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

[physicus]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

Thanks for any help!

physicus

 

[fzero]

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

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 D_i = i\phi_i.

 

[physicus]

Thanks! Why can we write D_M=\partial_M+iA_M if for some field X in the adjoint D_M X=\partial_M X +i[A_M,X] (which is given in the exercise)? I don't see where the commutator should come from.

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

Where am I wrong?

 

[fzero]

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.

 

[physicus]

Thanks, there seems to be an error in the exercise then.

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

All the operators have gauge group indices:
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

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