Spectral integral form of tensor products












3












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$T_{1}otimes T_{2}$ is the operator acting on $mathcal{H}otimes mathcal{H}$ then where $T_{1},T_{2}$ are self adjoint operators. What is the spectral resolution of identity for $T_{1}otimes T_{2}$, further what is the spectral measure of $T_{1}otimes T_{2}$?






operator-theory von-neumann-algebras







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$endgroup$

















    3












    $begingroup$


    $T_{1}otimes T_{2}$ is the operator acting on $mathcal{H}otimes mathcal{H}$ then where $T_{1},T_{2}$ are self adjoint operators. What is the spectral resolution of identity for $T_{1}otimes T_{2}$, further what is the spectral measure of $T_{1}otimes T_{2}$?






    operator-theory von-neumann-algebras







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      $T_{1}otimes T_{2}$ is the operator acting on $mathcal{H}otimes mathcal{H}$ then where $T_{1},T_{2}$ are self adjoint operators. What is the spectral resolution of identity for $T_{1}otimes T_{2}$, further what is the spectral measure of $T_{1}otimes T_{2}$?






      operator-theory von-neumann-algebras







      share|cite|improve this question











      $endgroup$




      $T_{1}otimes T_{2}$ is the operator acting on $mathcal{H}otimes mathcal{H}$ then where $T_{1},T_{2}$ are self adjoint operators. What is the spectral resolution of identity for $T_{1}otimes T_{2}$, further what is the spectral measure of $T_{1}otimes T_{2}$?






      operator-theory von-neumann-algebras




      operator-theory von-neumann-algebras






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 12:17







      mathlover

















      asked Jan 26 at 10:57









      mathlovermathlover

      152110




      152110






















          1 Answer
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          $begingroup$

          Let $E^1$ and $E^2$ be the spectral measures of $T_1$ and $T_2$. Let $E^1 otimes E^2$ be the spectral measure on $mathbb{R}times mathbb{R}$ defined by
          $$
          E^1 otimes E^2 (B_1 times B_2)=E^1 (B_1) otimes E^2 (B_2),
          $$
          where $B_1, B_2$ are Borel subsets of $mathbb{R}$.



          Remark: The above projection-valtued set function $E^1 otimes E^2$ defined on rectangles $B_1 times B_2$ has an unique extension to a spectral measure.



          Let us denote by $E$ the spectral measure on $mathbb{R}$ given by
          $$
          E(B)=(E^1 otimes E^2)({ (x,y):xy in B }).
          $$
          Then $E$ is the spectral measure of $T_1 otimes T_2$.



          Sketch of the proof:



          First, we prove the equality $T_1 otimes T_2 = int_{-infty}^{infty} int_{-infty}^{infty} xy d(E^1 otimes E^2)(x,y)$ (theorem 8.2 in W. F. Stinespring, "Integration Theorems For Gages and Duality for Unimodular Groups", Trans. Amer. Math. Soc. 90 (1959), 15–56).



          Second, we use the above equality to prove $T_1 otimes T_2 = int_{-infty}^{infty} x d E (x)$.






          share|cite|improve this answer









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            1 Answer
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            active

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            active

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            active

            oldest

            votes









            1





            +50







            $begingroup$

            Let $E^1$ and $E^2$ be the spectral measures of $T_1$ and $T_2$. Let $E^1 otimes E^2$ be the spectral measure on $mathbb{R}times mathbb{R}$ defined by
            $$
            E^1 otimes E^2 (B_1 times B_2)=E^1 (B_1) otimes E^2 (B_2),
            $$
            where $B_1, B_2$ are Borel subsets of $mathbb{R}$.



            Remark: The above projection-valtued set function $E^1 otimes E^2$ defined on rectangles $B_1 times B_2$ has an unique extension to a spectral measure.



            Let us denote by $E$ the spectral measure on $mathbb{R}$ given by
            $$
            E(B)=(E^1 otimes E^2)({ (x,y):xy in B }).
            $$
            Then $E$ is the spectral measure of $T_1 otimes T_2$.



            Sketch of the proof:



            First, we prove the equality $T_1 otimes T_2 = int_{-infty}^{infty} int_{-infty}^{infty} xy d(E^1 otimes E^2)(x,y)$ (theorem 8.2 in W. F. Stinespring, "Integration Theorems For Gages and Duality for Unimodular Groups", Trans. Amer. Math. Soc. 90 (1959), 15–56).



