Is there a purely topological definition of $\text{Spin}(p,q)$?

Question

I'm cross-posting this question from Math.SE, as it didn't get much attention there (even after a bounty).

A common way to define the group $\text{Spin}(p,q)$ is via Clifford algebras. However, $\text{Spin}(p,q)$ is always a connected double cover of $\text{SO}^+(p,q)$, so it seems we might be able to define it as "the unique (up to isomorphism) double cover such that...". Is there a purely topological definition of this form?

As I pointed out in the linked post, Wikipedia claims that there is always a unique connected double cover, period. When I asked about this here, Qiaochu Yuan pointed out that the statement is false. Hence, my question here could rephrased as, "how can Wikipedia's claim be fixed?"

There are two suggestions in the comments of the original post that might be right, but I don't know how to investigate them. I should add that I don't know much algebraic topology, so I probably won't easily understand the correct answer, even if there is one.

Answer 1

Here is a copy of my answer at MathStackExchange.

Almost nobody does this. There is a discussion in the Wikipedia article on spinor groups but it has too many mistakes and too few references. The only solid math reference (with a proof that I am not going to reproduce here) I know is in Chapter 5 (freely available from author's webpage here) in

Varadarajan, V. S., Supersymmetry for mathematicians: an introduction., Courant Lecture Notes in Mathematics 11. Providence, RI: American Mathematical Society (AMS); New York, NY: Courant Institute of Mathematical Sciences (ISBN 0-8218-3574-2/pbk). vi, 300 p. (2004). ZBL1142.58009.

It is useful if you read this answer in conjunction with my answer here.

To describe $Spin(p,q)$ as a 2-fold cover of $SO^+(p,q)$ ($p>1, q>1$) one has to look at the maximal compact subgroups since they carry all the homotopy information. The group $SO^+(p,q)$ has maximal compact subgroup $SO(p)\times SO(q)$ and $$ \pi_1(SO^+(p,q))\cong \pi_1(SO(p)\times SO(q))\cong H_1\times H_2, $$ where $H_1, H_2$ are cyclic groups (either infinite cyclic or ${\mathbb Z}_2$). In particular, if $h_1, h_2$ are generators of $H_1, H_2$, then there is a homomorphism $$ \phi: H_1\times H_2\to {\mathbb Z}_2, $$ sending both $h_1, h_2$ to the generator of ${\mathbb Z}_2$. It is clear that these homomorphisms are independent of the choices of generators. Let $H< H_1\times H_2$ denote the kernel of $\phi$, it is an index 2 subgroup in $\pi_1(SO(p)\times SO(q)\cong \pi_1(SO^+(p,q))$. Then $Spin(p,q)$ is the 2-fold cover of $SO^+(p,q)$ associated with the subgroup $H$ of the fundamental group. This is your topological description of the spinor group.

In Varadarajan's terminology, here is what's going on. The (unique up to conjugation) maximal compact subgroup of $Spin(p,q)$ is $(Spin(p)\times Spin(q))/G$, where $G\cong {\mathbb Z}_2$ is generated by an element $(a,b)\in Spin(p)\times Spin(q)$, where $a, b$ are the nontrivial central elements of $Spin(p), Spin(q)$ respectively, such that $$ Spin(p)/\langle a\rangle = SO(p), Spin(q)/\langle b\rangle = SO(q). $$ Hence, $(a,b)$ corresponds to the "diagonal" element $(h_1, h_2)$ of $H_1\times H_2$. Varadarajan proves in his book that the "usual" Clifford algebra definition of the spinor groups is equivalent to the one above.

Answer 2

What about the complex point of view?

The complex analogue of $SO(p,q)$ (complex matrices preserving a nondegenerate bilinear form and having determinant $1$) depends only on $p+q$, since all complex nondegenerate bilinear forms in $n$ indeterminates are isomorphic.

The complex analogue of $SO(n)$ contains the real $SO(n)$ as a deformation retract (maximal compact subgroup). So the complex group has fundamental group of order $2$ (if $n=p+q\ge 2$), and its unique connected double cover may be pulled back to give what I presume is the correct double cover of the real $SO(p,q)$.