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?
functional-analysis operator-theory
asked Sep 10 at 18:43
Haluter
506
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up vote
1
down vote
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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?
functional-analysis operator-theory
asked Sep 10 at 18:43
Haluter
506
add a comment |Â
up vote
1
down vote
favorite
up vote
1
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?
functional-analysis operator-theory
asked Sep 10 at 18:43
Haluter
506
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
functional-analysis operator-theory
asked Sep 10 at 18:43
Haluter
506
asked Sep 10 at 18:43
Haluter
506
asked Sep 10 at 18:43
Haluter
506
asked Sep 10 at 18:43
Haluter
506
asked Sep 10 at 18:43
Haluter
506
506
add a comment |Â
add a comment |Â
1 Answer
1
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.
answered Sep 10 at 19:49
Martin Argerami
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
add a comment |Â
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.
answered Sep 10 at 19:49
Martin Argerami
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
add a comment |Â
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.
answered Sep 10 at 19:49
Martin Argerami
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
add a comment |Â
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.
answered Sep 10 at 19:49
Martin Argerami
117k1071165
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.
answered Sep 10 at 19:49
Martin Argerami
117k1071165
answered Sep 10 at 19:49
Martin Argerami
117k1071165
answered Sep 10 at 19:49
Martin Argerami
117k1071165
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
add a comment |Â
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
add a comment |Â
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