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Numerical range of selfadjoint elements in non-unital C*-algebras
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If $a$ is an element of a C*-algebra $A$ then $V(a)=varphi(a): varphitext is a state of A$ is the numerical range of $a$. If $a$ is selfadjoint and $A$ is unital then it is known that $V(a)=[minsigma(a),maxsigma(a)]$.
I am interested in a similar formula when $A$ is non-unital. More, precisely I would like to know the relationship between the numerical range and the spectrum of a selfadjoint (in particular, positive) element of a non-unital C*-algebra.
I see that the same formula cannot be true since $0=minsigma(a)notin V(a)$ if a is a strictly positive element of $A$.
operator-algebras c-star-algebras
edited Aug 6 at 19:48
Martin Argerami
116k1071164
asked Aug 6 at 10:41
aly
492312
add a comment |Â
up vote
2
down vote
favorite
If $a$ is an element of a C*-algebra $A$ then $V(a)=varphi(a): varphitext is a state of A$ is the numerical range of $a$. If $a$ is selfadjoint and $A$ is unital then it is known that $V(a)=[minsigma(a),maxsigma(a)]$.
I am interested in a similar formula when $A$ is non-unital. More, precisely I would like to know the relationship between the numerical range and the spectrum of a selfadjoint (in particular, positive) element of a non-unital C*-algebra.
I see that the same formula cannot be true since $0=minsigma(a)notin V(a)$ if a is a strictly positive element of $A$.
operator-algebras c-star-algebras
edited Aug 6 at 19:48
Martin Argerami
116k1071164
asked Aug 6 at 10:41
aly
492312
Is the set of all states connected if $A$ is unital?
– C.Ding
Aug 7 at 1:12
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $a$ is an element of a C*-algebra $A$ then $V(a)=varphi(a): varphitext is a state of A$ is the numerical range of $a$. If $a$ is selfadjoint and $A$ is unital then it is known that $V(a)=[minsigma(a),maxsigma(a)]$.
I am interested in a similar formula when $A$ is non-unital. More, precisely I would like to know the relationship between the numerical range and the spectrum of a selfadjoint (in particular, positive) element of a non-unital C*-algebra.
I see that the same formula cannot be true since $0=minsigma(a)notin V(a)$ if a is a strictly positive element of $A$.
operator-algebras c-star-algebras
edited Aug 6 at 19:48
Martin Argerami
116k1071164
asked Aug 6 at 10:41
aly
492312
If $a$ is an element of a C*-algebra $A$ then $V(a)=varphi(a): varphitext is a state of A$ is the numerical range of $a$. If $a$ is selfadjoint and $A$ is unital then it is known that $V(a)=[minsigma(a),maxsigma(a)]$.
I am interested in a similar formula when $A$ is non-unital. More, precisely I would like to know the relationship between the numerical range and the spectrum of a selfadjoint (in particular, positive) element of a non-unital C*-algebra.
I see that the same formula cannot be true since $0=minsigma(a)notin V(a)$ if a is a strictly positive element of $A$.
operator-algebras c-star-algebras
edited Aug 6 at 19:48
Martin Argerami
116k1071164
asked Aug 6 at 10:41
aly
492312
edited Aug 6 at 19:48
Martin Argerami
116k1071164
edited Aug 6 at 19:48
Martin Argerami
116k1071164
edited Aug 6 at 19:48
Martin Argerami
116k1071164
116k1071164
asked Aug 6 at 10:41
aly
492312
asked Aug 6 at 10:41
aly
492312
asked Aug 6 at 10:41
aly
492312
492312
Is the set of all states connected if $A$ is unital?
– C.Ding
Aug 7 at 1:12
add a comment |Â
Is the set of all states connected if $A$ is unital?
– C.Ding
Aug 7 at 1:12
Is the set of all states connected if $A$ is unital?
– C.Ding
Aug 7 at 1:12
Is the set of all states connected if $A$ is unital?
