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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.

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Angie38750
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2 Answers2

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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.

knzhou
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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.

Frederic Brünner
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