Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 50 (Supersymmetry and Extra Dimensions)

Positive-energy one-particle Poincaré group representations have labels $(m,s)$ for $m>0$, and $(0,h)$ for finite-helicity massless states; individual massive polarizations add $m_s=-s,\dots ,s$. An $\mathcal{N}=1$ massive supermultiplet is labelled by mass $m$ and superspin $j$, with constituent labelsA massless supermultiplet has constituent labelsA CPT completion of a supermultiplet additionally supplies $(0,-h),(0,-h+\frac{1}{2})$, with conjugated internal charges where applicable. Constituent spins/helicities are the labels here, not counts of polarizations.

TeV-scale softly broken supersymmetry addresses the technical hierarchy problem: equal bosonic and fermionic couplings cancel quadratic ultraviolet contributions to the Higgs mass. Soft mass splittings leave corrections of order $\delta m_H^2\sim g^2{M}_{\mathrm{soft}}^2\log (\Lambda /M_{\mathrm{soft}})/(16{\pi }^2)$, so a weak-scale soft spectrum protects the weak scale against a much higher cutoff. Unbroken four-dimensional extended supersymmetry is incompatible with the observed complex chiral gauge spectrum: charged hypermultiplets contain conjugate representations, while vector multiplets have real adjoint matter. It must therefore be broken to at most $\mathcal{N}=1$ before reproducing low-energy Standard Model chirality.

In a supersymmetric Higgs mechanism, an internal gauge symmetry breaks while supersymmetry can remain unbroken. A massless vector multiplet absorbs a chiral multiplet and becomes a massive multiplet with spins $(1,\frac{1}{2},\frac{1}{2},0)$: the gauge boson acquires its longitudinal particle polarization from the Goldstone mode. In the super-Higgs mechanism, local supersymmetry itself breaks: the spin-$3/2$ gravitino absorbs the spin-$1/2$ goldstino, gaining two longitudinal polarizations and becoming massive. With only global supersymmetry breaking, the goldstino instead remains a physical massless fermion.

Both Kaluza-Klein theory and the brane-world scenario use a higher-dimensional bulk. Ordinary Kaluza–Klein matter propagates in the compact directions, producing observable massive towers; absence of weak-scale charged modes requires roughly $R^{-1}\gtrsim \mathrm{TeV}$, or $R\lesssim 2\times 10^{-19}\mathrm{m}$, and traditional constructions often take Planck-size radii. Brane-world matter is localized while gravity explores the bulk, so large flat dimensions can reach the submillimetre order of magnitude discussed in the 2004 setting. For $n$ flat dimensions, $M_{\mathrm{Pl}}^2\sim M_D^{n+2}(2\pi R{)}^n$. These are model-dependent scales, not universal maxima: warped gravity-localizing models can have a noncompact extra dimension. The historical large-flat-dimension mechanism is described in arxiv.org/abs/hep-ph/9803315.

Work with the local Wilsonian effective action, retaining the elementary chiral superfield and using a nonzero infrared cutoff. Its superpotential is holomorphic in both the field and chiral spurions representing the couplings. A convenient pair of spurion symmetries assigns chargesBoth couplings and $\Phi$ transform nontrivially under each displayed symmetry. Each superpotential monomial has ordinary charge zero and R-charge two, so the integral $\int d^2\theta W$ is invariant. This R assignment differs from the conventional $(1,0,-1)$ assignment for $(\Phi ,m,g)$ by twice the ordinary charge, and describes the same two-dimensional charge space. These are formal spurion symmetries; fixing the couplings to numerical values need not leave them as actual symmetries of the fields alone.

For $m\ne 0$, $m{\Phi }^2$ has the required charges and mass dimension three, while $z=g\Phi /m$ has zero charge under both symmetries and is dimensionless. Thus the allowed holomorphic form isOne can see its generality directly. A holomorphic monomial ${\Phi }^u{m}^v{g}^w$ must obey $u-2v-3w=0$ and $3u-4v-7w=2$. Solving gives $u=w+2$ and $v=1-w$, hence precisely $m{\Phi }^2(g\Phi /m{)}^w$. A field-independent constant is irrelevant to this global supersymmetric action and can be dropped.

