12  The Group \(\mathrm{SU}(2)\)

The next compact group of interest is given by \[ \operatorname{SU}(2) \coloneqq \{ Q \in \operatorname{Mat}_2(\mathbb{C}): Q^*Q = \mathbb{I}\text{ and } \det Q = 1 \}, \] the special unitary group of \(2 \times 2\) matrices. In this section we develop the structure of this group, by way of establishing an important isomorphism between it and the unit sphere of quaternions, before characterizing its representation theory. Finally, we explore the special relationship between \(\operatorname{SU}(2)\) and the group \(\operatorname{SO}(3)\) of rotation matrices in \(\mathbb{R}^3\) which allows us to deduce the representation theory of the latter as a nice final remark for the course.

12.1 Quaternions

The complex numbers can be understood as an algebra of \(2 \times 2\) real matrices via the \(\mathbb{R}\)-linear one-to-one ring homomorphism \(\mathbb{C}\to \operatorname{Mat}_2(\mathbb{R})\) given by \[ a + b i \mapsto \begin{pmatrix} a & -b \\ b & a \end{pmatrix}. \] This representation enjoys many nice properties: for example, the trace and determinant encode twice the real part and the norm squared, respectively, of the underlying complex number, while taking the transpose is analogous to complex conjugation.

Similarly, the Hamilton quaternions \(\mathbb{H}= \{a+bi+cj+dk: a,b,c,d \in \mathbb{R}\},\) where \(i, j,\) and \(k\) satisfy the relations from the group \(\mathcal{Q}_{8}\), can be understood using complex matrices: \[ a+bi+cj+dk \mapsto \begin{pmatrix} a+bi & c+di \\ -c+di & a-bi \end{pmatrix}. \] If we write \(\alpha = a+bi\) and \(\beta = c+di\), then this correspondence is given by \[ \alpha + \beta j \mapsto \begin{pmatrix} \alpha & \beta \\ -\bar{\beta} & \bar{\alpha} \end{pmatrix}. \tag{12.1}\] Again, taking the trace and determinant return twice the real part, \(2a\), and norm squared, \(a^2+b^2+c^2+d^2\), of the quaternion being represented. In this case, Hermitian conjugation of the matrix corresponds to an involution of \(\mathbb{H}\): \[ (a+bi+cj+dk)^* = a-bi-cj-dk. \] Unlike the complex numbers, the algebra \(\mathbb{H}\) is non-commutative, e.g., \(ij = k\) but \(ji = -k\). However, since \((AB)^* = B^* A^*\) holds on the level of matrices, we see that \[ (qp)^* = p^* q^* \] for all \(q, p \in \mathbb{H}\). Moreover, the usual Euclidean norm of a quaternion can be given as \[ \|q\|^2 = qq^* = q^* q \] and, therefore, this norm is well-behaved with respect to multiplication: \[ \|qp\|^2 = (qp)(qp)^* = qpp^*q^* = q\|p\|^2q^* = \|p\|^2 \|q\|^2. \] It follows that every non-zero quaternion is invertible with respect to multiplication, since \[ q \frac{q^*}{\|q\|^2} = \frac{\|q\|^2}{\|q\|^2} = 1, \] i.e., \(q^{-1} = \frac{q^*}{\|q\|^2}\). As such, the set of norm-1 quaternions are a group under multiplication: \[ \begin{split} \mathbb{S}^3 & = \{ q : \|q\|^2 = 1 \} \\ & = \{ \alpha + \beta j: |\alpha|^2+|\beta|^2 = 1 \} \\ & = \{ a+bi+cj+dk : a^2+b^2+c^2+d^2 = 1 \}, \end{split} \] where the notation \(\mathbb{S}^3\) is used because this set is the hypersphere (i.e., the \(3\)-sphere) in \(\mathbb{R}^4\).

Theorem 12.1 The correspondence of Equation 12.1 gives an isomorphism \(\mathbb{S}^3 \cong \operatorname{SU}(2)\) of topological groups.

While the details are left as an exercise, the key lemma involves showing that every element in \(\operatorname{SU}(2)\) is of the form \(\begin{psmallmatrix} \alpha & \beta \\ -\bar{\beta} & \bar{\alpha} \end{psmallmatrix}\) for \(\alpha, \beta \in \mathbb{C}\) satisfying \(|\alpha|^2+|\beta|^2=1\).

In order to understand the group \(\mathbb{S}^3 \cong \operatorname{SU}(2)\), it behooves us to better understand the multiplicative structure of the quaternions. Fortunately, those sweethearts Stewart et al. (2020) of Calculus: Early Transcendentals (and many other authors of calculus textbooks) have us covered—we are already practiced in multiplying the symbols \(i\), \(j\), and \(k\), except we thought of them as the standard basis vectors in \(\mathbb{R}^3\) and the underlying multiplication rule as the cross product!

