15.10  The Exterior Algebra, and Clifford at Last

Proof. Fix ${\mathbf{w}}_1,\dots ,{\mathbf{w}}_l\in V$. The map $V^k\to {\mathrm{\Lambda }}^{k+l}V$ given by $$({\mathbf{u}}_1,\dots ,{\mathbf{u}}_k)\mapsto {\mathbf{u}}_1\wedge \cdots \wedge {\mathbf{u}}_k\wedge {\mathbf{w}}_1\wedge \cdots \wedge {\mathbf{w}}_l$$ is $k$-linear, since the wedge of $k+l$ vectors is linear in each slot and the last $l$ are frozen, and alternating, since a repeat among the first $k$ arguments is a repeat among all $k+l$. By Definition 15.8.2 it induces a unique linear map $$R({\mathbf{w}}_1,\dots ,{\mathbf{w}}_l):{\mathrm{\Lambda }}^{k}V\to {\mathrm{\Lambda }}^{k+l}V$$ with $R({\mathbf{w}}_1,\dots ,{\mathbf{w}}_l)({\mathbf{u}}_1\wedge \cdots \wedge {\mathbf{u}}_k)$ equal to the displayed wedge.

Now let the $\mathbf{w}$’s vary. The assignment $({\mathbf{w}}_1,\dots ,{\mathbf{w}}_l)\mapsto R({\mathbf{w}}_1,\dots ,{\mathbf{w}}_l)$, with values in the vector space $\mathcal{L}({\mathrm{\Lambda }}^{k}V,{\mathrm{\Lambda }}^{k+l}V)$, is $l$-linear and alternating: two linear maps on ${\mathrm{\Lambda }}^{k}V$ are equal as soon as they agree on the wedges ${\mathbf{u}}_1\wedge \cdots \wedge {\mathbf{u}}_k$, which span ${\mathrm{\Lambda }}^{k}V$ (Theorem 15.8.5), and on those the required identities are the linearity and the alternation of the wedge of $k+l$ vectors in its last $l$ slots. So Definition 15.8.2 applies again and gives a unique linear map $${\mathrm{\Lambda }}^{l}V\to \mathcal{L}({\mathrm{\Lambda }}^{k}V,{\mathrm{\Lambda }}^{k+l}V),{\mathbf{w}}_1\wedge \cdots \wedge {\mathbf{w}}_l\mapsto R({\mathbf{w}}_1,\dots ,{\mathbf{w}}_l).$$ Reading a linear map into a space of linear maps as a bilinear map gives the required product, with the stated values. It is unique because the wedges span both factors and a bilinear map is determined on a pair of spanning sets. This proves the proposition.

Proof.

Proof. Both sides are bilinear in $(\omega ,\eta )$, so by Theorem 15.8.5 it suffices to take $\omega ={\mathbf{u}}_1\wedge \cdots \wedge {\mathbf{u}}_k$ and $\eta ={\mathbf{w}}_1\wedge \cdots \wedge {\mathbf{w}}_l$, and the cases $k=0$ or $l=0$ are scalar multiplication and are clear. The permutation of $\{1,\dots ,k+l\}$ carrying the list $({\mathbf{w}}_1,\dots ,{\mathbf{w}}_l,{\mathbf{u}}_1,\dots ,{\mathbf{u}}_k)$ to $({\mathbf{u}}_1,\dots ,{\mathbf{u}}_k,{\mathbf{w}}_1,\dots ,{\mathbf{w}}_l)$ has exactly $kl$ inversions, one for each pair consisting of a $\mathbf{u}$ and a $\mathbf{w}$, so its sign is $(-1{)}^{kl}$ (Definition 7.3.2). By Lemma 15.8.1 (b) the two wedges differ by that factor, which is the claim.

