Question regarding motivation of spectral theorem for unitary operators
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$newcommand{mc}{mathcal}$
$newcommand{ab}[1]{langle #1rangle}$
Theorem B.4 in Einsiedler and Ward's [EW] Ergodic Theory with a view towards Number Theory states the following:
Theorem 1.
Spectral Theorem.
Let $U $be a unitary operator on a complex Hilbert space $H$.
$(1)$ For each element $fin H$, there is a unique Borel measure $mu_f$ on $S^1$ with the property that
$$
ab{U^nf, f} = int_{S^1} z^n dmu_f(z)
$$
(2) The map
$$
sum_{n=-N}^N c_nz^n mapsto sum_{n=-N}^N c_n U^n f
$$
extends by continuity to a unitary isomorphism between $L^2(S^1, mu_f)$ and the smallest $U$-invariant subspace in $H$ containing $f$.
(There was some text here. Thanks to @DisintegrationByParts for pointing our an error. I have thus removed the text.)
Question 1.
I do not see any motivation for the theorem.
The statement seems out of the blue. Can somebody provide some perspective here?
Lastly, let $(X, mc F, mu, T)$ be a measure-preserving system.
Thus $U_T:L^2(X, mu)to L^2(X, mu)$ defined as $fmapsto fcirc T$ is a unitary operator.
Question 2.
If $gin L^2(X, mu)$, what is the meaning of the phrase ``By item (2) of the theorem above, $L^2(S^1, mu_g)$ is unitarily isomorphic to the cyclic sub-representation of $L^2(X, mu)$ generated by $g$ under the unitary map $U_T$."
The statement in quotes appears on pg 184 of EW. I am not able to see what is the representation here. The group is probably $S^1$, and the vector space is $L^2(X, mu)$. But the representation itself is not clear.
Also, how does item (2) of the theorem relate with representations?
Thank you.
functional-analysis measure-theory representation-theory ergodic-theory
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$newcommand{mc}{mathcal}$
$newcommand{ab}[1]{langle #1rangle}$
Theorem B.4 in Einsiedler and Ward's [EW] Ergodic Theory with a view towards Number Theory states the following:
Theorem 1.
Spectral Theorem.
Let $U $be a unitary operator on a complex Hilbert space $H$.
$(1)$ For each element $fin H$, there is a unique Borel measure $mu_f$ on $S^1$ with the property that
$$
ab{U^nf, f} = int_{S^1} z^n dmu_f(z)
$$
(2) The map
$$
sum_{n=-N}^N c_nz^n mapsto sum_{n=-N}^N c_n U^n f
$$
extends by continuity to a unitary isomorphism between $L^2(S^1, mu_f)$ and the smallest $U$-invariant subspace in $H$ containing $f$.
(There was some text here. Thanks to @DisintegrationByParts for pointing our an error. I have thus removed the text.)
Question 1.
I do not see any motivation for the theorem.
The statement seems out of the blue. Can somebody provide some perspective here?
Lastly, let $(X, mc F, mu, T)$ be a measure-preserving system.
Thus $U_T:L^2(X, mu)to L^2(X, mu)$ defined as $fmapsto fcirc T$ is a unitary operator.
Question 2.
If $gin L^2(X, mu)$, what is the meaning of the phrase ``By item (2) of the theorem above, $L^2(S^1, mu_g)$ is unitarily isomorphic to the cyclic sub-representation of $L^2(X, mu)$ generated by $g$ under the unitary map $U_T$."
The statement in quotes appears on pg 184 of EW. I am not able to see what is the representation here. The group is probably $S^1$, and the vector space is $L^2(X, mu)$. But the representation itself is not clear.
Also, how does item (2) of the theorem relate with representations?
Thank you.
functional-analysis measure-theory representation-theory ergodic-theory
1
Your version of the spectral theorem for normal operators is not true. Perhaps you were studying compact operators at the time?
– DisintegratingByParts
yesterday
1
Ah! Let me go back and revise. Thanks.