Toward this end, consider the decomposition \(\mathbb{H}= \mathbb{R}\oplus \mathbb{V}\) where \(\mathbb{V}= \operatorname{Span}_{\mathbb{R}}\{i,j,k\}\) is the subspace of “pure quaternions” that have no real part. We can see immediately that, when restricted to vectors \(x, y \in \mathbb{V}\), multiplication is almost the exact same as taking cross products in \(\mathbb{R}^3\). Indeed, we have relations like \(e_1 \times e_2 = e_3\) mirroring \(ij = k\); the only difference is that \(e_1 \times e_1 = 0\), whereas \(i^2\) becomes the real number \(-1\) (and similarly for \(j^2\) and \(k^2\)). For all \(x,y \in \mathbb{V}\), one can show that: \[ xy = \underbrace{-x \cdot y}_{\in \mathbb{R}} + \underbrace{x \times y}_{\in \mathbb{V}}. \tag{12.2}\]

This notation might seem a bit odd at first, expressing \(a+bi+cj+dk\) via the formal sum \[ a + \begin{psmallmatrix} b \\ c \\ d \end{psmallmatrix}, \] though it is quite useful. Indeed, if \(q = t+x\) and \(p = s+y\) for \(s,t \in \mathbb{R}\) and \(x,y \in \mathbb{V}\), then \[ qp = (\underbrace{ts-x \cdot y}_{\in \mathbb{R}})+(\underbrace{sx+ty+x\times y}_{\in \mathbb{V}}), \] which is a lovely generalization for the formulas involving products of complex numbers. From this we can deduce other fun identities, e.g., for any \(\alpha \in \mathbb{C}\) we have \[ \alpha j = j \bar{\alpha}, \tag{12.3}\] which means that the automorphism \(\mathbb{C}\to \mathbb{C}\) of complex conjugation is secretly conjugation by \(j\). Indeed, we have the commutator relation \(\tfrac{1}{2} [q,p] = x \times y\) with \(q,p \in \mathbb{H}\) as above.

Because the quaternions are non-commutative, we no longer have the theorems of elementary abstract algebra that bound the number of solutions to a polynomial by its degree. Indeed, we will deduce the following in homework:

Proposition 12.1 If \(u \in \mathbb{S}^3\) is a pure quaternion, then \(u^2 = -1\). In other words, \[ \mathbb{S}^2 \coloneqq \mathbb{S}^3 \cap \mathbb{V}= \left\{ bi+cj+dk \mid b^2+c^2+d^2 = 1 \right\} \] is exactly the set of solutions to the equation \(x^2+1 = 0\).

In other words, the quaternions contain a whole sphere’s worth of elements that behave like a square root of \(-1\). Note, however, that this sphere is not a multiplicative subgroup of \(\mathbb{S}^3\) since it is not closed under multiplication. All this motivates a quaternionic version of the Euler formula—if we define the exponential of \(q \in \mathbb{H}\) via \[ e^q \coloneqq \sum_{n=0}^\infty \frac{q^n}{n!}, \] where issues of convergence could be attended to carefully but are ignored here, then we can replicate the proof of the usual Euler identity to deduce \[ e^{\theta u} = \cos \theta + u \sin \theta \tag{12.4}\] for all \(\theta \in \mathbb{R}\) and \(u \in \mathbb{S}^2\). Note that, in general, one should be careful when exponentiating quaternions due to their non-commutative structure, but we can be somewhat sloppy here because \(u\) commutes with all its powers.

Theorem 12.2 Every \(q \in \mathbb{S}^3\) can be written as \(q = e^{\theta u}\), where \(\theta \in [0,\pi]\) and \(u \in \mathbb{S}^2\).

The proof is left as an exercise, though we again caution the reader that, because elements \(u,v \in \mathbb{S}^2\) need not commute, familiar formulas like \(e^{u+v} = e^u e^v\) do not hold in general. Indeed, this difficulty inspires an entirely separate regime of representation theory—that of Lie algebras.

Notice that \(\mathbb{T}\) is a subgroup of \(\mathbb{S}^3\), simply because \(\mathbb{C}\subset \mathbb{H}\). Indeed, as will be explored in homework, every element \(q \in \mathbb{S}^3\) is conjugate to a unique \(z\) in the top half of the unit circle. The reason we only need this half is because of Equation 12.4; we see that \(z \in \mathbb{T}\) and \(\bar{z} = z^{-1}\) are conjugate via \(j \in \mathbb{S}^3\). Translating these results to \(\operatorname{SU}(2)\), we have:

Theorem 12.3 Each element of \(\operatorname{SU}(2)\) is conjugate to an element in the subgroup \[ \left\{ \begin{pmatrix} e^{\theta i} & 0 \\ 0 & e^{-\theta i} \end{pmatrix} \mid \theta \in [0,2\pi) \right\} \cong \mathbb{T}. \] Moreover, these conjugacy classes are parameterized over \(\theta \in [0,\pi]\) in the above.

