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Dimensional reduction of 10D N=1 Super Yang Mills to 4D

  1. Jul 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Given [itex]N=1 SYM[/itex] in 10 dimensions (all fields in the adjoint representation):
    [itex]\int d^{10}x\, Tr\,\left( F_{MN}F^{MN}+\Psi\Gamma^M D_M\Psi\right)[/itex]
    [itex]D_M\Psi=\partial_M \Psi+i[A_M,\Psi][/itex] is the gauge covariant derivative.
    Reduce to 4 dimensions [itex]A_M=(A_\mu,\phi_i), \mu=0,\ldots,3, i=4,\ldots 9, \partial_i=0 \,\forall \,i[/itex] and show:
    [itex]F_{MN}F^{MN}=F_{\mu\nu}F^{\mu\nu}+D_\mu \phi_i D^\mu \phi^i + [\phi_i,\phi_j][\phi^i,\phi^j][/itex]

    2. Relevant equations

    3. The attempt at a solution

    My ansatz:
    [itex]F_{MN}F^{MN}=F_{\mu\nu}F^{\mu\nu}+F_{\mu i}F^{\mu i}+F_{i\nu}F^{i\nu}+F_{ij}F^{ij}[/itex]
    I have some trouble writing down what [itex]F_{\mu i}[/itex] and [itex]F^{\mu i}[/itex] are. I think my problem is that I only have the definition of the gauge covariant derivative acting on a field in the adjoint.
    [itex]F_{\mu i}=\frac{1}{i}[D_\mu,D_i]=\frac{1}{i}(D_\mu D_i - D_i D_\mu) = ?[/itex]
    Is [itex]D_i = i\phi_i[/itex]?

    Thanks for any help!

  2. jcsd
  3. Jul 25, 2013 #2


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    Up to sign and other conventions, yes. Remember that the ##\phi_i## are also in the adjoint representation. You seem to take conventions so that ##D_M = \partial_M +i A_M##. Then dimensional reduction tells to take ##\partial_i=0## when acting on any fields, so we recover [itex]D_i = i\phi_i[/itex].
  4. Jul 26, 2013 #3
    Thanks! Why can we write [itex]D_M=\partial_M+iA_M[/itex] if for some field [itex]X[/itex] in the adjoint [itex]D_M X=\partial_M X +i[A_M,X][/itex] (which is given in the exercise)? I don't see where the commutator should come from.

    I tried to determine [itex]F_{\mu i}=\frac{1}{i}[D_\mu,D_i][/itex] by letting it act on some [itex]X[/itex] in the adjoint:
    [itex][D_\mu,D_i]X=D\mu D_i X - D_i D_\mu X[/itex]
    [itex]=D_\mu(i[\phi_i,X])-D_i(\partial_\mu X+i[A_\mu,X])[/itex]
    [itex]=i\partial_\mu[\phi_i,X]-[A_\mu,[\phi_i,X]]-i[\phi_i,\partial_\mu X]+[\phi_i,[A_\mu,X]][/itex]
    Use the Jacobi identity in the last term and : [itex]\partial_\mu[\phi_i,X]=[\partial_\mu\phi_i,X]+[\phi_i,\partial_\mu X][/itex]
    The 2nd and 3rd term cancel.
    [itex]=i[\partial_\mu \phi_i+i[A\mu,\phi_i],X][/itex]
    I would have expected the result to be [itex]i(D_\mu\phi_i)X[/itex] without commutator. But even if that was the result i would get an unwanted factor 2:
    [itex]F_{MN}F^{MN}=F_{\mu\nu}F^{\mu\nu}+\color{red}{2} D_\mu \phi_i D^\mu \phi^i + [\phi_i,\phi_j][\phi^i,\phi^j][/itex]

    Where am I wrong?
  5. Jul 26, 2013 #4


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    The factor of 2 is correct. See pages 9-10 of http://arxiv.org/abs/hep-th/9801182. You can rescale the ##\phi_i##, which just puts a factor of 1/4 in front of the potential term.
  6. Jul 30, 2013 #5
    Thanks, there seems to be an error in the exercise then.

    I still have another question. I found for a field [itex]X[/itex] in the adjoint of the gauge group:
    [itex]F_{\mu i}F^{\mu i}X = \ldots = [D_\mu \phi_i,[D^\mu \phi^i,X]] = D_\mu \phi_i D^\mu \phi^i X+X D_\mu \phi_i D^\mu \phi^i[/itex]
    The calculation is equivalent to my previous post.
    What can i conclude from that for [itex]Tr\,(F_{\mu i}F^{\mu i})[/itex]?

    All the operators have gauge group indices:
    [itex]F_{\mu i}{}_a{}^b F^{\mu i}{}_b{}^c X_c{}^d = (D_\mu \phi_i)_a{}^b (D^\mu \phi^i)_b{}^c X_c{}^d+X_a{}^b (D_\mu \phi_i)_b{}^c (D^\mu \phi^i)_c{}^d[/itex]

    I need [itex]Tr\,(F_{\mu i}F^{\mu i})=F_{\mu i}{}_a{}^b F^{\mu i}{}_b{}^a[/itex]
    Can I get that from what i have done?
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