For the last sentence let $k$ be odd and $\omega \in {\mathrm{\Lambda }}^{k}V$. Write $\omega =\sum _S{c}_S{\mathbf{e}}_S$ in the basis of Theorem 15.10.2, the sum over the $S$ with $|S|=k$. Then $$\omega \wedge \omega =\sum _S\sum _T{c}_S{c}_T{\mathbf{e}}_S\wedge {\mathbf{e}}_T.$$ A term with $S=T$ has a repeated vector, so ${\mathbf{e}}_S\wedge {\mathbf{e}}_S=\mathbf{0}$. A pair $S\ne T$ occurs twice, and the two occurrences are $c_S{c}_T{\mathbf{e}}_S\wedge {\mathbf{e}}_T$ and $c_T{c}_S{\mathbf{e}}_T\wedge {\mathbf{e}}_S=-c_S{c}_T{\mathbf{e}}_S\wedge {\mathbf{e}}_T$, since $(-1{)}^{kl}=(-1{)}^{k^2}=-1$ for odd $k$. So the two cancel, and nothing divides by $2$ anywhere.

Proof. Suppose $\omega =\mathbf{u}\wedge \mathbf{w}$ for some $\mathbf{u},\mathbf{w}\in V$. Since ${\mathbf{e}}_1\wedge {\mathbf{e}}_2$ and ${\mathbf{e}}_3\wedge {\mathbf{e}}_4$ are distinct basis vectors of ${\mathrm{\Lambda }}^{2}V$ (Theorem 15.8.5), $\omega \ne \mathbf{0}$, so $(\mathbf{u},\mathbf{w})$ is linearly independent by Theorem 15.8.7; in particular $\mathbf{u}\ne \mathbf{0}$.

Consider $\mathbf{u}\wedge \omega =\mathbf{u}\wedge \mathbf{u}\wedge \mathbf{w}=\mathbf{0}$, the wedge having a repeated argument. We show this is impossible. Let $\mathbf{v}=\sum _{i=1}^n{x}_i{\mathbf{e}}_i$ be any vector and compute $\mathbf{v}\wedge \omega$, using bilinearity and dropping every term with a repeated basis vector: $$\begin{array}{rl}\mathbf{v}\wedge \omega =& x_3{\mathbf{e}}_3\wedge {\mathbf{e}}_1\wedge {\mathbf{e}}_2+x_4{\mathbf{e}}_4\wedge {\mathbf{e}}_1\wedge {\mathbf{e}}_2\\ & +x_1{\mathbf{e}}_1\wedge {\mathbf{e}}_3\wedge {\mathbf{e}}_4+x_2{\mathbf{e}}_2\wedge {\mathbf{e}}_3\wedge {\mathbf{e}}_4\\ & +\sum _{i\ge 5}{x}_i({\mathbf{e}}_i\wedge {\mathbf{e}}_1\wedge {\mathbf{e}}_2+{\mathbf{e}}_i\wedge {\mathbf{e}}_3\wedge {\mathbf{e}}_4).\end{array}$$ Sorting each wedge by Lemma 15.8.1 (b) turns this into a combination of the distinct basis vectors $${\mathbf{e}}_{123},{\mathbf{e}}_{124},{\mathbf{e}}_{134},{\mathbf{e}}_{234},{\mathbf{e}}_{12i},{\mathbf{e}}_{34i}(i\ge 5)$$ of ${\mathrm{\Lambda }}^{3}V$, with coefficients $x_3,x_4,x_1,x_2$ and $x_i,x_i$ respectively, up to signs. Since these basis vectors are distinct and the ${\mathbf{e}}_S$ are independent (Theorem 15.8.5), $\mathbf{v}\wedge \omega =\mathbf{0}$ forces $x_1=x_2=\cdots =x_n=0$, that is, $\mathbf{v}=\mathbf{0}$. This contradicts $\mathbf{u}\ne \mathbf{0}$ with $\mathbf{u}\wedge \omega =\mathbf{0}$. Hence $\omega$ is not decomposable.

Proof.

Proof. By Lemma 15.10.5 (a), $\mathrm{Cl}(q)$ is an $F$-algebra and $\pi$ is a surjective algebra homomorphism with $\ker \pi ={\mathcal{I}}_q$. The map $\iota =\pi {|}_V$ is linear, being a restriction of a linear map.

