Regarding the dimension of irreducible (finite-dimensional) group representations
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Ok, I admit it. I'm confused. I'm a physics student attempting to learn some group theory and topology in my spare time. I was reading about group representations. For example I get that the set of spherical harmonics $Y_{lm}(theta,phi)$ form a set of irreducible representations of $SO(3)$. What I don't get is their dimension. For example here (page 144 as it reads on the paper heading) it is stated:
The $Y_{lm}(theta,phi)$ form a $(2l+1)$ -dimensional representation of
$SO(3)$.
Now, in utilizing the spherical harmonics in physics, I know that I'm working in a three dimensional space. I further know that I can represent any “well behaved” function $f(theta ,phi)$ on the unit sphere in $R^3$ in terms of a series of these spherical harmonics (properly weighted with coefficients) like so:
$$f(theta,phi)=sum_{l=0}^{infty}sum_{m=-l}^{m=l}a_{lm}Y_{lm}(theta,phi)$$
It's not lost on me that the dimensionality for a given representation is the same as the number of $m$ values (ie the second summation). I know the function "lives" in a $2$-dimensional space (the unit sphere). So what is going on? Is there a mapping or reference to another space I'm missing? is this a physicist's notational/dictionarial clash with the mathematician's?? Thank you in advance.
group-theory representation-theory lie-groups spherical-harmonics
edited 2 days ago
Zvi
3,125217
asked 2 days ago
R. Rankin
281213
add a comment |
up vote
2
down vote
favorite
Ok, I admit it. I'm confused. I'm a physics student attempting to learn some group theory and topology in my spare time. I was reading about group representations. For example I get that the set of spherical harmonics $Y_{lm}(theta,phi)$ form a set of irreducible representations of $SO(3)$. What I don't get is their dimension. For example here (page 144 as it reads on the paper heading) it is stated:
The $Y_{lm}(theta,phi)$ form a $(2l+1)$ -dimensional representation of
$SO(3)$.
Now, in utilizing the spherical harmonics in physics, I know that I'm working in a three dimensional space. I further know that I can represent any “well behaved” function $f(theta ,phi)$ on the unit sphere in $R^3$ in terms of a series of these spherical harmonics (properly weighted with coefficients) like so:
$$f(theta,phi)=sum_{l=0}^{infty}sum_{m=-l}^{m=l}a_{lm}Y_{lm}(theta,phi)$$
It's not lost on me that the dimensionality for a given representation is the same as the number of $m$ values (ie the second summation). I know the function "lives" in a $2$-dimensional space (the unit sphere). So what is going on? Is there a mapping or reference to another space I'm missing? is this a physicist's notational/dictionarial clash with the mathematician's?? Thank you in advance.
group-theory representation-theory lie-groups spherical-harmonics
edited 2 days ago
Zvi
3,125217
asked 2 days ago
R. Rankin
281213
1
The space of functions on the unit sphere is infinite-dimensional, so the functions do not "live" in a 2-dimensional space (other than in the trivial sense that each individual function spans a 1-dimensional space). Maybe if you include the definition of the various symbols used here.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft I suppose I'm just used to two variables=two dimensions. please see my penultimate sentence in the question and elaborate then if willing.
– R. Rankin
2 days ago
1
I don't see any clash, other than most of the notation not being defined. The unit sphere is 2-dimensional (as a manifold), but the space of all (smooth) maps on the unit sphere (presumably with values in the reals) is an infinite-dimensional vector space.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft Ok I get it. i'm not sure why but I'd never seen the harmonics for a particular l written as a $(2l+1)x(2l+1)$ array (matrix) but putting the various entries for m into the diagonal of a matrix means that the $a_{lm}$ are similarly written in an array. In that sense I see how they're said to have dimension $(2l+1)$ In that sense, a function does have dimension 1.
– R. Rankin
yesterday
i think the only formal definition of dimension I'd come across was one for defining the dimension of a fractal in an old non-euclidean geometry class i'd had. but that doesn't apply here.
– R. Rankin
yesterday
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Ok, I admit it. I'm confused. I'm a physics student attempting to learn some group theory and topology in my spare time. I was reading about group representations. For example I get that the set of spherical harmonics $Y_{lm}(theta,phi)$ form a set of irreducible representations of $SO(3)$. What I don't get is their dimension. For example here (page 144 as it reads on the paper heading) it is stated:
The $Y_{lm}(theta,phi)$ form a $(2l+1)$ -dimensional representation of
$SO(3)$.
