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Motivation of Haar Measure [closed]
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What is the motivation of Haar measure? Where does a measure of abstract topological groups play a role? And why would one come to the conclusion to introduce a measure of these abstract topological groups?
analysis measure-theory
asked Jul 26 at 13:32
user109871
414217
closed as too broad by uniquesolution, amWhy, max_zorn, Isaac Browne, Lord Shark the Unknown Jul 27 at 5:24
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-2
down vote
favorite
What is the motivation of Haar measure? Where does a measure of abstract topological groups play a role? And why would one come to the conclusion to introduce a measure of these abstract topological groups?
analysis measure-theory
asked Jul 26 at 13:32
user109871
414217
closed as too broad by uniquesolution, amWhy, max_zorn, Isaac Browne, Lord Shark the Unknown Jul 27 at 5:24
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
"Where does a measure of abstract topological groups play a role?" Pretty much everywhere in harmonic analysis...
– Lord Shark the Unknown
Jul 26 at 13:35
1
But why would one want Fourier Analysis on an abstract group? I understand that one might want Fourier Analysis on euclidean spaces, the circle and discrete groups.
– user109871
Jul 26 at 13:54
1
Read amazon.co.uk/Fourier-Analysis-Number-Fields-Mathematics/dp/… and be amazed. @user109871
– Lord Shark the Unknown
Jul 26 at 14:10
1
@Shark Is there motivation outside of Number Theory?
– user109871
Jul 26 at 14:16
2
@user109871 "One of the most telling proofs of the worth of an abstract concept is what it, and the results about it, tells us in familiar situations" Israel Nathan Herstein
– Bob
Jul 26 at 14:20
 |Â
show 2 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
What is the motivation of Haar measure? Where does a measure of abstract topological groups play a role? And why would one come to the conclusion to introduce a measure of these abstract topological groups?
analysis measure-theory
asked Jul 26 at 13:32
user109871
414217
What is the motivation of Haar measure? Where does a measure of abstract topological groups play a role? And why would one come to the conclusion to introduce a measure of these abstract topological groups?
analysis measure-theory
asked Jul 26 at 13:32
user109871
414217
asked Jul 26 at 13:32
user109871
414217
asked Jul 26 at 13:32
user109871
414217
asked Jul 26 at 13:32
user109871
414217
414217
closed as too broad by uniquesolution, amWhy, max_zorn, Isaac Browne, Lord Shark the Unknown Jul 27 at 5:24
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by uniquesolution, amWhy, max_zorn, Isaac Browne, Lord Shark the Unknown Jul 27 at 5:24
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
"Where does a measure of abstract topological groups play a role?" Pretty much everywhere in harmonic analysis...
– Lord Shark the Unknown
Jul 26 at 13:35
1
But why would one want Fourier Analysis on an abstract group? I understand that one might want Fourier Analysis on euclidean spaces, the circle and discrete groups.
– user109871
Jul 26 at 13:54
1
Read amazon.co.uk/Fourier-Analysis-Number-Fields-Mathematics/dp/… and be amazed. @user109871
– Lord Shark the Unknown
Jul 26 at 14:10
1
@Shark Is there motivation outside of Number Theory?
– user109871
Jul 26 at 14:16
2
@user109871 "One of the most telling proofs of the worth of an abstract concept is what it, and the results about it, tells us in familiar situations" Israel Nathan Herstein
– Bob
Jul 26 at 14:20
 |Â
show 2 more comments
2
"Where does a measure of abstract topological groups play a role?" Pretty much everywhere in harmonic analysis...
– Lord Shark the Unknown
Jul 26 at 13:35
1
But why would one want Fourier Analysis on an abstract group? I understand that one might want Fourier Analysis on euclidean spaces, the circle and discrete groups.
– user109871
Jul 26 at 13:54
1
Read amazon.co.uk/Fourier-Analysis-Number-Fields-Mathematics/dp/… and be amazed. @user109871
– Lord Shark the Unknown
Jul 26 at 14:10
1
@Shark Is there motivation outside of Number Theory?
– user109871
Jul 26 at 14:16
2
@user109871 "One of the most telling proofs of the worth of an abstract concept is what it, and the results about it, tells us in familiar situations" Israel Nathan Herstein
– Bob
Jul 26 at 14:20
2
2
"Where does a measure of abstract topological groups play a role?" Pretty much everywhere in harmonic analysis...
