7 SL2

In this section we discuss the finite-dimensional representation theory of the Lie algebra 𝔰⁢𝔩2,ℂ. We then use that to study the representation theory of SL2⁡(ℂ), 𝔰⁢𝔩2,ℝ, SL2⁡(ℝ), and SU⁢(2), which are all closely related.

7.1 Weights

We fix the following ’standard’ basis of 𝔰⁢𝔩2,ℂ:

H =(100−1),
X =(0100),
and
Y =(0010).

These satisfy the following commutation relations, which are fundamental (check them!):

[H,X] =2⁢X,
[H,Y] =−2⁢Y,
and
[X,Y] =H.

Our strategy is to decompose representations of 𝔰⁢𝔩2,ℂ into eigenspaces for the action of H. The elements X and Y will then move vectors between these eigenspaces, and this will let us analyze the representation theory of 𝔰⁢𝔩2,ℂ.

Let (ρ,V) be a finite-dimensional complex-linear representation of 𝔰⁢𝔩2,ℂ. Since ρ⁢(H) is an endomorphism of a finite-dimensional complex vector space, it has at least one eigenvalue. For α∈ℂ, write

Vα={v∈V:ρ⁢(H)⁢v=α⁢v}

for the α-eigenspace of ρ⁢(H).

Definition 7.1.
  1. 1.

    An eigenvalue α of ρ⁢(H) is called a weight (more precisely, an H-weight) for the representation ρ.

  2. 2.

    Each Vα is called a weight space for ρ.

  3. 3.

    The nonzero vectors in Vα are called weight vectors for ρ.

We will see later (Remark 7.18) that ρ⁢(H) is in fact diagonalizable with integer eigenvalues, so that V decomposes as a direct sum of its weight spaces. However, the classification of irreducible representations does not require this.

Example 7.2.

The set of weights of the zero representation is empty, while the trivial representation has a single weight, 0.

Example 7.3.

Let V=ℂ2 be the standard representation. Write e1,e−1 for the standard basis. Then H⁢e1=e1 and H⁢e−1=−e−1. Thus the set of weights of V is {±1}.

Example 7.4.

We consider the adjoint representation ad of 𝔤=𝔰⁢𝔩2,ℂ on itself. By the commutation relations, we see directly that adH has eigenvalues 0 ([H,H]=0), 2 ([H,X]=2⁢X), and −2 ([H,Y]=−2⁢Y), so the set of weights is

{−2,0,2}.

The non-zero weights 2 and −2 are called the roots of 𝔰⁢𝔩2,ℂ and their weight spaces are the root spaces 𝔤2 and 𝔤−2. The weight vectors are called root vectors.

Thus we have the root space decomposition

𝔰⁢𝔩2,ℂ =𝔤0⊕𝔤2⊕𝔤−2
=⟨H⟩⊕⟨X⟩⊕⟨Y⟩.
Example 7.5.

We consider ℂ2⊗ℂ2 where ℂ2 is the standard representation. Then

H⁢(e1⊗e1)=(H⁢e1)⊗e1+e1⊗H⁢e1=2⁢e1⊗e1

and similarly

H⁢(e1⊗e−1)=H⁢(e−1⊗e1)=0,H⁢(e−1⊗e−1)=−2⁢e−1⊗e−1

so that the weights are {−2,0,0,2}. Note that this is a multiset — a set with repeated elements — and we say that the weight 0 has ‘multiplicity two’ (in general, the multiplicity of a weight is the dimension of the weight space).

Example 7.6.

Take V=Symk⁡(ℂ2). A set of basis vectors is

{e1a⁢e−1b:a,b≥0,a+b=k}.

We calculate

H⁢(e1a⁢e−1b) =a⁢(H⁢e1)⁢e1a−1⁢e−1b+b⁢(H⁢e−1)⁢e1a⁢e−1b−1
=(a−b)⁢e1a⁢e−1b.

Thus the weights are (writing b=k−a):

{−k,2−k,4−k,…,k−4,k−2,k}.

We will soon see an explanation for this pattern.

7.1.1 Highest weights

The following is our first version of the fundamental weight calculation.

Lemma 7.7.

Let (ρ,V) be a complex-linear representation of 𝔰⁢𝔩2,ℂ. Let α be a weight of V and let v∈Vα. Then

X⁢(v)∈Vα+2

and

Y⁢(v)∈Vα−2.

