In the Lie theory notes on my website it is claimed (Example 7.3.3.5) that $\mathrm{Pin}(4,0)$ and $\mathrm{Pin}(0,4)$ are not isomorphic. As Nigel Ray pointed out to me, this claim is not quite correct. I will begin by recalling some background, and then state my question.

Recall that the Clifford algebra $\mathrm{Cliff}(p,q)$ is the unital associative algebra over $\mathbb R$ with generators $x_1,\dots,x_{p+q}$ and relations $x_ix_j = -x_jx_i$ for $i\neq j$ and $x_i^2 = 1$ for $i\leq p$ and $x_i^2 = -1$ for $i > p$. I will let $V \subseteq \mathrm{Cliff}(p,q)$ denote the vector subspace spanned by the generators. A PBW-type theorem verifies that $V \cong \mathbb R^{p+q}$. Inspection of the relations reveals that $\mathrm{Cliff}(p,q)$ inherits a $\mathbb Z/2$ grading from the tensor algebra of $V$; the $\mathbb Z$-grading gets broken to a $\mathbb Z$-filtration. The $\mathbb Z/2$-grading is the same as an algebra involution $\alpha : \mathrm{Cliff}(p,q) \to \mathrm{Cliff}(p,q)$ determined by $\alpha|_V = -\mathrm{id}_V$. There is also an algebra *anti*involution $\tau$ determined by $\tau|_V = \mathrm{id}_V$; it reverses the order of monomials. For any $g \in \mathrm{Cliff}(p,q)$, its *norm* is $N(g) = \tau(g)g \in \mathrm{Cliff}(p,q)$. On $V$, the norm restricts to the usual norm of signature $(p,q)$ on $\mathbb R^{p+q}$.

The *Clifford group* $\Gamma(p,q) \subseteq \mathrm{Cliff}^\times(p,q)$ consists of those invertible $g\in \mathrm{Cliff}^\times(p,q)$ such that $gV\tau(g^{-1}) \subseteq V$. In particular, it acts on $V$ by definition, and it obviously preserves the norm. Thus there is a map $\Gamma(p,q) \to \mathrm{O}(p,q)$ (the latter being the subset of $\mathrm{GL}(V)$ preserving $N$). The kernel is a copy of $\mathbb R^\times$, and I believe the kernel is central. The *Pin group* $\mathrm{Pin}(p,q)$ is the subgroup of $\Gamma(p,q)$ consisting of those $g$ with $N(g) = \pm 1$. The map $\mathrm{Pin}(p,q) \to \mathrm{O}(p,q)$ is a double cover.

In any case, note that $\mathrm{O}(4,0)$ and $\mathrm{O}(0,4)$ are equal as subgroups of $\mathrm{GL}(V)$. But $\mathrm{Pin}(4,0)$ and $\mathrm{Pin}(0,4)$ are not isomorphic as double covers of $\mathrm{O}(4,0) = \mathrm{O}(0,4)$. Indeed, choose any reflection, and look at its lifts. In $\mathrm{Pin}(4,0)$, those lifts each have order $2$; in $\mathrm{Pin}(0,4)$, they have order $4$.

Nevertheless, there is an isomorphism $\mathrm{Pin}(4,0) \cong \mathrm{Pin}(0,4)$. The only way I know to construct it uses the interesting isomorphism $\mathrm{Cliff}(4,0) \cong \mathrm{Cliff}(0,4)$. Then some calculations check that this isomorphism identifies Clifford and Pin groups. What's going on downstairs? $\mathrm{O}(4)$ admits an interesting outer automorphism called "multiply by the determinant", which acts trivially on $\mathrm{SO}(4)$. This automorphism intertwines the two double covers.

I can come now to my question:

QuestionFor every $(p,q)$, the group $\mathrm{O}(p,q) \cong \mathrm{O}(q,p)$ admits an outer automorphism called "multiply by the determinant". When $p+q$ is even, this automorphism acts trivially on $\mathrm{SO}(p,q)$. In general, $\mathrm{Cliff}(p,q) \not\cong \mathrm{Cliff}(q,p)$; that happens only when $p-q \equiv 0 \pmod 4$. (For example, $\mathrm{Cliff}(2,0) \cong \mathrm{Mat}_2(\mathbb R)$ whereas $\mathrm{Cliff}(0,2) \cong \mathbb H$.) But does the automorphism on $\mathrm O$ nevertheless lift to an isomorphism $\mathrm{Pin}(p,q) \cong \mathrm{Pin}(q,p)$?

My expectation is something like the following. Central extensions are classified by classes of degree-$2$ in group cohomology; characters (like the determinant) are classified by degree-$0$ cohomology; you can multiply cohomology classes, and multiplication by "determinant" exchanges the two $\mathrm{Pin}$ extensions.

But I couldn't manage the calculations to check this expectation.