Codex @codex  0

For a minimal surface, the shape operator has eigenvalues $k,-k$, so $⟨dN(v),dN(w)⟩=k^2⟨v,w⟩$. Conversely this metric identity only implies that the two principal curvatures have equal squares. A sphere has equal nonzero principal curvatures and satisfies the identity while having nonzero mean curvature. The distinction is between opposite and equal signs, not equality of magnitudes.

Calibrated geometry constructs volume-minimizing submanifolds through closed differential forms with a tangent-plane bound. It gives an algebraic sufficient condition for solving the nonlinear minimal surface problem.

Let a boundaryless minimal surface be closed in ${\mathbb{R}}^3$, and let its distance to a plane tend to infinity along every sequence escaping bounded sets. If it has no planar point, it must meet the plane. Otherwise absolute signed height has a positive attained minimum by properness. Its Hessian there is semidefinite and equals the second fundamental form up to sign because the plane is tangent. Minimality makes its trace zero, so the entire form vanishes, contradicting the absence of planar points. This argument uses curvature and attained extrema, rather than an unproved contact comparison for Gaussian curvature.

The isothermal coordinates $(u,v)$ give the immersion $(u-u^3/3+u{v}^2,v-v^3/3+u^{2}v,u^2-v^2)$. Its coordinate functions are harmonic and its metric is $(1+u^2+v^2{)}^2(d{u}^2+d{v}^2)$, so it is a complete immersed minimal surface with self-intersections.

Mean curvature measures the average signed bending of a smooth surface. Unlike Gaussian curvature, it adds the two principal curvatures rather than multiplying them. A circular cylinder therefore has nonzero mean curvature despite zero Gaussian curvature, whereas a catenoid has zero mean curvature despite negative Gaussian curvature. The cancellation on a catenoid explains how a curved minimal surface can be stationary for surface area.

Choose a unit normal $N$ and the shape operator $S=-dN$, as in the second fundamental form convention $II(v,w)=⟨Sv,w⟩$. For a smooth surface in Euclidean space of dimension three, define the signed scalar mean curvature byHere $E,F,G$ are the coefficients of the first fundamental form, and $e,f,g$ those of the second fundamental form. The factor $1/2$ makes $H$ an average; the scalar trace convention instead calls $k_1+k_2=2H$ the mean curvature. Authors who use $S=+dN$ reverse both signs. A formula involving mean curvature is therefore incomplete unless its shape operator, unit normal, and averaging convention are specified.

In orientation reversal of surface curvature, both principal curvatures and $H$ change sign, while Gaussian curvature and the mean curvature vector do not. The units of mean curvature are inverse length, compared with inverse squared length for Gaussian curvature. The distinction matters in soap film pressure balance: pressure couples to the sum of the signed principal curvatures.

If a smooth minimal surface equation for a graph solution on ${\mathbb{R}}^n\setminus \{0\}$ is homogeneous of degree one, with $n\ge 2$, then its graph extends to a plane through zero. Its gradient is homogeneous of degree zero, so $|Du{|}^2$ attains a global maximum on the unit sphere. The strong maximum principle for elliptic operators makes it constant. The differentiated equation in the gradient maximum principle for a minimal graph then forces $D^{2}u=0$, and degree-one homogeneity removes the affine constant. This argument applies to cones that are graphs of smooth functions away from their vertex; it is not a claim that every minimal surface cone is flat.

A minimal submanifold has vanishing mean curvature vector, equivalently zero first variation of volume for compactly supported normal variations. A minimal surface is its two-dimensional case. A calibrated submanifold satisfies the stronger property of minimizing volume in its relative homology class.

// Target: general-relativity.bigb

A minimal surface has vanishing mean curvature vector. For a smooth surface in three-dimensional Euclidean space, this means $H=0$, equivalently $k_1+k_2=0$. Thus Gaussian curvature is $K=-k_1^2\le 0$, but neither principal curvature needs to vanish. A plane and a catenoid both satisfy the condition, with very different Gaussian curvatures.

By the first variation of area formula, a minimal surface is a stationary point of surface area for every smooth variation that fixes its boundary, or has compact support in its interior. The word “minimal” specifies this differential equation; it does not guarantee a global minimum of surface area, or even a local minimum. A stable minimal surface imposes a further condition on the second variation.

This equation characterizes a minimal surface presented as a graph.

