Geodesics in a surface of revolution

Consider a surface of revolution with respect to $z$-axis and suppose that the generating curve $s\mapsto (f(t),0,g(t))$ is parametrized by the arc-length. Let $\alpha(s)=X(u(s),v(s))$ be a geodesic where $X$ is the parametrization of the surface given by

$$X(u,v)=(f(u)\cos(v),f(u)\sin(v),g(u)).$$Suppose that $\alpha$ is a geodesic. Then we know that the functions $u(s)$, and $v(s)$ satisfy  two equations involving the Christoffel symbols. We pay attention in the second one, which is $$v''+\Gamma_{11}^2 u'^2+2\Gamma_{12}^2u'v'+\Gamma_{22}^2v'^2=0.$$

The computation of these symbols are: $\Gamma_{11}^2=\Gamma_{22}^2=0$ and $\Gamma_{12}^2=f'/f$. Thus the equation writes as

$$v''+\frac{f'}{2f}u'v'=0.$$ We deduce that $$(f^2v')'=2ff'v'+f^2v''=0.$$

If we compute the angle $\theta(s)$ that makes $\alpha$ with each parallel that meets, we have

$$\cos\theta=\frac{\langle u'X_u+v' X_v,X_v\rangle}{1}=v'$$ because $\alpha$ is parametrized by the arc-lengt since it is a geodesic. As a conclusion

Theorem: In a surface of revolution, we have that the angle that makes a geodesic with every parallel that intersects satisfies

$$f(u(s))^2\cos\theta(s)=\mbox{constant}.\quad (*)$$

Application. A sailor what to find/keep a specific path along a trip on the Earth. He needs to know at every time what is its position. The knowledge of the parallel is easy measuring with respect to the Polar star and this gives the value of $f$ by the radius of the Earth. Thus if the sailor want to follow a way doing a given angle $\theta$ with the parallel, he only has to maintain the value of the constant at (*): the value of $v(s)$ indicates which is the meridian where he is positioned.

Geodesics in a right cylinder

Consider the right cylinder $X(u,v)=\alpha(u)+v\vec{a}$, where $\alpha$ is a regular curve contained in a orthogonal plane to the vector $\vec{a}$. A curve $\alpha(t)=X(u(t),v(t))$ is a geodesic if the tangent part of $\alpha''(t)$ vanishes for every $t$. We have

$$\alpha''(t)=u''X_u+v''X_v+u'^2 X_{uu}+2u'v' X_{uv}+v'^2X_{vv},$$

where

$$X_u=\alpha'(u),\ X_v= \vec{a}$$

$$X_{uu}=\alpha''(u),\ X_{uv}=X_{vv}=0.$$

Thus the tangent part of $\alpha''$ is 

$$\alpha''(t)^T=u''\alpha'(u)+v''\vec{a}+u'^2\alpha''(t)^T.$$

Since $\alpha''(t)=\kappa(t)n(t)$ is a vector orthogonal to the surface, $\alpha''(t)^T=0$. Thus $\alpha$ is a geodesic if and only if $$u''=0,\ v''=0\Leftrightarrow u(t)=at+b, v(t)=ct+d,$$

for some constant $a,b,c,d$. Then the preimage of $\alpha$ in the domain of the parametrization is a straight-line so when we carry into the surface by $X$ we obtain a helix. For example, when $u=ct$, the geodesic is a vertical line in the cylinder and when $v=ct$, the geodesic is a vertical translation of the base curve $\alpha$. In the picture, and a for a circular cylinder $x^2+y^2=1$,  we have the lines in the domain of $X$, the geodesics and the cylinder with the geodesics.

The exponential map in the plane and in the sphere

A geodesic in a plane $P$ writes as $\gamma(s)=p+s v$, $p\in P$ and $v\in T_pP=P$. Then the exponential map is simply $$exp(v)=p+v$$

that is, the exponential map at $p$ is a translation on $P$ with translation vector $p$. Let us observe that the $\mbox{exp}$ is a diffeomorphism in the whole $T_pP$ and thus the normal neighborhood of $p$ is the very plane $P$.