In general, the element \(e^{\theta u}\) passes to a matrix in \(\operatorname{SU}(2)\) whose characteristic polynomial is given by \(t^2 - 2\cos \theta t + 1\) and thus has the eigenvalues \(e^{\theta i}\) and \(e^{-\theta i}\), i.e., it is conjugate to the diagonal form in the above theorem. In other words, for any \(u \in \mathbb{S}^2\) the element \(e^{\theta u}\) is conjugate to \(e^{\theta i} \in \mathbb{T}\) and so this quaternionic polar form is particularly handy. We note that \(\theta = 0\) and \(\theta = \pi\) correspond to the singleton conjugacy classes \(1\) and \(-1\), respectively.

Theorem 12.4 For any \(\theta \in [0,\pi]\) and \(u \in \mathbb{S}^2 \subset \mathbb{V}\), the element \(e^{\theta u} \in \mathbb{S}^3\) is conjugate to \(e^{\theta i}\).

12.2 Aside on Other Spheres

Recall that the unit \(n\)-sphere is defined as the set of points \[ \mathbb{S}^n = \left\{ (x_1,\dots,x_{n+1}) \in \mathbb{R}^{n+1} \mid x_1^2+\cdots+x_{n+1}^2 = 1 \right\}, \] so that the \(2\)-sphere in \(\mathbb{R}^3\) is the usual object we expect when someone says “sphere.” We can recall from past experience that the spheres \(\mathbb{S}^0\) and \(\mathbb{S}^1\) are groups, \[ \mathbb{S}^0 = \left\{ x \in \mathbb{R}\mid x^2 = 1 \right\} = \{ 1, -1 \} = \mathcal{C}_{2} \quad \text{ and } \quad \mathbb{S}^1 = \mathbb{T}\subset \mathbb{C}, \] and now we have also seen that the norm-1 quaternions \(\mathbb{S}^3 \subseteq \mathbb{H}\) is just the group \(\operatorname{SU}(2)\). A very natural question to ask is: “What about the \(2\)-sphere—is it a group?” In fact, no other sphere besides these three can be made into a topological group. While we will not prove this fact, we will at least construct an argument regarding the \(2\)-sphere that can be made rigorous using the apparatus of differential geometry—one which I, at least, find somewhat viscerally satisfying.

Note that the cases \(\mathbb{S}^0,\) \(\mathbb{S}^1\), and \(\mathbb{S}^3\) correspond to the multiplicative structure of \(\mathbb{R}\), \(\mathbb{C}\), and \(\mathbb{H}\), where \(\mathbb{C}= \mathbb{R}^2\) and \(\mathbb{H}= \mathbb{R}^4\) as Euclidean spaces, by restricting to vectors of length 1. It turns out these algebras are the only finite-dimensional division algebras over \(\mathbb{R}\)—which perhaps isn’t the most surprising, given our alternative proof of Schur’s Lemma that involved showing that \(\mathbb{C}\) is the only finite-dimensional division algebra over \(\mathbb{C}\)—and somehow this is deeply related to which spheres admit a topological group structure. We will give a casual presentation of this connection here.

Suppose that \(\mathbb{R}^n\) is a division algebra—along with the usual operation of vector addition, we assume that we can also multiply vectors to produce new vectors in an associative (but not necessarily commutative) way that distributes nicely, enjoys an identity and invertibility, and plays well with the scalar multiplication from \(\mathbb{R}\). Accordingly, the multiplicative identity of \(\mathbb{R}^n\) can be identified with the scalar \(1 \in \mathbb{R}\), so without loss of generality we can identity the first standard basis vector \(e_1=1\) in this way (this is analogous to decomposing \(\mathbb{C}= \mathbb{R}\oplus i \mathbb{R}\) and \(\mathbb{H}= \mathbb{R}\oplus \mathbb{V}\)). Moreover, one can show that the compatibility with scalar multiplication in \(\mathbb{R}\) guarantees that the maps \(\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n\) and \(\mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n \setminus \{0\}\) given by \((q,p) \mapsto qp\) and \(q \mapsto q^{-1}\) are continuous.

