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 $p$.

Starshaped surfaces

Given a positive differentiable function $f:{\mathbb S}^2\rightarrow{\mathbb R}$, define $$S(f)=\{f(p)p:p\in {\mathbb S}^2\}.$$ This surface is called a starshaped surface because the half-line starting from the origin of coordinates meets only at one point of $S(f)$. For example, if $f=2$, then $S(2)$ is the sphere cetered at the origin of radius $2$. In order to give more explicit examples, consider the parametrization of the sphere $X(t,\theta)=(\cos(t)\cos\theta,\cos(t)\sin\theta,\sin(t))$. Then define two functions $f$ by $f(X(t,\theta))=1+\cos(t)^2$ and $f(t)=1+\cos(s)^2$. The pictures are: 

We observe in the second surface that there appears 'strange point'. Exactly, points where the tangent plane is not well-defined. This occurs because $X$ is not a parametrization that cover the whole sphere, but only a part. Exactly, $s\in (0,2\pi)$, that is, except a meridian, exactly the points where appear the problem.

We give some properties of these surfaces.

1. The set $S(f)$ is, indeed, a surface. The map $$\phi:{\mathbb S}^2\rightarrow {\mathbb R}^3,\ \phi(p)=f(p)p$$ is differentiable and $d\phi_p$ is one-to-one. The proof is as follows. If $v\in T_p{\mathbb S}^2$, then if $0=d\phi_p(v)=(df_p(v))p+f(p)v$, we have a linear combination of $p$ and $v$. We know that $T_p{\mathbb S}^2=<p>^\bot$. If $v\not=0$, then $f(p)=0$, a contradiction. This proves that $\phi({\mathbb S}^2)=S(f)$ is a surface.

2. The surface $S(f)$ is compact and connected because  $S(f)=\phi({\mathbb S}^2)$.

3. The map $\phi$ is one-to-one. If $\phi(p)=\phi(q)$, then $f(p)p=f(q)q$. Taking the modulus, we have $f(p)=f(q)$, so $p=q$. 

4. The map $\phi:{\mathbb S}^2\rightarrow S(f)$ is a diffeomorphism because the inverse is   is $\phi^{-1}(p)=p/|p|$, so it is differentiable. 

Connected components in a surface

A connected component of a topological space is a closed set because the closure of a connected set is connected again. Consider now $S_i$ a connected component of a surface $S$. Then we prove that $S$ is an open set, in particular, $S_i$ is a surface. This result holds because any surface is locally connected. In order to avoid the use of this terminology, take $p\in S_i$ and we prove that $p$ is an interior point of $S_i$. Consider $X:U\rightarrow V\subset S$ a parametrization around $p$, where $V\subset S$ is an open set around $p$. Since $U$ is an open set of ${\mathbb R}^2$, there exists a ball $B$ centered at $q=X^{-1}(p)$. In particular $B$ is connected and consequently, $X(B)$ is a connected open set around $p$. Since $S_i$ is the connected component of $p$, then $X(B)\subset S_i$, q.e.d.

A consequence: any component of a compact surface is a compact surface. Indeed, if $S_i$ is the component, then $S_i$ is closed and clearly bounded, so $S_i$ is a compact. And by the above paragraph, $S_i$ is an open set of $S$, so it is a surface.

A compact surface

Consider the set $$S^\{(x,y,z)\in{\mathbb R}^3: e^{x^2}+e^{y^2}+e^{z^2}=4\}.$$

We prove that $S$ is a compact surface diffeomorphic to the unit sphere ${\mathbb S}^2$.

Here it is the picture of $S$

First  $S$ is a surface because $S=f^{-1}(\{4\})$ and $4$ is a regular value of the function $f:{\mathbb R}^3\rightarrow{\mathbb R}$ given by $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$. To prove that $S$ is compact, the same fact $S=f^{-1}(\{0\}$ proves that $S$ is closed. On the other hand, $S$ is bounded because for any $(x,y,z)\in S$, we have

$e^{x^2}\leq 4$ so $x^2\leq\log(4)$. This shows that $|(x,y,z)|^2\leq 3\log(4)$.

Finally we prove that $\phi:S\rightarrow {\mathbb S}^2$, $\phi(p)=p/|p|$ is a diffeomorphism. We point out that $|p|\not=0$ for any $p\in S$. It is immediate that $\phi$ is differentiable.

