Identification of Gelfand spectrum, $Sigma(mathcalA)$, with $sigma(A)$, spectrum of Banach Algebra element

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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.










share|cite|improve this question

























    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.










    share|cite|improve this question























      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.










      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 31 at 6:38









      Jagerber48

      1577




      1577




















          1 Answer
          1






          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 $sigma(A)$.






          share|cite|improve this answer




















            Your Answer




            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2900387%2fidentification-of-gelfand-spectrum-sigma-mathcala-with-sigmaa-spe%23new-answer', 'question_page');

            );

            Post as a guest






























            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 $sigma(A)$.






            share|cite|improve this answer
























              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 $sigma(A)$.






              share|cite|improve this answer






















                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 $sigma(A)$.






                share|cite|improve this answer












                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 $sigma(A)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 31 at 6:53









                Eric Wofsey

                166k12195308




                166k12195308



























                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2900387%2fidentification-of-gelfand-spectrum-sigma-mathcala-with-sigmaa-spe%23new-answer', 'question_page');

                    );

                    Post as a guest













































































                    這個網誌中的熱門文章

                    How to combine Bézier curves to a surface?

                    Mutual Information Always Non-negative

                    Why am i infinitely getting the same tweet with the Twitter Search API?