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  Intepreting Fermions as Differential Forms?

+ 2 like - 0 dislike
1212 views

In this paper,

https://projecteuclid.org/euclid.cmp/1104248307

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 xi. For any differential form aΩ(M), one has the operations

ψi:adxia,

and 

χj:aa(j).

One has the Clifford algebra 

{ψi,χj}=δij,{ψi,ψj}={χi,χj}=0

Define the Witten-index (1)F as 

(1)F:ω(1)qw,forωΩq(M).

Then one has the relation (equation 4.17)

ψi=(1)Fχi,χi=ψi(1)F

where must be a Hodge star operator (I will assume that there is a Riemannian metric on M so that 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 ωΩq(M) be a differential form on M. In local coordinates, one has

ω=1q!ωi1iqdxi1dxiq

 

Hodge star operator is defined as 

:Ωq(M)Ωnq(M)

such that 2=1.

One has

(ω)j1jnq=1q!ϵi1iqj1jnqωi1iq

where the ϵ symbol is raised by the metric tensor. Therefore, one has

ω=1(nq)!(1q!ϵi1iqj1jnqωi1iq)dxj1dxjnq

Then, one has 

ψiω=dxiω

=1(nq+1)!((nq+1)!(nq)!q!ϵi1iqj1jnqωi1iq)dxidxj1dxjnq

Applying the Hodge star operator again, one has

(ψiω)k1kq1=1(nq+1)!ϵij1jnqk1kq1(ψiω)ij1jnq

Thus, one has 
(ψiω)k1kq1=1(nq)!q!ϵij1jnqk1kq1ϵi1iqj1jnqωi1iq

Rearranging indices of ϵ tensors, one has

ϵij1jnqk1kq1ϵi1iqj1jnq=(1)(q1)(nq)ϵik1kq1j1jnqϵi1iqj1jnq

Using contraction rules of ϵ tensor, one has

ψi=(1)(q1)(nq)χi

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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