Clifford Algebras and Bott Periodicity — Clifford Algebras and Representations

1 Clifford Algebras and Representations

We begin by defining Clifford algebras.

Definition

Let VV be a vector space over R\mathbb {R} and Q:VRQ: V \to \mathbb {R} a quadratic form. The Clifford algebra C(V;Q)C(V; Q) is defined by the quotient

C(V;Q)=k=0Vk(xxQ(x)). C(V; Q) = \frac{\bigoplus _{k = 0}^\infty V^{\otimes k}}{(x \otimes x - Q(x))}.

Example

Let QkQ_ k be the standard negative-definite quadratic form on Rk\mathbb {R}^ k, so that

Q(x1,,xk)=xi2. Q(x_1, \ldots , x_ k) = - \sum x_ i^2.

We define Ck=C(Rk;Qk)C_ k = C(\mathbb {R}^ k; Q_ k) and Ck=C(Rk;Qk)C_ k' = C(\mathbb {R}^ k; -Q_ k). These are the Clifford algebras we are interested in.

Observe that our Clifford algebras are non-commutative algebras. For many purposes, it is convenient to treat them as Z2\mathbb {Z}_2-graded algebras 1 , with VV being in degree 11. In particular, it gives the correct tensor product.

Recall that if A,BA, B are Z2\mathbb {Z}_2-graded algebras, then the graded tensor product A^BA \mathbin {\hat\otimes }B is defined by the algebra with underlying vector space ABA \otimes B and multiplication

(xy)(vw)=(1)yvxvyw. (x \otimes y) \cdot (v \otimes w) = (-1)^{|y| |v|} xv \otimes yw.

Lemma

If V=V1V2V = V_1 \oplus V_2 and Q(x+y)=Q1(x)+Q2(y)Q(x + y) = Q_1(x) + Q_2(y) for all xV1x \in V_1 and yV2y \in V_2, then

C(V;Q)=C(V1;Q1)^C(V2;Q2). C(V; Q) = C(V_1; Q_1) \mathbin {\hat\otimes }C(V_2; Q_2).\fakeqed
Here the graded product is needed because we want

Q(a+b)=(a1+1b)2=a21+1b2=Q1(a)+Q2(b). Q(a + b) = (a \otimes 1 + 1 \otimes b)^2 = a^2 \otimes 1 + 1 \otimes b^2 = Q_1(a) + Q_2(b).

From this lemma, we immediately deduce that

Corollary

We have

Ck=C1^^C1k times. C_ k = \underbrace{C_1 \mathbin {\hat\otimes }\cdots \mathbin {\hat\otimes }C_1}_{k\text{ times}}.

Explicitly, CkC_ k is generated by e1,,eke_1, \ldots , e_ k subject to the relations

ei2=1,eiej+ejei=0. e_ i^2 = -1,\quad e_ i e_ j + e_ j e_ i = 0.

In particular, dimCk=2k\dim C_ k = 2^ k, and the dimension in each degree is 2k12^{k - 1}. Similar statements can be made for CkC_ k'.

Ultimately, we wish to understand the representations of CkC_ k. To do so, it is convenient to explicitly identify the Clifford algebras CkC_ k. While graded tensor products are the “morally correct” things to consider, they are not computationally helpful. Instead, we want to rephrase everything in terms of ordinary tensor products. We have the following result:

Proposition

We have

Ck+2CkRC2Ck+2CkRC2\begin{aligned} C_{k + 2} & \cong C_ k' \otimes _\mathbb {R}C_2\\ C_{k + 2}' & \cong C_ k \otimes _\mathbb {R}C_2' \end{aligned}

Proof
Let e1,,eke_1', \ldots , e_ k' be the generators of CkC_ k'. Consider the map of algebras Ck+2CkRC2C_{k + 2}' \to C_ k \otimes _\mathbb {R}C_2' defined by e11e1e21e2eiei2e1e2 for i3.\begin{aligned} e_1' & \mapsto 1 \otimes e_1'\\ e_2' & \mapsto 1 \otimes e_2'\\ e_ i' & \mapsto e_{i - 2} \otimes e_1' e_2'\text{ for }i \geq 3. \end{aligned} One checks that this is well-defined, and surjects onto the generators of CkRC2C_ k \otimes _\mathbb {R}C_2'. So this map is surjective, and since both sides have the same dimension, this is an isomorphism. The other isomorphism is obtained by swapping eie_ i' with eie_ i.

