Irreducible Representations of SO(3) and the Laplacian

Introduction

In our previous post on the irreducible representations of and , the irreducible representations of has been determined explicitly: , and irreducible representations of correspond to .

The result is satisfying for but not for . We hope it has something to do with but has not. In this post, we are delivering a much clear characterisation of .

This post would be relatively easier to read. Other than the basic language of representation theory (of Lie groups), only multivariable calculus is needed.

General Strategy & Playground

Like in the previous post, we first determine a good playground and then show that this is all we need. The playground here is

The reason for the symmetric product of is simple: we will be working on homogeneous polynomials. We complexified this space to make sure that we will not worry about eigenvalues (of ). In other words, is the complex vector space of homogeneous polynomials in three variables, viewed as functions on .

Recall that

Therefore , as a -vector space.

We will extract what we want from the spaces of this form.

The Representation of SO(3) And the Laplacian

The action of or in general on is defined in a similar way. For any and , we define

Here, , and is a product of and in sense of matrix multiplication. It is easy to verify that this indeed gives rise to a group representation.

To study this representation, we need to find some morphisms . The most obvious choice is the Laplacian, which is given by

In other words, is the trace of the Hessian matrix of . Trace is used in representation theory to define character so there is a chance to find its good connection to the representation of .

We shall also not forget the kernel of the Laplacian, which is called harmonic polynomials of degree in this context:

Since functions in are homogeneous, the value of at a point is determined by the value at , the unit sphere. Therefore we also call the spherical harmonics of degree . And we certainly need to know the nullity of .

Lemma 1. The dimension of is .

Proof. First of all we perform a Taylor expansion of with respect to the first variable :

Here, is homogeneous of degree in . Therefore we only need to study one term of the right hand side.

Now we can put them together naturally:

Let's try to explore the last term a little bit more. If or , then is of order and and consequently the second order derivative is . Therefore we write

Therefore if and only if

Therefore, once and is determined, all of are determined and so is itself. Therefore

where is the space of homogeneous polynomials with two variables, hence is isomorphic to , and we therefore have

Hence

Recall that . This should not be a coincidence, and we shall dive into it right now. To do this we immediately establish the connection between and .

Lemma 2. The action of the Laplacian on (which contains for all ) commutes with the action of , i.e. is -equivariant.

Proof. Really routine verification.

As a result, we have an very important result:

Theorem 1. The space is an -invariant subspace of .

Characters of SO(3)

We start with a direct observation of matrices in , that 'downgrades' the rotation to a plane:

Lemma 2. Every element in is conjugate to , where

Proof. Pick any . First of all we show that has an eigenvalue . Note

we therefore have . Hence we can pick with norm . Pick pedicular to with norm and . Then is an orthonormal basis and is in . is therefore conjugate to

In particular, also implies

Solving this equation system we must have , so that we can assign and , and the result follows.

Since characters are invariant under conjugation, the study of the character of is reduced to , the subgroup generated by matrices of the form . But direct computation is a nightmare so we try our best to do it elegantly. To do this, we return to the irreducible representations of (there are only two variables, anyway). The canonical map has a specific value:

One can refer to this document for the map above. Our study of characters is now reduced to , because , using the facts that character is invariant under isomorphism and that . We can compute that

Now we are ready for the irreducible representations of .

Determining All the Irreducible SO(3)-modules

Since we basically have , it is natural to believe that , in the sense of -modules, and the following theorem answers this question affirmatively.

Theorem 2. The space is isomorphic to . In other words, irreducible -modules are determined by spherical harmonics.

Proof. We will use the fact every compact Lie group is completely reducible. (First of all, is compact as it is a closed subgroup of this document. On the other hand, the fact that every compact Lie group is completely reducible can be found in section 3 of this document).

Therefore we have

where each is an irreducible representation of . Applying dimension on both sides yields

To prove that , it suffices to show that for some . On the other hand, applying characters on both side we see

In other words, the character is a linear combination of with . Each appeared above is an eigenvalue of the action of . We don't know about the distribution of yet but our job is done if we can show that has an eigenvalue .

To do this, we can consider vector . This is because for this vector we have

Ending Remarks

• We find the eigenvalue because it shows that appears in the summand of , hence . Since is finite, the maximum can be attained, and therefore our argument on dimension is done.

• The representation of can be deduced algebraically, for one only need to notice that , where . One will also need an odd-even argument just like we did to .

• Likewise, since , we can deduce the irreducible representations of in a same fashion.