Thus we have three maps:

H :Vα→Vα
X :Vα→Vα+2
Y :Vα→Vα−2.
Proof.

We have, for v∈Vα,

ρ⁢(H)⁢ρ⁢(X)⁢v =[ρ⁢(H),ρ⁢(X)]⁢v+ρ⁢(X)⁢ρ⁢(H)⁢v
=ρ⁢([H,X])⁢v+ρ⁢(X)⁢ρ⁢(H)⁢v
=2⁢ρ⁢(X)⁢v+α⁢ρ⁢(X)⁢v
=(α+2)⁢ρ⁢(X)⁢v.

So ρ⁢(X)⁢v∈Vα+2 as required.

The claim about the action of Y is proved similarly. ∎

Definition 7.8.

A vector v∈Vα is a highest weight vector if it is a weight vector and if

X⁢v=0.

In this case we call the weight of v a highest weight.

Lemma 7.9.

Any nonzero finite-dimensional complex linear representation V of 𝔰⁢𝔩2,ℂ has a highest weight vector.

Proof.

Since V is finite-dimensional, ρ⁢(H) has at least one eigenvalue. Among all weights of V, choose one, α, with maximal real part and let v be a weight vector of weight α. Then X⁢v has weight α+2 by the fundamental weight calculation. Since α+2 has strictly greater real part than α, it cannot be a weight of V, so X⁢v=0. ∎

Example 7.10.

Let V=ℂ2⊗ℂ2. Then the highest weight vectors are e1⊗e1 and e1⊗e−1−e−1⊗e1.

These are easily checked to be highest weight vectors — the first is killed by X since X⁢e1=0, the second becomes

e1⊗e1−e1⊗e1=0.

It is left to you to check that there are no further highest weight vectors.

7.2 Classification of representations of sl2,C.

The key point here is that a highest weight vector must have non-negative integer weight n, and it then generates an irreducible representation of dimension n+1 whose isomorphism class is determined by n and which has a very natural basis of weight vectors.

Let (ρ,V) be a complex-linear representation of 𝔰⁢𝔩2,ℂ.

Lemma 7.11.

Suppose that V has a highest weight vector v of weight n. Then the subspace W spanned by the vectors

v,Y⁢(v),Y2⁢(v)=Y⁢(Y⁢(v)),…

is an 𝔰⁢𝔩2,ℂ-invariant subspace of V.

Moreover, n≥0, the dimension of W is n+1, v,Y⁢(v),…,Yn⁢(v) are a basis for W, with Yn+1⁢(v)=0.

Proof.

Let W be the span of the Yk⁢(v). Since the Yk⁢(v) are weight vectors, their span is H-invariant. It is clearly also Y-invariant. So we only need to check the invariance under the X-action. I claim that for all m≥1,

X⁢Ym⁢(v)=m⁢(n−m+1)⁢Ym−1⁢(v). (7.1)

The proof is by induction. The case m=1 is:

X⁢Y⁢(v)=([X,Y]+Y⁢X)⁢v=H⁢v+Y⁢(X⁢v)=H⁢v=n⁢v.

If the formula holds for m, then

X⁢Ym+1⁢(v) =([X,Y]+Y⁢X)⁢Ym⁢(v)
=H⁢Ym⁢v+Y⁢(X⁢Ym⁢v)
=(n−2⁢m)⁢Ym⁢v+m⁢(n−m+1)⁢Ym⁢v (induction hypothesis)
=(m+1)⁢(n−m)⁢Ym⁢v

as required.

If n is not a nonnegative integer, then m⁢(n−m+1)≠0 for all m. So

Ym−1⁢v≠0⟹X⁢Ym⁢v≠0⟹Ym⁢v≠0,

whence Ym⁢v≠0 for all m. As these are weight vectors with distinct weights, they are linearly independent and so span an infinite dimensional subspace.

If Yi⁢v=0 for some 0<i≤n, then

0=X⁢Yi⁢v=i⁢(n−i+1)⁢Yi−1⁢v

and so Yi−1⁢v=0. Repeating gives that v=Y0⁢v=0, a contradiction.