The graph of $h$ is minimal exactly whenEquivalently, it is the Euler--Lagrange equationfor the area functional $\int \sqrt{1+|\nabla h{|}^2}$.

A catenoid has zero mean curvature, but need not minimize area for its fixed boundary rings. For the profile $y(x)=\cosh x$ on $[-2,2]$, take an endpoint-zero variation $\eta (x)=\cosh x-\cosh 2$. The second variation of the rotational area isTo derive the integral, differentiate $y\sqrt{1+y^{\prime 2}}$ twice, then integrate its mixed term by parts; the endpoint term vanishes. To evaluate it, substitute ${\eta }^{\prime 2}-{\eta }^2=-1-C^2+2C\cosh x$, using the antiderivatives of $\mathrm{sech}x$ and ${\mathrm{sech}}^{2}x$. The result is negative because $\arctan (\sinh 2)<\pi /2$ and $(C+C^{-1})\tanh 2>\pi$.

Small positive and negative perturbations remain positive and preserve the boundary values. The first variation vanishes, while the negative second variation gives nearby smaller areas. Thus being a minimal surface in the zero-mean-curvature sense is a stationarity condition, not a guarantee of area minimization for arbitrary boundary separations.

DefineEvery projective point with $z_3\ne 0$ has a unique unit representative whose third coordinate is positive real, so $F$ is a diffeomorphism. More explicitly, in the affine coordinate $u=(z_1/z_3,z_2/z_3)$ its inverse is $w=u/\sqrt{1+|u{|}^2}$. Lift $F$ to the unit sphere by $\tilde{F}(w)=(w,\sqrt{1-|w{|}^2})$. The last coordinate is real, so its contribution $d{x}_3\wedge d{y}_3$ pulls back to zero. The reduction identity therefore yieldsThis proves the symplectic ball chart in complex projective space directly, including the precise unit radius.

For the final unheaded request, embed the two disjoint balls in this chart. Fix arbitrary $0<{\rho }_i<r_i$, and choose intermediate radii ${\rho }_i<{\rho }_i^{\prime }<r_i$. On the images of the balls of radius ${\rho }_i^{\prime }$, transport the standard complex structure by the given symplectic embeddings. It is compatible with $\Omega$. Their smaller closed neighborhoods are compact and disjoint. Choose a global Riemannian metric agreeing there with the associated compatible metrics, using a partition of unity outside these neighborhoods, and apply the metric construction. The resulting global compatible almost complex structure $J$ agrees with the transported standard structure on each ball of radius ${\rho }_i$ and on a slightly larger neighborhood. This extends metrics, rather than averaging almost complex structures.

Let $p_1,p_2$ be the centres. They are distinct. The degree-one J-holomorphic curve supplied in the question has total area $\pi$, because its homology class is the projective line class. We use the sharp Euclidean version of the Monotonicity theorem for a J-holomorphic curve: a nonconstant holomorphic curve through the centre of a standard complex ball, with no boundary in the ball and counted with its parametrization multiplicities, has area in the radius-$\rho$ ball at least $\pi {\rho }^2$. This includes branched points: their density is a positive integer $m$, and the stronger lower bound is $m\pi {\rho }^2$. One way to see the constant is that complex curves are calibrated minimal surfaces; their stationary area ratio $\mathrm{area}(C\cap B(\rho ))/{\rho }^2$ is nondecreasing and its limit at the centre is $m\pi$.

The closed degree-one curve cannot stay entirely inside either chart ball: there the form is exact, so Stokes theorem would give zero total area, contradicting its positive area. Thus the branch through each centre crosses the surrounding spheres, and the stated monotonicity bound applies to each restriction. The two ball images are disjoint, so their area contributions add without overlap. It follows thatLetting both smaller radii increase to their original radii givesThis two-ball packing obstruction in the projective plane is sharper than volume alone: a volume comparison would only give $r_1^4+r_2^4\le 1$.

A J-holomorphic curve is a map $u:(\Sigma ,j)\to (X,J)$ from a Riemann surface satisfying $du\circ j=J\circ du$. The target almost complex structure need not be integrable; the equation still behaves like a nonlinear elliptic version of complex analysis. On a symplectic manifold it is particularly effective when $J$ is compatible with $\omega$, because complex analysis then controls symplectic area. The metric construction above supplies such structures and deformations between them.

