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Write $C_U=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes U$ for the controlled unitary gate. For a normalized eigenvector $|u⟩$, the first Hadamard gate and quantum phase kickback giveThe final Hadamard gate therefore produces the unentangled outputThe auxiliary system is unchanged, while its eigenphase becomes accessible through interference of the control qubit. The special unitary group assumption is not needed for this calculation; unitarity and the normalized eigenvector suffice.

Interpret triviality as holomorphic triviality. Choose a global holomorphic bundle frame $s_1,\dots ,s_k$ for $E$. Let $H_{\alpha }$ have as its columns the coordinates of these sections in the prescribed local frame over $U_{\alpha }$, with the convention that a fiber coordinate vector transforms as $s_{\beta }=F_{\alpha \beta }{s}_{\alpha }$. ThenEach $H_{\alpha }$ is holomorphic and invertible because its columns form a basis in every fiber. This proves the requested holomorphic splitting of transition functions on a trivial bundle. Smooth or topological triviality alone would not justify holomorphic splitting.

For the line bundle on projective twistor space, pull its patching function back to the twistor line $L_x$ using the incidence convention of this problem, ${\omega }^A=x^{A{A}^{\prime }}{\pi }_{A^{\prime }}$, without an extra factor of $i$. Triviality on each line gives nonvanishing linewise holomorphic functions $H_0,H_1$ with $F_{01}=H_1{H}_0^{-1}$. The factors can be chosen as holomorphic functions of $x$ locally: a degree-zero scalar transition function has zero winding on the overlap annulus, hence admits a logarithm there; splitting its positive and negative Laurent series parts gives factors whose coefficients are Cauchy integrals that are holomorphic functions of $x$. The explicit splitting is worked out below. Their only common linewise ambiguity is multiplication by a nonzero function $g(x)$: the ratio is a global holomorphic function on the compact complex projective line, hence constant on that line.

Take ${ϵ}_{0^{\prime }{1}^{\prime }}=1$, ${ϵ}^{0^{\prime }{1}^{\prime }}=-1$ and raise a two-component spinor by left multiplication with ${ϵ}^{A^{\prime }{B}^{\prime }}$. The derivative along an alpha-plane is ${\ell }_A={\pi }^{A^{\prime }}{\partial }_{A{A}^{\prime }}$. It annihilates the unsplit twistor function, becauseThusis holomorphic on the whole line and has projective weight one. A section of $\mathcal{O}(1)$ is linear in the homogeneous coordinates: in an affine coordinate, holomorphic regularity at infinity bounds its growth to at most first order, so the Cauchy integral formula excludes all higher powers. ThereforeThe inverse splitting functions obey $({\ell }_A+G_A)H_{\alpha }^{-1}=0$. Commuting these two equations provesThe spinor decomposition of gauge curvature iswhere both $\phi$ two-component spinors are symmetric. Vanishing of the contraction for every $\pi$ forces ${\phi }_{A^{\prime }{B}^{\prime }}=0$. This is exactly the anti-self-dual Maxwell equations. Since the Abelian gauge curvature is $dA$, it is closed; its duality condition also makes it co-closed, giving the source-free Maxwell equations. Changing both splitting factors by $g(x)$ changes $A$ by $d\log g$, the expected Abelian gauge transformation. This derives the rank-one Penrose-Ward correspondence in the needed local form.

For the exponential patching function, choose a representative $f$ of the Čech cohomology class, of homogeneous degree zero. On the overlap of the two line patches write $f_x(\zeta )=f(x\pi ,\pi )={\sum }_{n\in \mathbb{Z}}{f}_n(x){\zeta }^n$. To compute the gauge potential explicitly, use a constant two-component spinor basis in which ${\iota }_{A^{\prime }}=(1,0)$ and the finite patch ${\pi }_{0^{\prime }}=\zeta ,{\pi }_{1^{\prime }}=1$. Then ${\pi }^{0^{\prime }}=-1$, ${\pi }^{1^{\prime }}=\zeta$ andFor a positively oriented contour in the overlap annulus, split its Laurent series asThe first is holomorphic inside the contour, the second outside including infinity; $H_{\alpha }=e^{h_{\alpha }}$ split $e^f$. Since $(-{\partial }_{B{0}^{\prime }}+\zeta {\partial }_{B{1}^{\prime }})f_x=0$, coefficient comparison gives ${\partial }_{B{0}^{\prime }}{f}_n={\partial }_{B{1}^{\prime }}{f}_{n-1}$. Applying that operator separately to $h_0,h_1$ cancels all nonconstant terms and leavesThus $-A_{B{0}^{\prime }}+\zeta A_{B{1}^{\prime }}=G_B$ yields $A_{B{1}^{\prime }}=0$ and $A_{B{0}^{\prime }}=-{\partial }_{B{1}^{\prime }}{f}_{-1}$. MoreoverSubstitution gives $A_{B{0}^{\prime }}=-(2\pi i{)}^{-1}\oint f_{,B}d\zeta$, which in homogeneous form is preciselyThis is the Abelian Ward splitting by Laurent series. The factors have weights $-1,-1,2$, respectively, so the integrand is invariant under a change of homogeneous representative. The contour avoids ${\iota }_{C^{\prime }}{\pi }^{C^{\prime }}=0$; the basis chosen above places that point at infinity. One can keep the contour fixed for $x$ in a sufficiently small domain, allowing differentiation under the integral.

