BRS operator, ghost fields, Grassmann numbers

  • #1
Hi, I am learning what is going on with ghost fields in string theory. Does anyone know where I can find the basics for the brs (brst) operator? Specifically I need help with the following calculation. I have for the action of the brs operator on a ghost field and on the metric
[tex]s \xi^\alpha = \xi^\beta \partial_\beta \xi^\alpha [/tex]

[tex] s g_{\alpha\beta}=\Omega g_{\alpha\beta} + L_{\xi} g_{\alpha\beta}[/tex]

Omega is the Weyl field and L_xi the killing derivative.

I am trying to show explicitly that s^2 is zero when acting on the fields. I cannot show it for the ghost field or for the Weyl field Omega for which being a scalar the brs operator acts as

[tex]s \Omega = \xi^{\alpha} \partial_{\alpha} \Omega[/tex]

I believe it has to do with considering the ghost field as grassmann numbers and exploiting some property but I could not see what is going on only from the basic property x^2=0 and x1 x2=-x2 x1.
 
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  • #2
dextercioby
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<Quantization of gauge systems> by Henneaux and Teitelboim is the best reference on BRST symmetry.

Why can't you show that [itex] s^2 \Omega =0 [/itex] ?
 
  • #3
bigubau please bare with me I do not want spoon feeding.

This is my first time with grassmann numbers but so far it goes like this

[tex]s \Omega = \xi^\alpha \Omega_{,\alpha} [/tex]
[tex]s^2 \Omega = s (\xi^\alpha) \Omega_{,\alpha} + \xi^\alpha (s \Omega )_{,\alpha}[/tex]

I am assuming here that the partial differentiation operator and s commutes (do you agree?) as with any other differential operators I know, I don't have a formal definition of s at hand so I am not sure what is going on.

[tex]s^2 \Omega = \xi^\beta \xi^\alpha_{,\beta} \Omega_{,\alpha} + \xi^\alpha ( \xi^\beta \Omega_{,\beta} )_{,\alpha} [/tex]

[tex]s^2 \Omega = \xi^\beta \xi^\alpha_{,\beta} \Omega_{,\alpha} + \xi^\alpha \xi^\beta_{,\alpha} \Omega_{,\beta} +\xi^\alpha \xi^\beta \Omega_{,\beta \alpha} [/tex]

I tried several manipulations to cancel out the terms but no luck. So I believe it has to do with some property of grassmann numbers.

I believe I have to use the property I came up with that since
x1x2=-x2x1

x1x2+x2x1=0

if I have something like x^a x^b A_ab with the components of A_ab real numbers and x^a grassmann numbers then x^a x^b A_ab=0 if A_ab is symmetric. In 2d I get

x^a x^b A_ab= x1x1A11 + x2x2A22 + x1x2A12 + x2x1A21.

The first two terms are zero because xx=0 but for the next two I need A12=A21. This is ok for the last term above with Omega_,ab because partial differentiation commutes. But for the other terms I am not sure how to justify things.
 
  • #4
dextercioby
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Omega and xi are fermionic variables, which mean that their ANTIcommutator is 0. You've made a mistake in your second equation. The BRST differential is a fermionic derivation, let's choose it acting from the left. That's why:

[tex] s^2 \Omega = s (\xi^\alpha) \Omega_{,\alpha} - \xi^\alpha (s \Omega )_{,\alpha} [/tex]

That minus makes the world of difference, because it allows to establish the desired nillpotency.
 
  • #5
Thanks, so let me see if I got it right. When I wrote

[tex] s^2 \Omega = \xi^\beta \xi^\alpha_{,\beta} \Omega_{,\alpha} + \xi^\alpha \xi^\beta_{,\alpha} \Omega_{,\beta} +\xi^\alpha \xi^\beta \Omega_{,\beta \alpha} [/tex]

the two plus signs would be minus signs. So the first two terms cancel out (renaming the indices) and the third is zero because

[tex]\xi_\alpha \xi_\beta \Omega_{,\beta \alpha} = -\xi_\beta \xi_\alpha \Omega_{\beta \alpha}=-\xi_\beta \xi_\alpha \Omega_{,\alpha \beta}= -\xi_\alpha \xi_\beta \Omega_{,\beta \alpha}[/tex]

hence this term is zero. would you think of it in a different way?