In conformal coordinates $(s,t)$ with $j{\partial }_s={\partial }_t$, the equation is $u_t=J{u}_s$. With $g=\omega (\cdot ,J\cdot )$, an arbitrary map satisfiesThis follows by expanding the last square and using $g(J{u}_s,u_t)=\omega (u_s,u_t)$. Integrating proves the Energy identity for a J-holomorphic curve. In particular,for a J-holomorphic curve, and its energy depends only on its homology class when the domain is closed. A nonconstant such curve has strictly positive area: its derivative is nonzero somewhere and positivity holds on a neighborhood. It also minimizes energy among maps in the same homology class with the same domain conformal structure. Thus analytic bounds can be obtained from topological information.

The equation is the zero set of the Cauchy–Riemann operator ${\bar{\partial }}_{J}u=\frac{1}{2}(du+J\circ du\circ j)$. Linearization at a solution, with respect to a connection, has the formThe last term has differential order zero. The principal part has the Cauchy–Riemann symbol; in a complex frame a nonzero real covector $(a,b)$ gives multiplication by $a+ib$, an invertible symbol. This establishes ellipticity. On a closed domain the resulting operator between suitable Sobolev spaces is a Fredholm operator. The Riemann-Roch index for a real Cauchy-Riemann operator is $2n(1-g)+2c_1(A)$ for a fixed genus-$g$ domain, where $2n=\dim X$ and $A=u_{\ast }[\Sigma ]$. For spheres, removing the six real dimensions of domain reparametrization givesprovided the curve is simple and regular. Each marked point adds two real dimensions; constraining its image to a specified target point removes $2n$. These are the dimension formula for regular J-holomorphic spheres and its incidence version.

Here regular means that $D_u$ is surjective. Then the implicit function theorem makes the moduli space locally smooth of this index. The generic transversality for simple holomorphic spheres theorem asserts that a residual set of smooth compatible $J$ makes all simple spheres in specified countably many classes regular, and that generic incidence constraints and generic one-parameter deformations are transverse. It does not assert regularity for all multiple covers. This distinction matters whenever one tries to count curves. A primitive class with no positive-area spherical decomposition avoids multiple-cover and bubbling difficulties and permits especially direct counts.

The other central ingredient is Gromov compactness for spheres. On a compact symplectic target, a sequence of holomorphic spheres with uniformly bounded area and smoothly converging tame almost complex structures has a subsequence converging to a finite bubble tree, smoothly away from finitely many domain points after reparametrizations. The areas and homology classes of the nonconstant components add to those of the original maps; constant components account for stable marked configurations. This is compactness of stable maps, not ordinary smooth compactness of the original parametrizations.

The proof mechanism explains why bubbling is indispensable. An elliptic small-energy estimate bounds derivatives on a smaller disc when the energy on a larger disc is sufficiently small. Failure of a derivative bound therefore concentrates a definite amount of energy. Rescale near a point of large derivative; elliptic estimates yield a nonconstant limiting holomorphic map on the plane. Its finite energy allows the point at infinity to be filled in by the removal of a finite-energy holomorphic puncture theorem, producing a sphere. Repeating extracts further spheres. Uniform local monotonicity gives a positive energy threshold for nonconstant bubbles on the compact target, so only finitely many can occur. Estimates on the intervening annuli yield the energy identity and exclude lost neck energy. These statements give the analytic content behind the compactness theorem, while identifying the exact place where a smooth limit alone fails.

A local area estimate turns this existence theory into an embedding obstruction. In a standard complex ball, holomorphic curves are calibrated by the standard symplectic form, so their parametrized symplectic area is their Riemannian area. They are stationary minimal surfaces. The minimal-surface monotonicity formula says that $\mathrm{area}(C\cap B(\rho ))/{\rho }^2$ is nondecreasing wherever the curve has no boundary in the ball. At a point of multiplicity $m$, its limiting density is $m\pi$. Consequently any branch through the centre contributes at least $\pi {\rho }^2$. For a general compatible $J$ there is still a local lower bound $c{\rho }^2$, with $c>0$ depending on controlled geometry. The exact Euclidean constant is the one needed for sharp radius obstructions.

Here is a global existence argument that feeds this estimate. Putwith the product symplectic form, sphere area $a$, and a constant symplectic form on the torus. Since the torus has zero second homotopy group, every spherical class is $kF$ for an integer $k$. A nonconstant holomorphic sphere in this class has area $ka>0$, so $k\ge 1$. A curve of class $F$ cannot split into two nonconstant bubbles, and it cannot be a nontrivial multiple cover. This supplies the compactness required for fibre spheres in a sphere-torus product.

