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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...2012-04-20
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    What are you reading?2012-04-20
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    Check Sternbeg's notes on Lie algebra. There is explicit treatment on this.2012-04-23
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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?2012-04-23
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    @YBL : No, you're not the only one. My quarrel with the original question is basically the same as yours.2012-04-24
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    @ Buble I find your equation line incomprehensible, and "antisymmetric" does not seem to make any sense here... I'm not sure why the previous commentor did not like the form being symmetric.2012-04-24
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    @all, I was basing this on some seminar notes from quite a while ago. Perhaps I wrote down the wrong stuff, and have since lost sense of what I meant. I apologize for the poor language. I'm sorry for the confusion, and it may be best just to disregard this question for now.2012-04-25
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    No sorry, I didn't mean to be discouraging! I just thought if you looked at the line again you might see a typo that you could fix :)2012-04-25
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    I think this $Cl(V,\omega)$ is the same as the Weyl algebra. I'm guessing the defining relation is $vw - wv = \omega(v,w)$, unlike in Clifford algebras for symmetric forms where you have $vw + wv = g(v,w)$.2012-04-26
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    Actually, this can't be because the Weyl algebra is infinite dimensional whereas $End(\Lambda V)$ is finite-dimensional. You really should clarify what you mean by Clifford algebra in this context.2012-04-26
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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.2012-04-26
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    Judging from the first post, it sounded like "symmetric" was the OP's original question. Since then it looks like editors may have been bumping it back and forth. I tried to edit it to symmetric because I have never heard of a Clifford algebra using an anti-symmetric form.2012-04-26
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    I have now edited back to symmetric form, as was originally.2012-04-26

2 Answers 2

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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.2012-04-25
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    @Buble : You're welcome. Actually, it seems that rschwieb has found a better (simpler + shorter) answer.2012-04-26
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    @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 ?2012-04-26
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    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 post2012-04-26
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    @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.2012-04-26
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    Surely it doesn't take 40 lines and 10 equations to show?2012-04-26
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    @ rschwieb : the discussion about your answer is best located in the comments to your answer I guess, so I'll try to answer you there.2012-04-26
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    @ On third thoughts, rschwieb's neat argument for bijectivity makes all the final part in my proof complicated and unnecessary. On the other hand, I still think the first part of my proof (the "unhelpful" definition of the Clifford algebra action as rschwieb would say) is unavoidable.2012-04-26
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    @Ewan: it is not MathJax, but the StackExchange engine. It is a general community norm on the StackExchange sites that posts do not begin with "Dear XXX" and are not signed at the end (except by the automatically inserted name card you can see in your post). The software enforces this norm by automatically removing such salutations.2012-04-26
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    @Willie : thanks for the explanation2012-04-26
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    @EwanDelanoy It is unhelpful for computation, but *necessary* to make things work :) I'm trying to contact you via chat to ask a specific question. I am trying to decipher what has been written about the extension. Is $Y$ the proposed extension: $Y_i(a \wedge b)=Y_i(a)\wedge b + (-1)^{|I|} a \wedge Y_i(b)$? Plugging $a=1$ into this it looks like you get nonsense: $Y(b)=b+Y(b)$ but I could be making a silly mistake.2012-04-26
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    @rschwieb : sorry, I'm not used to the chat here, I tried but it wouldn't work. To answer your question : in my solution I define $Y_i$ by the "action on the basis" rather2012-04-26
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    (continued) than by extension. The formula for $Y(a\wedge b)$ is a property in this viewpoint, not a definition. And no, you don't get nonsense because $Y_i(1)=0$ so that $Y_i(a) \wedge b+a \wedge Y_i(b)=0 \wedge b+ 1 \wedge Y_i(b)=Y_i(b)$.2012-04-26
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    @EwanDelanoy OK thanks, so I wasn't using Y correctly. I'm not sure what you mean by "in this viewpoint". I have a hard time following the notation in your solution. Is it possible to write what the extension would be for $\sigma_{w+w'}(a\wedge b)$ for $a,b\in W$? This is vital for confirming $\sigma_v\circ\sigma_v=\omega(v,v)$, a step I haven't been able to reproduce.2012-04-26
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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.2012-04-25
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    @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.2012-04-26
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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, $End(\bigwedge W)=C\ell(Im(\sigma),\omega)$.2012-04-26
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    I will be devastated if I find out I did something stupid, but no matter what, thanks for looking carefully at my attempt.2012-04-26
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    @rschwieb : I am completely convinced now by your argument for injectivity/surjectivity once $\sigma$ is defined properly, but2012-04-26
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    I still cannot understand what you mean by "THE linear extension of $\sigma$ to $\Lambda (W)$". There are several extensions, and the natural definition $\sigma(a\wedge b)=\sigma(a) \wedge \sigma (b)$ does not work here.2012-04-26
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    @EwanDelanoy I should have typed *the given extension* rather than *the extension*. The given extension (that resembles a derivation) is the one I see the most used with Clifford algebras. I did not intend it to be taken that any extension would do. If there is a discrepancy between the given extension and the map I had in mind, please let me know. I have been trying to verify things by relying on bases, so there are chances to misstep.2012-04-26