Pure isometries are unitarily equivalent to a shift (Wold decomposition). Do the corresponding intertwining relations also hold for the adjoints?

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Let $T in L(H)$ be an isometry on a Hilbert space. Further assume that $T$ is pure, i.e. $T^*m xrightarrow m rightarrow infty 0$ in the strong operator topology, or equivalently, there is no unitary part in the Wold decomposition of $T$. In other words, by the Wold decomposition Theorem, there is a Hilbert space $D$ and a unitary $V: H rightarrow H^2(mathbbD, D)$ such that $VT = M_z V$ holds.



(Here, $H^2(mathbbD, D)$ denotes the $D$-valued Hardy space on $mathbbD$ and $$M_z: H^2(mathbbD, D) rightarrow H^2(mathbbD, D), (M_zf)(z) = z f(z)$$ the shift operator on it)



Is it true that we also have $VT^* = M_z^* V$ in this case?










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    Let $T in L(H)$ be an isometry on a Hilbert space. Further assume that $T$ is pure, i.e. $T^*m xrightarrow m rightarrow infty 0$ in the strong operator topology, or equivalently, there is no unitary part in the Wold decomposition of $T$. In other words, by the Wold decomposition Theorem, there is a Hilbert space $D$ and a unitary $V: H rightarrow H^2(mathbbD, D)$ such that $VT = M_z V$ holds.



    (Here, $H^2(mathbbD, D)$ denotes the $D$-valued Hardy space on $mathbbD$ and $$M_z: H^2(mathbbD, D) rightarrow H^2(mathbbD, D), (M_zf)(z) = z f(z)$$ the shift operator on it)



    Is it true that we also have $VT^* = M_z^* V$ in this case?










    share|cite|improve this question























      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      Let $T in L(H)$ be an isometry on a Hilbert space. Further assume that $T$ is pure, i.e. $T^*m xrightarrow m rightarrow infty 0$ in the strong operator topology, or equivalently, there is no unitary part in the Wold decomposition of $T$. In other words, by the Wold decomposition Theorem, there is a Hilbert space $D$ and a unitary $V: H rightarrow H^2(mathbbD, D)$ such that $VT = M_z V$ holds.



      (Here, $H^2(mathbbD, D)$ denotes the $D$-valued Hardy space on $mathbbD$ and $$M_z: H^2(mathbbD, D) rightarrow H^2(mathbbD, D), (M_zf)(z) = z f(z)$$ the shift operator on it)



      Is it true that we also have $VT^* = M_z^* V$ in this case?










      share|cite|improve this question













      Let $T in L(H)$ be an isometry on a Hilbert space. Further assume that $T$ is pure, i.e. $T^*m xrightarrow m rightarrow infty 0$ in the strong operator topology, or equivalently, there is no unitary part in the Wold decomposition of $T$. In other words, by the Wold decomposition Theorem, there is a Hilbert space $D$ and a unitary $V: H rightarrow H^2(mathbbD, D)$ such that $VT = M_z V$ holds.



      (Here, $H^2(mathbbD, D)$ denotes the $D$-valued Hardy space on $mathbbD$ and $$M_z: H^2(mathbbD, D) rightarrow H^2(mathbbD, D), (M_zf)(z) = z f(z)$$ the shift operator on it)



      Is it true that we also have $VT^* = M_z^* V$ in this case?







      functional-analysis operator-theory






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      asked Sep 10 at 18:43









      Haluter

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          Yes, because $V$ is a unitary. This allows you to write the intertwinning relation as
          $$
          VTV^*=M_z.
          $$
          Now you can take adjoints and multiply by $V$ on the right.






          share|cite|improve this answer




















          • Thank you so much! Seeing how immediate the argument is, I feel quite stupid for not seeing that by myself before.
            – Haluter
            Sep 10 at 22:13










          • Don't feel too bad about it. The lesson to be learned is that intertwinning is a way to write "unitary equivalence" without necessarily involving invertibility. Of course, when $V$ is not a unitary, you get different properties.
            – Martin Argerami
            Sep 10 at 22:50










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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Yes, because $V$ is a unitary. This allows you to write the intertwinning relation as
          $$
          VTV^*=M_z.
          $$
          Now you can take adjoints and multiply by $V$ on the right.






          share|cite|improve this answer




















          • Thank you so much! Seeing how immediate the argument is, I feel quite stupid for not seeing that by myself before.
            – Haluter
            Sep 10 at 22:13










          • Don't feel too bad about it. The lesson to be learned is that intertwinning is a way to write "unitary equivalence" without necessarily involving invertibility. Of course, when $V$ is not a unitary, you get different properties.
            – Martin Argerami
            Sep 10 at 22:50














          up vote
          1
          down vote



          accepted










          Yes, because $V$ is a unitary. This allows you to write the intertwinning relation as
          $$
          VTV^*=M_z.
          $$
          Now you can take adjoints and multiply by $V$ on the right.






          share|cite|improve this answer




















          • Thank you so much! Seeing how immediate the argument is, I feel quite stupid for not seeing that by myself before.
            – Haluter
            Sep 10 at 22:13










          • Don't feel too bad about it. The lesson to be learned is that intertwinning is a way to write "unitary equivalence" without necessarily involving invertibility. Of course, when $V$ is not a unitary, you get different properties.
            – Martin Argerami
            Sep 10 at 22:50












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Yes, because $V$ is a unitary. This allows you to write the intertwinning relation as
          $$
          VTV^*=M_z.
          $$
          Now you can take adjoints and multiply by $V$ on the right.






          share|cite|improve this answer












          Yes, because $V$ is a unitary. This allows you to write the intertwinning relation as
          $$
          VTV^*=M_z.
          $$
          Now you can take adjoints and multiply by $V$ on the right.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 10 at 19:49









          Martin Argerami

          117k1071165




          117k1071165











          • Thank you so much! Seeing how immediate the argument is, I feel quite stupid for not seeing that by myself before.
            – Haluter
            Sep 10 at 22:13










          • Don't feel too bad about it. The lesson to be learned is that intertwinning is a way to write "unitary equivalence" without necessarily involving invertibility. Of course, when $V$ is not a unitary, you get different properties.
            – Martin Argerami
            Sep 10 at 22:50
















          • Thank you so much! Seeing how immediate the argument is, I feel quite stupid for not seeing that by myself before.
            – Haluter
            Sep 10 at 22:13










          • Don't feel too bad about it. The lesson to be learned is that intertwinning is a way to write "unitary equivalence" without necessarily involving invertibility. Of course, when $V$ is not a unitary, you get different properties.
            – Martin Argerami
            Sep 10 at 22:50















          Thank you so much! Seeing how immediate the argument is, I feel quite stupid for not seeing that by myself before.
          – Haluter
          Sep 10 at 22:13




          Thank you so much! Seeing how immediate the argument is, I feel quite stupid for not seeing that by myself before.
          – Haluter
          Sep 10 at 22:13












          Don't feel too bad about it. The lesson to be learned is that intertwinning is a way to write "unitary equivalence" without necessarily involving invertibility. Of course, when $V$ is not a unitary, you get different properties.
          – Martin Argerami
          Sep 10 at 22:50




          Don't feel too bad about it. The lesson to be learned is that intertwinning is a way to write "unitary equivalence" without necessarily involving invertibility. Of course, when $V$ is not a unitary, you get different properties.
          – Martin Argerami
          Sep 10 at 22:50

















           

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