– caffeinemachine
yesterday
Do you know the classical Spectral Theorem for normal operators involving a projection-valued measure $P$, where $U = int_{sigma(U)}zdP(z)$?
– DisintegratingByParts
22 hours ago
I do not. Can you please articulate it in an answer and if possible give a reference. Thanks.
– caffeinemachine
22 hours ago
The spectral theorem involving spectral measures is not easily explained in a post. But it is the motivation for other versions, which is why other versions often seem unmotivated.
– DisintegratingByParts
14 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$newcommand{mc}{mathcal}$
$newcommand{ab}[1]{langle #1rangle}$
Theorem B.4 in Einsiedler and Ward's [EW] Ergodic Theory with a view towards Number Theory states the following:
Theorem 1.
Spectral Theorem.
Let $U $be a unitary operator on a complex Hilbert space $H$.
$(1)$ For each element $fin H$, there is a unique Borel measure $mu_f$ on $S^1$ with the property that
$$
ab{U^nf, f} = int_{S^1} z^n dmu_f(z)
$$
(2) The map
$$
sum_{n=-N}^N c_nz^n mapsto sum_{n=-N}^N c_n U^n f
$$
extends by continuity to a unitary isomorphism between $L^2(S^1, mu_f)$ and the smallest $U$-invariant subspace in $H$ containing $f$.
(There was some text here. Thanks to @DisintegrationByParts for pointing our an error. I have thus removed the text.)
Question 1.
I do not see any motivation for the theorem.
The statement seems out of the blue. Can somebody provide some perspective here?
Lastly, let $(X, mc F, mu, T)$ be a measure-preserving system.
Thus $U_T:L^2(X, mu)to L^2(X, mu)$ defined as $fmapsto fcirc T$ is a unitary operator.
Question 2.
If $gin L^2(X, mu)$, what is the meaning of the phrase ``By item (2) of the theorem above, $L^2(S^1, mu_g)$ is unitarily isomorphic to the cyclic sub-representation of $L^2(X, mu)$ generated by $g$ under the unitary map $U_T$."
The statement in quotes appears on pg 184 of EW. I am not able to see what is the representation here. The group is probably $S^1$, and the vector space is $L^2(X, mu)$. But the representation itself is not clear.
Also, how does item (2) of the theorem relate with representations?
Thank you.
functional-analysis measure-theory representation-theory ergodic-theory
$newcommand{mc}{mathcal}$
$newcommand{ab}[1]{langle #1rangle}$
Theorem B.4 in Einsiedler and Ward's [EW] Ergodic Theory with a view towards Number Theory states the following:
Theorem 1.
Spectral Theorem.
Let $U $be a unitary operator on a complex Hilbert space $H$.
$(1)$ For each element $fin H$, there is a unique Borel measure $mu_f$ on $S^1$ with the property that
$$
ab{U^nf, f} = int_{S^1} z^n dmu_f(z)
$$
(2) The map
$$
sum_{n=-N}^N c_nz^n mapsto sum_{n=-N}^N c_n U^n f
$$
extends by continuity to a unitary isomorphism between $L^2(S^1, mu_f)$ and the smallest $U$-invariant subspace in $H$ containing $f$.
(There was some text here. Thanks to @DisintegrationByParts for pointing our an error. I have thus removed the text.)
Question 1.
I do not see any motivation for the theorem.
The statement seems out of the blue. Can somebody provide some perspective here?
Lastly, let $(X, mc F, mu, T)$ be a measure-preserving system.
Thus $U_T:L^2(X, mu)to L^2(X, mu)$ defined as $fmapsto fcirc T$ is a unitary operator.
Question 2.
If $gin L^2(X, mu)$, what is the meaning of the phrase ``By item (2) of the theorem above, $L^2(S^1, mu_g)$ is unitarily isomorphic to the cyclic sub-representation of $L^2(X, mu)$ generated by $g$ under the unitary map $U_T$."
