What is the Gelfand-Naimark representation of functions that don't vanish at infinity?












3














The Gelfand-Naimark theorem says that every commutative C*-algebra is isometrically isomorphic to $C_0(X)$, the set of continuous functions $f:Xrightarrowmathbb{C}$ that vanish at infinity, for some locally compact Hausdorff space $X$.



What happens when we apply this construction to a commutative C*-algebra that looks almost like $C_0(X)$, but we relax one of the requirements? For example, we can consider the C*-algebra $C_b(Y)$, the set of bounded continuous functions $f:Yrightarrowmathbb{C}$, with no requirement they vanish at infinity. Is there any relation between the the space $Y$ of the original C*-algebra and the space $X$ that we generate using Gelfand-Naimark?



Bonus followup: What if Y is not Hausdorff? Or not locally compact?






abstract-algebra functional-analysis algebras gelfand-representation







share|cite|improve this question




















  • 2




    I think you need $C_b(Y)$ since $C(Y)$ is not necessarily a $C^*$-algebra if $Y$ is not compact. And if $Y$ is compact, then $C(Y)=C_0(Y)$.
    – SmileyCraft
    yesterday










  • I believe Stone Cech compactification has something to do with this en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification
    – SmileyCraft
    yesterday










  • @SmileyCraft You mean the functions need to be bounded to have a c*-compatible norm?
    – Jahan Claes
    yesterday










  • Exactly. At least for the standard norm (the supremum norm) to be $C^*$-compatible.
    – SmileyCraft
    yesterday
















3














The Gelfand-Naimark theorem says that every commutative C*-algebra is isometrically isomorphic to $C_0(X)$, the set of continuous functions $f:Xrightarrowmathbb{C}$ that vanish at infinity, for some locally compact Hausdorff space $X$.



What happens when we apply this construction to a commutative C*-algebra that looks almost like $C_0(X)$, but we relax one of the requirements? For example, we can consider the C*-algebra $C_b(Y)$, the set of bounded continuous functions $f:Yrightarrowmathbb{C}$, with no requirement they vanish at infinity. Is there any relation between the the space $Y$ of the original C*-algebra and the space $X$ that we generate using Gelfand-Naimark?



Bonus followup: What if Y is not Hausdorff? Or not locally compact?






abstract-algebra functional-analysis algebras gelfand-representation







share|cite|improve this question




















  • 2




    I think you need $C_b(Y)$ since $C(Y)$ is not necessarily a $C^*$-algebra if $Y$ is not compact. And if $Y$ is compact, then $C(Y)=C_0(Y)$.
    – SmileyCraft
    yesterday










  • I believe Stone Cech compactification has something to do with this en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification
    – SmileyCraft
    yesterday










  • @SmileyCraft You mean the functions need to be bounded to have a c*-compatible norm?
    – Jahan Claes
    yesterday










  • Exactly. At least for the standard norm (the supremum norm) to be $C^*$-compatible.
    – SmileyCraft
    yesterday














3












3








3







The Gelfand-Naimark theorem says that every commutative C*-algebra is isometrically isomorphic to $C_0(X)$, the set of continuous functions $f:Xrightarrowmathbb{C}$ that vanish at infinity, for some locally compact Hausdorff space $X$.



What happens when we apply this construction to a commutative C*-algebra that looks almost like $C_0(X)$, but we relax one of the requirements? For example, we can consider the C*-algebra $C_b(Y)$, the set of bounded continuous functions $f:Yrightarrowmathbb{C}$, with no requirement they vanish at infinity. Is there any relation between the the space $Y$ of the original C*-algebra and the space $X$ that we generate using Gelfand-Naimark?



Bonus followup: What if Y is not Hausdorff? Or not locally compact?






abstract-algebra functional-analysis algebras gelfand-representation







share|cite|improve this question















The Gelfand-Naimark theorem says that every commutative C*-algebra is isometrically isomorphic to $C_0(X)$, the set of continuous functions $f:Xrightarrowmathbb{C}$ that vanish at infinity, for some locally compact Hausdorff space $X$.



