Bicommutant theorem for commutative operator algebras












3












$begingroup$


Let $mathcal{B}(H)$ denote the space of bounded linear operators on a complex Hilbert space $H$. The von Neumann bicommutant theorem says:



Theorem. Suppose that $mathcal{A}$ is a $C^*$-subalgebra of $mathcal{B}(H)$ and that $mathcal{A}$ contains the identity operator. If $mathcal{A}$ is closed with respect to the strong operator topology, then $mathcal{A}$ coincides with its bicommutant $mathcal{A}''$.



[Note: The commutant $mathcal{S}'$ of a subset $mathcal{S} subseteq mathcal{B}(H)$ is defined as the set of all operators in $mathcal{B}(H)$ that commute with all operators in $mathcal{S}$.]



Question. Does this theorem remain true if we replace the assumption "$mathcal{A}$ is a $C^*$-subalgebra of $mathcal{B}(H)$" with the assumption "$mathcal{A}$ is a commutative subalgebra of $mathcal{B}(H)$"?



The question seems to be a bit bold at first glance, but it is motivated by the following simple finite-fimensional observation:



Motivation. The bicommutant theorem fails if we merely assume $mathcal{A}$ to be a subalgebra of $mathcal{B}(H)$ that is strongly closed and contains the identity operator. As a simple counterexample we can choose $mathcal{A}$ to be the set of all upper triangular matrices in $mathbb{C}^{2times 2} = mathcal{B}(mathbb{C}^2)$.



On the other hand, the algebra $mathcal{A}$ is not commutative in this example, and if we try to "adjust" the example by choosing $mathcal{A}$ as the commutative algebra
begin{align*}
mathcal{A} := {
begin{pmatrix}
a & b newline
0 & a
end{pmatrix}: ;
a, b in mathbb{C}
},
end{align*},
then we indeed have $mathcal{A} = mathcal{A}''$.






fa.functional-analysis matrices oa.operator-algebras operator-theory banach-algebras







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let $mathcal{B}(H)$ denote the space of bounded linear operators on a complex Hilbert space $H$. The von Neumann bicommutant theorem says:



    Theorem. Suppose that $mathcal{A}$ is a $C^*$-subalgebra of $mathcal{B}(H)$ and that $mathcal{A}$ contains the identity operator. If $mathcal{A}$ is closed with respect to the strong operator topology, then $mathcal{A}$ coincides with its bicommutant $mathcal{A}''$.



    [Note: The commutant $mathcal{S}'$ of a subset $mathcal{S} subseteq mathcal{B}(H)$ is defined as the set of all operators in $mathcal{B}(H)$ that commute with all operators in $mathcal{S}$.]



    Question. Does this theorem remain true if we replace the assumption "$mathcal{A}$ is a $C^*$-subalgebra of $mathcal{B}(H)$" with the assumption "$mathcal{A}$ is a commutative subalgebra of $mathcal{B}(H)$"?



    The question seems to be a bit bold at first glance, but it is motivated by the following simple finite-fimensional observation:



    Motivation. The bicommutant theorem fails if we merely assume $mathcal{A}$ to be a subalgebra of $mathcal{B}(H)$ that is strongly closed and contains the identity operator. As a simple counterexample we can choose $mathcal{A}$ to be the set of all upper triangular matrices in $mathbb{C}^{2times 2} = mathcal{B}(mathbb{C}^2)$.



    On the other hand, the algebra $mathcal{A}$ is not commutative in this example, and if we try to "adjust" the example by choosing $mathcal{A}$ as the commutative algebra
    begin{align*}
    mathcal{A} := {
    begin{pmatrix}
    a & b newline
    0 & a
    end{pmatrix}: ;
    a, b in mathbb{C}
    },
    end{align*}    ,
    then we indeed have $mathcal{A} = mathcal{A}''$.






    fa.functional-analysis matrices oa.operator-algebras operator-theory banach-algebras







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $mathcal{B}(H)$ denote the space of bounded linear operators on a complex Hilbert space $H$. The von Neumann bicommutant theorem says:



      Theorem. Suppose that $mathcal{A}$ is a $C^*$-subalgebra of $mathcal{B}(H)$ and that $mathcal{A}$ contains the identity operator. If $mathcal{A}$ is closed with respect to the strong operator topology, then $mathcal{A}$ coincides with its bicommutant $mathcal{A}''$.