            Second, we use the above equality to prove $T_1 otimes T_2 = int_{-infty}^{infty} x d E (x)$.






            share|cite|improve this answer









            $endgroup$


















              1





              +50







              $begingroup$

              Let $E^1$ and $E^2$ be the spectral measures of $T_1$ and $T_2$. Let $E^1 otimes E^2$ be the spectral measure on $mathbb{R}times mathbb{R}$ defined by
              $$
              E^1 otimes E^2 (B_1 times B_2)=E^1 (B_1) otimes E^2 (B_2),
              $$
              where $B_1, B_2$ are Borel subsets of $mathbb{R}$.



              Remark: The above projection-valtued set function $E^1 otimes E^2$ defined on rectangles $B_1 times B_2$ has an unique extension to a spectral measure.



              Let us denote by $E$ the spectral measure on $mathbb{R}$ given by
              $$
              E(B)=(E^1 otimes E^2)({ (x,y):xy in B }).
              $$
              Then $E$ is the spectral measure of $T_1 otimes T_2$.



              Sketch of the proof:



              First, we prove the equality $T_1 otimes T_2 = int_{-infty}^{infty} int_{-infty}^{infty} xy d(E^1 otimes E^2)(x,y)$ (theorem 8.2 in W. F. Stinespring, "Integration Theorems For Gages and Duality for Unimodular Groups", Trans. Amer. Math. Soc. 90 (1959), 15–56).



              Second, we use the above equality to prove $T_1 otimes T_2 = int_{-infty}^{infty} x d E (x)$.






              share|cite|improve this answer









              $endgroup$
















                1





                +50







                1





                +50



                1




                +50



                $begingroup$

                Let $E^1$ and $E^2$ be the spectral measures of $T_1$ and $T_2$. Let $E^1 otimes E^2$ be the spectral measure on $mathbb{R}times mathbb{R}$ defined by
                $$
                E^1 otimes E^2 (B_1 times B_2)=E^1 (B_1) otimes E^2 (B_2),
                $$
                where $B_1, B_2$ are Borel subsets of $mathbb{R}$.



                Remark: The above projection-valtued set function $E^1 otimes E^2$ defined on rectangles $B_1 times B_2$ has an unique extension to a spectral measure.



                Let us denote by $E$ the spectral measure on $mathbb{R}$ given by
                $$
                E(B)=(E^1 otimes E^2)({ (x,y):xy in B }).
                $$
                Then $E$ is the spectral measure of $T_1 otimes T_2$.



                Sketch of the proof:



                First, we prove the equality $T_1 otimes T_2 = int_{-infty}^{infty} int_{-infty}^{infty} xy d(E^1 otimes E^2)(x,y)$ (theorem 8.2 in W. F. Stinespring, "Integration Theorems For Gages and Duality for Unimodular Groups", Trans. Amer. Math. Soc. 90 (1959), 15–56).



                Second, we use the above equality to prove $T_1 otimes T_2 = int_{-infty}^{infty} x d E (x)$.






                share|cite|improve this answer









                $endgroup$



                Let $E^1$ and $E^2$ be the spectral measures of $T_1$ and $T_2$. Let $E^1 otimes E^2$ be the spectral measure on $mathbb{R}times mathbb{R}$ defined by
                $$
                E^1 otimes E^2 (B_1 times B_2)=E^1 (B_1) otimes E^2 (B_2),
                $$
                where $B_1, B_2$ are Borel subsets of $mathbb{R}$.



                Remark: The above projection-valtued set function $E^1 otimes E^2$ defined on rectangles $B_1 times B_2$ has an unique extension to a spectral measure.



                Let us denote by $E$ the spectral measure on $mathbb{R}$ given by
                $$
                E(B)=(E^1 otimes E^2)({ (x,y):xy in B }).
                $$
                Then $E$ is the spectral measure of $T_1 otimes T_2$.



                Sketch of the proof:



                First, we prove the equality $T_1 otimes T_2 = int_{-infty}^{infty} int_{-infty}^{infty} xy d(E^1 otimes E^2)(x,y)$ (theorem 8.2 in W. F. Stinespring, "Integration Theorems For Gages and Duality for Unimodular Groups", Trans. Amer. Math. Soc. 90 (1959), 15–56).



                Second, we use the above equality to prove $T_1 otimes T_2 = int_{-infty}^{infty} x d E (x)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 4 at 22:50









                RafałRafał

                1847




                1847






























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