David Tong Standard Model Solutions — Sheet 3: Anomalies

These prompts paraphrase David Tong's The Standard Model, Example Sheet 3 (February 2026). Problem numbers, starred status, equations, and guidance follow the official sheet.

1 Sheet 3, Problem 1: Chiral $U(1)$ Charge Assignments

Prove that four left-handed Weyl fermions with integer charges cannot form a consistent chiral $U(1)$ gauge theory that can also couple to gravity. Exhibit a consistent example with five such fermions.

2 Sheet 3, Problem 2*: Gauge and 't Hooft Anomalies in a Chiral Theory

Couple $SU(N)$ to one left-handed Weyl fermion $\chi$ in the two-index symmetric representation and to $p$ left-handed Weyl fermions $\psi_i$ in the anti-fundamental. Determine $p$ for gauge anomaly cancellation.

Classically, $SU(p)$ rotates the $\psi_i$, while $U(1)_\chi$ rotates only $\chi$ and $U(1)_\psi$ rotates only the $\psi_i$. Show that each Abelian symmetry is anomalous and find a surviving linear combination. Calculate its $SU(p)^3$, $SU(p)^2U(1)$, $U(1)^3$, and mixed $U(1)$--gravity 't Hooft anomalies.

The proposed confined spectrum consists of gauge-singlet fermions \[\lambda_{ij}=\psi_{[i}(\chi\psi_{j]}),\tag{2.1}\] in the antisymmetric representation of the global $SU(p)$. Find their surviving $U(1)$ charge and match all the anomalies computed above.

Useful source data: With $I(\bar R)=I(R)$ and $A(\bar R)=-A(R)$, the $SU(N)$ representation data are

$R$	fundamental	adjoint	symmetric	antisymmetric
$\dim R$	$N$	$N^2-1$	$N(N+1)/2$	$N(N-1)/2$
$I(R)$	$1$	$2N$	$N+2$	$N-2$
$A(R)$	$1$	$0$	$N+4$	$N-4$

For the theory to be consistent, we must have \[\sum_i A(R_i) = A(\mathrm{sym}) + pA(\bar{\mathbf{N}}) = N+4 -p = 0 \implies p = N+4.\tag{2.2}\]

Then, for the $U(1)_\chi SU(N)^2$ mixed anomaly, we have \[\mathcal{A}_{U(1)_\chi SU(N)^2} = \sum q_i I(R_i) = I(\mathrm{sym}) = N+2,\tag{2.3}\] while for the $U(1)_\psi SU(N)^2$ \[\mathcal{A}_{U(1)_\psi SU(N)^2} = \sum q_i I(R_i) = pI(\bar{\mathrm{fund}}) = N+4,\tag{2.4}\] which are both anomalous. For a general $U(1)$ transformation, let us take the charge of $\chi$ to be $Q$ and the charges of $\psi_i$ to be $q$. For the symmetry to be anomaly-free, we must have \[\mathcal{A}_{U(1) SU(N)^2} = pq + Q(N+4) = 0.\tag{2.5}\] A convenient choice of charge is $ q= -(N+2)$ and $Q = N+4$.

Now, we turn to the computation of the 't Hooft anomalies: \[\mathcal{A}_{SU(p)^3} = \sum_{\text{gauge index}} A(\mathbf{p}) = N,\tag{2.6}\] \[\mathcal{A}_{SU(p)^2U(1)} = \sum_{\text{gauge index}}q I(\mathbf{p}) = -N(N+2),\tag{2.7}\] \[\mathcal{A}_{U(1)^3} = \sum_{\text{gauge index}} \sum_i q^3_i = Npq^3 + \dim(\mathrm{sym}) Q^3 = -\frac{N^3(N+3)(N+4)}{2},\tag{2.8}\] \[\mathcal{A}_{U(1)} = \sum_{\text{gauge index}} \sum_i q_i = Npq + \dim(\mathrm{sym}) Q = -\frac{N(N+4)(N+3)}{2}.\tag{2.9}\]

Under $U(1)$ the $\lambda$ fermion transform as \[U(1):\quad \lambda^{ij} \to e^{i(Q+2q)\alpha}\lambda^{ij},\tag{2.10}\] which has a charge $q_\lambda = -N$.