In the unit sphere ${\mathbb S}^2$  a geodesic is $\gamma(s)=\cos(s)p+\sin(s) v/|v|$, where $p\in  {\mathbb S}^2$ and $v\in T_p{\mathbb S}^2$. Then $$exp(v)=\cos(1)p+\sin(1)v/|v|.$$ We want to study when $\mbox{exp}$ is one-to-one. By the property $\gamma(t;p,\lambda v)=\gamma(\lambda t; p,v)$, we have $\mbox{exp}(\lambda v)=\gamma(t)$. If $|v|=1$, for $\lambda=\pi$, $\gamma(\pi;p,\lambda v)=-p$. This implies that the exponential map $\mbox{exp}$ is one-to-one in the ball $B_\pi(0)$ of radius $\pi$.

Surfaces of revolution with constant mean curvature (II)

Following the above entry, the case $H=0$ is known: the plane and the catenoid are the only rotational minimal surfaces. If $H=0$, then we have 

$$\frac{f(z)}{\sqrt{1+f'(z)^2}}=c,\ c>0,$$ that is, $$\frac{f'}{\sqrt{f^2-c^2}}=\frac{1}{c}.$$ Then the solution is $$f(z)=c\cosh(\frac{1}{c}z+d),\ d\in{\mathbb R} (*).$$

Minimal surfaces are models of soap films. In this particular case of the catenoid, the surface is the soap film formed by two coaxial circles, that is, two circles $C_1\cup C_2$ in parallel planes and the straight-line joining their centers is orthogonal to the planes containing the circles. It is natural to ask if there exists a soap film joining two given circles in parallel planes. In order to simplify the arguments, we suppose 

1. the radii of the circles are identical, namely, $r>0$. 
2. the circles $C_1$ and $C_2$  are contained in the planes of equation $z=-h$ and $z=h>0$, respectively. 

Then we pose the next:

Problem. Under what conditions on $r$ and $h$ does exist a catenoid $S$ joining $C_1$ and $C_2$? In such a case, how many catenoids do exist?

By the symmetry of the hyperbolic cosine, and since $f(h)=f(-h)$, we conclude $d=0$ in (*). Thus, the problem reduces to find $c>0$ such that $$c\cosh(\frac{1}{c}h)=r (**).$$

It is natural to think that if the circles lie very close, then there do exists a catenoid, that is, if $h$ is small, then there exists a solution of (**). 

We propose the problem in the next direction. We suppose that the circles are given (the radius $r$, which we suppose $r=1$). If they are close, there exists a catenoid, but if we separate far then the catenoid is destroyed, that is, there do no exist a catenoid between both circles.

In order to simplify the problem, we do a homothety of the ambient space from the origin and we suppose that the value $r$ of the radius is $r=1$. Consider the function $$g(c)=c\cosh(\frac{1}{c}h).$$ Our idea is using the mean value theorem. It is not difficult to see that $$\lim_{c\rightarrow 0}g(c)= \lim_{c\rightarrow \infty}g(c)=\infty,$$

so we have to study carefully the monotonicity intervals of $g$. We calculate the critical points of $g$. We have

$$g'(c)=\cosh(h/c)-\frac{h}{c}\sinh(h/c).$$ By letting $y=h/c$, this is equivalent to find $y>0$ such that $$\frac{1}{y}=\tanh(y).$$ The function $1/y$ is decreasing from $-\infty$ to $0$ and $\tanh(y)$ is increasing from $0$ to $1$, s   there is only one critical point $c_0$ (with $h>c_0$). See the next figure:

Because the limits are $\infty$, then this critical point is a minimum, $c=c_0$. This the graphic of $g$ when $h=0,5$, $h=1$ and $h=4$, and the graphic of $y=1$.

Let us observe that the minimum increases with $h$! so we have to find that it is possible to choose $h$ so the value of this minimum is $1$ at more.

We compute the value of the minimum, that is, $g(c_0)$. We know that $c_0/h=\tanh(h/c_0)$, and numerically we obtain, $h/c_0=1.19968$. Then 

$$g(c_0)=\frac{hc_0}{\sqrt{h^2-c_0^2}}=\frac{h}{\sqrt{(h/c_0)^2-1}}=1,50888 h.$$

When $h$ is close to $0$, $g(c_0)<1$, thus the graphic of $g$ has points under the line $y=1$, proving that there exists two catenoids spanning $C_1\cup C_2$. For a certain height $h=h_0$, this minimum is exactly $1$, so there exists only one catenoid and when $h>h_0$ there do not exist a catenoid joining $C_1$ and $C_2$. The value of $h_0$ is

$$h_0=\frac{1}{1,50888}=0,6627.$$
Thus, and after a homothety, we obtain:

Theorem. Let $d$ be the distance $d$ between two coaxial circles of radii $r>0$.

