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.










share|cite|improve this question




















  • 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















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.










share|cite|improve this question




















  • 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













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.










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 22 hours ago









Zvi

3,005217




3,005217










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














  • 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















active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














 

draft saved


draft discarded


















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






























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes
















 

draft saved


draft discarded



















































 


draft saved


draft discarded














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





















































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







Popular posts from this blog

AnyDesk - Fatal Program Failure

How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

QoS: MAC-Priority for clients behind a repeater