Rotations

Any element of the orthogonal group $${O\left(r,s\right)}$$ is a rotation and/or reflection and the Cartan–Dieudonné theorem states that any such transformation can be obtained as a product of at most $${r+s}$$ reflections. Thus every element of $${O\left(r,s\right)}$$ corresponds to the Clifford product of some $${k}$$ unit vectors via

$$\displaystyle R_{u_{k}\dotsm u_{1}}\left(v\right)\equiv R_{u_{k}}\left(\dotsm\left(R_{u_{1}}\left(v\right)\right)\dotsm\right)=\left(-1\right)^{k}u_{k}\dotsm u_{1}vu_{1}^{-1}\dotsm u_{k}^{-1}.$$

The elements of $${C\left(r,s\right)}$$ that have the form of a Clifford product of unit vectors $${U=u_{k}\dotsm u_{1}}$$ form a Lie group denoted $${\textrm{Pin}\left(r,s\right)}$$ and called a Clifford group (AKA Pin group). In terms of the reverse operation of geometric algebra, the elements $${U\in\textrm{Pin}\left(r,s\right)}$$ are products of invertible vectors which satisfy $${U\widetilde{U}=\pm1}$$. $${R}$$ forms a homomorphism from $${\textrm{Pin}\left(r,s\right)}$$ to $${O\left(r,s\right)}$$ defined by $${U\mapsto R_{U}}$$, where

$$\displaystyle R_{U}(v)=\left(-1\right)^{k}UvU^{-1}.$$

This homomorphism is two-to-one, since $${R_{U}}$$ and $${R_{-U}}$$ map to the same transformation, so $${\textrm{Pin}(r,s)}$$ is a double covering of $${O(r,s)}$$.

Now, elements of $${SO\left(r,s\right)}$$ are pure rotations, i.e. they are obtained as a product of an even number of reflections. Therefore the special Clifford group (AKA Spin group) $${\textrm{Spin}\left(r,s\right)\equiv\textrm{Pin}\left(r,s\right)\cap C_{0}\left(r,s\right)}$$ is a double covering of $${SO\left(r,s\right)}$$, using the restriction of $${R_{U}}$$ to even elements:

$$\displaystyle R_{U}^{S}(v)=UvU^{-1}$$

 Δ It is important to remember that rotations in 4 dimensions and higher do not follow many intuitive ideas from 3 dimensions. In particular, a rotation can have more than one plane of rotation (where rotated vectors in the plane stay in the plane), and therefore can require more than two reflections.
 Δ Distinctions are sometimes made between (special) Clifford groups and (S)Pin groups.

These relationships are depicted in the following diagram. The above depicts how the Clifford group $${\textrm{Pin}}$$ and its even subgroup $${\textrm{Spin}}$$ are generated by the unit elements of the Clifford algebra $${C}$$. $${C_{1}}$$ and $${C_{0}}$$ both have dimension $${2^{n-1}}$$ as manifolds, and $${\textrm{Pin}}$$, $${\textrm{Spin}}$$, and $${\textrm{Spin}^{e}}$$ all have dimension $${n(n-1)/2}$$.

 Δ Some potential sources of confusion can be avoided by remembering that $${\textrm{Pin}\left(r,s\right)}$$ is the group generated by Clifford multiplication on the unit elements of the algebra $${C\left(r,s\right)}$$. Thus the elements of $${\textrm{Pin}\left(r,s\right)}$$ can only be multiplied with each other and always have inverses, while the elements of $${C\left(r,s\right)}$$ can be multiplied by scalars and added, but may not have multiplicative inverses.