The symmetries alone leave the function arbitrary. To determine its perturbative form, use regularity as well as the two limits. A regular weak-coupling expansion at $g=0$ hasThe $g\to 0$ limit is a free massive theory, so $c_0=1/2$. Since the elementary field has not been integrated out, the Wilsonian superpotential must remain regular term by term as $m\to 0$ with the infrared cutoff fixed. Every term with $r\ge 2$ would introduce a pole in $m$ and is excluded. Negative powers of $g$ were already excluded by perturbative regularity at $g=0$. Consequently $F=c_0+c_{1}z$.

At $m=0$, the only surviving allowed interaction is proportional to $g{\Phi }^3$. Its coefficient is first order in $g$, and matching the tree three-field vertex gives $c_1=1/3$. No closed-loop three-field diagram can be made from a single cubic vertex; extra interaction vertices would produce higher coupling order. Equivalently free superspace propagators join chiral to antichiral fields, so a closed loop would require additional vertices or conjugate couplings, which the holomorphic charge constraints exclude for this term. This fixes the coefficient without assuming the desired conclusion. ThereforeThe regularity condition is essential: it concerns the local Wilsonian superpotential with the field retained, not a nonlocal one-particle-irreducible functional at zero infrared cutoff or a superpotential obtained by eliminating a massive field. Such different operations can generate inverse masses or infrared singularities without contradicting this proof.

The Kähler potential is a real D-term, not a holomorphic F-term, and is not protected by this argument. For example its quadratic part becomes $Z(g,\bar{g},m,\bar{m};\mu ){\Phi }^{†}\Phi$, with higher real operators also allowed. Terms involving $|g{|}^2$ are compatible with the symmetries and produce wave-function renormalization. After the canonical rescaling ${\Phi }_c=Z^{1/2}\Phi$, the parameters in the superpotential expressed using canonical fields are $m_c=m/Z$ and $g_c=g/Z^{3/2}$, so physical couplings can run despite the holomorphic non-renormalization statement.

Holomorphy is crucial. Without it, neutral factors such as $1+c|g{|}^2$ could multiply either superpotential monomial while respecting both spurion symmetries. Excluding conjugate couplings, together with regularity and tree matching, is what turns the charge argument into the non-renormalization theorem.

For global four-dimensional $\mathcal{N}=1$ supersymmetry, useLet $\Pi =(-1{)}^F$ be fermion parity. On a finite one-particle supermultiplet at fixed positive-energy momentum, the supercharges act within the multiplet and $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$. Cyclicity of the finite-dimensional trace givesThus the trace with $\Pi$ of every displayed diagonal anticommutator vanishes. Summing the two components, ${\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}=4E1$, soSince $E>0$, this proves $N_B=N_F$, for both massive and massless one-particle multiplets. The restriction matters: a zero-momentum supersymmetric vacuum singlet, annihilated by every supercharge, need not contain a fermionic partner. The algebra does not imply equal counts for such a trivial zero-energy representation or for an arbitrary unregulated infinite-dimensional trace.

The same Super-Poincaré algebra gives the operator identityFor a normalized physical state $|v⟩$ in a positive-norm Hilbert space,An unbroken supersymmetric vacuum must be annihilated by every supercharge and its adjoint, hence has zero energy. Conversely, zero expectation of this positive sum forces each norm separately to vanish. Therefore a vacuum with strictly positive energy necessarily breaks supersymmetry. For an infinite-volume field theory the corresponding statement is about vacuum energy density, obtained with an appropriate finite-volume regulator. The zero of energy is fixed by the algebra; it cannot be arbitrarily shifted while preserving this identity. This is the global-supersymmetry result and does not assert positivity of the general local-supergravity scalar potential.