(C1). Let $\mathbf{v}\in V$. In $\mathrm{T}(V)$ the product of $\mathbf{v}$ with itself is $\mathbf{v}\otimes \mathbf{v}$ (Definition 15.6.2), and $\mathbf{v}\otimes \mathbf{v}-q(\mathbf{v})1$ is a generator of ${\mathcal{I}}_q=\ker \pi$. Since $\pi$ is multiplicative and linear, $$\iota (\mathbf{v}{)}^2=\pi (\mathbf{v})\pi (\mathbf{v})=\pi (\mathbf{v}\otimes \mathbf{v})=\pi (q(\mathbf{v})1)=q(\mathbf{v})1_{\mathrm{Cl}(q)}.$$

(C2). Let $A$ be an $F$-algebra and let $f:V\to A$ be linear with $f(\mathbf{v}{)}^2=q(\mathbf{v})1_A$ for every $\mathbf{v}\in V$. By Theorem 15.6.6 there is a unique algebra homomorphism $\tilde{f}:\mathrm{T}(V)\to A$ with $\tilde{f}{|}_V=f$. For every $\mathbf{v}\in V$, $$\tilde{f}(\mathbf{v}\otimes \mathbf{v}-q(\mathbf{v})1)=f(\mathbf{v})f(\mathbf{v})-q(\mathbf{v})1_A=0,$$ so $\ker \tilde{f}$ contains every generator of ${\mathcal{I}}_q$. By Lemma 15.10.5 (c), $\ker \tilde{f}$ is a two-sided ideal, and ${\mathcal{I}}_q$ is the smallest one containing those generators (Definition 15.10.3), so ${\mathcal{I}}_q\subseteq \ker \tilde{f}$. By Lemma 15.10.5 (b) there is exactly one algebra homomorphism $\overline{f}:\mathrm{Cl}(q)\to A$ with $\overline{f}\circ \pi =\tilde{f}$, and restricting to $V$ gives $\overline{f}\circ \iota =\tilde{f}{|}_V=f$.

For the uniqueness clause of (C2), let $g:\mathrm{Cl}(q)\to A$ be any algebra homomorphism with $g\circ \iota =f$. Then $g\circ \pi$ is an algebra homomorphism $\mathrm{T}(V)\to A$ whose restriction to $V$ is $g\iota =f$, so $g\pi =\tilde{f}$ by the uniqueness clause of Theorem 15.6.6. Hence $g\pi =\overline{f}\pi$, and $\pi$ is surjective, so $g=\overline{f}$. This proves the theorem.

Proof. First record what $E_i$ does on the basis. If $i\in S$ then ${\mathbf{e}}_i\wedge {\mathbf{e}}_S$ repeats ${\mathbf{e}}_i$ and is $\mathbf{0}$. If $i\notin S$, then sorting ${\mathbf{e}}_i\wedge {\mathbf{e}}_S$ moves ${\mathbf{e}}_i$ past the $\sigma (i,S)$ members of $S$ smaller than $i$, so by Lemma 15.8.1 (b) $$E_i({\mathbf{e}}_S)=(-1{)}^{\sigma (i,S)}{\mathbf{e}}_{S\cup \{i\}}.$$

First identity. For every $\omega \in \mathrm{\Lambda }V$, $E_i{E}_j(\omega )={\mathbf{e}}_i\wedge {\mathbf{e}}_j\wedge \omega$ and $E_j{E}_i(\omega )={\mathbf{e}}_j\wedge {\mathbf{e}}_i\wedge \omega$, and ${\mathbf{e}}_j\wedge {\mathbf{e}}_i=-{\mathbf{e}}_i\wedge {\mathbf{e}}_j$ by Theorem 15.10.3. Adding gives $0$. (For $i=j$ each term is separately $0$.)

Third identity, case $i=j$. If $i\notin S$, then $D_i({\mathbf{e}}_S)=\mathbf{0}$ and $E_i({\mathbf{e}}_S)=(-1{)}^{\sigma (i,S)}{\mathbf{e}}_{S\cup \{i\}}$; since $\sigma (i,S\cup \{i\})=\sigma (i,S)$, applying $D_i$ returns $(-1{)}^{2\sigma (i,S)}{\mathbf{e}}_S={\mathbf{e}}_S$. If $i\in S$, then $E_i({\mathbf{e}}_S)=\mathbf{0}$ and $D_i({\mathbf{e}}_S)=(-1{)}^{\sigma (i,S)}{\mathbf{e}}_{S\setminus \{i\}}$; since $\sigma (i,S\setminus \{i\})=\sigma (i,S)$, applying $E_i$ returns ${\mathbf{e}}_S$. Either way the sum is ${\mathbf{e}}_S$.

