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A table of the Cli¬ord groups Clp,q for 0 ¤ p, q ¤ 7 can be found in Kirillov [18], and for
0 ¤ p, q ¤ 8, in Lawson and Michelsohn [20] (but beware that their Clp,q is our Clq,p ). It can
also be shown that
Clp+1,q ≈ Clq+1,p
and that
Clp,q ≈ Cl0 ,

frow which it follows that
Spin(p, q) ≈ Spin(q, p).

1.7 The Complex Cli¬ord Algebras Cl(n, C)
One can also consider Cli¬ord algebras over the complex ¬eld C. In this case, it is well-known
that every nondegenerate quadratic form can be expressed by

¦C (x1 , . . . , xn ) = x2 + · · · + x2
n 1 n

in some orthonormal basis. Also, it is easily shown that the complexi¬cation C —R Clp,q of
the real Cli¬ord algebra Clp,q is isomorphic to Cl(¦C ). Thus, all these complex algebras are
isomorphic for p+q = n, and we denote them by Cl(n, C). Theorem 1.15 yields the following
periodicity theorem:

Theorem 1.18 The following isomorphisms hold:

Cl(n + 2, C) ≈ Cl(n, C) —C Cl(2, C),

with Cl(2, C) = C(2).

Proof . Since Cl(n, C) = C —R Cl0,n = C —R Cln,0 , by Lemma 1.15, we have

Cl(n + 2, C) = C —R Cl0,n+2 ≈ C —R (Cln,0 —R Cl0,2 ) ≈ (C —R Cln,0 ) —C (C —R Cl0,2 ).

However, Cl0,2 = H, Cl(n, C) = C —R Cln,0 , and C —R H ≈ C(2), so we get Cl(2, C) = C(2)
Cl(n + 2, C) ≈ Cl(n, C) —C C(2),
and the theorem is proved.
As a corollary of Theorem 1.18, we obtain the fact that

Cl(2k, C) ≈ C(2k ) and Cl(2k + 1, C) ≈ C(2k ) • C(2k ).

The table of the previous section can also be completed as follows:
n 0 1 2 3 4 5 6 7 8
H • H H(2) R(8) R(8) • R(8) R(16)
Cl0,n R C H C(4)
R R • R R(2) C(2) H(2) H(2) • H(2) H(4)
Cln,0 C(8) R(16)
Cl(n, C) C 2C C(2) 2C(2) C(4) 2C(4) 2C(8)
C(8) C(16).
where 2C(k) is an abbrevation for C(k) • C(k).

1.8 The Groups Pin(p, q) and Spin(p, q) as double covers
of O(p, q) and SO(p, q)
It turns out that the groups Pin(p, q) and Spin(p, q) have nice topological properties w.r.t.
the groups O(p, q) and SO(p, q). To explain this, we review the de¬nition of covering maps

and covering spaces (for details, see Fulton [14], Chapter 11). Another interesting source is
Chevalley [9], where is is proved that Spin(n) is a universal double cover of SO(n) for all
n ≥ 3.
Since Cp,q is an algebra of dimension 2p+q , it is a topological space as a vector space
p+q —
isomorphic to V = R2 . Now, the group Cp,q of units of Cp,q is open in Cp,q , because
x ∈ Cp,q is a unit if the linear map y ’ xy is an isomorphism, and GL(V ) is open in

End(V ), the space of endomorphisms of V . Thus, CP,q is a Lie group, and since Pin(p, q)

and Spin(p, q) are clearly closed subgroups of Cp,q , they are also Lie groups.