      [Note: The commutant $mathcal{S}'$ of a subset $mathcal{S} subseteq mathcal{B}(H)$ is defined as the set of all operators in $mathcal{B}(H)$ that commute with all operators in $mathcal{S}$.]



      Question. Does this theorem remain true if we replace the assumption "$mathcal{A}$ is a $C^*$-subalgebra of $mathcal{B}(H)$" with the assumption "$mathcal{A}$ is a commutative subalgebra of $mathcal{B}(H)$"?



      The question seems to be a bit bold at first glance, but it is motivated by the following simple finite-fimensional observation:



      Motivation. The bicommutant theorem fails if we merely assume $mathcal{A}$ to be a subalgebra of $mathcal{B}(H)$ that is strongly closed and contains the identity operator. As a simple counterexample we can choose $mathcal{A}$ to be the set of all upper triangular matrices in $mathbb{C}^{2times 2} = mathcal{B}(mathbb{C}^2)$.



      On the other hand, the algebra $mathcal{A}$ is not commutative in this example, and if we try to "adjust" the example by choosing $mathcal{A}$ as the commutative algebra
      begin{align*}
      mathcal{A} := {
      begin{pmatrix}
      a & b newline
      0 & a
      end{pmatrix}: ;
      a, b in mathbb{C}
      },
      end{align*}      ,
      then we indeed have $mathcal{A} = mathcal{A}''$.






      fa.functional-analysis matrices oa.operator-algebras operator-theory banach-algebras







      share|cite|improve this question









      $endgroup$




      Let $mathcal{B}(H)$ denote the space of bounded linear operators on a complex Hilbert space $H$. The von Neumann bicommutant theorem says:



      Theorem. Suppose that $mathcal{A}$ is a $C^*$-subalgebra of $mathcal{B}(H)$ and that $mathcal{A}$ contains the identity operator. If $mathcal{A}$ is closed with respect to the strong operator topology, then $mathcal{A}$ coincides with its bicommutant $mathcal{A}''$.



      [Note: The commutant $mathcal{S}'$ of a subset $mathcal{S} subseteq mathcal{B}(H)$ is defined as the set of all operators in $mathcal{B}(H)$ that commute with all operators in $mathcal{S}$.]



      Question. Does this theorem remain true if we replace the assumption "$mathcal{A}$ is a $C^*$-subalgebra of $mathcal{B}(H)$" with the assumption "$mathcal{A}$ is a commutative subalgebra of $mathcal{B}(H)$"?



      The question seems to be a bit bold at first glance, but it is motivated by the following simple finite-fimensional observation:



      Motivation. The bicommutant theorem fails if we merely assume $mathcal{A}$ to be a subalgebra of $mathcal{B}(H)$ that is strongly closed and contains the identity operator. As a simple counterexample we can choose $mathcal{A}$ to be the set of all upper triangular matrices in $mathbb{C}^{2times 2} = mathcal{B}(mathbb{C}^2)$.



      On the other hand, the algebra $mathcal{A}$ is not commutative in this example, and if we try to "adjust" the example by choosing $mathcal{A}$ as the commutative algebra
      begin{align*}
      mathcal{A} := {
      begin{pmatrix}
      a & b newline
      0 & a
      end{pmatrix}: ;
      a, b in mathbb{C}
      },
      end{align*}      ,
      then we indeed have $mathcal{A} = mathcal{A}''$.






      fa.functional-analysis matrices oa.operator-algebras operator-theory banach-algebras




      fa.functional-analysis matrices oa.operator-algebras operator-theory banach-algebras






