For every normal operator, is there a unitary operator with the same spectral measure?
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Given a (bounded) normal operator $N$ (satisfying $N^*N=NN^*$) in a von Neumann algebra $A$, does $A$ always contain a unitary operator $U$ having the same spectral (projection-valued) measure as $N$?
If $N$ has a purely discrete spectrum, then the answer is clearly yes, because then $N$ has the form $N = sum_n z_n P_n$ where the $P_n$ are mutually orthogonal projection operators and the $z_n$ are distinct complex coefficients. Replacing the original coefficients with any list of distinct unit-magnitude coefficients gives a unitary operator $U$ with the desired property.
Is the answer always yes? If so, is there a simple proof (or a reference)? If not, then what is a counterexample?
I browsed other questions on math.stackexchange with similar keywords but didn't notice any that answered this particular question.
operator-algebras spectral-theory
asked Nov 16 at 20:10
Dan Yand
1116
|
show 2 more comments
up vote
2
down vote
favorite
Given a (bounded) normal operator $N$ (satisfying $N^*N=NN^*$) in a von Neumann algebra $A$, does $A$ always contain a unitary operator $U$ having the same spectral (projection-valued) measure as $N$?
If $N$ has a purely discrete spectrum, then the answer is clearly yes, because then $N$ has the form $N = sum_n z_n P_n$ where the $P_n$ are mutually orthogonal projection operators and the $z_n$ are distinct complex coefficients. Replacing the original coefficients with any list of distinct unit-magnitude coefficients gives a unitary operator $U$ with the desired property.
Is the answer always yes? If so, is there a simple proof (or a reference)? If not, then what is a counterexample?
I browsed other questions on math.stackexchange with similar keywords but didn't notice any that answered this particular question.
operator-algebras spectral-theory
asked Nov 16 at 20:10
Dan Yand
1116
2
What if you take a 1-1 Borel function $f : mathbb{C} to S^1$ and let $U = f(N)$ using the functional calculus?
– Nate Eldredge
Nov 16 at 21:52
1
What do you mean by "having the same spectral measure"? The spectral measure of a unitary operator is supported in $S^1$ while the spectral measure of $N$ need not be so. It seems that your are using an equivalence of spectral measures that factors the equivalence of the underlying subsets of $mathbb C$.
– Adrián González-Pérez
Nov 17 at 14:48
@AdriánGonzález-Pérez I'm not a mathematician, and I'm probably not using the right words. I'm trying to ask if, given $N$, there is always a $U$ that implicitly defines the same set of mutually commuting projection operators via the spectral decomposition theorem. I understand that $N$ and $U$ will typically have different spectra (numbers $z$ for which $N-z$ or $U-z$ is not invertible), but can the projection-valued measure in the spectral decomposition theorem be defined independently of the spectrum? Sorry if my vocabulary is mixed up. I know that makes deciphering my question difficult.
– Dan Yand
Nov 17 at 15:08
1
@DanYand: Heuristically, a Borel function is anything you can describe explicitly without using the axiom of choice. For example, to get a 1-1 Borel function from the unit square to the unit interval, you can do something like mapping $(0.123456dots, 0.789789dots)$ to $0.172839475869dots$ by interleaving the digits. One could use the same sort of trick to get a 1-1 Borel function from $mathbb{C}$ to $S^1$.
– Nate Eldredge
Nov 17 at 15:42
2
@DanYand: More generally, between any two uncountable complete separable metric spaces, there is a bijective Borel function. This fact is found in lots of measure theory and descriptive set theory books, e.g. Kechris, Classical Descriptive Set Theory.
– Nate Eldredge
Nov 17 at 15:44
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given a (bounded) normal operator $N$ (satisfying $N^*N=NN^*$) in a von Neumann algebra $A$, does $A$ always contain a unitary operator $U$ having the same spectral (projection-valued) measure as $N$?
If $N$ has a purely discrete spectrum, then the answer is clearly yes, because then $N$ has the form $N = sum_n z_n P_n$ where the $P_n$ are mutually orthogonal projection operators and the $z_n$ are distinct complex coefficients. Replacing the original coefficients with any list of distinct unit-magnitude coefficients gives a unitary operator $U$ with the desired property.
Is the answer always yes? If so, is there a simple proof (or a reference)? If not, then what is a counterexample?
I browsed other questions on math.stackexchange with similar keywords but didn't notice any that answered this particular question.
operator-algebras spectral-theory
asked Nov 16 at 20:10
Dan Yand
1116
Given a (bounded) normal operator $N$ (satisfying $N^*N=NN^*$) in a von Neumann algebra $A$, does $A$ always contain a unitary operator $U$ having the same spectral (projection-valued) measure as $N$?
If $N$ has a purely discrete spectrum, then the answer is clearly yes, because then $N$ has the form $N = sum_n z_n P_n$ where the $P_n$ are mutually orthogonal projection operators and the $z_n$ are distinct complex coefficients. Replacing the original coefficients with any list of distinct unit-magnitude coefficients gives a unitary operator $U$ with the desired property.
