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Let $V$ be a real or complex finite dimensional vector space with nondegenerate quadratic form $Q$. According to the spin representation Wikipedia article,

Up to group isomorphism, SO$(V, Q)$ has a unique connected double cover, the spin group Spin$(V, Q)$.

There is no proof or citation for this claim, and I don't see this exact claim being made on the spin group or spinor articles, though it is often implied by use of the phrase "the" double cover. Is this true, and if so, why? If this follows from standard well known theorems, I would be happy to just be pointed to these theorems.

I am more familiar with universal covers than more general $n$-fold covers, and I know of the following results.

  • Every connected manifold [e.g., SO$(V, Q)$] has a universal cover, which is the unique (up to equivalence) simply-connected covering space.

  • The universal cover of SO$(n,\mathbb R)$ happens to be a double cover for $n>2$. As far as I know, this breaks down for quadratic forms of indefinite signature.

It isn't clear to me whether either of these facts (or anything about universal covers) is relevant to proving the above claim.

WillG
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1 Answers1

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This is false in indefinite signature. The Wikipedia article uses nonstandard notation here and I will use standard notation: for me $SO$ refers to the subgroup of the orthogonal group of elements with determinant $1$, which is not connected in indefinite signature. The connected component of the identity can be denoted $SO^{+}$.

Now for the counterexample: $SO^{+}(2, 2)$ has fundamental group $\mathbb{Z} \times \mathbb{Z}$, so it has three connected double covers corresponding to the three subgroups of index $2$. This follows from the general fact that $O(p, q)$ has maximal compact subgroup $O(p) \times O(q)$ and deformation retracts onto its maximal compact, so $SO^{+}(p, q)$ has maximal compact subgroup $SO(p) \times SO(q)$ and deformation retracts onto this. This shows more generally that $SO^{+}(p, q)$ has three connected double covers for $p, q \ge 2$.

(Of these three double covers, the spin group should be the "diagonal" one, but I don't know a proof or reference for this off the top of my head.)

Qiaochu Yuan
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  • And these three groups are not isomorphic? It seems like the three index-2 subgroups of $\mathbb Z\times\mathbb Z$ are isomorphic, but I'm not sure if this carries over to the double covers. – WillG Jan 02 '23 at 17:43
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    @WillG: I don't have a proof but I don't think they are. In any case they aren't isomorphic as covers. And we can switch the example to $SO^{+}(2, 3)$, which has fundamental group $\mathbb{Z} \times C_2$, for an example where the double covers have non-isomorphic fundamental groups (either $\mathbb{Z}$ or $\mathbb{Z} \times C_2$). – Qiaochu Yuan Jan 02 '23 at 21:09
  • Very late follow-up: What did you mean by the "diagonal" double cover? – WillG Nov 04 '23 at 19:43
  • @WillG: the fundamental group is $A \times B$ where $A, B$ are either $\mathbb{Z}$ or $C_2$. Either way there are exactly three (connected) double covers because there are exactly three nontrivial homomorphisms $A \times B \to C_2$. The "diagonal" cover is the one where the restriction to both $A$ and $B$ is nontrivial; the other two are the one factoring through $A$ and the one factoring through $B$. When $A = B = C_2$ this is the double cover corresponding to the diagonal copy of $C_2$ in $C_2 \times C_2$. – Qiaochu Yuan Nov 04 '23 at 22:53
  • Yes, indeed, these are "diagonal" 2-fold covers, see my answer here. – Moishe Kohan Feb 20 '24 at 02:03