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

      126k1182181




      126k1182181










      asked Dec 8 '18 at 12:38









      sirjoesirjoe

      505




      505






















          1 Answer
          1






          active

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          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 ?
            $endgroup$
            – user42761
            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











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          $begingroup$

          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 ?
            $endgroup$
            – user42761
            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 ?
            $endgroup$
            – user42761
            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

          126k1182181




          126k1182181












          • $begingroup$
            The $p_i$ must be pairwise orthogonal, right ?
            $endgroup$
            – user42761
            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 ?
            $endgroup$
            – user42761
            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 ?
          $endgroup$
          – user42761
          Dec 9 '18 at 12:31




          $begingroup$
          The $p_i$ must be pairwise orthogonal, right ?
          $endgroup$
          – user42761
          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


















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