For the unheaded extended-algebra request, fix the central-charge normalization explicitly:The central charges are scalar numbers on an irreducible massive representation. A unitary skew-diagonalization of an antisymmetric matrix puts $Z$ into two-by-two blocks $\left(\begin{array}{cc}0& z_r\\ -z_r& 0\end{array}\right)$ and, for odd $\mathcal{N}$, a leftover zero entry. Since this change of supercharge basis is unitary, it preserves the mixed anticommutator. Go to the rest frame $P^{\mu }=(M,0,0,0)$.

For one block set $q_{\alpha }=Q_{\alpha }^1$ and $r_{\alpha }={ϵ}_{\alpha \gamma }(Q_{\gamma }^2{)}^{†}$, with ${ϵ}_{12}=1$. Direct use of the algebra givesWrite $z=|z|e^{i\phi }$ and form the combinations suggested by the hint:ThenEach diagonal anticommutator has nonnegative expectation because it is the sum of squared norms of an operator and its adjoint. Therefore $M-|z_r|\ge 0$ for every block, provingIf the algebra is instead written with $\{Q,Q\}=ϵZ$ without the factor two, the same proof gives $M\ge \frac{1}{2}{\max }_r|Z_r|$. This is the normalization used by some definitions of the BPS bound in supersymmetry; the physical bound is unchanged by the naming of the central charge.

A massive BPS state saturates at least one block bound. At saturation $\{B_{\alpha },B_{\alpha }^{†}\}=0$, so both $B_{\alpha }$ and $B_{\alpha }^{†}$ annihilate every state in its multiplet. The two complex inactive oscillators are four preserved real supersymmetries per saturated block. If $r$ blocks saturate, the preserved fraction is $r/\mathcal{N}$ and only $2\mathcal{N}-2r$ complex fermionic oscillators remain active. For a spin-$j$ Clifford vacuum,compared with $(2j+1)2^{2\mathcal{N}}$ for a long massive multiplet. Thus BPS states have an exact mass-charge relation, preserve part of the supersymmetry, and belong to shortened representations. A physical CPT-invariant spectrum may additionally require the conjugate-charge multiplet. The short representation protects the saturated mass relation while it exists; it does not prohibit a decay at a threshold where central-charge phases align and the sum of daughter masses equals the bound.

Here the sphere's “volume” is its intrinsic $(N-1)$-dimensional area, not the volume of the enclosed ball. Write ${\Omega }_{N-1}$ for the area of the unit sphere. The Gaussian integral givesFor completeness the one-dimensional integral is positive and its square is $2\pi {\int }_0^{\infty }\rho e^{-{\rho }^2}d\rho =\pi$. In hyperspherical coordinates, the same integral isThe substitution is $u={\rho }^2$, and the last integral defines the Gamma function. Equating the two evaluations and scaling lengths by $r$ proves

Take $D$ to mean spacetime dimension, so the number of spatial dimensions is $d=D-1$. This convention is needed when comparing five-dimensional spacetime with four-dimensional spacetime. For a point charge $q$, Gauss's law in $d$ spatial dimensions givesUsing $E_r=-{\varphi }^{\prime }(r)$ and choosing $\varphi (\infty )=0$ for $d>2$ givesThe Green function of the Laplacian with $-{\nabla }^2{G}_d={\delta }^{(d)}$ is consequently $G_d(r)=1/[(d-2){\Omega }_{d-1}{r}^{d-2}]$. Define the gravitational Poisson coupling ${\kappa }_D$ by ${\nabla }^2\Psi ={\kappa }_{D}M{\delta }^{(d)}$, and use acceleration $\mathbf{g}=-\nabla \Psi$. The same flux calculation givesThis explicitly fixes the normalization of the gravitational coupling: in four dimensions ${\kappa }_4=4\pi G_4$, so ${\Psi }_4=-G_{4}M/r$. If a higher-dimensional Newton constant is instead defined by the force law $|\mathbf{g}|=G_{D}M/r^{D-2}$, then ${\kappa }_D={\Omega }_{D-2}{G}_D$ in the displayed expression. In all conventions the force scales as $r^{-(D-2)}$ and the potential as $r^{-(D-3)}$. If “$D$ dimensions” denotes spatial dimensions instead, replace $d$ by $D$ and the potential exponent is $D-2$.