Using this, we are now in a position to explicitly describe the Clifford algebras inductively. We first identify the Clifford algebras in low dimensions explicitly.

Notation

For any algebra RR, we write R(n)R(n) for the algebra of n×nn \times n matrices with entries in RR.

Proposition
C1CC1RRC2HC2R(2).\begin{aligned} C_1 & \cong \mathbb {C}& C_1' \cong \mathbb {R}\oplus \mathbb {R}\\ C_2 & \cong \mathbb {H}& C_2' \cong \mathbb {R}(2). \end{aligned}

Proof
The descriptions of C1C_1 and C2C_2 are essentially by definition. The identification of C1C_1' is obtained by picking basis vectors 1±e12\frac{1 \pm e_1'}{2}. The identification of C2C_2' is obtained by setting e1=12(1111),e2=12(1111). e_1' = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},\quad e_2' = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}.\qedhere

We will also need the following observations:

Lemma

We have

CRCCC \mathbb {C}\otimes _\mathbb {R}\mathbb {C}\cong \mathbb {C}\oplus \mathbb {C}

and

HRCEndC(H)C(2)HRHEndR(H)R(4)pz(qzqpˉ)pr(qpqrˉ)\begin{aligned} \mathbb {H}\otimes _\mathbb {R}\mathbb {C}& \cong \operatorname{End}_\mathbb {C}(\mathbb {H}) \cong \mathbb {C}(2) & \mathbb {H}\otimes _\mathbb {R}\mathbb {H}& \cong \operatorname{End}_\mathbb {R}(\mathbb {H}) \cong \mathbb {R}(4)\\ p \otimes z & \mapsto (q \mapsto zq\bar{p}) & p \otimes r & \mapsto (q \mapsto p q \bar{r})\fakeqed \end{aligned}

We can then write down the table of Clifford algebras

kk

CkC_ k

CkC_ k'

1

C\mathbb {C}

RR\mathbb {R}\oplus \mathbb {R}

2

H\mathbb {H}

R(2)\mathbb {R}(2)

3

HH\mathbb {H}\oplus \mathbb {H}

C(2)\mathbb {C}(2)

4

H(2)\mathbb {H}(2)

H(2)\mathbb {H}(2)

5

C(4)\mathbb {C}(4)

H(2)H(2)\mathbb {H}(2) \oplus \mathbb {H}(2)

6

R(8)\mathbb {R}(8)

H(4)\mathbb {H}(4)

7

R(8)R(8)\mathbb {R}(8) \oplus \mathbb {R}(8)

C(8)\mathbb {C}(8)

8

R(16)\mathbb {R}(16)

R(16)\mathbb {R}(16)

Moreover, Ck+8=CkRR(16)C_{k + 8} = C_ k \otimes _\mathbb {R}\mathbb {R}(16).

Note that all our Clifford algebras are semi-simple, and most of them are in fact simple. One might think the representations are therefore quite uninteresting, but that would be wrong.

Definition

Let AA be an Z2\mathbb {Z}_2-graded algebra. We define M(A)M(A) to be the Grothendieck group of graded representations of AA, and N(A)N(A) be the Grothendieck group of ungraded representations of AA. We also write A0A^0 for the degree zero part of AA.