– Lord Shark the Unknown
Jul 26 at 13:35
"Where does a measure of abstract topological groups play a role?" Pretty much everywhere in harmonic analysis...
– Lord Shark the Unknown
Jul 26 at 13:35
1
1
But why would one want Fourier Analysis on an abstract group? I understand that one might want Fourier Analysis on euclidean spaces, the circle and discrete groups.
– user109871
Jul 26 at 13:54
But why would one want Fourier Analysis on an abstract group? I understand that one might want Fourier Analysis on euclidean spaces, the circle and discrete groups.
– user109871
Jul 26 at 13:54
1
1
Read amazon.co.uk/Fourier-Analysis-Number-Fields-Mathematics/dp/… and be amazed. @user109871
– Lord Shark the Unknown
Jul 26 at 14:10
Read amazon.co.uk/Fourier-Analysis-Number-Fields-Mathematics/dp/… and be amazed. @user109871
– Lord Shark the Unknown
Jul 26 at 14:10
1
1
@Shark Is there motivation outside of Number Theory?
– user109871
Jul 26 at 14:16
@Shark Is there motivation outside of Number Theory?
– user109871
Jul 26 at 14:16
2
2
@user109871 "One of the most telling proofs of the worth of an abstract concept is what it, and the results about it, tells us in familiar situations" Israel Nathan Herstein
– Bob
Jul 26 at 14:20
@user109871 "One of the most telling proofs of the worth of an abstract concept is what it, and the results about it, tells us in familiar situations" Israel Nathan Herstein
– Bob
Jul 26 at 14:20
 |Â
show 2 more comments
2 Answers
2
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3
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The comments provided very good theoretical insights to the question, but I have the feeling that the author of the post is looking for a more "practical" approach.
My answer is certainly not addressing the most important aspects of Haar measure, but it shows an example how it can sometimes be used as a proof technique.
Assume first that $G$ is a finite group.
Then every action of $G$ (compatible with the additive structure) on a finite real vector space, or a $G$-invariant convex subset of it, has a fixed point. (Cf. a finite group of congruences of a convex subset of a Euclidean space $mathbbR^n$ has a fixed point that serves as some kind of "center" of the subset.) Namely, pick any vector $v$, and "average out": $frac1sumlimits_gin G gcdot v$ is a fixed point.
Now if $G$ is not finite but compact, a similar idea works. Instead of averaging out, we integrate with respect to the Haar measure $mu$:
$$intlimits_G gcdot v ,dmu$$
This makes sense, since if you focus on a given coordinate, then $gcdot v$ yields a function $Grightarrow mathbbR$, so there is no formal obstruction of integration. (OK, maybe the function is not integrable, but often it is.)
More generally, you can obtain a $G$-invariant object this way, preserving some properties of the original object.
Check out this really nice application of the idea, which is one of the variants of the so-called Weyl trick (see Theorem 4.1.):
https://sites.math.washington.edu/~morrow/336_17/papers17/thomas.pdf
It is used to show that under some conditions, representations can be decomposed into a sum of irreducible representations (see Theorem 4.2.).
Of course you can keep saying "OK, but why are representations useful", but at a point, you will lose your audience...
answered Jul 26 at 15:40
A. Pongrácz
1,584115
add a comment |Â
up vote
2
down vote
Even if one does not find the abstraction and generality appealing in itself, such general existence-and-uniqueness results are reassuring: even if we anticipate being able to exhibit an explicit or formulaic invariant measure on a tangible topological group of interest, it is comforting to know _in_advance_, _with_certainty_, that there exists such... and that up to scalar multiples it is unique. In practice, in explicit situations, it is indeed often possible to construct an invariant measure... but far less clear that there are no alternatives. (Further, there is the potential confusion between alternative constructions and alternative outcomes...)
The essential uniqueness also means that we can often manipulate Haar measures and measures on quotients $G/H$ without reference to formulaic or construction details, but, rather, just using properties which uniquely characterize them. For example, for unimodular $G$ and $H$, there is a unique measure $ddotg$ on $G/H$ such that, for $fin C^o_c(G)$
$$
int_G f(g); dg ;=; int_G/HBig( int_H f(dotgh);dhBig);ddot g
$$
with Haar measures $dg$ and $dh$... This also applies when $H$ is a discrete subgroup of $G$, and, significantly, shows that we would not need to determine a fundamental domain for the action of $H$ on $G$ to have this integration formula.
answered Jul 26 at 18:14
paul garrett
30.8k360116
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The comments provided very good theoretical insights to the question, but I have the feeling that the author of the post is looking for a more "practical" approach.
