SUSY $SU(5)$ GUT chiral adjoint repersentations

when you have a chiral superfield in the adjoint representation in your theory (i.e the Higgs $\Sigma_{24}$ in $SU(5)$), you can write it as a $5\times5$ Matrix: $(\Sigma_{24})_i^j=\Sigma^a (t^a)_i^j$ (sum over a), where the $t^a$ are the generatorns of the $SU(5)$.

My question is: is it possible that the $\Sigma_a$ are real, or do they have to be complex for $SU(5)$ SUSY GUT?

I'm asking because some steps in the calculation of the minimum (vev) seem to need real $\Sigma_a$'s, since that would make $\Sigma_{24}$ Hermitian. But on the other side I'm not sure if real scalars would make problems with supersymmetry (degrees of freedom).

thanks

Update:

The problem with real $\Sigma_a$ is the following:

Suppose you want to write down the superfield $S$ in components. For one complex scalar $\Phi$ it would look as follows: $S=\Phi+\sqrt{2} \theta \Psi+\theta\theta F$. Looking at the degrees of freedome on-shell $\rightarrow dof(\Phi)=2, dof(\Psi)=2,dof(F)=0$

Now looking at the adjoint representation of $\Sigma$, wich is a 24-dimesional multiplett ($\Sigma_1,...,\Sigma_{24}$)(this are the $\Sigma_a$ from above, (btw. $\Sigma_a=\Sigma^a$, sorry for mixing notation)). Now each field in this supermultiplett has to be a superfield, like above $\Sigma_a=\Phi_a+\sqrt{2} \theta \Psi_a+\theta\theta F_a$. Counting now the degress of freedome reveals the problem: $\rightarrow dof(\Phi_a)=1, dof(\Psi_a)=2,dof(F_a)=0$. That shows that the fermionic dof. are not equal to the bosonic dof. wich contradicts SUSY.

Edit:

• $i,j$ in $\Sigma_i^j$ are matrix indices and run from 1 to 5.

• $a$ in $t^a$ and $\Sigma^a$ runs from 1 to 24 (24 = dimension of adjoint representation of $SU(5)$)

• Could you explain your notation in a little more detail? For instance, to what are the characters $j$, $i$, and $a$ referring? It would help the reader (though I imagine Susy crowd probably understands this already)... – honeste_vivere Sep 11 '16 at 18:11
• sure I added a bit of detail ;) – johnny1557 Sep 11 '16 at 18:50
• @Johnny1557, are you asking that is it possible to choose Higgs in the adjoint representation $SU(5)$ to be real? – AMS Sep 11 '16 at 19:07
• Yes. I know that it is possible in non-susy models, but I'm not sure about SUSY. Since Wikipedia tells me "For particles that are real scalars (such as an axion), there is a fermion superpartner as well as a second, real scalar field",(en.wikipedia.org/wiki/Superpartner). But what if I have 24, real scalars (i.e the adjoint of $SU(5)$), do I need another 24 ??? – johnny1557 Sep 11 '16 at 19:44
• your question is edited. please check whether it reflects your actual query or not. – AMS Sep 11 '16 at 20:19

In any theory in which supersymmetry is relevant down to 1 TeV, the supersymmetric partners of the leptons are spinless bosons with the right $SU(2)\times U(1)$ properties to be Higgs mesons. However, for various reasons, they cannot be.
The most serious objection to using the lepton partners as Higgs mesons is that the partners of the charge $-1/3$ quarks then mediate baryon number changing interactions. To eliminate this difficulty, the Higgs doublets must be part of a different $SU(5)$ multiplet whose color triplet components are superheavy. To ensure that baryon number is conserved except by the usual $SU(5)$ baryon number changing interactions, we require that the Higgs multiplets couple to pairs of quark and lepton multiplets just as in the usual $SU(5)$ model.
This in turns saying that taking the representations of Higgs in the adjoint (for real) just like minimal $SU(5)$, is acceptable.
• Consider the case that one have to show that $SU(5)$ SUSY GUT in anomaly free. Which representation of Higgs you gonna take, in order to make sure the anomaly cancellation? Answer of this question will fix the possible representations (real, pseudo real or complex) of particle. – AMS Sep 11 '16 at 21:46
• Another point is, anomaly coefficient for the adjoint representation of $SU(N)$ gauge group is zero. – AMS Sep 12 '16 at 16:53