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In reading Sternberg's notes on Clifford algebras and spin representations (page 148) I encountered the following:

"...Consider the linear map $$C(\mathbf p)\rightarrow \wedge \mathbf p, x\mapsto x1$$ where $1\in \wedge^{0}\mathbf p$ under the identification of $\wedge^{0}\mathbf p$ with the ground field. The element $x1$ on the extreme right means the image of 1 under the action of $x\in C(\mathbf p)$. "

I am wondering how to derive the explicit form of this map. Sternberg give a homomorphism $C(\mathbf p)\rightarrow \operatorname{End}{(\wedge \mathbf p)}$ by extending the map $\mathbf p\rightarrow \operatorname{End}{(\wedge \mathbf p)}$, which is defined by $v\mapsto \epsilon(v)+\iota(v)$. In here $\epsilon(v)$ denotes the exterior mulplication by $v$ and $\iota(v)$ be the the adjoint of $\epsilon(v)$ relative to the biinear form given by $(x_{1}\wedge\cdots\wedge x_{k},y_{1}\wedge\cdots\wedge y_{k})=\det((x_{i},y_{j}))$.

Hence we have $ \epsilon(v)+\iota(v)(u)=v\wedge u+Au$, with $(v\wedge u, w)=(u,Aw),\forall u,v,w\in \wedge\mathbf p$. Sternberg argues that $(\epsilon(v)+\iota(v))^{2}=(v,v)_{\mathbf p}\operatorname{id}$, therefore we may extend it via universal property to a map $C(\mathbf p)\rightarrow \wedge(\mathbf p)$.

This relationship is not clear to me because I do not see how $\epsilon(v)(\iota(v)(u))+\iota(v)(\epsilon(v)(u))=(v,v)_{\mathbf p}u$, namely LHS is in exterior product while RHS is in a nice closed form.

This confusion hindered me to understand the nature of the linear map he gave. For example, Sternberg wrote: $$v_{1}v_{2}\rightarrow v_{1}\wedge v_{2}+(v_{1},v_{2})1$$ and $$v_{1}v_{2}v_{3}\rightarrow v_{1}\wedge v_{2}\wedge v_{3}+(v_{1},v_{2})v_{3}-(v_{1},v_{3})v_{2}+(v_{2},v_{3})v_{1}$$ I do not know how to derive these formulas explicitly, I believe they should be elementary in nature and easy to work out by hand. So I must have missed something.

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Edited, thanks for pointing out. Really embarrassing mistake. – Kerry Nov 20 '11 at 7:02
Use \wedge ($\wedge$) instead of \vee ($\vee$) for the wedge product. – wildildildlife Nov 20 '11 at 13:11
updated. Thanks. – Kerry Nov 21 '11 at 1:11
You can WLOG assume that $u$ is a "pure wedge", i. e., of the form $u=u_1\wedge u_2\wedge ...\wedge u_n$ for some $u_1,u_2,...,u_n\in\mathbf p$ (because such "pure wedges" generate the vector space $\wedge\left(\mathbf p\right)$). Then both $\epsilon(v)(\iota(v)(u))$ and $\iota(v)(\epsilon(v)(u))$ can be expanded into sums, and you will easily see that the addends of these sums cancel out pairwise except of one which is exactly $(v,v)_{\mathbf p}u$. – darij grinberg Nov 29 '11 at 17:17
Note: Working with $\iota$ becomes much easier once you explicitly write down $\iota$ as follows: $\left(\iota\left(v\right)\right)\left(u_1\wedge u_2\wedge ...\wedge u_n\right) = \sum\limits_{i=1}^n \left(-1\right)^{i-1}\left(v,u_i\right)_{\mathbf p} u_1\wedge u_2\wedge ...\wedge \hat{u_i}\wedge ...\wedge u_n$. – darij grinberg Nov 29 '11 at 17:18
up vote 1 down vote accepted

It is enough to consider action of $\epsilon_v \equiv \epsilon(v)$ and $\iota_v \equiv \iota(v)$, $v \in {\bf p}$ on a basis of $\wedge {\bf p}$.

$\epsilon_v(1) \equiv v$; $\epsilon_v(v_1 \wedge \cdots \wedge v_k) \equiv v \wedge v_1 \wedge \cdots \wedge v_k$.

$\iota_v(1) \equiv 0$; $\iota_v(v_1 \wedge \cdots \wedge v_k) \equiv \sum_{j=1}^k (-1)^{j-1} (v,v_j)_{\bf p} v_1 \wedge \cdots \hat{v}_j \cdots \wedge v_k$,

where $\hat{v}_j$ means that $v_j$ is omitted from product.

Then for any $v, w \in {\bf p}$ and $x = x_1 \wedge \cdots \wedge x_k \in \wedge {\bf p}$

$\iota_v (\epsilon_w (x)) = \iota_v (w \wedge x) = (v,w)_{\bf p} x - w \wedge \iota_v (x)$

So $\iota_v (\epsilon_w (x)) + \epsilon_w (\iota_v (x)) = (v,w)_{\bf p} x$

after omitting $x$ we have $\iota_v \epsilon_w + \epsilon_w \iota_v = (v,w)_{\bf p} {\rm id}$.

E.g. see Gilbert J.E., Murray M.A.M. Clifford algebras and Dirac operators in harmonic analysis (CUP, 1991)

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