Now,

X⁢Yn+1⁢(v)=(n+1)⁢(n−n)⁢v=0

and so Yn+1⁢v is either zero or a highest weight vector of weight −(n+2)<0. We have already seen that the second possibility cannot happen, so Yn+1⁢v=0.

Thus W is spanned by the (nonzero) weight vectors v,Y⁢v,…,Yn⁢v with distinct weights, which are therefore linearly independent and so a basis for W. ∎

Remark 7.12.

If we didn’t assume that V was finite-dimensional, then the first part of the previous lemma would still be true.

Corollary 7.13.

In the situation of the previous lemma, W is irreducible.

Proof.

Suppose that W′⊂W is a nonzero subrepresentation. Then it has a highest weight vector, which must be (proportional to) Yi⁢v for some 0≤i≤n. But then

X⁢Yi⁢v=i⁢(n−i+1)⁢Yi−1⁢v≠0

if i>0 and so i=0, meaning that v∈W′. But then Yi⁢v∈W′ for all i, so W′=W as required. ∎

The weights of W are illustrated in Figure 7.2.

[Uncaptioned image]
Theorem 7.14.

Suppose V is an irreducible finite-dimensional complex-linear representation of 𝔰⁢𝔩2,ℂ. Then:

  1. 1.

    There is a unique (up to scalar) highest weight vector, with highest weight n≥0.

  2. 2.

    The weights of V are n,n−2,…,2−n,−n.

  3. 3.

    All weight spaces Vα are one-dimensional (we say that V is ‘multiplicity free’).

  4. 4.

    The dimension of V is n+1.

For every n≥0, there is a unique irreducible complex-linear representation of 𝔰⁢𝔩2,ℂ (up to isomorphism) with highest weight n.

Proof.

Let V be an irreducible finite-dimensional complex-linear representation of 𝔰⁢𝔩2,ℂ. Let v∈V be a highest weight vector. By Lemma 7.11 its weight is a nonnegative integer n, and by Corollary 7.13 the vectors v,Y⁢v,…,Yn⁢v span an irreducible subrepresentation of V of dimension n+1, which must therefore be the whole of V. The claims all follow immediately.

Moreover, the actions of X, Y and H on this basis are given by explicit matrices depending only on n, so the isomorphism class of V is determined by n.

It remains only to show that a representation with highest weight n exists for all n≥0. Consider Symn⁡(ℂ2). The vector e1n is a highest weight vector of weight n, so we are done. In fact, in this case Ya⁢e1n is proportional to e1n−a⁢e−1a, so the irreducible representation generated by e1n is in fact the whole of Symn⁡(ℂ2). ∎

In fact, if (ρ,V) is the irreducible representation with highest weight n, highest weight vector v, then the matrices of H, Y, and X with respect to the basis

{v,Y⁢(v),⋯,Yn⁢(v)}

are respectively

ρ⁢(H) =(nn−2⋱−n),
ρ⁢(Y) =(010⋱⋱10),
and
ρ⁢(X) =(0n02⁢(n−1)03⁢(n−2)⋱⋱⋱n0).

Since SL2⁡(ℂ) is simply connected, we have:

Proposition 7.15.

Every finite-dimensional complex-linear representation of 𝔰⁢𝔩2,ℂ is the derivative of a unique representation of SL2⁡(ℂ).

Theorem 7.16.

Every finite-dimensional irreducible complex-linear representation of SL2⁡(ℂ) or 𝔰⁢𝔩2,ℂ is isomorphic to Symn⁡(ℂ2), the symmetric power of the standard representation.

Proof.

We already proved this for 𝔰⁢𝔩2,ℂ. By Proposition 7.15, every representation of SL2⁡(ℂ) arises from one of 𝔰⁢𝔩2,ℂ, and since SL2⁡(ℂ) is connected the result follows. ∎

7.3 Decomposing representations

Theorem 7.17.

Every finite-dimensional complex-linear representation of 𝔰⁢𝔩2,ℂ is completely reducible, that is, splits into a direct sum of irreducible representations.

Proof.

This is a special case of Theorem 7.27 below. ∎

Remark 7.18.

