Lie group properties

Although the Lie groups $${\textrm{Pin}\left(r,s\right)}$$ and $${\textrm{Spin}\left(r,s\right)}$$ are double covers of $${O(r,s)}$$ and $${SO(r,s)}$$ (except for $${r=s=1}$$, see table below), they are in general not simply connected and so are not universal covering groups. However, it turns out that $${\textrm{Spin}\left(n\right)\equiv\textrm{Spin}\left(n,0\right)\cong\textrm{Spin}\left(0,n\right)}$$ is simply connected for $${n>2}$$, and thus is the universal covering group of $${SO(n)}$$. In addition, for $${n>2}$$ the identity component $${\textrm{Spin}\left(n,1\right)^{e}\cong\textrm{Spin}\left(1,n\right)^{e}}$$ is simply connected and is the double cover of $${SO\left(1,n\right)^{e}\cong SO\left(n,1\right)^{e}}$$. $${\textrm{Spin}\left(r,s\right)}$$ does not always have a simple description in terms of common matrix groups. Specific isomorphisms for the identity component of the first few $${\mathrm{Spin}}$$ groups in Euclidean and Lorentzian signatures are listed in the following table.

$${n}$$$${\textrm{Spin}\left(n,0\right)\cong \textrm{Spin}\left(0,n\right)}$$$${\textrm{Spin}\left(n-1,1\right)^{e}\cong \textrm{Spin}\left(1,n-1\right)^{e}}$$
2$${U\left(1\right)\cong SO\left(2\right)}$$$${SO\left(1,1\right)^{e}\cong\mathbb{R}^{+}}$$
3$${Sp\left(1\right)\cong SU\left(2\right)}$$$${Sp\left(2,\mathbb{R}\right)\cong SL\left(2,\mathbb{R}\right)}$$
4$${Sp\left(1\right)\oplus Sp\left(1\right)\cong SU\left(2\right)\oplus SU\left(2\right)}$$$${Sp\left(2,\mathbb{C}\right)\cong SL\left(2,\mathbb{C}\right)}$$
5$${Sp\left(2\right)}$$$${Sp\left(1,1\right)}$$
6$${SU\left(4\right)}$$$${SL\left(2,\mathbb{H}\right)}$$

Notes: $${\textrm{Spin}(2)}$$ is isomorphic to $${SO(2)}$$ as an abstract group, but is nevertheless a double cover of $${SO(2)}$$ in terms of its action on vectors; however it is not simply connected. $${\textrm{Spin}(1,1)^{e}}$$ is not a double cover of $${SO(1,1)^{e}}$$, which is instead obtained by imposing the restriction $${U\widetilde{U}=1}$$ on $${\textrm{Spin}(1, 1)}$$ to obtain $${GL\left(1,\mathbb{R}\right)\cong\left\{ R-0\right\} }$$. $${\textrm{Spin}(2,1)^{e}}$$ is not simply connected, but is a double cover of $${SO(2,1)^{e}}$$. In all higher dimensions $${\textrm{Spin}(n,0)}$$ and $${\textrm{Spin}(n,1)^{e}}$$ are always the simply connected double covers of $${SO(n)}$$ and $${SO(n,1)^{e}}$$. $${Sp(1,1)}$$ is defined in a straightforward way, but $${SL(2,\mathbb{H})}$$ (not defined here) is more tricky due to non-commutativity.

Since $${\textrm{Spin}(r,s)}$$ is a cover of $${SO(r,s)}$$, its Lie algebra is $${so(r,s)}$$, which turns out to be the space of bivectors $${\Lambda^{2}\mathbb{R}^{n}}$$ under the Lie commutator using Clifford multiplication: $${\left[A,B\right]\equiv AB-BA}$$. A more general construct sometimes seen is the Lie group generated by all invertible elements of $${C(r,s)}$$, called the Lipschitz group (AKA Clifford group); its Lie algebra is $${C(r,s)}$$ itself, also under the Lie commutator using Clifford multiplication.

We consequently have an equivalent definition of $${\textrm{Spin}\left(r,s\right)^{e}}$$ as the group generated by the exponentials of bivectors under Clifford multiplication. For the compact Lie groups $${\textrm{Spin}\left(n,0\right)\:\left(n>1\right)}$$, every element is then the exponential of a bivector; as it turns out, this is also true for $${\textrm{Spin}\left(n,1\right)^{e}}$$ for $${n>3}$$, and every element of $${\textrm{Spin}\left(3,1\right)^{e}}$$ is plus or minus such an exponential.