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 $\lambda\in{\mathbb R}$.

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.

Parametrizations of a surface of revolution

Consider $S$ the surface of revolution obtained by rotating the planar curve $\alpha(t)=(f(t),0,g(t))$, $t\in I$ which it is contained in the halfplane $y=0, x\geq 0$. We know that $\alpha$ is regular and we have two possible types of curves, i) $\alpha$ is an embedding or ii) $\alpha$ is a simple closed curve. The surface $S$ is $X(I\times{\mathbb R})$, where $$X(t,\theta)=(f(t)\cos\theta,f(t)\sin\theta,g(t)).$$ In order to prove that $S$ is a surface, it appears the problem about how many of parametrizations are needed and how to prove that they are homeomorphisms.

First consider that $\alpha$ is an embedding. We compute the inverse function of $X$ so we will discover what is the right domain of $X$. By the injectivity, it is necessary that $\theta$ moves in an interval of length $2\pi$ at most. Thus a possibility is $X:U_1:=I\times (0,2\pi)\rightarrow X(U_1)$. The set $X(U_1)$ is an open set of $S$ because $$X(U_1)=S-\alpha(I)=S-S\cap(\{y=0,x\geq 0\}).$$ Here we use that $f(t)>0$. Because we need to cover the curve $\alpha(I)$, then the other parametrization is $Y:U_2:=I\times (-\pi,\pi)\rightarrow X(U_2)$. For $Y$ we have to remove `the other side' of $S$, that is $\Phi_\pi(\alpha(I))$, where $\Phi_\theta$ is the rotation of angle $\theta$. In other words, $Y(U_2)=S-(S\cap(\{y=0,x\leq 0\})$.

We compute the inverse of $X$ (or $Y$). We have to write $t$ and $\theta$ in terms of $x,y,z$ from the next equations $$\left\{\begin{array}{l}x=f(t)\cos\theta\\ y=f(t)\sin\theta\\ z=g(t)\end{array}\right.$$ From $x^2+y^2=f(t)^2$, we obtain $f(t)=\sqrt{x^2+y^2}$, and using that $\alpha$ is an embedding, then $t=\alpha^{-1}(\sqrt{x^2+y^2},0,z)$. For $\theta$ we have two possibilities:

1. If we want to write `something' with the arc tangent, then it would be $\theta=\mbox{arc tan}(y/x)$. But in such a case, $\theta$ is defined in $(-\pi/2,\pi/2)$. Because the initial interval is $(0,2\pi)$, we have to change the domain of $X$. We write now $X:U_1:=I\times(-\pi/2,\pi/2)\rightarrow X(U_1)$. The only difference is that we have to prove that $X(U_1)$ is an open set of $S$. But by the picture, $X(U_1)=S-(S\cap \{x\leq 0\})$. Other parametrization is with $I\times (\pi/2,3\pi/2)$ where it is possible to define the inverse of the tangent. And what about the points with $\theta=\pi/2$ or $\theta=3\pi/2$? Here, the $x$-coordinate of the point $(x,y,z)\in S$ vanishes. Now we consider that inverse of the cotangent and taking $\theta={\mbox arc cot}(x/y)$. We need two more domains, namely, $I\times (0,\pi)$ and $I\times (\pi,2\pi)$. Finally, we observe that, using the inverse functions of the tangent of the cotangent function, we need 4 parametrizations: all them write 'very similar', but  the domain goes changing.
2.  If we do not want to use trigonometric functions, we can do the following. If one looks the picture of a surface of revolution, it is clear that the parametrization $X$ can be defined in $I\times (0,2\pi)$, because the angle $\theta$ is well defined. The problem appeared in how to catch the variable $\theta$. The trigonometric functions have added a bit of confusion. Other way is the following.                                                                                                                        Consider $\beta:(0,2\pi)\rightarrow {\mathbb S}^1-\{(0,0\})$ a parametrization of the circle minus one point. The key is the following: the map $\beta$ is a homeomorphism! It is clear that $\beta$ is one-to-one and continuous. One could do the following argument. It is well known that  ${\mathbb S}^1$ minus one point is homeomorphic to the real line ${\mathbb R}$, which it is homeomorphic to the interval $(0,2\pi)$. But the problem is if $\beta$ is, indeed, a homeomorphism. The only trouble is about the inverse. But $\beta$ is an open map because the image of an interval of $(0,2\pi)$   is an open set of ${\mathbb S}^1$.             Once proved that $\beta$ is a homeomorphism, from $x=f(t)\cos\theta$ and $y=f(t)\sin\theta$ we conclude $$\theta=\beta^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right).$$ Then $$X^{-1}(x,y,z)=\left(\alpha^{-1}(\sqrt{x^2+y^2}),\beta^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)\right).$$ Thus we need 2 parametrizations.