Complete reducibility shows that every finite-dimensional representation of 𝔰⁢𝔩2,ℂ is a direct sum of irreducibles. Each irreducible has integer weights and a basis of weight vectors. Thus these properties hold for any finite-dimensional representation: ρ⁢(H) is diagonalizable with integer eigenvalues and we get a weight space decomposition

V=⨁n∈ℤVn. (7.2)

One can also see this more directly: by Proposition 7.15, ρ exponentiates to SL2⁡(ℂ). The subgroup U⁡(1)⊂SL2⁡(ℂ) (embedded via ei⁢t↦(ei⁢te−i⁢t)) acts completely reducibly by Maschke’s theorem (section 6.5), and its irreducible representations are ei⁢t↦ei⁢n⁢t for n∈ℤ (Theorem 6.18). Taking the derivative shows that ρ⁢(i⁢H) is diagonalizable with eigenvalues in i⁢ℤ and so ρ⁢(H) is diagonalizable with integer eigenvalues. The integrality of the weights thus reflects the fact that the characters of U⁡(1) are indexed by integers.

It is easy to decompose a representation V of 𝔰⁢𝔩2,ℂ into irreducibles by looking at the weights. Firstly, look at the maximal weight k of V. Then there must be a weight vector v of weight k, which is necessarily a highest weight vector, and so V must contain a copy of Symk⁡(ℂ2) — namely, the subspace ⟨v,Y⁢v,…,Yk⁢v⟩. By complete reducibility we have

V≅Symk⁡(ℂ2)⊕W.

The weights of W are then obtained by removing the weights of Symk⁡(ℂ2) from the weights of V, and we repeat the process.

In particular, this shows that a finite-dimensional representation of 𝔰⁢𝔩2,ℂ is determined, up to isomorphism, by its multiset of weights.

Proposition 7.19.

If V, W are representations of 𝔰⁢𝔩2,ℂ then:

  • •

    {weights of V⊗W}={weights of V}+{weights of W}.

  • •

    {weights of Symk(V)}={sums of unordered k-tuples of weights of V}.

  • •

    {weights of Λk(V)}={sums of unordered ‘distinct’ k-tuples of weights of V}.

Proof.

If v1,…,vm is a basis of weight vectors of V with H⁢vi=αi⁢vi, and w1,…,wn is a basis of weight vectors of W with H⁢wi=βi⁢vi, then

{vi⊗wj:1≤i≤m,1≤j≤n}

is a basis of weight vectors of V⊗W and

H⁢(vi⊗wj)=(H⁢vi)⊗wj+vi⊗H⁢Wj=(αi+βj)⁢(vi⊗wj).

This shows part (1), and the other parts are very similar.

See also section 8.5 below. ∎

Example 7.20.

We should illustrate what is meant by ‘distinct’: it is ‘distinct’ as elements of the multiset. Suppose that the weights of V are {2,0,0,−2}. Then to obtain the weights of Λ2⁢(V) we add together unordered, distinct, pairs of these in every possible way, getting:

{2+0,2+0,2+−2,0+0,0+−2,0+−2}={2,2,0,0,−2,−2}.
Example 7.21.

Let ℂ2 be the standard representation of SL2⁡(ℂ) with weight basis e1, e−1. Consider V=Sym2⁡(ℂ2)⊗Sym2⁡(ℂ2).

The weights of Sym2⁡(ℂ2) are {−2,0,2} and so the weights of V are

{−2,0,2}+{−2,0,2}={−4,−2,−2,0,0,0,2,2,4}.

This is the same as the set of weights of

Sym4⁡(ℂ2)⊕Sym2⁡(ℂ2)⊕ℂ

and so this is the required decomposition into irreducibles.

Refer to caption
Figure 6: Decomposing Sym2⁡(ℂ2)⊗Sym2⁡(ℂ2).

We can go further, and decompose V into irreducible subrepresentations. This means finding irreducible subrepresentations of V such that V is their direct sum. To start with we write down the weight vectors:

Let v2=e12, v0=e1⁢e−1 and v−2=e−12 be weight vectors in Sym2⁡(ℂ2) corresponding to the weights 2, 0 and −2. Then the weights of V are

  • •

    4, multiplicity one, weight vector: v2⊗v2.

  • •

    2, multiplicity two, weight space: ⟨v2⊗v0,v0⊗v2⟩.

  • •

    0, multiplicity three, weight space: ⟨v2⊗v−2,v0⊗v0,v−2⊗v2⟩.

  • •

    −2, multiplicity two, weight space: ⟨v0⊗v−2,v−2⊗v0⟩.

