arbitrary $n$-dimensional matrix algebras in II$_1$ factor












1












$begingroup$


I'm struggling to show that in a type II$_1$ and any $n$ $exists$ a subfactor $M$ such that $M cong M_n$ .



I suppose it should follow from the isomorphism between the equivalence classes of projections and the interval $[0,operatorname{tr}(1)]$ but i can not figure out the details






functional-analysis operator-algebras von-neumann-algebras







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm struggling to show that in a type II$_1$ and any $n$ $exists$ a subfactor $M$ such that $M cong M_n$ .



    I suppose it should follow from the isomorphism between the equivalence classes of projections and the interval $[0,operatorname{tr}(1)]$ but i can not figure out the details






    functional-analysis operator-algebras von-neumann-algebras







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm struggling to show that in a type II$_1$ and any $n$ $exists$ a subfactor $M$ such that $M cong M_n$ .



      I suppose it should follow from the isomorphism between the equivalence classes of projections and the interval $[0,operatorname{tr}(1)]$ but i can not figure out the details






      functional-analysis operator-algebras von-neumann-algebras







      share|cite|improve this question











      $endgroup$




      I'm struggling to show that in a type II$_1$ and any $n$ $exists$ a subfactor $M$ such that $M cong M_n$ .



      I suppose it should follow from the isomorphism between the equivalence classes of projections and the interval $[0,operatorname{tr}(1)]$ but i can not figure out the details






      functional-analysis operator-algebras von-neumann-algebras




      functional-analysis operator-algebras von-neumann-algebras






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 21:40









      Martin Argerami

      125k1177177




      125k1177177










      asked Dec 8 '18 at 12:38









      sirjoesirjoe

      284




      284






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Say your II$_1$-factor is $N$. You can always find projections $p_1,ldots,p_n$ with $sum_jp_j=I $ and $tau(p_j)=1/n$. Because you are in a factor and they have equal trace, these projections are pairwise equivalent. In particular there exist partial isometries $v_1,ldots,v_n$ such that $v_j^*v_j=p_j$, $v_jv_j^*=p_1$.



          Next you define
          $$
          e_{kj}=v_k^*v_j, k,j=1,ldots,n.
          $$
          You have
          $$tag1
          e_{kj}e_{st}=v_k^*v_jv_s^*v_t=v_k^*v_jp_jp_sv_s^*v_t=delta_{j,s},v_k^*v_jv_j^*v_t^*
          =delta_{j,s},v_k^*p_1v_t=delta_{j,s},v_k^*v_t=delta_{j,s},e_{kt}.
          $$
          Now let $M=operatorname{span}{e_{kj}, k,j=1,ldots,n}subset N$. Then the map $pi:Mto M_n(mathbb C)$ given by $pi(e_{kj})=E_{kj}$ is a $*$-isomorphism. This is easily checked using $(1)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The $p_i$ must be pairwise orthogonal, right ?
            $endgroup$
            – André S.
            Dec 9 '18 at 12:31






          • 1




            $begingroup$
            Yes, they have to add to the identity if we want the subfactor to have the same identity (and this guarantees pairwise orthogonal). If we don't mind the identity, they just need to be pairwise orthogonal with equal trace.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 13:36










          • $begingroup$
            How can you always find such a sum of projections? It is obvious you can find any projection of size 1/n but im not sure how you could find a sum of such projections equal to the identity
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:31










          • $begingroup$
            Because if $tau(p_1)=1/n$, then $tau(1-p)=1-1/n$ and you can find a $p_2leq 1-p_1$ with $tau(p_2)=1/n$. After $n$ steps you have $p_1,ldots,p_n$, pairwise orthogonal, and the sum has trace 1.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 21:32












          • $begingroup$
            Oh okay! I was struggling to see why under any projection you can choose a projection of any given trace and i suppose this is because in your proof above $(1-p_1)N(1-p_1)$ is a $Pi_1$ factor itself so there exists $p_2$ inside it of trace $1/n$. Thank you!!
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:46











          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%2f3031061%2farbitrary-n-dimensional-matrix-algebras-in-ii-1-factor%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









          1












          $begingroup$

          Say your II$_1$-factor is $N$. You can always find projections $p_1,ldots,p_n$ with $sum_jp_j=I $ and $tau(p_j)=1/n$. Because you are in a factor and they have equal trace, these projections are pairwise equivalent. In particular there exist partial isometries $v_1,ldots,v_n$ such that $v_j^*v_j=p_j$, $v_jv_j^*=p_1$.



