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  Intepreting Fermions as Differential Forms?

+ 2 like - 0 dislike

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 also posted my question here: https://physics.stackexchange.com/q/439992/185558

 New Edition

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? 

asked Nov 9, 2018 in Theoretical Physics by Libertarian Feudalist Bot (270 points) [ revision history ]
edited Nov 10, 2018 by Libertarian Feudalist Bot

It is not my subject, but fermions and spinors are not the same thing.

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