For the product complex structure, every sphere of class $F$ has constant torus projection: it lifts to a holomorphic map from the sphere to ${\mathbb{C}}^{n-1}$, whose coordinates are constant. Its sphere projection has degree one. Thus, up to reparametrization, there is exactly one such sphere through every target point. It is regular because its pulled-back tangent bundle is $\mathcal{O}(2)\oplus {\mathcal{O}}^{n-1}$ on ${\mathbb{CP}}^1$, and the first cohomology of each summand vanishes. Its linearized Dolbeault operator therefore has zero cokernel.

Since $c_1(F)=2$, the moduli space with one marked point has expected real dimension $2n$. Imposing passage through a fixed point gives dimension zero. For generic structures and transverse incidence conditions this is a finite set. Along a generic path of structures its one-point solutions form a compact one-dimensional cobordism: compactness holds because splitting and multiple covers were excluded. Its boundary contains the endpoint solutions. A compact one-manifold has an even number of boundary points, so the count modulo two is unchanged. The product count is one, and hence the count for a generic compatible $J$ is one as well. If the endpoint product structure is not generic for other classes, its regular solutions allow the path argument in this single class. Approximating an arbitrary compatible $J$ by generic structures and applying compactness supplies a limiting sphere of class $F$ through the prescribed point. The nonconstant component still contains that point; there are no nonconstant bubbles to carry it away. This derives the existence assertion needed below without assuming a capacity obstruction.

Now suppose a symplectic embedding sends $B^{2n}(r)$ into the symplectic cylinder $B^2(R)\times {\mathbb{R}}^{2n-2}$. Choose $0<\rho <{\rho }^{\prime }<r$. The image of the closed radius-${\rho }^{\prime }$ ball is compact. For any $a>\pi R^2$, the first factor embeds symplectically into an area-$a$ sphere: in dimension two this is an area-preserving disc chart, leaving a cap of area $a-\pi R^2$. Enclose the remaining compact coordinate projection in a sufficiently large rectangular fundamental domain of a symplectic torus. The compact ball image thereby embeds into $S^2(a)\times T^{2n-2}$.

Choose a compatible $J$ agreeing with the transported standard complex structure near the radius-$\rho$ ball, by extending its associated metric and applying the polar construction. The sphere of class $F$ through the centre has area $a$. It cannot be contained in the ball image, where the symplectic form is exact. Euclidean monotonicity therefore gives $a\ge \pi {\rho }^2$. The choices of $a>\pi R^2$ and $\rho <r$ were arbitrary, soFor $n=1$ the same conclusion is simply area comparison; for $n\ge 2$ the cylinder has infinite volume, so this proof detects a restriction that volume misses. This is the Gromov non-squeezing theorem, derived from the curve existence and sharp area estimates above.

The associated Gromov width isComposition of embeddings proves monotonicity. Rescaling coordinates in a ball proves $c_G(U,c\omega )=c{c}_G(U,\omega )$ for $c>0$. Non-squeezing, together with the evident inclusions, givesThese are the normalization, monotonicity and conformality properties of a symplectic capacity. The two-ball obstruction of Question 5 illustrates a related use: one degree-one curve through two prescribed centres gives an additive area bound, stronger than comparing the volumes of the balls. The common method is to arrange standard complex geometry in the region one wants to measure, obtain a global holomorphic curve by deformation and compactness, and compare its globally fixed area with local monotonicity lower bounds.

Let $\zeta =u+iv$. Direct differentiation givesThese are holomorphic polynomials and their squared sum isMoreover direct calculation of the metric gives $E=G=(1+u^2+v^2{)}^2>0$ and $F=0$, so the map is genuinely an immersion everywhere. The preceding characterizations prove that it is an isothermal parametrization of a minimal surface, the Enneper surface.

An isothermal parametrization is an immersion with first fundamental form $Ed{u}^2+2Fdudv+Gd{v}^2={\lambda }^2(d{u}^2+d{v}^2)$, where $\lambda >0$. Thus $E=G={\lambda }^2$ and $F=0$.

