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In mostly-minus signature, adding the boundary term ${\partial }_{\mu }{K}^{\mu }$ with $K^{\mu }=\frac{1}{2}(A_{\nu }{\partial }^{\nu }{A}^{\mu }-A^{\mu }{\partial }_{\nu }{A}^{\nu })$ to the subtracted Feynman gauge density changes ${\tilde{\pi }}^{\mu }=-{\dot{A}}^{\mu }$ to the displayed canonical momenta. For fields with vanishing spatial boundary terms, $\int K^0{d}^{3}x=\int A_0{\partial }_i{A}_i{d}^{3}x$, whose functional derivatives are these momentum shifts. They depend only on coordinates, so $[A_{\mu },{\pi }^{\nu }]=i{\delta }_{\mu }{}^{\nu }{\delta }^3$ is unchanged. The only potentially nonzero mixed momentum bracket is $[{\pi }^0(x),{\pi }^i(y)]=i({\partial }_{y^i}+{\partial }_{x^i}){\delta }^3(x-y)=0$. Thus both density representatives define canonical commutation relations, although their momenta must not be equated componentwise.

For a real sequence $(a_n)$, the partial sum of ${\sum }_{k=1}^{\infty }(a_k-a_{k-1})$ is exactly $a_n-a_0$. Consequently it is a convergent series if and only if $a_n$ has a finite limit $L$, and its sum is $L-a_0$. Cancellation takes place inside each finite partial sum; taking a limit is a separate step.^[ref]

The partial fraction telescoping example has $S_N=N/(3N+1)$ and therefore limit $1/3$. In contrast, let $a_n=(-1{)}^n$. Its first entry is $a_0=1$, so $S_N=(-1{)}^N-1$. The even partial sums are zero and the odd partial sums are negative two. Both finite cancellations are correct, but the sequence has no single limit, giving a divergent series.

It is also essential to keep the two contributions paired. For example, ${\sum }_{k=1}^{\infty }(1/k-1/(k+1))$ is a convergent series with sum one, as its finite partial sum is $1-1/(N+1)$. Writing it instead as ${\sum }_{k=1}^{\infty }1/k-{\sum }_{k=1}^{\infty }1/(k+1)$ asks to subtract two divergent series. The permitted finite cancellation does not define that separate subtraction. Working through partial sums keeps the indices aligned and the operation meaningful.

The Euler-Lagrange field equation implements the principle of stationary action for a Lagrangian density depending on scalar fields and their first derivatives. For a single scalar field $\Phi$,

Vary $S=\int d^{4}x\mathcal{L}$. Integration by parts transfers the derivative from $\delta \Phi$ to $\partial \mathcal{L}/\partial ({\partial }_{\mu }\Phi )$. The boundary term vanishes for variations supported away from the boundary; arbitrary interior variations then force the displayed coefficient to vanish. For $\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\Phi {\partial }^{\mu }\Phi -\frac{1}{2}{m}^2{\Phi }^2$, the result is the Klein-Gordon equation $({\partial }_{\mu }{\partial }^{\mu }+m^2)\Phi =0$, using natural units and Minkowski metric signature $(+---)$. ^[ref]

A functional derivative represents the linear first-order response of a functional to a perturbation of its input function. With a specified integration measure and admissible perturbations,For $F[\varphi ]=\int \varphi (x{)}^{2}dx$, direct expansion gives $\delta F/\delta \varphi =2\varphi$. For $F[\varphi ]=\frac{1}{2}\int |{\varphi }^{\prime }{|}^{2}dx$, integration by parts gives $-{\varphi }^{\prime \prime }$ when the boundary term vanishes under the chosen perturbation conditions. The boundary conditions and function space are therefore part of identifying the derivative, rather than optional details.

Integration by parts transfers a derivative from one factor of a product to the other, while keeping an endpoint contribution. Its usefulness comes from choosing which factor becomes simpler when differentiated, and which has an accessible antiderivative. The integral is then replaced by a boundary term minus another integral.