Non-trivially, take \(n>1\) and consider the sphere \(\mathbb{S}^{n-1} \subset \mathbb{R}^n\) of length \(1\) vectors. In the parlance of differential geometry, our claim is that the tangent bundle of \(\mathbb{S}^{n-1}\) must be trivial (this is, in fact, true for every Lie group by a similar argument). For those unfamiliar with this language, our goal is to show that we can establish continuously-varying bases for the space of tangent vectors at every point on \(\mathbb{S}^{n-1}\) due to this multiplication law.

Let’s begin at the point \(e_1 = 1 \in \mathbb{S}^{n-1}\). At this location, the vector \(e_1\) is normal to the sphere whereas the other vectors \(e_2, \dots, e_n\) are tangent to it; indeed, the tangent plane to the sphere at this point is described by the equation \(x_1 = 1\). At any other point \(q \in \mathbb{S}^{n-1}\), because of the topological group structure, we have a canonical way of moving points near \(1\) to points near \(q\) simply by multiplying everything by \(q\) (say, on the left). This process of translating from \(1\) to \(q\) is very nice; because \(\mathbb{R}^n\) is presumed to be a division algebra, we can also multiply by \(q^{-1}\) to undo the process. It follows that the vectors \(\{q, q e_2, \dots, q e_n\}\) are linearly independent, as a relation \[ \alpha_1 q + \alpha_2 q e_2 + \cdots + \alpha_n q e_n = 0 \] for \(\alpha_i \in \mathbb{R}\) becomes \(\alpha_1 e_1 + \cdots + \alpha_n e_n = 0\) after multiplying on the left by \(q^{-1}\), i.e., we must have each \(\alpha_i = 0\) by the linear independence of the standard basis.

Moreover, since the vector \(q\) is normal to \(\mathbb{S}^{n-1}\) at \(q\), orthogonally projecting \(qe_2, \dots, qe_n\) onto the tangent plane \(q \cdot (x-q) = 0\) at \(q\) gives a basis for the space of tangent vectors (denoted \(T_q\mathbb{S}^{n-1}\)). In this way, we get continuous vector fields \(X_2, \dots, X_n: \mathbb{S}^{n-1} \to \mathbb{R}^n\) via \[ X_i(q) = \operatorname{Proj}_{T_q\mathbb{S}^{n-1}}(q e_i) \] so that, for all \(q \in \mathbb{S}^{n-1}\), the \(X_i(q)\) give a basis for the space of all tangent vectors to \(\mathbb{S}^{n-1}\) at \(q\). Students familiar with the basics of differential geometry can modify this argument if we assume from the get-go that \(\mathbb{S}^{n-1}\) is a group, rather than assuming the ambient \(\mathbb{R}^n\) is a division algebra, whose operations are continuously differentiable.¹

In this sense, infinitesimal beings living on \(\mathbb{S}^{n-1}\) can use the presumed algebraic structure of the underlying \(\mathbb{R}^n\) to maintain a globally-understood sense of cardinal direction—in the \(n=2\) case, an ant walking around a circle will always have a notion of which way is clockwise and which is counterclockwise. This is in contrast to our earthly lives, where our notions of direction break down at the ends of the planet—which direction is East at the North pole? A particularly delightful mathematical result that gets at this phenomenon is known as the , a.k.a. the hedgehog theorem, which states that any continuous vector field \(X: \mathbb{S}^2 \to \mathbb{R}^3\) such that \(X(q) \in T_q\mathbb{S}^2\) for all \(q \in \mathbb{S}^2\) must vanish for at least one \(p \in \mathbb{S}^2\). More colloquially, this theorem is understood as saying something equivalent to: “No matter how you comb the hair on a hedgehog, it’ll always have a cowlick.”

The situation outlined by the hairy ball theorem is much worse than what we have presented for the sphere \(\mathbb{S}^{n-1}\) living in an ambient division algebra \(\mathbb{R}^n\). Not only do we not have a global sense of direction on \(\mathbb{S}^2\), which would involve writing down a continuously varying pair of vectors spanning the tangent plane \(T_q\mathbb{S}^{n-1}\) at every \(q \in \mathbb{S}^{n-1}\), but we can’t even write down a single vector field that is non-zero everywhere. Hence, if you are willing to accept that you cannot perfectly comb the hair of your imaginary hedgehog Emilio, you can sleep soundly knowing that \(\mathbb{S}^2\) can never be a topological group.