1. If $d<1,3254 r$, there exists exactly two catenoids spanning $C_1\cup C_2$.
2. If $d=1,3254 r$, there exists exactly two catenoids spanning $C_1\cup C_2$.
3. If $d>1,3254 r$, there do not  exist a catenoid spanning $C_1\cup C_2$.

Now we give one example of two circles that bound two catenoids. Take $r=1$ and we choose $h=0.5$. The solutions of $g(c)=1$ are: $c_1=0,235095$ and $c_2=0,848338$. The picture of the two catenoids is

Finally a remark: when one dips two coaxial circles in a soapy water container, only one catenoid is formed. In the above case, it would be the blue catenoid. Among the two catenoids, the physical systems chooses that catenoid with minimum area (minimum energy) and in this case, is the 'exterior' catenoid.

Surfaces of revolution with constant mean curvature (I)

We calculate the equation of a surface of revolution with constant mean curvature $H$. Without loss of generality, we suppose that the profile curve is a planar curve in the $xz$-plane and that the $z$-axis is the rotational axis. Also, suppose that the curve is a graph on the $z$-axis, that is, a parametrization of the profile curve is $(f(z),0,z)$, $z\in I$. Then a parametrization of the rotational surface is $$X(t,s)=(f(t)\cos(s),f(t)\sin(s),t),\ t\in I,s\in [0,2\pi].$$ Thus $H$ satisfies 

$$\frac{-f''}{(1+f'^2)^{3/2}}+\frac{1}{f\sqrt{1+f'^2}}=2H.$$

The key of this equation is that because $H$ is constant, it is possible to obtain a first integral of this equation (which is of second order). Indeed, multiplying by $ff'$ we have $$\frac{-ff'f''}{(1+f'^2)^{3/2}}+\frac{f'}{\sqrt{1+f'^2}}=2Hff',$$which can be written as 

$$\left(\frac{f}{\sqrt{1+f'^2}}\right)'=(Hf^2)'.$$ Therefore there exists $c\in {\mathbb R}$ such that 

$$\frac{f(z)}{\sqrt{1+f'(z)^2}}=Hf(z)^2+c.$$

For example, the sphere and the cylinder can be obtained from (*). For the sphere, take $c=0$. Then we have 

$$f'=\frac{1}{H}\sqrt{\frac{1}{f^2}-H^2},$$

or

$$\frac{f'}{\sqrt{\frac{1}{f^2}-H^2}}=\frac{1}{H}.$$ By integrating, we have

$$\frac{1}{H^2}\sqrt{1-H^2 f^2}=\frac{1}{H}{x}.$$

Definitively, $$f(z)=\sqrt{\frac{1}{H^2}-z^2}$$ which is a circle of radius $1/|H|$, and the surface is a sphere of radius $1/|H|$.

For the cylinder, we have to come back to the initial equation for $H$. If  $f(z)=r$, then  $H=1/(2r)$.  

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.

Asymptotic curves and lines of curvature in a hyperbolic paraboloid

The explicit computation of the asymptotic curves and lines of a curvature in a given surface uses parametrizations. Thus their computations depend what is the chosen parametrization. We illustrate this problem with the hyperbolic paraboloid. The usual way to work with this surface is as $z=x^2-y^2$. Then $X(x,y)=(x,y,x^2-y^2)$ is a parametrization of the surface and the coefficients of $I$ are:

$$E=1+4x^2,\ F=-4xy,\ G=1+4y^2.$$

For the coefficients of the second fundamental form suffices to consider the numerators in the expressions of these coefficients, which are:

$$en=2,\ fn=0,\ gn=-2.$$

Then $\alpha(t)=X(x(t),y(t))$ is an asymptotic curve if $2x'^2-2y'^2=0$, that is, $y(t)=x(t)+c$ or $y(t)=-x(t)+c$, $c\in{\mathbb R}$. If $x(t)=t$, then the curves $X^{-1}\alpha$ are straight-lines of slope $1$ and $-1$. 

For the lines of curvature, $\alpha(t)=X(x(t),y(t))$ is line of curvature if 

$$\left|\begin{array}{ccc}y'^2&-x'y'&x'^2\\ 1+4x^2& -4xy& 1+4y^2\\ 2&0&-2\end{array}\right|=0.$$

If $x(t)=t$, then $$4ty(1-y'^2)=y'(2+4t^2+4y^2),$$ which it is very difficult to solve. 

