10 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Classification of Clifford algebras | 5/7 | https://en.wikipedia.org/wiki/Classification_of_Clifford_algebras | reference | science, encyclopedia | 2026-05-05T09:08:16.280956+00:00 | kb-cron |
Cl
1
,
0
(
R
)
≅
R
⊕
R
,
Cl
0
,
1
(
R
)
≅
C
,
{\displaystyle \operatorname {Cl} _{1,0}(\mathbf {R} )\cong \mathbf {R} \oplus \mathbf {R} ,\qquad \operatorname {Cl} _{0,1}(\mathbf {R} )\cong \mathbf {C} ,}
while
Cl
2
,
0
(
R
)
≅
M
2
(
R
)
,
Cl
0
,
2
(
R
)
≅
H
.
{\displaystyle \operatorname {Cl} _{2,0}(\mathbf {R} )\cong M_{2}(\mathbf {R} ),\qquad \operatorname {Cl} _{0,2}(\mathbf {R} )\cong \mathbf {H} .}
So the failure of symmetry appears already in the first few low-dimensional cases. There are two different mechanisms behind this asymmetry. In odd dimension, the distinction is visible in the center. If p − q ≡ 1 (mod 4), then ω2 = +1 and the algebra splits as a direct sum of two simple ideals, so its center is R⊕R. If instead p − q ≡ −1 (mod 4), then ω2 = −1 and the center is C. Thus swapping p and q can change the center from split real to complex. In even dimension, both algebras are central simple, so the distinction is instead in their Brauer classes. For example, when p − q ≡ 2 (mod 8) the algebra is a split matrix algebra over R, while when q − p ≡ 2 (mod 8)—equivalently p − q ≡ 6 (mod 8)—the algebra is a matrix algebra over H. So swapping p and q can also change a split algebra into a quaternionic one. Equivalently, one has
Cl
p
,
q
(
R
)
≅
Cl
q
,
p
(
R
)
{\displaystyle \operatorname {Cl} _{p,q}(\mathbf {R} )\cong \operatorname {Cl} _{q,p}(\mathbf {R} )}
if and only if p − q ≡ 0 or 4 (mod 8). This is simply the fixed-point condition for the involution
d
↦
−
d
{\displaystyle d\mapsto -d}
on the real classification table. This is one reason sign conventions matter in the literature: authors using the opposite convention for Clifford multiplication often write
Cl
p
,
q
{\displaystyle \operatorname {Cl} _{p,q}}
for what this article denotes by
Cl
q
,
p
{\displaystyle \operatorname {Cl} _{q,p}}
. The non-symmetry under
(
p
,
q
)
↦
(
q
,
p
)
{\displaystyle (p,q)\mapsto (q,p)}
is a property of real Clifford algebras, not just a notational artefact. This asymmetry belongs to the full Clifford algebra, not to the spin group. Let (V,q) be a real quadratic space. The spin group is defined inside the even Clifford algebra by
Spin
(
V
,
q
)
=
Pin
(
V
,
q
)
∩
Cl
0
(
V
,
q
)
,
{\displaystyle \operatorname {Spin} (V,q)=\operatorname {Pin} (V,q)\cap \operatorname {Cl} ^{0}(V,q),}
where Pin(V,q) is generated by the unit vectors v with q(v)=± 1. Under the standard twisted-adjoint action, such a vector acts on V by reflection in the hyperplane orthogonal to v, so products of an even number of unit vectors act by orientation-preserving orthogonal transformations. Now replacing q by −q does not change the orthogonal group: the same linear maps preserve q and −q, so O(V,q)=O(V,−q) and hence SO(V,q)=SO(V,−q). This is why one has SO(p,q) ≅ SO(q,p) and correspondingly Spin(p,q) ≅ Spin(q,p). The point is that although the Clifford algebras need not be the same, the spin group is built from even products of the same reflections, inside the even Clifford algebra. The lowest-dimensional examples already show the distinction. In the sign convention used in this article,
Cl
1
,
0
(
R
)
≅
R
⊕
R
,
Cl
0
,
1
(
R
)
≅
C
,
{\displaystyle \operatorname {Cl} _{1,0}(\mathbf {R} )\cong \mathbf {R} \oplus \mathbf {R} ,\qquad \operatorname {Cl} _{0,1}(\mathbf {R} )\cong \mathbf {C} ,}
so the full algebras are different, but in both cases the even subalgebra is just R. Hence
Spin
(
1
,
0
)
≅
Spin
(
0
,
1
)
≅
{
±
1
}
.
{\displaystyle \operatorname {Spin} (1,0)\cong \operatorname {Spin} (0,1)\cong \{\pm 1\}.}
A more instructive example is
Cl
2
,
0
(
R
)
≅
M
2
(
R
)
,
Cl
0
,
2
(
R
)
≅
H
.
{\displaystyle \operatorname {Cl} _{2,0}(\mathbf {R} )\cong M_{2}(\mathbf {R} ),\qquad \operatorname {Cl} _{0,2}(\mathbf {R} )\cong \mathbf {H} .}
Here the full algebras, and therefore their irreducible real modules, are of different types: in the first case the irreducible module is real 2-dimensional, whereas in the second it is quaternionic 1-dimensional. But the spin group only sees the even subalgebra. In both signatures the even subalgebra is generated by 1 and the bivector e1e2, and
(
e
1
e
2
)
2
=
−
e
1
2
e
2
2
=
−
1.
{\displaystyle (e_{1}e_{2})^{2}=-e_{1}^{2}e_{2}^{2}=-1.}
Therefore
Cl
2
,
0
0
(
R
)
≅
Cl
0
,
2
0
(
R
)
≅
C
,
{\displaystyle \operatorname {Cl} _{2,0}^{0}(\mathbf {R} )\cong \operatorname {Cl} _{0,2}^{0}(\mathbf {R} )\cong \mathbf {C} ,}
and in either case the spin group is the circle group
{
cos
θ
+
sin
θ
e
1
e
2
:
θ
∈
R
}
≅
U
(
1
)
.
{\displaystyle \{\cos \theta +\sin \theta \,e_{1}e_{2}:\theta \in \mathbf {R} \}\cong U(1).}
So the full Clifford algebra can distinguish real and quaternionic module types even when the associated spin group cannot: after passing to the even subalgebra, both cases are governed by the same complex structure. The same phenomenon persists in higher dimensions. For example, although
Cl
1
,
3
(
R
)
{\displaystyle \operatorname {Cl} _{1,3}(\mathbf {R} )}
and
Cl
3
,
1
(
R
)
{\displaystyle \operatorname {Cl} _{3,1}(\mathbf {R} )}
are different entries in the real classification table, the associated spin groups are both the double cover of the Lorentz group; in particular
Spin
(
1
,
3
)
≅
SL
2
(
C
)
,
{\displaystyle \operatorname {Spin} (1,3)\cong \operatorname {SL} _{2}(\mathbf {C} ),}