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 7 hours ago









      Jochen GlueckJochen Glueck

      2,94611323




      2,94611323






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          The answer is negative. Consider the Hardy space $H^{infty}(mathbb{T})$ consisting of holomorphic functions admitting bounded extension to the unit circle and view it is a subalgebra of $B(L^{2}(mathbb{T})$. Let us compute the commutant. Denote the operator of multiplication by $z$ by $M_z$. If $T$ belongs to the commutant of $H^{infty}(mathbb{T})$, it commutes with $M_z$. But it also has to commute with its inverse $M_{overline{z}}$. But these two generate $L^{infty}(mathbb{T})$, so the commutant is equal to $L^{infty}(mathbb{T})$, hence also the bicommutant, as this a maximal abelian subalgebra of $B(L^{2}(mathbb{T}))$. It remains to see that $H^{infty}(mathbb{T})$ is strongly closed, but this is simple: a function $fin L^{infty}(mathbb{T})$ belongs to $H^{infty}(mathbb{T})$ iff all of its negative Fourier coefficients vanish and this property is clearly preserved by strong convergence (even by weak convergence).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's great, thanks!
            $endgroup$
            – Jochen Glueck
            4 hours ago











          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: "504"
          };
          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%2fmathoverflow.net%2fquestions%2f323585%2fbicommutant-theorem-for-commutative-operator-algebras%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









          5












          $begingroup$

          The answer is negative. Consider the Hardy space $H^{infty}(mathbb{T})$ consisting of holomorphic functions admitting bounded extension to the unit circle and view it is a subalgebra of $B(L^{2}(mathbb{T})$. Let us compute the commutant. Denote the operator of multiplication by $z$ by $M_z$. If $T$ belongs to the commutant of $H^{infty}(mathbb{T})$, it commutes with $M_z$. But it also has to commute with its inverse $M_{overline{z}}$. But these two generate $L^{infty}(mathbb{T})$, so the commutant is equal to $L^{infty}(mathbb{T})$, hence also the bicommutant, as this a maximal abelian subalgebra of $B(L^{2}(mathbb{T}))$. It remains to see that $H^{infty}(mathbb{T})$ is strongly closed, but this is simple: a function $fin L^{infty}(mathbb{T})$ belongs to $H^{infty}(mathbb{T})$ iff all of its negative Fourier coefficients vanish and this property is clearly preserved by strong convergence (even by weak convergence).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's great, thanks!
            $endgroup$
            – Jochen Glueck
            4 hours ago
















          5












          $begingroup$

          The answer is negative. Consider the Hardy space $H^{infty}(mathbb{T})$ consisting of holomorphic functions admitting bounded extension to the unit circle and view it is a subalgebra of $B(L^{2}(mathbb{T})$. Let us compute the commutant. Denote the operator of multiplication by $z$ by $M_z$. If $T$ belongs to the commutant of $H^{infty}(mathbb{T})$, it commutes with $M_z$. But it also has to commute with its inverse $M_{overline{z}}$. But these two generate $L^{infty}(mathbb{T})$, so the commutant is equal to $L^{infty}(mathbb{T})$, hence also the bicommutant, as this a maximal abelian subalgebra of $B(L^{2}(mathbb{T}))$. It remains to see that $H^{infty}(mathbb{T})$ is strongly closed, but this is simple: a function $fin L^{infty}(mathbb{T})$ belongs to $H^{infty}(mathbb{T})$ iff all of its negative Fourier coefficients vanish and this property is clearly preserved by strong convergence (even by weak convergence).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's great, thanks!
            $endgroup$
            – Jochen Glueck
            4 hours ago














          5












          5








          5





          $begingroup$

          The answer is negative. Consider the Hardy space $H^{infty}(mathbb{T})$ consisting of holomorphic functions admitting bounded extension to the unit circle and view it is a subalgebra of $B(L^{2}(mathbb{T})$. Let us compute the commutant. Denote the operator of multiplication by $z$ by $M_z$. If $T$ belongs to the commutant of $H^{infty}(mathbb{T})$, it commutes with $M_z$. But it also has to commute with its inverse $M_{overline{z}}$. But these two generate $L^{infty}(mathbb{T})$, so the commutant is equal to $L^{infty}(mathbb{T})$, hence also the bicommutant, as this a maximal abelian subalgebra of $B(L^{2}(mathbb{T}))$. It remains to see that $H^{infty}(mathbb{T})$ is strongly closed, but this is simple: a function $fin L^{infty}(mathbb{T})$ belongs to $H^{infty}(mathbb{T})$ iff all of its negative Fourier coefficients vanish and this property is clearly preserved by strong convergence (even by weak convergence).