          Next you define
          $$
          e_{kj}=v_k^*v_j, k,j=1,ldots,n.
          $$
          You have
          $$tag1
          e_{kj}e_{st}=v_k^*v_jv_s^*v_t=v_k^*v_jp_jp_sv_s^*v_t=delta_{j,s},v_k^*v_jv_j^*v_t^*
          =delta_{j,s},v_k^*p_1v_t=delta_{j,s},v_k^*v_t=delta_{j,s},e_{kt}.
          $$
          Now let $M=operatorname{span}{e_{kj}, k,j=1,ldots,n}subset N$. Then the map $pi:Mto M_n(mathbb C)$ given by $pi(e_{kj})=E_{kj}$ is a $*$-isomorphism. This is easily checked using $(1)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The $p_i$ must be pairwise orthogonal, right ?
            $endgroup$
            – André S.
            Dec 9 '18 at 12:31






          • 1




            $begingroup$
            Yes, they have to add to the identity if we want the subfactor to have the same identity (and this guarantees pairwise orthogonal). If we don't mind the identity, they just need to be pairwise orthogonal with equal trace.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 13:36










          • $begingroup$
            How can you always find such a sum of projections? It is obvious you can find any projection of size 1/n but im not sure how you could find a sum of such projections equal to the identity
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:31










          • $begingroup$
            Because if $tau(p_1)=1/n$, then $tau(1-p)=1-1/n$ and you can find a $p_2leq 1-p_1$ with $tau(p_2)=1/n$. After $n$ steps you have $p_1,ldots,p_n$, pairwise orthogonal, and the sum has trace 1.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 21:32












          • $begingroup$
            Oh okay! I was struggling to see why under any projection you can choose a projection of any given trace and i suppose this is because in your proof above $(1-p_1)N(1-p_1)$ is a $Pi_1$ factor itself so there exists $p_2$ inside it of trace $1/n$. Thank you!!
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:46
















          1












          $begingroup$

          Say your II$_1$-factor is $N$. You can always find projections $p_1,ldots,p_n$ with $sum_jp_j=I $ and $tau(p_j)=1/n$. Because you are in a factor and they have equal trace, these projections are pairwise equivalent. In particular there exist partial isometries $v_1,ldots,v_n$ such that $v_j^*v_j=p_j$, $v_jv_j^*=p_1$.



          Next you define
          $$
          e_{kj}=v_k^*v_j, k,j=1,ldots,n.
          $$
          You have
          $$tag1
          e_{kj}e_{st}=v_k^*v_jv_s^*v_t=v_k^*v_jp_jp_sv_s^*v_t=delta_{j,s},v_k^*v_jv_j^*v_t^*
          =delta_{j,s},v_k^*p_1v_t=delta_{j,s},v_k^*v_t=delta_{j,s},e_{kt}.
          $$
          Now let $M=operatorname{span}{e_{kj}, k,j=1,ldots,n}subset N$. Then the map $pi:Mto M_n(mathbb C)$ given by $pi(e_{kj})=E_{kj}$ is a $*$-isomorphism. This is easily checked using $(1)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The $p_i$ must be pairwise orthogonal, right ?
            $endgroup$
            – André S.
            Dec 9 '18 at 12:31






          • 1




            $begingroup$
            Yes, they have to add to the identity if we want the subfactor to have the same identity (and this guarantees pairwise orthogonal). If we don't mind the identity, they just need to be pairwise orthogonal with equal trace.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 13:36










          • $begingroup$
            How can you always find such a sum of projections? It is obvious you can find any projection of size 1/n but im not sure how you could find a sum of such projections equal to the identity
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:31










          • $begingroup$
            Because if $tau(p_1)=1/n$, then $tau(1-p)=1-1/n$ and you can find a $p_2leq 1-p_1$ with $tau(p_2)=1/n$. After $n$ steps you have $p_1,ldots,p_n$, pairwise orthogonal, and the sum has trace 1.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 21:32