We prove that $d\phi_p$ is an isomorphism. Since

$$d\phi_p(v)=\frac{v}{|p|}-\frac{\langle p,v\rangle}{|p|^3}p$$

if $d\phi_p(v)=0$, then $v$ is proportional to $p$, exactly $v=\langle p,v\rangle p/|p|^2$. If $\langle v,p\rangle=0$, then $v=0$. On the contrary, $p=\lambda v$ with $\lambda\not=0$. In particular, $p$ is a tangent vector. Because the tangent plane $T_pS$ is orthogonal to $\nabla f(p)=2(x e^{x^2},y e^{y^2},z e^{z^2})$,  $p=(x,y,z)$, then  $\langle p,\nabla f (p)\rangle=2(x^2 e^{x^2}+y^2 e^{y^2}+z^2 e^{z^2})=0$, which is not possible. Then the inverse function theorem proves that $\phi$ is a local diffeomorphism.

We prove that $\phi$ is tsurjective. Given $p\in {\mathbb S}^2$, we have to find $q\in S$ such that $q/|q|=p$. Then we take the half-straightline starting from the origin across $p$ until that we intersect with $S$ at one point. Then it is clear that $q$ is the desired point. Thus we have to find $\lambda\in {\mathbb R}$ such that if $(x,y,z)\in {\mathbb S}^2$, $$e^{\lambda^2 x^2}+e^{\lambda^2 y^2}+e^{\lambda^2 z^2}=4.$$ If we see the left hand-side as continuous function on $\lambda$, namely, $g(\lambda)$, we use the intermediate value theorem: if $\lambda\rightarrow 0$, $g(\lambda)\rightarrow 3$ and if $\lambda\rightarrow\infty$, then $g(\lambda)\rightarrow\infty$, obtaining the result.

Finally, and because $S$ is compact, it suffices to see that $\phi$ is injective. Suppose $\phi(p)=\phi(q)$. Then $p$ and $q$ are proportional, that is, $q=m p$, $m>0$. If $p=(x,y,z)$, then

$$ e^{x^2}+e^{y^2}+e^{z^2}= e^{m^2 x^2}+e^{ m^2y^2}+e^{m^2 z^2}.$$But it is clear that the function $g(m)=e^{m^2 x^2}+e^{ m^2y^2}+e^{m^2 z^2}$ is one-to-one because $g'(m)\not=0$ for $m>0$. This proves that in (*) $m$ is necessarily $1$.

From curves to cylinders

We know that from a planar curve $\alpha:I\rightarrow{\mathbb R}^2$ we can define the generalized cylinder as $S=\{(\alpha(s),t): s\in I, t\in{\mathbb R}\}$. This can also view as follows. 

Consider a smooth function $f:O\subset{\mathbb R}^2\rightarrow {\mathbb R}$ and $a\in {\mathbb R}$ a regular value. Then we know that $C=f^{-1}(\{a\})$ is a regular curve: the $1$-dimensional case than a surface. With this function $f$, we define a new function $F$ as $$F:O\times{\mathbb R}\rightarrow{\mathbb R},\ F(x,y,z)=f(x,y),$$

which is differentiable. We prove that $a$ is a regular value of $F$. Indeed, $$\nabla F=(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z})=(\nabla f,0).$$ Consequently, $(x,y,z)$ is a critical point of $F$ if and only if $(x,y)$ is a critical point of $f$. In particular, if $F(x,y,z)=a$, then $f(x,y)=a$ and $(x,y)$ is not a critical point of $f$, and thus, $(x,y,z)$ is not a critical point of $F$. This proves that $a$ is a regular value of $F$. 

As a consequence, $S=F^{-1}(\{a\})$ is a surface. We describe this surface: $$S=\{(x,y,z)\in O\times {\mathbb R}: f(x,y)=a\}=C\times {\mathbb R}.$$ This shows that $S$ is the generalized cylinder over the planar curve $C$.

The next picture is a generalized cylinder based on the cardioid.