Since the $\lambda$ fermion is a gauge singlet, the 't Hooft anomalies of this theory are given by \[\mathcal{A}_{SU(p)^3}' = A(\mathrm{antisym}) = p-4 = N,\tag{2.11}\] \[\mathcal{A}_{SU(p)^2U(1)}' = q_\lambda I(\mathrm{antisym}) = -N(p-2) = -N(N+2),\tag{2.12}\] \[\mathcal{A}_{U(1)^3}' = \sum_{\text{fermion index}} q_\lambda^3 = -\frac{1}{2}p(p-1)N^3 = -\frac{N^3(N+3)(N+4)}{2},\tag{2.13}\] \[\mathcal{A}_{U(1)}' = \sum_{\text{fermion index}} q_\lambda = -\frac{1}{2}p(p-1)N = -\frac{N(N+3)(N+4)}{2}.\tag{2.14}\] Therefore, we see that the anomalies indeed match.

3 Sheet 3, Problem 3*: Real Hypercharges and Anomaly Cancellation

Use one Standard Model generation, omitting a right-handed neutrino, with arbitrary real hypercharges: \[Q_L:(\mathbf2,\mathbf3)_q,\quad L_L:(\mathbf2,\mathbf1)_l,\quad u_R:(\mathbf1,\mathbf3)_u,\quad d_R:(\mathbf1,\mathbf3)_d,\quad e_R:(\mathbf1,\mathbf1)_x.\tag{3.1}\] The entries specify $SU(2)\times SU(3)$ representations and the subscript is the $U(1)$ charge.

Write every gauge anomaly-cancellation condition, including the mixed $U(1)$--gravity condition.
Show that the real solutions consist of two branches: one with $u=-d$ and another with the usual Standard Model hypercharges up to overall rescaling.

We start with the $SU(3)^3$ gauge anomaly, which only involves the quarks: \[\sum_\text{left} A(\mathbf{3}) = \sum_\text{right} A(\mathbf{3}),\tag{3.2}\] which is indeed true. For $SU(2)^3$, we simply have $A(\mathbf{2}) = 0$, and hence is always anomaly-free. For $U(1)^3$, we have \[\sum_\text{left}Y_i^3 = \sum_\text{right}Y_i^3 \implies 6q^3 + 2l^3 = 3u^3+ 3d^3+x^3.\tag{3.3}\] Now, we turn to the mixed gauge anomalies. For $SU(2)^2U(1)$, which only involves left-handed particles, we have \[\sum_\text{left}Y_iI(\mathbf{2}) = \sum_\text{right}Y_iI(\mathbf{2}) \implies 3q + l = 0.\tag{3.4}\] For $SU(3)^2U(1)$, we find \[\sum_\text{left}Y_iI(\mathbf{3}) = \sum_\text{right}Y_iI(\mathbf{3}) \implies 2q = u+d.\tag{3.5}\] Finally, $\text{grav}^2U(1)$ gives \[\sum_\text{left} Y_i = \sum_\text{right} Y_i \implies 6q + 2l = 3u + 3d + x.\tag{3.6}\]
The 4 equations boil down to \[6u^3 +6d^3 + 21u^2d+21ud^2 = 0.\tag{3.7}\] Clearly, this has the solution $u/d = -1 \equiv y$. The polynomial factorises to \[(y+1)(6y^2+15y+6) = 0,\tag{3.8}\] which has solutions, $y= -1, -2, -1/2$. Since $u_R$ and $d_R$ have the same representation, we can always relabel them. Therefore, we actually only have two possible classes of solutions: \[u=-d, \quad q=0,\quad l=0,\quad x=0,\tag{3.9}\] \[u=-2d,\quad q=\frac{d}{2},\quad l=\frac{3d}{2},\quad x =3d.\tag{3.10}\]

4 Sheet 3, Problem 4: Integer Hypercharges

Repeat Problem 3 with $q,l,u,d,x\in\mathbb Z$.

Impose all gauge anomaly conditions, without imposing the mixed gravitational condition.
Show that the solution is unique up to overall scaling and automatically cancels the mixed gravitational anomaly. You may quote a familiar pure-mathematics result without proving it.

Hint: First write $u-d=2y$ for integer $y$. Then use the variables $v,w$ related by \[x=-\frac6{v+w}, \qquad y=\frac{3(v-w)}{v+w}.\tag{4.1}\]

Since $u+d = 2q \in 2\mathbb{Z}$, we must have $u-d = 2y \in 2\mathbb{Z}$. Substituting this and $3q = -l$ into the $U(1)^3$ anomaly-free condition, we find \[48q^3 +3(q+y)^3+3(q-y)^3 +x^3 = 0,\tag{4.2}\] which simplifies to \[54q^3 +18qy^2 +x^3 = 0.\tag{4.3}\] Absorbing $q$ into $x,y$, so that $x,y \in \mathbb{Q}$ and \[54 +18y^2 +x^3 = 0.\tag{4.4}\] Making the redefinition of variables and working through a bit of algebra, we find \[v^3 +w^3 =1,\tag{4.5}\] which is the famous Fermat's last theorem with $N=3$. This, famously enough, has only trivial solutions, namely $v = 1, w=0$ or $v=0, w=1$. Again, these are equivalent solutions since we can always relabel $u$ and $d$. It is not hard to see that this is indeed the hypercharge of our world up to a scaling, and hence free of gravitational anomaly.