Is the answer always yes? If so, is there a simple proof (or a reference)? If not, then what is a counterexample?
I browsed other questions on math.stackexchange with similar keywords but didn't notice any that answered this particular question.
operator-algebras spectral-theory
operator-algebras spectral-theory
asked Nov 16 at 20:10
Dan Yand
1116
asked Nov 16 at 20:10
Dan Yand
1116
edited Nov 16 at 20:44
asked Nov 16 at 20:10
Dan Yand
1116
asked Nov 16 at 20:10
Dan Yand
1116
asked Nov 16 at 20:10
Dan Yand
1116
1116
2
What if you take a 1-1 Borel function $f : mathbb{C} to S^1$ and let $U = f(N)$ using the functional calculus?
– Nate Eldredge
Nov 16 at 21:52
1
What do you mean by "having the same spectral measure"? The spectral measure of a unitary operator is supported in $S^1$ while the spectral measure of $N$ need not be so. It seems that your are using an equivalence of spectral measures that factors the equivalence of the underlying subsets of $mathbb C$.
– Adrián González-Pérez
Nov 17 at 14:48
@AdriánGonzález-Pérez I'm not a mathematician, and I'm probably not using the right words. I'm trying to ask if, given $N$, there is always a $U$ that implicitly defines the same set of mutually commuting projection operators via the spectral decomposition theorem. I understand that $N$ and $U$ will typically have different spectra (numbers $z$ for which $N-z$ or $U-z$ is not invertible), but can the projection-valued measure in the spectral decomposition theorem be defined independently of the spectrum? Sorry if my vocabulary is mixed up. I know that makes deciphering my question difficult.
– Dan Yand
Nov 17 at 15:08
1
@DanYand: Heuristically, a Borel function is anything you can describe explicitly without using the axiom of choice. For example, to get a 1-1 Borel function from the unit square to the unit interval, you can do something like mapping $(0.123456dots, 0.789789dots)$ to $0.172839475869dots$ by interleaving the digits. One could use the same sort of trick to get a 1-1 Borel function from $mathbb{C}$ to $S^1$.
– Nate Eldredge
Nov 17 at 15:42
2
@DanYand: More generally, between any two uncountable complete separable metric spaces, there is a bijective Borel function. This fact is found in lots of measure theory and descriptive set theory books, e.g. Kechris, Classical Descriptive Set Theory.
– Nate Eldredge
Nov 17 at 15:44
|
show 2 more comments
2
What if you take a 1-1 Borel function $f : mathbb{C} to S^1$ and let $U = f(N)$ using the functional calculus?
– Nate Eldredge
Nov 16 at 21:52
1
What do you mean by "having the same spectral measure"? The spectral measure of a unitary operator is supported in $S^1$ while the spectral measure of $N$ need not be so. It seems that your are using an equivalence of spectral measures that factors the equivalence of the underlying subsets of $mathbb C$.
– Adrián González-Pérez
Nov 17 at 14:48
@AdriánGonzález-Pérez I'm not a mathematician, and I'm probably not using the right words. I'm trying to ask if, given $N$, there is always a $U$ that implicitly defines the same set of mutually commuting projection operators via the spectral decomposition theorem. I understand that $N$ and $U$ will typically have different spectra (numbers $z$ for which $N-z$ or $U-z$ is not invertible), but can the projection-valued measure in the spectral decomposition theorem be defined independently of the spectrum? Sorry if my vocabulary is mixed up. I know that makes deciphering my question difficult.
– Dan Yand
Nov 17 at 15:08
1
@DanYand: Heuristically, a Borel function is anything you can describe explicitly without using the axiom of choice. For example, to get a 1-1 Borel function from the unit square to the unit interval, you can do something like mapping $(0.123456dots, 0.789789dots)$ to $0.172839475869dots$ by interleaving the digits. One could use the same sort of trick to get a 1-1 Borel function from $mathbb{C}$ to $S^1$.
– Nate Eldredge
Nov 17 at 15:42
2
@DanYand: More generally, between any two uncountable complete separable metric spaces, there is a bijective Borel function. This fact is found in lots of measure theory and descriptive set theory books, e.g. Kechris, Classical Descriptive Set Theory.
– Nate Eldredge
Nov 17 at 15:44
2
2
What if you take a 1-1 Borel function $f : mathbb{C} to S^1$ and let $U = f(N)$ using the functional calculus?
– Nate Eldredge
Nov 16 at 21:52
What if you take a 1-1 Borel function $f : mathbb{C} to S^1$ and let $U = f(N)$ using the functional calculus?
– Nate Eldredge
Nov 16 at 21:52
1
1
What do you mean by "having the same spectral measure"? The spectral measure of a unitary operator is supported in $S^1$ while the spectral measure of $N$ need not be so. It seems that your are using an equivalence of spectral measures that factors the equivalence of the underlying subsets of $mathbb C$.
