Regarding the dimension of irreducible (finite-dimensional) group representations
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 22 hours ago
Zvi
3,005217
asked yesterday
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 22 hours ago
Zvi
3,005217
asked yesterday
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
yesterday
@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
yesterday
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
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
11 hours ago
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
11 hours ago
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 22 hours ago
Zvi
3,005217
asked yesterday
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 22 hours ago
Zvi
3,005217
asked yesterday
R. Rankin
281213
edited 22 hours ago
Zvi
3,005217
asked yesterday
R. Rankin
281213
edited 22 hours ago
Zvi
3,005217
edited 22 hours ago
Zvi
3,005217
edited 22 hours ago
Zvi
3,005217
3,005217
asked yesterday
R. Rankin
281213
asked yesterday
R. Rankin
281213
asked yesterday
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
yesterday
@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
yesterday
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
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
11 hours ago
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
11 hours ago
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
yesterday
@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
yesterday
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
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
11 hours ago
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
11 hours ago
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
yesterday
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
yesterday
@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
yesterday
@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
yesterday
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
yesterday
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
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
11 hours 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
11 hours ago
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
11 hours ago
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
11 hours ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999372%2fregarding-the-dimension-of-irreducible-finite-dimensional-group-representation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
yesterday
@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
yesterday
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
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
11 hours ago
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
11 hours ago