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When looking into the definition of a Pin group, it turns out that there are - at least - three different ones in the literature, and they do not agree --- but thankfully all yield the same Spin groups.

In detail, let $q:V\to K$ be a nonsingular=regular quadratic form on a finite dimensional vector space $V$ over a field $K$ of characteristic not $2$, and $C = Cl(q)$ its Clifford algebra. With $\alpha$ the principal automorphism of $C$, determined by $\alpha|_V =-\mathsf{id}_V$, and $t$ the principal antiautomorphism of $C$, determined by $t|_V=\mathsf{id}_V$, the Clifford norm is defined as $N(x) =t(\alpha(x))x$, for $x\in C$, in most sources, but by $N'(x)=t(x)x$ in Scharlau's 1985 book (see below.)

With $C^*_{hom}$ the multiplicative group of $\mathbb Z/2\mathbb Z$--homogeneous invertible elements in $C$, the Clifford group $\Gamma(q)$ consists of those $u\in C^*_{hom}$ for which $\alpha(u)Vu^{-1}\subseteq V$. The Clifford group maps naturally onto the orthogonal group $\mathsf O(q)$, due to the Cartan--Dieudonné Theorem.

Restricting $N, N'$, or $N^2$ to $\Gamma(q)$, each map defines a group homomorphism to $K^*$, and "the" Pin group has been defined as the kernel of either in various sources (In what follows I made up the notation to keep the notions apart, except for the first one.):

$\mathsf{Pin}(q) = \mathsf{ker}(N|_{\Gamma(q)})$ is the definition given in

  • Scharlau, Winfried: Quadratic forms. Queen's Papers in Pure and Applied Mathematics, No. 22 Queen's University, Kingston, Ont., 1969 iii+162 pp. and in

  • Knus, Max-Albert: Quadratic and Hermitian forms over rings. Grundlehren der Mathematischen Wissenschaften, 294. Springer-Verlag, Berlin, 1991. xii+524 pp.

It is also the definition in the original paper by Atiyah-Bott-Shapiro on Clifford Modules. However, these authors only deal with negative definite real forms so that $\mathsf{Pin}(q)$ coincides with $\mathsf{PIN}(q)$ in the notation proposed below.

$\mathsf{Pin}'(q) = \mathsf{ker}(N'|_{\Gamma(q)})$ is the definition put forward in

  • Scharlau, Winfried: Quadratic and Hermitian forms. Grundlehren der Mathematischen Wissenschaften, 270. Springer-Verlag, Berlin, 1985. x+421 pp.

$\mathsf{PIN}(q) = \mathsf{ker}(N^2|_{\Gamma(q)})=N^{-1}(\{\pm1\})$ is the definition one finds in sources more concerned with real Clifford algebras and their applications in Geometry or Physics, such as

  • Lawson, H. Blaine, Jr.; Michelsohn, Marie-Louise: Spin geometry. Princeton Mathematical Series, 38. Princeton University Press, Princeton, NJ, 1989. xii+427 pp.

  • Choquet-Bruhat, Yvonne; DeWitt-Morette, Cécile: Analysis, manifolds and physics. Part II. 92 applications. North-Holland Publishing Co., Amsterdam, 1989. xii+449 pp.

  • Lounesto, Pertti: Clifford algebras and spinors. London Mathematical Society Lecture Note Series, 239. Cambridge University Press, Cambridge, 1997. x+306 pp.

Note that $\mathsf{Pin}(q), \mathsf{Pin}'(q)$ are subgroups of $\mathsf{PIN}(q)$ of index $1$ or $2$.

To make things even more confusing, sometimes competing definitions are used side by side, such as on the wikipedia page on Clifford algebras or on an earlier page here on mathoverflow.

The advantage of the "small" Pin groups is that they map onto the kernel of the Spinor norm in $\mathsf O(q)$, but whether one uses $N$ or $N'$ changes the sign of the Spinor norm...

The advantage of the "large" PIN group is that it maps onto the orthogonal group whenever $K^*/(K^*)^2 =\{\pm1\}$.

So, what should be the authoritative definition, or should we live with small and large Pin groups, in which case an author should specify very clearly which one is meant?

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  • $\begingroup$ Personally I read only some works of Pertti Lounesto from mentioned above. For me the definition of things is less important. More important is what would you like to investigate on. Then take your favourite definition and go ahead. For me Spin group is interesting enough, because it is double cover of $SO_n$. I don't know what is the advantage of Pin groups. Logically it should be double cover of $O_n$. That would give the same reason for adding letter $S$ in front of the name. That explanation is given in wikipedia article for Pin group and I like this. "This joke is due to J-P. Serre" ... $\endgroup$
    – user21230
    Commented Oct 3, 2016 at 9:36
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    $\begingroup$ @MarekMitros: "For me the definition of things is less important." -- What? How can you communicate about mathematics if you don't have precise definitions? The OP points out exactly that having different definitions for the same notion can be very confusing, and should be avoided as much as possible. $\endgroup$ Commented Oct 3, 2016 at 15:03
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    $\begingroup$ @TomDeMedts: Certainly definitions are very important, but one's choice of definitions are best guided by what one wishes to do with them. Marek Mitros is pointing that the author doesn't give any indication about what (s)he wishes to do with Pin groups, and that consequently it is not possible to adequately answer the question (beyond that one has to be clear on one's choice). The literature has multiple conventions because there are many purposes for Pin groups (and allowing algebraic groups or group schemes rather than Lie or abstract groups makes it even more abundant!). $\endgroup$
    – nfdc23
    Commented Oct 3, 2016 at 16:15
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    $\begingroup$ @TomDeMedts: Probably, I am more intuitionist. When you publish something, surely you should use definitions understandable for others. But when you're doing research you can leave some details to clarify later on. For example it is enough to understand that Pin is double cover of $O_n$. Besides it is not the first neither the last situation in mathematics that there are several definitions of the same thing. On school level, I have learned that 0 is not natural number. Now I hear that in most western countries 0 is natural number. In Poland it is not settled, so on exams 0 is treated separate $\endgroup$
    – user21230
    Commented Oct 3, 2016 at 16:49
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    $\begingroup$ From what I can tell, $\operatorname{Pin'}(q) = \operatorname{Pin}(-q)$ and vice versa $\endgroup$
    – Eric
    Commented Aug 4, 2022 at 21:35

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