A surface is locally the graph of a function

We know that any surface is locally the graph of a function. In fact, we proved two things. First, the graph is defined in one of the three coordinate planes and, second, we know exactly what plane is. The proof says that if $p\in S$ and $X$ is a parametrization around $p$, then the rank of the derivative $dX_q$, $q=X^{-1}(p)$, is $2$. In particular, a $2\times 2$ matrix of $$dX_q=\left(\begin{array}{ccc}x_u& y_u&z_u\\ x_v& y_v&z_v\end{array}\right)$$ has non zero determinant, where $X(u,v)=(x(u,v),y(u,v),z(u,v))$. If we choose $$\left|\begin{array}{cc}x_u& y_u\\ x_v&y_v\end{array}\right|(q)\not=0,$$ then it is a graph on the $xy$-plane. 

It is clear that if $S$ is a graph on the $xy$-plane around $p$, then $T_pS$ can not be vertical, that is, $T_pS$ can not be orthogonal to the $xy$-plane. This can also proved in terms of the projection map. Let $\pi:{\mathbb R}^3\rightarrow{\mathbb R}^2$ be the projection $\pi(x,y,z)=(x,y)$ onto the $xy$-plane. This map is smooth, so when we restrict to $S$ is a differentiable map, which we denote by $\pi$ again. If $S$ is a graph on the $xy$-plane, then $\pi$ would be a local diffeomorphism. Indeed, the differential of $\pi$ is $$(d\pi)_p(v)=(\pi\circ\alpha)'(0)=(\alpha_1'(0),\alpha_2'(0))=(v_1,v_2)$$ where $\alpha$ is a curve that represents $v\in T_pS$ and $\alpha(t)=(\alpha_1(t),\alpha_2(t),\alpha_3(t))$. We ask when $(d\pi)_p$ is an isomorphism. Thus if $v\in T_pS$ where $(d\pi)_p(v)=(0,0)$ then $v_1=v_2=0$. This means that $v=(0,0,v_3)$ is a vertical vector. We conclude that $(d\pi)_p$ is an isomorphism if and only if  the tangent plane is not orthogonal to the $xy$-plane. In such a case, the inverse function theorem asserts the existence of an open set $V\subset S$ around $p\in S$ and and open set $W\subset{\mathbb R}^2$ around $\pi(p)$ such that $\pi:V\rightarrow W$ is a diffeomorphism. If $\phi=\pi^{-1}:W\rightarrow V$ and we stand for $\phi=(\phi_1,\phi_2,\phi_3)$, then for any $(x,y,z)\in V$ we have $$(x,y,z)=\phi(\pi(x,y,z))=\phi(x,y)=(\phi_1(x,y),\phi_2(x,y),\phi_3(x,y)).$$ This proves that $$V=\mbox{graph}(f)=\{(x,y,f(x,y)):(x,y)\in W\},$$ where $f=\phi_3:W\rightarrow\mathbb{R}$.

Ruled surfaces

A ruled surface is a surface constructed moving a straight-line along a given curve. If $\alpha=\alpha(s)$ is this curve and $w(s)$ is the direction of the straight-line at $\alpha(s)$, the straight-line is the set $\{\alpha(s)+t w(s):t\in{\mathbb R}\}$. Thus the surface $S=\{\alpha(s)+t w(s): s\in I,t\in{\mathbb R}\}$ and the parametrization is $X(s,t)=\alpha(s)+t w(s)$. Since $X_s=\alpha'(s)+tw'(s)$ and $X_t=w(s)$, then we have to assume that they are independent linearly. Then $X$ is a parametrized surface and so, locally, $S$ is a surface.   

We show some examples. Suppose $\alpha$ is a planar curve included in the plane $z=0$. If we take $w(s)=a=(0,0,1)$, we have the right cylinder over the curve $\alpha$. In the next figure, $\alpha$ is the parabola $\alpha(s)=(s,s^2,0)$ and $w(s)=(0,0,1)$.

We can take $w(s)$ to be tilted at each point of $s$. If $\{T(s), N(s), B(s)\}$ is the Frenet trihedron along $\alpha$,  then up to reparametrizations, $B(s)=(0,0,1)$. If we take $w(s)=\cos(m) N(s)+\sin(m) B(s)$, with $m\in{\mathbb R}$,  we obtain a cone along $\alpha$. In the next picture,  $\alpha$ is the parabola again.

