Motivation for differential Manifolds

Question

I've an epistemological question about Manifolds. According to the text I'm reading the motivation for introducing differential manifolds is the following:

"In general terms, smooth manifolds are objects that locally look like $\mathbb{R}^n$, and on which it is possible to define operations that extend those ones used in classic Calculus. The most familiar examples are spheres, ellipsoid, etc".

So to extend Calculus to curved surfaces. Considering for example the sphere as the subset of points in $\mathbb{R}^3$ for which $x^2 + y^2 + z^2 = r_0$ holds, and which can be parametrized by the vector function

$$(\theta,\phi) ⟼ r_0 \cos \theta \cos \phi \hat{i} + r_0 \sin \theta \cos \phi \hat{j} + r_0 \sin φ \hat{k} \quad(1)$$

in the parameter domain, the rectangle $D = [0, \pi] \times [0, 2\pi].$

If the motivation for Manifold is doing calculus as in $\mathbb{R}^n$, I don't see why I cannot consider the sphere as an object in $\mathbb{R}^3$, for example mapping each point on the surface (with the function (1) defined above) of the sphere with a 3D vector in $\mathbb{R}^3$ and do calculus in $\mathbb{R}^3$; Instead of considering the manifold $M$ a set of points with a continuous 1-1 map from each open neighborhood onto an open set of $\mathbb{R}^n$, which associates with each point $P$ of $M$ an $n$-tupel $(x_1(P) , ..., x_n(P))$. It seems there exists already a natural mapping (1) which maps points on the surface of the sphere to 3-tuples, i.e. vectors in $\mathbb{R}^3$.

Answer 1

First, let me give you an example of a manifold which is not so obviously part of $\mathbb{R}^3$. Take the compactified complex plane $$ \hat{\mathbb{C}}:=\mathbb{C} \cup \{\infty \} $$ It is basically a plane with a poinf at infinity added to it. But you cannot easily realize it as the plane in $\mathbb{R}^2$, due to the point at infinity. So - how do we see this is still a surface?
Let us first note that $\mathbb{C}\cong \mathbb{R}^2$. By using the convention $\frac{1}{\infty}=0$, you can see that for $\mathbb{C} \subset \hat{\mathbb{C}}$, we can simply use the map $$ \phi_1:\mathbb{C} \to \mathbb{C},z \mapsto z=x+iy $$ and see that it indeed locally "looks like" $\mathbb{R}^2$.
For the point infinity, take an open neighbourhood of $\infty$, that is $ \mathbb{C} \setminus \overline{B_R(0)}$ for some $R>0$ together with the point at infinity. Then we can use the map $$ \phi_2:\mathbb{C} \setminus \overline{B_R(0)} \to \overline{B_R(0)}, z \mapsto \frac{1}{z}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}. $$ (Here I assume that $\infty$ gets mapped to $0$).
Now on the overlap these maps are also well defined in the sense that $\phi_i^{-1}\circ \phi_j$ is smooth. You can even show they are not just smooth, but also biholomorphic in this case.

Anyway, this shows you that $\hat{\mathbb{C}}$ looks like $\mathbb{C}\cong\mathbb{R}^2$ locally, but none of those two maps above can be defined globally on $\hat{\mathbb{C}}$. You can use those $\phi_i$ to interpret maps from, to and between $\hat{\mathbb{C}}$ locally as maps from some subset of the complex plane. This is a set for which you know how to do calculus! Any local statements can be deduced using those maps! This includes, for example, finding solutions of ODE on manifolds.
This is what you refer to as instrinstic geometry - none of the definitions above rely on the fact that $\hat{\mathbb{C}}$ can be embedded into $\mathbb{R}^3$.

On the other hand, extrinstic geometry relies on the fact that surfaces can in fact be seen as object in, for example, $\mathbb{R}^3$. In the above example, you have no meaningful notion of unit normal vector, unless you specify some "bigger space" $\hat{\mathbb{C}}$ belongs to.

