= Solution
A canonical <Wess–Zumino model> with a cubic <superpotential> already supplies a microscopic membrane-like object. Choose positive real $\mu,g$ and
$$
W(\Phi)=\frac{\mu^2}{g}\Phi-\frac g3\Phi^3,
\qquad U(\phi)=\left|\frac{\mu^2}{g}-g\phi^2\right|^2.
$$
There are two isolated <supersymmetric vacua>, $\phi_\pm=\pm\mu/g$. 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 is
$$
T=\int dz\,(|\phi'|^2+|W'|^2)
=\int dz\,|\phi'-e^{i\alpha}\overline{W'}|^2
+2\operatorname{Re}\bigl(e^{-i\alpha}\Delta W\bigr).
$$
Choosing $\alpha=\arg\Delta W$ gives $T\geq2|\Delta W|$. The real increasing solution saturates the bound:
$$
\phi'=\frac{\mu^2}{g}-g\phi^2,
\qquad
\boxed{\phi(z)=\frac\mu g\tanh(\mu(z-z_0)),\qquad
T=\frac{8\mu^3}{3g^2}.}
$$
The tension follows either from integrating $2(\phi')^2$ 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 $1/\mu$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-53-domain-wall.png]
{title=Derived Wess-Zumino double-well potential, domain-wall profile and localized energy density}
The wall breaks transverse translations: changing $z_0$ gives a normalizable scalar zero mode proportional to $\phi'(z)$. It also preserves only half the four real <supercharges>. In the chiral-fermion transformation, $F=-\overline{W'}=-\phi'$ and the derivative term combine into a rank-half condition on the constant <supersymmetry> parameter, for example $\epsilon=i\sigma^3\bar\epsilon$ in a compatible Weyl convention. The orthogonal two real parameters are broken and generate localized <fermion> zero modes. One such profile satisfies $(\partial_z+2g\phi)f=0$, giving $f\propto\operatorname{sech}^2(\mu(z-z_0))$. 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 action>
$$
\boxed{S_{\mathrm{mem}}=-T\int d^3\xi\sqrt{-\det h_{ij}},
\qquad h_{ij}=\partial_iX^m\partial_jX^n\eta_{mn}.}
$$
It is invariant under <worldvolume> reparametrizations and the target <Poincaré group>. In static gauge $X^i=\xi^i$, $X^3=Y$, the <matrix determinant lemma> gives
$$
S_{\mathrm{mem}}=-T\int d^3\xi\sqrt{1+\partial_iY\partial^iY}
=-T\int d^3\xi\left(1+\tfrac12\partial_iY\partial^iY+\cdots\right).
$$
Directly substituting the leading collective-coordinate profile into the scalar kinetic term gives the same coefficient, since $\int dz\,(\phi')^2=T/2$. 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 $\partial_i(\sqrt{-h}\,h^{ij}\partial_jX^m)=0$.
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:
$$
\Pi_i^m=\partial_iX^m+i\bar\theta\gamma^m\partial_i\theta,
\qquad h_{ij}=\Pi_i^m\Pi_j^n\eta_{mn}.
$$
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,
$$
S=-T\int d^3\xi\sqrt{-h}+T\int_{\Sigma_3}Z^*C_3,
\qquad
G_4=dC_3=\frac i2\Pi^m\wedge\Pi^n\wedge d\bar\theta\gamma_{mn}d\theta.
$$
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 $dG_4=0$. 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> $\gamma_i=\Pi_i^m\gamma_m$ and
$$
\Gamma_\kappa=\frac{\epsilon^{ijk}}{3!\sqrt{-h}}\gamma_{ijk},
\qquad
\boxed{\delta_\kappa\theta=(1+\Gamma_\kappa)\kappa(\xi),
\quad\delta_\kappa X^m=-i\bar\theta\gamma^m\delta_\kappa\theta.}
$$
With mostly-plus target signature, a local orthonormal <worldvolume> frame gives $\Gamma_\kappa=\gamma_0\gamma_1\gamma_2$, so $\Gamma_\kappa^2=1$ and $\operatorname{tr}\Gamma_\kappa=0$. 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 $\epsilon^{ijk}\gamma_{ij}=2\sqrt{-h}\Gamma_\kappa\gamma^k$ gives the opposite Wess-Zumino contribution with an extra $\Gamma_\kappa$. Their sum is proportional to
$$
\delta\bar\theta(1-\Gamma_\kappa)\gamma^i\partial_i\theta.
$$
For 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. \b[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.
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