Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 307 / 1

Choose $P_{\mu }=-i{\partial }_{\mu }$ and left Grassmann derivatives. A consistent differential realization of four-dimensional N=1 supersymmetry isThe signs can change with the definitions of $P_{\mu }$, Grassmann differentiation and the finite transformation, but these operators obey the convention-independent content of the Super-Poincaré algebra,They follow by expanding the finite transformation to first order and requiring a supersymmetry translation of the superspace coordinates.

A vector superfield is a real superfield,For a non-Abelian gauge theory it is also Lie-algebra valued, $V=V^a{T}_a$. The reality condition permits the component expansion to be reduced to Wess-Zumino gauge, where the physical fields are a gauge field and gaugino together with a real auxiliary field.

A chiral superfield is annihilated by the antichiral supersymmetric covariant derivative:Equivalently, it depends on $(y^{\mu },\theta )$ with $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$, and has the finite expansion $\Phi =\varphi +\sqrt{2}\theta \psi +{\theta }^{2}F$.

The gauge-transformation superfield $\Lambda ={\Lambda }^a{T}_a$ is a dimensionless chiral superfield:It takes values in the complexified $\mathfrak{su}(3)$ Lie algebra. It need not be real; its Hermitian conjugate is antichiral. This complexification is necessary because a general superspace gauge transformation must preserve chirality.

With canonical normalization, the renormalizable supersymmetric gauge theory hasThe first term is the gauge-covariant Kähler potential and the second is the Supersymmetric Yang-Mills action. A holomorphic superpotential would also be integrated as $\int d^2\theta \mathcal{W}(\Phi )+h.c.$, but one commuting field in the fundamental representation of $SU(3)$ admits no nonconstant renormalizable gauge-invariant superpotential: the apparent cubic ${ϵ}_{ijk}{\Phi }^i{\Phi }^j{\Phi }^k$ vanishes.

Under ${\Phi }^{\prime }=e^{-2i\Lambda }\Phi$, the conjugate field transforms as ${\Phi }^{\prime }{}^{†}={\Phi }^{†}{e}^{2i{\Lambda }^{†}}$. Invariance of the canonical Kähler term therefore requiresorThis is the finite non-Abelian supergauge transformation of the vector superfield.

The relation in part vi givesso the full-superspace term has gauge invariance. The chiral field-strength superfield transforms covariantly,Cyclicity of the matrix trace then gives $\mathrm{Tr}(W^{\prime }{}^{\alpha }{W}_{\alpha }^{\prime })=\mathrm{Tr}(W^{\alpha }{W}_{\alpha })$. Both superspace integrals, and hence the entire Lagrangian density, are invariant.