Answer 2

Yes, the sphere sits inside of $\Bbb{R}^3$ very nicely, and there are other examples of surfaces which sit nicely inside of an ambient $\Bbb{R}^n$ (e.g parabolas, hyperbolas, or their revolutions about an axis, graphs of smooth functions etc). But the problem is that these are not open subsets of $\Bbb{R}^n$. If you are on the usual unit sphere $S^2\subset\Bbb{R}^3$ at the north pole $N=(0,0,1)$, and if you decide to move directly south, then your position will be $p=(0,0,z)$ for some $0<z<1$. However, this point $p$ does not lie on the sphere: $p\notin S^2$. This is what it means to not be open: you can find some direction such that if you move a small amount in that direction then you leave the original subset. Why are non-open subsets problems? Well for continuity it's not really an issue, but if you want to talk about differentiability or even just about partial derivatives, it gets much more troublesome. Imagine you have a very "nicely behaved" function $f: S^2\to\Bbb{R}$. If you naively think that just because $S^2\subset\Bbb{R}^3$, and hence it makes sense to do 'calculus as usual', then you'd be wrong. For example, partial derivatives like $\frac{\partial f}{\partial z}\bigg|_{(0,0,1)}$ don't even make sense because by definition it ought to be \begin{align} \lim_{h\to 0}\frac{f(0,0,1+h)-f(0,0,1)}{h}, \end{align} but this expression is completely meaningless since for small non-zero $h$, the point $(0,0,1+h)\notin S^2$ so it doesn't even make sense to evaluate $f:S^2\to\Bbb{R}$ on this point. Even if I decide to define $f:S^2\to\Bbb{R}$ to be the constant zero function $f=0$, it would still not make sense to talk about its partial derivative in the $z$-direction while at the north pole. This is just because the function is not defined on an open set (sure, the zero function has an obvious extension to all of $\Bbb{R}^3$, but that's a separate matter). The sphere is such a nice simple shape, and just visually so smooth looking, and about as perfect as things get. Even the constant zero function has troubles with partial derivatives. Thus, the obstruction to doing 'calculus as usual' is really the non-openness.

Ok, so this immediately brings a halt to 'doing calculus as usual', but should we give up? No! Calculus is about studying changes in objects, and the sphere is so nice and smooth looking that we should definitely be able to talk about how functions defined on the sphere change. Intuitively, it is '2-dimensional', and more precisely we know from experience that we can locally write the various hemispheres as graphs of functions of two variables. Alternatively, we know how to use spherical polar coordinates to parameterize a huge portion of the sphere etc. These 2-tuples of parameters now come from an open subset of $\Bbb{R}^2$, so by composing functions $f$ on the sphere, we now get functions $\tilde{f}$ defined on open subsets of $\Bbb{R}^2$, and it is here that we can do 'calculus as usual'.

The only subtlety (which is really a recurring theme) we have to keep track of is that our definitions/theorems do not depend on one specific choice of coordinates/parameters (e.g polar coordinates $(\theta,\phi)\mapsto (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ is no better than a graph-parametrization $(x,y)\mapsto (x,y,\sqrt{1-x^2-y^2})$ from a logical standpoint). This is where the definitions of charts and atlases comes from. More intuitively: suppose you're a tourist looking for directions. The map provided to you on an iPhone vs the map provided to you by Google are both good enough, they'll get you from point A to point B. Sure, the phones/apps might look and feel different but they're just as good (roughly speaking) in their ultimate objective of helping you navigate.

Of course, the next issue is what happens when your sets are not in any obvious manner given to you as embedded in some $\Bbb{R}^n$. This was discussed in the other answer, and other examples include things like the projective planes $\Bbb{RP}^n$, which are better thought of as quotients rather than subsets of $\Bbb{R}^{n+1}$. Well, this motivates the study of abstract manifolds. As a rough guideline, here's the setting of various topics:

• Linear Algebra: study vector spaces over fields, and linear maps between them
• Differential Calculus: given normed vector spaces $V,W$, an open subset $A\subset V$, and you study properties of functions $f:A\to W$. The key to studying such functions is to approximate $f$, near each point, by a linear map; this 'reduces' many problems down to linear algebra.
• Differential Manifolds: now you have smooth manifolds $M,N$, so you don't assume a vector space structure, and you're given functions $f:M\to N$. The main idea here is that you 'approximate' $M,N$ by the so called tangent spaces (i.e approximate non-linear spaces near each point by a linear space), and you approximate the non-linear $f$ by a linear map (the tangent map/pushforward between tangent spaces). This 'reduces' problems to differential calculus

As you can see, at each stage, we're generalizing one of the items, either the space, or the functions, going from linear in both to non-linear in both.