My answer is certainly not addressing the most important aspects of Haar measure, but it shows an example how it can sometimes be used as a proof technique.
Assume first that $G$ is a finite group.
Then every action of $G$ (compatible with the additive structure) on a finite real vector space, or a $G$-invariant convex subset of it, has a fixed point. (Cf. a finite group of congruences of a convex subset of a Euclidean space $mathbbR^n$ has a fixed point that serves as some kind of "center" of the subset.) Namely, pick any vector $v$, and "average out": $frac1sumlimits_gin G gcdot v$ is a fixed point.
Now if $G$ is not finite but compact, a similar idea works. Instead of averaging out, we integrate with respect to the Haar measure $mu$:
$$intlimits_G gcdot v ,dmu$$
This makes sense, since if you focus on a given coordinate, then $gcdot v$ yields a function $Grightarrow mathbbR$, so there is no formal obstruction of integration. (OK, maybe the function is not integrable, but often it is.)
More generally, you can obtain a $G$-invariant object this way, preserving some properties of the original object.
Check out this really nice application of the idea, which is one of the variants of the so-called Weyl trick (see Theorem 4.1.):
https://sites.math.washington.edu/~morrow/336_17/papers17/thomas.pdf
It is used to show that under some conditions, representations can be decomposed into a sum of irreducible representations (see Theorem 4.2.).
Of course you can keep saying "OK, but why are representations useful", but at a point, you will lose your audience...
answered Jul 26 at 15:40
A. Pongrácz
1,584115
add a comment |Â
up vote
3
down vote
The comments provided very good theoretical insights to the question, but I have the feeling that the author of the post is looking for a more "practical" approach.
My answer is certainly not addressing the most important aspects of Haar measure, but it shows an example how it can sometimes be used as a proof technique.
Assume first that $G$ is a finite group.
Then every action of $G$ (compatible with the additive structure) on a finite real vector space, or a $G$-invariant convex subset of it, has a fixed point. (Cf. a finite group of congruences of a convex subset of a Euclidean space $mathbbR^n$ has a fixed point that serves as some kind of "center" of the subset.) Namely, pick any vector $v$, and "average out": $frac1sumlimits_gin G gcdot v$ is a fixed point.
Now if $G$ is not finite but compact, a similar idea works. Instead of averaging out, we integrate with respect to the Haar measure $mu$:
$$intlimits_G gcdot v ,dmu$$
This makes sense, since if you focus on a given coordinate, then $gcdot v$ yields a function $Grightarrow mathbbR$, so there is no formal obstruction of integration. (OK, maybe the function is not integrable, but often it is.)
More generally, you can obtain a $G$-invariant object this way, preserving some properties of the original object.
Check out this really nice application of the idea, which is one of the variants of the so-called Weyl trick (see Theorem 4.1.):
https://sites.math.washington.edu/~morrow/336_17/papers17/thomas.pdf
It is used to show that under some conditions, representations can be decomposed into a sum of irreducible representations (see Theorem 4.2.).
Of course you can keep saying "OK, but why are representations useful", but at a point, you will lose your audience...
answered Jul 26 at 15:40
A. Pongrácz
1,584115
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The comments provided very good theoretical insights to the question, but I have the feeling that the author of the post is looking for a more "practical" approach.
My answer is certainly not addressing the most important aspects of Haar measure, but it shows an example how it can sometimes be used as a proof technique.
Assume first that $G$ is a finite group.
Then every action of $G$ (compatible with the additive structure) on a finite real vector space, or a $G$-invariant convex subset of it, has a fixed point. (Cf. a finite group of congruences of a convex subset of a Euclidean space $mathbbR^n$ has a fixed point that serves as some kind of "center" of the subset.) Namely, pick any vector $v$, and "average out": $frac1sumlimits_gin G gcdot v$ is a fixed point.
