Naïve question about manifolds

Question

On this youtube video by XylyXylyX explaining curves on differentiable manifolds the following drawing is presented:

with $(X,\mathcal T_X,\mathcal A)$ denoting a topological space, i.e. $(X,\mathcal T_X),$ with Housdorff, second countability and paracompactness, and an atlas, $\mathcal A;$ $f(\lambda)$ plotting the real line to the manifold: $\mathbb R \to X$, so as to parametrize the line in black on the manifold; and $\gamma$ and $\phi$ representing charts $X\to \mathbb R^2$ (or $\mathbb R^d)$ for different chart regions $U$ and $V.$

It is clear how after we land safely in Euclidean space through $\gamma$ and $\phi$ we can apply calculus; however, and before we get there (or to change coordinates) we have to go through $X$. And if $X$ is not in Euclidean space,

What mathematical form does $f$ assume? It can't be $y = f(\lambda),$ which would imply coordinates.

Can I have an example (other than the sterographic world map)?

Answer 1

We define differentiability of maps between manifolds (and many other things) in coordinate representations because that is precisely the only way we can "concretely" deal with them.

If you want to explicitly give a curve $\gamma:\mathbb{R}\rightarrow M$ you will always give $(\varphi\circ\gamma)(t)=(x^1(t),...,x^n(t))$. If you want to explicitly give a function $f:M\rightarrow\mathbb{R}$, you will always give $(f\circ\varphi^{-1})(x^1,...,x^n)$.

The construction holds because differentiability (and other properties) do not depend on the coordinates. You see, if we only allow $C^{\infty}$ coordinate transformations, then if $\varphi$ and $\psi$ are both chart functions, then if, say, $f:M\rightarrow \mathbb{R}$ is a $C^k$ function, because $f\circ\varphi^{-1}$ is $C^k$, then $f\circ\psi^{-1}=f\circ\varphi^{-1}\circ\varphi\circ\psi^{-1}=(f\circ\varphi^{-1})\circ(\varphi\circ\psi^{-1})$, where the expression in the first parentheses is $C^k$ by our postulate (but it can be explicitly checked, as this is just an $\mathbb{R}^n\rightarrow \mathbb{R}$ function), and the expression in the second parentheses is a change-of-coordinates function, which is $C^{\infty}$ because both charts are part of the manifold's maximal smooth atlas.

Now, the composition of a $C^k$ function with a $C^{\infty}$ function is $C^{k}$, so if the coordinate representation $f\circ\varphi^{-1}$ is $C^{k}$ for any one chart, then the coordinate representation $f\circ\psi^{-1}$ will also be $C^{k}$ automatically.

The coordinate-free forms are there only to make 'theoretical' or 'abstract' statements. When you want to calculate things explicitly, the only way to do it is by working directly with the representations $f\circ\varphi^{-1}$ and the likes.

Answer 2

$f$ is a function that takes a real number and returns a point in $X$. If you don't have a concrete definition of $X$ (perhaps as a subset of $R^n$) and you don't use coordinates, then you can't write a formula for $f$, because the elements of $X$ are just abstract points.

Answer 3

However, sometimes there actually is an explicit function f. Consider a set of, say, for ordered numbers (x,y,z,w) such that the sum of their squares equals the constant A. That set is “M” in the diagram above. You can easily construct a function from [0,1] into that set of order quadruples. M, it turns out, is a manifold, and can be given coordinates (in an almost obvious way) and an atlas. “F” can be explicitly constructed and so can the chart mappings.

You forbade the stereographic projection as an example, that would be one way of constructing charts in this case. There are others however. Lot’s of others.

Regarding spacetime, we are dealing with the abstract idea of spacetime as the set M. That is why it is an axiom of GR that spacetime is a 4-D manifold. This assumption guarantees that we can do everything in terms of some set of coordinates. The next assumption is that the laws of nature do not depend on the choice of coordinates and that leads to all laws being in tensorial form. So the “actual set M of spacetime” is there to help us understand that we are modeling spacetime via the mathematics of a manifold.