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Suppose you have a vector space $V$ of dimension $2n$. I know that there exists a basis $x_1,\dots,x_n,y_1,\dots,y_m$ such that $\omega(x_i,x_j)=\omega(y_i,y_j)=0$ and $\omega(x_i,y_j)=\delta_{ij}$, where $\omega$ is a bilinear symmetric nondegenerate form.

Now let $W=\operatorname{span}\{x_1,\dots,x_n\}$, and $W'=\operatorname{span}\{y_1,\dots,y_n\}$. If $Cl(V,\omega)$ denotes the corresponding Clifford algebra, then one can define a $Cl(V,\omega)$-module structure on the exterior algebra $\Lambda(W)$ as follows. For $x\in W$, let $\sigma(x)\in\operatorname{End}(\Lambda(W))$ as wedge multiplication by $x$ on the left. Also, for $y\in W'$, one can define $\sigma(y)x=\omega(y,x)$ for $x\in W$, which then extends to $\sigma(y)\in\operatorname{End}(\Lambda(W))$ by $$ y(a\wedge b)=y(a)\wedge b+(-1)^{\deg a}a\wedge y(b). $$

My question is, why does the $\sigma$ defined above in fact give an isomorphism between $Cl(V,\omega)$ and $\operatorname{End}(\Lambda(W))$? I'm viewing the endomorphisms as over the underlying field $k$, by the way.

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I assume you mean $\omega$ is a bilinear nondegenerate antisymmetric form? If not, I don't understand your basis... –  Jason DeVito Apr 20 '12 at 2:45
    
What are you reading? –  Parsa Apr 20 '12 at 23:20
    
Check Sternbeg's notes on Lie algebra. There is explicit treatment on this. –  Kerry Apr 23 '12 at 3:27
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Am I the only one confused by the fact the we are talking about the Clifford algebra of an antisymmetric bilinear form? Aren't Clifford algebra associated with quadratic forms? Are we talking about the zero form? –  YBL Apr 23 '12 at 20:53
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Please find out what you wanted to ask... No one seems to know what is the Clifford algebra of an antisymmetric form, and currently one answer decided to drop the anti and the other seems to have simply ignored the matter. –  Mariano Suárez-Alvarez Apr 26 '12 at 7:39
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2 Answers 2

up vote 3 down vote accepted
+100

I do not think that your question is all that bad, and in fact I think I have found a correct statement and a correct proof, see below. To summarize the proof, we have a basis indexed by subsets of $\lbrace 1,2, \ldots ,n \rbrace$ and to show the isomorphism property, we need to show we can go freely from any subset to any other. To achieve this, we use the $x_i$ to delete elements and the $y_i$ to insert elements.

There are two things to be corrected in your original question :

1) The form $w$ is symmetric, not antisymmetric (nobody ever constructed Clifford algebras from antisymmetric forms)

2) The formula for $\sigma(y)(x)$ should be $2w(y,x)$ instead of $w(y,x)$.

To be more specific : $\lbrace x_1, \ldots ,x_n \rbrace$ is a basis for $W$. So $\Lambda (W)$ has a basis ${\cal B}=\big({\lambda_I}\big)_{I \subseteq E}$ indexed by the subsets $I$ of $E=\lbrace 1,2, \ldots ,n \rbrace$, where for $I=\lbrace i_1<i_2< \ldots <i_t \rbrace$ we put

$$ \lambda_I=w_{i_1} \wedge w_{i_2} \wedge \ldots w_{i_t} $$

(in particular, $\lambda_{\emptyset}=1$, the unity element in $\Lambda (W)$). Note that the wedge product by $x_i$ (let us denote that operation by $X_i$) acts on this basis by

$$ X_i(\lambda_I)=\left\lbrace \begin{array}{ll} 0 , & {\rm if } \ i \in I, \\ (-1)^s \lambda_{I \cup \lbrace i \rbrace}, & {\rm if } \ i \not\in I, {\rm where} \ s \ {\rm is \ the \ number \ of \ elements \ in \ } I {\rm \ below \ } i. \end{array} \right. $$

It is natural to reverse this process and define a new action $Y_i$ on $W$ by

$$ Y_i(\lambda_I)=\left\lbrace \begin{array}{ll} 0 , & {\rm if } \ i\not\in I, \\ (-1)^s \lambda_{I \setminus \lbrace i \rbrace}, & {\rm if } \ i \in I, {\rm where} \ s \ {\rm is \ the \ number \ of \ elements \ in \ } I {\rm \ below \ } i. \end{array} \right. $$