12.3 Haar Integral

We begin with a more geometrically intuitive scenario: finding a Haar integral on \(\mathbb{S}^3\). This analysis is more difficult than in the case for \(\mathbb{T}\) because we do not have a simple formula for products of exponentials! In particular, we have: \[ \begin{split} e^{\theta u} e^{\phi v} & = (\cos \theta - u \sin \theta)(\cos \phi + v \sin \phi) \\ & = (\underbrace{\cos \theta \cos \phi + u \cdot v \sin \theta \sin \phi}_{\in \mathbb{R}}) + (\underbrace{\cos \theta \sin \phi v - \sin \theta \cos \phi u - u \times v \sin \theta \sin \phi}_{\in \mathbb{V}}) \end{split} \] and it is unclear how to write this new result in polar form as a function of \(\theta, \phi, u,\) and \(v\).

Fortunately, it turns out that multiplying by a fixed element \(e^{\theta u}\) preserves volume. To see this, consider the linear map \(L: \mathbb{H}\to \mathbb{H}\) given by \(L(q) = e^{\theta u}q\). Thinking of \(\mathbb{V}\) as \(\mathbb{R}^3\), we can take \(v \in \mathbb{S}^2 \subset \mathbb{V}\) orthogonal to \(u\) (there is a planes’ worth of choices) and set \(w = u \times v\). In light of our formula (Equation 12.2), we can see that \(u^2 = v^2 = w^2 = 1\) and \[ uv = -u \cdot v + u \times v = w, \quad vw = u, \quad \text{ and } \quad wu = v. \] Given \(q = a + b u + c v + d w\), we have \[ \begin{split} L(q) & = (\cos \theta + u \sin \theta)(a + b u + c v + d w) \\ & = (a \cos \theta - b \sin \theta) + (a \sin \theta + b \cos \theta) u + (c \cos \theta - d \sin \theta) v + (c \sin \theta + d \cos \theta) w. \end{split} \] So in the basis \(\mathscr{B}=(1,i,j,k)\) we see that \(L\) is given by \[ [L]_{\mathscr{B}} = \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \\ - \sin \theta & \cos \theta & 0 & 0 \\ 0 & 0 & \cos \theta & \sin \theta \\ 0 & 0 & - \sin \theta & \cos \theta \\ \end{pmatrix}, \] i.e., \(L\) rotates the planes \(\operatorname{Span}\{1,u\}\) and \(\operatorname{Span}\{v,w\}\) in the by \(\theta\). Rotations are famous for not distorting length—\(\det L = 1\), after all—and so in particular multiplying by \(e^{\theta u}\) describes a transformation that preserves the tradition notion of volume. So, since rotating a sphere does not alter its size, we are interested in the usual integration inherited from \(\mathbb{H}= \mathbb{R}^4\).

To actually write down this Haar integral, we should change variables to spherical polar coordinates. We can specify any \(u \in \mathbb{S}^2\) in terms of \(\psi \in [0,\pi]\) and \(\phi \in [0,2\pi]\): \[ u = \cos \psi \, i + \sin \psi \cos \phi \, j + \sin \psi \sin \phi \, k. \] Hence we will change from the standard Euclidean coordinates \(a + b i + c j + d k\) via the rule \[ a = r \cos \theta, \quad b = r \sin \theta \cos \psi, \quad c = r \sin \theta \sin \psi \cos \phi, \quad \text{ and } \quad d = r \sin \theta \sin \psi \sin \phi. \] In order to compute the change of variables we write down the Jacobian matrix \[ \begin{pmatrix} \cos \theta & \sin \theta \cos \psi & \sin \theta \sin \psi \cos \phi & \sin \theta \sin \psi \sin \phi \\ -r \sin \theta & r \cos \theta \cos \psi & r \cos \theta \sin \psi \cos \phi & r \cos \theta \sin \psi \sin \phi \\ 0 & -r \sin \theta \sin \psi & r \sin \theta \cos \psi \cos \phi & r \sin \theta \cos \psi \sin \phi \\ 0 & 0 & -r \sin \theta \sin \psi \sin \phi & r \sin \theta \sin \psi \cos \phi \end{pmatrix} \] which has determinant \(r^3 \sin^2 \theta \sin \psi\). Taking \(r=1\), we see that \[ \int_{\mathbb{S}^3} f(a+bi+cj+dk) \mathop{}\!\mathrm{d}{V} = \int_0^{2\pi} \int_0^\pi \int_0^\pi f(\theta,\psi,\phi) \sin^2 \theta \sin \psi \mathop{}\!\mathrm{d}{\theta} \mathop{}\!\mathrm{d}{\psi} \mathop{}\!\mathrm{d}{\phi}. \] If we integrate the constant function \(f=1\), we see that the volume of \(S^3\) is \(2\pi^2\). Therefore, the Haar integral on the unit quaternions, and therefore \(\operatorname{SU}(2)\), is given by \[ f \mapsto \frac{1}{2\pi^2} \int_0^{2\pi} \int_0^\pi \int_0^\pi f(\theta,\psi,\phi) \sin^2 \theta \sin \psi \mathop{}\!\mathrm{d}{\theta} \mathop{}\!\mathrm{d}{\psi} \mathop{}\!\mathrm{d}{\phi}. \tag{12.5}\] In particular, since every element of \(\mathbb{S}^3\) is conjugate to \(e^{\theta i}\)—which is parallel to each matrix in \(\operatorname{SU}(2)\) being conjugate to \(\operatorname{Diag}(e^{\theta i}, e^{-\theta i})\)—class functions are only a function of \(\theta\). We leave it to Homework to verify that simplifying Equation 12.5 leads to the inner product \[ \langle f_1, f_2 \rangle = \frac{2}{\pi} \int_0^\pi \overline{f_1(\theta)} f_2(\theta) \sin^2 \theta \mathop{}\!\mathrm{d}{\theta}. \tag{12.6}\] for all \(f_1, f_2 \in C^{}(\operatorname{SU}(2))\).