What happens when we apply this construction to a commutative C*-algebra that looks almost like $C_0(X)$, but we relax one of the requirements? For example, we can consider the C*-algebra $C_b(Y)$, the set of bounded continuous functions $f:Yrightarrowmathbb{C}$, with no requirement they vanish at infinity. Is there any relation between the the space $Y$ of the original C*-algebra and the space $X$ that we generate using Gelfand-Naimark?



Bonus followup: What if Y is not Hausdorff? Or not locally compact?






abstract-algebra functional-analysis algebras gelfand-representation




abstract-algebra functional-analysis algebras gelfand-representation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









Jahan Claes

889310




889310








  • 2




    I think you need $C_b(Y)$ since $C(Y)$ is not necessarily a $C^*$-algebra if $Y$ is not compact. And if $Y$ is compact, then $C(Y)=C_0(Y)$.
    – SmileyCraft
    yesterday










  • I believe Stone Cech compactification has something to do with this en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification
    – SmileyCraft
    yesterday










  • @SmileyCraft You mean the functions need to be bounded to have a c*-compatible norm?
    – Jahan Claes
    yesterday










  • Exactly. At least for the standard norm (the supremum norm) to be $C^*$-compatible.
    – SmileyCraft
    yesterday














  • 2




    I think you need $C_b(Y)$ since $C(Y)$ is not necessarily a $C^*$-algebra if $Y$ is not compact. And if $Y$ is compact, then $C(Y)=C_0(Y)$.
    – SmileyCraft
    yesterday










  • I believe Stone Cech compactification has something to do with this en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification
    – SmileyCraft
    yesterday










  • @SmileyCraft You mean the functions need to be bounded to have a c*-compatible norm?
    – Jahan Claes
    yesterday










  • Exactly. At least for the standard norm (the supremum norm) to be $C^*$-compatible.
    – SmileyCraft
    yesterday








2




2




I think you need $C_b(Y)$ since $C(Y)$ is not necessarily a $C^*$-algebra if $Y$ is not compact. And if $Y$ is compact, then $C(Y)=C_0(Y)$.
– SmileyCraft
yesterday




I think you need $C_b(Y)$ since $C(Y)$ is not necessarily a $C^*$-algebra if $Y$ is not compact. And if $Y$ is compact, then $C(Y)=C_0(Y)$.
– SmileyCraft
yesterday












I believe Stone Cech compactification has something to do with this en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification
– SmileyCraft
yesterday




I believe Stone Cech compactification has something to do with this en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification
– SmileyCraft
yesterday












@SmileyCraft You mean the functions need to be bounded to have a c*-compatible norm?
– Jahan Claes
yesterday




@SmileyCraft You mean the functions need to be bounded to have a c*-compatible norm?
– Jahan Claes
yesterday












Exactly. At least for the standard norm (the supremum norm) to be $C^*$-compatible.
– SmileyCraft
yesterday




Exactly. At least for the standard norm (the supremum norm) to be $C^*$-compatible.
– SmileyCraft
yesterday










1 Answer
1






active

oldest

votes


















2














Since $C_b(Y)$ has a unit, if it is isometrically $^*$-isomorphic to $C_0(X)$ then $C_0(X)$ must also have a unit, so $X$ is compact. This is intuitively why $X$ will need to be some sort of compactification of $Y$. To be exact, we will get the Stone Cech compactification.



The Stone Cech compactification $beta Y$ of $Y$ comes with an embedding $Delta:Ytobeta Y$ such that $Delta(Y)$ is dense in $beta Y$. The important property is that $fmapsto fcircDelta:C(beta Y)to C_b(Y)$ is an isometric $^*$-isomorphism. Since $beta Y$ is by definition also compact, we have $C_0(beta Y)=C(beta Y)$, so $C_b(Y)$ is isometrically $^*$-isomorphic to $C_0(beta Y)$.



Do note that $Y$ is required to be completely regular for the Stone Cech compactification to be well-defined. I do not know about the case where $Y$ is not completely regular.






