Meaning of the differential of a map between manifolds as an approximation

The differential $df_x$ of a map $f$ between two Euclidean spaces is the linear approximation which satisfies $f(x + v) = df_x(v) + o(|v|)$. The meaning of the word "approxiamtion" is clear here. I'm struggling to get what is the meaning of the differential of a map between two smooth manifolds as an approximation. In general there is no concept of a distance between two points in a manifold or a length of a vector in a tangent space of a manifold. Then how could we say about approximation precisely? In what sense does the tangent space at a point of a manifold locally approximate the manifold? In what sense does the differential of a map between smooth manifolds approximate the map?

Given $f: M \to N$, we use charts $(U, \phi)$ and $(V, \psi)$ for $U,V$ respectively such that $\phi(U) \cap V \not = \emptyset$. If you wish to do calculus in the traditional sense then we use $\tilde{f}: \phi(U) \subset \mathbb{R}^n \to \phi(V) \subset \mathbb{R}^m$ defined by,

$$\tilde{f} = \psi \circ f \circ \phi^{-1}$$

If you like, you could just write that $\tilde{f}$ is just $f$, how? Well, $\psi(\textbf{x})$ is just the coordinate representation of $\textbf{x}$ in $U$ and if we let $\psi = (y^1,...,y^m)$ then $\psi(f(\textbf{x})) = (y^1(\textbf{x}),...,y^m(\textbf{x}))$ is just the coordinate rep. of $f$ in $V$. Hence, if the use of a chart is understood then we just write,

$$f(x^1,...,x^n) = (y^1,...,y^m)$$

The moral is, in order to make sense of smoothness, we use charts. The idea behind smooth manifold theory is that if you always assume that you manifold is embedded in some ambient space then you're probably deducing things about your space as a result of being in another i.e you've lost focus of the primary object, your manifold.

To counteract this though, you consider your manifold as an abstract space with just a differentiable structure and a few other things, but you never mention a metric or anything of that kind. You now run into a problem if you want to do traditional calculus, but if you make the right identifications there may be a chance. The above is an illustration of some of these identifications.

• Thank you for the answer. However, I think that actually you didn't answer to my questions. I'm familiar with such coordinate thinking. But the coordinate representation has nothing to do with my questions, or I don't understand its implications on my questions. Would you please elaborate the answer to show more concrete connection to my questions?. Nov 15 '17 at 4:30

If you zoom in close enough on a smooth curve, it's a line. Likewise for a smooth manifold, we approximate it at each point by a tangent vector space. A map between two manifolds is then approximated by the map derivative acting between the two tangent spaces.

• So we approximate both the domain and range as well as the function itself? Jul 23 '19 at 10:17
• @user1, yes, and a linear map is only defined between vector spaces (modules) as one that preserves addition and scalar multiplication. An infinitesimal displacement on a smooth manifold is a vector in the tangent vector space and is mapped to a vector in the target manifold's tangent vector space.
– rych
Jul 23 '19 at 13:29

Consider the real line $\mathbb{R}$. It is a topological ring in a familiar way; you can add points, multiply points and so forth.

There is a nice algebraic description of its tangent bundle $T\mathbb{R}$: it is constructed like the complex numbers as the space of all pairs $a + b \epsilon$, but the multiplication is defined by $\epsilon^2 = 0$. This ring is called the (real) dual numbers, and is denoted $\mathbb{R}[\epsilon]$.

Since $\epsilon^2 = 0$ but $\epsilon \neq 0$, it captures some of our intuition of what it means to be infinitesimal. More precisely, we call it a "nilpotent infinitesimal".

For any smooth function $f : \mathbb{R} \to \mathbb{R}$, you have that its derivative is given by the formula

$$f_*(a + b \epsilon) = f(a) + f'(a) b \epsilon$$

which is formally of the same shape as the differential approximation, but even stronger: it is exactly true.

A similar algebraic picture applies to $\mathbb{R}^n$; its tangent bundle is $\mathbb{R}[\epsilon]^n$.

• How can I apply this idea to general smooth manifolds? Nov 15 '17 at 4:31