For real continuously differentiable functions $u,v$ on $[a,b]$, the precise definite integral formula isIn the shorter notation, $du=u^{\prime }(x)dx$ and $dv=v^{\prime }(x)dx$, giving $\int udv=uv-\int vdu$ for indefinite integrals, with a constant of integration understood. The product rule says $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$; the fundamental theorem of calculus turns its integral into the endpoint difference. Thus the formula packages a differentiation rule into a tool for integration.^[ref]

For example, choose $u=x$ and $dv=e^{x}dx$. Then $du=dx$ and $v=e^x$, since the exponential function is its own derivative:The boundary term is $e\approx 2.718282$; the remaining integral is $e-1\approx 1.718282$. Their difference is exactly one. The endpoint product alone would overcount, because its derivative includes $u^{\prime }v=e^x$ as well as the desired $u{v}^{\prime }=x{e}^x$.

In the figure, the area under $(x{e}^x{)}^{\prime }=(1+x)e^x$ is the endpoint contribution $e$. Its lower band has area $e-1$, and its upper band has thickness $x{e}^x$ and area one. This decomposition makes the subtraction visible. The endpoint contribution is a difference of product values; it need not equal the area under either factor separately.

The same choice repeatedly lowers the degree of a polynomial when integrating a polynomial times an exponential function, while choosing $u=\ln x$ gives $\int \ln xdx=x\ln x-x+C$ for $x>0$. More broadly, moving a derivative onto a convenient test function is the mechanism behind a weak derivative. Always evaluate the boundary term: it disappears only when the relevant endpoint values justify doing so, often through a boundary condition.

With the curvature sign convention in which $\delta R_{bc}={\nabla }_c\delta {\Gamma }^a{}_{ba}-{\nabla }_a\delta {\Gamma }^a{}_{bc}$, the inverse-metric variation of $\int \sqrt{-g}\Phi R$ is $\int \sqrt{-g}{E}_{ab}\delta g^{ab}$ plus a boundary term, for a scalar multiplier held fixed in that variation. The Palatini identity expresses the curvature variation as divergences of $\delta \Gamma$. Two integrations by parts move the derivatives onto $\Phi$, yielding the displayed terms. For constant $\Phi$ they vanish. The opposite curvature convention instead has $E_{ab}=\Phi G_{ab}+g_{ab}\square \Phi -{\nabla }_a{\nabla }_b\Phi$.

Put $\mathcal{T}=\beta \hslash >0$, the interval of imaginary time, and assume $m,\omega >0$. For a variation $x_0+ϵ\eta$ with $\eta (0)=\eta (\mathcal{T})=0$, the first variation of the Euclidean action isThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationIts sign differs from the real-time harmonic oscillator equation because the Euclidean action contains positive kinetic energy and positive potential energy. The general solution is $A\cosh (\omega \tau )+B\sinh (\omega \tau )$. Applying the two endpoint conditions gives $A=x$ and $B=-x\tanh (\omega \mathcal{T}/2)$. Equivalently, the equal-endpoint Euclidean oscillator path isThis form displays reflection about the midpoint and satisfies both endpoints immediately. On this path,Thus the Euclidean action reduces to a boundary term. Since ${\dot{x}}_0(\mathcal{T})=\omega x\tanh (\omega \mathcal{T}/2)$ and ${\dot{x}}_0(0)=-\omega x\tanh (\omega \mathcal{T}/2)$,The positive quadratic form $m{\int }_0^{\mathcal{T}}({\dot{\eta }}^2+{\omega }^2{\eta }^2)d\tau$ is the second variation, so this stationary path is the unique minimum, rather than merely an arbitrary stationary point.

For fixed-endpoint variations, the second variation iswith $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$ evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$. The Jacobi equation, written without differentiating $r$ separately, isWhen $r$ is differentiable this is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$.

Positive definiteness means $q(\phi )>0$ for every nonzero admissible variation. Uniform or coercive positivity of a second variation is the stronger assertion that $q(\phi )\ge c\int {\phi }^{\prime 2}$ for some fixed $c>0$. On a fixed-endpoint interval this is equivalent to a positive lower bound in the $W^{1,2}$ norm.

