Let $E$ be a finite dimensional real vector space with $E^*$ its dual, and let $\langle \; , \; \rangle$ be an inner product on $E$. For any $e \in E$, denote by $e^* = \langle e, \; \rangle \in E^*$ the element dual to $e$ via $\langle \; , \; \rangle$.
We define a homomorphism
$C: E \to \text{End}(\Lambda^* (E^*)) \; , \; e \mapsto e^* \wedge i_{e},$
where $i_e: \Lambda^*(E^*) \to \Lambda^{*-1}(E^*)$ denotes the contraction (interior multiplication) induced by $e$. Observe that $C(e)$ is a degree-preserving endomorphism of $\Lambda^*(E^*)$. It is claimed in this article on page 65 that one has
$$ C(e)^2 = \langle e,e \rangle \cdot \mathbb 1_{\Lambda^*(E)}$$
To me, this equality seems absurd. For example, if $e_1,e_2$ are two orthonormal vectors, then one has $i_{e_1}(e_2^*) = \langle e_2,e_1 \rangle = 0$, so that in particular $C(e_1)^2 e_2^* = 0 \neq e_2^*$. I believe that there must be something fundamental that I misunderstand, but I can't see what it is. Any help is appreciated.
You are interpreting the definition (4.15) incorrectly. $\hat{c}(e)$ is the sum of $e^*\wedge(-)$ and $i_e(-)$, similarly $c(e)$ is the difference.