The statement in quotes appears on pg 184 of EW. I am not able to see what is the representation here. The group is probably $S^1$, and the vector space is $L^2(X, mu)$. But the representation itself is not clear.
Also, how does item (2) of the theorem relate with representations?
Thank you.
functional-analysis measure-theory representation-theory ergodic-theory
functional-analysis measure-theory representation-theory ergodic-theory
edited yesterday
asked yesterday
caffeinemachine
6,36721248
6,36721248
1
Your version of the spectral theorem for normal operators is not true. Perhaps you were studying compact operators at the time?
– DisintegratingByParts
yesterday
1
Ah! Let me go back and revise. Thanks.
– caffeinemachine
yesterday
Do you know the classical Spectral Theorem for normal operators involving a projection-valued measure $P$, where $U = int_{sigma(U)}zdP(z)$?
– DisintegratingByParts
22 hours ago
I do not. Can you please articulate it in an answer and if possible give a reference. Thanks.
– caffeinemachine
22 hours ago
The spectral theorem involving spectral measures is not easily explained in a post. But it is the motivation for other versions, which is why other versions often seem unmotivated.
– DisintegratingByParts
14 hours ago
add a comment |
1
Your version of the spectral theorem for normal operators is not true. Perhaps you were studying compact operators at the time?
– DisintegratingByParts
yesterday
1
Ah! Let me go back and revise. Thanks.
– caffeinemachine
yesterday
Do you know the classical Spectral Theorem for normal operators involving a projection-valued measure $P$, where $U = int_{sigma(U)}zdP(z)$?
– DisintegratingByParts
22 hours ago
I do not. Can you please articulate it in an answer and if possible give a reference. Thanks.
– caffeinemachine
22 hours ago
The spectral theorem involving spectral measures is not easily explained in a post. But it is the motivation for other versions, which is why other versions often seem unmotivated.
– DisintegratingByParts
14 hours ago
1
1
Your version of the spectral theorem for normal operators is not true. Perhaps you were studying compact operators at the time?
– DisintegratingByParts
yesterday
Your version of the spectral theorem for normal operators is not true. Perhaps you were studying compact operators at the time?
– DisintegratingByParts
yesterday
1
1
Ah! Let me go back and revise. Thanks.
– caffeinemachine
yesterday
Ah! Let me go back and revise. Thanks.
– caffeinemachine
yesterday
Do you know the classical Spectral Theorem for normal operators involving a projection-valued measure $P$, where $U = int_{sigma(U)}zdP(z)$?
– DisintegratingByParts
22 hours ago
Do you know the classical Spectral Theorem for normal operators involving a projection-valued measure $P$, where $U = int_{sigma(U)}zdP(z)$?
– DisintegratingByParts
22 hours ago
I do not. Can you please articulate it in an answer and if possible give a reference. Thanks.
– caffeinemachine
22 hours ago
I do not. Can you please articulate it in an answer and if possible give a reference. Thanks.
– caffeinemachine
22 hours ago
The spectral theorem involving spectral measures is not easily explained in a post. But it is the motivation for other versions, which is why other versions often seem unmotivated.
– DisintegratingByParts
14 hours ago
The spectral theorem involving spectral measures is not easily explained in a post. But it is the motivation for other versions, which is why other versions often seem unmotivated.
– DisintegratingByParts
14 hours ago
add a comment |
1 Answer
1
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You have a unitary representation $pi:mathbb{Z} to Unit(L^2(X))$. This induces a *-homomorphism $pi:L^1(mathbb{Z}) to Lin(L^2(X)): (pi(f)u,v):= int f(n)(pi(n)u,v)dn$ (Folland's harmonic analysis theorem $3.9$).
$L^1(mathbb{Z})$ is a unital, commutative, Banach algebra whose spectrum can be identified with $S^1$, the dual of $mathbb{Z}$ (Folland theorem 4.2).
Theorem $1.54$ of the same book then shows how to obtain part $(1)$ of the theorem in EW.
