Definition of a spin group

Question

$\DeclareMathOperator\Pin{Pin}\DeclareMathOperator\Spin{Spin}$This follows on from Definition of Pin groups?, which notes there are three different definitions of the Pin group; thankfully, all of which "yield the same definition of the spin group".

However, it seems there are even multiple definitions of spin groups:

1. Elements with spinor norm 1:

   • The wikipedia page on Clifford Algebras:

     The pin group $\Pin_V(K)$ is the subgroup of the Lipschitz group $Γ$ of elements of spinor norm 1, and similarly the spin group $\Spin_V(K)$ is the subgroup of elements of Dickson invariant 0 in $\Pin_V(K)$.

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

     The Pin group of $(V,q)$ is the subgroup $\operatorname{Pin}(V,q)$ of $P(V,q)$ generated by the elements $v \in V$ with $q(v) = \pm 1$. The associated spin group of $(V,q)$ is then defined by $$\operatorname {Spin} (V,q)=\operatorname {Pin} (V,q)\cap \operatorname {Cl} ^{0}(V,q)$$

2. Products of elements of norm 1:

   • The ncatlab page on spin group

     The Pin group $\Pin(V;q)$ of a quadratic vector space, def. 2.1, is the subgroup of the group of units in the Clifford algebra $Cl(V,q)$ $$\Pin(V,q) \hookrightarrow \mathrm{GL}_1(Cl(V,q))$$ on those elements which are multiples $v_1 \cdots v_{n}$ of elements $v_i∈V$ with $q(v_i)=1$.

   • The wikipedia page on Spin groups:

     The pin group $\operatorname {Pin} (V)$ is a subgroup of $\operatorname{Cl} (V)$'s Clifford group of all elements of the form $$v_{1}v_{2}\cdots v_{k}$$ where each $v_{i}\in V$ is of unit length:$q(v_{i})=1$. The spin group is then defined as $$\operatorname{Spin} (V)=\operatorname{Pin} (V)\cap \operatorname{Cl}^{\text{even}}$$

These two definitions seem equivalent on positive (or complex) quadratic forms; but over the space $\mathbb{R}^2$ spanned by $e_1, e_2$ with quadratic form $q(e_1) = q(e_2) = -1$, it seems to me that the two definitions disagree on whether $e_1e_2 \in \operatorname{Spin}(V,q)$:

• Definition 1 permits it since the spinor norm of $e_1e_2$ is either $N(e_1e_2) = (-e_2)(-e_1)e_1e_2 = 1$ or $N'(e_1e_2) = e_2e_1e_1e_2 = 1$ (using the two possible meanings of spinor norm from Definition of Pin groups?)
• Definition 2 excludes it because there is no $v\in V$ with $q(v) = 1$. In fact, definition 2 permits only one element of the spin group, $1$!

Am I right to claim that definition 2 is just incorrect? Or are there really two competing notions of spin groups which intentionally disagree on whether $e_1e_2 \in \operatorname{Spin}(V,q)$?

Answer 1

Here is the status of the definitions (as of Oct. 2023):

1. The definition in the Wikipedia page on Clifford algebras makes sense for any nondegenerate quadratic space over any field, and has the properties that it claims to hold.

2. The Lawson-Michelsohn definition does not agree with the Wikipedia Clifford algebra definition, because it allows elements with spinor norm -1. This is convenient for getting a double cover of $SO(V,q)$ for real vector spaces when the signature is indefinite, but makes a mess over other fields. I haven't read their book, but I would guess they work over $\mathbb{R}$.

3. The ncatlab page uses the general language of quadratic spaces over fields, but it is likely the definition of Pin group is lifted from a source that is limited to real quadratic spaces that are positive definite or of signature $(n,1)$, $n \geq 1$. As you mention, it yields trivial groups when it is not supposed to. Oddly enough, both this article and the article on Pin groups cite Lawson-Michelsohn while producing a different definition.

4. The wikipedia page on Spin groups has more than one definition! The "Construction" section explicitly assumes the quadratic form is definite, but as you note, it needs to further assume the form is positive definite for that definition to yield a nontrivial group. The "Double covering" section gives a definition of Pin group in terms of spinor norm, but doesn't include the condition that twisted conjugation preserves vectors.

In summary, the answer to your first question is "yes".