Notes on The Quaternions and \(SU(2)\)

The following short write-up is largely a summary of the blog post by Qiaochu Yuan of similar title. In this I will focus on a few different aspect of Qiaochu's write up as well as provide a little clarification on some details.

The quaternions \(\mathbb{H}\) are the familiar 4-dimensional \(\mathbb{R}\)-algebra with basis \(1,\bf{i},\bf{j},\bf{k}\), with \(1\) as our unit and the multiplication of \(\bf{i,j,k}\) given by the ``cross-product'' chart.

We can define an adjoint operation \(\dagger\) on \(\mathbb{H}\) in a nice way generalizing complex conjugation, given by $$ q = t1 + x\textbf{i} + y\textbf{j} + z\textbf{k} \quad\quad q^\dagger \equiv t1 - x\textbf{i} - y\textbf{j} - z\textbf{k}$$ This allows us to define a norm, namely \( |q|^2 \equiv qq^\dagger \), which gives the familiar $$ |q|^2 = t^2 + x^2 + y^2 + z^2$$ It is not difficult but tedious to check that our norm is multiplicative, as we would expect. Now we are ready to discuss how \(SU(2)\) fits into the picture.

The group \(SU(2)\) is the set of \(2\times 2\) unitary (that is, \(MM^\dagger = M^\dagger M = I\)) \(\mathbb{C}\) matrices with determinant 1, namely $$ SU(2) \equiv \left\{ \left( \begin{array}{cc} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{array} \right) : |\alpha|^2 + |\beta|^2 = 1 \right\}$$ We now embed \(SU(2)\) into the quaternions. Picking an element of \(SU(2)\) is equivalent to picking the two complex numbers \(\alpha,\beta\) above, and given, \(\alpha = t1+zi, \beta = y1 + xi\), we can simply consider this to be a 4-tuple of real numbers and then map them \((x,y,z,t) \mapsto t1 + x\textbf{i} + y\textbf{j} + z\textbf{k} \). We could, but won't verify that this is a bijective homomorphism of \(SU(2)\) onto the norm-1 quaternions. Thus herein when we speak of an element of \(SU(2)\), we will regard it as a norm 1 quaternion.

Turning our attention to the quaternions themselves, we make some observations. First, \(\mathbb{H}\) can be decomposed \(\mathbb{H} = \mathop{span}(1) \oplus \mathop{span}(\textbf{i,j,k})\). We note that this is a specific case of a general fact about \(C^*\) algebras, namely that any \(C^*\) algebra can be decomposed as the direct sum of the Hermitian subspace and the anti-Hermitian subspace. This is seen at once by writing any element $$ q = \frac{q+q^\dagger}{2} + \frac{q-q^\dagger}{2} $$ Next, given some \(q \in \mathbb{H}\), we consider its "adjoint" action on \(\mathbb{H}\) $$ \mathrm{Ad}\,q : \mathbb{H} \to \mathbb{H}, \quad\quad \zeta \mapsto q\zeta q^{-1}$$ A quick check reveals that if \(\zeta\) is Hermitian or anti-Hermitian, then \(q\zeta q^{-1}\) is still Hermitian or anti-Hermitian respectively. Thus the adjoint action of a given \(q\) maps the subspaces to themselves, and thus we might consider elements of \(\mathbb{H}\) or in particular \(SU(2)\) as acting on the space \(\mathrm{span}(\textbf{i,j,k})\). Since the norm is multiplicative, we find that this action preserves norm and thus we have arrived at an action of \(SU(2)\) on \(\mathrm{span}(\textbf{i,j,k})= \mathbb{R}^3\), and even more an action of the 2-sphere to the 2-sphere, \(S^2\),that is, an element of \(SO(3)\) which we will find to be our rotation.

Given our \(q\), we can directly verify that \(\mathrm{Im}(q)\equiv (1/2)(q-q^\dagger)\) is fixed under the mapping \(\mathrm{Ad}\,q\), and thus the \(S^2\) point \( \textbf{h}_q \equiv \frac{\mathrm{Im}(q)}{|\mathrm{Im}(q)|}\) is our axis of rotation. Given our \(SU(2)\) element we express it as $$ q = \cos\theta 1 + \sin\theta \textbf{h}_q $$ for some \(\theta\). A quick check reveals that \(\textbf{h}_q^2 = -1\), and with this we can simply observe that our quaternion norm coincides with the complex norm, making valid the exponential formula $$ q = \cos\theta 1 + \sin\theta \textbf{h}_q = e^{\textbf{h}_q \theta} $$ This is wonderful since we can now define $$ \eta(\theta) \equiv \cos\theta 1 + \sin\theta \textbf{h}_q = e^{\textbf{h}_q \theta} $$ and be assured that it is a continuous homomorphism in \(\theta\). In effect what this is is a homomorphism of the copy of \(S^1 \subset SU(2)\) associated with \(q\), explicitly it is \(e^{it}q\) where \(t \in [0,2\pi]\).