= Solution
For bosonic embedding coordinates $X^M(\sigma)$, the <supermembrane action> truncates to
$$
\boxed{S=-T_2\int d^3\sigma\,\sqrt{-\det h}
+\frac{qT_2}{3!}\int d^3\sigma\,\varepsilon^{ijk}
\partial_iX^M\partial_jX^N\partial_kX^P A_{MNP}(X),\qquad
h_{ij}=\partial_iX^M\partial_jX^NG_{MN}.}
$$
The first term is its induced <worldvolume> volume and the second is its <Wess-Zumino brane coupling>. A membrane of the opposite orientation reverses $q$.
The <M2-brane supergravity solution> admits the <harmonic function>
$$
H(y)=1+\sum_a\frac{Q_a}{|y-y_a|^6},\qquad Q_a>0,
$$
with arbitrary centre positions $y_a$ in eight transverse dimensions. Away from the sources, the <Laplacian> of each term is zero. The unrestricted positions describe static separated membranes carrying aligned charges, so their separations have no potential. Physically, their gravitational attraction is cancelled by their four-form charge repulsion. This is the <M2-brane probe no-force identity>, which can also be checked directly without relying only on the existence of the multicentre solution.
Keep the field-strength ordering printed in the paper and use $F=dA$. If $\operatorname{vol}_{012}=dx^0\wedge dx^1\wedge dx^2$, a gauge vanishing at infinity is
$$
A=(1-H^{-1})\operatorname{vol}_{012},\qquad
dA=\operatorname{vol}_{012}\wedge dH^{-1}.
$$
The last equality uses the sign acquired by moving a one-form past a three-form. For this convention the parallel probe has $q=-1$. In <static gauge>, at a fixed transverse position, its <induced worldvolume metric> is $h_{ij}=H^{-2/3}\eta_{ij}$, hence $\sqrt{-\det h}=H^{-1}$ and
$$
\boxed{\mathcal L_{\mathrm{parallel}}=-T_2H^{-1}-T_2(1-H^{-1})=-T_2,
\qquad\nabla_yV_{\mathrm{parallel}}=0.}
$$
Using the opposite convention for $F$, $A$ and orientation together yields the same cancellation. Changing only the probe charge produces an anti-membrane, with nonconstant potential $T_2(2H^{-1}-1)$. Thus the orientation cannot be dropped from the argument.
The cancellation also survives an expansion for slow transverse motion. With $v^2=|\dot y|^2$,
$$
\mathcal L=-T_2H^{-1}\sqrt{1-Hv^2}-T_2(1-H^{-1})
=-T_2+\frac{T_2}{2}v^2+O(Hv^4).
$$
There is no static potential and no position dependence in the leading kinetic coefficient.
Near an isolated positive membrane charge, write $H\sim R^6/r^6$ with $r=|y-y_a|$ and $R^6=Q_a$. The regular membrane <near-horizon limit> removes the asymptotically flat constant and gives
$$
ds^2=\frac{r^4}{R^4}\eta_{\mu\nu}dx^\mu dx^\nu
+R^2\frac{dr^2}{r^2}+R^2d\Omega_7^2.
$$
Set $z=R^3/(2r^2)$. Then
$$
\boxed{ds^2=\frac{(R/2)^2}{z^2}
\left(\eta_{\mu\nu}dx^\mu dx^\nu+dz^2\right)+R^2d\Omega_7^2,
\qquad\text{geometry }\mathrm{AdS}_4(R/2)\times S^7(R).}
$$
Thus the <Anti-de Sitter spacetime> radius is half the sphere radius. In orthonormal units the four-form has magnitude $|F|=6/R$ along the AdS volume form, the <Freund-Rubin compactification> flux.
The full asymptotically flat solution has spinors $\epsilon=H^{-1/6}\epsilon_0$ subject to one oriented membrane projector, leaving sixteen real parameters. In the throat, the <Killing spinor> equation separates into the standard AdS and sphere equations with correlated signs fixed by the flux. There are four real AdS$_4$ solutions and eight real $S^7$ solutions for the required signs, so
$$
\boxed{N_{\mathrm{susy}}=4\times8=32.}
$$
This is <maximal supersymmetry of AdS4 times S7>: the additional sixteen spinors need not extend through the asymptotically flat region. A multicentre geometry therefore remains globally sixteen-supersymmetric even though each regular local throat has this enhancement. The classical supergravity description is controlled when $R$ is large compared with the eleven-dimensional <Planck length>; singular harmonic multipoles not representing regular positive membrane centres do not automatically have the same throat.
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