For each $I\subseteq E$, denote by $F_I$ and $G_I$, respectively, the subspace of ${\Lambda}(W)$ generated by the $\lambda_J$ such that $I\subseteq J$ ($ I \not\subseteq J$, respectively). Then $F_I$ and $G_I$ are complementary subspaces in ${\Lambda}(W)$. Let us also write $F_i$ instead of $F_{\lbrace i \rbrace}$ and $G_i$ instead of $G_{\lbrace i \rbrace}$ for short. Then by contruction, $Z_i=X_iY_i$ is the projector onto $F_i$ according to $G_i$, and $Y_iX_i$ is the projector onto $G_i$ according to $F_i$. Similarly we have

$$ X_i^2=0, Y_i^2=0, X_iY_j+X_jY_i=2\delta_{ij} $$

so that we have in fact defined an action of the Clifford algebra. Note also that if $a$ and $b$ are two elements of ${\sf End}(\Lambda(W))$, and $a$ is decomposable, say $a=\lambda_{I}$ for some $I \subseteq E$, then it is easily checked that

$$ Y_i(a \wedge b)=Y_i(a)\wedge b + (-1)^{|I|} a \wedge Y_i(b) $$

(a property you mention in your original post).

Let us now show the isomorphism property. First note that the two spaces, ${\sf Cl}(V,w)$ and ${\sf End}({\lambda}(W))$ have the same dimensionality, namely $2^{2n}$. So it suffices to show that this homorphism is surjective, i.e. that its image $\cal I$ is the full space ${\sf End}({\Lambda}(W))$. Now the basis we have given for ${\Lambda}(W)$ provides a standard basis $(\gamma_{I,J})_{I,J \subseteq E}$ for ${\sf End}({\Lambda}(W))$, where

$$ \gamma_{I,J} (\lambda_K)=\delta_{KJ}\lambda_I $$ So all we need to show is that all the $\gamma_{I,J}$ are in $\cal I$.

Now $\cal I$ contains all the $Z_i$. So $\cal I$ contains $Z_I$ for any $I \subseteq E$, where we put $Z_{ \lbrace i_1 < i_2< \ldots < i_t \rbrace }=Z_{i_1}Z_{i_2} \ldots Z_{i_t}$. So $\cal I$ also contains all the $Z_{I,t}$ for $I \subseteq E$ and $0 \leq t \leq n$, where we put

$$ Z_{I,t}=\sum_{J \subseteq I, |J|=t } Z_J $$

The action of all those elements on the basis $\cal B$ can be written as follows : for any $K,I \subseteq E, i \in E, 0 \leq t \leq n$, we have

$$ Z_i(\lambda_{K})=|K \cap \lbrace i \rbrace| \lambda_K, \ Z_{I,t}(\lambda_{K})=\binom{|K \cap I|}{t} \lambda_K $$

Note that the polynomials $1,x,\frac{x(x-1)}{2},\binom{x}{3}, \ldots , \binom{x}{n}$ form a basis of the space of polynomials of degree $\leq n$ in $x$. For $i\in E$, we know that there a polynomial $\Delta_i(x)$ such that $\Delta_i(x)=\delta_{ix}$ for all $x\in E$ (explicitly, $\Delta_i(x)=\prod_{j \neq i}{\frac{x-j}{i-j}}$). So for any $i\in E$ there are coefficients $a_{i0},a_{i1}, \ldots ,a_{in}$ such that

$$ \delta_{ix}=\sum_{t=0}^{n} a_{it} \binom{x}{t} \ (x \in E) $$

In fact, all the coefficients $a_{it}$ can be computed explicitly, but it is not necessary here and I am too lazy to do it. If we put $$ {Z'}_I=\sum_{t=0}^{n} a_{|I|t} Z_{I,t} $$ then by construction $Z'_I$ is in $\cal I$, and is the projector onto the line ${\sf span} (\lambda_I)$ according to the hyperplane ${\sf span} (\lambda_J)_{J \neq I}$.

Let $I,J \subseteq E$. We claim that there is a $f\in {\cal I}$ such that $f(\lambda_I)=\lambda_{J}$. Indeed, if we put $I \setminus J=\lbrace j_1,j_2, \ldots j_t \rbrace$ and $J \setminus I = \lbrace i_1,i_2, \ldots i_s \rbrace$ and $f=X_{i_1}X_{i_2} \ldots X_{i_t}Y_{j_1}Y_{j_2} \ldots Y_{j_t}$, then $f\in {\cal I}$ and $f(\lambda_I)=\lambda_{J}$. It follows that $\gamma_{I,J}=fZ'_I$ is also in $\cal I$, as wished.

