Extended supersymmetry and chiral gauge theories

Question

I'm trying to understand the argument that extended supersymmetry cannot produce the chiral structure of the Standard Model, as explained on page 25 of these notes. My impression of the argument goes like this:

• All particles in a supermultiplet must transform the same way under internal symmetries.
• All massless particles with helicity $|\lambda| = 1$ must be created by gauge fields, because gauge symmetry is the only way to get rid of the extra degrees of freedom.
• Gauge fields must transform under the adjoint representation of any gauge group, and for all matrix Lie groups the adjoint representation is real.
• In extended SUSY, with one exception, all multiplets containing $|\lambda| = 1/2$ also contain $|\lambda| = 1$. Thus we can only get fermions in real representations of any gauge group.
• This is incompatible with the chiral structure of the Standard Model, which requires that fermions live in complex representations, such as left-handed quarks, which are in the doublet representation of $SU(2)_L$.

Granted that I paraphrased the argument correctly, I'm fine with every step except for the last step, which just doesn't seem right to me.

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Here are the preliminary thoughts I've had about this:

• It's ambiguous whether we're talking about the representations of fields or representations of particles. These are different but I haven't been able to make the argument work either way.
• It's ambiguous whether 'complex' means $\text{complex}_1$, i.e. the base field is the complex numbers, or $\text{complex}_2$, i.e. the representation is $\text{complex}_1$ and not equivalent to its conjugate. I think the only sensible option is $\text{complex}_2$.
• The doublet representation of $SU(2)_L$ is not $\text{complex}_2$! There is only one two-dimensional representation of $SU(2)_L$ so it must necessarily be equivalent to its conjugate.
• In this answer it is argued that if $q_L$ transforms in a representation $R$, then $q_R$ transforms in the conjugate representation $\bar{R}$. Therefore if $R$ is real, $q_L$ and $q_R$ transform the same way, which contradicts observation. But this seems clearly wrong, it's not true for $q_L$ and $q_R$ in the Standard Model!

Answer 1

1. You must be talking about the representations of the fields, since the particles are quantum states that always live in infinite-dimensional complex Hilbert space with corresponding unitary representations on it.

2. Let us make clear what we really mean by "chirality" of the Standard Model. Given a Dirac fermion $\psi$ with its chiral parts/Weyl fermions $\psi_L,\psi_R$, chirality means that $\psi_L$ transforms under a different representation of at least one gauge group (the weak one in the case of the SM) than $\psi_R$. For how exactly this translates to our ordinary natural language of left/right-handed electrons and positron, see this answer of mine to a previous question of yours.

3. The meaning of "complex" here is "not real", not "complex" as in "not real or pseudoreal". For the difference between real, pseudoreal and complex, see this answer of mine to another previous question of yours.

4. When we say that "a Weyl spinor" $\psi_{R/L}$ transforms in the representation $\rho$ of a gauge group, what we really formally would have to say that the object transforms in the combined representation $(1/2,0)\otimes \rho$ where $(1/2,0)$ is the usual notation for the Weyl spinor representation of the Lorentz group.

5. Finally, we are in a position to observe the following: Given a massless multiplet with a Weyl fermion of any handedness in it and a real representation of the gauge group, we can split the result of the tensor product since we can split the real representation $\rho$ into two representations on real vector spaces as $\rho_\mathbb{R} + J\rho_\mathbb{R}$, where $J$ is basically a complex conjugation map: $$ (1/2,0) \otimes \rho = \left( (1/2,0) \otimes \rho_\mathbb{R} \right) \oplus \left( (1/2,0) \otimes J\rho_\mathbb{R}\right)$$

All that is left to do is to observe that $(1/2,0)\otimes J\rho_\mathbb{R}$ is equivalent to $J(1/2,0)\otimes \rho_\mathbb{R}$, and that $J(1/2,0) = (0,1/2)$ - the conjugate representation of a left-handed Weyl fermion is a right-handed Weyl fermion. Therefore, choosing a real representation $\rho$ of the gauge group for a Weyl fermion actually results in the Weyl fermion transforming in $\rho_\mathbb{R}$ and its opposite-chirality version also transforming in $\rho_\mathbb{R}$, explaining why real representations can never lead to chirally symmetric theories.

Observe also that this argument fails for a pseudoreal representation, which also resolves the apparent contradiction with the doublet of $\mathrm{SU}(2)_L$ being "not complex".