Sign of $K$ does not imply local convexity

We know that if $K(p)>0$ at one pint, then the surface locally in one side of its affine tangent plane. On the other hand, if the surface lies in one side around a point, then $K(p)\geq 0$. However, there are surfaces that at one point $p\in S$, $K(p)=0$, $K>0$ around $p$, BUT the surface lies in both sides of $T_pS$. An example is the surface $z=f(x,y)=x^3(1+y^2)$. Let $p=(0,0,0)$. Take the parametrization $$X(x,y)=(x,y,x^3(1+y^2)).$$ Then $X^{-1}(p)=(0,0)$ and as $X_x(0,0)=(1,0,0)$ and $X_y(0,0)=(0,1,0)$, then $T_pS$ is the plane $z=0$. Since $1+y^2\geq 0$ and $x^3$ changes of sign at $x=0$, then $f$ change of sign around $p$, that is, the surface has points in both sides of $T_pS$, as it appears in the next figure:

We now compute the Gauss curvature using the formula $$K(X(x,y))=\frac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}.$$ Then $$K(X(x,y))=\frac{12x^4(1-2y^2)}{(1+f_x^2+f_y^2)^2}.$$ In the open set of $S$ gieven by $V=X(U)$, where $U=\{(x,y): x\in{\mathbb R},|y|<1/\}$, $K>0$ in $V-\{p\}$ and $K(p)=0$. The next figure is the numerator of $K$ in $|x|<1$, $|y|<1/2$ that hows that the sign of $K$ is, indeed, positive.

Surface with only one parabolic point

We have proved that at an elliptic point, the surface lies in one side of the tangent plane at that point. Exactly, if $K(p)>0$, then there exists a neighborhood $V\subset S$ of $p$ such that $V\cap T_pS=\{p\}$ and $V-\{p\}$ lies in one of the two open halfspaces determined by $T_pS$. 

Here we show a surface with the same property but $K(p)=0$. The surface is obtained by rotating the curve $z=x^4$ around the $z$-axis. At the point $p=(0,0,0)$, $K(p)=0$, the tangent plane $T_pS$ is the $xy$-plane and $S-\{p\}$ lies in the halfspace $z>0$. Exactly, with the usual parametrization $X(x,s)=(x\cos(s), x\sin(s), x^4)$, we have 

$$K(x,s)=\frac{36 x^4}{(1+16x^6)^2},$$

so $K(p)=0$ and the rest of points are elliptic, that is, $p$ is the only parabolic point of $S$.

Surfaces of revolution with positive constant Gauss curvature

We know that when we write $K=c$ in the family of rotational surfaces, then this equation is an ordinary differential equation, so there is a unique solution for each initial conditions. We show this phenomenon when $K=1$. Suppose that the profile curve is locally a graph on the rotation axis, that is, $z\mapsto (f(z),0,z)$ for $z\in I$, $f(z)>0$. The parametrization of the surface is $X(z,s)=(f(z)\cos(s),f(z)\sin(s),z)$. Equation $K=1$ writes as 

$$-\frac{f''}{f(1+f'^2)^2}=1.$$

Thus we have $f''+f(1+f'^2)^2=0$. This is differential equation is not possible to integrate, up to special cases. We think that sphere should easily solve. The initial conditions are put on $z=0$, that is $f(0)=xo$ and $f'(0)=0$. With this last condition, we are imposing that the tangent line at $z=0$ is vertical. Moreover, by this condition, we can suppose that the solution is symmetric with respect to $z=0$. 

We use Mathematica to solve numerically the initial value problem $$(*) \left\{\begin{array}{l} f''+f(1+f'^2)^2=0\\ f(0)=xo\\ f'(0)=0\end{array}\right.$$ We study the solutions depending on the initial value $xo$, that is, the intersection point of the profile curve with the $x$-axis.

When $xo=1$, we know that the solution is the sphere, exactly, $f(z)=\sqrt{1-z^2}$ is a solution of (*). 

In order to study with Mathematica (*) we write here the sentences: 

profile =  NDSolve[{F''[z] + F[z] (1 + F'[z]^2)^2 == 0, F[0] == xo, F'[0] == 0}, F[z], {z, -Zo, Zo}]

f[z_] := F[z] /. profile[[1]]

ParametricPlot[{{z, 0}, {f[z], z}}, {z, -Zo, Zo}, PlotRange -> All]

The first line numerically solves the ODE with initial conditions as we have presented. Here $Zo$ is the width of the interval when the solution $f$ is defined. The second line `takes' the numerical value f in order to manage in the next line, where we plot the profile curve. In fact, the last line indicates that we also draw the $x$-line. I write this because Mathematica `reduces' the picture to the interval where is defined the solution and we want to compare the profile curve with its position with respect to the rotation axis. Finally, we use

ParametricPlot3D[{f[z] Cos[s], f[z] Sin[s], z}, {s, 0, 2 Pi}, {z, -Zo, Zo}]

for drawing the surface.