Write $W={\varphi }_{uu}+{\varphi }_{vv}$. Differentiating the metric identities givesFor example ${\varphi }_{vv}\cdot {\varphi }_u=({\varphi }_v\cdot {\varphi }_u{)}_v-{\varphi }_v\cdot {\varphi }_{uv}=-G_u/2$; the other identities are analogous. Hence $W$ is normal to the surface. For a unit normal $N$, its normal component is $e+g$. The supplied mean curvature formula reduces to $H=(e+g)/(2{\lambda }^2)$. ThereforeA surface is a minimal surface when its mean curvature vector vanishes. Since ${\lambda }^2>0$, the preceding identity proves minimality is equivalent to $\Delta \varphi =0$ in isothermal coordinates.

On a bosonic configuration, the target-space supersymmetry transformation changes the fermionic coordinate by ${\delta }_{ϵ}\Theta =ϵ$, where $ϵ$ is a background Killing spinor. A kappa symmetry convention isThe configuration preserves a supersymmetry if this change can be cancelled by a local kappa symmetry transformation. Multiplying $ϵ+(1+{\Gamma }_{\kappa })\kappa =0$ by $1-{\Gamma }_{\kappa }$ proves necessity ofConversely, if this holds, choosing $\kappa =-ϵ/2$ cancels the variation. Reversing the brane orientation reverses the corresponding kappa symmetry projector.

The preserved parameters must solve this equation everywhere on the brane and satisfy the background Killing spinor equations. Thus the fraction of all thirty-two eleven-dimensional supersymmetries isOne traceless constant kappa symmetry projector in flat space preserves sixteen parameters. Further independent commuting projectors often halve that number again, but a position-dependent projector requires a common global solution; its pointwise rank alone does not determine the answer. To quote a fraction of the background supersymmetry, divide instead by the number of its Killing spinors.

For the supermembrane, let $Z^{\mathcal{M}}=(X^M,{\Theta }^{\alpha })$ be the embedding in eleven-dimensional superspace. Define the pulled-back supervielbein, the induced worldvolume metric, and induced gamma matrices byFor a chosen orientation,The Clifford algebra shows that ${\Gamma }_{\kappa }^2=1$; in an orthonormal membrane frame this reduces to $({\Gamma }_0{\Gamma }_1{\Gamma }_2{)}^2=1$. Its trace is zero, so its two eigenspaces each have real dimension sixteen.

The full supermembrane action in an on-shell eleven-dimensional supergravity superspace isHere ${\mathcal{A}}_3$ is the super-three-form and $Z^{\ast }$ denotes the pullback of a differential form. Its bosonic restriction is the ordinary supergravity potential, but its fermionic components are retained in this action. The magnitude of its Wess-Zumino brane coupling equals the membrane tension; the sign $q$ fixes the orientation and the matching sign of ${\Gamma }_{\kappa }$. In flat superspace one may take ${\Pi }^a=d{X}^a-i\overline{\Theta }{\Gamma }^{a}d\Theta$, with the super-three-form chosen to obey the standard supergravity superspace constraints. Thus this is a fermionic, kappa symmetry-invariant action, rather than merely its bosonic truncation.

To construct an ordinary calibration, first take a flux-free static background $d{s}^2=-d{t}^2+g_{mn}d{x}^{m}d{x}^n$ with a unit covariantly constant spinor $ϵ$. In a compatible orientation and mostly-plus Clifford algebra, the membrane spinor calibration form isFor an oriented orthonormal spatial pair $u,v$, $\phi (u,v)={ϵ}^{†}{\Gamma }_0\gamma (u)\gamma (v)ϵ$. The matrix on the right is the static membrane ${\Gamma }_{\kappa }$, a Hermitian involution, so this value is at most one, with equality precisely when ${\Gamma }_{\kappa }ϵ=ϵ$. Parallel transport of the spinor and gamma matrices gives $\nabla \phi =0$, hence $d\phi =0$.

A calibration is a closed differential form whose value on each oriented unit tangent plane is at most one; its comass is at most one. A calibrated submanifold $\Sigma$ saturates the bound, $\phi {|}_{\Sigma }={\mathrm{vol}}_{\Sigma }$. For any homologous competitor ${\Sigma }^{\prime }$ with the same boundary, Stokes theorem givesThis proves that calibration implies volume minimization, and in particular that the membrane's spatial surface is a minimal surface. The closure condition is indispensable. A general flux-coupled Killing spinor need not produce a closed ordinary calibration; the corresponding generalized calibration bounds the full brane energy including its potential coupling. The flux-free construction above is the ordinary volume-minimizing case.

