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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.

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.

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

The physical argument for the Penrose inequality combines weak cosmic censorship conjecture, the dominant energy condition, and relaxation to a stationary black hole. Work in geometrized units. Let $M_f$ and $A_f$ be the final Kerr black hole mass and horizon area. Positive energy radiated to infinity gives $E_{\mathrm{ADM}}\ge M_f$, where $E_{\mathrm{ADM}}$ is the initial ADM energy. For a Kerr black hole with $a_f=J_f/M_f$,If the initial apparent horizon obeys the necessary apparent-horizon area comparison with the enclosing event horizon, and Hawking's area theorem applies during the evolution, thenConsequently the anticipated answer, under those additional hypotheses, isThe bound is saturated by a nonrotating Schwarzschild black hole with no energy loss. Rotation or outgoing radiation makes the argument's inequalities stricter.

There is an essential qualification: inclusion inside an event horizon does not by itself compare areas. An arbitrary apparent horizon on general, non-time-symmetric initial data need not satisfy the displayed apparent-horizon area comparison; the unqualified version with its area is not universally true, even with the dominant energy condition. On time-symmetric data the relevant outermost minimal surface is an outer area-minimizing surface, as used in the Riemannian Penrose inequality, with nonnegative scalar curvature. In more general formulations an appropriate enclosing-area quantity is needed. The physical expectation is conditional on this area comparison, as well as on censorship, predictability, settling, and the energy assumptions; the mere presence of a trapped surface does not supply every step.

An immersion $\varphi :U\to {\mathbb{R}}^3$ has differential of rank $2$, equivalently linearly independent ${\varphi }_u,{\varphi }_v$. It has isothermal coordinates when its first fundamental form has $E=G>0$, $F=0$. A minimal surface has zero mean curvature.

For an isothermal immersion, differentiation of $E=G$ and $F=0$ givesand similarly its scalar product with ${\varphi }_v$ is zero. Its normal component is $e+g$. The given mean curvature formula reduces to $H=(e+g)/(2E)$, soThus an isothermal immersion is minimal exactly when each coordinate is a harmonic function.

An explicit example is the Enneper surface, parametrized on ${\mathbb{R}}^2$ byEvery coordinate has zero Laplacian. Direct differentiation gives ${\varphi }_u\cdot {\varphi }_v=0$ and $|{\varphi }_u{|}^2=|{\varphi }_v{|}^2=(1+u^2+v^2{)}^2>0$. Thus it is an isothermal immersion and is minimal everywhere.

It is not an open subset of any of the three listed surfaces. The parameter points $(\sqrt{3},0)$ and $(-\sqrt{3},0)$ both map to $(0,0,3)$, but their tangent planes differ: ${\varphi }_u=(-2,0,\pm 2\sqrt{3})$ while ${\varphi }_v=(0,4,0)$. A plane, catenoid, or helicoid is an embedded smooth surface with a unique tangent plane at each point. If our whole image lay in any one of them, both rank-$2$ tangent planes at this point would have to equal its tangent plane, a contradiction.

A regular surface is a minimal surface when its mean curvature vanishes identically, equivalently when its two principal curvatures sum to zero everywhere.

The catenoidis an example. In part (a), take $f(v)=\cosh v$ and $g(v)=v$, so $q=\cosh v$. The formula there givesA plane and a circular cylinder have zero Gaussian curvature, as does a flat cone away from its vertex. Because a local isometry preserves Gaussian curvature, the catenoid is minimal and is not locally isometric to any of those surfaces; this is the catenoid is a minimal surface.

The map $\varphi$ is an immersion when $d\varphi$ has rank two, equivalently ${\varphi }_u$ and ${\varphi }_v$ are linearly independent. It is isothermal when its first fundamental form hasIn general the mean curvature isThe surface is minimal when $H=0$. In isothermal coordinates the Laplace-Beltrami operator identity isThus $H=0$ exactly when each coordinate function $x,y,z$ is harmonic.

Forthe isothermal conditions areand harmonicity givesConsequentlyand the isothermal equation reduces toThese are all the solutions for which $f$ never vanishes. When $AB>0$ they parameterize a catenoid, up to translations, reflection and scaling. The limiting cases $AB=0$, $c=0$, with exactly one of $A,B$ nonzero, parameterize a punctured plane.

The Gauss map sends $p$ to the chosen unit normal $N(p)$. Differentiating $⟨N,N⟩=1$ shows that $d{N}_p$ maps $T_{p}S$ into itself. The shape operator $S_p=-d{N}_p$ is self-adjoint because its associated second fundamental form is symmetric. Its eigenvalues $k_1,k_2$ are the principal curvatures. A point is umbilical when $k_1=k_2$, and the surface is minimal when its mean curvature $H=(k_1+k_2)/2$ vanishes everywhere.

In an orthonormal principal basis, $d{N}_p$ has diagonal matrix $\mathrm{diag}(-k_1,-k_2)$. It is conformal exactly whenwhich is equivalent to $k_1^2=k_2^2\ne 0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.

The Ryu–Takayanagi formula gives the leading entropy of a static boundary region:On the $t=0$ slice of Poincaré ${\mathrm{AdS}}_4$,The minimal surface anchored on a boundary circle of radius $R_1$ is the hemisphereCutting it off at $z=ϵ$, its area isThereforeThe universal requirement was the perimeter-law divergence $\pi R_1/(2Gϵ)$; the finite constant depends on the stated pure ${\mathrm{AdS}}_4$ vacuum geometry and equals $-\pi /(2G)$ here. Since $G\sim N^{-3/2}$, both terms are of order $N^{3/2}$.

Use the conventionAn embedded submanifold is minimal when its mean curvature vector $\mathbf{H}$ vanishes identically. For a variation $M_t$ with velocity field $V$, the first variation of area iswhere $\eta$ is the outward unit conormal. For a boundaryless $M$, only the first integral remains. This sign convention is consistent with the outward variation of a round sphere increasing its area, because its $\mathbf{H}$ points inward.