Identification of Gelfand spectrum, $Sigma(mathcalA)$, with $sigma(A)$, spectrum of Banach Algebra element
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I am reading Mathematical Structure of Quantum Mechanics A short Course for Mathematicians by F. Strocchi.
My question is about the text after Proposition 1.5.3. The proposition is that
For a commutative Banach algebra $mathcalA$ there is a one to one correspondence between the set $Sigma(mathcalA)$ of multiplicative linear functionals and proper maximal ideals of $mathcalA$.
Furthermore, given $Ain mathcalA$, $lambda in sigma(A)$ iff there exists a multiplicative linear functional $m in Sigma(mathcalA)$ such that $m(A) = lambda$.
Afterwards Strocchi calls $Sigma(mathcalA)$ the Gelfand spectrum of $mathcalA$ and claims
Indeed, if $mathcalA$ is generated by a single element $A$, i.e. the linear span of the powers of $A$ is dense in $mathcalA$, then $Sigma(mathcalA)= sigma(A)$.
I don't understand this very last statement that $Sigma(mathcalA)=sigma(A)$. I see that there might be a correspondence between elements of $Sigma(mathcalA)$ and elements of $sigma(A)$ but I can't see how these two sets could be equal. My issue is that elements of $sigma(A)$ are complex numbers, $lambda$, (such that $A-lambda mathbb1$ is not invertible) whereas elements of $Sigma(mathcalA)$ are multiplicative linear functionals on $mathcalA$, i.e. not necessarily complex numbers.
Is the statement that $Sigma(mathcalA)=sigma(A)$ just making the point that there is a one to one correspondence between the two sets so we might as well call them equal. Perhaps since there is a homomorphism involved it is justified/conventional? I work in physics so I don't know what might be conventional in mathematics texts. My stronger suspicion is that I'm missing something.
notation banach-algebras gelfand-representation
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I am reading Mathematical Structure of Quantum Mechanics A short Course for Mathematicians by F. Strocchi.
My question is about the text after Proposition 1.5.3. The proposition is that
For a commutative Banach algebra $mathcalA$ there is a one to one correspondence between the set $Sigma(mathcalA)$ of multiplicative linear functionals and proper maximal ideals of $mathcalA$.
Furthermore, given $Ain mathcalA$, $lambda in sigma(A)$ iff there exists a multiplicative linear functional $m in Sigma(mathcalA)$ such that $m(A) = lambda$.
Afterwards Strocchi calls $Sigma(mathcalA)$ the Gelfand spectrum of $mathcalA$ and claims
Indeed, if $mathcalA$ is generated by a single element $A$, i.e. the linear span of the powers of $A$ is dense in $mathcalA$, then $Sigma(mathcalA)= sigma(A)$.
I don't understand this very last statement that $Sigma(mathcalA)=sigma(A)$. I see that there might be a correspondence between elements of $Sigma(mathcalA)$ and elements of $sigma(A)$ but I can't see how these two sets could be equal. My issue is that elements of $sigma(A)$ are complex numbers, $lambda$, (such that $A-lambda mathbb1$ is not invertible) whereas elements of $Sigma(mathcalA)$ are multiplicative linear functionals on $mathcalA$, i.e. not necessarily complex numbers.
Is the statement that $Sigma(mathcalA)=sigma(A)$ just making the point that there is a one to one correspondence between the two sets so we might as well call them equal. Perhaps since there is a homomorphism involved it is justified/conventional? I work in physics so I don't know what might be conventional in mathematics texts. My stronger suspicion is that I'm missing something.
notation banach-algebras gelfand-representation
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading Mathematical Structure of Quantum Mechanics A short Course for Mathematicians by F. Strocchi.
My question is about the text after Proposition 1.5.3. The proposition is that
For a commutative Banach algebra $mathcalA$ there is a one to one correspondence between the set $Sigma(mathcalA)$ of multiplicative linear functionals and proper maximal ideals of $mathcalA$.
Furthermore, given $Ain mathcalA$, $lambda in sigma(A)$ iff there exists a multiplicative linear functional $m in Sigma(mathcalA)$ such that $m(A) = lambda$.
Afterwards Strocchi calls $Sigma(mathcalA)$ the Gelfand spectrum of $mathcalA$ and claims
Indeed, if $mathcalA$ is generated by a single element $A$, i.e. the linear span of the powers of $A$ is dense in $mathcalA$, then $Sigma(mathcalA)= sigma(A)$.
I don't understand this very last statement that $Sigma(mathcalA)=sigma(A)$. I see that there might be a correspondence between elements of $Sigma(mathcalA)$ and elements of $sigma(A)$ but I can't see how these two sets could be equal. My issue is that elements of $sigma(A)$ are complex numbers, $lambda$, (such that $A-lambda mathbb1$ is not invertible) whereas elements of $Sigma(mathcalA)$ are multiplicative linear functionals on $mathcalA$, i.e. not necessarily complex numbers.
