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.