability works in the case of locally compact groups but with a modification. In that case $tau$ is not a trace but a tracial weight (in the case of unimodular groups) or a non-tracial weight in the case of general non-unimodular groups. In either case it still holds that the unbounded functional $tau: C_lambda^ast(G) _+ to [0,infty]$ (now is only well-defined in the positive cone of the algebra) is a faithful weight iff $G$ is amenable.
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:13
1
$begingroup$
For the first question: It also holds that you can extend $tau$ to a weight -not a conditional expectation- (although some people think of weights as "unbounded conditional expectations"). Of course, it would not be given by a vector state but by a sum of vector states. In the case in which the group is second countable you can take that sum to be countable. The construction of this is a little bit more technical. It is described in detail in Pedersen's book "$C^ast$-algebras and their automorphism groups".
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:17
1
$begingroup$
For the first question, if you want $tau$ to be a conditional expectation you need to impose that $G$ is discrete. Otherwise $tau(mathbb{1})$ would be infinity since, intuitively, $lambda_e$ is given by $f$ being a Dirac delta over the unit $e in G$. And evaluation in $e$ would give $infty$. This argument can be made rigorous by taking an increasing approximate unit on $C_lambda^ast(G)$
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:22
add a comment |
$begingroup$
The answer to the second is negative. In fact, it is a well known characterization of amenability.
Indeed, a group $G$ is amenable iff the trace $tau:C^ast(G) to mathbb C$, that you define as $$f mapsto f(e)$$ is faithful. To see that, just use that if $tau$ is faithful so shall its associated GNS representation be. But the image of the associated GNS is $C_lambda^ast(G)$ and $C_lambda^ast(G)$ and $C^ast(G)$ are isomorphic iff $G$ is amenable.
The first is true since $tau$ is given by composing $q:C^ast(G) to C^ast_lambda(G)$ with the trace of $C_lambda^ast(G)$. But you can express that trace as a vector state $tau(x) = langle delta_e, x , delta_e rangle$.
answered Jan 26 at 23:59
Adrián González-PérezAdrián González-Pérez
1,206139
$endgroup$
2
$begingroup$
Is the answer also for general locally compact groups? or just discrete?
$endgroup$
– Shirly Geffen
Jan 27 at 15:07
1
$begingroup$
Perhaps I should add a clarification to the answer. The characterization of amenability works in the case of locally compact groups but with a modification. In that case $tau$ is not a trace but a tracial weight (in the case of unimodular groups) or a non-tracial weight in the case of general non-unimodular groups. In either case it still holds that the unbounded functional $tau: C_lambda^ast(G) _+ to [0,infty]$ (now is only well-defined in the positive cone of the algebra) is a faithful weight iff $G$ is amenable.
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:13
1
$begingroup$
For the first question: It also holds that you can extend $tau$ to a weight -not a conditional expectation- (although some people think of weights as "unbounded conditional expectations"). Of course, it would not be given by a vector state but by a sum of vector states. In the case in which the group is second countable you can take that sum to be countable. The construction of this is a little bit more technical. It is described in detail in Pedersen's book "$C^ast$-algebras and their automorphism groups".
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:17
1
$begingroup$
For the first question, if you want $tau$ to be a conditional expectation you need to impose that $G$ is discrete. Otherwise $tau(mathbb{1})$ would be infinity since, intuitively, $lambda_e$ is given by $f$ being a Dirac delta over the unit $e in G$. And evaluation in $e$ would give $infty$. This argument can be made rigorous by taking an increasing approximate unit on $C_lambda^ast(G)$
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:22
add a comment |
$begingroup$
The answer to the second is negative. In fact, it is a well known characterization of amenability.
Indeed, a group $G$ is amenable iff the trace $tau:C^ast(G) to mathbb C$, that you define as $$f mapsto f(e)$$ is faithful. To see that, just use that if $tau$ is faithful so shall its associated GNS representation be. But the image of the associated GNS is $C_lambda^ast(G)$ and $C_lambda^ast(G)$ and $C^ast(G)$ are isomorphic iff $G$ is amenable.
The first is true since $tau$ is given by composing $q:C^ast(G) to C^ast_lambda(G)$ with the trace of $C_lambda^ast(G)$. But you can express that trace as a vector state $tau(x) = langle delta_e, x , delta_e rangle$.
answered Jan 26 at 23:59
Adrián González-PérezAdrián González-Pérez
1,206139
$endgroup$
The answer to the second is negative. In fact, it is a well known characterization of amenability.
Indeed, a group $G$ is amenable iff the trace $tau:C^ast(G) to mathbb C$, that you define as $$f mapsto f(e)$$ is faithful. To see that, just use that if $tau$ is faithful so shall its associated GNS representation be. But the image of the associated GNS is $C_lambda^ast(G)$ and $C_lambda^ast(G)$ and $C^ast(G)$ are isomorphic iff $G$ is amenable.
The first is true since $tau$ is given by composing $q:C^ast(G) to C^ast_lambda(G)$ with the trace of $C_lambda^ast(G)$. But you can express that trace as a vector state $tau(x) = langle delta_e, x , delta_e rangle$.
answered Jan 26 at 23:59
Adrián González-PérezAdrián González-Pérez
1,206139
answered Jan 26 at 23:59
Adrián González-PérezAdrián González-Pérez
1,206139
answered Jan 26 at 23:59
Adrián González-PérezAdrián González-Pérez
1,206139
answered Jan 26 at 23:59
Adrián González-PérezAdrián González-Pérez
1,206139
1,206139
2
$begingroup$
Is the answer also for general locally compact groups? or just discrete?