We change of parametrization of the hyperbolic paraboloid. After a change of variables $u=x-y$, $v=x+y$ and a rotation about the $z$-axis, the surface is $z=xy$. Let $X(x,y)=(x,y,xy)$. Now we have $$E=1+y^2,\ F=xy,\ G=1+x^2$$ and $en=0,\ fn=1,\ gn=0$. Then $\alpha(t)=X(x(t),y(t))$ is asymptotic curve if $x'y'=0$, that is, $x(t)=c$ or $y(t)=c$, or in other words, they are the coordinate curves. For the lines of curvatures, the determinant to solve simplifies into $$(1+y^2)x'^2=(1+x^2)y'^2.$$ Letting $x(t)=t$, then 

$$\frac{y'}{1+ y^2}=\pm\frac{t}{1+t^2}\Rightarrow {\mbox arc}\sinh(y(t))={\mbox arc} \sinh(t)+c.$$

Then $$y(t)=\sinh(\pm {\mbox arc} \sinh(t)+c)=\pm\cosh(c)t+\sinh(c)\sqrt{1+t^2}.$$

Thus with this parametrization we have found the lines of curvature of the surfaces, in contrast to the initial parametrization. 

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 $S$.

Metrics and Gauss curvature

The first and second fundamental form in a surface are symmetric bilinear forms, in particular they are metrics. Of course, the first fundamental form is positive definite because is the Euclidean metric in the tangent plane. We consider the second fundamental form $$\sigma_p:T_pS\times T_pS\rightarrow{\mathbb R}$$ $$\sigma_p(v,v)=\langle -dN_p(v),v\rangle.$$

If $X=X(u,v)$ is a parametrization of the surface, the matrix of $\sigma_p$ with respect to the basis $\{X_u,X_v\}$ is $$\left(\begin{array}{cc}e&f\\ f&g\end{array}\right).$$

We study the type of $\sigma_p$. However, it is better to choose a more suitable basis of $T_pS$, indeed, a basis of principal directions because in such a case, $$\sigma_p\rightarrow \left(\begin{array}{cc}\kappa_1(p)&0\\ 0&\kappa_2(p)\end{array}\right).$$ Because this basis diagonalizes the metric $\sigma_p$, we conclude:

1.  The point is elliptic ($K(p)>0$) is equivalent to $\kappa_i(p)>0$ for $i=1,2$, or $\kappa_i(p)<0$ for $i=1,2$ and this means that   $\sigma_p$ is  definite. 
2. The point is hyperbolic ($K(p)<0$) is equivalent to $\kappa_1(p)<0<\kappa_2(p)$. Then $\sigma_p$ is a non-degenerate indefinite metric with signature $(1,1)$.
3. $K(p)=0$, that is, some principal curvature is $0$. We have two subcases. 
• If the point is parabolic, then the non-zero principal curvature is positive or negative. This is equivalent to say that $\sigma_p$ is positive semidefinite or negative semidefinite and also, the metric is degenerate. Here the radical of $\sigma_p$ is the vector subspace spanned by the principal direction of the zero principal curvature.
• If the point is flat, then the metric is null.  

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.

A surface whose all points are hyperbolic

Consider the hyperbolic paraboloid, that is,  the surface $S$ given by the graph of $z=f(x,y)=x^2-y^2$. By the formula of the Gauss curvature for a surface which is a graph, we have $$K=\frac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}=-\frac{4}{(1+4x^2+4y^2)^2}<0.$$ 

Then all points are negative. Consider the point $p=(0,0,0)\in S$, whose tangent plane is the plane $z=0$. Let $c>0$. Then the intersection of $S$ with planes parallel to $T_pS$ is $$S\cap \{z=c\}=\{(x,y,c): x^2-y^2=c\}=\{(x,y,c):\left(\frac{x}{\sqrt{c}}\right)^2-\left(\frac{y}{\sqrt{c}}\right)^2=1\},$$

that is, an ellipse, for each height $c>0$. Similarly, for $c<0$.

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.

Comparison of curves by curvatures (II)

The definition of the mean curvature is the average of the principal curvatures. But it is also the average of the normal curvatures in two orthogonal directions, that is, 

$$H(p)=\frac12\left(\kappa_n(\theta)+\kappa_n(\theta+\frac{\pi}{2})\right),\ \ (*)$$ where $\kappa_n(\theta)$ is the normal curvature in the direction that makes an angle $\theta$ with a fix direction of the tangent plane.

