Clifford algebra

Given a field $k$. Consider a $k$-vector field $V$ endowed with a quadratic form $q$. Recall that a quadratic form $q$ is a map

$$q:V\rightarrow k$$

such that $q(\lambda v)=\lambda^2q(v)$, for all $\lambda\in k$ and $v\in V$, and that

$$q(u,v):=\frac{1}{2}(q(u+v)-q(u)-q(v))$$

is a bilinear map. In particular, $q$ is symmetric.

Def. The Clifford algebra of $(V,q)$ is defined as

$$Cl(V,q):=\mathscr{T}(V)/\mathscr{I}_q(V)$$

where

$$\mathscr{T}(V):=\sum_{i=0}^\infty V^{\otimes i}$$

is the tensor algebra of $V$, and $\mathscr{I}_q(V)$ is the ideal generated by $v\otimes v+q(v)1_k$​. Denote the multiplication of elements in $Cl(V,q)$ by

$$v\cdot w=[v\otimes w].$$

Remark. The vector space $V$ can be thought of as a subspace in $Cl(V,q)$, as $V=V^{\otimes 1}$.

---

As an algebraic object, or regarding $Cl(V,q)$ as a quotient space, we can think of its universal property.

Prop. The Clifford algebra of $(V,q)$ has the following universal property: given an associative algebra $A$ and a $k$-linear map $f:V\rightarrow A$, satisfying

$$f(v)^2=-q(v)1_A$$

there exists a unique map $\tilde{f}: Cl(V,q)\rightarrow A$ that extends $f$.

With this property, once given a morphism $f:(V,q)\rightarrow (V',q')$, then this morphism induces a morphism between Clifford algebra

$$f_*:Cl(V,q)\rightarrow Cl(V',q')$$

such that $f_*(v\otimes w)=f(v)\otimes f(w)$. Moreover, one can verify that $*$ is a covariant functor.

---

Clifford algebra as a filtered algebra

Clifford algebra has a filtered structure as a

Def. A filtered algebra $A$ is an associative algebra with a sequence of subsets

$$\{0\}\subset\mathcal{F}^0A\subset \mathcal{F}^1A\subset\cdots\subset A$$

such that $\bigcup\mathcal{F}^iA=A$, and $\mathcal{F}^iA\cdot\mathcal{F}^jA\subset\mathcal{F}^{i+j}A$. The associated graded algebra $\mathscr{G}$ is defined by

$$\mathcal{G}^iA:=\mathcal{F}^iA/\mathcal{F}^{i-1}A.$$

e.g. The tensor algebra $\mathscr{T}(V)$ of a $k$-vector space is a filtered algebra, with filtration

$$\mathcal{F}^iA:=\sum_{k\leq i} V^{\otimes k}$$

Remark. There is a strong relation between the filtered structures on $\mathscr{T}(V)$ and $Cl(V,q)$​​.

---

Structure on $Cl(V,q)$ and $\Lambda^* V$

Prop. The associated graded algebra of $Cl(V,q)$ is naturally isomorphic to the exterior algebra $\Lambda^*V$.

Proof. Consider a map

$$\bigotimes^rV\rightarrow\mathscr{F}^r\rightarrow\mathscr{F}^r/\mathscr{F}^{r-1}$$

defined by $v_1\otimes\cdots\otimes v_r\mapsto[v_1\cdots v_r]$. This map could be descended to

$$\Lambda^rV\rightarrow\mathscr{F}^r/\mathscr{F}^{r-1}$$

as we have the Clifford equation

$$vw+wv=-2q(v,w)\in k$$

has degree $0$, which vanishes in $\mathscr{F}^{r-1}$. This map is surjective, as Clifford algebra is a quotient of tensor algebra. To see that this map is injective, notice that element in the kernel of

$$\bigotimes^r V\rightarrow \mathscr{G}^r$$

can be written as $\sum a_i\otimes(v_i\otimes v_i)\otimes b_i$, where $v_i\in V$ and $a_i,b_i$ are of pure degree such that $\deg(a_i)+\deg(b_i)\leq r-2$. Such element descends to $0$ in $\Lambda^*V$, which implies that this map is an injection. $\square$

---

Prop. There is a canonical vector space isomorphism

$$f:\Lambda^*V\rightarrow Cl(V,q)$$

compatible with the filtration, i.e.

