Dependence of operator topologies in a $C^*$ algebra on the representation












4














Let $A$ be a $C^*$ algebra. Given a faithful representation $pi:Ato mathcal{B}(H)$, we can define the weak operator topology with respect to $pi$ as initial with respect to the maps $amapsto langle pi(a) x,yrangle$ for each $x,yinmathcal{H}$. However, this definition depends on our representation $pi$.



Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?










share|cite|improve this question



























    4














    Let $A$ be a $C^*$ algebra. Given a faithful representation $pi:Ato mathcal{B}(H)$, we can define the weak operator topology with respect to $pi$ as initial with respect to the maps $amapsto langle pi(a) x,yrangle$ for each $x,yinmathcal{H}$. However, this definition depends on our representation $pi$.



    Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?










    share|cite|improve this question

























      4












      4








      4







      Let $A$ be a $C^*$ algebra. Given a faithful representation $pi:Ato mathcal{B}(H)$, we can define the weak operator topology with respect to $pi$ as initial with respect to the maps $amapsto langle pi(a) x,yrangle$ for each $x,yinmathcal{H}$. However, this definition depends on our representation $pi$.



      Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?










      share|cite|improve this question













      Let $A$ be a $C^*$ algebra. Given a faithful representation $pi:Ato mathcal{B}(H)$, we can define the weak operator topology with respect to $pi$ as initial with respect to the maps $amapsto langle pi(a) x,yrangle$ for each $x,yinmathcal{H}$. However, this definition depends on our representation $pi$.



      Are these operator topologies intrinsic to $A$, or can different faithful representations induce different weak operator, strong operator, etc. topologies?







      functional-analysis operator-algebras c-star-algebras






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Sep 25 '18 at 22:56









      Ashwin TrisalAshwin Trisal

      1,2441516




      1,2441516






















          1 Answer
          1






          active

          oldest

          votes


















          2














          No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.



          For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).



          Then we have
          $$
          overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
          $$

          a II$_1$-factor. While
          $$
          overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
          $$

          a type I factor.



          The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.






          share|cite|improve this answer























          • It would be really interesting to know what the downvoter things is wrong with this answer.
            – Martin Argerami
            Sep 30 '18 at 18:57











          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%2f2930886%2fdependence-of-operator-topologies-in-a-c-algebra-on-the-representation%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














          No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.



          For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).



          Then we have
          $$
          overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
          $$

          a II$_1$-factor. While
          $$
          overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
          $$

          a type I factor.



          The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.






          share|cite|improve this answer























          • It would be really interesting to know what the downvoter things is wrong with this answer.
            – Martin Argerami
            Sep 30 '18 at 18:57
















          2














          No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.



          For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).



          Then we have
          $$
          overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
          $$

          a II$_1$-factor. While
          $$
          overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
          $$

          a type I factor.



          The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.






          share|cite|improve this answer























          • It would be really interesting to know what the downvoter things is wrong with this answer.
            – Martin Argerami
            Sep 30 '18 at 18:57














          2












          2








          2






          No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.



          For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).



          Then we have
          $$
          overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
          $$

          a II$_1$-factor. While
          $$
          overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
          $$

          a type I factor.



          The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.






          share|cite|improve this answer














          No, they are not intrinsic. That's why we need things like Sakai's Theorem to tell us which C$^*$-algebras are von Neumann algebras in an appropriate representation.



          For instance, let $A$ be any simple infinite-dimensional C$^*$-algebra; like $A=C^*_r(mathbb F_2)$, say. Since $A$ is simple, any representation is faithful. Take $pi_1:Ato B(H_1)$ be the representation induced by the left-regular representation. Take $pi_2:Ato B(H_2)$ be an irreducible representation (which is faithful, since $A$ is simple).



          Then we have
          $$
          overline{pi_1(A)}^{rm sot, wot, etc.}=L(mathbb F_2),
          $$

          a II$_1$-factor. While
          $$
          overline{pi_2(A)}^{rm sot, wot, etc.}=B(H_2),
          $$

          a type I factor.



          The example is a bit more dramatic if you take $Asubset B(H_1)$ to be a II$_1$-factor (only that now $A$ is not separable as a C$^*$-algebra). Take $pi_1$ to be the identity, and $pi_2$ to be an irreducible representation. Then $pi_1(A)$ is sot/wot closed, while $overline{pi_2(A)}^{rm sot, wot}=B(H_2)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 21:43

























          answered Sep 26 '18 at 15:04









          Martin ArgeramiMartin Argerami

          124k1176175




          124k1176175












          • It would be really interesting to know what the downvoter things is wrong with this answer.
            – Martin Argerami
            Sep 30 '18 at 18:57


















          • It would be really interesting to know what the downvoter things is wrong with this answer.
            – Martin Argerami
            Sep 30 '18 at 18:57
















          It would be really interesting to know what the downvoter things is wrong with this answer.
          – Martin Argerami
          Sep 30 '18 at 18:57




          It would be really interesting to know what the downvoter things is wrong with this answer.
          – Martin Argerami
          Sep 30 '18 at 18:57


















          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%2f2930886%2fdependence-of-operator-topologies-in-a-c-algebra-on-the-representation%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

          Bressuire

          Cabo Verde

          Gyllenstierna