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

1. Jul 25, 2013

physicus

1. The problem statement, all variables and given/known data

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:
$F_{MN}F^{MN}=F_{\mu\nu}F^{\mu\nu}+D_\mu \phi_i D^\mu \phi^i + [\phi_i,\phi_j][\phi^i,\phi^j]$

2. Relevant equations

3. 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

2. Jul 25, 2013

fzero

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 $D_i = i\phi_i$.

3. Jul 26, 2013

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?

4. Jul 26, 2013

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.

5. Jul 30, 2013

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?