Now, in utilizing the spherical harmonics in physics, I know that I'm working in a three dimensional space. I further know that I can represent any “well behaved” function $f(theta ,phi)$ on the unit sphere in $R^3$ in terms of a series of these spherical harmonics (properly weighted with coefficients) like so:
$$f(theta,phi)=sum_{l=0}^{infty}sum_{m=-l}^{m=l}a_{lm}Y_{lm}(theta,phi)$$
It's not lost on me that the dimensionality for a given representation is the same as the number of $m$ values (ie the second summation). I know the function "lives" in a $2$-dimensional space (the unit sphere). So what is going on? Is there a mapping or reference to another space I'm missing? is this a physicist's notational/dictionarial clash with the mathematician's?? Thank you in advance.
group-theory representation-theory lie-groups spherical-harmonics
edited 2 days ago
Zvi
3,125217
asked 2 days ago
R. Rankin
281213
Ok, I admit it. I'm confused. I'm a physics student attempting to learn some group theory and topology in my spare time. I was reading about group representations. For example I get that the set of spherical harmonics $Y_{lm}(theta,phi)$ form a set of irreducible representations of $SO(3)$. What I don't get is their dimension. For example here (page 144 as it reads on the paper heading) it is stated:
The $Y_{lm}(theta,phi)$ form a $(2l+1)$ -dimensional representation of
$SO(3)$.
Now, in utilizing the spherical harmonics in physics, I know that I'm working in a three dimensional space. I further know that I can represent any “well behaved” function $f(theta ,phi)$ on the unit sphere in $R^3$ in terms of a series of these spherical harmonics (properly weighted with coefficients) like so:
$$f(theta,phi)=sum_{l=0}^{infty}sum_{m=-l}^{m=l}a_{lm}Y_{lm}(theta,phi)$$
It's not lost on me that the dimensionality for a given representation is the same as the number of $m$ values (ie the second summation). I know the function "lives" in a $2$-dimensional space (the unit sphere). So what is going on? Is there a mapping or reference to another space I'm missing? is this a physicist's notational/dictionarial clash with the mathematician's?? Thank you in advance.
group-theory representation-theory lie-groups spherical-harmonics
group-theory representation-theory lie-groups spherical-harmonics
edited 2 days ago
Zvi
3,125217
asked 2 days ago
R. Rankin
281213
edited 2 days ago
Zvi
3,125217
asked 2 days ago
R. Rankin
281213
edited 2 days ago
Zvi
3,125217
edited 2 days ago
Zvi
3,125217
edited 2 days ago
Zvi
3,125217
3,125217
asked 2 days ago
R. Rankin
281213
asked 2 days ago
R. Rankin
281213
asked 2 days ago
R. Rankin
281213
281213
1
The space of functions on the unit sphere is infinite-dimensional, so the functions do not "live" in a 2-dimensional space (other than in the trivial sense that each individual function spans a 1-dimensional space). Maybe if you include the definition of the various symbols used here.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft I suppose I'm just used to two variables=two dimensions. please see my penultimate sentence in the question and elaborate then if willing.
– R. Rankin
2 days ago
1
I don't see any clash, other than most of the notation not being defined. The unit sphere is 2-dimensional (as a manifold), but the space of all (smooth) maps on the unit sphere (presumably with values in the reals) is an infinite-dimensional vector space.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft Ok I get it. i'm not sure why but I'd never seen the harmonics for a particular l written as a $(2l+1)x(2l+1)$ array (matrix) but putting the various entries for m into the diagonal of a matrix means that the $a_{lm}$ are similarly written in an array. In that sense I see how they're said to have dimension $(2l+1)$ In that sense, a function does have dimension 1.
– R. Rankin
yesterday
i think the only formal definition of dimension I'd come across was one for defining the dimension of a fractal in an old non-euclidean geometry class i'd had. but that doesn't apply here.
– R. Rankin
yesterday
add a comment |
1
The space of functions on the unit sphere is infinite-dimensional, so the functions do not "live" in a 2-dimensional space (other than in the trivial sense that each individual function spans a 1-dimensional space). Maybe if you include the definition of the various symbols used here.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft I suppose I'm just used to two variables=two dimensions. please see my penultimate sentence in the question and elaborate then if willing.
