Representations of SU(2) & su(2)

su(2) ≅ so(3) and the representation of so(3) is important in the computation of angular momentum. By studying the representation theory of su(2) we can know:

(1) how to commutation relations to determine the representations of a Lie algebra.

(2) how to determine the representations of semisimple Lie algebras, e.g., su(3).

Some Representations of SU(2)

By definition, an element of U of SU(2) is a linear transformation of C². Let z denotes the (z₁, z₂) pair in C². Then, we may define a linear transformation Π_m(U) on the space V_m by the formula [Π_m(U)f] (z) = f(U^-1z). The inverse is necessary in order to make Π_m a representation.  And V_m is the space of functions of the form f(z₁, z₂) = a₀z₁^m + a₁z₁^m-1z₂  + a₂z₁^m-2z₂²  + ... + a_mz₂^m .
 

Therefore, [Π_m(U)f] (z₁, z₂) =  ∑ a_k(U₁₁^-1z₁ + U₁₂^-1z₂)^m-k (U₂₁^-1z₁ + U₂₂^-1z₂)^k for k = 0 .. m.  Π_m(U)f is a homogeneous polynomial of degree m. Thus, Π_m(U) maps V_m into V_m. Moreover, Π_m(U₁) [Π_m(U₂)f] (z) = Π_m(U₁U₂) f(z). So Π_m is a (finite-dimensional complex) representation of SU(2).

The Lie algebra representation of Π_m can be computed as π_m(X) = d/dt Π_m(e^tX) |_t=0. So (π_m(X)f) (z) = d/dt f(e^-tXz) |_t=0. Let z(t) be the curve e^-tXz. We have z(0) = z and (dz/dt) |_t=0 = -Xz. Since z(t) can also be written as z(t) = (z₁(t), z₂(t)), with z_i(t) ∈ C. By chain rule, π_m(X)f = ∂f / ∂z₁(dz₁/dt) |_t=0 + ∂f / ∂z₂(dz₂/dt) |_t=0 . We have

π_m(X)f = -∂f / ∂z₁(X₁₁z₁ + X₁₂z₂)  - ∂f / ∂z₂(X₂₁z₁ + X₂₂z₂). 

Because every finite-dimensional complex representation of the Lie algebra su(2) extends uniquely to a complex-linear representation of the complexification of su(2). And the complexification of su(2) is sl(2;C). Therefore, the representation π_m of su(2) given by above extends to a representation of sl(2;C). 

Consider H = (1,0 : 0,-1)

(π_m(H)f) (z) = -(∂f / ∂z₁)z₁ + (∂f / ∂z₂)z₂ .

 π_m(H) = -z₁(∂f / ∂z) + z₂(∂f / ∂z₂). Apply this to a basis element z₁^kz₂^m-k , we have

π_m(H) z₁^kz₂^m-k = -k z₁^kz₂^m-k  + (m-k)z₁^kz₂^m-k  = (m-2k) z₁^kz₂^m-k .

Thus, z₁^kz₂^m-k is an eigenvector for π_m(H) with eigenvalue (m-2k). In particular, π_m(H) is diagonalizable. 

Let X and Y be the elements, X = (0,0 : 1,0) , Y = (0,1 : 0,0) in sl(2;C). We have π_m(X) = -z₂(∂f / ∂z₁) and π_m(X) = -z₁(∂f / ∂z₂). Apply these to the basis element 

π_m(X) z₁^kz₂^m-k = -k z₁^k-1z₂^m-k+1  

π_m(Y) z₁^kz₂^m-k = -(m-k) z₁^k+1z₂^m-k-1  

It suffices to show that every nonzero invariant subspace of V_m is equal to V_m. Let W be such a space. Since W is assumed nonzero, there is at least one nonzero element w in W. Then w can be written uniquely in the form 

w =  a₀z₁^m + a₁z₁^m-1z₂  + a₂z₁^m-2z₂²  + ... + a_mz₂^m

with at least one of the a_k's nonzero. Let k₀ be the smallest value of k for which  a_k ≠ 0 and consider π_m(X)^m-k0 w. Since π_m(X) lowers the power of z₁ by 1, it will kill all the terms in w except a_k0 z₁^m-k0 z₂^k0 . So we have  π_m(X)^m-k0 z₁^m-k0 z₂^k0 = (-1)^m-k0(m-k₀)! z₂^m . Since W is assumed invariant, W must contain z₂^m . Furthermore, π_m(Y)^k z₂^m is a nonzero multiple of z₁^kz₂^m-k for all 0 ≤ k ≤ m. Because these elements form a basis for V_m . In fact W = V_m. Therefore, representation π_m is an irreducible representation of sl(2;C). 