  • •

    −4, multiplicity one, weight vector: ⟨v−2⊗v−2⟩.

The copy of Sym4⁡(ℂ2) in V has highest weight vector v2⊗v2. We can find a basis by repeatedly hitting this with Y (writing ∝ for ‘equal up to a nonzero scalar’):

Y⁢(v2⊗v2) =2⁢(v2⊗v0+v0⊗v2)∝v2⊗v0+v0⊗v2
Y2⁢(v2⊗v2) ∝v2⊗v−2+4⁢v0⊗v0+v−2⊗v2
Y3⁢(v2⊗v2) ∝6⁢(v0⊗v−2+v−2⊗v0)∝v0⊗v−2+v−2⊗v0
Y4⁢(v2⊗v2) ∝v−2⊗v−2.

These vectors are a basis for the copy of Sym4⁡(ℂ2) in V.

Next, we find the copy of Sym2⁡(ℂ2) in V. We start by looking for a highest weight vector of weight 2:

v2⊗v0−v0⊗v2

does the trick. Hitting this with Y gives v2⊗v−2−v−2⊗v2, and doing so again gives v−2⊗v0−v0⊗v−2 (up to scalar). These vectors are a basis for the copy of Sym2⁡(ℂ2) in V.

Finally, we find the trivial representation ℂ in V. We need only find a weight vector of weight 0 which is killed by X, and

v2⊗v−2−2⁢v0⊗v0+v−2⊗v2

does the job: this vector spans a copy of the trivial representation.

7.4 Real forms and complete reducibility

We use our understanding of the representation theory of 𝔰⁢𝔩2,ℂ to understand the representation theory of SL2⁡(ℝ) and SU⁢(2), and prove complete reducibility.

Definition 7.22.

A real form of a complex Lie algebra 𝔤 is a real Lie algebra 𝔥⊂𝔤 such that every element Z of 𝔤 can be written uniquely as X+i⁢Y for X,Y∈𝔥.

For dimension reasons, necessary and sufficient conditions are that 𝔥∩i⁢𝔥=0 and dimℝ𝔥=dimℂ𝔤.

Example 7.23.

The Lie algebra 𝔰⁢𝔩n,ℝ is a real form of 𝔰⁢𝔩n,ℂ.

Example 7.24.

The Lie algebra 𝔰⁢𝔲n is a real form of 𝔰⁢𝔩n,ℂ. Indeed, dimℝ𝔰⁢𝔲n=n2−1=dimℂ𝔰⁢𝔩n,ℂ and if X,i⁢X∈𝔰⁢𝔲n then

i⁢X=−(i⁢X)†=i⁢X†=−i⁢X

so X=0.

Explicitly, we may write A∈𝔰⁢𝔩n,ℂ as X+i⁢Y with

X=12⁢(A−A†)

and

Y=−i2⁢(A+A†)

in 𝔰⁢𝔲n.

Proposition 7.25.

Let 𝔥 be a real form of 𝔤.

There is a one-to-one correspondence between representations of 𝔥 and complex-linear representations of 𝔤 under which irreducible representations correspond to irreducible representations.

Proof.

Given a ℂ-linear representation of 𝔤, it is a representation of 𝔥 by restriction. Conversely, if (ρ,V) is a representation of 𝔥, then it extends to a unique ℂ-linear representation of 𝔤 given by the formula (forced by ℂ-linearity)

ρ⁢(X+i⁢Y)⁢v=ρ⁢(X)⁢v+i⁢ρ⁢(Y)⁢v.

It is easy to see that this preserves the Lie bracket. The proposition follows (the final statement is left as an exercise). ∎

As a corollary we immediately obtain

Theorem 7.26.

The representation theories of 𝔰⁢𝔩n,ℝ and 𝔰⁢𝔲n are ‘the same’ as the complex-linear representation theory of 𝔰⁢𝔩n,ℂ. All finite-dimensional irreducible representations of 𝔰⁢𝔩2,R, SL2⁡(ℝ), 𝔰⁢𝔲2, or SU⁢(2) are of the form Symn⁡(ℂ2).

Proof.