Suppose again  $S_1$ and $S_2$  two surfaces with $p\in S_1\cap S_2$, $T_pS_1=T_pS_2$ and $N_1(p)=N_2(p)$, where $N_i$ are the orientation in each surface $S_i$. Take all normal sections through $p$, that is, the intersection of the planes formed by $N_i(p)$ and the tangent vectors. The set of these planes in both surfaces is the same because $N_1(p)=N_2(p)$. Fix $v\in T_p S_i$ and $\Pi_v$ the corresponding normal section. This plane meets $S_i$ in two curves $\alpha_i^v$. We are computing the normal curvature with respect to the normal (of the curve) coincides with $N_i(p)$ at $p$ and this normal curvature is the curvature of the curve

Suppose that $S_1$ lies above $S_2$ around $p$. Then for each $v\in T_pS_i$, the curve $\alpha_1^v$ lies above $\alpha_2^v$ at $p$. But we know by the theory of curves that the curvature of $\alpha_1^v$ is greater or equal to the one of $\alpha_2^v$ at $p$. Thus, all normal curvatures of $S_1$ at $p$ are greater or equal to the ones of $S_2$ at $p$. We state this remarkable result:

Theorem. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then for avery $v\in T_pS_i$, we have the inequality
$$\kappa_n^1(v)\geq\kappa_n^2(v).$$

As a consequence,  it is immediate from (*) 

Corollary. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then $H_1(p)\geq H_2(p)$.

Comparison of surfaces by curvatures (I)

We address the following problem: suppose two surfaces tangent at a point $p$ and a surface lies above the other one around $p$. Could we say something about the curvatures of the surfaces at $p$?

Since the sign of the principal curvatures depend on the orientation, suppose that $S_1$ and $S_2$ are two surfaces with $p\in S_1\cap S_2$, $T_pS_1=T_pS_2$ and $N_1(p)=N_2(p)$, where $N_i$ are the orientation in each surface $S_i$. After a rigid motion, we suppose that $p=(0,0,0)$,  $T_pS_i$ is the horizontal plane of equation $z=0$ and $N_i(p)=(0,0,1)$. By this condition, $S_i$ is locally the graph of a function on the tangent plane $T_pS_i$ (exercise!). Let $S_1$ and $S_2$ be the graph of $u$ and $v$ respectively.

Because the tangent plane is $z=0$, then with the usual parametrization $X^i(x,y)$ of the surface, 

$X^1_x(0,0)=(1,0,u_x(0,0))$, $X^1_y(0,0)=(0,1,u_y(0,0))$. Since they belong to the $xy$-plane then $u_x(0,0)=u_y(0,0)=0$. Similarly, $v_x(0,0)=v_y(0,0)=0$. By the expression of the mean curvature $H$, we obtain

$$H_1(p)=\frac12(u_{xx}+u_{yy})(0,0),\ H_2(p)=\frac12(v_{xx}(0,0)+v_{yy}(0,0)\ \ (*).$$

Suppose that $S_1$ lies above $S_2$ around $p$ and with respect to the direction that indicates $N_i(p)$. This is equivalent to $u\geq v$ around $q=(0,0)$. Thus the function $u-v$ attains a minimum at $q$. In particular, the Hessian of $(u-v)$ at $q$ is positive semidefinite. But the Hessian is

$$\left(\begin{array}{cc} (u-v)_{xx}&(u-v)_{xy}\\ (u-v)_{xy} &(u-v)_{yy}\end{array}\right)(q).$$

Thus $$(u-v)_{xx}(q)\geq 0,\ \ (u-v)_{yy}(q)\geq 0$$

and (*) implies $H_1(p)\geq H_2(p)$. Summarizing, the result is:

Theorem. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then $H_1(p)\geq H_2(p)$.

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 $g(y)=\sin(y)$.

On surfaces of revolution (II)

The idea about the concept of a surface of revolution is as 'something that rotates'. However there is a characterization of this class of surfaces in terms of the tangent planes. It is not difficult to see that in a surface of revolution the normal lines through any point meets the rotation axis. Now, and it is here the surprise, this property characterizes a surface of revolution. Thus the result is the following.

Theorem. If all normal lines in a surface meet a given straight-line $L$, then the surface is included in a surface of revolution and $L$ is the rotation axis.

The proof consists into prove that the intersection of any orthogonal plane to $L$ with the surface $S$ is a circle centered at $L$. Then it suffices to finish the result: the surface is formed by the union of (arcs of)  circles centered at $L$ and this is just the definition of a surface of revolution. Denote by $N$ the unit normal vector to $S$.

