# SUSY gauge theories and representations

Hi,

I'm currently reading "Supersymmetry demystifed" by Patrick Labelle, chapter 10, about SUSY non-Abelian gauge theories.

We have a Lagrangian with SU(N)-gauge fields, and gaugino's. What puzzles me are the following claims of Labelle about the representations. In the SUSY-transformation of the gauge fields A he remarks (and it doesn't seem to be a typo) "that the gauge fields A don't transform in the adjoint representation" (page 221).

Maybe my knowledge of gauge theories is a bit rusty, but gauge fields transform in de adjoint representation and SUSY doesn't change this fact, right? Is it some sort of convention? Maybe there are more people who read this part of Labelle's book and are puzzled?

I haven't looked at the book, but maybe he's just pointing out that the field strengths transform in the adjoint, while the potentials transform inhomogeneously.

tom.stoer
I haven't looked at the book, but maybe he's just pointing out that the field strengths transform in the adjoint, while the potentials transform inhomogeneously.
The reason why gauge potentials live in the adjoint rep. is that you get the adjoint rep. by taking the product fund rep. * fund rep'; as the fermions live in the fund. rep. and the lagrangian is biliear in the fermions, the gauge fields must be in the adjoint.

Gauge fields must always transform under the adjoint rep, this is completely independent from whether there is SUSY or not, or any other fields like fermions. If the theory is supersymmetric, then there are extra fermionic partners of the gauge fields, which also must transform in the adjoint rep.

@Tomstoer: I'm not sure what you mean by your product of two rep's; could you elaborate on that?

@Surprised: Yes, that's what i also thought, that's why I'm confused by his section on gauge theories. These gaugino's indeed transform in de adjoint, and these also appear in the susy transformation of the gauge potential. Labelle then remarks that potentially there could be a problem, because "the gaugino's transform in de adjoint while the gauge potential A does not".

. Labelle then remarks that potentially there could be a problem, because "the gaugino's transform in de adjoint while the gauge potential A does not".
Well perhaps he meant that the gauge field transforms also with a shift, ie, as a connection and not as an ordinary tensor in the adjoint rep?

tom.stoer
@Tomstoer: I'm not sure what you mean by your product of two rep's; could you elaborate on that?
It's about coupling of multiplets. The most famous example is the coupling of one 3-dim. fundamental triplet rep. of SU(3) containing the quarks (u,d,s) times the conjugate fund. rep. containing the anti-quarks(anti-u,anti-d, anti-s). This product contains two irreducible reps., namely the 8-dim octet containing the pseudo-scalar mesons (3 pions, 4 K, 1 eta) and one 1-dim. singulet which (the eta-prime). The octet is the adjoint rep. of the flavor-SU(3).

See http://en.wikipedia.org/wiki/Quark_model

tom.stoer
Gauge fields must always transform under the adjoint rep, this is completely independent from ... or any other fields like fermions.
Why? I mean I know that this is the case, but why?

tom.stoer
Are you happy with the explanations

?
No, not really :-)

This well known construction just shows that (and how) it works, but it does neither exclude other possibilities nor does it give us a deeper reason for this specific construction.

There must be some deep connection between the Lorentz symmetry and the gauge symmetry which forces the gauge potentials to transform inhomogeniously and which excludes other representations. Perhaps SUSY and / or SUGRA provide a way to understand it but from gauge symmetry itself it looks accidental.

If a certain Lie group G is gauged, then by definition the gauge fields take values in the Lie algebra of G, and thus must transform in the adjoint rep. This can easily be seen as follows.

Gauge fields are supposed to compensate for local (space-time dependent) gauge transformations, which mess up when there are derivatives acting on the fields. These derivatives must be changed to covariant derivatives in which the gauge field enters as connection:

d_mu -> D_mu = d_mu + A_mu

Then under local gauge transformations U=exp(lambda(x)^a T^a) the connection transforms with an inhomogenous piece

delta(A_mu) = U^(-1) d_mu U = d_mu lambda(x)^a T^a

so that D_mu transforms covariantly as a tensor (ie without a shift).
Now the variation, being proportional to the generators T^a, takes values in the Lie algebra of the gauge group (defined to be the tangent space of the group, and this is precisely what is expressed by writing U=1+lambda^a T^a+…. near the origin). So this means that the gauge fields necessarily transform like the generators T^a, which form the adjoint rep of the group G.

What seems to be confusing, especially also in that cited thread and in the last posts there,
https://www.physicsforums.com/showpost.php?p=1526666
is how this is related to the representation of the fields to which the connection couples; for example, what if the matter fields are not in the fundamental rep, etc.

This is elementary!... the gauge fields taking values in the adjoint rep can be represented by matrices _in_ the representation of the matter field. In other words, apart from the lorentz index mu, a gauge field can be written as

(A)^i_j = A^a (T^a)^i_j

where "a" runs over the number of generators, T^a. The generators act on arbitrary other representations R by matrix multplication, so that i,j run from 1 to dim(R), the dimension of the rep R. More precisely, i corresponds to R and j to the complex conjugate representation.

R can be any rep, say 3 or 27 or adjoint of SU(3), even the singlet rep for which the generators vanish. This means that the combination of fields

psi*^j (A)^i_j psi_i

is invariant, which corresponds to the singlet in the tensor product R* x adjoint x R.

Even if there are no matter fields around, this applies as well to the gauge fields themselves, whose kinetic terms contain derivatives acting on the gauge fields. For these couplings, the gauge fields are represented by matrices

(A^a)^b_c

where all indices run over the same range, ie, the adjoint rep. (For abelian gauge theories, the self-couplings of the gauge fields vanish of course).

tom.stoer
@suprised: I agree with everythign you are writung here, but I am questioning the starting point

$$\partial_\mu \to D_\mu$$

and

$$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu +i[A_\mu, A_\nu]$$

If you take this as input it's (nearly) obvious that you will always arrive at the adjoint rep. But the starting point already is more or less identical to what you want to prove. The last equation says that by construction the gauge fields transform in the adjoint rep. as the adjoint rep. uses the commutator for it's very definition.

The question remains: why must a spin-1 field sit in the adjoint rep. (whereas fermions can sit in the fund rep. or in the adjoint rep.)? why does the Dirac operator not enforce an additional condition on the fermions (whereas the construction of the field strength does)? what are the reasons to start with exactly the equations we have just written down - except for the fact that we know that they are working and produce viable and theories?

There is a relation between spacetime and gauge symmetry which is described by the equations, but not explained.

@suprised: I agree with everythign you are writung here, but I am questioning the starting point

$$\partial_\mu \to D_\mu$$
Well it depends what you want to do - if you like to right down a theory that is invariant under local gauge transformations, and which contains derivatives acting non-trivially on those transformations, then you need a connection in order to cancel the effect of those derivatives.

And as I was saying above, such connections lie in the tangent space of the gauge group (if we talk about cancelling infinitesimal gauge transformations), and that is per def the adjoint rep.

If you change the starting point, eg by considering a theory without space-time derivatives in the lagrangian, then you don't need gauge connections for having local gauge invariance.