Finally differentiate the contour potential for an anti-self-dual Maxwell field. By the twistor incidence relation, ${\partial }_{A{A}^{\prime }}{f}_{,B}={\pi }_{A^{\prime }}{f}_{,AB}$. ConsequentlyThe second equality uses the elementary two-spinor identity ${\pi }_{A^{\prime }}{\iota }_{B^{\prime }}-{\pi }_{B^{\prime }}{\iota }_{A^{\prime }}={ϵ}_{A^{\prime }{B}^{\prime }}({\iota }_{C^{\prime }}{\pi }^{C^{\prime }})$ in the declared convention. The remaining two-component spinor is symmetric in $A,B$, and no primed symmetric gauge curvature part survives. This directly verifies $dA$ is anti-self-dual. In complexified Lorentzian signature duality has Hodge star eigenvalues $\pm i$; here the vanishing primed two-component spinor fixes the stated ASD convention rather than incorrectly imposing the Euclidean equation $\ast F=-F$ on that signature.

Let $F_{+}=(F+\ast F)/2$ and $F_{-}=(F-\ast F)/2$, using the oriented Euclidean metric. For a self-dual two-form $\omega$,Apply this componentwise in the Lie algebra. Since the three ${\omega }_i$ span the self-dual subspace, their vanishing wedges force $F_{+}=0$. Thus $\ast F=-F$. The gauge-theory Bianchi identity is $D_{A}F=0$, so $D_A\ast F=-D_{A}F=0$, the Euclidean Yang-Mills equations. This proves the implication locally; finite action would additionally be needed to call the solution an instanton.

Use the complex orientation for $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The real and imaginary parts of $dw\wedge dz$, and $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. Accordingly the three gauge curvature equations areFor a spectral parameter $\lambda \in {\mathbb{CP}}^1$, consider the Lax pairTheir commutator isIt vanishes for every $\lambda$ exactly when all three coefficients vanish. The derivative directions ${\partial }_w-\lambda {\partial }_{\bar{z}}$ and ${\partial }_z+\lambda {\partial }_{\bar{w}}$ are mutually orthogonal null vectors for the complexified metric. Flatness on the corresponding two-planes makes the linear parallel-transport equations locally compatible. Conversely compatibility for an invertible fundamental matrix $\Psi$ forces the commutator to vanish; a single zero solution would not establish it. This is the Euclidean complex-coordinate anti-self-dual Lax pair.

For an Abelian U(1) connection, all commutators of the connection components vanish, and the gauge curvature equations therefore readLocally the first two equations say that $A^{1,0}$ is $\partial$-closed and $A^{0,1}$ is $\bar{\partial }$-closed. The Dolbeault-Poincaré lemma, and its complex conjugate, give scalar functions $u,v$ withSubstituting in the third equation givesIn a complex gauge transformation, use $g=e^{-u}$ and the derivative-plus-connection convention $A\mapsto A+g^{-1}dg$. ThenThis is the scalar Laplace equation, since for the stated metric $\Delta =2({\partial }_z{\partial }_{\bar{z}}+{\partial }_w{\partial }_{\bar{w}})$. The complex gauge transformation need not remain in $U(1)$.

A reduction using only a unitary gauge transformation is also available. Since $A$ is anti-Hermitian, $A_{\bar{w}}=-\overline{A_w}$ and $A_{\bar{z}}=-\overline{A_z}$, so the potential for the barred components can be chosen as $v=-\bar{u}$. Write $u=p+iq$ with real $p,q$. Then $A=\partial p-\bar{\partial }p+idq$, and $g=e^{-iq}$ removes the last term. With $f=-2p$ this givesIts gauge curvature is $\partial \bar{\partial }f$. Thus the Abelian anti-self-dual connections from harmonic scalar potentials reduction is local and does not require silently enlarging the real gauge group.

