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

$$\psi^{i}:a\rightarrow dx^{i}\wedge a,$$


$$\chi_{j}: a\rightarrow a(\partial_{j}).$$

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


Then one has the relation (equation 4.17)


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


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


$$=\frac{1}{(n-q+1)!}\left(\frac{(n-q+1)!}{(n-q)!q!}\epsilon^{i_{1}\cdots i_{q}}_{\qquad j_{1}\cdots j_{n-q}}\,\,\omega_{i_{1}\cdots i_{q}}\right)dx^{i}\wedge dx^{j_{1}}\wedge\cdots\wedge dx^{j_{n-q}}$$

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


I expect to have $(-1)^{q}$. Where did I make mistakes?

  • $\begingroup$ What's the definition of $*$? $\endgroup$ Commented Nov 9, 2018 at 22:03
  • $\begingroup$ I think it's just the usual Hodge star. $\endgroup$
    – Valac
    Commented Nov 9, 2018 at 22:11
  • $\begingroup$ The usual Hodge star acts on $\Omega$, not on $\mathrm{End}(\Omega)$. So there must be some explicit definition somewhere. $\endgroup$ Commented Nov 9, 2018 at 22:14
  • $\begingroup$ This hodge star here is acting on forms, not on the fermionic operators. $\endgroup$
    – Valac
    Commented Nov 9, 2018 at 22:16
  • $\begingroup$ Oh, so $*\psi*$ is actually $*\circ\psi\circ*$? That makes more sense. Have you tried acting with both sides on an arbitrary form? It shouldn't be too difficult. $\endgroup$ Commented Nov 9, 2018 at 22:18

1 Answer 1


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:

$$*(\nu \wedge *\omega)= (-1)^q i_{\nu}\omega$$


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