We begin with the study and sphere is our starting point: sphere appears when $xo=1$ and the domain of $f$ is for $Zo=1$. We now increase $xo$, for example $xo=1.5$. If we put $Zo=1$, Mathematica says that the solution is not defined in the interval $(-Zo,Zo)$ because appear errors. In fact, Mathematica says what is the maximum interval. In this example, the output is

NDSolve::ndsz: At z == -0.559099, step size is effectively zero; singularity or stiff system suspected.

This means that we have to take $Zo=0.5590$, obtaining the profile curve in its maximum domain, namely:

If we increase $xo$, that is, we move far the point $(f(xo),0,0)$, the profile moves far from the rotation axis: let us observe that the profile curve does not meet the rotation axis. In the figure, it indicates that the tangent plane at the boundary circles is horizontal, and the surface one `hole'.

Now we let $xo\rightarrow 0$. If $xo=0.7$, and for $Zo=2$ we see that the profile curve meets the $z$-axis, which is not possible.

Then, and after some trials, we see that for $Zo=1.35$, the profile meets exactly the $z$-axis. The figures are:

Now the surface presents two `singularities' exactly in the intersection points with the $z$-axis.

Elliptic and parabolic points.

Two days ago, we have shown a surface where all its points are hyperbolic. Now, we consider the elliptic paraboloid $z=x^2+y^2$. Now $$K=\frac{4}{(1+4x^2+4y^2)^2},$$proving that all its points are elliptic. 

On the other hand, the parabolic paraboloid $z=x^2$ has as Gauss curvature $K=0$ so all its points are parabolic or flat. In order to distinguish, we have to compute the principal curvatures. Since $K=0$, the principal curvatures are $\kappa_1=H$ and $\kappa_2=0$ by the relation $$\kappa_i=H\pm\sqrt{H^2-K}.$$ The mean curvature is given by $$H=\frac12\frac{(1+f_y^2)f_{xx}-2f_xf_yf_{xy}+(1+f_x^2)f_{yy}}{(1+f_x^2+f_y^2)^{3/2}}=\frac{1}{(1+4x^2)^{3/2}}\not=0$$ for any $x$. This proves that all its points are parabolic.

In the next pictures we have a elliptic paraboloid (left) and a parabolic paraboloid (right).

Surfaces with constant Gauss curvature

Consider surfaces where the Gauss curvature $K$ is a 'simplest' function, namely, $K$ is constant on the surface. For example, in a plane $K=0$ and in a sphere of radius $r$, $K=1/r^2$. Other surface with constant Gauss curvature is the cylinder, where $K=0$ again. If one want to obtain more examples, one may consider this problem in the family of surfaces of revolution. In such a case, the equation $K=c$ is an ordinary differential equation, and by the general theory, for each initial conditions, there is a solution. Thus this provides us a huge family of examples. 

If the profile curve is a graph on the $x$-line, that is, $\alpha(x)=(x,0,f(x))$, then 

$$K(X(x,\theta))=\frac{f'f''}{x(1+f'^2)^2}.$$

We know that when $K=0$, then $f''=0$, obtaining planes, circular cylinders and cones. On the other hand, the equation $$\frac{f'f''}{x(1+f'^2)^2}=c$$ is not possible to integrate completely and only some particular cases can solved: for example, the sphere and the pseudosphere. 

However, if we assume that the surface is compact, then panorama changes drastically, because we have

Theorem (Hilbert): Sphere is the only compact surface with constant Gauss curvature.

We observe that the topological assumption on compactness is essential in the result. It is known that in a compact surface there exists elliptic points, so the value of the constant $K$ must be positive. Then the key is based in the following result:

Lemma (Hilbert): Let $S$ be a surface and $p\in S$ an elliptic point. Consider $\kappa_1\geq \kappa_2$ the principal curvatures on $S$. If $p$ is a global maximum for $\kappa_1$ and it is a global minimum for $\kappa_2$, then $p$ is an umbilical point.

The proof of the Hilbert's theorem is then as follows. Since $K=\kappa_1\kappa_2>0$, the sign of the principal curvature is the same. After a change of orientation, we suppose $\kappa_i>0$. Let $p\in S$ a point where the (continuous) function $\kappa_2$ attains a global minimum. Since $\kappa_1=\frac{c}{\kappa_2}$, then $p$ is a global maximum for $\kappa_1$. By the lemma, $\kappa_1(p)=\kappa_2(p)$. Then we have for any $x\in  S$, $$\kappa_2(p)\leq\kappa_2(x)\leq\kappa_1(x)\leq\kappa_1(p)$$ and thus, $\kappa_1=\kappa_2$ on $S$. Then $S$ is an open of a plane or a sphere. Since $S$ is closed, by connectedness, $S$ is a plane or it is a sphere. But sphere is the only one that is compact.

Comparison surfaces (IV): elliptic points

We prove the analogous result that was showed for curves about the position of the surface with respect to the tangent plane in terms of its Gauss curvature.