For $d=2$, integration gives $\varphi =-q\log (r/r_0)/(2\pi {\epsilon }_3)$ and $\Psi ={\kappa }_{3}M\log (r/r_0)/(2\pi )$ for the Poisson-normalized gravitational model; these logarithmic potentials have no zero-at-infinity convention. For $d=1$, a fundamental solution of the negative Laplacian is $G_1(x)=-|x|/2$, giving linear potentials up to an additive homogeneous solution.

Now compactify the fourth spatial coordinate $y$ on a circle of circumference $L=2\pi R$, retaining three noncompact coordinates with radial distance $r$. In uncompactified five-dimensional spacetime, $G_4(r,y)=1/[4{\pi }^2(r^2+y^2)]$. A periodic point source is therefore represented by its images:For $r/L\to \infty$, write the sum as $r^{-2}{\sum }_n[1+(nL/r{)}^2{]}^{-1}$ on the source slice $y=0$. The spacing $L/r$ tends to zero, so its Riemann sum approaches the integrable function's integral:Thus the compactified Green function is asymptotic to $1/(4\pi Lr)$, already proving the four-dimensional inverse-distance law.

An exact expression makes the crossover transparent. The Fourier transform of $(r^2+y^2{)}^{-1}$ is $\pi e^{-r|k|}/r$, obtained by closing the integration contour around $y=ir$ or $-ir$. Applying the Poisson summation formula then gives, at $y=0$,The last equality sums a geometric series. Hence the point-source potential with one compact spatial dimension isAt short distances $r\ll L$, $\coth (\pi r/L)\sim L/(\pi r)$ and these reproduce the five-dimensional $1/r^2$ potentials. At long distances $r\gg L$, the hyperbolic factor is $1+O(e^{-2\pi r/L})$, giving the four-dimensional $1/r$ potentials with effective Poisson couplings ${\epsilon }_4=L{\epsilon }_5$ and ${\kappa }_4={\kappa }_5/L$. Equivalently only the massless Kaluza-Klein mode survives at long distance, while the massive modes contribute exponentially suppressed terms.

Start with the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$. For the dimension count, complexify and rescale the timelike matrix so that the generators obey the Euclidean complex relation $\{{\gamma }_a,{\gamma }_b\}=2{\delta }_{ab}$. This changes neither the complex representation dimension nor its decomposition under the complexified even algebra. In $D=2k$, pair the generators intoDirect substitution gives $\{a_i,a_j\}=\{b_i,b_j\}=0$ and $\{a_i,b_j\}={\delta }_{ij}$. The commuting operators $n_j=b_j{a}_j$ satisfy $n_j^2=n_j$, so their occupation eigenvalues are zero or one. An irreducible module contains an all-zero occupation state $|0⟩$, obtained by lowering the occupied entries of a simultaneous eigenvector. The statesare nonzero and independent by their distinct joint occupation eigenvalues. They span an invariant module; irreducibility makes this the full module. Conversely this Fock construction supplies a representation on exactly these $2^k$ states. Products of occupation projectors and raising/lowering operators give its matrix units, so the generated algebra is the full matrix algebra. Therefore the minimal complex Dirac spinor has

This module is irreducible under the full Clifford algebra, but the Spinor representation of the Lorentz group is generated by the even bilinears ${\Sigma }^{\mu \nu }=\frac{1}{4}[{\gamma }^{\mu },{\gamma }^{\nu }]$. The chirality matrix in the Euclideanized basis isIts two eigenspaces therefore remain invariant under Lorentz transformations. Each has half the Fock states, since any gamma matrix bijects them. The even algebra connects every pair of occupation states of the same parity and has no further invariant subspaces within either one. HenceThese are the irreducible Weyl spinors. Globally they are ordinary representations of $\mathrm{Spin}(D-1,1)$, the double cover; calling them representations of $SO(D-1,1)$ uses the conventional projective spinor terminology.