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Dear Ewan, thanks for this interesting answer! I will do my best to read through it. –  Buble Apr 25 '12 at 20:27
    
@Buble : You're welcome. Actually, it seems that rschwieb has found a better (simpler + shorter) answer. –  Ewan Delanoy Apr 26 '12 at 4:33
    
@Buble : by the way, my full answer starts with "Dear Buble, I do not think" ... But I edited it several times and mathjax always edits out the "Dear Buble" part. Does anyone know why this happens ? –  Ewan Delanoy Apr 26 '12 at 4:33
    
On second thoughts, I find rschwieb's answer unsatisfying : he does not give any indication about how $\sigma$ is extended, which was the major problem that caused confusion in the original post –  Ewan Delanoy Apr 26 '12 at 6:53
    
@EwanDelanoy I thought the linear extension of a map defined on $W$ to $\bigwedge W$ was well known, and looked like the one above. I did not think this was the issue so much as the question of how to verify the map was one-to-one. –  rschwieb Apr 26 '12 at 12:39
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I would like to begin similiarly to a previous post with the comment that both algebras have the same dimension, but I find it easier to show that the map is injective by using properties of the Clifford algebra (because I find writing out the extension to be unhelpful). Let me also add that I'm working under the assumption the form is symmetric. (I have never heard of a Clifford algebra using anything but.)

I think $^\dagger$ the suggested map is :$$ \sigma_{w+w'}(x):=w\wedge x+\omega(w',x)\wedge 1 $$ for all x in $W$ and $w+w'\in W+W'=V$, and then it is extended to have domain all of $\bigwedge W$. So, $\sigma$ is defined from $V$ into $End(\bigwedge W)$.

If $\sigma_{w+w'}(x)=0$ for all $x\in V$, then both $w\wedge x=0$ and $\omega(w',x)=0$ for all $x$ in $V$, but then by the definition of $\omega$, both $w$ and $w'$ must be zero. Hence $\sigma$ is injective on V, and its image has dimension $2n$.

It is (or at least seemed) routine to verify that $\sigma_v\circ\sigma_v=\omega(v,v)$ for all $v\in V$, and so the universal mapping property would kick in and extend the map $$ v\mapsto \sigma_v $$ to an algebra homomorphism from $C\ell(V,\omega)$ to $End(\bigwedge W)$.

Since $\sigma$ injected $V$ into $End(\bigwedge W)$, the extension is an injection of $C\ell(V,\omega)$ into $End(\bigwedge W)$ as well. Since these two algebras have the same finite dimension, this would be an isomorphism.

$^\dagger$ I'm not 100% confident though :(

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Many thanks rschwieb, I appreciate the alternate argument for injectivity. –  Buble Apr 25 '12 at 20:28
    
@rshcwieb : how is $\sigma_{w+w'}$ extended to all of $\Lambda (W)$ ? And why is it that the injectivity of $\sigma$ on $V$ implies the injectivity of its extension ? It seems to me that you cannot avoid writing out the extension somehow here. –  Ewan Delanoy Apr 26 '12 at 4:47
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@EwanDelanoy Yes, giving the extension formula to extend $\bigwedge(W)$ is necessary. (When I said I didn't find the formula helpful, I was only thinking of verifying injectivity/surjectivity). For your second question, let $\sigma$ denote the map. The condition $\sigma_v\circ\sigma_v=\omega(v,v)$ says that $Im(\sigma)$ generates $C\ell(Im(\sigma),\omega)$ inside $End(\bigwedge W)$. Since $V$ and $\sigma(V)$ are isometric quadratic spaces, their Clifford algebras are isomorphic, too. So, $End(\bigwedge W)=C\ell(Im(\sigma),\omega)$. –  rschwieb Apr 26 '12 at 12:28
    
I will be devastated if I find out I did something stupid, but no matter what, thanks for looking carefully at my attempt. –  rschwieb Apr 26 '12 at 12:32
    
@rschwieb : I am completely convinced now by your argument for injectivity/surjectivity once $\sigma$ is defined properly, but –  Ewan Delanoy Apr 26 '12 at 13:32
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