The claims about Lie algebras follow from the above discussion. Every (finite-dimensional) irreducible representation of 𝔰⁢𝔩2,ℝ is of the form Symn⁡(ℂ2), and these clearly exponentiate to representations of SL2⁡(ℝ), despite this not being a simply connected group! Similarly for SU⁢(2) (which is simply connected). Since SL2⁡(ℝ) and SU⁢(2) are connected, every representation of them is determined by its derivative, so we have a complete list of the irreducible representations. ∎

Theorem 7.27.

Every finite-dimensional complex-linear representation of 𝔰⁢𝔩n,ℂ is completely reducible.

Proof.

Let V be a finite-dimensional complex-linear representation of 𝔰⁢𝔩n,ℂ and let W⊂V be a subrepresentation. Then W is an 𝔰⁢𝔲n-subrepresentation. As SU⁢(n) is simply-connected, V and W exponentiate to representations of SU⁢(n). Since SU⁢(n) is compact, by Maschke’s theorem there is a complementary SU⁢(n)-subrepresentation W′ with

V=W⊕W′.

Then W′ is a 𝔰⁢𝔲n-subrepresentation, and so (since 𝔰⁢𝔲n is a real form of 𝔰⁢𝔩n,ℂ) a ℂ-linear 𝔰⁢𝔩n,ℂ-subrepresentation. Complete reducibility follows. ∎

The argument in this proof is called Weyl’s unitary trick.

7.5 SO(3)

7.5.1 Classification of irreducible representations

We have already (see the first problem set this term) seen that 𝔰⁢𝔲2 and 𝔰⁢𝔬3 are isomorphic. We therefore have:

Theorem 7.28.

There is an irreducible two-dimensional representation V of 𝔰⁢𝔬3 (coming from 𝔰⁢𝔬3→∼𝔰⁢𝔲2⊂𝔤⁢𝔩2,ℂ) such that the irreducible complex representations of 𝔰⁢𝔬3 are exactly

Symn⁡(V)

for n≥0.

We want to know which of these representations exponentiate to an irreducible representation of SO⁢(3). For this, we revisit the connection with SU⁢(2) from section 6.4.

There is a surjective homomorphism π:SU⁢(2)→SO⁢(3), given by the adjoint action of SU⁢(2) on its Lie algebra endowed with a certain symmetric positive definite bilinear form. The kernel of π is {±I}.

Theorem 7.29.

For each ℓ≥0, there is a unique irreducible representation V(ℓ) of SO⁢(3) of dimension 2⁢ℓ+1.

The derivative of this representation is isomorphic to the representation Sym2⁢ℓ⁡(V) of 𝔰⁢𝔬3.

This gives the complete list of irreducible representations of SO⁢(3) up to isomorphism.

Proof.

We simply have to work out which representations Symk⁡(V) of 𝔰⁢𝔬3 exponentiate to a representation of SO⁢(3). Since we have

SU⁢(2)/{±I}→∼SO⁢(3)

and each Symk⁡(V) exponentiates to a unique representation of SU⁢(2) — which we also call Symk⁡(V) — this is equivalent to asking for which k the centre {±I} of SU⁢(2) acts trivially on Symk⁡(V). But we see that −I acts as (−1)k, so the answer is: for even k only.

Thus Symk⁡(V) exponentiates to a representation of SO⁢(3) if, and only if, k=2⁢ℓ is even, and we obtain the result. ∎

There is an orthonormal basis ℐ,𝒥,𝒦 of 𝔰⁢𝔲2 given by

12⁢(01−10),12⁢(0ii0),12⁢(i00−i)

respectively. We have computed in 6.4 that, under the (adjoint) map D⁢π, with respect to this basis,

ℐ↦Jx =(00000−1010)
𝒥↦Jy =(001000−100)
𝒦↦Jz =(0−10100000).

It will be useful to know where this isomorphism takes the elements H,X,Y of 𝔰⁢𝔩2,ℂ=𝔰⁢𝔲2,ℂ. For example, as H=−2⁢i⁢𝒦, we see that it goes to −2⁢i⁢Jz. Or, for the lowering operator Y, we have

Y=(0010)=−(ℐ+i⁢𝒥)↦−(Jx+i⁢Jy)

and similarly X↦Jx−i⁢Jy.

