### 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 $$t \in [0,2\pi]$$.