For $D=2k+1$, adjoin ${\gamma }_{2k+1}=\pm {\gamma }_{\ast }$. It anticommutes with the old matrices and gives a Clifford module of the same dimension $2^k$. The new even generators $[{\gamma }_a,{\gamma }_{2k+1}]$ interchange the two old chirality eigenspaces, so the odd-dimensional Lorentz spinor is irreducible of complex dimension $2^k$, with no Weyl splitting. The product of all odd-dimensional gamma matrices is central and acts as a scalar; it cannot be a new anticommuting chirality operator. Its two signs distinguish the two complex odd-Clifford modules, whose restrictions to Spin group are equivalent. Majorana spinor reality conditions, when the signature permits them, are additional conditions and should not be confused with complex chirality.

For the reduction, take the massless zero modes on a flat seven-torus with no flux or projections. The eleven-dimensional fields are the metric $G_{MN}$, a three-form $A_{MNP}$ and a Majorana gravitino ${\Psi }_M$. Split $M=(\mu ,i)$ with $\mu =0,\dots ,3$ and $i=1,\dots ,7$. The metric supplies $G_{\mu \nu }$, seven vectors $G_{\mu i}$ and 28 scalar fields $G_{ij}=G_{ji}$. The three-form supplies a four-dimensional three-form $A_{\mu \nu \rho }$, seven two-forms $A_{\mu \nu i}$, $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ vectors $A_{\mu ij}$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ scalars $A_{ijk}$.

A massless two-form in four dimensions has one physical particle polarization and can be dualized to a scalar by $H_3=\ast d\sigma$; the seven two-forms thus supply seven more scalars. The four-dimensional three-form has no local particle polarization, since its four-form field strength is locally a constant times the volume form. The total bosonic content is consequentlyThe eleven-dimensional Majorana spinor has 32 real components, decomposing into eight four-dimensional Majorana spinors. Accordingly ${\Psi }_{\mu }$ supplies eight gravitini, while ${\Psi }_i$ supplies $7\times 8=56$ spin-$1/2$ fields after diagonalizing the gravitino/matter kinetic terms. The resulting theory has 32 real supercharges, hence is four-dimensional N=8 supergravity, withThe on-shell counts check the reduction: the metric has $11(11-3)/2=44$ polarizations and the three-form has $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, totaling 128. In four dimensions $2+28\cdot 2+70=128$ bosonic polarizations match $8\cdot 2+56\cdot 2=128$ fermionic ones. Higher Kaluza–Klein levels, fluxes or projections would be additional choices beyond this massless toroidal reduction.

Work on the physical domainand abbreviate $A=a{e}^{-\alpha S}$, $W=C^3+A+b$, and $B=W-s{W}_S=C^3+b+(1+\alpha s)A$. There are no vector multiplets specified, hence no additional gauge D-term. In Planck units the supergravity F-term potential and upper-index supergravity auxiliary fields areThe overall auxiliary-field sign is a convention; its vanishing and norm are invariant statements.

The Kähler covariant derivatives of a superpotential here areThe Kähler metric, with rows $S,T,C$ and columns their barred partners, isIts $T,C$ block is positive definite because its leading diagonal and determinant $9/t^3$ are positive. In the inverse-tensor convention $K_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, the nonzero upper components areIn particular the conjugations in the off-diagonal components matter.

