The aim of this note is to compute the fundamental group of SO(3) using quaternions. More precisely, an isomorphism SU(2)/ \mathbb{Z}_2 \simeq SO(3) will be found leading to:

Property 1: SU(2) is the univeral covering of SO(3) and \pi_1(SO(3)) \simeq \mathbb{Z}_2.

Before introducing quaternions, recall the construction of \mathbb{C}. Usually, \mathbb{C} is defined as the splitting field of the polynomial X^2+1 over \mathbb{R}, namely \mathbb{R}[X]/(X^2+1). Taking the companion matrix A= \left( \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right) of X^2+1, we get an isomorphism \mathbb{C} \simeq \mathbb{R}[A] = \left\{ \left( \begin{matrix} a & -b \\ b & a \end{matrix} \right) \mid a,b \in \mathbb{R} \right\} with a subalgebra of M_2(\mathbb{R}). Thus, a possible construction of the quaternions \mathbb{H} is to replace real coefficients with complex ones, namely \mathbb{H} = \left\{ \left( \begin{matrix} a & -\overline{b} \\ b & \overline{a} \end{matrix} \right) \mid a,b \in \mathbb{C} \right\}. Notice that if 1= \left( \begin{matrix} 1&0 \\ 0&1 \end{matrix} \right), I= \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right), J= \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right) and K= \left( \begin{matrix} 0 & i \\ i & 0 \end{matrix} \right), then every element of \mathbb{H} can be uniquely written as a \mathbb{R}-linear combination of 1, I, J and K.

Formally, \mathbb{H} is a four-dimensional real algebra with a basis (1,i,j,k) satisfying i^2=j^2=k^2=ijk=-1; in particular, ij=-ijk^2=-(ijk)k=k and ji=i^2j(ij)j=i(ijk)j=-ij=-k, so \mathbb{H} is not commutative. In fact, it can be shown that \mathbb{H} is a noncommutative division algebra.

Viewing \mathbb{H} as a subalgebra of M_2(\mathbb{C}), there are a natural conjugaison \overline{ a+bi+cj+dk}= a-bi-cj-dk and a natural norm |a+bi+cj+dk|=a^2+b^2+c^2+d^2 corresponding respectively to the conjugaison and the determinant in M_2(\mathbb{C}). As a real vector space, the norm |\cdot | define a definite positive quadratic form on \mathbb{H}, whose bilinear form is \langle h_1,h_2 \rangle= \frac{1}{2}(h_1\overline{h_2} + \overline{h_1}h_2 ). In particular, notice that \mathbb{R}1 and \mathbb{I}= i \mathbb{R}+j \mathbb{R}+k \mathbb{R} are orthogonal.

Now let SU(2) act on \mathbb{H} by conjugaison, \varphi : \left\{ \begin{array}{ccc} SU(2) & \to & \text{Aut}(\mathbb{H}) \\ h & \mapsto & (\varphi_h : k \mapsto hkh^{-1}) \end{array} \right. (\text{Aut}(\mathbb{H}) is the group of automorphisms of \mathbb{H} viewed as a real vector space). Since \text{Aut}(\mathbb{H}) \simeq GL_4(\mathbb{C}) and each \varphi_h preserve the quadratic form \det, taking the restrictions of the \varphi_h‘s to \mathbb{I} we can suppose that \text{Im}(\varphi) \subset O(3) \simeq \mathbb{I}. Moreover, SU(2) is connected and \varphi is continuous so \text{Im}(\varphi) \subset O_0(3)=SO(3).

To show that \varphi : SU(2) \to SO(3) is onto, notice that \varphi is C^1 and  that d \varphi(I_2) : H \mapsto (K \mapsto HK-KH) induces a linear map between the Lie algebra \mathfrak{su}(2)=\mathfrak{u}(2)= \{ M \in M_2(\mathbb{C}) \mid M^*+M=0 \} and \mathfrak{so}(3)=\mathfrak{o}(3) = \{ M \in M_3(\mathbb{R}) \mid M^t +M=0 \}; by connectedness of SO(3) and using the inverse function theorem, it is sufficient to show that d\phi(I_2) is invertible. But \text{ker}(d \varphi(I_2))= \{ H \in \mathfrak{su}(2) \mid \forall K \in \mathbb{I}, HK=KH\}= \{I_2\} and \dim(\mathfrak{su}(2))=\dim(\mathfrak{so}(3))= 3. Therefore, \varphi : SU(2) \to SO(3) is onto.

Noticing finally that \text{ker}(\varphi)= \{\pm I_2\}, we get the isomorphism SU(2)/\mathbb{Z}_2 \simeq SU(2) / \{ \pm I_2\} \simeq SO(3). But the isomorphism is also a homeomorphis since the coordinate functions of \varphi are just rational functions, and identifying SU(2) with the element of \mathbb{H} \simeq \mathbb{R}^4 of norm 1, SU(2) is homeomorphic to the sphere \mathbb{S}^3; in particulier, SU(2) is simply connected. Moreover, the action of \mathbb{Z}_2 on SU(2) by left multiplication is properly discontinuously, so we can deduce property 1.

In fact, the isomorphism SU(2)/ \mathbb{Z}_2 \simeq SO(3) has a topological interpretation:

Property 2: SO(3) is homeomorphic to the projective space P^3\mathbb{R}.

By definition, P^3\mathbb{R} is the quotient of \mathbb{R}^4 \backslash \{0\} by the equivalence relation (x,y,z,t) \sim (x',y',z',t') \Leftrightarrow \exists \lambda \in \mathbb{R}^*, (x',y',z',t')= \lambda (x,y,z,t).  In fact, all points of \mathbb{R}^4 \backslash \{0\} are not needed, and P^3 \mathbb{R} is also homeomorphic to the quotient of X:=\{ (x,y,z,t) \in \mathbb{R}^4 \backslash \{0\} \mid x^2+y^2+z^2+t^2=1, t \geq 0 \} by \sim; since the projection \text{pr}_{\mathbb{R}^3 \times \{0\}} induces a bijection from X to B^3:= \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2+z^2 \leq 1 \}, we get that P^3\mathbb{R} is also homeomorphic to B^3 quotiented by the symmetry with respect to (0,0,0).

Now let \psi : \left\{ \begin{array}{ccc} B^3 & \to & SO(3) \\ x & \mapsto & r(x,||x||\pi) \end{array} \right. where r(x,\theta) denotes the rotation of axis x and angle \theta. Then \psi is a continuous surjection such that for all x,x' \in B^3, \psi(x)=\psi(x') is equivalent to x= \pm x'. Therefore, \psi induces a continuous bijection \tilde{\psi} : P^3\mathbb{R} \to SO(3); since P^3\mathbb{R} is compact we deduce that \tilde{\psi} is in fact a homeomorphism.