– C.Ding
Aug 7 at 1:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
It is well-known that, since $A$ is non-unital,
$$tag1
sigma(a)=tau(a): tau text is a character of C^*(a)cup0.
$$
Note that characters are states. This shows that $sigma(a)setminus0subset V(a)$. Take any state $varphi$ on $A$ and extend it (uniquely!) to $tilde A$. In $tilde A$, we have $$(minsigma(a),Ileq aleq (maxsigma(a)),I.$$ Applying the extension of $varphi$ to $(1)$, we get
$$
minsigma(a)leq varphi(a)leqmaxsigma(a).
$$
Thus
$$tag2
sigma(a)setminus0subset V(a)subset [minsigma(a),maxsigma(a)].
$$
As $V(a$ is convex, this inclusion is fairly tight: if $0notinminsigma(a),maxsigma(a)$, the convexity implies $V(a)=[minsigma(a),maxsigma(a)]$.
So the only pathological case is that where either $minsigma(a)=0$ or $maxsigma(a)=0$. Both cases are similar, since they are switched by considering $-a$. Say $minsigma(a)=0$. If there exists a state $varphi$ with $varphi(a)=0$, then $V(a)=[0,maxsigma(a)]$. Otherwise, $0notin V(a)$; in that case $0$ cannot be an isolated point in $sigma(a)$ (if $0$ is isolated, there exists a nonzero projection $p$ with $pa=0$; then any state $psi$ on $pAp$ induces a state $tildepsi$ on $A$ by $tildepsi(x)=psi(pxp)$, and $tildepsi(a)=0$). So we have that $[t,maxsigma(a)]subset V(a)$ for all $t>0$. Thus
$$
V(a)=(0,maxsigma(a)].
$$
answered Aug 6 at 15:42
Martin Argerami
116k1071164
Maybe I am missing something, but an element of $A$ cannot be invertible in $tildeA$ if $A$ is non-unital. The computation is simple: $(a,0)(b,lambda)=(0,1)$ iff $(ab+lambda a,0)=(0,1)$, which is impossible.
– aly
Aug 6 at 17:06
Probably the formula should be $[minsigma(a),maxsigma(a)]=V(a)cup0$.
– aly
Aug 6 at 17:07
I have rewritten the answer. And yes, your formula holds.
– Martin Argerami
Aug 7 at 2:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It is well-known that, since $A$ is non-unital,
$$tag1
sigma(a)=tau(a): tau text is a character of C^*(a)cup0.
$$
Note that characters are states. This shows that $sigma(a)setminus0subset V(a)$. Take any state $varphi$ on $A$ and extend it (uniquely!) to $tilde A$. In $tilde A$, we have $$(minsigma(a),Ileq aleq (maxsigma(a)),I.$$ Applying the extension of $varphi$ to $(1)$, we get
$$
minsigma(a)leq varphi(a)leqmaxsigma(a).
$$
Thus
$$tag2
sigma(a)setminus0subset V(a)subset [minsigma(a),maxsigma(a)].
$$
As $V(a$ is convex, this inclusion is fairly tight: if $0notinminsigma(a),maxsigma(a)$, the convexity implies $V(a)=[minsigma(a),maxsigma(a)]$.
So the only pathological case is that where either $minsigma(a)=0$ or $maxsigma(a)=0$. Both cases are similar, since they are switched by considering $-a$. Say $minsigma(a)=0$. If there exists a state $varphi$ with $varphi(a)=0$, then $V(a)=[0,maxsigma(a)]$. Otherwise, $0notin V(a)$; in that case $0$ cannot be an isolated point in $sigma(a)$ (if $0$ is isolated, there exists a nonzero projection $p$ with $pa=0$; then any state $psi$ on $pAp$ induces a state $tildepsi$ on $A$ by $tildepsi(x)=psi(pxp)$, and $tildepsi(a)=0$). So we have that $[t,maxsigma(a)]subset V(a)$ for all $t>0$. Thus
$$
V(a)=(0,maxsigma(a)].
$$
answered Aug 6 at 15:42
Martin Argerami
116k1071164
Maybe I am missing something, but an element of $A$ cannot be invertible in $tildeA$ if $A$ is non-unital. The computation is simple: $(a,0)(b,lambda)=(0,1)$ iff $(ab+lambda a,0)=(0,1)$, which is impossible.
– aly
Aug 6 at 17:06
Probably the formula should be $[minsigma(a),maxsigma(a)]=V(a)cup0$.
– aly
Aug 6 at 17:07
I have rewritten the answer. And yes, your formula holds.
