Spin group

Example

Let $n \ge 2$ and let $\mathrm{Cl}_n$ be the real Clifford algebra of $\R^n$ with the relations $v^2 = -\abs{v}^2$ for $v \in \R^n$. The spin group $\mathrm{Spin}(n)$ is the set of products $u_1 u_2 \cdots u_{2k}$ in $\mathrm{Cl}_n$ of an even number of unit vectors $u_i \in \R^n$, with the Clifford product. It is a group: the product of two such elements is again one. The inverse of $u_1 \cdots u_{2k}$ is $u_{2k} \cdots u_1$, because $u_i u_i = -1$ and the number of factors is even.

For $x \in \mathrm{Spin}(n)$ and $v \in \R^n$, the element $x v x^{-1}$ lies again in $\R^n$. Check for one unit vector $u$: $u^{-1} = -u$, and $uv = -vu$ when $v \perp u$. So $-u v u^{-1}$ equals $v$ for $v \perp u$ and $-u$ for $v = u$, and $v \mapsto -u v u^{-1}$ is the reflection $r_u$ of $\R^n$ in the hyperplane $u^\perp$. For $x = u_1 \cdots u_{2k}$ the signs cancel, and $v \mapsto x v x^{-1}$ is the product $r_{u_1} \cdots r_{u_{2k}}$ of $2k$ reflections, which lies in SO(n).

This gives a homomorphism $\rho \colon \mathrm{Spin}(n) \to SO(n)$, $\rho(x)(v) = x v x^{-1}$. It is surjective, because every element of $SO(n)$ is a product of an even number of reflections, and its kernel is $\set{1, -1}$.