# 2D anomaly-free condition for a gauge theory

+ 4 like - 0 dislike
246 views

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

+ 3 like - 0 dislike

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

$$N_L = N_R$$

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

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
answered Jul 19, 2018 by (1,925 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.