Spin(2,C) - Knowlpedia

$\operatorname{Spin}(2,\mathbb C)$ is the spin group in the even Clifford algebra of $\mathbb C^{2}$ 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 $2$ , complex dimension $1$ . Conjugation on the generating vector space gives a two-to-one homomorphism to $\mathrm{SO}(2,\mathbb C)$ with kernel $\{\pm1\}$ . In dimension two it is isomorphic to $\mathbb C^\times$ . The double cover is a squaring map after identifying both groups with that torus; this cover is not universal.

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 $\operatorname{SO}(2,\mathbb C)$ .

The even Clifford algebra in dimension two identifies its norm-one subgroup with the circle in the real case, and with C* in the complex case. Under the same identification of SO(2), the covering map is squaring. See $\mathbb C^\times$ .

Two-dimensional calculation

Writing $e_1^2=e_2^2=-1$ , the even Clifford algebra has basis $1,e_1e_2$ and $(e_1e_2)^2=-1$ . Its spin elements have the form $s=a+b e_1e_2$ with $a^2+b^2=1$ . Over the reals, $s\mapsto a+ib$ identifies the group with the circle. Over the complex numbers the same formula identifies it with $\mathbb C^\times$ : its inverse uses $a=(z+z^{-1})/2$ and $b=(z-z^{-1})/(2i)$ . Conjugating the two basis vectors shows that the orthogonal action corresponds to $z\mapsto z^2$ .

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.