De¬nition 1.7 Given two topological spaces X and Y , a covering map is a continuous
surjective map, p: Y ’ X, with the property that for every x ∈ X, there is some open
subset, U ⊆ X, with x ∈ U , so that p’1 (U ) is the disjoint union of open subsets, V± ⊆ Y ,
and the restriction of p to each V± is a homeomorphism onto U . We say that U is evenly
covered by p. We also say that Y is a covering space of X. A covering map p: Y ’ X is
called trivial if X itself is evenly covered by p (i.e., Y is the disjoint union of open subsets
subsets Y± each homeomorphic to X), and nontrivial , otherwise. When each ¬ber, p’1 (x),
has the same ¬nite cardinaly n for all x ∈ X, we say that p is an n-covering (or n-sheeted

Note that a covering map, p: Y ’ X, is not always trivial, but always locally trivial (i.e.,
for every x ∈ X, it is trivial in some open neighborhood of x). A covering is trivial i¬ Y
is isomorphic to a product space of X — T , where T is any set with the discrete topology.
Also, if Y is connected, then the covering map is nontrivial.

De¬nition 1.8 An isomorphism • between covering maps p: Y ’ X and p : Y ’ X is a
homeomorphism, •: Y ’ Y , so that p = p —¦ •.

Typically, the space X is connected, in which case it can be shown that all the ¬bers
p’1 (x) has the same cardinality.
One of the most important properties of covering spaces is the path“lifting property, a
property that we will use to show that Spin(n) is path-connected.

Proposition 1.19 (Path lifting) Let p: Y ’ X be a covering map, and let γ: [a, b] ’ X
be any continuous curve from xa = γ(a) to xb = γ(b) in X. If y ∈ Y is any point so that
p(y) = xa , then there is a unique curve, γ: [a, b] ’ Y , so that y = γ(a) and
p —¦ γ(t) = γ(t) for all t ∈ [a, b].

Proof . See Fulton [15], Chapter 11, Lemma 11.6.
Many important covering maps arise from the action of a group G on a space Y . If Y
is a topological space, an action (on the left) of a group G on Y is a map ±: G — Y ’ Y
satisfying the following conditions, where, for simplicity of notation, we denote ±(g, y) by
g · y:

(1) g · (h · y) = (gh) · y, for all g, h ∈ G and y ∈ Y ;

(2) 1 · y = y, for all ∈ Y , where 1 is the identity of the group G;

(3) The map y ’ g · y is a homeomorphism of Y for every g ∈ G.

We de¬ne an equivalence relation on Y as follows: x ≡ y i¬ y = g · x for some g ∈ G
(check that this is an equivalence relation). The equivalence class G · x = {g · x | g ∈ G} of
any x ∈ Y is called the orbit of x. We obtain the quotient space Y /G and the projection
map p: Y ’ Y /G sending every y ∈ Y to its orbit. The space Y /G is given the quotient
topology (a subset U of Y /G is open i¬ p’1 (U ) is open in Y ).
Given a subset V of Y and any g ∈ G, we let

g · V = {g · y | y ∈ V }.

We say that G acts evenly on Y if for every y ∈ Y there is an open subset V containing y
so that g · V and h · V are disjoint for any two distinct elements g, h ∈ G.
The importance of the notion a group acting evenly is that such actions induce a covering

Proposition 1.20 If G is a group acting evenly on a space Y , then the projection map,
p: Y ’ Y /G, is a covering map.

Proof . See Fulton [15], Chapter 11, Lemma 11.17.
The following proposition shows that Pin(p, q) and Spin(p, q) are nontrivial covering
spaces unless p = q = 1.

Proposition 1.21 For all p, q ≥ 0, the groups Pin(p, q) and Spin(p, q) are double covers of
O(p, q) and SO(p, q), respectively. Furthermore, they are nontrivial covers unless p = q = 1.