If we replace the constant $m$ by a function $\theta(s)$, then $w(s)$ goes changing at each point. Here we take $\alpha$ the circle $\alpha(s)=(\cos(s),\sin(s),0)$. If $w(s)$ is a $2\pi$-periodic function, then $w$ is also $2\pi$-periodic. This occurs for example if $\theta(s)=s$. The parametrization is $X(s,t)=(\cos (s)-t \cos ^2(s),\sin (s)-t \sin (s) \cos (s),t \sin (s))$ and the surface is:

But if $w$ is $4\pi$-periodic, then we obtain a Möbius strip. For this, we take $\theta(s)=s/2$. Then $$X(s,t)=\left(\cos (s) \left(1-t \cos \left(\frac{s}{2}\right)\right),\sin (s) \left(1-t \cos \left(\frac{s}{2}\right)\right),t \sin \left(\frac{s}{2}\right)\right).$$

Differential of the restriction of a function

We know that if $F:{\mathbb R}^3\rightarrow{\mathbb R}^m$ is a smooth function, and $S$ is a surface, then $F_{|S}:S\rightarrow{\mathbb R}^m$ is differentiable. Moreover, if $p\in S$, then $(dF_{|S})_p={(dF)_p}{\Big |}_{T_pM}$. In the right side, we have the differential of the restriction of $F$ and in the right side, we have the derivative of $F$ which is only viewed in the tangent plane $T_pS$. Denote $f=F_{|S}$. From now, and to simplify, we suppose $m=1$.

It is usual that we work with the Jacobian matrix of $dF$, which is noting the matrix expression of the linear map $dF$ with respect to the usual basis of Euclidean spaces, here, of ${\mathbb R}^3$ and ${\mathbb R}$. By definition, the elements are the partial derivatives of $f$, namely, $$\mbox{Jac F}(p)=M(dF_p,B_u,B_u')=(f_x\ f_y\ f_z)(x,y,z),$$ where $p=(x,y,z)$.

If we want to write the matrix expression of the linear map $df_p$, we need to fix basis in both spaces, that is, in $T_pS$ and in ${\mathbb R}$. Although for ${\mathbb R}$ we choose the basis $B_u'=\{1\}$, there is not a natural basis on $T_pS$. This only appears when we fix previously a parametrization $X=X(u,v)$ around $p$. Then the basis could be $B=\{X_u(q),X_v(q)\}$, with $X(q)=p$ and the matrix would be $M(df_p,B,B_u')$. Therefore, there is not a relation between this matrix and $\mbox{Jac F}(p)$.

We show an example. In order to avoid 'simple numbers', consider the elliptic paraboloid $z=x^2+y^2$ and the function $F(x,y,z)=z$. Then $\mbox{Jac F}(x,y,z)=(0\ 0\ 0)$. 

Consider now $f=F_{|S}$. Then $(df)_p(v)=\alpha_3'(0)=v_3$, where $v=(v_1,v_2,v_3)$. Let $p=(x,y,z)\in S$ and the parametrization $X(x,y)=(x,y,x^2+y^2)$. Then $X_x=(1,0,2x)$ and $X_y=(0,1,2y)$. Moreover, using the chain-rule, we have

$$(df)_{X(x,y)}(X_x)=(f\circ X)_x=2x,\  (df)_{X(x,y)}(X_y)=(f\circ X)_y=2y.$$

If $B=\{X_x,X_y\}$, we have  $$M((df)_{X(x,y)},B,B_u')=(2x\ 2y),$$

a matrix completely different than ${\mbox Jac F}(p)$.

Vector structure of the tangent plane

We know that the tangent plane $T_pS$ of $S$ at $p$ is a vector space of dimension $2$. We ask here what is the vector structure of $T_pS$. Indeed, let $v,w\in T_pS$ and let $\alpha, \beta: I\rightarrow S$ be two curves such that $\alpha(0)=\beta(0)=p$ and $\alpha'(0)=v$ and $\beta'(0)=w$. Since $T_pS$ is a vector space, then $v+w\in T_pS$. We ask: 

Question. What is the curve on $S$ whose tangent vector is $v+w$?