With this calculation in hand, we can consider the weights of the representations in 7.29. Under the isomorphism

𝔰⁢𝔲2,ℂ=𝔰⁢𝔩2,ℂ→𝔰⁢𝔬3,ℂ

the element H=−2⁢i⁢𝒦 maps to −2⁢i⁢Jz. Since the H-weights of V(ℓ) are −2⁢ℓ,−2⁢(ℓ−1),…,2⁢(ℓ−1),2⁢ℓ, we must divide these by −2⁢i to find the weights of Jz acting on V(ℓ):

−i⁢ℓ,−i⁢(ℓ−1),…,i⁢(ℓ−1),i⁢ℓ.
Example 7.30.

Consider the standard three-dimensional representation of SO⁢(3) on ℂ3. The weights of Jz are simply its eigenvalues as a 3×3 matrix, which are −i,0,i. We see that this representation is isomorphic to V(1).

7.5.2 Harmonic functions

Remark 7.31.

We went through this very quickly in the lecture; I have marked as nonexaminable the proofs that we skipped.

We can use our understanding of the representation theory of SO⁢(3) to shed light on the classical theory of spherical harmonics.

We let 𝒫ℓ be the subspace of ℂ⁢[x,y,z] consisting of homogeneous polynomials of degree ℓ. This has an action of SO⁢(3) given by

(g⁢f)⁢(𝐱)=f⁢(gT⁢𝐱)

where 𝐱=(xyz) is a vector in ℂ3. We therefore get a representation of SO⁢(3) and hence also of 𝔰⁢𝔬3.

Lemma 7.32.

The elements Jx,Jy, and Jz of 𝔰⁢𝔬3 act on 𝒫ℓ according to the following formulae:

Jx =z⁢∂∂y−y⁢∂∂z
Jy =x⁢∂∂z−z⁢∂∂x
Jz =y⁢∂∂x−x⁢∂∂y.
Proof.

(nonexaminable) Exercise. In fact, prove that the action of A=(ai⁢j)∈𝔰⁢𝔬3 is given by

∑i,jai⁢j⁢xi⁢∂∂xj

by considering

dd⁢t⁢f⁢(exp⁡(t⁢AT)⁢𝐱)=dd⁢t⁢f⁢((I+t⁢AT)⁢𝐱)

at t=0 (where we rewrite x,y,z as x1,x2,x3). ∎

The representation 𝒫ℓ is not irreducible. Let r2∈ℂ⁢[x,y,z] be the polynomial

r2=x2+y2+z2.

Note that r2 is clearly invariant under the action of SO⁢(3).

Lemma 7.33.

The map 𝒫ℓ→𝒫ℓ+2 defined by

f↦r2⁢f

is an injective homomorphism of SO⁢(3)-representations.

Proof.

We have, for g∈SO⁢(3),

g⁢(r2⁢f)=g⁢(r2)⁢g⁢(f)=r2⁢g⁢(f)

as required. ∎

Next, we consider the Laplace operator:

Δ⁢f=∂2∂x2+∂2∂y2+∂2∂z2.

This is map from 𝒫ℓ→𝒫ℓ−2.

Lemma 7.34.

The map Δ is a map of SO⁢(3) representations.

Proof.

(Nonexaminable.) We must show that, for g=(gi⁢j)∈SO⁢(3),

(Δf)(gT𝐱)=Δ(gf)(𝐱)).

We have

∂∂xi⁢(g⁢f)⁢(x)=∑jgi⁢j⁢∂f∂xj⁢(gT⁢x)

and so

∂∂xi⁢∂∂xi⁢(g⁢f)⁢(x)=∑j,kgi⁢j⁢gi⁢k⁢∂∂xk⁢∂∂xj⁢(gT⁢x).

We sum over i, for fixed j and k:

∑igi⁢j⁢gi⁢k=δj⁢k

since g is orthogonal. We therefore obtain

∑i∂2∂xi2⁢(g⁢f)⁢(x)=∑j∂2∂xj2⁢(f)⁢(gT⁢x),

and therefore

Δ⁢(g⁢f)=g⁢Δ⁢(f).

∎

An element f∈𝒫ℓ is harmonic if Δ⁢f=0. Since dim(𝒫ℓ)>dim(𝒫ℓ−2), nonzero harmonic polynomials must exist for all ℓ. We write ℋℓ⊂𝒫ℓ for the space of harmonic polynomials.

Lemma 7.35.

On 𝒫ℓ, we have

r2⁢Δ=Jx2+Jy2+Jz2+ℓ2+ℓ.
Proof.

