Intepreting Fermions as Differential Forms? In this paper on path-integral quantization of Chern-Simons theory, on page 434 (equation 4.17), the authors used fermions to interpret wedge product and contractions of differential forms.
Let $M$ be a manifold, with local coordinate $x^{i}$. For any differential form $a\in\Omega(M)$, one has the operations
$$\psi^{i}:a\rightarrow dx^{i}\wedge a,$$
and 
$$\chi_{j}: a\rightarrow a(\partial_{j}).$$
One has the Clifford algebra 
$$\left\{\psi^{i},\chi_{j}\right\}=\delta^{i}_{j},\quad \left\{\psi^{i},\psi^{j}\right\}=\left\{\chi_{i},\chi_{j}\right\}=0$$
Define the Witten-index $(-1)^{F}$ as 
$$(-1)^{F}:\omega\rightarrow(-1)^{q}w,\,\,\,\,\mathrm{for}\,\,\,\forall\omega\in\Omega^{q}(M).$$
Then one has the relation (equation 4.17)
$$\ast\psi^{i}\ast=(-1)^{F}\chi^{i},\quad\ast\chi^{i}\ast=\psi^{i}(-1)^{F}$$
where $\ast$ must be a Hodge star operator (I will assume that there is a Riemannian metric on $M$ so that $\ast^{2}=1$.)
Can anybody explain to me how to derive the relations (4.17)

I calculated this by myself but I cannot obtain the correct $(-1)^{F}$ factor.
Let $\omega\in\Omega^{q}(M)$ be a differential form on $M$. In local coordinates, one has
$$\omega=\frac{1}{q!}\omega_{i_{1}\cdots i_{q}}dx^{i_{1}}\wedge\cdots\wedge dx^{i_{q}}$$ 
Hodge star operator is defined as 
$$\ast:\Omega^{q}(M)\rightarrow\Omega^{n-q}(M)$$
such that $\ast^{2}=1$.
One has
$$(\ast\omega)_{j_{1}\cdots j_{n-q}}=\frac{1}{q!}\epsilon^{i_{1}\cdots i_{q}}_{\qquad j_{1}\cdots j_{n-q}}\,\,\omega_{i_{1}\cdots i_{q}}$$
where the $\epsilon$ symbol is raised by the metric tensor. Therefore, one has
$$\ast\omega=\frac{1}{(n-q)!}\left(\frac{1}{q!}\epsilon^{i_{1}\cdots i_{q}}_{\qquad j_{1}\cdots j_{n-q}}\,\,\omega_{i_{1}\cdots i_{q}}\right)dx^{j_{1}}\wedge\cdots\wedge dx^{j_{n-q}}$$
Then, one has 
$$\psi^{i}\ast\omega=dx^{i}\wedge\ast\omega$$
$$=\frac{1}{(n-q+1)!}\left(\frac{(n-q+1)!}{(n-q)!q!}\epsilon^{i_{1}\cdots i_{q}}_{\qquad j_{1}\cdots j_{n-q}}\,\,\omega_{i_{1}\cdots i_{q}}\right)dx^{i}\wedge dx^{j_{1}}\wedge\cdots\wedge dx^{j_{n-q}}$$
Applying the Hodge star operator again, one has
$$(\ast\psi^{i}\ast\omega)_{k_{1}\cdots k_{q-1}}=\frac{1}{(n-q+1)!}\epsilon^{ij_{1}\cdots j_{n-q}}_{\qquad\quad\,k_{1}\cdots k_{q-1}}(\psi^{i}\ast\omega)_{ij_{1}\cdots j_{n-q}}$$
Thus, one has 
$$(\ast\psi^{i}\ast\omega)^{k_{1}\cdots k_{q-1}}=\frac{1}{(n-q)!q!}\epsilon^{ij_{1}\cdots j_{n-q}\,k_{1}\cdots k_{q-1}}\,\epsilon_{i_{1}\cdots i_{q}j_{1}\cdots j_{n-q}}\,\omega^{i_{1}\cdots i_{q}}$$
Rearranging indices of $\epsilon$ tensors, one has
$$\epsilon_{ij_{1}\cdots j_{n-q}\,k_{1}\cdots k_{q-1}}\epsilon^{i_{1}\cdots i_{q}j_{1}\cdots j_{n-q}}=(-1)^{(q-1)(n-q)}\epsilon_{ik_{1}\cdots k_{q-1}\,j_{1}\cdots j_{n-q}}\,\epsilon^{i_{1}\cdots i_{q}\,j_{1}\cdots j_{n-q}}$$
Using contraction rules of $\epsilon$ tensor, one has
$$\ast\psi^{i}\ast=(-1)^{(q-1)(n-q)}\chi^{i}$$
I expect to have $(-1)^{q}$. Where did I make mistakes? 
 A: The calculation can be performed in a coordinate free manner. 
I am following here: Walter Thirring: A course in mathematical physics 2: Classical field theory, especially page 14 section 1.2.16, where he defines contraction with respect to differential forms. (I am assuming flat Euclidean metric, this $**=1$)
In Thirring's notation the $\psi$ field is a wedge product with a one-form which I'll denote by $\nu\in \Omega^1$  to conform with the Thirring's notation; the $\chi$ field will be the contraction with $\nu$ denoted by $i_{\nu}$, which is the same as the contraction with the vector $g^{ij}\nu_j$ in coordinate notation.
Given $\omega \in \Omega^q$; then by using contractions with respect to forms, we have
$$*\omega = i_{\omega} \epsilon$$
where $\epsilon$ is the completely anti-symmetric tensor.
Thus
$$i_{\nu}*\omega = i_{\nu}i_{\omega} \epsilon = i_{\omega \wedge \nu} \epsilon =  (-1)^q i_{\nu\wedge \omega}\epsilon = (-1)^q *(\nu \wedge \omega) $$
Taking the hodge star of both sides, we obtain
$$*i_{\nu}*\omega = (-1)^q  \nu \wedge \omega$$
Replacing $\omega$ by $*\omega$ in the original equality, we obtain:
$$*(\nu \wedge *\omega)= (-1)^q i_{\nu}\omega$$