A first attempt would be $\alpha+\beta$, because $(\alpha(t)+\beta(t)'(0)=v+w$. Of course, this is completely wrong because $\alpha(t)+\beta(t)$ is a curve in ${\mathbb R}^3$ which it is not included at $S$! To find the right curve $\gamma:I\rightarrow S$ such that $\gamma(0)=p$ and $\gamma'(0)=v+w$ we have to come back to the moment where we proved that $T_pS$ is a vector space. Recall that $$T_pS=(dX)_q({\mathbb R}^2),\ q=X^{-1}(p),$$ where $X$ is a parametrization around $p$, and $(dX)_q:{\mathbb R}^2\rightarrow {\mathbb R}^3$ is the derivative of the map $X:U\subset {\mathbb R}^2\rightarrow{\mathbb R}^3$. The above identity says us that $v+w$ is the sum of two vectors in ${\mathbb R}^2$. Thus the idea to find $\gamma$ is: first, compute the preimages of $v$ and $w$, namely, $\bar{v}$, $\bar{w}$, compute $\bar{v}+\bar{w}$, then take a curve passing $q$ with tangent vector $\bar{v}+\bar{w}$, and finally, consider the image of this curve by the parametrization $X$.

Let $\bar{\alpha}(t)=X^{-1}(\alpha(t))$, $\bar{\beta}(t)=X^{-1}(\beta(t))$. Then  $\alpha(t) = X(\bar{\alpha}(t))$, $\beta(t)=X(\bar{\beta}(t))$. Thus $$v=(dX)_q(\bar{\alpha}'(0)),\ w=(dX)_q(\bar{\beta}'(0)).$$ Here we obtain $\bar{v}= \bar{\alpha}'(0)$ and $\bar{w}= \bar{\beta}'(0)$. Consider $\bar{v}+\bar{w}$ placed at $q$ and consider a curve passing $q$ with this tangent vector: if suffices $$\sigma:I\rightarrow U,\ \sigma(t)=q+t(\bar{v}+\bar{w}),$$ where it is immediate $\sigma'(0)=\bar{v}+\bar{w}$.

Here we find the key of the vector structure of the tangent plane $T_pS$: of course, we can not sum the curves $\alpha$ and $\beta$, neither the curves $\bar{\alpha}$ and $\bar{\beta}$ (in this case, outside of $U$). We take another curve that represents the (tangent) vector $\bar{v}+\bar{w}$. In this case, it is enough a straight-line and here we use strongly that $U$ is an open set of ${\mathbb R}^2$, because at least around $t=0$, the straight-line  $\sigma$ is included in $U$. 

The last steps are now easy. Consider $\gamma(t)=X(\sigma(t))$, which is a curve on $S$ with $\gamma(0)=X(\sigma(0))=X(q)=p$. Then using the chain rule, we have $$\gamma'(0)=(dX)_{\sigma(0)}(\sigma'(0))= (dX)_q(\bar{v}+\bar{w})=(dX)_q(\bar{v})+(dX)_q(\bar{w})=v+w.$$

Similarly, we can find the curve that represents the tangent vector $\lambda v$, where $\lambda\in{\mathbb R}$.

Writing differentiable maps on surfaces

In calculus it is usual to work with smooth functions in terms of `variables', I mean, something as $f(x,y,z)=x^2+\sin(z)+e^y$, in terms of `x', `y' and `z'. However working on surfaces, sometimes (or many), we prefer do not use `variables', specially when we need to compute the derivative of the function. The next example clarifies this issue. 

If $S$ is a surface, define $$f:S\rightarrow{\mathbb R},\ f(p)=|p|^2=\langle p,p\rangle.$$ Here we use `p' instead of the variables. This function measures the square of the distance of the point $p$ to the origin of ${\mathbb R}^3$. We observe that if $p=(x,y,z)$, then $f(x,y,z)=x^2+y^2+z^2$, which is a known differentiable function in ${\mathbb R}^3$, but now $f$ is defined on a surface. If we want to prove that $f$ is differentiable on $S$, first we consider $F:{\mathbb R}^3\rightarrow{\mathbb R}$ the function $F(p)=\langle p,p\rangle$. Since $p\mapsto p$ is the identity, which is differentiable, then $F$ is noting the scalar product of a differentiable map by itself. Then $F$ is differentiable. Finally, $f=F_{|S}$, that is, the restriction on $S$ of a differentiable map of ${\mathbb R}^3$. This proves definitively that $f$ is differentiable.