          • $begingroup$
            Oh okay! I was struggling to see why under any projection you can choose a projection of any given trace and i suppose this is because in your proof above $(1-p_1)N(1-p_1)$ is a $Pi_1$ factor itself so there exists $p_2$ inside it of trace $1/n$. Thank you!!
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:46














          1












          1








          1





          $begingroup$

          Say your II$_1$-factor is $N$. You can always find projections $p_1,ldots,p_n$ with $sum_jp_j=I $ and $tau(p_j)=1/n$. Because you are in a factor and they have equal trace, these projections are pairwise equivalent. In particular there exist partial isometries $v_1,ldots,v_n$ such that $v_j^*v_j=p_j$, $v_jv_j^*=p_1$.



          Next you define
          $$
          e_{kj}=v_k^*v_j, k,j=1,ldots,n.
          $$
          You have
          $$tag1
          e_{kj}e_{st}=v_k^*v_jv_s^*v_t=v_k^*v_jp_jp_sv_s^*v_t=delta_{j,s},v_k^*v_jv_j^*v_t^*
          =delta_{j,s},v_k^*p_1v_t=delta_{j,s},v_k^*v_t=delta_{j,s},e_{kt}.
          $$
          Now let $M=operatorname{span}{e_{kj}, k,j=1,ldots,n}subset N$. Then the map $pi:Mto M_n(mathbb C)$ given by $pi(e_{kj})=E_{kj}$ is a $*$-isomorphism. This is easily checked using $(1)$.






          share|cite|improve this answer











          $endgroup$



          Say your II$_1$-factor is $N$. You can always find projections $p_1,ldots,p_n$ with $sum_jp_j=I $ and $tau(p_j)=1/n$. Because you are in a factor and they have equal trace, these projections are pairwise equivalent. In particular there exist partial isometries $v_1,ldots,v_n$ such that $v_j^*v_j=p_j$, $v_jv_j^*=p_1$.



          Next you define
          $$
          e_{kj}=v_k^*v_j, k,j=1,ldots,n.
          $$
          You have
          $$tag1
          e_{kj}e_{st}=v_k^*v_jv_s^*v_t=v_k^*v_jp_jp_sv_s^*v_t=delta_{j,s},v_k^*v_jv_j^*v_t^*
          =delta_{j,s},v_k^*p_1v_t=delta_{j,s},v_k^*v_t=delta_{j,s},e_{kt}.
          $$
          Now let $M=operatorname{span}{e_{kj}, k,j=1,ldots,n}subset N$. Then the map $pi:Mto M_n(mathbb C)$ given by $pi(e_{kj})=E_{kj}$ is a $*$-isomorphism. This is easily checked using $(1)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 13:33

























          answered Dec 8 '18 at 18:27









          Martin ArgeramiMartin Argerami

          125k1177177




          125k1177177












          • $begingroup$
            The $p_i$ must be pairwise orthogonal, right ?
            $endgroup$
            – André S.
            Dec 9 '18 at 12:31






          • 1




            $begingroup$
            Yes, they have to add to the identity if we want the subfactor to have the same identity (and this guarantees pairwise orthogonal). If we don't mind the identity, they just need to be pairwise orthogonal with equal trace.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 13:36










          • $begingroup$
            How can you always find such a sum of projections? It is obvious you can find any projection of size 1/n but im not sure how you could find a sum of such projections equal to the identity
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:31










          • $begingroup$
            Because if $tau(p_1)=1/n$, then $tau(1-p)=1-1/n$ and you can find a $p_2leq 1-p_1$ with $tau(p_2)=1/n$. After $n$ steps you have $p_1,ldots,p_n$, pairwise orthogonal, and the sum has trace 1.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 21:32












          • $begingroup$
            Oh okay! I was struggling to see why under any projection you can choose a projection of any given trace and i suppose this is because in your proof above $(1-p_1)N(1-p_1)$ is a $Pi_1$ factor itself so there exists $p_2$ inside it of trace $1/n$. Thank you!!
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:46


















          • $begingroup$
            The $p_i$ must be pairwise orthogonal, right ?
            $endgroup$
            – André S.
            Dec 9 '18 at 12:31