To verify the no-scale supergravity cancellation explicitly, temporarily write $u=D_{T}W=-3W/t$ and $v=D_{C}W=W_C+3\bar{C}W/t$. Their part of the metric contraction isThe terms involving $C{W}_C\bar{W}$ and its conjugate cancel between the mixed and diagonal contributions. The remaining $3|W{|}^2$ cancels the universal negative term. Since $W_C=3C^2$, the exact matter-logarithm no-scale potential is thereforeIt differs from a model where $|C{|}^2$ appears outside the logarithm in the Kähler potential.

Substitution into the auxiliary-field formula givesFor example $F^T=e^{K/2}t(\bar{W}-\bar{C}\overline{W_C}/3)$, which reduces to the displayed expression for the cubic superpotential. To test unbroken supersymmetry at any finite point, $F^C=0$ first forces $C=0$. Then $F^T=0$ forces $A+b=0$, whereas $F^S=0$ forces $(1+\alpha s)A+b=0$. Subtracting gives $\alpha sA=0$. Because $s>0$, $\alpha >0$ and $e^{-\alpha S}\ne 0$ at a finite point, all auxiliaries can vanish only if $a=b=0$. Thus every nontrivial choice of $(a,b)$ breaks supersymmetry at finite field values.

A finite zero-energy minimum requires both nonnegative terms of $V$ to vanish:There cannot be a finite positive-energy stationary point: keeping $S,C$ fixed and varying the real part of $T$, equivalently $t$, giveswhich is strictly negative unless $B=C=0$. Hence these zero-energy solutions are all the possible finite minima.

Because $a,b$ were allowed to be arbitrary, existence must be checked. If both are nonzero, put $x=\alpha s>0$ and $r=|b/a|$. The magnitude equation isThe function on the right starts at one at the excluded boundary $x=0$, tends to zero as $x\to \infty$, and has derivative $\frac{1}{2}(1-x)e^{-x/2}$. Its maximum is $2e^{-1/2}$ at $x=1$. Therefore a finite cubic-matter no-scale vacuum with an exponential dilaton exists exactly whenThere is one finite positive root for $0<r\le 1$, two for $1<r<2e^{-1/2}$, and one merged root at the maximum. The complex phase equation fixes $\mathrm{Im}S$ modulo $2\pi /\alpha$. If exactly one of $a,b$ vanishes, or the ratio exceeds the bound, no finite minimum exists; the potential instead has infimum zero along the runaway $t\to \infty$. A zero infimum at an infinite-field boundary is not an attained finite vacuum.

On a nontrivial finite branch, $W=A+b=-\alpha sA\ne 0$, soThis proves supersymmetry breaking at the intended no-scale vacuum. Its zero energy is consistent with the negative gravitational term: $K_{T\bar{T}}|F^T{|}^2=3e^K|W{|}^2$, which cancels $3e^K|W{|}^2$ in $V$. It is not a counterexample to the global-supersymmetry positivity argument in question 3.

Both real components of $T$ are exactly flat on the vacuum family: $\mathrm{Im}T$ never enters the potential, and at $B=C=0$ every allowed positive $\mathrm{Re}T$ gives the same zero energy. Generically both components of $S$ are fixed. Although the matter field has no quadratic mass there, it is not an exact flat direction: for $S$ fixed at its root, $B=C^3$ and the leading positive potential is $3|C{|}^4/(s{t}^2)$. At the special merged root $x=1$, the real-$S$ quadratic curvature also vanishes, but higher orders lift it; this does not add a continuous flat vacuum direction. If $a=b=0$, by contrast, $C=0$ is a supersymmetric zero-energy family with both complex $S$ and $T$ flat and all auxiliaries zero. This exception prevents an unconditional breaking assertion for arbitrary constants.

Finally the gravitino mass from a superpotential isAt a nontrivial finite minimum it becomesIts value varies along the flat volume modulus; the classical potential does not select a unique supersymmetry-breaking or gravitino-mass scale. On the trivial supersymmetric branch $a=b=C=0$, it is zero.