If $\alpha$ is a simple closed curve, for covering the curve $\alpha$ by embeddings, we need two parametrizations, namely, $\alpha:(0,T)\rightarrow \alpha(0,T)$ and $\alpha:(T/2,3T/2)\rightarrow \alpha(T/2,3T/2)$. Then for the surface, we need 4 parametrizations.

There is another elegant argument. We observe that when one has proved that $X$ is a parametrization, then we think $Y$ as a `rotation' of $X$. Recall that if $S$ is a parametrization and $\psi$ is a diffeomorphism of ${\mathbb R}^3$, then $\psi(S)$ is a surface. But the parametrizations of $\psi(S)$ are of type  $\psi\circ X$, where $X$ is a parametrization of $S$. With this idea in mind, consider now $\Phi_\theta$ the rotation about the $z$-axis of angle $\theta$. By the definition of $S$, we have $S=\Phi_\theta(S)$. Suppose that we have proved that $X:U_1:=I\times (-\pi/2,\pi/2)\rightarrow {\mathbb R}^3$ is a parametrization of $S$: it is only of a part of $S$, exactly, $V_1=X(U_1)$. Now we take $\theta\in{\mathbb R}$. Because $\Phi_\theta$ is a homeomorphism, it is clear that $\Phi_\theta\circ X$ satisfies the properties of a parametrization, where now the coordinate open is $\Phi_\theta(V_1)$. We point out that $\Phi_\theta(V_1)$ is an open set of $S$. Therefore, if we are going taking many $\theta$, we cover all the surface $S$ by coordinate opens of type $\Phi_\theta(U_1)$. In fact, it is enough only one rotation!, namely, $\Phi_\pi$. With this argument, the effort lies only in the first parametrization $X$: the other ones are obtained `by rotating' $X$.

Finally, we conclude that it is clear that the parametrizations $X$ hold to prove that $S$ is a surface of revolution, for example, taking very small domains, or rotations of $X$. However, the proof is a bit tedious if one wants to write precise arguments.

Curves-maps; surfaces-sets

I remarked in the classroom the differences between the definition of a curve and a surface: a curve is a differentiable map and a surface is a subset of Euclidean space where there do exist parametrizations. I return again with it.

If a curve $\alpha:I\rightarrow{\mathbb R}^3$ is regular $t_0$, then $\alpha'(t_0)\not=0$. If we write in terms of the differential map of $\alpha$, it means that $(d\alpha)_t:{\mathbb R}\rightarrow {\mathbb R}^3$ is a non-zero linear map. This is equivalent to say that $\mbox{rank}(d\alpha)_t=1$, because $$(d\alpha)_{t_0}(1)=\frac{d}{ds}{\Big |}_{s=0}\alpha(t_0+s)=(x'(t_0),y'(t_0),z'(t_0))\not=(0,0,0).$$ Thus the rank  of $(d\alpha)_t$ is the maximum possible (it would be $0$ or $1$). Furthermore, by using the inverse function theorem, ``the curve is a graph locally around $t_0$''. In fact, it was proved that there exists $\epsilon>0$ such that $$\alpha:J=(t_0-\epsilon,t_0+\epsilon)\rightarrow \alpha(t_0-\epsilon,t_0+\epsilon)$$ coincides with the graph of a function, that is, there exists a differentiable function $f:K\subset {\mathbb R}\rightarrow {\mathbb R}$ such that $\{(x,f(x)):x\in K\}=\alpha(J)$. As a consequence, $\alpha:J\rightarrow\alpha(J)$ is homeomorphic to an interval of ${\mathbb R}$.

Then the map $\alpha$ would play (almost) the same role of parametrizations in a surface. If we want to give the definition of the  analogous $1$-dimensional case of a surface, then a subset $C\subset{\mathbb R}^3$ is a $1$-surface (=curve) if for each point $p\in C$ there exists $I\subset {\mathbb R}$ and a map $X:I\rightarrow V\subset C$ a homeomorphism, where $V$ is an open of $C$ around $p$, $X:I\rightarrow {\mathbb R}^3$ is differentiable and $X'(t)\not=(0,0,0)$.

The question is the definition of curve given in chapter $1$ is now a $1$-surface, more precisely, if the trace $\alpha(I)$ is a such $1$-surface. Then one would think `yes' by taking around each point $p\in C$ the corresponding restriction of $\alpha$ to the suitable interval $J$. However, the only problem is the following: Is $\alpha(J)$ an open set of $C$? because the other properties have been showed. We find the answer in the curve $\alpha(t)=(\cos(t),\sin(2t))$, $t\in {\mathbb R}$.

This curve self-intersects at the origin. Thus it can not be a $1$-surface because this point has not a neighbourhood which is homeomorphic to ${\mathbb R}$. By the inverse function theorem, around $t_0=0$, $\alpha(J)\cong J$ is a graph, but $\alpha(J)$ is not an open set of $\alpha({\mathbb R})$, which it happens exactly in our example, as one can see in the next picture: the red color line is $\alpha(J)$ is not an open set in $C$.