The strict Legendre implication uses the uniform interpretation of positivity. If $q\ge c\int {\phi }^{\prime 2}$, localized rapidly oscillating variations show $p(x)\ge c$: otherwise support them in an interval where $p-c<0$, so the derivative term of $q-c\int {\phi }^{\prime 2}$ is negative of order frequency squared, overwhelming its mixed and zero-order terms. Continuity extends the bound to the endpoints, giving $p>0$ on $[a,b]$. Under mere positive definiteness this strict conclusion is false: $f=u^2$ at $\bar{u}=0$ gives $q=2\int {\phi }^2>0$ but $p=0$. In that weaker convention only the ordinary Legendre necessary condition $p\ge 0$ follows. This distinction makes the implication precise.

Under the stated regular condition $p>0$, a conjugate point $c\in (a,b]$ to $a$ is an endpoint admitting a nonzero Jacobi solution with $z(a)=z(c)=0$. The disconjugacy form of the Jacobi positivity criterion says that no such point in $(a,b]$ exists if and only if there is a Jacobi solution strictly positive on all of $[a,b]$.

For the required identity, let that positive solution be $z$ and write $\phi =zh$. Set $A=pz{z}^{\prime }+r{z}^2$. The Jacobi equation givesExpanding the second-variation integrand therefore yieldsFor compactly supported $\phi$, the boundary term vanishes, provingThis is the Jacobi identity for the second variation. The divergence form is also appropriate under the original $C^1$ regularity; with $p>0$, the Euler-Lagrange equation gives the usual further regularity along the extremal.

Multiplication by an integrating factor givessince $p^{\prime }=ap$ and hence $(p{w}^{\prime }{)}^{\prime }=p(w^{\prime \prime }+a{w}^{\prime })$.

For the singular-endpoint equation, take $p=x^{-1/2}$, $r=x^{1/2}/2$, $q=0$. The indicial equation is $\alpha (2\alpha -3)=0$. The constant branch violates the zero endpoint value unless its coefficient vanishes; the nontrivial admissible branch therefore has $\boxed{\alpha =3/2}$. More precisely, an eigenfunction has $w=C{x}^{3/2}+O(x^{9/2})$ and $w^{\prime }=(3C/2)x^{1/2}+O(x^{7/2})$ near zero.

Integrate the self-adjoint equation after multiplying by $w$, initially on $[a,1]$ with $a>0$:At zero the boundary expression is $O(x^{3/2})$, the first integrand is $O(x^{1/2})$, and the second is $O(x^{7/2})$. Both integrals converge and the boundary term vanishes. Division by the positive weighted norm gives $\boxed{\lambda =I(w)}$.

The change of coordinate $z=x^{3/2}$, $w(x)=W(z)$ gives exactlyThus ${\lambda }_n=9n^2{\pi }^2/2$ and $w_n=\sin (n\pi x^{3/2})$. To prove a minimum over trial functions, rather than just identify the least eigenvalue, transform the quotient itself:The Dirichlet Poincaré inequality gives the last step, with equality precisely for a nonzero multiple of $W=\sin \pi z$. The trial class is the nonzero finite-energy class with zero Sobolev traces; endpoint values alone without an integrability requirement do not define a finite quotient. This is the Power-coordinate reduction of a singular Sturm-Liouville quotient.

On each interval $[k,k+1]$, $k\in \mathbb{Z}$, the function is $d(x)=x-k$ on $[k,k+\frac{1}{2}]$ and $d(x)=k+1-x$ on $[k+\frac{1}{2},k+1]$. Its ordinary derivative exists off $\frac{1}{2}\mathbb{Z}$ and is the locally integrable functionAssign arbitrary values at the endpoints. For a test function $\varphi$, split $-\int d{\varphi }^{\prime }$ at these corners and apply integration by parts on each piece. The adjacent boundary terms cancel because $d$ is continuous, and only finitely many pieces meet the compact support of $\varphi$. HenceThe distributional derivative is exactly the regular distribution of the ordinary derivative defined almost everywhere. There are no Dirac delta distributions in this first distributional derivative: corners change slope, while a jump of the function itself would create a point mass.

