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.
Best Answer
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.)