Lemma

We have an isomorphism of groups

\begin{useimager} 
    \begin{align*}
      M(C_k) &\cong N(C_k^0)\\
      M^0 \oplus M^1 &\mapsto M^0\\
      C_k \otimes_{C_k^0} M &\leftmapsto M.\qedhere
    \end{align*}
  \end{useimager}

Moreover, we can also easily identify Ck0C_ k^0:

Lemma

We have an isomorphism of algebras

ϕ:Ck1Ck0eieiek.\begin{aligned} \phi : C_{k - 1} & \cong C_ k^0\\ e_ i & \mapsto e_ i e_ k. \end{aligned}

This immediately allows us to write down the Grothendieck group of representations of CkC_ k, as well as the dimensions of the representations.

kk

CkC_ k

M(Ck)M(C_ k)

dimR(M0)\dim _\mathbb {R}(M^0)

1

C\mathbb {C}

Z\mathbb {Z}

11

2

H\mathbb {H}

Z\mathbb {Z}

22

3

HH\mathbb {H}\oplus \mathbb {H}

Z\mathbb {Z}

44

4

H(2)\mathbb {H}(2)

ZZ\mathbb {Z}\oplus \mathbb {Z}

44

5

C(4)\mathbb {C}(4)

Z\mathbb {Z}

88

6

R(8)\mathbb {R}(8)

Z\mathbb {Z}

88

7

R(8)R(8)\mathbb {R}(8) \oplus \mathbb {R}(8)

Z\mathbb {Z}

88

8

R(16)\mathbb {R}(16)

ZZ\mathbb {Z}\oplus \mathbb {Z}

88

Now there is always an inclusion i:CkCk+1i: C_ k \to C_{k + 1}, which induces a map i:M(Ck+1)M(Ck)i^*: M(C_{k + 1}) \to M(C_ k).

Definition

We define Ak=M(Ck)/iM(Ck+1)A_ k = M(C_ k) / i^* M(C_{k + 1}).

When k̸=4nk \not= 4n, computing AkA_ k is trivial, since there is only one irreducible representation. Hence we know what the image of ii^* is simply by looking at the dimension of the pulled back representation. We claim that A4n=ZA_{4n} = \mathbb {Z} for all nn.

One can prove this using some more direct arguments, but it is convenient for us to prove this by understanding the module structure better. Given a graded CkC_ k module M=M0M1M = M^0 \oplus M^1, we can obtain a new CkC_ k module M=M1M0M^* = M^1 \oplus M^0. If k̸=4nk \not= 4n, then for purely dimensional reasons, we must have MMM \cong M^*.

Proposition

Let k=4nk = 4n, and let MM and NN be the irreducible CkC_ k-modules. Then M=NM = N^* and N=MN = M^*.

Proof

We wish to reduce this problem to a problem of ungraded modules.

Note that in general, for an algebra AA and an automorphism α\alpha , if MM is an AA-module, then we can construct another AA-module MαM^\alpha whose underlying set is the same, but the action is given by

xm=α(x)m. x \cdot m = \alpha (x)m.

With this in mind, consider the automorphisms

α:CkCkβ:Ck1Ck1xekxek1x(1)xx.\begin{aligned} \alpha : C_ k & \mapsto C_ k & \beta : C_{k - 1} & \mapsto C_{k - 1}\\ x & \mapsto e_ k x e_ k^{-1} & x & \mapsto (-1)^{|x|} x. \end{aligned}

Check that these are related by the commutative diagram

\begin{useimager} 
    \[
      \begin{tikzcd}
        C_{k - 1} \ar[r, "\phi"] \ar[d, "\beta"] & C_k \ar[d, "\alpha"]\\
        C_{k - 1} \ar[r, "\phi"] & C_k.
      \end{tikzcd}
    \]
  \end{useimager}

Since multiplication by eke_ k gives an isomorphism of vector spaces M0M1M^0 \cong M^1, we know that MMαM^* \cong M^{\alpha }. Hence under the correspondence M(Ck)N(Ck1)M(C_ k) \cong N(C_{k - 1}), the operation MMM \mapsto M^* corresponds to M0(M0)βM^0 \mapsto (M^0)^{\beta }.

Thus, it suffices to show that β\beta swaps the two irreducible ungraded representations of Ck1C_{k - 1}. Since Ck1Ck2Ck2C_{k - 1} \cong C_{k - 2} \oplus C_{k - 2}, we want to find the central idempotents that project to the two ideals, and show that β\beta swaps the two. We can just write this down.