First step. Let $P$ be a orthogonal plane to $L$ that meets $S$. In particular, for any $p\in S\cap P$, $P\not= T_pS$: on the contrary, the normal line is parallel to $L$ so it does not meet $L$. Thus  $S$ and $P$ meet transversally and $S\cap P$ can parametrized as a regular curve, namely, $\alpha=\alpha(s)$. 

Second step. The normal line of $\alpha$, as curve of ${\mathbb R}^3$, meets $L$. First, recall that the normal line is included in $P$. Furthermore, the normal vector $n(s)$ of $\alpha$ at $s$ is orthogonal to $\alpha'(s)$, which lies in $P$. But the orthogonal projection $\pi(N(\alpha(s)))$ of $N(\alpha(s))$ on $P$ is also a vector orthogonal to $\alpha'(s)$. Thus $n(s)$ and $N(\alpha(s))$ are collinear. Since the normal line through $\alpha(s)$ meets $L$, the same occurs for the line through $\alpha(s)$ and with direction $\pi(N(\alpha(s)))$. 

Third step. The only planar curve whose normal lines meet at one point $p_0$ is a circle centered at $p_0$. Indeed, for each $s\in I$, there exists $\lambda(s)$ such that $p_0=\alpha(s)+\lambda n(s)$, where we are assuming that $\alpha$ is parametrized by the length-arc. If we differentiate with respect to $s$ and using the Frenet equations, we obtain, $$0=\alpha'(s)+\lambda'(s)n(s)-\lambda(s)\kappa(s)\alpha'(s).$$ This proves that $\lambda'=0$ on $I$, that is, $\lambda$ is a non-zero constant and $1-\lambda(s)\kappa(s)=0$, so $\kappa(s)=1/\lambda$, that is, $\alpha$ is included in a circle. 

On surfaces of revolution

There are two ways to define a surface of revolution in Euclidean space ${\mathbb R}^3$.

1. A surface $S$ is a surface of revolution with respect to the line $L$ is $\phi(S)=S$ for any rotation of axis $L$ (type I)

2. A surface of revolution is a surface constructed as follows. Fix $L$ a straight-line and let $P$ be a plane containing $L$. Consider a curve $C$ contained in $P$. Then the surface of revolution generated by $C$ is the set of points obtained when we rotate $C$ about the axis $L$ (type II). We denote this surface as $S(C)$

It is clear that any surface of type II is a surface of type I by the definition given in 2. It is less clear if a surface of type I is of type II, that is, if $S$ satisfies I, is there exists a curve $C$ in a plane $P$ containing $L$ such that $S=S(C)$? After a rigid motion of ${\mathbb R}^3$, we suppose that $L$ is the $z$-axis. 

Take $P$ a plane containing $L$. Then at any point   $p\in S\cap P$, the surfaces $S$ and $P$ are transversal, that is, $T_pS\not=T_pP$. Indeed, if $p=(x,y,z)\in S$ with $x^2+y^2\not=0$ (we are assuming that $S$ does not intersect the axis $L$), the rotation about the $z$ axis is the curve $$\alpha(\theta)=  \left(\cos\theta x-\sin\theta y,\sin\theta x+\cos\theta y,z \right).$$ Since $\alpha(0)=p$, then $\alpha'(0)\in T_pS$, that is, $(-y,x,0)$. The plane $P$ containing $p$ and the $z$ axis is the plane orthogonal to $(-y,x,0)$ through $p$. This proves $T_pS\not=T_pP$.

As a consequence $S\cap P$ defines a regular curve $C_p$ around $p$. Since $C_p\subset S$, then $\phi_\theta(C_p)\subset S$ for any rotation $\phi_\theta$ about the $z$-axis. In particular, the surface of revolution $S(C_p)\subset S$. 

Therefore we have prove that if $C=S\cap P$ is the intersection curve (with possible many components), then $S(C)\subset S$.

For the other inclusion, if $(x,y,z)\in S$, it is immediate that the circle $\alpha(\theta)$ defined previously intersects $P$. For example, if $P$ is the $xz$-plane, we are asking if there exists $\theta$ such that  $$\sin\theta x+\cos\theta y=0.$$ It suffices by taking $\theta$ such that $\tan\theta=-y/x$ if $x\not=0$ and $\theta=\pi/2$ if $x=0$ (it is not possible $x=y=0$). If $q=\alpha(\theta)$, then it is immediate that a suitable rotation of $q$ (exactly that rotation with angle $-\theta$) gives $p$.