Third identity, case $i\ne j$. If $j\in S$ then $E_j({\mathbf{e}}_S)=\mathbf{0}$, and $D_i({\mathbf{e}}_S)$ is either $\mathbf{0}$ or a multiple of ${\mathbf{e}}_{S\setminus \{i\}}$, which still contains $j$, so $E_j$ kills it; both terms vanish. If $j\notin S$ and $i\notin S$ then $D_i({\mathbf{e}}_S)=\mathbf{0}$ and $D_i({\mathbf{e}}_{S\cup \{j\}})=\mathbf{0}$ as well, since $i\ne j$; again both terms vanish. So let $j\notin S$ and $i\in S$. Then $$\begin{array}{rl}{D}_i{E}_j({\mathbf{e}}_S)& =(-1{)}^{\sigma (j,S)+\sigma (i,S\cup \{j\})}{\mathbf{e}}_{(S\cup \{j\})\setminus \{i\}},\\ E_j{D}_i({\mathbf{e}}_S)& =(-1{)}^{\sigma (i,S)+\sigma (j,S\setminus \{i\})}{\mathbf{e}}_{(S\setminus \{i\})\cup \{j\}},\end{array}$$ and the two index sets agree. Now $\sigma (i,S\cup \{j\})=\sigma (i,S)+[j<i]$ and $\sigma (j,S\setminus \{i\})=\sigma (j,S)-[i<j]$, writing $[P]$ for $1$ if $P$ holds and $0$ otherwise. So the two exponents differ by $[j<i]+[i<j]=1$, which is odd because $i\ne j$. The two terms therefore cancel.

Second identity. Both $D_i{D}_j({\mathbf{e}}_S)$ and $D_j{D}_i({\mathbf{e}}_S)$ vanish unless $i,j\in S$ and $i\ne j$, and in that case they are $(-1{)}^{\sigma (j,S)+\sigma (i,S\setminus \{j\})}$ and $(-1{)}^{\sigma (i,S)+\sigma (j,S\setminus \{i\})}$ times the same ${\mathbf{e}}_{S\setminus \{i,j\}}$. The same two identities for $\sigma$ show the exponents differ by an odd number, so the terms cancel. For $i=j$, $D_i^2({\mathbf{e}}_S)=\mathbf{0}$ because $i\notin S\setminus \{i\}$. This proves the lemma.

Proof. An orthogonal basis exists by Theorem 14.4.2, since $\mathrm{char}F\ne 2$. Write $d_i=q({\mathbf{e}}_i)$ and $c_i=E_i+d_i{D}_i\in \mathcal{L}(\mathrm{\Lambda }V)$, with $E_i,D_i$ as in Lemma 15.10.7. By Proposition 14.12.3 the sorted products span $C$ and ${\dim }_{F}C\le 2^n$, so only independence is left.

Step 1: the $c_i$ satisfy the Clifford relations. Expanding the product and collecting the four brackets, $$\begin{array}{rl}{c}_i{c}_j+c_j{c}_i=& (E_i{E}_j+E_j{E}_i)+d_j(E_i{D}_j+D_j{E}_i)\\ & +d_i(D_i{E}_j+E_j{D}_i)+d_i{d}_j(D_i{D}_j+D_j{D}_i).\end{array}$$ By Lemma 15.10.7 the first and last brackets are $0$ and each middle bracket is ${\delta }_{ij}\mathrm{id}$. So $c_i{c}_j=-c_j{c}_i$ for $i\ne j$. Setting $i=j$ in the display would give $2c_i^2$, so expand $c_i^2$ directly instead; with $E_i^2=D_i^2=0$, which were checked separately in the lemma, $$c_i^2=E_i^2+d_i(E_i{D}_i+D_i{E}_i)+d_i^2{D}_i^2=d_i\mathrm{id},$$ with no division by $2$ anywhere. Now $\mathcal{L}(\mathrm{\Lambda }V)$ is an $F$-algebra under composition (Definition 14.12.1). Let $f:V\to \mathcal{L}(\mathrm{\Lambda }V)$ be the linear map with $f({\mathbf{e}}_i)=c_i$, which exists and is unique by Theorem 2.1.6. For $\mathbf{v}=\sum _i{x}_i{\mathbf{e}}_i$, $$f(\mathbf{v}{)}^2=\sum _i{x}_i^2{c}_i^2+\sum _{i<j}{x}_i{x}_j(c_i{c}_j+c_j{c}_i)=(\sum _i{d}_i{x}_i^2)\mathrm{id},$$ the cross terms vanishing because $i\ne j$ there. Since the basis is orthogonal, $\beta ({\mathbf{e}}_i,{\mathbf{e}}_j)=0$ for $i\ne j$, so $q(\mathbf{v})=\beta (\mathbf{v},\mathbf{v})=\sum _i{d}_i{x}_i^2$. Hence $f(\mathbf{v}{)}^2=q(\mathbf{v})\mathrm{id}$, which is (C1) for $f$.

