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For the standard catenoid, the two principal curvatures are opposite, so its mean curvature vanishes. Its Gaussian curvature is $K=-{\mathrm{sech}}^{4}u<0$, which also shows that no neighbourhood is locally isometric to a plane, cylinder, or regular part of a cone, all of which have zero Gaussian curvature.

Curvature depresses the equilibrium melting temperature of solid convex into liquid. Here $\mathcal{K}$ is the sum of principal curvatures, equal to twice a common convention for mean curvature, and $T_m$ is absolute temperature. The coefficient uses solid-liquid surface energy, mass density and latent heat. This stabilizes fine corrugations of a solidification front.

For a proper Euclidean motion of Euclidean three-space $E(x)=Ax+b$, transport the unit normal by $\tilde{N}(E(p))=AN(p)$. The corresponding shape operators are orthogonally conjugate:Their determinant and trace, hence Gaussian curvature and mean curvature, are unchanged.

A circular cylinder of radius $\rho$ has induced metric locally equal to $d{u}^2+d{z}^2$ with $u=\rho \phi$. Its Riemann curvature tensor therefore vanishes, despite its nonzero extrinsic curvature. With outward unit normal and convention $K_{ij}=e_i^a{e}_j^b{\nabla }_a{n}_b$, its trace is $1/\rho$ and its principal curvatures are $1/\rho$ and zero. Its averaged mean curvature is $1/(2\rho )$. The angular identification changes global topology, not local flatness.

A minimal submanifold has vanishing mean curvature vector. For a surface in ${\mathbb{R}}^3$, this says that the scalar mean curvature is zero, equivalently that its two principal curvatures sum to zero.

Replacing the chosen unit normal $N$ by $-N$ replaces the shape operator $S$ by $-S$. On a surface, Gaussian curvature $K=\det S$ is unchanged, while mean curvature $H=\mathrm{tr}(S)/2$ changes sign.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\mathrm{grad}$. A Riemannian submersion is a surjective smooth submersion $\pi :(M,g)\to (N,h)$ for which, at every $p$, the restriction of $d{\pi }_p$ to $H_p=(\ker d{\pi }_p{)}^{\perp }$ is a linear isometry onto $T_{\pi (p)}N$. The spaces $V_p=\ker d{\pi }_p$ and $H_p$ are its vertical and horizontal spaces. Its fibres are totally geodesic submanifolds precisely when ${\nabla }_{U}V$ is vertical for vertical vector fields $U,V$: their second fundamental form vanishes. Equivalently, a geodesic initially tangent to a fibre remains in that fibre while defined.

For a smooth $f:N\to \mathbb{R}$, its basic function $F=f\circ \pi$ is constant along each fibre. The Riemannian gradient of $F$ is the horizontal lift of a vector field through a submersion of ${\mathrm{grad}}_{N}f$, sincefor horizontal $X$, and $dF(U)=0$ for vertical $U$. In particular no derivative of $f$ in a vertical direction occurs.

Here is the needed connection fact, which also follows directly from the Koszul formula: for horizontal lifts $X,Y$ of vector fields $\bar{X},\bar{Y}$ on $N$, the horizontal component of ${\nabla }_{X}Y$ projects to ${\nabla }_{\bar{X}}^N\bar{Y}$. To see this, pair the Koszul formula with a third horizontal lift $Z$. The horizontal inner products are pulled back from $N$, and the horizontal components of their Lie brackets of vector fields project to the brackets on $N$. Thus all six terms are the pullbacks of the corresponding terms on $N$.

Choose an adapted Riemannian orthonormal frame $X_1,\dots ,X_n,U_1,\dots ,U_r$, with the $X_a$ horizontal lifts. Using $\mathrm{Hess}F(A,A)=A(AF)-({\nabla }_{A}A)F$, the connection fact givesFor a vertical $U_b$, both $U_{b}F=0$ and $({\nabla }_{U_b}{U}_b)F=0$, the latter because the fibres are totally geodesic submanifolds. Taking the negative metric trace of the Riemannian Hessian therefore proves the basic-function Laplacian identityThis identity is local and does not require compactness. Vanishing mean curvature of the fibres would already suffice; total geodesicity makes each vertical summand vanish separately.

For the discrete eigenspace assertion, assume the two Riemannian manifolds are closed manifolds. Without a discrete spectral realization, an unrestricted noncompact version need not have an eigenbasis. The projections of the Riemannian product $M\times N$ are Riemannian submersions with totally geodesic submanifolds as fibres. Its Levi-Civita connection splits into the two factor connections. Consequently its positive Laplace-Beltrami operator isThe cross term in the product rule for the positive Laplace-Beltrami operator is zero because the two factor Riemannian gradients are orthogonal.

We use the standard compact elliptic compact elliptic spectral theorem: the positive Laplace-Beltrami operator on a closed manifold is self-adjoint, has compact resolvent, and has a complete orthonormal eigenbasis of smooth eigenfunctions, with finite-dimensional eigenspaces and eigenvalues tending to infinity. Let ${\Delta }_M{u}_i={\mu }_i{u}_i$ and ${\Delta }_N{v}_j={\nu }_j{v}_j$. Fubini's theorem and completeness on each factor show that $u_i(x)v_j(y)$ form a complete orthonormal basis of $L^2(M\times N)$. For example, a function orthogonal to all these products has, for each $i$, zero $u_i$-coefficient as an $L^2(N)$ function, hence is zero.

