$C^*$ algebra, existence of particular state
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If we have $ain A$ be arbitrary element of $C^*$ algebra $A$. Can we find a faithful state $phi$ such that $phi(a) = k$ for $k$ in $spec(a)$?
banach-algebras
asked Nov 17 at 13:22
Sushil
989925
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up vote
1
down vote
favorite
If we have $ain A$ be arbitrary element of $C^*$ algebra $A$. Can we find a faithful state $phi$ such that $phi(a) = k$ for $k$ in $spec(a)$?
banach-algebras
asked Nov 17 at 13:22
Sushil
989925
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If we have $ain A$ be arbitrary element of $C^*$ algebra $A$. Can we find a faithful state $phi$ such that $phi(a) = k$ for $k$ in $spec(a)$?
banach-algebras
asked Nov 17 at 13:22
Sushil
989925
If we have $ain A$ be arbitrary element of $C^*$ algebra $A$. Can we find a faithful state $phi$ such that $phi(a) = k$ for $k$ in $spec(a)$?
banach-algebras
banach-algebras
asked Nov 17 at 13:22
Sushil
989925
asked Nov 17 at 13:22
Sushil
989925
asked Nov 17 at 13:22
Sushil
989925
asked Nov 17 at 13:22
Sushil
989925
asked Nov 17 at 13:22
Sushil
989925
989925
add a comment |
add a comment |
1 Answer
1
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1
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First, not every C $^*$-algebra has a separable state.
But even then, it may fail, and even when $ageq0$. Take $A=C_0 (mathbb R)+mathbb C,1$, and $a $ the function $a (t)=tfrac1{1+t^2} $. The spectrum of $a $ is its range, $[0,1] $. So take $1insigma (a) $ and suppose there is a state $phi $ with $phi (a)=1$. Then $$phi(1-a)=phi (1)-phi (a)=0. $$ As $phi $ is faithful, this would imply $1-a=0$, a contradiction. So no such $phi $ exists.
answered Nov 18 at 3:10
Martin Argerami
122k1173172
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First, not every C $^*$-algebra has a separable state.
But even then, it may fail, and even when $ageq0$. Take $A=C_0 (mathbb R)+mathbb C,1$, and $a $ the function $a (t)=tfrac1{1+t^2} $. The spectrum of $a $ is its range, $[0,1] $. So take $1insigma (a) $ and suppose there is a state $phi $ with $phi (a)=1$. Then $$phi(1-a)=phi (1)-phi (a)=0. $$ As $phi $ is faithful, this would imply $1-a=0$, a contradiction. So no such $phi $ exists.
answered Nov 18 at 3:10
Martin Argerami
122k1173172
add a comment |
up vote
1
down vote
accepted
First, not every C $^*$-algebra has a separable state.
But even then, it may fail, and even when $ageq0$. Take $A=C_0 (mathbb R)+mathbb C,1$, and $a $ the function $a (t)=tfrac1{1+t^2} $. The spectrum of $a $ is its range, $[0,1] $. So take $1insigma (a) $ and suppose there is a state $phi $ with $phi (a)=1$. Then $$phi(1-a)=phi (1)-phi (a)=0. $$ As $phi $ is faithful, this would imply $1-a=0$, a contradiction. So no such $phi $ exists.
answered Nov 18 at 3:10
Martin Argerami
122k1173172
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First, not every C $^*$-algebra has a separable state.
But even then, it may fail, and even when $ageq0$. Take $A=C_0 (mathbb R)+mathbb C,1$, and $a $ the function $a (t)=tfrac1{1+t^2} $. The spectrum of $a $ is its range, $[0,1] $. So take $1insigma (a) $ and suppose there is a state $phi $ with $phi (a)=1$. Then $$phi(1-a)=phi (1)-phi (a)=0. $$ As $phi $ is faithful, this would imply $1-a=0$, a contradiction. So no such $phi $ exists.
answered Nov 18 at 3:10
Martin Argerami
122k1173172
First, not every C $^*$-algebra has a separable state.
But even then, it may fail, and even when $ageq0$. Take $A=C_0 (mathbb R)+mathbb C,1$, and $a $ the function $a (t)=tfrac1{1+t^2} $. The spectrum of $a $ is its range, $[0,1] $. So take $1insigma (a) $ and suppose there is a state $phi $ with $phi (a)=1$. Then $$phi(1-a)=phi (1)-phi (a)=0. $$ As $phi $ is faithful, this would imply $1-a=0$, a contradiction. So no such $phi $ exists.
answered Nov 18 at 3:10
Martin Argerami
122k1173172
answered Nov 18 at 3:10
Martin Argerami
122k1173172
answered Nov 18 at 3:10
Martin Argerami
122k1173172
answered Nov 18 at 3:10
Martin Argerami
122k1173172
122k1173172
add a comment |
add a comment |
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