Now if $G$ is not finite but compact, a similar idea works. Instead of averaging out, we integrate with respect to the Haar measure $mu$:
$$intlimits_G gcdot v ,dmu$$
This makes sense, since if you focus on a given coordinate, then $gcdot v$ yields a function $Grightarrow mathbbR$, so there is no formal obstruction of integration. (OK, maybe the function is not integrable, but often it is.)
More generally, you can obtain a $G$-invariant object this way, preserving some properties of the original object.
Check out this really nice application of the idea, which is one of the variants of the so-called Weyl trick (see Theorem 4.1.):
https://sites.math.washington.edu/~morrow/336_17/papers17/thomas.pdf
It is used to show that under some conditions, representations can be decomposed into a sum of irreducible representations (see Theorem 4.2.).
Of course you can keep saying "OK, but why are representations useful", but at a point, you will lose your audience...
answered Jul 26 at 15:40
A. Pongrácz
1,584115
The comments provided very good theoretical insights to the question, but I have the feeling that the author of the post is looking for a more "practical" approach.
My answer is certainly not addressing the most important aspects of Haar measure, but it shows an example how it can sometimes be used as a proof technique.
Assume first that $G$ is a finite group.
Then every action of $G$ (compatible with the additive structure) on a finite real vector space, or a $G$-invariant convex subset of it, has a fixed point. (Cf. a finite group of congruences of a convex subset of a Euclidean space $mathbbR^n$ has a fixed point that serves as some kind of "center" of the subset.) Namely, pick any vector $v$, and "average out": $frac1sumlimits_gin G gcdot v$ is a fixed point.
Now if $G$ is not finite but compact, a similar idea works. Instead of averaging out, we integrate with respect to the Haar measure $mu$:
$$intlimits_G gcdot v ,dmu$$
This makes sense, since if you focus on a given coordinate, then $gcdot v$ yields a function $Grightarrow mathbbR$, so there is no formal obstruction of integration. (OK, maybe the function is not integrable, but often it is.)
More generally, you can obtain a $G$-invariant object this way, preserving some properties of the original object.
Check out this really nice application of the idea, which is one of the variants of the so-called Weyl trick (see Theorem 4.1.):
https://sites.math.washington.edu/~morrow/336_17/papers17/thomas.pdf
It is used to show that under some conditions, representations can be decomposed into a sum of irreducible representations (see Theorem 4.2.).
Of course you can keep saying "OK, but why are representations useful", but at a point, you will lose your audience...
answered Jul 26 at 15:40
A. Pongrácz
1,584115
answered Jul 26 at 15:40
A. Pongrácz
1,584115
answered Jul 26 at 15:40
A. Pongrácz
1,584115
answered Jul 26 at 15:40
A. Pongrácz
1,584115
1,584115
add a comment |Â
add a comment |Â
up vote
2
down vote
Even if one does not find the abstraction and generality appealing in itself, such general existence-and-uniqueness results are reassuring: even if we anticipate being able to exhibit an explicit or formulaic invariant measure on a tangible topological group of interest, it is comforting to know _in_advance_, _with_certainty_, that there exists such... and that up to scalar multiples it is unique. In practice, in explicit situations, it is indeed often possible to construct an invariant measure... but far less clear that there are no alternatives. (Further, there is the potential confusion between alternative constructions and alternative outcomes...)
The essential uniqueness also means that we can often manipulate Haar measures and measures on quotients $G/H$ without reference to formulaic or construction details, but, rather, just using properties which uniquely characterize them. For example, for unimodular $G$ and $H$, there is a unique measure $ddotg$ on $G/H$ such that, for $fin C^o_c(G)$
$$
int_G f(g); dg ;=; int_G/HBig( int_H f(dotgh);dhBig);ddot g
$$
with Haar measures $dg$ and $dh$... This also applies when $H$ is a discrete subgroup of $G$, and, significantly, shows that we would not need to determine a fundamental domain for the action of $H$ on $G$ to have this integration formula.
answered Jul 26 at 18:14
paul garrett
30.8k360116
add a comment |Â
up vote
2
down vote
Even if one does not find the abstraction and generality appealing in itself, such general existence-and-uniqueness results are reassuring: even if we anticipate being able to exhibit an explicit or formulaic invariant measure on a tangible topological group of interest, it is comforting to know _in_advance_, _with_certainty_, that there exists such... and that up to scalar multiples it is unique. In practice, in explicit situations, it is indeed often possible to construct an invariant measure... but far less clear that there are no alternatives. (Further, there is the potential confusion between alternative constructions and alternative outcomes...)