The displayed operator identity makes $u_i{v}_j$ an eigenfunction with eigenvalue ${\mu }_i+{\nu }_j$. Conversely, if $\Delta w=\lambda w$, self-adjointness of the positive Laplace-Beltrami operator givesAll other coefficients vanish. Only finitely many pairs can have ${\mu }_i+{\nu }_j=\lambda$, since both spectra are nonnegative and have finitely many eigenvalues below any fixed bound. Thus the product Laplacian eigenspace decomposition isThe tensor product summands are mutually orthogonal; their elements are actual smooth eigenfunctions, so this is an equality of eigenspaces, not just a formal expansion.

Let $\xi =z-Vt$ increase into the liquid, with the planar solidification front at $\xi =0$. Denote the common thermal conductivity by $K=\rho c_p\kappa$. A bounded travelling temperature profile satisfies $\kappa T^{\prime \prime }+V{T}^{\prime }=0$. In the solid, boundedness as $\xi \to -\infty$ forces $T_s=T_m$; in the liquid,The Stefan condition then givesThus the travelling solidification front in a pure supercooled melt releases just enough latent heat to warm the initially supercooled liquid to $T_m$. There is no separate imposed heat flux at infinity in this bounded travelling-wave problem.

For the interface perturbation write $\xi =\zeta (y,t)=\hat{\zeta }{e}^{i\alpha y+\sigma t}$. Use the Gibbs--Thomson relation for curvature-induced melting-temperature depressionHere $\gamma$ is the solid-liquid surface energy, $T_m$ in this expression is absolute temperature in kelvin, and $\mathcal{K}$ is the sum of principal curvatures, positive for solid convex into liquid. For this graph, $\mathcal{K}=-{\zeta }_{yy}={\alpha }^2\zeta$. If mean curvature is defined as half this sum, $\mathcal{K}=2H$; this fixes the capillary factor unambiguously.

Under the small-wavelength thermal approximation for a solidification front, take harmonic perturbation temperatures that decay away from the interface:This approximates their temperature equation by Laplace's equation, neglecting perturbation time dependence and the moving-frame advection. Expanding the interfacial temperature at the displaced surface givesThe second expression includes displacement through the negative base liquid gradient. Linearizing the Stefan condition also requires evaluation of the base gradient at the displaced surface:Using $S=1$ gives the requested reduced dispersion relation:orThe destabilizing $V\alpha$ term describes a protrusion entering colder liquid. The cubic term expresses stabilization by the curvature-induced temperature decrease.

There is an important accuracy qualification. The harmonic approximation gives the leading short-wave thermal exponents, but the retained constant term is not a uniformly controlled next-order correction to the full moving-boundary problem. If perturbation diffusion is retained, its exponents areSubstitution into exactly the same linearized boundary balances gives the full thermal stability of a travelling solidification front:For example, with capillarity removed, $\sigma =V\alpha$ solves this relation exactly, whereas the harmonic reduction gives $V\alpha -V^2/\kappa$. The difference is smaller than the leading term when $\alpha \gg V/\kappa$, but demonstrates why the reduced formula should not be asserted as an exact dispersion law or a precise onset calculation. To neglect perturbation time dependence one also needs $|\sigma |\ll \kappa {\alpha }^2$; extremely large capillary damping is outside that approximation.

To analyse the displayed cubic, defineThen $\kappa \sigma /V^2=g(q)=-1+q-2q^3/R_c$. It is negative at zero and at sufficiently large $q$. Its positive stationary point is the unique maximum:Thus the capillary threshold of a reduced solidification dispersion iswith equality giving a neutral maximum and strict inequality giving a band of unstable positive wavenumbers. At equality, however, $q_{\ast }=3/2$, not $q_{\ast }\gg 1$. The number $27/2$ is therefore the threshold of the stated reduced cubic, not a controlled threshold of the full thermal problem under the stated short-wave ordering. Far above this threshold its fastest-growing scale does satisfy $q_{\ast }\gg 1$, making the short-wave mechanism meaningful.

The analogy for a snowflake is dendritic growth of a snowflake. A protrusion has better access to the driving supersaturation or undercooling and grows faster than a sheltered region; this amplifies tips and promotes branching into crystal dendrites. Interfacial capillarity suppresses arbitrarily fine tips. In an actual snowflake, the main growth is deposition from water vapour, so the diffusion field and its driving variable differ from this pure-melt thermal example. The hexagonal ice crystal anisotropy supplies six equivalent preferred arm directions; an isotropic planar stability calculation alone cannot predict six arms. Nonlinear tip selection, vapour transport and side branching are needed for the later detailed morphology. The analysis explains the competition between diffusion-driven protrusion growth and capillary smoothing, with this physically necessary qualification.

Choose the outward normal vector $n=(\cos \phi ,\sin \phi ,0)$ and the positive conventionThe ambient connection vanishes in Cartesian coordinates. Since ${\partial }_{\phi }n=e_{\phi }/\rho$ and ${\partial }_{\hat{z}}n=0$, the extrinsic curvature components are $K_{\phi \phi }=\rho$, $K_{\phi \hat{z}}=K_{\hat{z}\hat{z}}=0$. Taking the trace using the induced metric,Reversing the normal or using the negative extrinsic-curvature convention gives $-1/\rho$. The two principal curvatures are $1/\rho$ and zero in the chosen convention; the averaged mean curvature would be $K/2$, so it must not be confused with the requested trace.

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.