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 $a=(0,0,1)$. 

Surfaces by implicit equations

We have proved that the inverse of a regular value of a function $f:O\subset{\mathbb R}^3\rightarrow{\mathbb R}$ is a surface. Consider the prototype of function $f(x,y,z)=x^n+y^n+z^n$ where $n\in {\mathbb N}$ and let $S_a=f^{-1}(\{a\})$. The gradient of $f$ is $$\nabla f(x,y,z)=n(x^{n-1},y^{n-1},z^{n-1}).$$

First we compute the critical points, that is,  $\nabla f(x,y,z)=(0,0,0)$, obtaining  that there do no exist if $n=1$ and for $n>1$, the only critical point is $(0,0,0)$. Then 

1. For $n=1$, any value $a\in {\mathbb R}$ is regular. Here $S_a$ is a plane.
2. For $n>1$, any value $a\not=0$ is regular and $S_a\not=\emptyset$ if $a>0$. Of course, for $n=2$, we have the sphere, but if $n$ is large, $S_a$ seems a cube, but a smooth cube! 

Some pictures for $n=4$ and $n=10$:

[Remark. We observe that the set of regular value is 'big', in the above case, ${\mathbb R}-\{0\}$. There is a result that informs us about the size of $f(A)$, where $A\subset O$ is the set of critical points: the Sard's theorem says that $f(A)$ has Lebesgue measure $0$. For example, the compliment of the set of regular value has not interior points.]

Consider now the function $g(x,y,z)=x^n+y^n-z^n$ that gives the hyperboloid of one and two sheets. The set of regular values is ${\mathbb R}-\{0\}$ again. Using Mathematica, I put some pictures when $a=1$ and $n=4$ and $n=10$ 

and when $a=-1$ and $n=4$ and $n=10$