I don't think I understand how to derive correctly
[tex] s^2 \Omega = s (\xi^\alpha) \Omega_{,\alpha} - \xi^\alpha (s \Omega )_{,\alpha} [/tex]
with a minus from the anticommutativity of omega and xi alone. Is this a matter of definition? Can you give me a defining property of fermionic derivation operators so that I can derive these things?

Thanks again
 
  • #6
ok I am getting really confused.

A is grassmann (fermionic field). So A^2=0. For a derivation operator acting "normally"

d(A^2)=dA A + A dA =0 thus dA A=-A dA which is a property of grassmann numbers (see http://www.mathematics.thetangentbundle.net/wiki/Grassmann_number" [Broken] )

but if I make the derivation using the rule you gave me for a fermionic operator

s(A^2)=sA A - A sA =0 thus sA A=A sA and so
s(A^2)=2 sA A=0

thus sA A=A sA =0

??
 
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  • #7
dextercioby
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Thanks, so let me see if I got it right. When I wrote

[tex] s^2 \Omega = \xi^\beta \xi^\alpha_{,\beta} \Omega_{,\alpha} + \xi^\alpha \xi^\beta_{,\alpha} \Omega_{,\beta} +\xi^\alpha \xi^\beta \Omega_{,\beta \alpha} [/tex]

the two plus signs would be minus signs. So the first two terms cancel out (renaming the indices) and the third is zero because

[tex]\xi_\alpha \xi_\beta \Omega_{,\beta \alpha} = -\xi_\beta \xi_\alpha \Omega_{\beta \alpha}=-\xi_\beta \xi_\alpha \Omega_{,\alpha \beta}= -\xi_\alpha \xi_\beta \Omega_{,\beta \alpha}[/tex]

hence this term is zero. would you think of it in a different way?

No, excellent.

christodouloum said:
I don't think I understand how to derive correctly
[tex] s^2 \Omega = s (\xi^\alpha) \Omega_{,\alpha} - \xi^\alpha (s \Omega )_{,\alpha} [/tex]
with a minus from the anticommutativity of omega and xi alone. Is this a matter of definition? Can you give me a defining property of fermionic derivation operators so that I can derive these things?

I don't have Henneaux's book, nor the notes from the courses I took on BRST, so I can't give you the full definitions/axioms. So take them like that and I will update you on this at a certain point in the future.
 
  • #8
dextercioby
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ok I am getting really confused.

A is grassmann (fermionic field). So A^2=0. For a derivation operator acting "normally"

d(A^2)=dA A + A dA =0 thus dA A=-A dA which is a property of grassmann numbers (see http://www.mathematics.thetangentbundle.net/wiki/Grassmann_number" [Broken] )

but if I make the derivation using the rule you gave me for a fermionic operator

s(A^2)=sA A - A sA =0 thus sA A=A sA and so
s(A^2)=2 sA A=0

No, it's wrong, where did you get that ? s(A^2) = s(0) = 0 = (sA)A-A(sA), because both s and A are fermionic => (sA)A=A(sA) which brings nothing new, because it's obvious from sA being a bosonic variable.
 
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  • #9
Forget about my last post I was getting tired... thanks for all the help I am much better off now. I will look for the book you proposed
 
  • #10
Firstly after some studying the fermionic operators are Z2-graded (even-odd rank for a differential form) differential operators, and s here was a left such derivative. That means that depending on the grade of the two forms a,b in s(ab) the sign is decided as + if a is of even rank and - if odd.

Second, I noticed something interesting related to our discussion before. A p-form is defined as fully antisymmetric in its arguments, which for a 2-form translates as antisymmetry in the two indices of the components:

[tex]/omega_{/alpha /beta}=-/omega_{/beta /alpha}[\tex]

and so I noticed that the contraction of a 2-form with any two vector fields would vanish because

[tex]/omega_{a b} u^a v^b=-/omega_{ b a} u^a v^b=-/omega_{b a} u^b v^a=-/omega_{a b} u^a v^b[\tex]

which is the same think tha happened two posts above
 
  • #11
dextercioby
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it doesn't vanish, the 2 form picks up the antisymmetrized product of the 2 vectors.

Please, use [/tex]
 

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