In this John Preskill et al paper, a statement is made in page 1:

We will refer to a gauge theory with fermions transforming as a complex representation of the gauge group as a chiral gauge theory, because the gauged symmetry is a chiral symmetry, rather than a vector-like symmetry (such as QCD).

But my question is: why does a Complex Representation of gauge group imply a Chiral Gauge Theory?

If fundamental representation of SU(3) is a complex representation (with complex conjugate anti-fundamental Rep), then isn't QCD with fundamental representation of SU(3) a perfect counter example where the gauge symmetry is vector-like, instead of chiral???

ps. See this page, or learn that from Wiki:

In physics, a complex representation is a group representation of a group (or Lie algebra) on a complex vector space that is neither real nor pseudoreal. In other words, the group elements are expressed as complex matrices, and the complex conjugate of a complex representation is a different, non-equivalent representation. For compact groups, the Frobenius-Schur indicator can be used to tell whether a representation is real, complex, or pseudo-real.

For example, the N-dimensional fundamental representation of SU(N) for N greater than two is a complex representation whose complex conjugate is often called the antifundamental representation.


This answer is coming extremely late, but hopefully it's helpful to someone.

For simplicity let's consider one quark flavor. There are two distinct Weyl spinors involved here: a left-chiral Weyl spinor $q_L$ and a right-chiral Weyl spinor $q_R$, both of which transform in the fundamental representation $3$. Intuitively, the theory is not chiral, because both chiralities are being treated the same way, but how is this equivalent to Preskill's definition, when $3 \oplus 3$ is complex?

Note a left-chiral Weyl spinor in a representation $R$ is exactly the same thing as a right-chiral Weyl spinor in the conjugate representation $R^*$ (as I explain here). So it is inherently ambiguous what representation "the fermions" transform under. Depending on whether I want to use left-chiral or right-chiral spinors, it could be $3 \oplus 3$, $3 \oplus \bar{3}$, or $\bar{3} \oplus \bar{3}$.

The convention used here is to make everything left-chiral for consistency, just as done in GUT model building. Then the representation for the quark is $3 \oplus \bar{3}$ which is perfectly real. Contrast this with the electroweak theory with $SU(2)_L \times U(1)_Y$, where one quark generation would be in $(2, 1/6) \oplus (1, -2/3) \oplus (1, 1/3)$, which is complex.


The statement you cited does not imply that a complex representation of a gauge group implies a chiral gauge theory in general. This only holds true if the gauge group corresponds to a chiral symmetry in the first place. A chirally symmetric theory contains massless fermions.

Regarding your counterexample: it is true that QCD contains fermions in the complex representation of the gauge group. However, the gauge group in this case is not chiral symmetry, but $SU(N_c)$ colour symmetry. Hence it is possible that chiral symmetry is broken and fermions acquire mass. This can happen through spontaneous, explicit and anomalous symmetry breaking.


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