Spin(1,C) - Knowlpedia

$\operatorname{Spin}(1,\mathbb C)$ is the spin group in the even Clifford algebra of $\mathbb C^{1}$ with quadratic form $Q(v)=\sum_jv_j^2$ and convention $v^2=-Q(v)1$ . It is generated by products of an even number of vectors satisfying $Q(v)=1$ .

Dimensions and structure

This group has real dimension $0$ , complex dimension $0$ . Conjugation on the generating vector space gives a two-to-one homomorphism to $\mathrm{SO}(1,\mathbb C)$ with kernel $\{\pm1\}$ . In dimension one this is the discrete group $\{\pm1\}$ , and SO(1) is trivial. It is not connected.

Its complex Lie group structure uses holomorphic multiplication and inversion. Forgetting the complex structure gives the underlying real Lie group; real and holomorphic homomorphisms belong to different categories.

Clifford conjugation gives a two-sheeted covering with kernel {±1}. See $\operatorname{SO}(1,\mathbb C)$ .

In the one-dimensional Clifford algebra, even products of unit vectors are exactly {±1}; identify each scalar with the same one-by-one orthogonal matrix. See $\operatorname{O}(1,\mathbb C)$ .

One-dimensional calculation

Let $e$ be the basis vector with $e^2=-1$ . The only vectors satisfying $Q(v)=1$ are $e$ and $-e$ , and their even products are exactly $1,-1$ . They commute with $e$ , so both act trivially on the one-dimensional vector space.

References

Pavel Etingof, Lie Groups and Lie Algebras (MIT lecture notes), §31.2, Example 31.3 and Proposition 31.4; §31.3, Example 31.10, pp. 166–168.