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In the classification of simple Lie algebras one has the familiar picture of 4 families, $A_n$, $B_n$, $C_n$ and $D_n$, and 5 exceptional groups, $F_4,$ $G_2,$ $E_6$, $E_7$ and $E_8$. The $D_n$ family has the unique feature that it contains, among all the corresponding Lie groups, groups whose center is non-cyclic: for $Spin(4n+2)$ the center is $\mathbb{Z}_4$, but one has $$Z(Spin(4n))=\mathbb{Z}_2\times \mathbb{Z}_2.$$ One can take the quotient of $Spin(4n)$ by any of the 3 subgroups of its center isomorphic to $\mathbb{Z}_2$ - one of these quotients is $SO(4n)$, the other are two are isomorphic to each other and are sometimes referred to as the half-spin or semi-spin groups, denoted by $SSpin(4n)$ (the notation may not be completely standard). They are a little bit the forgotten Lie groups - simple groups that are neither exceptional nor a quotient or covering of a classical group.

My question is: other than occuring in the classification, are there any places where these half-spin groups show up (naturally so to speak)?

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up vote 5 down vote accepted

Even if you are only interested in proving theorems about simply connected groups, you can naturally run into quotients like the half-spin groups. For example, you might try to prove something about your simply connected group by considering a reductive subgroup $G$ (maybe a centralizer of a rank 1 torus, for example). To prove the result for $G$, you may need to reduce to the case of the semisimple group $G/T$, where $T$ denotes the connected center of $G$ (a torus). Even when $G$ has derived subgroup $Spin(4n)$, the quotient $G/T$ could be $HSpin(4n)$, because $T$ can overlap the center of $Spin(4n)$.

Here is a specific example of this that some people care about, where $HSpin(12)$ arises from $E_7$. (This can be viewed as a follow-on to Christian's answer--this example is less "natural" than his.) One of the very few remaining cases of the Kneser-Tits Conjecture [google] concerns a simply connected group of type E7 with rank 1--deleting the distinguished vertex leaves a subdiagram of type $D_6$. If you attempt to prove rationality of the $E_7$ by standard reductions as in Chernousov-Platonov, then you end up doing the procedure I outlined in the previous paragraph, and you are stuck with proving rationality for a form of $HSpin(12)$.

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There's a "natural" $SSpin(16)\subset E_8$ (cf. Adam's "Lectures on the exceptional Lie groups").

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The term "half-spin" seems to be more widespread in the literature. Anyway, the groups have serious uses in the study of $E_8$ and other exceptional Lie algebras (by people including Skip Garibaldi), as well as uses in mathematical physics (explained by people like Jose Figueroa-O'Farrill). Anyone with access to MathSciNet will find it worthwhile to search for "half-spin" and browse the many papers listed under subject numbers like 17, 22, 81. But there doesn't seem to be a completely standard notation for these groups.

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$SSpin(32)$ appears in two superficially unrelated places that I know of.

  1. There are two heterotic string theories: the one based on the root lattice of $E_8 \oplus E_8$ and the one based on the weight lattice of $SSpin(32)$. This is often (but incorrectly!) called the $SO(32)$ heterotic string theory.

  2. Milnor gave the first example of isospectral manifolds which are not isometric: the flat tori obtained by quotienting $\mathbb{R}^{16}$ by the above two lattices.

Another factoid is the following. Although they are isomorphic, it is convenient for this to distinguish between, say, $SSpin(4n)$ and $S'Spin(4n)$, which are the quotients of $Spin(4n)$ by different $\mathbb{Z}_2$ subgroups of the centre, which are not $SO(4n)$. Then $SSpin(8n)$ and $S'Spin(8n)$ are self-dual (under Langlands duality), whereas $SSpin(8n+4)$ is the Langlands (a.k.a. magnetic) dual of $S'Spin(8n+4)$.

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Nice observations! – Skip Dec 2 '10 at 18:58

The case of $SSpin(8)$ is very special, in that all three groups are isomorphic to $SO(8)$. The isomorphisms are provided by triality, namely the outer automorphism group of $Spin(8)$, which acts on $Z(Spin(8))\simeq (\mathbb{Z}/2)^2$ as $GL_2(\mathbb{Z}/2)\simeq S_3$, in particular transitively on the order 2 subgroups. This also has to do with the octonions, since $Spin(8)$ may be realized as the subgroup of $(A,B,C) \in SO(8)^3$ such that $A(x)B(y)=C(xy)$ where multiplication in $\mathbb{R}^8$ is that of octonions. For more details, see John Baez great survey on octonions.

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