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In mathematics the spin group Spin(n) is the double cover of the special orthogonal group SO(n), such that there exists a short exact sequence of Lie groups

1 \to \mathbb{Z}_2 \to \operatorname{Spin}(n) \to \operatorname{SO}(n) \to 1.

For n > 2, Spin(n) is simply connected and so coincides with the universal cover of SO(n). As a Lie group Spin(n) therefore shares its dimension, n(n - 1)/2, and its Lie algebra with the special orthogonal group.

Spin(n) can be constructed as a subgroup of the invertible elements in the Clifford algebra Cl(n).

Contents

Accidental isomorphisms

In low dimensions, there are isomorphisms among the classical Lie groups called accidental isomorphisms. For instance, there are isomorphisms between low dimensional spin groups and certain classical Lie groups, due to low dimensional isomorphisms between the Root systems of the different families of simple Lie algebras. Specifically, we have

Spin(1) = O(1)
Spin(2) = U(1) = SO(2)
Spin(3) = Sp(1) = SU(2)
Spin(4) = Sp(1) × Sp(1)
Spin(5) = Sp(2)
Spin(6) = SU(4)

There are certain vestiges of these isomorphisms left over for n = 7,8 (see Spin(8) for more details). For higher n, these isomorphisms disappear entirely.

Indefinite signature

In indefinite signature, the spin group Spin(p,q) is constructed through Clifford algebras in a similar way to standard spin groups. It is a connected double cover of SO0(p,q), the connected component of the identity of the indefinite orthogonal group SO(p,q) (there are a variety of conventions on the connectedness of Spin(p,q); in this article, it is taken to be connected for p+q>2). As in definite signature, there are some accidental isomorphisms in low dimensions:

Spin(1,1) = GL(1,R)
Spin(2,1) = SL(2,R)
Spin(3,1) = SL(2,C)
Spin(2,2) = SL(2,R) × SL(2,R)
Spin(4,1) = Sp(1,1)
Spin(3,2) = Sp(4,R)
Spin(5,1) = SL(2,H)
Spin(4,2) = SU(2,2)
Spin(3,3) = SL(4,R)

Note that Spin(p,q) = Spin(q,p).

Topological considerations

Connected and simply connected Lie groups are classified by their Lie algebra. So if G is a connected Lie group with a simple Lie algebra, with G' the universal cover of G, there is an inclusion

 \pi_1 (G) \subset Z(G'),

with Z(G') the centre of G'. This inclusion and the Lie algebra \mathfrak{g} of G determine G entirely (note that it is not the fact that \mathfrak{g} and p1(G) determine G entirely; for instance SL(2,R) and PSL(2,R) have the same Lie algebra and same fundamental group \mathbb{Z}, but are not isomorphic).

The definite signature Spin(n) are all simply connected for (n>2), so they are the universal coverings for SO(n). In indefinite signature, the maximal compact connected subgroup of Spin(p,q) is

(\mbox{Spin}(p) \times \mbox{Spin}(q))/ \{(1,1),(-1,-1)\}.

This allows us to calculate the fundamental groups of Spin(p,q):

\pi_1(\mbox{Spin}(p,q)) = \begin{cases}
\{0\} & (p,q)=(1,1) \mbox{ or } (1,0) \\
\{0\} & p > 2, q = 0,1 \\
\mathbb{Z} & (p,q)=(2,0) \mbox{ or } (2,1) \\
\mathbb{Z} \times \mathbb{Z} & (p,q) = (2,2) \\
\mathbb{Z} & p > 2, q=2 \\
\mathbb{Z}_2 & p >2, q >2 \\
\end{cases}

For p,q > 2, this implies that the map \pi_1(\mbox{Spin}(p,q)) \to \pi_1(SO(p,q)) is given by  1 \in \mathbb{Z}_2 going to (1,1) \in \mathbb{Z}_2\times \mathbb{Z}_2. For p=2, q>2, this map is given by 1 \in \mathbb{Z} \to (1,1) \in \mathbb{Z} \times \mathbb{Z}_2. And finally, for p=q=2, (1,0) \in \mathbb{Z} \times \mathbb{Z} is sent to (1,1) \in \mathbb{Z} \times \mathbb{Z} and (0,1) is sent to (1, - 1).

See also



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