5 Sheet 3, Problem 5: Global Symmetries of the Standard Model

Determine the global symmetries of the classical Standard Model Lagrangian in each case:

there are no right-handed neutrinos;
right-handed neutrinos exist and every allowed relevant and marginal interaction is included.

For each case, reassess the result after accounting for anomalies.

Naively, there seems to be a $U(1)^5$ global symmetry, which is \[\begin{aligned}U(1):\quad Q_L &\to e^{i\alpha}Q_L,\\ d_R &\to e^{i\beta}d_R,\\ u_R &\to e^{i\gamma}u_R,\\ L_L &\to e^{i\delta}L_L,\\ e_R &\to e^{i\lambda}e_R.\end{aligned}\tag{5.1}\] However, the Yukawa couplings impose additional restrictions on the charges, which are \[\alpha = \beta, \quad \alpha = \gamma, \quad \delta = \lambda.\tag{5.2}\] This reduces the global symmetry to $U(1)^2$. In particular, we choose this to be $U(1)_B \times U(1)_L$.
After accounting for the anomalies, the mixed anomaly with $SU(2)$ and $SU(3)$ vanishes for both lepton and baryon number symmetry. However, the mixed anomaly with hypercharge survives: \[\mathcal{A}_{U(1)_BU(1)^2} = -\frac{1}{2} = -\mathcal{A}_{U(1)_LU(1)^2}.\tag{5.3}\] Therefore, the only surviving global symmetry in the quantum theory is $U(1)_B - U(1)_L$.
The situation is almost identical after including right-handed neutrinos, since they have no hypercharge.

6 Sheet 3, Problem 6: Surviving Discrete Symmetries

Part (a) is optional on the source sheet.

Couple $SU(N)$ Yang--Mills theory to a single massless adjoint left-handed Weyl fermion $\lambda$. Explain quantum consistency. Although the theory has $\mathcal N=1$ supersymmetry, that fact is not needed here. Classically $\lambda\mapsto e^{i\alpha}\lambda$ is a $U(1)$ symmetry. Follow the shift of the theta angle to show that only $\mathbb Z_{2N}$ survives quantum mechanically. If confinement produces $\langle\operatorname{Tr}\lambda\lambda\rangle \sim\Lambda_{QCD}^3$, determine the action of $\mathbb Z_{2N}$ on this condensate and count the vacua.
Now couple $SU(N)$, with $N>2$, to one Dirac fermion in the two-index symmetric representation. Determine its classical and quantum global symmetries. Assuming confinement and a condensate $\langle\bar\psi\psi\rangle \sim\Lambda_{QCD}^3$, count the vacua. Repeat for the two-index antisymmetric representation. The representation data in Problem 2 may help.

Since $A(\mathrm{adj}) =0$, the theory is free of gauge anomaly, and hence consistent. For the $U(1)$ global symmetry, the ABJ anomaly is \[\mathcal{A}_{U(1)SU(N)^2} = I(\mathrm{adj}) = 2N.\tag{6.1}\] Hence, the theta term transform as \[\theta \to \theta + 2\alpha N.\tag{6.2}\] Therefore, only a discrete $\mathbb{Z}_{2N}$ survives. Under the discrete symmetry, the condensate transforms as \[\langle\operatorname{Tr}\lambda\lambda\rangle \to e^{2in\pi/N}\langle\operatorname{Tr}\lambda\lambda\rangle.\tag{6.3}\] Therefore, the symmetry is further broken to $\mathbb{Z}_{2}$, which means that there are $N$ vacuum states for this theory.
In this case the classical global symmetry is $U(1)_V \times U(1)_A$, while the quantum global symmetry reduces to $U(1)_V$ and a surviving discrete group of $U(1)_A$. In particular, the theta angle transform as \[\theta \to \theta +2\beta(N\pm 2)\tag{6.4}\] ($+$ for symmetric representation and $-$ for antisymmetric representation). Therefore, the surviving discrete symmetry is $\mathbb{Z}_{N\pm 2}$. Under the discrete symmetry, the condensate transforms as \[\langle\bar\psi\psi\rangle \to e^{2\pi ni/(N\pm 2)}\langle\bar\psi\psi\rangle.\tag{6.5}\] Therefore, there are $N\pm 2$ vacuum for the theory.