Step 2: the induced homomorphism. By the universal property (C2) of Definition 14.12.2 there is an algebra homomorphism $\mathrm{\Psi }:C\to \mathcal{L}(\mathrm{\Lambda }V)$ with $\mathrm{\Psi }\iota =f$. Define the linear map $$\mathrm{\Phi }:C\to \mathrm{\Lambda }V,\mathrm{\Phi }(x)=\mathrm{\Psi }(x)(1),$$ where $1\in {\mathrm{\Lambda }}^{0}V=F\subseteq \mathrm{\Lambda }V$. It is linear because $\mathrm{\Psi }$ is linear and evaluation at a fixed element is linear.

Step 3: $\mathrm{\Phi }$ sends sorted products to basis vectors. Let $S=\{i_1<\cdots <i_k\}\subseteq [n]$. Since $\mathrm{\Psi }$ is multiplicative and $\mathrm{\Psi }\iota =f$, $$\mathrm{\Psi }(\iota ({\mathbf{e}}_{i_1})\cdots \iota ({\mathbf{e}}_{i_k}))=c_{i_1}{c}_{i_2}\cdots c_{i_k}.$$ We claim $c_{i_r}{c}_{i_{r+1}}\cdots c_{i_k}(1)={\mathbf{e}}_{T_r}$ with $T_r=\{i_r,\dots ,i_k\}$, by downward induction on $r$. For $r=k+1$ the product is empty and the value is $1={\mathbf{e}}_{\mathrm{\varnothing }}$. Assume the claim for $r+1$. Because $i_r<i_{r+1}$ and the indices increase, $i_r\notin T_{r+1}$, so $D_{i_r}({\mathbf{e}}_{T_{r+1}})=\mathbf{0}$; and $i_r$ is smaller than every element of $T_{r+1}$, so $\sigma (i_r,T_{r+1})=0$ and $E_{i_r}({\mathbf{e}}_{T_{r+1}})={\mathbf{e}}_{T_r}$. Hence $c_{i_r}({\mathbf{e}}_{T_{r+1}})={\mathbf{e}}_{T_r}$, completing the induction. Taking $r=1$, $$\mathrm{\Phi }(\iota ({\mathbf{e}}_{i_1})\cdots \iota ({\mathbf{e}}_{i_k}))={\mathbf{e}}_S,$$ and for $S=\mathrm{\varnothing }$, $\mathrm{\Phi }(1_C)=\mathrm{id}(1)=1={\mathbf{e}}_{\mathrm{\varnothing }}$.

Step 4: conclusion. Suppose a linear combination of the sorted products is $0$. Applying $\mathrm{\Phi }$ gives the same combination of the ${\mathbf{e}}_S$, which are linearly independent by Theorem 15.10.2 (b); so all coefficients vanish. The sorted products are therefore linearly independent, and being also a spanning list they form a basis of $C$, of size $2^n$. Thus ${\dim }_{F}C=2^n$.

Since $1_C$ and $\iota ({\mathbf{e}}_1),\dots ,\iota ({\mathbf{e}}_n)$ are among the basis elements just found, the list $(\iota ({\mathbf{e}}_1),\dots ,\iota ({\mathbf{e}}_n))$ is independent, so $\iota$ carries a basis of $V$ to an independent list and is injective. Finally $\mathrm{\Phi }$ carries the basis of $C$ bijectively onto the basis $({\mathbf{e}}_S)$ of $\mathrm{\Lambda }V$, so $\mathrm{\Phi }$ is a vector space isomorphism. This proves the theorem.