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',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    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%2f3061990%2fwhat-is-the-gelfand-naimark-representation-of-functions-that-dont-vanish-at-inf%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Since $C_b(Y)$ has a unit, if it is isometrically $^*$-isomorphic to $C_0(X)$ then $C_0(X)$ must also have a unit, so $X$ is compact. This is intuitively why $X$ will need to be some sort of compactification of $Y$. To be exact, we will get the Stone Cech compactification.



    The Stone Cech compactification $beta Y$ of $Y$ comes with an embedding $Delta:Ytobeta Y$ such that $Delta(Y)$ is dense in $beta Y$. The important property is that $fmapsto fcircDelta:C(beta Y)to C_b(Y)$ is an isometric $^*$-isomorphism. Since $beta Y$ is by definition also compact, we have $C_0(beta Y)=C(beta Y)$, so $C_b(Y)$ is isometrically $^*$-isomorphic to $C_0(beta Y)$.



    Do note that $Y$ is required to be completely regular for the Stone Cech compactification to be well-defined. I do not know about the case where $Y$ is not completely regular.






    share|cite|improve this answer


























      2














      Since $C_b(Y)$ has a unit, if it is isometrically $^*$-isomorphic to $C_0(X)$ then $C_0(X)$ must also have a unit, so $X$ is compact. This is intuitively why $X$ will need to be some sort of compactification of $Y$. To be exact, we will get the Stone Cech compactification.



      The Stone Cech compactification $beta Y$ of $Y$ comes with an embedding $Delta:Ytobeta Y$ such that $Delta(Y)$ is dense in $beta Y$. The important property is that $fmapsto fcircDelta:C(beta Y)to C_b(Y)$ is an isometric $^*$-isomorphism. Since $beta Y$ is by definition also compact, we have $C_0(beta Y)=C(beta Y)$, so $C_b(Y)$ is isometrically $^*$-isomorphic to $C_0(beta Y)$.



      Do note that $Y$ is required to be completely regular for the Stone Cech compactification to be well-defined. I do not know about the case where $Y$ is not completely regular.






      share|cite|improve this answer
























        2












        2








        2






        Since $C_b(Y)$ has a unit, if it is isometrically $^*$-isomorphic to $C_0(X)$ then $C_0(X)$ must also have a unit, so $X$ is compact. This is intuitively why $X$ will need to be some sort of compactification of $Y$. To be exact, we will get the Stone Cech compactification.



        The Stone Cech compactification $beta Y$ of $Y$ comes with an embedding $Delta:Ytobeta Y$ such that $Delta(Y)$ is dense in $beta Y$. The important property is that $fmapsto fcircDelta:C(beta Y)to C_b(Y)$ is an isometric $^*$-isomorphism. Since $beta Y$ is by definition also compact, we have $C_0(beta Y)=C(beta Y)$, so $C_b(Y)$ is isometrically $^*$-isomorphic to $C_0(beta Y)$.



        Do note that $Y$ is required to be completely regular for the Stone Cech compactification to be well-defined. I do not know about the case where $Y$ is not completely regular.






        share|cite|improve this answer












        Since $C_b(Y)$ has a unit, if it is isometrically $^*$-isomorphic to $C_0(X)$ then $C_0(X)$ must also have a unit, so $X$ is compact. This is intuitively why $X$ will need to be some sort of compactification of $Y$. To be exact, we will get the Stone Cech compactification.



        The Stone Cech compactification $beta Y$ of $Y$ comes with an embedding $Delta:Ytobeta Y$ such that $Delta(Y)$ is dense in $beta Y$. The important property is that $fmapsto fcircDelta:C(beta Y)to C_b(Y)$ is an isometric $^*$-isomorphism. Since $beta Y$ is by definition also compact, we have $C_0(beta Y)=C(beta Y)$, so $C_b(Y)$ is isometrically $^*$-isomorphic to $C_0(beta Y)$.



        Do note that $Y$ is required to be completely regular for the Stone Cech compactification to be well-defined. I do not know about the case where $Y$ is not completely regular.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        SmileyCraft

        2,971416




        2,971416






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061990%2fwhat-is-the-gelfand-naimark-representation-of-functions-that-dont-vanish-at-inf%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            Understanding the size os this class of aleatory events

            Image
            0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
            Read more

            Partial Derivative Guidance.

            Image
            0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
            Read more