Clifford Algebras and Bott PeriodicityClifford Algebras and Representations

# 1 Clifford Algebras and Representations

We begin by defining Clifford algebras.

Definition

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

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

Example

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

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

We define $C_ k = C(\mathbb {R}^ k; Q_ k)$ and $C_ k' = C(\mathbb {R}^ k; -Q_ k)$. 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 $\mathbb {Z}_2$-graded algebras 1 , with $V$ being in degree $1$. In particular, it gives the correct tensor product.

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

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

Lemma

If $V = V_1 \oplus V_2$ and $Q(x + y) = Q_1(x) + Q_2(y)$ for all $x \in V_1$ and $y \in V_2$, then

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

$Q(a + b) = (a \otimes 1 + 1 \otimes b)^2 = a^2 \otimes 1 + 1 \otimes b^2 = Q_1(a) + Q_2(b).$

From this lemma, we immediately deduce that

Corollary

We have

$C_ k = \underbrace{C_1 \mathbin {\hat\otimes }\cdots \mathbin {\hat\otimes }C_1}_{k\text{ times}}.$

Explicitly, $C_ k$ is generated by $e_1, \ldots , e_ k$ subject to the relations

$e_ i^2 = -1,\quad e_ i e_ j + e_ j e_ i = 0.$

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

Ultimately, we wish to understand the representations of $C_ k$. To do so, it is convenient to explicitly identify the Clifford algebras $C_ k$. 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

\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 $e_1', \ldots , e_ k'$ be the generators of $C_ k'$. Consider the map of algebras $C_{k + 2}' \to C_ k \otimes _\mathbb {R}C_2'$ defined by \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 $C_ k \otimes _\mathbb {R}C_2'$. So this map is surjective, and since both sides have the same dimension, this is an isomorphism. The other isomorphism is obtained by swapping $e_ i'$ with $e_ i$.

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 $R$, we write $R(n)$ for the algebra of $n \times n$ matrices with entries in $R$.

Proposition
\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 $C_1$ and $C_2$ are essentially by definition. The identification of $C_1'$ is obtained by picking basis vectors $\frac{1 \pm e_1'}{2}$. The identification of $C_2'$ is obtained by setting $e_1' = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},\quad e_2' = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}.\qedhere$

We will also need the following observations:

Lemma

We have

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

and

\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

 $k$ $C_ k$ $C_ k'$ 1 $\mathbb {C}$ $\mathbb {R}\oplus \mathbb {R}$ 2 $\mathbb {H}$ $\mathbb {R}(2)$ 3 $\mathbb {H}\oplus \mathbb {H}$ $\mathbb {C}(2)$ 4 $\mathbb {H}(2)$ $\mathbb {H}(2)$ 5 $\mathbb {C}(4)$ $\mathbb {H}(2) \oplus \mathbb {H}(2)$ 6 $\mathbb {R}(8)$ $\mathbb {H}(4)$ 7 $\mathbb {R}(8) \oplus \mathbb {R}(8)$ $\mathbb {C}(8)$ 8 $\mathbb {R}(16)$ $\mathbb {R}(16)$

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

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

Lemma

We have an isomorphism of groups

Moreover, we can also easily identify $C_ k^0$:

Lemma

We have an isomorphism of algebras

\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 $C_ k$, as well as the dimensions of the representations.

 $k$ $C_ k$ $M(C_ k)$ $\dim _\mathbb {R}(M^0)$ 1 $\mathbb {C}$ $\mathbb {Z}$ $1$ 2 $\mathbb {H}$ $\mathbb {Z}$ $2$ 3 $\mathbb {H}\oplus \mathbb {H}$ $\mathbb {Z}$ $4$ 4 $\mathbb {H}(2)$ $\mathbb {Z}\oplus \mathbb {Z}$ $4$ 5 $\mathbb {C}(4)$ $\mathbb {Z}$ $8$ 6 $\mathbb {R}(8)$ $\mathbb {Z}$ $8$ 7 $\mathbb {R}(8) \oplus \mathbb {R}(8)$ $\mathbb {Z}$ $8$ 8 $\mathbb {R}(16)$ $\mathbb {Z}\oplus \mathbb {Z}$ $8$

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

Definition

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

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

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 $C_ k$ module $M = M^0 \oplus M^1$, we can obtain a new $C_ k$ module $M^* = M^1 \oplus M^0$. If $k \not= 4n$, then for purely dimensional reasons, we must have $M \cong M^*$.

Proposition

Let $k = 4n$, and let $M$ and $N$ be the irreducible $C_ k$-modules. Then $M = N^*$ and $N = M^*$.

Proof

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

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

$x \cdot m = \alpha (x)m.$

With this in mind, consider the automorphisms

\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

Since multiplication by $e_ k$ gives an isomorphism of vector spaces $M^0 \cong M^1$, we know that $M^* \cong M^{\alpha }$. Hence under the correspondence $M(C_ k) \cong N(C_{k - 1})$, the operation $M \mapsto M^*$ corresponds to $M^0 \mapsto (M^0)^{\beta }$.

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

Set

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

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

Corollary

$A_{4n} \cong \mathbb {Z}$.

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

This $w$ will be useful later on. Observe that, when restricted to degree $0$ components, $w$ acts as $1$ on one of the irreducible representations and $-1$ on the other. So if we are given some representation of $C_{4n}$, if we want to know how many of each irrep we have got, we simply have to take the degree $0$ component and count the eigenvectors of $\phi (w)$.

We can now tabulate

 $k$ $C_ k$ $M(C_ k)$ $\dim _\mathbb {R}(M^0)$ $A_ k$ 1 $\mathbb {C}$ $\mathbb {Z}$ $1$ $\mathbb {Z}_2$ 2 $\mathbb {H}$ $\mathbb {Z}$ $2$ $\mathbb {Z}_2$ 3 $\mathbb {H}\oplus \mathbb {H}$ $\mathbb {Z}$ $4$ $0$ 4 $\mathbb {H}(2)$ $\mathbb {Z}\oplus \mathbb {Z}$ $4$ $\mathbb {Z}$ 5 $\mathbb {C}(4)$ $\mathbb {Z}$ $8$ $0$ 6 $\mathbb {R}(8)$ $\mathbb {Z}$ $8$ $0$ 7 $\mathbb {R}(8) \oplus \mathbb {R}(8)$ $\mathbb {Z}$ $8$ $0$ 8 $\mathbb {R}(16)$ $\mathbb {Z}\oplus \mathbb {Z}$ $8$ $\mathbb {Z}$

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

Observe that if $M \in M(C_ k)$ and $N \in M(C_ m)$, then since $C_{k + m} \cong C_ k \mathbin {\hat\otimes }C_ m$, we can view $M \mathbin {\hat\otimes }N$ as an $C_{k + m}$ module. Then this turns $M_* = M(C_*)$ into a ring, and since the image of $i_*: M_* \to M_*$ is an ideal ($M \cdot i^* N = i^* (M \cdot N)$), this implies $A_*$ is a ring as well.

Theorem (Bott Periodicity)

Let $\lambda$ be the irreducible module of $C_8$. Then multiplication by $\lambda$ induces an isomorphism $M(C_ k) \to M(C_{k + 8})$, hence an isomorphism $A_ k \cong A_{k + 8}$.

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

This is interesting, since we observe that we have $A_ k \cong KO^{-k}(*)$, 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 $KO^{-k}(*)$, and this would give us both Bott periodicity, as well as explicit computations of $KO^{-k}(*)$.