12.4 Irreducible Representations

We begin by remarking that any subgroup \(G \leq \operatorname{GL}_m(\mathbb{C})\) acts naturally on the vector space of polynomials in \(m\) variables: if \(M \in \operatorname{GL}_m(\mathbb{C})\) and \(p \in \mathbb{C}[x_1,\dots,x_m]\), then we define \[ M \cdot p(x_1,\cdots,x_n) = p( M^{-1} (x_1,\cdots,x_n)), \] where \(M^{-1}\) acts formally on \((x_1,\cdots,x_n)\) as though it were a column vector. Notice that this action behaves particularly nicely with respect to degree; for each \(d \in \mathbb{N}\) we obtain a subrepresentation, denoted \[ \mathbb{C}^{(d)}[x_1,\dots,x_n] \coloneqq \operatorname{Sym}^{d}({(\mathbb{C}^n)^*}), \] of the homogeneous degree \(d\) polynomials (c.f. Exercise E.3). This is historically motivated by a branch of mathematics known as invariant theory and is familiar also to physicists and applied mathematicians interested in solutions to equations that are invariant under certain symmetries. We are particularly interested in the \(m=2\) case, where elements \(Q \in \operatorname{SU}(2)\) act on polynomials in two variables by \[ \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \cdot p(x,y) = p(\delta x - \beta y,\alpha y - \gamma x). \] We will use the notation \[ V_d \coloneqq \mathbb{C}^{(d)}[x,y] = \operatorname{Span}\{x^d,x^{d-1}y,\dots,y^d\} \] for the subrepresentation of degree \(d\) homogeneous polynomials in two variables.

Lemma 12.1 The representation \(\rho_d: \operatorname{SU}(2) \to \operatorname{GL}(V_d)\) is irreducible for all \(d \in \mathbb{N}\).

Proof (Sketch). The character \(\chi_d\) of this action is given by \[ \chi_d(\theta) = e^{-d\theta i} + e^{-(d-2)\theta i} + \cdots + e^{d \theta i} = \frac{\sin(d+1)\theta}{\sin \theta}, \] where \(\theta\) parameterizes the conjugacy class \(\begin{psmallmatrix} e^{\theta i} & 0 \\ 0 & e^{-\theta i} \end{psmallmatrix} \in \operatorname{SU}(2)\). That these representations are irreducible follows from evaluating \(\langle \chi_d, \chi_e \rangle\) under Equation 12.6 and appealing to Corollary 7.4, which is left as an exercise.

Theorem 12.5 Any irreducible representation \(W\) of \(\operatorname{SU}(2)\) is isomorphic to some \(V_d\).

Proof. In some sense, this proof is exploring the Fourier series of functions \(f: \mathbb{T}\to \mathbb{C}\) that are symmetric, in the sense that \(f(z) = f(\overline{z})\) for all \(z \in \mathbb{T}\). If we choose to parameterize \(z = e^{\theta i}\) with \(\theta \in (-\pi,\pi]\), i.e., imagining \(f: \mathbb{R}\to \mathbb{C}\) as \(2\pi\)-periodic, this is just the fact that the Fourier series of an even function only includes cosine terms.

We proceed by thinking of \(W\) as a representation of \(\mathbb{T}\) via the natural inclusion \[ \begin{aligned} \mathbb{T}& \hookrightarrow S^3 \stackrel{\cong}{\longrightarrow} \operatorname{SU}(2) \\ z & \mapsto z+0j \mapsto \begin{pmatrix} z & 0 \\ 0 & z^{-1} \end{pmatrix} \end{aligned} \] So, given a representation \(W\) of \(\operatorname{SU}(2)\), we can obtain a character \(\chi_W: \mathbb{T}\to \mathbb{C}\). Given the apparatus of Fourier series, i.e., the representation theory of \(\mathbb{T}\), we can therefore write \[ \chi_W(z) = \sum_{k \in \mathbb{Z}} b_k z^k, \] where only finitely many of the multiplicities \(b_k \in \mathbb{Z}\) are non-zero since \(W\) is a finite-dimensional representation of \(\mathbb{T}\) and thus decomposes into only finitely many irreducibles.