          share|cite|improve this answer









          $endgroup$



          The answer is negative. Consider the Hardy space $H^{infty}(mathbb{T})$ consisting of holomorphic functions admitting bounded extension to the unit circle and view it is a subalgebra of $B(L^{2}(mathbb{T})$. Let us compute the commutant. Denote the operator of multiplication by $z$ by $M_z$. If $T$ belongs to the commutant of $H^{infty}(mathbb{T})$, it commutes with $M_z$. But it also has to commute with its inverse $M_{overline{z}}$. But these two generate $L^{infty}(mathbb{T})$, so the commutant is equal to $L^{infty}(mathbb{T})$, hence also the bicommutant, as this a maximal abelian subalgebra of $B(L^{2}(mathbb{T}))$. It remains to see that $H^{infty}(mathbb{T})$ is strongly closed, but this is simple: a function $fin L^{infty}(mathbb{T})$ belongs to $H^{infty}(mathbb{T})$ iff all of its negative Fourier coefficients vanish and this property is clearly preserved by strong convergence (even by weak convergence).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          Mateusz WasilewskiMateusz Wasilewski

          2,905920




          2,905920












          • $begingroup$
            That's great, thanks!
            $endgroup$
            – Jochen Glueck
            4 hours ago


















          • $begingroup$
            That's great, thanks!
            $endgroup$
            – Jochen Glueck
            4 hours ago
















          $begingroup$
          That's great, thanks!
          $endgroup$
          – Jochen Glueck
          4 hours ago




          $begingroup$
          That's great, thanks!
          $endgroup$
          – Jochen Glueck
          4 hours ago


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to MathOverflow!


          • 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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f323585%2fbicommutant-theorem-for-commutative-operator-algebras%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

          Paul Cézanne

          Image
          Pour l’article ayant un titre homophone, voir Sézanne. Paul Cézanne.mw-parser-output .entete.artiste{background-image:url("//upload.wikimedia.org/wikipedia/commons/f/f3/Picto_infobox_artiste.png")} Autoportrait vers 1861 Naissance 19 janvier 1839 Aix-en-Provence (France) Décès 22 octobre 1906 (à 67 ans) Aix-en-Provence (France) Nationalité Française Activité Peintre Formation Académie de Charles Suisse Maître Joseph Gilbert, Antoine Guillemet, Camille Pissarro Mouvement Impressionnisme postimpressionnisme Influencé par Eugène Delacroix, Gustave Courbet, Véronèse A influencé Braque, Picasso, Matisse Conjoint Hortense Fiquet (depuis 1886) Enfant Paul Cézanne  ( d ) Œuvres principales La Montagne Sainte-Victoire, Portrait de madame Cézanne signature modifier - modifier le code - modifier Wikidata Paul Cézanne , né le 19 janvier 1839 à Aix-en-Provence et mort le 22 octobre 1906 dans la mê...
          Read more

          UIScrollView CustomStickyHeader Resize height generates problems when scroll is too fast

          Image
          0 I have been trying for a while to create a UIViewController that has a CustomStickyHeader and a UITableView. The idea I have been following is use the UITableView scrolling for changing the header view with the value of the contentOffset. The layout for the ViewController (AlbumsViewController) looks like: The UItableView Scrolling functions from the delegate are implemented like that: func scrollViewDidScroll(_ scrollView: UIScrollView) { let endReached = contentTableView.contentOffset.y >= (contentTableView.contentSize.height - contentTableView.frame.size.height) let isUP = lastContentOffset > scrollView.contentOffset.y if scrollView.contentOffset.y < 0 { headerHeightConstraint.constant += abs(scrollView.contentOffset.y) } else{ if !endReached{ ...
          Read more

          Angular material date-picker (MatDatepicker) auto completes the date on focus out

          Image
          3 3 I am using angular material/mat-datepicker in my project and whenever the user enters one digit inside of it, and goes out of focus - the date picker is auto completing with a random date. Example Just enter 1 and go out of focus, or enter 5/5/20 The date picker is adding date automatically Update : Also I am using reactive forms and I'm validating the input with that approach Can I disable this ? angular datepicker angular-material angular6 angular-material-datetimepicker share | improve this question edited Nov 20 '18 at 13:21 ...
          Read more