Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 305 / 3

A gauge anomaly breaks a redundancy required to remove unphysical polarizations. It violates the Ward identities and makes the quantum gauge theory inconsistent, so all gauge anomalies must cancel. A chiral anomaly, or Adler-Bell-Jackiw anomaly, instead breaks a classically conserved global axial symmetry; it is physically allowed and explains effects such as anomalous pseudoscalar decays and instanton-induced charge violation. A 't Hooft anomaly is an obstruction to gauging a global symmetry. It is invariant under renormalization-group flow, so 't Hooft anomaly matching constrains the infrared theory to reproduce it through massless fields, symmetry breaking, or a topological sector.

Count every right-handed field as a left-handed conjugate, which reverses its $U(1)$ charge and conjugates its non-Abelian representation. The nontrivial local anomaly-cancellation conditions areThe pure $SU(3{)}^3$ anomaly cancels because the two triplets in $Q_L$ balance the conjugates of $u_R$ and $d_R$. There is no perturbative $SU(2{)}^3$ anomaly, and the Witten SU(2) anomaly also vanishes because the three colored quark doublets plus one lepton doublet make four. The mixed gauge-gravitational anomaly is excluded at this stage as requested.

The $[SU(3){]}^{2}U(1)$ condition gives $u+d=2q$, which is even because all hypercharges are integers. The sum $u+d$ and difference $u-d$ have the same parity, so $u-d$ is even. ThereforeSolving the sum and difference equations gives

The $[SU(2){]}^{2}U(1)$ condition gives $l=-3q$. Substituting this and $u=q+y$, $d=q-y$ into the cubic anomaly givesUsing $(q+y{)}^3+(q-y{)}^3=2q^3+6q{y}^2$ and multiplying by minus one yields

If $q\ne 0$, divide the cubic equation by $q^3$ and define the rational numbersThen every integer anomaly-free assignment supplies a rational solution of

Use the stated rational change of variablesDirect substitution turns the anomaly equation intoWriting $v=a/c$ and $w=b/c$ in lowest common denominator gives the integer equation $a^3+b^3=c^3$. The supplied special case of Fermat's Last Theorem says that one of $a,b$ must vanish. Thus $(v,w)=(1,0)$ or $(0,1)$, which givesThe two signs merely exchange the names of the up- and down-type singlets. Choosing $q=1$ and the conventional sign $y=3$ gives the unique assignment up to overall scaling and that exchange:These are six times the conventional Standard Model hypercharges.

The coefficient of the mixed gauge-gravitational anomaly is the sum of all left-handed $U(1)$ charges, with right-handed fields counted as conjugates:For the solution above,More invariantly, $l=-3q$, $u+d=2q$, and $x=-6q$ make the expression vanish identically. Thus the unique cubic-anomaly solution automatically cancels the mixed gauge-gravitational anomaly.