I'm guessing that for $(2)$, theorem $1.47$ (and perhaps also theorem $4.44$) of Folland's book are relevant.
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1 Answer
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1 Answer
1
active
oldest
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You have a unitary representation $pi:mathbb{Z} to Unit(L^2(X))$. This induces a *-homomorphism $pi:L^1(mathbb{Z}) to Lin(L^2(X)): (pi(f)u,v):= int f(n)(pi(n)u,v)dn$ (Folland's harmonic analysis theorem $3.9$).
$L^1(mathbb{Z})$ is a unital, commutative, Banach algebra whose spectrum can be identified with $S^1$, the dual of $mathbb{Z}$ (Folland theorem 4.2).
Theorem $1.54$ of the same book then shows how to obtain part $(1)$ of the theorem in EW.
I'm guessing that for $(2)$, theorem $1.47$ (and perhaps also theorem $4.44$) of Folland's book are relevant.
add a comment |
up vote
0
down vote
You have a unitary representation $pi:mathbb{Z} to Unit(L^2(X))$. This induces a *-homomorphism $pi:L^1(mathbb{Z}) to Lin(L^2(X)): (pi(f)u,v):= int f(n)(pi(n)u,v)dn$ (Folland's harmonic analysis theorem $3.9$).
$L^1(mathbb{Z})$ is a unital, commutative, Banach algebra whose spectrum can be identified with $S^1$, the dual of $mathbb{Z}$ (Folland theorem 4.2).
Theorem $1.54$ of the same book then shows how to obtain part $(1)$ of the theorem in EW.
I'm guessing that for $(2)$, theorem $1.47$ (and perhaps also theorem $4.44$) of Folland's book are relevant.
add a comment |
up vote
0
down vote
up vote
0
down vote
You have a unitary representation $pi:mathbb{Z} to Unit(L^2(X))$. This induces a *-homomorphism $pi:L^1(mathbb{Z}) to Lin(L^2(X)): (pi(f)u,v):= int f(n)(pi(n)u,v)dn$ (Folland's harmonic analysis theorem $3.9$).
$L^1(mathbb{Z})$ is a unital, commutative, Banach algebra whose spectrum can be identified with $S^1$, the dual of $mathbb{Z}$ (Folland theorem 4.2).
Theorem $1.54$ of the same book then shows how to obtain part $(1)$ of the theorem in EW.
I'm guessing that for $(2)$, theorem $1.47$ (and perhaps also theorem $4.44$) of Folland's book are relevant.
You have a unitary representation $pi:mathbb{Z} to Unit(L^2(X))$. This induces a *-homomorphism $pi:L^1(mathbb{Z}) to Lin(L^2(X)): (pi(f)u,v):= int f(n)(pi(n)u,v)dn$ (Folland's harmonic analysis theorem $3.9$).
$L^1(mathbb{Z})$ is a unital, commutative, Banach algebra whose spectrum can be identified with $S^1$, the dual of $mathbb{Z}$ (Folland theorem 4.2).
Theorem $1.54$ of the same book then shows how to obtain part $(1)$ of the theorem in EW.
I'm guessing that for $(2)$, theorem $1.47$ (and perhaps also theorem $4.44$) of Folland's book are relevant.
answered 7 hours ago
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1
Your version of the spectral theorem for normal operators is not true. Perhaps you were studying compact operators at the time?
– DisintegratingByParts
yesterday
1
Ah! Let me go back and revise. Thanks.
– caffeinemachine
yesterday
Do you know the classical Spectral Theorem for normal operators involving a projection-valued measure $P$, where $U = int_{sigma(U)}zdP(z)$?
– DisintegratingByParts
22 hours ago
I do not. Can you please articulate it in an answer and if possible give a reference. Thanks.
– caffeinemachine
22 hours ago
The spectral theorem involving spectral measures is not easily explained in a post. But it is the motivation for other versions, which is why other versions often seem unmotivated.
– DisintegratingByParts
14 hours ago