As we have seen, in the context of the unit quaternions \(\mathbb{S}^3\), every \(z \in \mathbb{T}\) is conjugate (in the group-theoretic sense) to its complex conjugate \(\bar{z} = z^{-1} \in \mathbb{T}\) via \(j \in \mathbb{S}^3\). Because \(\chi_W\) is a class function on \(\operatorname{SU}(2)\), this means that we must have \(\chi_W(z) = \chi_W(z^{-1})\), i.e., \[ \chi_W(z) = \sum_{k \in \mathbb{N}} c_k (z^k + z^{-k}) \] for \(c_k \in \mathbb{N}\) and with only finitely many terms non-zero.

But recall that the character of \(\mathbb{T}\hookrightarrow\operatorname{SU}(2) \to \operatorname{GL}(V_d)\) is given by \[ \chi_d(z) = z^{-d} + z^{-(d-2)} + \cdots + z^{d-2} + z^d. \] If the highest degree non-zero term of \(\chi_W\) is \(c_\ell z^\ell\) then its lowest order term is \(c_\ell z^{-\ell}\). Hence \[ \chi_W - c_\ell \chi_\ell \] is also symmetric in \(z\) and \(z^{-1}\) with non-zero terms of degree strictly between \(\ell\) and \(-\ell\). In this way we can write \(\chi_W\) as a linear combination of the characters \(\chi_d\) for \(d \in \mathbb{N}\), which corresponds to a decomposition of \(\operatorname{SU}(2)\)-representations: \[ W = V_0^{a_0} \oplus \cdots \oplus V_\ell^{a_\ell}. \] In particular, if \(W\) is irreducible, we must have \(W \cong V_\ell\).

12.5 Double Cover of \(\operatorname{SO}(3)\)

We conclude by mentioning the representation theory of one more compact group.

Given \(q \in \mathbb{S}^3\) and \(x \in \mathbb{V}\), we can see that \(qxq^* \in Vb\). This follows because the conjugation on \(\mathbb{H}\) diagonalizes via the decomposition \(\mathbb{R}\oplus \mathbb{V}\), where \(*\) acts trivially on \(\mathbb{R}\) and by \(-1\) on \(\mathbb{V}\): \[ (qxq^*)^* = q x^* q^* = - qxq^*. \] This defines a representation \(\mathbb{S}^3 \to \operatorname{GL}(\mathbb{V})\) called the adjoint representation (for a much more general construction, see (Rossmann 2002, 14)) by \(q \mapsto \operatorname{Ad}_q\), where \(\operatorname{Ad}_q(x) \coloneqq qxq^*.\) We claim that the image of this map is isomorphic to the subgroup of special orthogonal matrices: \[ \operatorname{SO}(3) = \left\{ R \in \operatorname{Mat}_{3}(\mathbb{R}): R^\top R = \mathbb{I}\text{ and } \det R = 1 \right\}. \]

Theorem 12.6 In the ordered basis \(\mathscr{B} = (i,j,k)\) for \(\mathbb{V}\), the adjoint representation \[ q \mapsto [\operatorname{Ad}_q]_{\mathscr{B}} \] is a surjective homomorphism \(\mathbb{S}^3 \twoheadrightarrow\operatorname{SO}(3)\) with kernel \(\{ \pm 1\}\) and thus \(\operatorname{SU}(2) / \langle - \mathbb{I}\rangle \cong \operatorname{SO}(3)\).

Before investigating this, we first consider the structure of \(\operatorname{SO}(3)\). We will implicitly rely on the fact that \(\mathbb{T}\cong \operatorname{SO}(2)\) via the correspondence \[ e^{\theta i} \mapsto \begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}. \]

Lemma 12.2 Every \(R \in \operatorname{SO}(3)\) has \(1\) as an eigenvalue. If \(R \not = \mathbb{I}\), then the eigenspace \[ \ell \coloneqq \{ x \in \mathbb{R}^3 \mid Rx = x \} \] is \(1\)-dimensional, i.e., \(R\) is a rotation about the line \(\ell\).

