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The helicoid $H(s,v)=(\sinh s\cos v,\sinh s\sin v,v)$ and catenoid $C(s,v)=(\cosh s\cos v,\cosh s\sin v,s)$ have the same first fundamental form ${\cosh }^{2}s(d{s}^2+d{v}^2)$. Therefore matching equal parameter pairs gives a local isometry. The result is local: a full turn changes the height on the helicoid while the angular coordinate on the catenoid repeats. Equal intrinsic lengths do not require the same embedding into Euclidean space.

For the helicoid,Hence $E=1+v^2$, $F=0$, $G=1$. Since $r_{uu}=(-v\cos u,-v\sin u,0)$ and $r_{vv}=0$, both $L$ and $N$ vanish. The mean-curvature formula therefore givesFor comparison, $M=1/\sqrt{1+v^2}$, so $K=-1/(1+v^2{)}^2$; zero mean curvature does not mean zero Gaussian curvature.

Differentiate to getTheir squared lengths are $a^2$ and $a^2{u}^2+1$, and their inner product is zero. Hence the helicoid has first fundamental form

Using the two first fundamental forms calculated below, chooseThe surface of revolution is then a catenoid, with metric ${\cosh }^{2}s(d{s}^2+d{v}^2)$. On the helicoid, take $u=\sinh s/a$, assuming $a\ne 0$ so that its parametrization is regular. Substitution givesThus the coordinate mapis a local isometry on sufficiently small patches. The constant $a$ only changes the radial parametrization of the given helicoid, whose pitch is fixed at one. For $a=0$ the displayed helicoid parametrization is not a regular surface.

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.

Withthe coordinate tangent vectors satisfyThus the first fundamental form isGeometrically, $S$ is the half of a helicoid with positive radius, winding upwards indefinitely.

Consider the upper half of the catenoid, parametrized as the surface of revolutionIt has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a local isometry. By the invariance of Gaussian curvature under local isometry, the curvatures agree; directly, the conformal metric above gives