Is this exterior algebra identity correct? (Repost) I’m reading DJH Garling’s Clifford Algebras: An Introduction.
He defines creation and annihilation operators as follows.
Given a finite dimensional vector space $E$ (over $\mathbb{R}$, say), let $A^k(E)$ be the space of skew-symmetric functions on $\times^k E$ and $\bigwedge^k(E) := A^k(E)^*$ (dual). $\bigwedge(E) := \bigoplus \bigwedge^k(E)$ is the exterior algebra over $E$.
The creation operator defined by $x\in E$ is
\begin{align}
m_x: \bigwedge(E) \rightarrow \bigwedge(E)\\
a \mapsto x\wedge a
\end{align}
Next, if $\varphi\in E^*$,
\begin{align}
P_\varphi: &\,A^k(E) \rightarrow A^{k+1}(E) \ {} \\
P_\varphi (a)&(x_1, \dots, x_{k+1})\\
 &:= \frac{1}{(k+1)!}\sum_{\sigma\in S_{k+1}}\mathrm{sgn}\,\sigma\ \varphi(x_{\sigma(1)})\,a(x_{\sigma(2)}, \dots, x_{\sigma(k+1)})
\end{align}
the annihilation operator $\delta_\varphi: \bigwedge(E) \rightarrow \bigwedge(E)$ is defined to be the adjoint $P_\varphi^*$ and by varying $k$ is extended to $\bigwedge(E)$ ($\delta_\varphi(c) := 0$ for all $c \in \mathbb{R}$). There is the following expression for $\delta_\varphi$:
\begin{align*}
\delta_\varphi & (x_1 \wedge \dots \wedge x_{k}) \\
&= \frac{1}{k}\sum_{j=1}^{k} (-1)^{j-1} \varphi(x_j)(x_1 \wedge \dots \wedge \hat{x_j} \wedge \dots \wedge x_{k})
\end{align*}
$x_i$’s in $E$.

To finally get to my question, the author sets out (p.48) to prove the “anticommutation” relation:
$$
m_x\delta_\varphi + \delta_\varphi m_x= \varphi(x)I
$$
Now let $\{e_1, \dots, e_d\}$ be a basis of $E$. If I take $x = e_1$, $\varphi = e_1^*$ (from the dual basis), then $(m_{e_1}\delta_{e_1^*} + \delta_ {e_1^*} m_{e_1})(e_2) = (1/2) e_2$ whereas the right side is simply $e_2$.
 A: It turns out that the identity is correct, but the expression of the annihilation operator seems incorrect. The factor $\frac{1}{k}$ should not be there. The proof of Theorem 3.4.1 in Garling is also incorrect due to this. If we keep the factor, then
$$
\begin{multline}
\delta_\phi m_x(x_1\wedge \dots \wedge x_k) = \color{red}{\frac{1}{k+1}}\phi(x)(x_1\wedge\dots\wedge x_k) \\
+ \color{red}{\frac{1}{k+1}}\sum_{j=1}^k (-1)^j\phi(x_j)(x\wedge x_1\wedge \dots \wedge \hat{x}_j \wedge \dots \wedge x_k)\ .
\end{multline}
$$
The factors I marked are missing from the expressions in the book (page 48). Note how the second term no longer cancels out $m_x\delta_\phi$ due to the factor.
There are two remarks in John M. Lee's Introduction to Smooth Manifolds that might be relevant to you. On pp. 357-358 (2nd edition), at the end of the subsection on the wedge product, he points out that there are two conventions for the definition of wedge products, the difference being a factor of $k!l!/(k+l)!$ where $k$ and $l$ are the degrees of the factors. At the end of the following subsection on interior products, he notes that in the two conventions, the definition of interior products are also off by a factor of $k$ as a result. I realize that the annihilation operator is not the same as an interior product, but they are quite similar, so it's worth considering this point. For example, assuming there are no errors between the definition of $P_\phi$ and Theorem 3.4.1, if we modify the definition of $l_\phi$ on page 47 to $l_\phi(a)(x_1, \dots, x_{k+1}) =(k+1) \phi(x_1)a(x_2, \dots, x_{k+1})$, then this inconsistency might get fixed.
