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 \({SO(r,s)}\), 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.

Δ \({\textrm{Pin}(r,s)}\) is sometimes defined as the Clifford product of “unit” vectors \({u}\) with only positive \({\left\langle u,u\right\rangle =+1}\). Then \({\textrm{Spin}(r,s)}\) ends up being the double cover of \({SO(r,s)^{e}}\), i.e. it is what we call \({\textrm{Spin}(r,s)^{e}}\) for most cases: it is connected for \({r}\) or \({s}\) greater than one, and is simply connected if \({s=0,1}\) and \({r>2}\) (or vice versa). |

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.