– Martin Argerami
Aug 7 at 2:26
add a comment |Â
up vote
0
down vote
It is well-known that, since $A$ is non-unital,
$$tag1
sigma(a)=tau(a): tau text is a character of C^*(a)cup0.
$$
Note that characters are states. This shows that $sigma(a)setminus0subset V(a)$. Take any state $varphi$ on $A$ and extend it (uniquely!) to $tilde A$. In $tilde A$, we have $$(minsigma(a),Ileq aleq (maxsigma(a)),I.$$ Applying the extension of $varphi$ to $(1)$, we get
$$
minsigma(a)leq varphi(a)leqmaxsigma(a).
$$
Thus
$$tag2
sigma(a)setminus0subset V(a)subset [minsigma(a),maxsigma(a)].
$$
As $V(a$ is convex, this inclusion is fairly tight: if $0notinminsigma(a),maxsigma(a)$, the convexity implies $V(a)=[minsigma(a),maxsigma(a)]$.
So the only pathological case is that where either $minsigma(a)=0$ or $maxsigma(a)=0$. Both cases are similar, since they are switched by considering $-a$. Say $minsigma(a)=0$. If there exists a state $varphi$ with $varphi(a)=0$, then $V(a)=[0,maxsigma(a)]$. Otherwise, $0notin V(a)$; in that case $0$ cannot be an isolated point in $sigma(a)$ (if $0$ is isolated, there exists a nonzero projection $p$ with $pa=0$; then any state $psi$ on $pAp$ induces a state $tildepsi$ on $A$ by $tildepsi(x)=psi(pxp)$, and $tildepsi(a)=0$). So we have that $[t,maxsigma(a)]subset V(a)$ for all $t>0$. Thus
$$
V(a)=(0,maxsigma(a)].
$$
answered Aug 6 at 15:42
Martin Argerami
116k1071164
Maybe I am missing something, but an element of $A$ cannot be invertible in $tildeA$ if $A$ is non-unital. The computation is simple: $(a,0)(b,lambda)=(0,1)$ iff $(ab+lambda a,0)=(0,1)$, which is impossible.
– aly
Aug 6 at 17:06
Probably the formula should be $[minsigma(a),maxsigma(a)]=V(a)cup0$.
– aly
Aug 6 at 17:07
I have rewritten the answer. And yes, your formula holds.
– Martin Argerami
Aug 7 at 2:26
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is well-known that, since $A$ is non-unital,
$$tag1
sigma(a)=tau(a): tau text is a character of C^*(a)cup0.
$$
Note that characters are states. This shows that $sigma(a)setminus0subset V(a)$. Take any state $varphi$ on $A$ and extend it (uniquely!) to $tilde A$. In $tilde A$, we have $$(minsigma(a),Ileq aleq (maxsigma(a)),I.$$ Applying the extension of $varphi$ to $(1)$, we get
$$
minsigma(a)leq varphi(a)leqmaxsigma(a).
$$
Thus
$$tag2
sigma(a)setminus0subset V(a)subset [minsigma(a),maxsigma(a)].
$$
As $V(a$ is convex, this inclusion is fairly tight: if $0notinminsigma(a),maxsigma(a)$, the convexity implies $V(a)=[minsigma(a),maxsigma(a)]$.
So the only pathological case is that where either $minsigma(a)=0$ or $maxsigma(a)=0$. Both cases are similar, since they are switched by considering $-a$. Say $minsigma(a)=0$. If there exists a state $varphi$ with $varphi(a)=0$, then $V(a)=[0,maxsigma(a)]$. Otherwise, $0notin V(a)$; in that case $0$ cannot be an isolated point in $sigma(a)$ (if $0$ is isolated, there exists a nonzero projection $p$ with $pa=0$; then any state $psi$ on $pAp$ induces a state $tildepsi$ on $A$ by $tildepsi(x)=psi(pxp)$, and $tildepsi(a)=0$). So we have that $[t,maxsigma(a)]subset V(a)$ for all $t>0$. Thus
$$
V(a)=(0,maxsigma(a)].
$$
answered Aug 6 at 15:42
Martin Argerami
116k1071164
It is well-known that, since $A$ is non-unital,
$$tag1
sigma(a)=tau(a): tau text is a character of C^*(a)cup0.