Proof . We know that kernel of the homomorphism ρ: Pin(p, q) ’ O(p, q) is Z2 = {’1, 1}.
If we let Z2 act on Pin(p, q) in the natural way, then O(p, q) ≈ Pin(p, q)/Z2 , and the reader
can easily check that Z2 acts evenly. By Proposition 1.20, we get a double cover. The
argument for ρ: Spin(p, q) ’ SO(p, q) is similar.
Let us now assume that p = 1 and q = 1. In order to prove that we have nontrivial
covers, it is enough to show that ’1 and 1 are connected by a path in Pin(p, q) (If we had
Pin(p, q) = U1 ∪ U2 with U1 and U2 open, disjoint, and homeomorphic to O(p, q), then ’1
and 1 would not be in the same Ui , and so, they would be in disjoint connected components.
Thus, ’1 and 1 can™t be path“connected, and similarly with Spin(p, q) and SO(p, q).) Since
(p, q) = (1, 1), we can ¬nd two orthogonal vectors e1 and e2 so that ¦p,q (e1 ) = ¦p,q (e2 ) = ±1.
γ(t) = ± cos(2t) 1 + sin(2t) e1 e2 = (cos t e1 + sin t e2 )(sin t e2 ’ cos t e1 ),

for 0 ¤ t ¤ π, de¬nes a path in Spin(p, q), since

(± cos(2t) 1 + sin(2t) e1 e2 )’1 = ± cos(2t) 1 ’ sin(2t) e1 e2 ,

as desired.
In particular, if n ≥ 2, since the group SO(n) is path-connected, the group Spin(n) is
also path-connected. Indeed, given any two points xa and xb in Spin(n), there is a path
γ from ρ(xa ) to ρ(xb ) in SO(n) (where ρ: Spin(n) ’ SO(n) is the covering map). By
Proposition 1.19, there is a path γ in Spin(n) with origin xa and some origin xb so that
ρ(xb ) = ρ(xb ). However, ρ’1 (ρ(xb )) = {’xb , xb }, and so, xb = ±xb . The argument used in
the proof of Proposition 1.21 shows that xb and ’xb are path-connected, and so, there is a
path from xa to xb , and Spin(n) is path-connected. In fact, for n ≥ 3, it turns out that
Spin(n) is simply connected. Such a covering space is called a universal cover (for instance,
see Chevalley [9]).
This last fact requires more algebraic topology than we are willing to explain in detail,
and we only sketch the proof. The notions of ¬bre bundle, ¬bration, and homotopy sequence
associated with a ¬bration are needed in the proof. We refer the perseverant readers to Bott
and Tu [5] (Chapter 1 and Chapter 3, Sections 16“17) or Rotman [24] (Chapter 11) for a
detailed explanation of these concepts.
Recall that a topological space is simply connected if it is path connected and π1 (X) = (0),
which means that every closed path in X is homotopic to a point. Since we just proved that
Spin(n) is path connected for n ≥ 2, we just need to prove that π1 (Spin(n)) = (0) for all
n ≥ 3. The following facts are needed to prove the above assertion:
(1) The sphere S n’1 is simply connected for all n ≥ 3.
SU(2) is homeomorphic to S 3 , and thus, Spin(3) is simply
(2) The group Spin(3)
(3) The group Spin(n) acts on S n’1 in such a way that we have a ¬bre bundle with ¬bre
Spin(n ’ 1):
Spin(n ’ 1) ’’ Spin(n) ’’ S n’1 .

Fact (1) is a standard proposition of algebraic topology and a proof can found in many
books. A particularly elegant and yet simple argument consists in showing that any closed
curve on S n’1 is homotopic to a curve that omits some point. First, it is easy to see that
in Rn , any closed curve is homotopic to a piecewise linear curve (a polygonal curve), and
the radial projection of such a curve on S n’1 provides the desired curve. Then, we use the
stereographic projection of S n’1 from any point omitted by that curve to get another closed
curve in Rn’1 . Since Rn’1 is simply connected, that curve is homotopic to a point, and so is
its preimage curve on S n’1 . Another simple proof uses a special version of the Seifert”van
Kampen™s theorem (see Gramain [17]).
Fact (2) is easy to establish directly, using (1).

