I’m seeking for a straightforward construction of well-known $\mathrm{Spin}(4) = \mathrm{Spin}(3)\times\mathrm{Spin}(3)$ isomorphism using geometric algebra-based definition of “Spin”, without topological detours. Explicit description of the even part of $\mathrm{C\ell}_{4,0}$ would be also desirable. I know that $\mathrm{C\ell}_{3,0}^0$ is the same as quaternions, $\mathbb{H}$. Also I have following considerations about 4 dimensions but am not sure they are all correct.
- $\mathrm{C\ell}_{4,0}^0$ is a 8-dimensional algebra that contains one dimension of scalars, six dimensions generated by bivectors (or, the same, $\mathfrak{so}(4)$), and one dimension generated by $e_1 e_2 e_3 e_4$, where $\{e_1, e_2, e_3, e_4\}$ is an orthonormal basis.
- There is an explicit isomorphism $\mathrm{C\ell}_{4,0}^0 = \mathrm{C\ell}_{3,0}^0 \oplus \mathrm{C\ell}_{3,0}^0$ of associative algebras.
- One of terms, a 4-dimensional subalgebra, can be described with the basis $$ \{ \frac{1}{2}(1 - e_1 e_2 e_3 e_4),\ \frac{1}{2}(e_1 e_2 + e_3 e_4),\ \frac{1}{2}(e_1 e_3 - e_2 e_4),\ \frac{1}{2}(e_1 e_4 + e_2 e_3)\}$$ that shows its isomorphism to $\mathbb{H}$.
- Another term has a similar basis, with opposite signs.
- Bivectors from both instances of $\mathrm{C\ell}_{3,0}^0$ sum to 6-dimensional $\mathfrak{so}(4)$, whereas two “scalar” dimensions form the rest of $\mathrm{C\ell}_{4,0}^0$.
Would anyone evaluate this stuff and/or suggest an alternative approach?