The distributional derivative of $u$ is defined byIt is a distribution because $\varphi \mapsto {\varphi }^{\prime }$ is a continuous map of the space of test functions into itself. If $f\in C^1(a,b)$ represents a regular distribution, integration by parts givesThe boundary term vanishes because $\varphi$ has compact support. Thus the distributional derivative agrees with the ordinary derivative.

For $\mathrm{supp}\varphi \subseteq K$, the assumed order of a distribution estimate givesConsequently $\boxed{\mathrm{ord}(u^{\prime })\le m+1}$.

To repeat integration by parts across whole periods, choose periodic primitives with zero mean. Set $h_0=f$, and recursively defineEvery $h_r$ has zero mean. Part (i) gives $h_{r+1}^{\prime }=-h_r$ almost everywhere, and $h_{r+1}(0)=h_{r+1}(1)$ because $R{h}_r$ vanishes at both endpoints. Its periodic extension is consequently locally an absolutely continuous function, with no jump at the integers. Part (ii) gives the uniform estimateThe weak derivative identity on the whole line is $D{\overline{h}}_{r+1}=-{\overline{h}}_r$. After scaling by the integer $j$ and repeating integration by parts $N$ times,The boundary terms vanish because $\varphi$ has compact support in $(0,1)$. The interval contains exactly $j$ periods, soTaking $N=m+1$ therefore yieldsThis rapid distributional decay of periodic oscillations uses only integrability of $f$ and smoothness of the test function.

Write $z=x+iy$, $r=|z|$, and $L=(4\pi {)}^{-1}\log (x^2+y^2)$. The Wirtinger derivatives satisfy $4{\partial }_{\bar{z}}{\partial }_z=\Delta$. Away from zero, ordinary differentiation givesThis equality also holds for distributional derivatives. To check that no Dirac delta distribution was lost at zero, integrate by parts outside the disk $r\le \epsilon$ against a test function. The new boundary term has absolute value at most $C\epsilon |\log \epsilon |\Vert \varphi {\Vert }_{\infty }\to 0$. Both $L$ and its displayed first derivatives, of size $O(r^{-1})$, are locally integrable functions in two dimensions, so the remaining integrals converge to their full-plane values.

Using the supplied fundamental solution of the Laplace equation identity $\Delta L={\delta }_0$, and commutation of distributional derivatives, now givesThus the Cauchy kernel fundamental solution has exactly the required normalization.

Interchange the two Wirtinger derivatives in part (i). The same real fundamental solution of the Laplace equation $L$ givesThe calculation is valid for distributions by the same vanishing boundary term estimate. Any other fundamental solution differs by a distribution $H$ with ${\partial }_{z}H=0$. Then $\Delta H=0$, so the regularity argument of part (ii) makes $H$ smooth, and its Cauchy-Riemann equations make it an antiholomorphic function. On the whole plane it can be written $H(z)=h(\bar{z})$ with $h$ an entire function. Conversely every such $H$ has ${\partial }_{z}H=0$. Therefore

The in-plane equilibrium equations are ${\tau }_{xx,x}+{\tau }_{xy,y}=0$ and ${\tau }_{xy,x}+{\tau }_{yy,y}=0$. On a simply connected patch, each divergence-free row has a streamfunction: write ${\tau }_{xx}={\varphi }_y$, ${\tau }_{xy}=-{\varphi }_x=-{\psi }_y$, ${\tau }_{yy}={\psi }_x$. Symmetry gives ${\varphi }_x={\psi }_y$, so another potential exists with $A_y=\varphi$, $A_x=\psi$. HenceIf two such Airy stress functions give the same stress, their difference has zero Hessian and is therefore an arbitrary affine function $a+bx+cy$ on a connected domain. This is the full nonuniqueness for a fixed stress. The existence argument is local, or global on a simply connected domain; domains that are multiply connected regions require the associated force and moment period conditions for a single-valued global potential. Equilibrium alone should not silently erase that topological qualification.

