§ Spin groups
- Spin group is a 2 to 1 cover of $SO(n)$.
- We claim that for 3 dimensions, $Spin(3) \simeq SU(2)$. So we should have a 2 to 1 homomorphism $\rho: SU(2) \to SO(3)$.
- We want to write the group in some computational way. Let's use the adjoint action (how the lie group acts on its own lie algebra).
- What is the lie algebra $su(2)$? It's trace-free hermitian.
- Why? Physicist: $UU^\dagger = I$ expanded by epsilon gives us $(I + i \epsilon H)(I - i \epsilon H) = I$, which gives $H = H^\dagger$.
- Also the determinant condition gives us $det(1 + i \epsilon H) = 1$ which means $1 + tr(i \epsilon H) = 1$, or $tr(H) = 0$.
- The adjoint action is $SU(2) \to Aut(H)$ given by $U \mapsto \lambda x. ad_U x$ which is $\lambda x. U X U^{-1}$. By unitarry, this is $U \mapsto \lambda x. U X U^{\dagger}$.
- $SO(3)$ acts on $\mathbb R^3$. The trick is to take $\mathbb R^3$ and compare it to the lie algebra $su(2)$which has 3 dimensions, spanned by pauli matrices.
- Conjecture: There is an isomorphism $\mathbb R^3 \simeq H$ as an inner product space for a custom inner product $\langle, \rangle$ on $H$.
- Reference