Is the statement that $Sigma(mathcalA)=sigma(A)$ just making the point that there is a one to one correspondence between the two sets so we might as well call them equal. Perhaps since there is a homomorphism involved it is justified/conventional? I work in physics so I don't know what might be conventional in mathematics texts. My stronger suspicion is that I'm missing something.
notation banach-algebras gelfand-representation
I am reading Mathematical Structure of Quantum Mechanics A short Course for Mathematicians by F. Strocchi.
My question is about the text after Proposition 1.5.3. The proposition is that
For a commutative Banach algebra $mathcalA$ there is a one to one correspondence between the set $Sigma(mathcalA)$ of multiplicative linear functionals and proper maximal ideals of $mathcalA$.
Furthermore, given $Ain mathcalA$, $lambda in sigma(A)$ iff there exists a multiplicative linear functional $m in Sigma(mathcalA)$ such that $m(A) = lambda$.
Afterwards Strocchi calls $Sigma(mathcalA)$ the Gelfand spectrum of $mathcalA$ and claims
Indeed, if $mathcalA$ is generated by a single element $A$, i.e. the linear span of the powers of $A$ is dense in $mathcalA$, then $Sigma(mathcalA)= sigma(A)$.
I don't understand this very last statement that $Sigma(mathcalA)=sigma(A)$. I see that there might be a correspondence between elements of $Sigma(mathcalA)$ and elements of $sigma(A)$ but I can't see how these two sets could be equal. My issue is that elements of $sigma(A)$ are complex numbers, $lambda$, (such that $A-lambda mathbb1$ is not invertible) whereas elements of $Sigma(mathcalA)$ are multiplicative linear functionals on $mathcalA$, i.e. not necessarily complex numbers.
Is the statement that $Sigma(mathcalA)=sigma(A)$ just making the point that there is a one to one correspondence between the two sets so we might as well call them equal. Perhaps since there is a homomorphism involved it is justified/conventional? I work in physics so I don't know what might be conventional in mathematics texts. My stronger suspicion is that I'm missing something.
notation banach-algebras gelfand-representation
notation banach-algebras gelfand-representation
asked Aug 31 at 6:38
Jagerber48
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You're not really missing anything here. The statement $Sigma(mathcalA)=sigma(A)$ is just a sloppy way of saying there is a canonical bijection between these two sets (which in fact is a homeomorphism for their natural topologies). It is not exactly normal for mathematicians to write something like this (we would more properly write $Sigma(mathcalA)cong sigma(A)$ instead), but this kind of sloppiness is reasonably common.
Explicitly, the canonical bijection is defined as follows: given $minSigma(mathcalA)$, we map it to $m(A)$, which is an element of $sigma(A)$.
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
You're not really missing anything here. The statement $Sigma(mathcalA)=sigma(A)$ is just a sloppy way of saying there is a canonical bijection between these two sets (which in fact is a homeomorphism for their natural topologies). It is not exactly normal for mathematicians to write something like this (we would more properly write $Sigma(mathcalA)cong sigma(A)$ instead), but this kind of sloppiness is reasonably common.
Explicitly, the canonical bijection is defined as follows: given $minSigma(mathcalA)$, we map it to $m(A)$, which is an element of $sigma(A)$.
add a comment |Â
up vote
1
down vote
accepted
You're not really missing anything here. The statement $Sigma(mathcalA)=sigma(A)$ is just a sloppy way of saying there is a canonical bijection between these two sets (which in fact is a homeomorphism for their natural topologies). It is not exactly normal for mathematicians to write something like this (we would more properly write $Sigma(mathcalA)cong sigma(A)$ instead), but this kind of sloppiness is reasonably common.
Explicitly, the canonical bijection is defined as follows: given $minSigma(mathcalA)$, we map it to $m(A)$, which is an element of $sigma(A)$.
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You're not really missing anything here. The statement $Sigma(mathcalA)=sigma(A)$ is just a sloppy way of saying there is a canonical bijection between these two sets (which in fact is a homeomorphism for their natural topologies). It is not exactly normal for mathematicians to write something like this (we would more properly write $Sigma(mathcalA)cong sigma(A)$ instead), but this kind of sloppiness is reasonably common.
Explicitly, the canonical bijection is defined as follows: given $minSigma(mathcalA)$, we map it to $m(A)$, which is an element of $sigma(A)$.
You're not really missing anything here. The statement $Sigma(mathcalA)=sigma(A)$ is just a sloppy way of saying there is a canonical bijection between these two sets (which in fact is a homeomorphism for their natural topologies). It is not exactly normal for mathematicians to write something like this (we would more properly write $Sigma(mathcalA)cong sigma(A)$ instead), but this kind of sloppiness is reasonably common.
Explicitly, the canonical bijection is defined as follows: given $minSigma(mathcalA)$, we map it to $m(A)$, which is an element of $sigma(A)$.
answered Aug 31 at 6:53
Eric Wofsey
166k12195308
166k12195308
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