There are two possible ways to define the $\mathrm{Spin}(n)$ group of Euclidean $n$-space from $\mathrm{Pin}(n)$. First is that $\mathrm{Spin}(n)$ is the identity component of $\mathrm{Pin}(n)$. Second is that $\mathrm{Spin}(n)$ is the even part of $\mathrm{Pin}(n)$, i.e. $\mathrm{Spin}(n) = \mathrm{Pin}(n) ∩ \mathrm{Cℓ}^0(n)$. For $n ≥ 2$ we can see equivalence of two definitions directly from geometric algebra. Even and odd parts of $\mathrm{Pin}(n)$ are disjoint (follows from the spin norm), and we have to demonstrate connectedness of the even part of $\mathrm{Pin}(n)$ as:
- Each $w ∊ \mathrm{Pin}(n)$ has a decomposition $w = v_1 v_2 ⋯ v_k$ to unit vectors for some $k$ (one of definitions of “Pin”);
- If $w$ is even, then $k$ is even (follows from grading);
- Each of pair products $v_1 v_2$, $v_3 v_4$, …, $v_{k-1} v_k$ is, essentially, a $\mathrm{Spin}(2)$ “rotation”, each of which is homotopic to the identity element.
Now, the question:
Is a similar argument possible to demonstrate that $\mathrm{Spin}(n)$ is simply connected for $n > 2$ without resorting to knowledge of homotopic properties of $\mathrm{SO}(n)$?