Proof. Recall that any matrix \(R \in \operatorname{Mat}_n(\mathbb{R})\) satisfying \(R^\top R = \mathbb{I}\) has \(\det R = \pm 1\), since \[ \det(R)^2 = \det(R^\top)\det(R) = \det(R^\top R) = \det(\mathbb{I}) = 1. \] Similarly, \[ \det(\mathbb{I}- R) = \det(R^\top R - R) = \det(R^\top - \mathbb{I})\det(R) = \det(R - \mathbb{I})\det(R). \] If we also assume that \(\det(R) = 1\) and that \(n\) is odd, then \[ \det(R - \mathbb{I}) = \det(-(\mathbb{I}- R)) = -\det(\mathbb{I}- R), \] so \[ \det(\mathbb{I}- R) = -\det(\mathbb{I}- R). \] Therefore \(\det(\mathbb{I}- R) = 0\), which shows that \(1\) is an eigenvalue of \(R\).

Now let \[ U = \ker(\mathbb{I}- R) \leq \mathbb{R}^3. \] Choose a unit vector \(u \in U\), so that \(Ru = u.\) We claim that the line \[ \ell = \operatorname{Span}\{u\} \] is the axis of rotation.

Next consider the orthogonal complement \[ U^\perp = \{x \in \mathbb{R}^3 : x \cdot u = 0\}. \] We first show that \(U^\perp\) is preserved by \(R\). Indeed, if \(x \in U^\perp\), then \[ (Rx)\cdot u = (Rx)\cdot(Ru) = x\cdot u = 0, \] since orthogonal matrices preserve the Euclidean inner product. Hence \(Rx \in U^\perp\).

Now choose a unit vector \(v \in U^\perp\), and define \[ w = u \times v. \] Then \(w \in U^\perp\), and the set \[ \mathscr{B} = \{u,v,w\} \] is an orthonormal basis for \(\mathbb{R}^3\).

Since \(Ru = u\) and \(U^\perp\) is \(R\)-invariant, the matrix of \(R\) in the basis \(\mathscr{B}\) has the form \[ [R]_{\mathscr{B}} = \begin{pmatrix} 1 & 0 \\ 0 & S \end{pmatrix}, \] for some \(S \in \operatorname{Mat}_2(\mathbb{R})\).

Moreover, since \(R^\top R = \mathbb{I}\), the matrix \(S\) is orthogonal, and because \[ \det(S) = \det(R) = 1, \] we actually have \(S \in \operatorname{SO}(2)\). Therefore there exists some \(\omega \in [0,2\pi)\) such that \[ [R]_{\mathscr{B}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\omega & -\sin\omega \\ 0 & \sin\omega & \cos\omega \end{pmatrix}. \] Thus, if \(R \neq \mathbb{I}\), then \(\dim \ell = 1\) and \(R\) is a rotation through \(\ell\) by angle \(\omega\).

Proof (Sketch for Theorem 12.6). While the details are left as an exercise, the argument is as follows. Given \(q = e^{\theta u}\), we want to show that \([\operatorname{Ad}_q]_{\mathscr{B}}\) is a rotation through \(u \in S^2 \subset \mathbb{V}\) by angle \(2 \theta\); from here it follows that the kernel of the homomorphism \(S^3 \to \operatorname{SO}(3)\) is \(\pm 1\). To prove the desired claim regarding \([\operatorname{Ad}_q]_{\mathscr{B}}\), take \(v \in S^2\) orthogonal to \(u\) and \(w = uv = u \times v\) and write down \(\operatorname{Ad}_q\) in the basis \(\{u,v,w\}\).

Recall from the Homework that representations of the quotient \(G/N\) are exactly the representations of \(G\) whose kernel contains \(N\). In particular, by verifying which of the representations \(\operatorname{SU}(2) ⟳V_d\) of Theorem 12.5 have \(-\mathbb{I}\) in the kernel, we deduce

Corollary 12.1 The irreducible representations of \(\operatorname{SO}(3)\) are given by \(V_{2k}\) for each \(k \in \mathbb{N}\), i.e., there is exactly one irreducible representation with degree \(2n+1\) for every \(n \in \mathbb{N}\).

This correspondence is of special interest to topologists, who say things like “\(\operatorname{SU}(2)\) is the universal cover of \(\operatorname{SO}(3),\)” but also to the physicists, who speak of representations of \(\operatorname{SO}(3)\) as corresponding to “integer spin” while the faithful representations of \(\operatorname{SU}(2)\) correspond to “half-integer spin.” But these digressions are for another course!

Rossmann, Wulf. 2002. Lie Groups: An Introduction Through Linear Groups. Oxford University Press.

Stewart, J., D. K. Clegg, and S. Watson. 2020. Calculus: Early Transcendentals. 9th ed. Cengage Learning.

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1. In order to drop the differentiability condition and honestly show that \(\mathbb{S}^2\) cannot be a topological group, one needs access to some tools from an introductory course in algebraic topology.