Since the result is local, we suppose that the surface at the point $p$ is tangent to the plane $z=0$ and writes as $z=f(x,y)$ with $p=(q,0)=(0,0,0)$. In such a case, we know that $$K(p)=(f_{xx}f_{yy}-f_{xy}^2)(q).$$

If $K(p)>0$, then the determinant of the Hessian is positive. Since $f_{xx}(q)\not=0$ (on the contrary, $K(p)\leq 0$), then $f_{xx}(q)$ is positive or negative, that is, the Hessian is positive definite or negative definite, respectively. This proves that $q$ is a local minimum or a local maximum, respectively, proving:

Theorem. If $K(p)>0$, then the surface lies in one side of $T_pS$ around $p$.

Corollary. If in any neighbourhood around $p$, $S$ has points in both sides of $T_pS$, then $K(p)\leq 0$.

Comparison of curves by curvatures (III)

By the theorem of the previous entry, we prove:
Theorem. Any compact surface has points with positive Gauss curvature.

Proof. Take $p_0\in S$ the fairest point of $S$ from the origin of ${\mathbb R}^3$: this point exists because $S$ is compact and the distance function to a fixed point is a continuous function. We do the next steps.

Take ${\mathbb S}^2(r)$ the sphere centered at the origin and radius $r=|p_0|$: this number is positive because on the contrary is only one point. 

The surfaces $S$ and ${\mathbb S}^2(r)$ are tangent at $p_0$. For ${\mathbb S}^2(r)$ we know that the tangent plane is orthogonal to the position vector $p_0$. For $S$, consider the function $f(p)=|p|^2$. Because $p_0$ is a maximum, it is a critical point, so $df_{p_0}=0$. But it is is immediate that $df_{p_0}(v)=2\langle p_0,v\rangle$ for any $v\in T_{p_0}S$. Thus $T_{p_0}S$ is orthogonal to $p_0$. 

We orient ${\mathbb S}^2(r)$ according the orientation pointing inside, so the normal curvature for any tangent vector is $1/r$. Consider the orientation on $S$ so $N(p_0)=-p_0/|p_0|$, that is, the same than ${\mathbb S}^2(r)$. Moreover, $S$ lies above ${\mathbb S}^2(r)$ around $p_0$.

By the Theorem in the previous day, $\kappa_n(v)\geq 1/r$, in particular, in along the principal directions, $\kappa_i(p_0)\geq 1/r$, so $K(p_0)\geq 1/r^2$.

In particular, we have an estimate of the Gauss curvature at the fairest point  $p_0$ from the origin: $$K(p_0)\geq\frac{1}{|p_0|^2}.$$

As a consequence of the inequality $H^2\geq K$, we have:

Corollary. There are no compact minimal surfaces.

Translation surfaces with constant zero curvature

A translation surface is a surface that is the sum of two planar curves contained in orthogonal planes. Thus a parametrization of the surfaces is $$X(u,v)=(x,0,f(x))+(0,y,g(y))$$ where $f$ and $g$ are smooth functions defined in some intervals of ${\mathbb R}$. We are interesting in the translation surfaces with $H=0$ or $K=0$ on the whole surface. 

For the mean curvature, it is immediate that $H=0$ is equivalent to $$\frac{f''(x)}{1+f'(x)^2}+\frac{g''(y)}{1+g'(y)^2}=0.$$ Then necessarily we have that $$\frac{f''(x)}{1+f'(x)^2}=-\frac{g''(y)}{1+g'(y)^2}=c$$ for some real number $c$. If $c=0$, then $f''=g''=0$, obtaining $f(x)=ax+b$, $g(y)=cy+d$ and $z=ax+cy+b+d$, that is, the surface is a plane. If $c\not=0$, integrating $f$ and $g$ we obtain $$f(x)=-\frac{1}{c}\log\cos(cx+m),\ g(y)=\frac{1}{c}\log\cos(cy+n),\ m,n\in{\mathbb R}.$$ Thus we write $$z=\frac{1}{c}\log\left(\frac{\cos(cy+n)}{\cos(cx+m)}\right).$$ This surface is called the Scherk's surface. In order to study the domain of the $z(x,y)$, we take $c=1$ and $m=n=0$. Then $$z=\log\left|\frac{\cos(y)}{\cos(x)}\right|.$$ Then the domain is 

$$(x,y)\in(-\frac{\pi}{2},\frac{\pi}{2})\times (-\frac{\pi}{2},\frac{\pi}{2}).$$

It is clear that in the sides of this square, the function $z=z(x,y)$ takes $\infty$ or $-\infty$ values, as it is shown in the next picture.

If we now study translation surfaces with $K=0$, then this identity is equivalent to 

$$f''g''=0.$$

Then $f''=0$ identically or $g''=0$ identically. Without loss of generality, we suppose $f''=0$, that is, $f(x)=ax+b$ for some numbers $a,b$. Then the surface writes as $z=ax+g(y)+b$ or in terms of $X$, $$X(x,y)=x(1,0,a)+(0,y,g(y)).$$ This surface is a ruled surface whose base curve is any curve as $\alpha(y)=(0,y,g(y))$ and the rulings as parallel to the direction $(1,0,a)$. In the picture we consider $a=0$ and $g(y)=\sin(y)$.