For plane strain put $s=e_{xx}+e_{yy}$. The constitutive equation gives ${\tau }_{xx}+{\tau }_{yy}=2(\lambda +\mu )s$. ConsequentlyThe displacement Saint-Venant compatibility equations $e_{xx,yy}+e_{yy,xx}-2e_{xy,xy}=0$ then becomesFor stable elastic moduli the coefficient is nonzero, so $\boxed{{\Delta }^{2}A=0}$. If $D$ is bounded and smooth and $A={\partial }_{n}A=0$ on its boundary, integrating twice by parts gives $0={\int }_{D}A{\Delta }^{2}AdV={\int }_D(\Delta A{)}^{2}dV$. Thus $A$ is harmonic with zero boundary value and $\boxed{A=0}$ by uniqueness of the Dirichlet problem. The normal-derivative condition is what removes the extra fourth-order boundary terms.

For the pressurized annulus, substituting the radial constitutive equations into equilibrium gives $u^{\prime \prime }+u^{\prime }/r-u/r^2=0$, so $u=Cr+D/r$. The stresses areUsing gauge stresses relative to ambient, the boundaries are ${\tau }_{rr}(a)=-P$, ${\tau }_{rr}(b)=0$. ThereforeandThe maximum occurs at the bore. Once the chosen yield criterion is reached, Hooke's law cannot describe the total strain alone: write total strain as elastic plus plastic strain and supplement the elastic stress law with a yield condition, a plastic flow rule and any required work hardening law. A plastic region starts near the bore and must be matched to the remaining elastic region. The exact conversion between a principal-stress difference and the quoted yield stress depends on whether the material criterion is Tresca or von Mises. This is the elastic stress concentration in a pressurized thick cylinder.

Although connection coefficients are not tensor components, their variation $C^a{}_{bc}=\delta {\Gamma }^a{}_{bc}$ is tensorial: the difference of affine connections is a tensor. Varying the Levi-Civita connection at fixed coordinates givesIn the paper's curvature sign convention, varying the derivative and quadratic-connection terms and grouping them into covariant derivatives gives the Palatini identityOne may verify the grouping in normal coordinates at a point, where the affine connection vanishes and the expression is just the difference of two partial derivatives; both sides are tensors, so it holds in every coordinate frame. Reversing the definition of the Riemann curvature tensor reverses this identity's right-hand side.

For completeness, put $q_{ab}=\delta g_{ab}$ and $q=g^{ab}{q}_{ab}$. The contracted variation uses $C^a{}_{ba}={\nabla }_{b}q/2$ and $g^{bc}{C}^a{}_{bc}={\nabla }_b{q}^{ab}-{\nabla }^{a}q/2$. After two integrations by parts,up to the discarded boundary term. Substituting $q^{ab}=-\delta g^{ab}$ gives the differentiated-scalar terms in the metric variation of a scalar-curvature coupling.

The Brans-Dicke theory couples a scalar field $\Phi$ to the Ricci scalar. Assume $\Phi \ne 0$, constant $\omega$, matter independent of $\Phi$, and variations of compact support so boundary terms may be discarded. Use the paper's curvature sign convention and signature $(+---)$. Set $X=g^{ab}{\partial }_a\Phi {\partial }_b\Phi$ and $\square \Phi ={\nabla }_a{\nabla }^a\Phi$. Define the stress-energy tensor consistently byThis convention fixes the sign of the matter term.

Vary with respect to $g^{ab}$. The variation of metric volume density contributes $-g_{ab}/2$ times each scalar Lagrangian, while $\delta R=R_{ab}\delta g^{ab}+g^{ab}\delta R_{ab}$. Applying the Palatini identity followed by two rounds of integration by parts gives the metric variation of a scalar-curvature coupling,The differentiated $\Phi$ terms remain because the curvature multiplier is not constant. Varying the kinetic term at fixed $\Phi$ givesCombining these terms with the matter variation, the Euler-Lagrange equation isRaising both free indices reproduces the contravariant form, because ${\nabla }_a{g}_{bc}=0$.

