Definitions

Pin group

In mathematics, the pin group is a certain subgroup of the Clifford algebra associated to a quadratic space. It maps 2-to-1 to the orthogonal group, just as the spin group maps 2-to-1 to the special orthogonal group.

In general the map from the Pin group to the orthogonal group is not onto or a universal covering space, but if the quadratic form is definite, it is both.

Definite form

The pin group of a definite form maps onto the orthogonal group, and each component is simply connected: it double covers the orthogonal group. The pin groups for a positive definite quadratic form $Q$ and for its negative $-Q$ are not isomorphic, but the orthogonal groups are.

In terms of the standard forms, $O\left(n,0\right) = O\left(0,n\right)$, but $mbox\left\{Pin\right\}\left(n,0\right) notcong mbox\left\{Pin\right\}\left(0,n\right)$. Using the "+" sign convention for Clifford algebras (where $v^2=Q\left(v\right) in Cell\left(V,Q\right)$), one writes

$mbox\left\{Pin\right\}_+\left(n\right) := mbox\left\{Pin\right\}\left(n,0\right) qquad mbox\left\{Pin\right\}_-\left(n\right) := mbox\left\{Pin\right\}\left(0,n\right)$
and these both map onto $O\left(n\right) = O\left(n,0\right) = O\left(0,n\right)$.

By contrast, we have the isomorphism $mbox\left\{Spin\right\}\left(n,0\right) cong mbox\left\{Spin\right\}\left(0,n\right)$ and they are both the (unique) universal cover of the special orthogonal group SO(n).

As topological group

Every connected topological group has a unique universal cover as a topological space, which has a unique group structure as a central extension by the fundamental group. For a disconnected topological group, there is a unique universal cover of the identity component of the group, and one can take the same cover as topological spaces on the other components (which are principal homogeneous spaces for the identity component) but the group structure on other components is not uniquely determined in general.

The Pin and Spin groups are particular topological groups associated to the orthogonal and special orthogonal groups, coming from Clifford algebras: there are other similar groups, corresponding to other double covers or to other group structures on the other components, but they are not referred to as Pin or Spin groups, nor studied much.

Construction

The two pin groups correspond to the two central extensions
$1 to \left\{pm 1\right\} to mbox\left\{Pin\right\}_pm\left(V\right) to O\left(V\right) to 1$
The group structure on $mbox\left\{Spin\right\}\left(V\right)$ (the connected component of determinant 1) is already determined; the group structure on the other component is determined up to the center, and thus has a $pm 1$ ambiguity.

The two extensions are distinguished by whether the preimage of a reflection squares to $pm 1 in ker left\left(mbox\left\{Spin\right\}\left(V\right) to SO\left(V\right)right\right)$, and the two pin groups are named accordingly. Explicitly, a reflection has order 2 in $O\left(V\right)$, $r^2=1$, so the square of the preimage of a reflection (which has determinant one) must be in the kernel of $mbox\left\{Spin\right\}_pm\left(V\right) to SO\left(V\right)$, so $tilde r^2 = pm 1$, and either choice determines a pin group (since all reflections are conjugate by an element of $SO\left(V\right)$, which is connected, all reflections must square to the same value).

Concretely, in $mbox\left\{Pin\right\}_+$, $tilde r$ has order 2, and the preimage of a subgroup $\left\{1,r\right\}$ is $C_2 times C_2$: if one repeats the same reflection twice, one gets the identity.

In $mbox\left\{Pin\right\}_-$, $tilde r$ has order 4, and the preimage of a subgroup $\left\{1,r\right\}$ is $C_4$: if one repeats the same reflection twice, one gets "a rotation by 2π"—the non-trivial element of $mbox\left\{Spin\right\}\left(V\right) to SO\left(V\right)$ can be interpreted as "rotation by 2π" (every axis yields the same element).

Low dimensions

In 2 dimensions, the distinction between $mbox\left\{Pin\right\}_+$ and $mbox\left\{Pin\right\}_-$ mirrors the distinction between the dihedral group of a $2n$-gon and the dicyclic group of the cyclic group $C_\left\{2n\right\}$.

In $mbox\left\{Pin\right\}_+$, the preimage of the dihedral group of an $n$-gon, considered as a subgroup $mbox\left\{Dih\right\}_n < O\left(2\right)$, is the dihedral group of an $2n$-gon, $mbox\left\{Dih\right\}_\left\{2n\right\} < mbox\left\{Pin\right\}_+\left(2\right)$, while in $mbox\left\{Pin\right\}_-$, the preimage of the dihedral group is the dicyclic group $mbox\left\{Dic\right\}_n < mbox\left\{Pin\right\}_-\left(2\right)$.

In 1 dimension, the pin groups are congruent to the first dihedral and dicyclic groups:

begin\left\{align\right\}
mbox{Pin}_+(1) &cong C_2 times C_2 = mbox{Dih}_1 mbox{Pin}_-(1) &cong C_4 = mbox{Dic}_1 end{align}

Indefinite Pin groups

There are as many as eight different double covers of Spin(p,q), for $p,qneq 0$, which correspond to the extensions of the center (which is either $C_2 times C_2$ or $C_4$) by $C_2$. Only two of them are taken to be pin groups, namely, those which admit the Clifford algebra as a representation. They are called Pin(p,q) and Pin(q,p) respectively.

Name

The name was introduced in M.F. Atiyah, R. Bott, A. Shapiro: Clifford modules, Topology 3, suppl. 1 (1964), pp. 3-38, on page 3, line 17, where they state "This joke is due to J-P. Serre". It is a back-formation from Spin: "Pin is to O(n) as Spin is to SO(n)", hence dropping the "S" from "Spin" yields "Pin". Further, the word "Pin" sounds like vulgar French slang when pronounced in French, which is alluded to by the name originating with (or being attributed to) Serre.

References

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