– R. Rankin
2 days ago
1
I don't see any clash, other than most of the notation not being defined. The unit sphere is 2-dimensional (as a manifold), but the space of all (smooth) maps on the unit sphere (presumably with values in the reals) is an infinite-dimensional vector space.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft Ok I get it. i'm not sure why but I'd never seen the harmonics for a particular l written as a $(2l+1)x(2l+1)$ array (matrix) but putting the various entries for m into the diagonal of a matrix means that the $a_{lm}$ are similarly written in an array. In that sense I see how they're said to have dimension $(2l+1)$ In that sense, a function does have dimension 1.
– R. Rankin
yesterday
i think the only formal definition of dimension I'd come across was one for defining the dimension of a fractal in an old non-euclidean geometry class i'd had. but that doesn't apply here.
– R. Rankin
yesterday
1
1
The space of functions on the unit sphere is infinite-dimensional, so the functions do not "live" in a 2-dimensional space (other than in the trivial sense that each individual function spans a 1-dimensional space). Maybe if you include the definition of the various symbols used here.
– Tobias Kildetoft
2 days ago
The space of functions on the unit sphere is infinite-dimensional, so the functions do not "live" in a 2-dimensional space (other than in the trivial sense that each individual function spans a 1-dimensional space). Maybe if you include the definition of the various symbols used here.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft I suppose I'm just used to two variables=two dimensions. please see my penultimate sentence in the question and elaborate then if willing.
– R. Rankin
2 days ago
@TobiasKildetoft I suppose I'm just used to two variables=two dimensions. please see my penultimate sentence in the question and elaborate then if willing.
– R. Rankin
2 days ago
1
1
I don't see any clash, other than most of the notation not being defined. The unit sphere is 2-dimensional (as a manifold), but the space of all (smooth) maps on the unit sphere (presumably with values in the reals) is an infinite-dimensional vector space.
– Tobias Kildetoft
2 days ago
I don't see any clash, other than most of the notation not being defined. The unit sphere is 2-dimensional (as a manifold), but the space of all (smooth) maps on the unit sphere (presumably with values in the reals) is an infinite-dimensional vector space.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft Ok I get it. i'm not sure why but I'd never seen the harmonics for a particular l written as a $(2l+1)x(2l+1)$ array (matrix) but putting the various entries for m into the diagonal of a matrix means that the $a_{lm}$ are similarly written in an array. In that sense I see how they're said to have dimension $(2l+1)$ In that sense, a function does have dimension 1.
– R. Rankin
yesterday
@TobiasKildetoft Ok I get it. i'm not sure why but I'd never seen the harmonics for a particular l written as a $(2l+1)x(2l+1)$ array (matrix) but putting the various entries for m into the diagonal of a matrix means that the $a_{lm}$ are similarly written in an array. In that sense I see how they're said to have dimension $(2l+1)$ In that sense, a function does have dimension 1.
– R. Rankin
yesterday
i think the only formal definition of dimension I'd come across was one for defining the dimension of a fractal in an old non-euclidean geometry class i'd had. but that doesn't apply here.
– R. Rankin
yesterday
i think the only formal definition of dimension I'd come across was one for defining the dimension of a fractal in an old non-euclidean geometry class i'd had. but that doesn't apply here.
– R. Rankin
yesterday
add a comment |
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1
The space of functions on the unit sphere is infinite-dimensional, so the functions do not "live" in a 2-dimensional space (other than in the trivial sense that each individual function spans a 1-dimensional space). Maybe if you include the definition of the various symbols used here.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft I suppose I'm just used to two variables=two dimensions. please see my penultimate sentence in the question and elaborate then if willing.
– R. Rankin
2 days ago
1
I don't see any clash, other than most of the notation not being defined. The unit sphere is 2-dimensional (as a manifold), but the space of all (smooth) maps on the unit sphere (presumably with values in the reals) is an infinite-dimensional vector space.
– Tobias Kildetoft
2 days ago
@TobiasKildetoft Ok I get it. i'm not sure why but I'd never seen the harmonics for a particular l written as a $(2l+1)x(2l+1)$ array (matrix) but putting the various entries for m into the diagonal of a matrix means that the $a_{lm}$ are similarly written in an array. In that sense I see how they're said to have dimension $(2l+1)$ In that sense, a function does have dimension 1.
– R. Rankin
yesterday
i think the only formal definition of dimension I'd come across was one for defining the dimension of a fractal in an old non-euclidean geometry class i'd had. but that doesn't apply here.
– R. Rankin
yesterday