The independent scalar variation givesor $R+\omega {\Phi }^{-2}X-2\omega {\Phi }^{-1}\square \Phi =0$. The trace of the metric tensor equation in four dimensions isMultiplying the scalar equation by $\Phi$ eliminates the same combination $\Phi R+\omega {\Phi }^{-1}X$. Thus the scalar equation of Brans-Dicke theory isAt the degenerate coupling of Brans-Dicke theory, $\omega =-3/2$, the combined equations impose $T=0$ instead; division by $3+2\omega$ is unavailable. If matter depended directly on $\Phi$, its scalar variation would add a source and the displayed scalar equation would change.

The following lettered sections justify the supplied variation identities; they are supporting identities in the PDF, rather than three independent field-equation problems.

A canonical Wess–Zumino model with a cubic superpotential already supplies a microscopic membrane-like object. Choose positive real $\mu ,g$ andThere are two isolated supersymmetric vacua, ${\varphi }_{\pm }=\pm \mu /g$. A planar domain wall has a static scalar profile depending on the transverse coordinate $z=X^3$, with one vacuum at either end. Its energy per unit area isChoosing $\alpha =\arg \Delta W$ gives $T\ge 2|\Delta W|$. The real increasing solution saturates the bound:The tension follows either from integrating $2({\varphi }^{\prime }{)}^2$ or from $W(\mu /g)-W(-\mu /g)=4{\mu }^3/(3g^2)$. This is a supersymmetric domain wall, and the first-order equation also implies the second-order scalar equation. Its thickness is of order $1/\mu$.

The wall breaks transverse translations: changing $z_0$ gives a normalizable scalar zero mode proportional to ${\varphi }^{\prime }(z)$. It also preserves only half the four real supercharges. In the chiral-fermion transformation, $F=-\overline{W^{\prime }}=-{\varphi }^{\prime }$ and the derivative term combine into a rank-half condition on the constant supersymmetry parameter, for example $ϵ=i{\sigma }^3\overline{ϵ}$ in a compatible Weyl convention. The orthogonal two real parameters are broken and generate localized fermion zero modes. One such profile satisfies $({\partial }_z+2g\varphi )f=0$, giving $f\propto {\mathrm{sech}}^2(\mu (z-z_0))$. The scalar and a three-dimensional Majorana fermion therefore form the low-energy wall multiplet. The domain-wall charge in N=1 supersymmetry is a boundary/tensor charge contributing to the extended algebra; the preserved combinations saturate $T=2|\Delta W|$. This avoids falsely applying the unextended particle-algebra argument of Question 3 to a wall with nonzero tension.

Promote the translation collective coordinate to a slowly varying worldvolume field $Y({\xi }^0,{\xi }^1,{\xi }^2)$. At wavelengths long compared with the wall thickness, its leading relativistic geometric action is the Dirac membrane actionIt is invariant under worldvolume reparametrizations and the target Poincaré group. In static gauge $X^i={\xi }^i$, $X^3=Y$, the matrix determinant lemma givesDirectly substituting the leading collective-coordinate profile into the scalar kinetic term gives the same coefficient, since $\int dz({\varphi }^{\prime }{)}^2=T/2$. Target Lorentz symmetry fixes the leading first-derivative nonlinear completion to the invariant area; finite-thickness physics can add higher-curvature terms. The area action is therefore a long-wavelength effective field theory, not an exact replacement for the microscopic wall at arbitrary momenta. Its embedding equation is ${\partial }_i(\sqrt{-h}{h}^{ij}{\partial }_j{X}^m)=0$.

For the four-dimensional supermembrane, extend the embedding to $(X^m(\xi ),{\theta }^{\alpha }(\xi ))$ with a four-real-component Majorana spinor. Pull back the invariant coframe of Question 7:Replacing $\partial X$ by $\Pi$ makes the area term invariant under rigid target supersymmetry, but does not yet give the appropriate fermionic gauge symmetry. Add a Wess-Zumino brane coupling to a superspace three-form potential $C_3$. In one orientation and matching normalization,The supermembrane closed four-form uses the distinct four-dimensional Fierz identity $(C{\gamma }_{mn}{)}_{(\alpha \beta }(C{\gamma }^n{)}_{\gamma \delta )}=0$. Differentiating $G_4$ reduces its coefficient to this symmetric tensor, so $d{G}_4=0$. A local potential therefore exists. Since the curvature is supertranslation invariant, the potential can vary by an exact form; its pullback integral is invariant modulo boundary terms. This is why the three-form $H_3$ of Question 7 is not itself the membrane field strength: integrating a membrane potential requires three-form degree and hence four-form curvature.