To prove (3), we let Spin(n) act on S n’1 via the standard action: x · v = xvx’1 . Because
SO(n) acts transitively on S n’1 and there is a surjection Spin(n) ’’ SO(n), the group
Spin(n) also acts transitively on S n’1 . Now, we have to show that the stabilizer of any
element of S n’1 is Spin(n ’ 1). For example, we can do this for e1 . This amounts to some
simple calculations taking into account the identities among basis elements. Details of this
proof can be found in Mneimn´ and Testard [22], Chapter 4. It is still necessary to prove that
Spin(n) is a ¬bre bundle over S n’1 with ¬bre Spin(n ’ 1). For this, we use the following
results whose proof can be found in Mneimn´ and Testard [22], Chapter 4:

Lemma 1.22 Given any topological group G, if H is a closed subgroup of G and the pro-
jection π: G ’ G/H has a local section at every point of G/H, then

H ’’ G ’’ G/H

is a ¬bre bundle with ¬bre H.

Lemma 1.22 implies the following key proposition:

Proposition 1.23 Given any linear Lie group G, if H is a closed subgroup of G, then

H ’’ G ’’ G/H

is a ¬bre bundle with ¬bre H.

Now, a ¬bre bundle is a ¬bration (as de¬ned in Bott and Tu [5], Chapter 3, Section 16,
or in Rotman [24], Chapter 11). For a proof of this fact, see Rotman [24], Chapter 11, or
Mneimn´ and Testard [22], Chapter 4. So, there is a homotopy sequence associated with
the ¬bration (Bott and Tu [5], Chapter 3, Section 17, or Rotman [24], Chapter 11, Theorem
11.48), and in particular, we have the exact sequence

π1 (Spin(n ’ 1)) ’’ π1 (Spin(n)) ’’ π1 (S n’1 ).

Since π1 (S n’1 ) = (0) for n ≥ 3, we get a surjection

π1 (Spin(n ’ 1)) ’’ π1 (Spin(n)),

and so, by induction and (2), we get

π1 (Spin(n)) ≈ π1 (Spin(3)) = (0),

proving that Spin(n) is simply connected for n ≥ 3.
We can also show that π1 (SO(n)) = Z/2Z for all n ≥ 3. For this, we use Theorem 1.11
and Proposition 1.21, which imply that Spin(n) is a ¬bre bundle over SO(n) with ¬bre
{’1, 1}, for n ≥ 2:
{’1, 1} ’’ Spin(n) ’’ SO(n).

Again, the homotopy sequence of the ¬bration exists, and in particular, we get the exact
π1 (Spin(n)) ’’ π1 (SO(n)) ’’ π0 ({’1, +1}) ’’ π0 (SO(n)).
Since π0 ({’1, +1}) = Z/2Z, π0 (SO(n)) = (0), and π1 (Spin(n)) = (0), when n ≥ 3, we get
the exact sequence
(0) ’’ π1 (SO(n)) ’’ Z/2Z ’’ (0),
and so, π1 (SO(n)) = Z/2Z. Therefore, SO(n) is not simply connected for n ≥ 3.

Remark: Of course, we have been rather cavalier in our presentation. Given a topological
space, X, the group π1 (X) is the fundamental group of X, i.e., the group of homotopy
classes of closed paths in X (under composition of loops). But π0 (X) is generally not a
group! Instead, π0 (X) is the set of path-connected components of X. However, when X is
a Lie group, π0 (X) is indeed a group. Also, we have to make sense of what it means for the
sequence to be exact. All this can be made rigorous (see Bott and Tu [5], Chapter 3, Section
17, or Rotman [24], Chapter 11).

1.9 More on the Topology of O(p, q) and SO(p, q)
It turns out that the topology of the group, O(p, q), is completely determined by the topology
of O(p) and O(q). This result can be obtained as a simple consequence of some standard
Lie group theory. The key notion is that of a pseudo-algebraic group.
Consider the group, GL(n, C), of invertible n — n matrices with complex coe¬cients. If
A = (akl ) is such a matrix, denote by xkl the real part (resp. ykl , the imaginary part) of akl
(so, akl = xkl + iykl ).

