
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 surjective or a universal covering space, but if the quadratic form is definite (and dimension is greater than 2), it is both.
The non-trivial element of the kernel is denoted
-1,
-I.
Let
V
Q
\operatorname{Pin}(V,Q)
\operatorname{Cl}(V,Q)
v1v2 … vk
vi
Q(vi)=\pm1
\operatorname{Spin}(V,Q)
k
In this article,
V
V
e1,...,ep+q
| 2, | |
| e | |
| 1 |
...,
| 2 | |
| e | |
| p |
=1
| 2, | |
| e | |
| p+1 |
...,
| 2 | |
| e | |
| p+q |
=-1,
Geometrically, for vectors
v
Q(v)\ne0
-vwv-1
w
v
v1v2 … vk
w
(-1)kv1v2 … vkw(v1v2 …
| -1 | |
| v | |
| k) |
I\inO(p,q)
O(p,q)
\operatorname{Pin}(p,q)
The pin group of a definite form maps onto the orthogonal group, and each component is simply connected (in dimension 3 and higher): 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.[1]
In terms of the standard forms, O(n, 0) = O(0, n), but Pin(n, 0) and Pin(0, n) are in general not isomorphic. Using the "+" sign convention for Clifford algebras (where
v2=Q(v)\inCl(V,Q)
Pin+(n):=Pin(n,0) Pin-(n):=Pin(0,n)
By contrast, we have the natural isomorphism[2] Spin(n, 0) ≅ Spin(0, n) and they are both the (unique) non-trivial double cover of the special orthogonal group SO(n), which is the (unique) universal cover for n ≥ 3.
Depending on how one counts, and the values of q and p, there may be as many as thirty-two different double covers of O(p, q), for p, q ≠ 0. Only two of them are the standard pin groups—those that are contained within the standard Clifford algebra constructions. They are called Pin(p, q) and Pin(q, p) respectively. Not all of these extensions are 'spinoral': one of these double covers, for example, is the trivial extension, which topologically is merely two copies of O(q, p).
It is worth pointing out for the indefinite case that while the group (in particular here, the component connected to the identity) Spin(p, q) is by definition a double cover of the special orthogonal group SO(p, q), this double covering is unique only in the compact and Lorentzian cases SO(n) and SO(1, q) with q>2. For SO(p, q) when both p and q are greater than two, the group has 3 unique connected double covers (only one of which is the spin group). Its universal cover is none of these, and is a non-trivial extension of the special orthogonal group by the Klein group, and so is a four fold covering of SO(p, q). The case of p or q being 2 is complicated (but well understood) by the infinite homotopy group of SO(2).
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.
In 2001, Andrzej Trautman[3] found the set of all 32 inequivalent double covers of O(p) x O(q), the maximal compact subgroup of O(p, q) and an explicit construction of 8 double covers of the same group O(p, q).
The two pin groups correspond to the two central extensions
1\to\{\pm1\}\toPin\pm(V)\toO(V)\to1.
The two extensions are distinguished by whether the preimage of a reflection squares to ±1 ∈ Ker (Spin(V) → SO(V)), and the two pin groups are named accordingly. Explicitly, a reflection has order 2 in O(V), r2 = 1, so the square of the preimage of a reflection (which has determinant one) must be in the kernel of Spin±(V) → SO(V), so
\tilder2=\pm1
Concretely, in Pin+,
\tilder
In Pin−,
\tilder
In 1 dimension, the pin groups are congruent to the first dihedral and dicyclic groups:
\begin{align} Pin+(1)&\congC2 x C2=Dih1\\ Pin-(1)&\congC4=Dic1. \end{align}
In 2 dimensions, the distinction between Pin+ and Pin− mirrors the distinction between the dihedral group of a 2n-gon and the dicyclic group of the cyclic group C2n.
In Pin+, the preimage of the dihedral group of an n-gon, considered as a subgroup Dihn < O(2), is the dihedral group of a 2n-gon, Dih2n < Pin+(2), while in Pin−, the preimage of the dihedral group is the dicyclic group Dicn < Pin−(2).
The resulting commutative square of subgroups for Spin(2), Pin+(2), SO(2), O(2) – namely C2n, Dih2n, Cn, Dihn – is also obtained using the projective orthogonal group (going down from O by a 2-fold quotient, instead of up by a 2-fold cover) in the square SO(2), O(2), PSO(2), PO(2), though in this case it is also realized geometrically, as "the projectivization of a 2n-gon in the circle is an n-gon in the projective line".
In 3 dimensions the situation is as follows. The Clifford algebra generated by 3 anticommuting square roots of +1 is the algebra of 2×2 complex matrices, and Pin+(3) is isomorphic to
\{A\inU(2):\detA=\pm1\}
H ⊕ H
For real (as opposed to complex) compact Pin groups, the center depends upon the dimension
n
n=2k
Pin
C2
n=4k+1
Pin+(4k+1)
C2 x C2
Pin-(4k+1)
C4
n=4k+3
Pin+(4k+3)
C4
Pin-(4k+3)
C2 x C2
In the indefinite case for
Pin(p,q)
n=p+q
(p-q)\operatorname{mod}8
(p-q)\operatorname{mod}2=0
C2
p-q\operatorname{mod}8=4k+1
C2 x C2
p-q\operatorname{mod}8=4k+3
C4
C4
C2 x C2
Though it may seem the center should always be order four, as the orthogonal group always has center of order two, those Pin group elements which are mapped to the center of the orthogonal group need only commute with everything up to a sign, as the Pin group serves as a projective representation of the orthogonal group. That is to say, pre-images of the center of the orthogonal group, need only commute or anti-commute with every element of the Pin group.
The name was introduced in, 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".
Cl(n,0)\not\congCl(0,n)
Cl(n,0)=Cl(0,n)=Λ*Rn