The required additional local fermionic gauge invariance is kappa symmetry. Define the worldvolume gamma matrices ${\gamma }_i={\Pi }_i^m{\gamma }_m$ andWith mostly-plus target signature, a local orthonormal worldvolume frame gives ${\Gamma }_{\kappa }={\gamma }_0{\gamma }_1{\gamma }_2$, so ${\Gamma }_{\kappa }^2=1$ and $\mathrm{tr}{\Gamma }_{\kappa }=0$. Thus $(1\pm {\Gamma }_{\kappa })/2$ are rank-two kappa symmetry projectors. The embedding variation gives $\delta {\Pi }_i^m=2i\delta \bar{\theta }{\gamma }^m{\partial }_i\theta$. The area variation is consequently proportional to $-2iT\sqrt{-h}\delta \bar{\theta }{\gamma }^i{\partial }_i\theta$. Contracting the displayed $G_4$ with the fermionic variation and using ${ϵ}^{ijk}{\gamma }_{ij}=2\sqrt{-h}{\Gamma }_{\kappa }{\gamma }^k$ gives the opposite Wess-Zumino contribution with an extra ${\Gamma }_{\kappa }$. Their sum is proportional toFor the boxed kappa variation, $\delta \bar{\theta }=\bar{\kappa }(1+{\Gamma }_{\kappa })$, so it vanishes by $(1+{\Gamma }_{\kappa })(1-{\Gamma }_{\kappa })=0$. Reversing membrane orientation reverses the Wess-Zumino sign and the matching projector. The cancellation fixes the relative charge magnitude to the tension; rigid supersymmetry alone would not fix it.

Worldvolume reparametrizations remove three of the four embedding scalars, leaving one transverse mode. Kappa symmetry removes half the four real embedding spinor components, leaving a two-component three-dimensional Majorana field, whose first-order equation has one on-shell fermionic polarization. The physical worldvolume modes are one scalar and one Majorana fermion, matching the wall zero modes. Without kappa symmetry the fermionic degrees would be doubled. A planar configuration preserves the target supersymmetries that can be compensated by the kappa projector, reproducing the wall's half-supersymmetry condition. This is the four-dimensional effective domain-wall interpretation of a supermembrane, rather than the eleven-dimensional M2-brane particle-content problem.

In the ordinary local two-derivative matter class, the general ungauged supersymmetric action for a chiral superfield isThe Kähler potential $K$ is real, and the superpotential $W$ is a holomorphic function. Adding $f(Z)+\bar{f}(\bar{Z})$ to $K$ changes the action only by a boundary term: full superspace integration of a chiral or antichiral function vanishes modulo that term. This is a Kähler transformation. A positive Kähler metric $G=K_{z\bar{z}}$ is required for a nonsingular positive-norm kinetic term, not just for derivative counting.

To see the differential order, the bosonic component Lagrangian in mostly-plus signature containsThe auxiliary field equation is algebraic: $F=-G^{-1}\overline{W_z}$ on a purely bosonic configuration. It gives the scalar potential $U=G^{-1}|W_z{|}^2$. The scalar equations have at most two spacetime derivatives, the fermion equations from the sigma-model kinetic term have one, and the remaining interaction terms do not raise this order. The fermionic corrections to the auxiliary equation are still algebraic.

For four-dimensional renormalizability, assign engineering dimensions $[Z]=1$, $[d^4\theta ]=2$, and $[d^2\theta ]=1$. Hence, up to a Kähler transformation and a constant field normalization, the restrictions areA mixed higher-degree term in $K$ or a term of degree above three in $W$ introduces a negative-mass-dimension coupling. The constant $w_0$ has no local effect in global supersymmetry, as in the constant superpotential in global supersymmetry.