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.
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)