De¬nition 1.9 A subgroup, G, of GL(n, C) is pseudo-algebraic i¬ there is a ¬nite set of
polynomials in 2n2 variables with real coe¬cients, {Pi (X1 , . . . , Xn2 , Y1 , . . . , Yn2 )}t , so that

A = (xkl + iykl ) ∈ G i¬ Pi (x11 , . . . , xnn , y11 , . . . , ynn ) = 0, for i = 1, . . . , t.

Recall that if A is a complex n — n-matrix, its adjoint, A— , is de¬ned by A— = (A) .
Also, U(n) denotes the group of unitary matrices, i.e., those matrices A ∈ GL(n, C) so
that AA— = A— A = I, and H(n) denotes the vector space of Hermitian matrices, i.e., those
matrices A ∈ GL(n, C) so that A— = A. Then, we have the following theorem which is
essentially a re¬ned version of the polar decomposition of matrices:

Theorem 1.24 Let G be a pseudo-algebraic subgroup of GL(n, C) stable under adjunction
(i.e., we have A— ∈ G whenever A ∈ G). Then, there is some integer, d ∈ N, so that G is
homeomorphic to (G © U(n)) — Rd . Moreover, if g is the Lie algebra of G, the map
(U, H) ’ U eH ,
(U(n) © G) — (H(n) © g) ’’ G, given by
is a homeomorphism onto G.

Proof . A proof can be found in Knapp [19], Chapter 1, or Mneimn´ and Testard [22], Chapter
We now apply Theorem 1.24 to determine the structure of the space O(p, q). Let Jp,q be
the matrix
Ip 0
Jp,q = .
0 ’Iq
We know that O(p, q) consists of the matrices, A, in GL(p + q, R) such that

A Jp,q A = Jp,q ,

and so, O(p, q) is clearly pseudo-algebraic. Using the above equation, it is easy to determine
the Lie algebra, o(p, q), of O(p, q). We ¬nd that o(p, q) is given by

X1 X2
X1 = ’X1 , X3 = ’X3 , X2 arbitrary
o(p, q) =
X2 X3

where X1 is a p — p matrix, X3 is a q — q matrix and X2 is a p — q matrix. Consequently, it
immediately follows that

0 X2
o(p, q) © H(p + q) = X2 arbitrary ,
X2 0

a vector space of dimension pq.
Some simple calculations also show that

X1 0 ∼ O(p) — O(q).
O(p, q) © U(p + q) = X1 ∈ O(p), X2 ∈ O(q) =
0 X2

Therefore, we obtain the structure of O(p, q):

Proposition 1.25 The topological space O(p, q) is homeomorphic to O(p) — O(q) — Rpq .

Since O(p) has two connected components when p ≥ 1, we see that O(p, q) has four
connected components when p, q ≥ 1. It is also obvious that

X1 0
SO(p, q) © U(p + q) = X1 ∈ O(p), X2 ∈ O(q), det(X1 ) det(X2 ) = 1 .
0 X2

This is a subgroup of O(p) — O(q) that we denote S(O(p) — O(q)). Furthermore, it is easy
to show that so(p, q) = o(p, q). Thus, we also have

Proposition 1.26 The topological space SO(p, q) is homeomorphic to S(O(p)—O(q))—Rpq .

Note that SO(p, q) has two connected components when p, q ≥ 1. The connected
component of Ip+q is a group denoted SO0 (p, q). This latter space is homeomorphic to
SO(q) — SO(q) — Rpq .
As a closing remark observe that the dimension of all these space depends only on p + q:
It is (p + q)(p + q ’ 1)/2.

Acknowledgments. I thank Eric King whose incisive questions and relentless quest for the
“essence” of rotations eventually caused a level of discomfort high enough to force me to
improve the clarity of these notes. Rotations are elusive!

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