$endgroup$
– Shirly Geffen
Jan 27 at 15:07
1
$begingroup$
Perhaps I should add a clarification to the answer. The characterization of amenability works in the case of locally compact groups but with a modification. In that case $tau$ is not a trace but a tracial weight (in the case of unimodular groups) or a non-tracial weight in the case of general non-unimodular groups. In either case it still holds that the unbounded functional $tau: C_lambda^ast(G) _+ to [0,infty]$ (now is only well-defined in the positive cone of the algebra) is a faithful weight iff $G$ is amenable.
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:13
1
$begingroup$
For the first question: It also holds that you can extend $tau$ to a weight -not a conditional expectation- (although some people think of weights as "unbounded conditional expectations"). Of course, it would not be given by a vector state but by a sum of vector states. In the case in which the group is second countable you can take that sum to be countable. The construction of this is a little bit more technical. It is described in detail in Pedersen's book "$C^ast$-algebras and their automorphism groups".
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:17
1
$begingroup$
For the first question, if you want $tau$ to be a conditional expectation you need to impose that $G$ is discrete. Otherwise $tau(mathbb{1})$ would be infinity since, intuitively, $lambda_e$ is given by $f$ being a Dirac delta over the unit $e in G$. And evaluation in $e$ would give $infty$. This argument can be made rigorous by taking an increasing approximate unit on $C_lambda^ast(G)$
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:22
add a comment |
2
$begingroup$
Is the answer also for general locally compact groups? or just discrete?
$endgroup$
– Shirly Geffen
Jan 27 at 15:07
1
$begingroup$
Perhaps I should add a clarification to the answer. The characterization of amenability works in the case of locally compact groups but with a modification. In that case $tau$ is not a trace but a tracial weight (in the case of unimodular groups) or a non-tracial weight in the case of general non-unimodular groups. In either case it still holds that the unbounded functional $tau: C_lambda^ast(G) _+ to [0,infty]$ (now is only well-defined in the positive cone of the algebra) is a faithful weight iff $G$ is amenable.
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:13
1
$begingroup$
For the first question: It also holds that you can extend $tau$ to a weight -not a conditional expectation- (although some people think of weights as "unbounded conditional expectations"). Of course, it would not be given by a vector state but by a sum of vector states. In the case in which the group is second countable you can take that sum to be countable. The construction of this is a little bit more technical. It is described in detail in Pedersen's book "$C^ast$-algebras and their automorphism groups".
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:17
1
$begingroup$
For the first question, if you want $tau$ to be a conditional expectation you need to impose that $G$ is discrete. Otherwise $tau(mathbb{1})$ would be infinity since, intuitively, $lambda_e$ is given by $f$ being a Dirac delta over the unit $e in G$. And evaluation in $e$ would give $infty$. This argument can be made rigorous by taking an increasing approximate unit on $C_lambda^ast(G)$
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:22
2
2
$begingroup$
Is the answer also for general locally compact groups? or just discrete?
$endgroup$
– Shirly Geffen
Jan 27 at 15:07
$begingroup$
Is the answer also for general locally compact groups? or just discrete?
$endgroup$
– Shirly Geffen
Jan 27 at 15:07
1
1
$begingroup$
Perhaps I should add a clarification to the answer. The characterization of amenability works in the case of locally compact groups but with a modification. In that case $tau$ is not a trace but a tracial weight (in the case of unimodular groups) or a non-tracial weight in the case of general non-unimodular groups. In either case it still holds that the unbounded functional $tau: C_lambda^ast(G) _+ to [0,infty]$ (now is only well-defined in the positive cone of the algebra) is a faithful weight iff $G$ is amenable.
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:13
$begingroup$
Perhaps I should add a clarification to the answer. The characterization of amenability works in the case of locally compact groups but with a modification. In that case $tau$ is not a trace but a tracial weight (in the case of unimodular groups) or a non-tracial weight in the case of general non-unimodular groups. In either case it still holds that the unbounded functional $tau: C_lambda^ast(G) _+ to [0,infty]$ (now is only well-defined in the positive cone of the algebra) is a faithful weight iff $G$ is amenable.
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:13
1
1
$begingroup$
For the first question: It also holds that you can extend $tau$ to a weight -not a conditional expectation- (although some people think of weights as "unbounded conditional expectations"). Of course, it would not be given by a vector state but by a sum of vector states. In the case in which the group is second countable you can take that sum to be countable. The construction of this is a little bit more technical. It is described in detail in Pedersen's book "$C^ast$-algebras and their automorphism groups".
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:17
$begingroup$
For the first question: It also holds that you can extend $tau$ to a weight -not a conditional expectation- (although some people think of weights as "unbounded conditional expectations"). Of course, it would not be given by a vector state but by a sum of vector states. In the case in which the group is second countable you can take that sum to be countable. The construction of this is a little bit more technical. It is described in detail in Pedersen's book "$C^ast$-algebras and their automorphism groups".
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:17
1
1
$begingroup$
For the first question, if you want $tau$ to be a conditional expectation you need to impose that $G$ is discrete. Otherwise $tau(mathbb{1})$ would be infinity since, intuitively, $lambda_e$ is given by $f$ being a Dirac delta over the unit $e in G$. And evaluation in $e$ would give $infty$. This argument can be made rigorous by taking an increasing approximate unit on $C_lambda^ast(G)$
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:22
$begingroup$
For the first question, if you want $tau$ to be a conditional expectation you need to impose that $G$ is discrete. Otherwise $tau(mathbb{1})$ would be infinity since, intuitively, $lambda_e$ is given by $f$ being a Dirac delta over the unit $e in G$. And evaluation in $e$ would give $infty$. This argument can be made rigorous by taking an increasing approximate unit on $C_lambda^ast(G)$
$endgroup$
– Adrián González-Pérez
Jan 27 at 15:22
add a comment |
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