Other example is the height function. Let $a\in {\mathbb R}^3$ be a unit vector and define $$f:S\rightarrow{\mathbb R},\ f(p)= \langle p,a\rangle.$$ This function measures the square of the distance of the point $p$ to the vector plane $\Pi$ orthogonal to $a$. For this reason, it is named height function. If we write in coordinates and $p=(x,y,z)$, we have $f(x,y,z)=a_1 x+a_2 y+a_3 z$, where $a=(a_1,a_2,a_3)$.  If we want to prove that $f$ is differentiable on $S$ without the use of `x's', define $F:{\mathbb R}^3\rightarrow{\mathbb R}$ the function $F(p)=\langle p,a\rangle$. Since $p\mapsto p$ and $p\mapsto a$ are differentiable maps, then  $F$ is the scalar product of two differentiable vector maps, so $F$ is differentiable. Finally, $f=F_{|S}$, proving that $f$ is differentiable.  

Two possible definition of differentiability between two surfaces

The definition of a differentiable map between two surfaces given in the course is extrinsic. I explain it. Consider  $f:S_1\rightarrow S_2$ a map between two surfaces and $p\in S_1$. Then $f$ is differentiable at $p$ if $i\circ f\circ X:U\rightarrow{\mathbb R}^3$ is differentiable at $q=X^{-1}(q)$, where $i:S_2\rightarrow{\mathbb R}^3$ is the inclusion map (definition I). Here we use strongly that $S_2$ is included in Euclidean space ${\mathbb R}^3$. If one changes the viewpoint, one would request that the definition does not depend if $S_2$ is or is not included in ${\mathbb R}^3$, but only on $S_2$, that is, an intrinsic definition. Then the natural way to do it is by means of parametrizations and the definition would be: $f$ is differentiable at $p$ if $Y^{-1}\circ f\circ X:U\rightarrow W$ is smooth at $q=X^{-1}(p)$, where $X:U\rightarrow S_1$ and $Y_W\rightarrow S_2$ are parametrizations around $p$ and $f(p)$ respectively (definition II). Now it is not important if the surface is included in Euclidean space. 

This allows to extend the above definition to object with similar properties than surfaces, that is, objects with a set of parametrizations between open sets of ${\mathbb R}^n$ and open sets of the object. Then it will appear the concept of manifold of dimension $n$.

Returning, we prove that both definition are equivalents. 

1. (II) $\Rightarrow$ (I). Suppose a such $f$ which is differentiable at $p$ with definition II. When we consider a parametrization $X$ around $p$, then $i\circ f\circ X=(i \circ Y)\circ (Y^{-1} \circ f\circ X)$ and thus, it is the composition of two smooth maps between open sets of Euclidean spaces.
2. (I) $\Rightarrow$ (II). Suppose $f$ which is differentiable at $p$ with definition I, that is, we know $i\circ f\circ X$ is smooth at $q$ for any $X$. Without loss of generality, and fi $Y=(Y_1,Y_2,Y_3)$, we suppose that $$\left|\begin{array}{cc}\frac{\partial Y_1}{\partial u}&\frac{\partial Y_2}{\partial u}\\  \frac{\partial Y_1}{\partial v}& \frac{\partial Y_2}{\partial v}\end{array}\right|\not=0.$$ The Inverse function theorem asserts that the function $$(Y_1,Y_2):W'\rightarrow O', (u,v)\mapsto (Y_1(u,v),Y_2(u,v))$$ is a diffeomorphism between suitable open sets of ${\mathbb R}^2$. Let $\phi=(Y_1,Y_2)^{-1}$. If $(i\circ f\circ X)=(f_1,f_2,f_3)$, then $$Y^{-1} \circ f\circ X (u',v')=\phi^{-1} (f_1(u',v'),f_2(u',v'')),$$ which is differentiable because it is the composition of two differentiable maps.

Finally, we will adopt the definition I because it is more intuitive, although we are loosing `generality'.

Using the theory on differentiability for the properties of differentiability on surfaces

Almost all properties on the differentiability of maps on surfaces are proved by the analogous properties of differentiable maps between open sets of Euclidean spaces. I point out two of them

1. A parametrization of a surface is differentiable. Here we are saying that the parametrization $X_U\subset{\mathbb R}^2\rightarrow V\subset S$ is differentiable, where $V$ is an open set of a surface $S$. In order to clarify the notation, we stand for $Y$ the above map, and $X:U\rightarrow {\mathbb R}^3$ the parametrization. In fact, $Y$ is noting the restriction of $X$ into the codomain. Because $Y$ arrives to a surface, $Y$ is differentiable if $i\circ Y: U\rightarrow{\mathbb R}^3$ is smooth. But this map is just $X$, which it is smooth because is the second property of a parametrization.
2. The inverse of a parametrization is differentiable. Here we mean $X^{-1}:V\rightarrow U\subset{\mathbb R}^2$ is differentiable. Now $X^{-1}$ is a map whose domain is a surface, in fact, the open set $V$ of $S$, which is indeed a surface. By the definition, we have to prove that $X^{-1}\circ Z$ is smooth for some parametrization of $S$. Here we take $Z=X$. Then $X^{-1}\circ X$ is the identity map on the open set $U$, which is trivially smooth.

