2D anomaly-free condition for a gauge theory

Question

Take a $SU(2)$ gauge theory in 2d spacetime, say there are $n_1$ left-handed Weyl fermion in spin-1 written as

$$1_L,$$ and $n_0$ left-handed Weyl fermion in spin-0 written as $$0_L .$$ and $n_{1/2}$ right-handed Weyl fermion in spin-1/2 written as $$\frac{1}{2}{}_R .$$

There are total certain number of left-handed Weyl fermions and right-handed Weyl fermions.

Can we show the theory is anomaly free from $SU(2)^2$-anomaly? (2 pt function coupling to $SU(2)$ field)? and other anomalies?

This post imported from StackExchange Physics at 2020-11-08 17:29 (UTC), posted by SE-user annie marie heart

Answer 1

The anomaly cancellation condition for the gravitational part of the chiral anomaly (neglecting gauge charges) is simply that there are the same number of left and right moving fields

$$N_L = N_R$$

which in your case is

$$3 N(1_L) + N(0_L) = 2 N(1/2_R),$$

where $N(R)$ indicates the number of (complex) fields in each representation. The prefactors are the (complex) dimensions of the gauge multiplets.

To determine the gauge part of the chiral anomaly, note that the $SU(2)$ Chern-Simons level may be determined from its $U(1)$ subgroup, so it is equivalent to determine the $U(1)$ gauge anomaly. This anomaly cancellation condition is

$$\sum_i q_i^2 = \sum_j q_j^2,$$

where $i$ indexes left-moving charge carriers and $j$ indexes right-moving charge carriers.

The $SU(2)$ triplet has a charge $+2$ and a charge $-2$ carrier, the doublet has a $+1$ and a $-1$, and the singlet has no charge carriers. So the anomaly cancellation condition is

$$8 N(1_L) = 2 N(1/2_R).$$

Note that this technique works for any connected compact gauge group, since the Chern-Simons levels are determined by their maximal torus. Thus, the chiral anomaly in 1+1D always comes down to a number of $U(1)$ anomalies (plus the gravitational part) which must be checked.

This post imported from StackExchange Physics at 2020-11-08 17:29 (UTC), posted by SE-user Ryan Thorngren