The essential uniqueness also means that we can often manipulate Haar measures and measures on quotients $G/H$ without reference to formulaic or construction details, but, rather, just using properties which uniquely characterize them. For example, for unimodular $G$ and $H$, there is a unique measure $ddotg$ on $G/H$ such that, for $fin C^o_c(G)$
$$
int_G f(g); dg ;=; int_G/HBig( int_H f(dotgh);dhBig);ddot g
$$
with Haar measures $dg$ and $dh$... This also applies when $H$ is a discrete subgroup of $G$, and, significantly, shows that we would not need to determine a fundamental domain for the action of $H$ on $G$ to have this integration formula.
answered Jul 26 at 18:14
paul garrett
30.8k360116
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Even if one does not find the abstraction and generality appealing in itself, such general existence-and-uniqueness results are reassuring: even if we anticipate being able to exhibit an explicit or formulaic invariant measure on a tangible topological group of interest, it is comforting to know _in_advance_, _with_certainty_, that there exists such... and that up to scalar multiples it is unique. In practice, in explicit situations, it is indeed often possible to construct an invariant measure... but far less clear that there are no alternatives. (Further, there is the potential confusion between alternative constructions and alternative outcomes...)
The essential uniqueness also means that we can often manipulate Haar measures and measures on quotients $G/H$ without reference to formulaic or construction details, but, rather, just using properties which uniquely characterize them. For example, for unimodular $G$ and $H$, there is a unique measure $ddotg$ on $G/H$ such that, for $fin C^o_c(G)$
$$
int_G f(g); dg ;=; int_G/HBig( int_H f(dotgh);dhBig);ddot g
$$
with Haar measures $dg$ and $dh$... This also applies when $H$ is a discrete subgroup of $G$, and, significantly, shows that we would not need to determine a fundamental domain for the action of $H$ on $G$ to have this integration formula.
answered Jul 26 at 18:14
paul garrett
30.8k360116
Even if one does not find the abstraction and generality appealing in itself, such general existence-and-uniqueness results are reassuring: even if we anticipate being able to exhibit an explicit or formulaic invariant measure on a tangible topological group of interest, it is comforting to know _in_advance_, _with_certainty_, that there exists such... and that up to scalar multiples it is unique. In practice, in explicit situations, it is indeed often possible to construct an invariant measure... but far less clear that there are no alternatives. (Further, there is the potential confusion between alternative constructions and alternative outcomes...)
The essential uniqueness also means that we can often manipulate Haar measures and measures on quotients $G/H$ without reference to formulaic or construction details, but, rather, just using properties which uniquely characterize them. For example, for unimodular $G$ and $H$, there is a unique measure $ddotg$ on $G/H$ such that, for $fin C^o_c(G)$
$$
int_G f(g); dg ;=; int_G/HBig( int_H f(dotgh);dhBig);ddot g
$$
with Haar measures $dg$ and $dh$... This also applies when $H$ is a discrete subgroup of $G$, and, significantly, shows that we would not need to determine a fundamental domain for the action of $H$ on $G$ to have this integration formula.
answered Jul 26 at 18:14
paul garrett
30.8k360116
answered Jul 26 at 18:14
paul garrett
30.8k360116
answered Jul 26 at 18:14
paul garrett
30.8k360116
answered Jul 26 at 18:14
paul garrett
30.8k360116
30.8k360116
add a comment |Â
add a comment |Â
2
"Where does a measure of abstract topological groups play a role?" Pretty much everywhere in harmonic analysis...
– Lord Shark the Unknown
Jul 26 at 13:35
1
But why would one want Fourier Analysis on an abstract group? I understand that one might want Fourier Analysis on euclidean spaces, the circle and discrete groups.
– user109871
Jul 26 at 13:54
1
Read amazon.co.uk/Fourier-Analysis-Number-Fields-Mathematics/dp/… and be amazed. @user109871
– Lord Shark the Unknown
Jul 26 at 14:10
1
@Shark Is there motivation outside of Number Theory?
– user109871
Jul 26 at 14:16
2
@user109871 "One of the most telling proofs of the worth of an abstract concept is what it, and the results about it, tells us in familiar situations" Israel Nathan Herstein
– Bob
Jul 26 at 14:20