Surfaces constructed from curves (II)

Surfaces of revolutions are other type of surfaces constructed by curves. In the previous entry, a cylinder is noting a planar curve $\alpha$ moved along a fix direction $a$, that is, we translate $\alpha$ along a direction. If $a$ is the given direction, a translation in this direction is $T_t(x)=x+t a$, $x\in{\mathbb R}^3$ and $t\in{\mathbb R}$. Then the cylinder on basis $\alpha$ is $$\cup_{t\in {\mathbb R}}T_t(\alpha(s)):s\in I\}.$$

In order to  define a surface of revolution, we consider a   curve $\alpha$ contained in a plane $P$ and we rotate $\alpha$ about a line $L$ contained in the plane $P$. We know that the parametrization is $X(s,\theta)=(f(t)\cos\theta,f(t)\sin\theta,g(t))$, where $\alpha(t)=(f(t),0,g(t))$. Again, the difficulties appear when we prove that $X$ is an embedding. For this reason we assume again that $\alpha$ is an embedding or a simple closed curve. 

With the curve $\alpha(t)=( \sin(t),0,1+\cos(t)\cos(2t))$, with $t\in (0.5,2.5)$, we observe that there is a self-intersection, so it does not define a surface. 

Other example is the torus generated by the circle $\alpha(t)=(1+2\cos(t),0,2\sin(t))$ because it intersects the $z$-axis.

Definitively, we impose that the curve $\alpha$ is an embedding or a simple closed curve. In the next picture, the surface is generated by the simple closed curve  $\alpha(t)=(2+\sin(t),0,\cos(t)+\cos(2t))$.

Surfaces constructed from curves: cylinders

We have defined some types of surfaces from curves, for example, generalized cylinders. Let $\alpha:I\rightarrow{\mathbb R}^3$ a curve contained in a plane $P$, which we suppose it is the plane $z=0$ and let $a\in {\mathbb R}^3$ be a vector that is not contained in $P$. The cylinder on base $\alpha$ in the direction of $a$ is the set $$S=\{\alpha(s)+ta:s\in I,t\in{\mathbb R}\}.$$ Of course, the parametrization is $$X:I\times{\mathbb R}\rightarrow{\mathbb R}^3, X(s,t)=\alpha(s)+ta.$$ If we prove that $S$ is a surface, it is immediate that $X$ is differentiable, $X_s=\alpha'(s)$, $X_t=a$ and both vectors are independent linearly. The difficulty appears when we want to prove that $X$ is a parametrization. The sets $I\times{\mathbb R}$ and $S$ are open in ${\mathbb R}^2$ and $S$, respectively. Also, it is immediate that $X$ is continuous. It remains to prove that $X$ is biyective and $X^{-1}$ is continuous. Of course, if $\alpha$ is not one-to-one, then $X$ is not, as in the next pictures (here the vector $a$ is $a=(1,1,1)$.

1. For this reason, we suppose two cases: $\alpha:I\rightarrow{\mathbb R}^3$ is an embedding or 
2. $\alpha:{\mathbb R}\rightarrow {\mathbb R}^3$ is a simple closed curve.

In the first case, $X$ is one-to-one. If $(x,y,z)=(\alpha_1(s)+ta_1,\alpha_2(s)+t a_2,ta_3)$, then $t=z/a_3$ and so $$s=\alpha^{-1}(x-\frac{z}{a_3} a_1,y-\frac{z}{a_3} a_2).$$ It is immediate that $X^{-1}$ is continuous.

In the second case, $\alpha$ is an embedding in an interval of length less than $T$, where $T>0$ is the period of $\alpha$. 

We have the next pictures for the simple closed curve $\alpha(s)=(3 \cos (s),\sin(s)+\cos(s)+\cos(2s)$ and $a=(0,0,1)$.