– Adrián González-Pérez
Nov 17 at 14:48
What do you mean by "having the same spectral measure"? The spectral measure of a unitary operator is supported in $S^1$ while the spectral measure of $N$ need not be so. It seems that your are using an equivalence of spectral measures that factors the equivalence of the underlying subsets of $mathbb C$.
– Adrián González-Pérez
Nov 17 at 14:48
@AdriánGonzález-Pérez I'm not a mathematician, and I'm probably not using the right words. I'm trying to ask if, given $N$, there is always a $U$ that implicitly defines the same set of mutually commuting projection operators via the spectral decomposition theorem. I understand that $N$ and $U$ will typically have different spectra (numbers $z$ for which $N-z$ or $U-z$ is not invertible), but can the projection-valued measure in the spectral decomposition theorem be defined independently of the spectrum? Sorry if my vocabulary is mixed up. I know that makes deciphering my question difficult.
– Dan Yand
Nov 17 at 15:08
@AdriánGonzález-Pérez I'm not a mathematician, and I'm probably not using the right words. I'm trying to ask if, given $N$, there is always a $U$ that implicitly defines the same set of mutually commuting projection operators via the spectral decomposition theorem. I understand that $N$ and $U$ will typically have different spectra (numbers $z$ for which $N-z$ or $U-z$ is not invertible), but can the projection-valued measure in the spectral decomposition theorem be defined independently of the spectrum? Sorry if my vocabulary is mixed up. I know that makes deciphering my question difficult.
– Dan Yand
Nov 17 at 15:08
1
1
@DanYand: Heuristically, a Borel function is anything you can describe explicitly without using the axiom of choice. For example, to get a 1-1 Borel function from the unit square to the unit interval, you can do something like mapping $(0.123456dots, 0.789789dots)$ to $0.172839475869dots$ by interleaving the digits. One could use the same sort of trick to get a 1-1 Borel function from $mathbb{C}$ to $S^1$.
– Nate Eldredge
Nov 17 at 15:42
@DanYand: Heuristically, a Borel function is anything you can describe explicitly without using the axiom of choice. For example, to get a 1-1 Borel function from the unit square to the unit interval, you can do something like mapping $(0.123456dots, 0.789789dots)$ to $0.172839475869dots$ by interleaving the digits. One could use the same sort of trick to get a 1-1 Borel function from $mathbb{C}$ to $S^1$.
– Nate Eldredge
Nov 17 at 15:42
2
2
@DanYand: More generally, between any two uncountable complete separable metric spaces, there is a bijective Borel function. This fact is found in lots of measure theory and descriptive set theory books, e.g. Kechris, Classical Descriptive Set Theory.
– Nate Eldredge
Nov 17 at 15:44
@DanYand: More generally, between any two uncountable complete separable metric spaces, there is a bijective Borel function. This fact is found in lots of measure theory and descriptive set theory books, e.g. Kechris, Classical Descriptive Set Theory.
– Nate Eldredge
Nov 17 at 15:44
|
show 2 more comments
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2
What if you take a 1-1 Borel function $f : mathbb{C} to S^1$ and let $U = f(N)$ using the functional calculus?
– Nate Eldredge
Nov 16 at 21:52
1
What do you mean by "having the same spectral measure"? The spectral measure of a unitary operator is supported in $S^1$ while the spectral measure of $N$ need not be so. It seems that your are using an equivalence of spectral measures that factors the equivalence of the underlying subsets of $mathbb C$.
– Adrián González-Pérez
Nov 17 at 14:48
@AdriánGonzález-Pérez I'm not a mathematician, and I'm probably not using the right words. I'm trying to ask if, given $N$, there is always a $U$ that implicitly defines the same set of mutually commuting projection operators via the spectral decomposition theorem. I understand that $N$ and $U$ will typically have different spectra (numbers $z$ for which $N-z$ or $U-z$ is not invertible), but can the projection-valued measure in the spectral decomposition theorem be defined independently of the spectrum? Sorry if my vocabulary is mixed up. I know that makes deciphering my question difficult.
– Dan Yand
Nov 17 at 15:08
1
@DanYand: Heuristically, a Borel function is anything you can describe explicitly without using the axiom of choice. For example, to get a 1-1 Borel function from the unit square to the unit interval, you can do something like mapping $(0.123456dots, 0.789789dots)$ to $0.172839475869dots$ by interleaving the digits. One could use the same sort of trick to get a 1-1 Borel function from $mathbb{C}$ to $S^1$.
– Nate Eldredge
Nov 17 at 15:42
2
@DanYand: More generally, between any two uncountable complete separable metric spaces, there is a bijective Borel function. This fact is found in lots of measure theory and descriptive set theory books, e.g. Kechris, Classical Descriptive Set Theory.
– Nate Eldredge
Nov 17 at 15:44