Set

w=e1e2e4n1. w = e_1 e_2 \cdots e_{4n - 1}.

Then 11 and ww are in the center, and w2=1w^2 = 1. So the desired idempotents are 1±w2\frac{1 \pm w}{2}, and β(w)=w\beta (w) = -w.

Corollary

A4nZA_{4n} \cong \mathbb {Z}.

Proof
Let x,yx, y be the two irreducible modules in M(C4n)M(C_{4n}), and zz the irreducible module in M(C4n+1)M(C_{4n + 1}). Then z=zz^* = z, so (iz)=iz=iz(i^*z)^* = i^*z^* = i^*z. So we must have i(z)=x+yi(z) = x + y.

This ww will be useful later on. Observe that, when restricted to degree 00 components, ww acts as 11 on one of the irreducible representations and 1-1 on the other. So if we are given some representation of C4nC_{4n}, if we want to know how many of each irrep we have got, we simply have to take the degree 00 component and count the eigenvectors of ϕ(w)\phi (w).

We can now tabulate

kk

CkC_ k

M(Ck)M(C_ k)

dimR(M0)\dim _\mathbb {R}(M^0)

AkA_ k

1

C\mathbb {C}

Z\mathbb {Z}

11

Z2\mathbb {Z}_2

2

H\mathbb {H}

Z\mathbb {Z}

22

Z2\mathbb {Z}_2

3

HH\mathbb {H}\oplus \mathbb {H}

Z\mathbb {Z}

44

00

4

H(2)\mathbb {H}(2)

ZZ\mathbb {Z}\oplus \mathbb {Z}

44

Z\mathbb {Z}

5

C(4)\mathbb {C}(4)

Z\mathbb {Z}

88

00

6

R(8)\mathbb {R}(8)

Z\mathbb {Z}

88

00

7

R(8)R(8)\mathbb {R}(8) \oplus \mathbb {R}(8)

Z\mathbb {Z}

88

00

8

R(16)\mathbb {R}(16)

ZZ\mathbb {Z}\oplus \mathbb {Z}

88

Z\mathbb {Z}

Note that we have Ck+8=CkRR(16)C_{k + 8} = C_ k \otimes _\mathbb {R}\mathbb {R}(16), so we see that AkA_ k is 88-periodic. In fact, more is true.

Observe that if MM(Ck)M \in M(C_ k) and NM(Cm)N \in M(C_ m), then since Ck+mCk^CmC_{k + m} \cong C_ k \mathbin {\hat\otimes }C_ m, we can view M^NM \mathbin {\hat\otimes }N as an Ck+mC_{k + m} module. Then this turns M=M(C)M_* = M(C_*) into a ring, and since the image of i:MMi_*: M_* \to M_* is an ideal (MiN=i(MN)M \cdot i^* N = i^* (M \cdot N)), this implies AA_* is a ring as well.

Theorem (Bott Periodicity)

Let λ\lambda be the irreducible module of C8C_8. Then multiplication by λ\lambda induces an isomorphism M(Ck)M(Ck+8)M(C_ k) \to M(C_{k + 8}), hence an isomorphism AkAk+8A_ k \cong A_{k + 8}.

Proof
This is trivial except in the case k=4nk = 4n, by dimension counting. If M(C4n)M(C_{4n}) is generated by the irreps x,yx, y, then λx\lambda \cdot x is one of the irreps in M(C4n+8)M(C_{4n + 8}), and then λy=λx=(λx) \lambda \cdot y = \lambda \cdot x^* = (\lambda \cdot x)^* is the other irrep.

This is interesting, since we observe that we have AkKOk()A_ k \cong KO^{-k}(*), and after a bit of checking, this is in fact an isomorphism of rings! Of course, it would be terrible to just observe that we know what both sides are, and that they are the same. We should prove that they are the same without knowledge of KOk()KO^{-k}(*), and this would give us both Bott periodicity, as well as explicit computations of KOk()KO^{-k}(*).