$$f(\Lambda^n V)=\mathscr{G}^n(Cl(V,q))$$

Proof. Define $f_r:V\times\cdots V\rightarrow Cl(V,q)$, by setting

$$f_r(v_1,\cdots,v_r)=\frac{1}{r!}\sum_{\sigma\in Sym(r)}\mathrm{sgn}(\sigma)v_{\sigma(1)}\cdots v_{\sigma(r)}.$$

Since $f_r$ is anti-symmetric and multi-linear, then $f_r$ can be extended uniquely to $\tilde{f}_r:\Lambda^r V\rightarrow Cl(V,q)$. Moreover, by linear expansion, $\{\tilde{f}_r\}$ can be expanded to

$$\tilde{f}:\Lambda^*V\rightarrow Cl(V,q).$$

where $\tilde{f}(\Lambda^rV) \subset \mathscr{G}^r(Cl(V,q))$. Since $\tilde{f}_r$ is the map discussed in the above Prop, we know that their combination $\tilde{f}$ is an isomorphism. $\square$

Remark. This isomorphism is NOT an algebra isomorphism, as

$$f(v)^2=-q(v)\cdot 1\neq f(v\wedge v)=0,$$

unless the quadratic form $q=0$.

---

Divided into two algebras

Question: if one can divide $(V,q)$ into two separating parts, what are the behaviors of their corresponding Clifford algebra?

Prop. Suppose that $(V,q)$ is a $k$-vector space with quadratic form $q$, and $(V_1,q_1)$, $(V_2,q_2)$ are two subspaces of $V$ such that

$$V=V_1\oplus V_2$$

and $q(v_1+v_2)=q_1(v_1)+q_2(v_2)$, for all $v_1\in V_1$ and $v_2\in V_2$. Then

$$Cl(V,q)\simeq Cl(V_1,q_1)\hat{\otimes}Cl(V_2,q_2),$$

where $\hat{\otimes}$ is the tensor of two $\mathbb{Z}_2$-graded algebra.

Proof. Consider a map $f:V\rightarrow Cl(V_1,q_1)\hat{\otimes}Cl(V_2,q_2)$ defined by

$$f(v)=v_1\otimes 1+1\otimes v_2$$

and its linear expansion, where $v=v_1+v_2$ is the unique factorization of $v\in V_1\oplus V_2$. Then

$$\begin{aligned} f(v)\cdot f(v) &=(v_1\otimes 1+1\otimes v_2)^2\\ &=v_1^2\otimes 1+(-1)^{\deg 0\deg 0}v_1\otimes v_2+(-1)^{\deg v_1\deg v_2}v_1\otimes v_2+1\otimes v^2_2\\ &=-q_1(v_1) 1\otimes 1-q_2(v_2)1\otimes 1\\ &=-q(v)1\otimes 1. \end{aligned}$$

By the universal property of Clifford algebra, $f$ can be extended uniquely to

$$\tilde{f}:Cl(V,q)\rightarrow Cl(V_1,q_1)\hat{\otimes}Cl(V_2,q_2).$$

• Surjectivity: since the image of $\tilde{f}$ is the subalgebra containing $Cl(V_1,q_1)\otimes 1$and $1\otimes Cl(V_2,q_2)$, then $\tilde{f}$ is surjective.
• Injectivity: consider a basis $\{e_1,\cdots,e_n\}$ for the vector space $V$. Then a basis for the Clifford algebra is

$$e_{i_1}e_{i_2}\cdots e_{i_k}, \quad 1\leq k\leq n, \quad 1\leq i_1<\cdots<i_k\leq n.$$

by analyzing the behavior of $\tilde{f}$ on such basis, we know that $\tilde{f}$ is injective. $\square$

---

Involution

Since the tensor algebra $\mathscr{T}(V)$ has a natural involution defined by

$$v_1\otimes\cdots\otimes v_r\mapsto v_r\otimes\cdots\otimes v_1,$$

then this map also descend to a map on Clifford algebra

$$(\cdot)^t:Cl(V,q)\rightarrow Cl(V,q)$$

called a transpose. Note that $(\cdot)^t$ is an anti-automorphism, namely

$$(\phi\psi)^t=\psi^t\phi^t.$$

---