$$
Note that characters are states. This shows that $sigma(a)setminus0subset V(a)$. Take any state $varphi$ on $A$ and extend it (uniquely!) to $tilde A$. In $tilde A$, we have $$(minsigma(a),Ileq aleq (maxsigma(a)),I.$$ Applying the extension of $varphi$ to $(1)$, we get
$$
minsigma(a)leq varphi(a)leqmaxsigma(a).
$$
Thus
$$tag2
sigma(a)setminus0subset V(a)subset [minsigma(a),maxsigma(a)].
$$
As $V(a$ is convex, this inclusion is fairly tight: if $0notinminsigma(a),maxsigma(a)$, the convexity implies $V(a)=[minsigma(a),maxsigma(a)]$.
So the only pathological case is that where either $minsigma(a)=0$ or $maxsigma(a)=0$. Both cases are similar, since they are switched by considering $-a$. Say $minsigma(a)=0$. If there exists a state $varphi$ with $varphi(a)=0$, then $V(a)=[0,maxsigma(a)]$. Otherwise, $0notin V(a)$; in that case $0$ cannot be an isolated point in $sigma(a)$ (if $0$ is isolated, there exists a nonzero projection $p$ with $pa=0$; then any state $psi$ on $pAp$ induces a state $tildepsi$ on $A$ by $tildepsi(x)=psi(pxp)$, and $tildepsi(a)=0$). So we have that $[t,maxsigma(a)]subset V(a)$ for all $t>0$. Thus
$$
V(a)=(0,maxsigma(a)].
$$
answered Aug 6 at 15:42
Martin Argerami
116k1071164
edited Aug 7 at 2:25
answered Aug 6 at 15:42
Martin Argerami
116k1071164
answered Aug 6 at 15:42
Martin Argerami
116k1071164
answered Aug 6 at 15:42
Martin Argerami
116k1071164
116k1071164
Maybe I am missing something, but an element of $A$ cannot be invertible in $tildeA$ if $A$ is non-unital. The computation is simple: $(a,0)(b,lambda)=(0,1)$ iff $(ab+lambda a,0)=(0,1)$, which is impossible.
– aly
Aug 6 at 17:06
Probably the formula should be $[minsigma(a),maxsigma(a)]=V(a)cup0$.
– aly
Aug 6 at 17:07
I have rewritten the answer. And yes, your formula holds.
– Martin Argerami
Aug 7 at 2:26
add a comment |Â
Maybe I am missing something, but an element of $A$ cannot be invertible in $tildeA$ if $A$ is non-unital. The computation is simple: $(a,0)(b,lambda)=(0,1)$ iff $(ab+lambda a,0)=(0,1)$, which is impossible.
– aly
Aug 6 at 17:06
Probably the formula should be $[minsigma(a),maxsigma(a)]=V(a)cup0$.
– aly
Aug 6 at 17:07
I have rewritten the answer. And yes, your formula holds.
– Martin Argerami
Aug 7 at 2:26
Maybe I am missing something, but an element of $A$ cannot be invertible in $tildeA$ if $A$ is non-unital. The computation is simple: $(a,0)(b,lambda)=(0,1)$ iff $(ab+lambda a,0)=(0,1)$, which is impossible.
– aly
Aug 6 at 17:06
Maybe I am missing something, but an element of $A$ cannot be invertible in $tildeA$ if $A$ is non-unital. The computation is simple: $(a,0)(b,lambda)=(0,1)$ iff $(ab+lambda a,0)=(0,1)$, which is impossible.
– aly
Aug 6 at 17:06
Probably the formula should be $[minsigma(a),maxsigma(a)]=V(a)cup0$.
– aly
Aug 6 at 17:07
Probably the formula should be $[minsigma(a),maxsigma(a)]=V(a)cup0$.
– aly
Aug 6 at 17:07
I have rewritten the answer. And yes, your formula holds.
– Martin Argerami
Aug 7 at 2:26
I have rewritten the answer. And yes, your formula holds.
– Martin Argerami
Aug 7 at 2:26
add a comment |Â
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Is the set of all states connected if $A$ is unital?
– C.Ding
Aug 7 at 1:12