Left as a nonexaminable exercise. You may find the following useful:

Lemma 7.36.

(Euler’s formula) If f∈𝒫ℓ, then

x⁢∂∂x⁢f+y⁢∂∂y⁢f+z⁢∂∂z⁢f=ℓ⁢f.

∎

It follows that r2⁢Δ preserves every subrepresentation of 𝒫ℓ (since Jx,Jy,Jz do). Furthermore, by Schur’s lemma it must act on each irreducible subrepresentation as a scalar. We determine that scalar.

Lemma 7.37.

nonexaminable Suppose that V⊂𝒫ℓ is an irreducible subrepresentation with highest weight i⁢k. Then

(r2⁢Δ)⁢(f)=(ℓ2+ℓ−k2−k)⁢f=(ℓ−k)⁢(ℓ+k+1)⁢f.
Proof.

Since r2⁢Δ is a 𝔰⁢𝔬3-homomorphism and V is irreducible, by Schur’s lemma it acts as a scalar on V. It therefore suffices to compute the action on a highest weight vector v∈V. So

Jz⁢v=i⁢k⁢v,(Jx−i⁢Jy)⁢v=0.

It follows that Jz2⁢v=−k2⁢v and, as

Jx2+Jy2 =(Jx+i⁢Jy)⁢(Jx−i⁢Jy)+i⁢[Jx,Jy]
=(Jx+i⁢Jy)⁢(Jx−i⁢Jy)+i⁢Jz,

we have (Jx2+Jy2)⁢v=0⁢v−k⁢v.

Applying the previous lemma gives the result. ∎

Theorem 7.38.

For every ℓ≥2,

𝒫ℓ=ℋℓ⊕r2⁢𝒫ℓ−2.

The space ℋℓ is the irreducible highest weight representation of SO⁢(3) of dimension 2⁢ℓ+1, and the space 𝒫ℓ has the following decomposition into irreducible subrepresentations:

𝒫ℓ=ℋℓ⊕r2⁢ℋℓ−2⊕r4⁢ℋℓ−4⊕….
Proof.

(Nonexaminable.) We use induction on ℓ. The case ℓ=0 is clear (we just have the trivial representation). Suppose true for ℓ−1 with ℓ≥1.

By the previous lemma, the space ℋℓ is the sum of all the copies inside 𝒫ℓ of the irreducible representation with highest weight i⁢ℓ. Since 𝒫ℓ−2 does not contain this irreducible representation, by the inductive hypothesis, we have

ℋℓ∩r2⁢𝒫ℓ−2={0}.

Since ℋℓ≠0 as already discussed, its dimension is a positive multiple of 2⁢ℓ+1. However, its dimension is at most

dim𝒫ℓ−dim𝒫ℓ−2=(ℓ+22)−(ℓ2)=2⁢ℓ+1.

It follows that ℋℓ is irreducible, and that we have

ℋℓ⊕r2⁢𝒫ℓ−2=𝒫ℓ.

The statement about the decomposition into irreducibles follows. ∎

Thus every homogeneous polynomial has a unique decomposition as a sum of harmonic polynomials multiplied by powers of r2.

We can go further and give nice bases for the ℋℓ by taking weight vectors for Jz. First, we have

Lemma 7.39.

The function (x−i⁢y)ℓ∈𝒫ℓ is a highest weight vector of weight i⁢ℓ.

Proof.

Exercise! ∎

We then obtain a weight basis by repeatedly applying the lowering operator

Jx+i⁢Jy=(i⁢x−y)⁢∂∂z+z⁢(∂∂y−i⁢∂∂x).

The functions thus obtained are known as ’spherical harmonics’ (at least, up to normalization), and give a particularly nice basis for the space of functions on the sphere S2⊂ℝ3. The decomposition of a function into spherical harmonics is analogous to the Fourier decomposition of a function on the unit circle.

Example 7.40.

If ℓ=1, then 𝒫ℓ=ℋℓ, and the weight vectors are

x+i⁢y,z,x−i⁢y.

If ℓ=2, a basis of ℋℓ made up of weight vectors is

(x−i⁢y)2,z⁢(x−i⁢y),x2+y2−2⁢z2,z⁢(x+i⁢y),(x+i⁢y)2.