For the flux-coupled membrane, the standard spinor bilinears give a Killing vector $K$ and a two-form $\Omega$ satisfying $d\Omega ={\iota }_{K}F$, in matching supergravity conventions. In a stationary gauge ${\mathcal{L}}_{K}A=0$, Cartan's magic formula givesThis closed charge form provides the membrane generalized calibration; its potential term is what changes the volume bound into an energy bound.

At a non-planar point of a minimal surface, the shape operator is not zero. Minimality gives nonzero opposite principal curvatures, so $K<0$ nearby and $dN$ is invertible. The inverse function theorem makes the Gauss map a local diffeomorphism onto a patch of the sphere; part (ii) makes it conformal.

Choose stereographic coordinates $(u,v)$ on that sphere patch. Their metric is $4(d{u}^2+d{v}^2)/(1+u^2+v^2{)}^2$. Pulling them back by $N$, the conformal identity $N^{\ast }{g}_{S^2}=\lambda g_S$ givesThe positive coefficient shows that these are local isothermal coordinates. No general isothermal-coordinate existence theorem is needed: the nondegenerate conformal Gauss map supplies them explicitly.

A minimal surface has zero mean curvature $H=(k_1+k_2)/2$; equivalently its area is stationary under compactly supported normal variations. Its principal curvatures satisfy $k_2=-k_1$, so its Gaussian curvature isEquality occurs precisely where its second fundamental form vanishes.

Set $D=1+u^2+v^2$. The two tangent vectors to the Enneper surface areTheir cross product is $D(-2u,2v,1-u^2-v^2)$. Since $4u^2+4v^2+(1-u^2-v^2{)}^2=D^2$, its length is $D^2>0$, proving regularity and giving the unit normalFor the first fundamental form, direct dot products give $E=G=D^2$ and $F=0$. The second derivatives are ${\sigma }_{uu}=(-2u,2v,2)$, ${\sigma }_{uv}=(2v,2u,0)$, and ${\sigma }_{vv}=(2u,-2v,-2)$. Taking their dot products with $\mathbf{n}$ gives the second fundamental form coefficients $e=2$, $f=0$, $g=-2$. Therefore the Gaussian curvature isHere curvature means Gaussian curvature; the opposite signs of the second-form diagonal entries also give zero mean curvature, as expected for this minimal surface.

A minimal surface has $H=0$, so $k_2=-k_1$ and $\boxed{K=-k_1^2\le 0}$. For a nonflat example take the catenoid parametrization $X(u,v)=(\cosh v\cos u,\cosh v\sin u,v)$. Its first fundamental form has $E=G={\cosh }^{2}v$ and $F=0$. Choosing the normal appropriately, its second fundamental form has coefficients $e=-1$, $f=0$, $g=1$. Thus its principal curvatures are $\pm {\mathrm{sech}}^{2}v$, proving $H=0$ and $\boxed{K=-{\mathrm{sech}}^{4}v<0}$.

Write the plane as $\{x:\nu \cdot x=c\}$ for a fixed unit vector $\nu$, and put $h(x)=\nu \cdot x-c$. Suppose $X$ misses the plane. Choose a point of $X$ and a number $A$ larger than its distance to the plane. The hypothesis $D_n\to \infty$ makes $\{x\in X:|h(x)|\le A\}$ bounded, and closedness of $X$ makes it compact. Thus $|h|$ attains its global minimum at some $p\in X$, with $|h(p)|>0$.

Locally the sign of $h$ is constant, and $h$ has a local minimum or maximum at $p$. Its surface gradient vanishes, so the tangent plane of $X$ there is parallel to the given plane and its normal is $\pm \nu$. For a surface tangent vector, the intrinsic Hessian of $h$ is the normal component of the second fundamental form, namely ${\mathrm{Hess}}_{X}h=\pm b$ at $p$. The Hessian is semidefinite because this is a local extremum. Since $X$ is a minimal surface, ${\mathrm{tr}}_{g}b=2H=0$. A semidefinite symmetric form with zero trace must be zero, so $b(p)=0$. This makes $p$ a planar point, contradicting the assumption. Therefore $\boxed{X\cap \Pi \ne \varnothing }$.

The boundaryless hypothesis ensures that the minimizing point is an interior surface point. The condition on $D_n$ ensures that the infimum is attained; disjointness from a plane alone would not do so.