          • 1




            $begingroup$
            Yes, they have to add to the identity if we want the subfactor to have the same identity (and this guarantees pairwise orthogonal). If we don't mind the identity, they just need to be pairwise orthogonal with equal trace.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 13:36










          • $begingroup$
            How can you always find such a sum of projections? It is obvious you can find any projection of size 1/n but im not sure how you could find a sum of such projections equal to the identity
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:31










          • $begingroup$
            Because if $tau(p_1)=1/n$, then $tau(1-p)=1-1/n$ and you can find a $p_2leq 1-p_1$ with $tau(p_2)=1/n$. After $n$ steps you have $p_1,ldots,p_n$, pairwise orthogonal, and the sum has trace 1.
            $endgroup$
            – Martin Argerami
            Dec 9 '18 at 21:32












          • $begingroup$
            Oh okay! I was struggling to see why under any projection you can choose a projection of any given trace and i suppose this is because in your proof above $(1-p_1)N(1-p_1)$ is a $Pi_1$ factor itself so there exists $p_2$ inside it of trace $1/n$. Thank you!!
            $endgroup$
            – sirjoe
            Dec 9 '18 at 21:46
















          $begingroup$
          The $p_i$ must be pairwise orthogonal, right ?
          $endgroup$
          – André S.
          Dec 9 '18 at 12:31




          $begingroup$
          The $p_i$ must be pairwise orthogonal, right ?
          $endgroup$
          – André S.
          Dec 9 '18 at 12:31




          1




          1




          $begingroup$
          Yes, they have to add to the identity if we want the subfactor to have the same identity (and this guarantees pairwise orthogonal). If we don't mind the identity, they just need to be pairwise orthogonal with equal trace.
          $endgroup$
          – Martin Argerami
          Dec 9 '18 at 13:36




          $begingroup$
          Yes, they have to add to the identity if we want the subfactor to have the same identity (and this guarantees pairwise orthogonal). If we don't mind the identity, they just need to be pairwise orthogonal with equal trace.
          $endgroup$
          – Martin Argerami
          Dec 9 '18 at 13:36












          $begingroup$
          How can you always find such a sum of projections? It is obvious you can find any projection of size 1/n but im not sure how you could find a sum of such projections equal to the identity
          $endgroup$
          – sirjoe
          Dec 9 '18 at 21:31




          $begingroup$
          How can you always find such a sum of projections? It is obvious you can find any projection of size 1/n but im not sure how you could find a sum of such projections equal to the identity
          $endgroup$
          – sirjoe
          Dec 9 '18 at 21:31












          $begingroup$
          Because if $tau(p_1)=1/n$, then $tau(1-p)=1-1/n$ and you can find a $p_2leq 1-p_1$ with $tau(p_2)=1/n$. After $n$ steps you have $p_1,ldots,p_n$, pairwise orthogonal, and the sum has trace 1.
          $endgroup$
          – Martin Argerami
          Dec 9 '18 at 21:32






          $begingroup$
          Because if $tau(p_1)=1/n$, then $tau(1-p)=1-1/n$ and you can find a $p_2leq 1-p_1$ with $tau(p_2)=1/n$. After $n$ steps you have $p_1,ldots,p_n$, pairwise orthogonal, and the sum has trace 1.
          $endgroup$
          – Martin Argerami
          Dec 9 '18 at 21:32














          $begingroup$
          Oh okay! I was struggling to see why under any projection you can choose a projection of any given trace and i suppose this is because in your proof above $(1-p_1)N(1-p_1)$ is a $Pi_1$ factor itself so there exists $p_2$ inside it of trace $1/n$. Thank you!!
          $endgroup$
          – sirjoe
          Dec 9 '18 at 21:46




          $begingroup$
          Oh okay! I was struggling to see why under any projection you can choose a projection of any given trace and i suppose this is because in your proof above $(1-p_1)N(1-p_1)$ is a $Pi_1$ factor itself so there exists $p_2$ inside it of trace $1/n$. Thank you!!
          $endgroup$
          – sirjoe
          Dec 9 '18 at 21:46


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031061%2farbitrary-n-dimensional-matrix-algebras-in-ii-1-factor%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

          Berounka

          Image
          Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
          Read more

          Sphinx de Gizeh

          Image
          Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
          Read more

          Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

          Image
          3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
          Read more