Irreducible Representations of su(2)

Every finite-dimensional complex representation π of su(2) extends to a complex-linear representation of the complexification of su(2), namely sl(2;C). Studying the irreducible representations of su(2) is equivalent to studying the irreducible representation of sl(2;C). Passing to the complexified Lie algebra makes computations easier, in that there is a nice basis for sl(2;C) that has no counterpart among the bases of su(2).

We can use commutation relations to determine the representation of a Lie algebra. Consider the following basis for sl(2;C) and commutation relations:

H = (1,0 : 0,-1);  X = (0,0 : 1,0);  Y = (0,1 : 0,0)

[H, X] = 2X,  [H, Y] = -2Y,  [X, Y] = H.

If V is a (finite-dimensional complex) vector space and A, B, and C are operators on V satisfying

[A, B] = 2B,  [A, C] = -2C,  [B, C] = A, then

because of the skew symmetry and bilinearity of brackets, the linear map π : sl(2;C) --> gl(V) satisfying π(H) = A, π(X) = B, π(Y) = C will be a representation of sl(2;C).

We call π(X) the "raising operator", because it has the effect of raising the eigenvalue of π(H) by 2, and call π(Y) the "lowering operator". Since [π(H), π(X)] = π([H, X]) = 2π(X). Let u be an eigenvector of π(H) with eigenvalue α ∈ C. Thus,

π(H)π(X)u = π(X)π(H)u + 2π(X)u

= π(X)(αu) + 2π(X)u = (α + 2)π(X)u.

Either π(X)u = 0 or π(X)u is an eigenvector for π(H) with eigenvalue α+2. More general, π(H)π(X)ⁿu = (α + 2n)π(X)ⁿu. Similarly, for [π(H), π(Y)] = -2π(Y), we have π(H)π(Y)u = (α - 2)π(Y)u.

An operator on a finite-dimensional space can have only finitely many distinct eigenvalues. Therefore, there is some N ≥ 0 such that π(X)^N+1u = 0.

Define u₀ = π(X)^Nu and λ = α + 2N. Then,

π(H)u₀ = λu₀, π(X)u₀ = 0. 

Define u_k = π(Y)^ku₀ , for k ≥ 0. Thus, we have π(H)u_k = (λ - 2k)u_k . Since π(H) can have only finitely many eigenvalues, the u_k's cannot be all be nonzero.

For k = 1,

π(H)u₁ = π(H)π(Y)u₀ = (α - 2)π(Y)u₀ = (α - 2)u₁.

π(X)u₁ = π(X)π(Y)u₀ = (π(Y)π(X) + π(H))u₀ = π(H)u₀ = λu₀ (as π(X)u₀ = 0)

π(Y)u₁ = π(Y)π(Y)u₀  = π(Y)²u₀ = u₂  

If π(X)u_k = [kλ - k(k-1)]u_k-1. By induction, π(X)u_k+1= π(X)π(Y)u_k

= (π(Y)π(X) + π(H))u_k = π(Y)π(X)u_k + (λ - 2k)u_k

= π(Y)[kλ - k(k-1)]u_k-1 + (λ - 2k)u_k  = [kλ - k(k-1) + (λ - 2k)]u_k

= [(k+1)λ - (k+1)k]u_k  

Because π(H) can have only finitely many eigenvalues, the u_k's cannot all be nonzero. For all k ≤ m, u_m+1 = π(Y)^m+1u₀  = 0. If u_m+1 = 0. Then π(X)u_m+1  = (m+1)(λ - m)u_m  = 0. Since m ≠ 0 and m + 1 ≠ 0. So we have λ = m, where m is a non-negative integer. 

In summary, given a finite-dimensional irreducible representation π of sl(2;C) acting on a space V and putting λ = m, there exists an integer m ≥ 0 and nonzero vectors u₀, ..., u_m such that

π(H)u_k = (m - 2k)u_k, 

π(Y)u_k = u_k+1 (k < m),

π(Y)u_m = 0, 

π(X)u_k = [km - k(k-1)]u_k-1 (k > 0),

π(X)u₀ = 0.

The vectors u₀, ..., u_m must be linearly independent, since they are eigenvectors of π(H) with distinct eigenvalues. Moreover, the (m+1)-dimensional span u₀, ..., u_m is explicitly invariant under π(H), π(X), and π(Y). Hence under π(Z) for all Z ∈ sl(2;C). Since π is irreducible, this space must be all of V. 

The (m+1)-dimensional representation Π_m described above must be equivalent to π. This can be seen explicitly by introducing the following basis for V_m :

u_k = [π_m(Y)]^k (z₂)^m = (z₂)^m (m! / (m-k)!) z₁^kz₂^m-k  (k ≤ m).

In other words, π_m have a basis of the form π(H), π(X), and π(Y). π's have the right commutation relations to forma representation of sl(2;C) and that this representation is irreducible.