Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 52 / 4 / Solution

Take ${\eta }^{\alpha \beta }=\mathrm{diag}(-1,1)$ with coordinate zero equal to $\tau$ and coordinate one to $\sigma$. In the supplied representation setThey satisfy the Clifford algebra $\{{\rho }^{\alpha },{\rho }^{\beta }\}=2{\eta }^{\alpha \beta }1$. For anticommuting spinor components and an ordinary matrix $B$, relabelling the two summed indices givesNow $C^T=-C$, whereas $C{\rho }^0=-1$ and $C{\rho }^1=\mathrm{diag}(-1,1)$ are symmetric. Therefore the Majorana Grassmann bilinear interchange yieldsThese signs require fermionic, Grassmann-odd fields; commuting spinors would have the opposite interchange signs.

In flat conformal gauge, suppress the common factor $(4\pi {\alpha }^{\prime }{)}^{-1}$ and write the combined density as $\mathcal{L}=-{\partial }_{\alpha }X\cdot {\partial }^{\alpha }X+\bar{\psi }\cdot {\rho }^{\alpha }{\partial }_{\alpha }\psi$. The supersymmetry variation including its odd parameter is an even derivation. From $({\rho }^{\alpha }{)}^{T}C=-C{\rho }^{\alpha }$ we obtainAlso $(C{\rho }^{\alpha }{\rho }^{\beta }{)}^T=-C{\rho }^{\beta }{\rho }^{\alpha }$, so interchanging odd spinors gives $\bar{\chi }{\rho }^{\alpha }{\rho }^{\beta }ϵ=\overline{ϵ}{\rho }^{\beta }{\rho }^{\alpha }\chi$. The two density variations, with target contractions understood, are thereforeIn the second term the symmetric second derivative annihilates the antisymmetric part of ${\rho }^{\beta }{\rho }^{\alpha }$. Using $2{\eta }^{\alpha \beta }-{\rho }^{\beta }{\rho }^{\alpha }={\rho }^{\alpha }{\rho }^{\beta }$ now gives the exact off-shell identityThus the action is invariant when this boundary flux vanishes and the transformation preserves the field boundary conditions, for example with suitable falloff or compatible supersymmetric endpoint conditions. On a spatial circle, periodic fields admit the constant parameter; antiperiodic fermions with periodic bosons require the local superconformal treatment, since a nonzero constant parameter would make the boson variation antiperiodic. The constant parameter and ordinary derivatives describe rigid supersymmetry in flat gauge; a locally supersymmetric curved-worldsheet formulation includes the worldsheet gravitino. The result is the rigid worldsheet supersymmetry boundary term without using field equations.

For the commutator on $X$, keep the order of the odd parameters:For the fermion, the order is equally important. Moving ${ϵ}_2$ through both odd factors in $({\overline{ϵ}}_1{\partial }_{\alpha }\psi ){ϵ}_2$ incurs two minus signs. ConsequentlyTo check the spinor matrix explicitly, set ${ϵ}_1=(a,b{)}^T$, ${ϵ}_2=(c,d{)}^T$, with all four components anticommuting and monomials ordered $a,b,c,d$. ThenDirect multiplication gives ${\rho }^{\alpha }M+M{\rho }^{\alpha }=2V^{\alpha }1$. HenceVariation of the fermion action gives the Dirac equation ${\rho }^{\alpha }{\partial }_{\alpha }\psi =0$. On this equation the second term vanishes, provingThis is on-shell closure of rigid worldsheet supersymmetry; the action's invariance was off shell, but its fermionic transformation algebra closes in this field content only on shell.

The associated superconformal current is proportional to $\psi \cdot \partial X$, of conformal weight $3/2$. Its fermionic modes $G_r$ extend the stress-tensor generators to the N=1 super-Virasoro algebra:The modes are half-integral in the Neveu–Schwarz sector and integral in the Ramond sector. Their anticommutator giving a Virasoro generator is the conformal-mode counterpart of the translation commutator just derived. Ten coordinate bosons and ten Majorana fermions give matter central charge $c=10+10/2=15$, canceled by the usual reparametrization and superconformal ghosts in the critical RNS string.