Use the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$. On the Adjoint representation, the adjoint covariant derivative is $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Set $a_{\mu }=\delta A_{\mu }$ and $\chi =\delta \Phi$. ThenThe trace obeys $\mathrm{Tr}(X[Y,Z])=\mathrm{Tr}([Z,X]Y)$, and covariant integration by parts has the ordinary boundary term because the trace of a commutator is zero. Thus compactly supported variations giveThe trace pairing on $\mathfrak{su}(2)$ is nondegenerate, so the Euler-Lagrange field equations areIn particular the sign of the scalar mass term follows from the plus sign in the supplied trace Lagrangian density; it should not be guessed from a different metric convention. This is the adjoint Yang-Mills-Higgs trace variation.

For the static Bogomolny-Prasad-Sommerfield monopole, take the usual purely magnetic extension ${\partial }_0{A}_i={\partial }_0\Phi =0$ and temporal gauge $A_0=0$. At $m=0$, the equations reduce towith the $\nu =0$ equation identically zero. We now verify both spatial equations from $B_i=D_i\Phi$ without assuming them. The gauge-theory Bianchi identity gives $D_i{B}_i=0$, hence immediately $D_i{D}_i\Phi =0$. Also $F_{ji}={\epsilon }_{jik}{B}_k$, soHere ${\epsilon }_{jik}{\epsilon }_{jk\ell }=-2{\delta }_{i\ell }$. This proves that the Bogomolny equations imply all static zero-mass Yang-Mills equations and the scalar equation. The specification of $A_0$ matters: the spatial Bogomolny equations by themselves do not constrain an arbitrary extra electric potential, so the implication is for their standard purely magnetic static extension.

The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and total energy areThe Euler-Lagrange field equation is ${\varphi }_{tt}-{\varphi }_{xx}+U^{\prime }(\varphi )=0$. Multiplication by ${\varphi }_t$ gives the local conservation of energy identityThus $E$ is conserved when the energy flux vanishes at spatial infinity. For finite energy with asymptotically constant fields, the endpoints ${\varphi }_{\pm }$ must be scalar-field vacua, where $U=0$.

Choose an antiderivative $G^{\prime }(\varphi )=\sqrt{2U(\varphi )}$. For either sign $s=\pm 1$, completing the square givesChoosing $s$ to make the boundary term nonnegative proves the Bogomolny boundThese Bogomolny equations obtained by square completion for a one-dimensional kink describe minima within a fixed endpoint sector when a saturating configuration exists. Differentiating the spatial equation gives ${\varphi }_{xx}=U^{\prime }(\varphi )$ wherever $U>0$, with smooth extension to its limiting scalar-field vacua. The absolute minimum across all sectors is a constant zero-potential scalar-field vacuum when one exists; the nontrivial minima are sector-specific kinks.

For $\beta \ne 0$, write $b=|\beta |$. The scalar-field vacua are $-b,0,b$. A static finite-energy solution has the first integralbecause differentiating its left side gives ${\varphi }_x({\varphi }_{xx}-U^{\prime })=0$, and its value is zero at infinity. In either adjacent interval, $\sqrt{2U}=|\varphi |(b^2-{\varphi }^2)$. Set $y={\varphi }^2/b^2$. The first-order equation becomes $y_x=\eta 2b^{2}y(1-y)$, with $\eta =\pm 1$, and separation gives $\log [y/(1-y)]=2\eta b^2(x-x_0)$. HenceFor $\eta =1$ the endpoints are $0\to \sigma b$; for $\eta =-1$ they are $\sigma b\to 0$. Thus there are four oriented static kink families, up to translation, namely two increasing kinks and two decreasing antikinks. Each has rest energyA static field joining $-b$ directly to $b$ would have to pass through $\varphi =0$ at finite $x$. The first integral then gives ${\varphi }_x=0$, and the Picard-Lindelöf theorem applied to the smooth second-order equation with those initial conditions forces $\varphi \equiv 0$. Therefore there is no such additional kink; this is the intermediate-vacuum obstruction to a kink. The Bogomolny classification of a rescaled phi-six kink also shows that if $\beta =0$, only the scalar-field vacuum $0$ remains and there is no nontrivial finite-energy static kink. Indeed the first integral gives a monotone field wherever it is nonzero, precluding a return to the same scalar-field vacuum.

A Lorentz boost of the $0\to b$ profile gives a moving solution with arbitrary $|v|<1$:If $K$ denotes the static profile and $\xi =\gamma (x-vt-x_0)$, then ${\varphi }_{tt}-{\varphi }_{xx}={\gamma }^2(v^2-1)K^{\prime \prime }=-K^{\prime \prime }$, proving the Euler-Lagrange field equation directly. Its energy is $\gamma b^4/4$: since $K^{\prime 2}=2U(K)$, changing variable to $\xi$ in the energy integral gives the factor $\gamma$. Reversing $\sigma$ or $\eta$ yields the other moving elementary profiles.

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.

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.

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.