In the Lie theory notes on my website it is claimed (Example 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 antiinvolution $\tau$ 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:

Question For 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.

  • 1
    $\begingroup$ Credit where it's due: Nige asked me to mention that he learned about the isomorphism $Pin(4,0) \cong Pin(0,4)$, and about the error in my notes, from his student Yumi Boote. $\endgroup$ Feb 9, 2015 at 15:42

2 Answers 2


In "Analysis, Manifolds and Physics. Part II" by Y. Choquet-Bruhat and C. De Witt-Morette (Elsevier, 2000), there is a claim (example 3 in 1.7, pp. 25-27) that while $\mathrm{Pin}(p,q)\simeq\mathrm{Pin}(q,p)$ for $p\equiv q\pmod{4}$ (which, as you note, follows from the mod 8 "Bott periodicity" for Clifford algebras), for other values of $p$ and $q$ there is generally no isomorphism. While I think they only refer to $(p,q)=(1,0)$ and $(p,q)=(2,0)$ as examples of non-isomorphisms, I think that the case of $(p,q)=(2,0)$ is already instructional enough.

  • $\begingroup$ Great, I'll take a look. $\endgroup$ Oct 24, 2014 at 14:46

Your initial claim about $Pin(4,0)\not\simeq Pin(0,4)$ seems to be correct. In fact let $\phi:Pin(4,0)\rightarrow Pin(0,4)$ be an (abstract) group isomorphism. When $p+q$ is even, every element $g$ of the Clifford-Lipschitz group $\Gamma(p,q)$ is a product of some anisotropic elements $v_i, (i=1,\cdots,n)$ of $V$ (this follows from the Cartan-Dieudonn\'e theorem). Now let $g=v_1\cdots v_n\in Pin(4,0)$. By scaling we can still have $g=v_1\cdots v_n$ with $N(v_i)=\pm1$, hence we may assume that $v_i\in Pin(4,0)$. So $\phi(g)=\Pi_{i=1}^n\phi(v_i)$. We then obtain \begin{eqnarray*} N(\phi(g))&=&\tau(\phi(g))\phi(g)\\ & =&\tau(\phi(\Pi_{i=1}^nv_i))\phi(\Pi_{i=1}^nv_i)\\ & =&\tau(\Pi_{i=1}^n\phi(v_i))\phi(\Pi_{i=1}^nv_i)\\ & =&\Pi_{i=1}^n\phi(v_{n-i})\phi(\Pi_{i=1}^nv_i)\\ & =&\phi(\Pi_{i=1}^nv_{n-i})\phi(\Pi_{i=1}^nv_i)\\ & =&\phi(\Pi_{i=1}^nv_{n-i}\cdot\Pi_{i=1}^nv_i)\\ & =&\phi(\tau(g)g)\\ & =&\phi(N(g)) \end{eqnarray*} But this is contradiction because considering the fact that $(4,0)$ is positive definite, $N$ can take only positive values on $\Gamma(4,0)$. But we have many elements of $Pin(0,4)$ with norm $-1$.

I edited my answer, I agree with your comment but I think I don't use it. Please tell me if something is still wrong.

  • $\begingroup$ Why does $\phi$ commute with $\tau$? Indeed, let $\phi: \mathrm{Cliff}(4,0) \to \mathrm{Cliff}(0,4)$ be the isomorphism. It identifies a generator $x$ with a cubic $yzw$ (for some arbitrary ortho(normal up to sign) bases). Then $\phi(\tau(x)) = \phi(x) = yzw$, whereas $\tau(\phi(x)) = \tau(yzw) = wzy = -yzw$. $\endgroup$ Oct 24, 2014 at 0:53
  • $\begingroup$ If your argument were correct, it would show en passant that $\tau\bigl(\phi\bigl(\prod_{i = 1}^n v_i\bigr)\bigr) = \tau\bigl(\prod_{i = 1}^n \phi(v_i)\bigr) = \phi\bigl(\prod_{i = 1}^n v_{n - i}\bigr) = \phi\bigl(\tau\bigl(\prod_{i = 1}^n v_i\bigr)\bigr)$, which is why @TheoJohnson-Freyd mentioned $\phi$ not commuting with $\tau$. The problem is that your equality $\tau\bigl(\prod_{i = 1}^n \phi(v_i)\bigr) = \prod_{i = 1}^n \phi(v_i)$ assumes that $\phi(v_i)$ are still anisotropic elements of $V$. $\endgroup$
    – LSpice
    Jun 13, 2020 at 23:18

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