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This is something that's been bugging me for a while now, I understand that one of the most important facts about differential geometry is that it doesn't require embeddings into higher dimensional spaces. My problem is how we are then able to meaningfully define coordinates without making reference to an external coordinate system in which our manifold is embedded.

For example, when we define a coordinate chart on the circle, we can use the projection onto the axes as: $$(x,y)\mapsto x\quad (x,y)\mapsto y \tag{1}.$$ This basically takes a circle centred at the origin and provides a four separate coordinate charts which cover the circle. This is fine, but it seems to require us to specify the $(x,y)$ coordinates in the first place. I've seen a similar construction where we begin with the polar angle $\theta$ and use this to define the coordinates in a similar way. But this again seems to require us to have the circle originally embedded in $\Bbb R^2$?

How can we provide coordinates for a simple manifold like the circle without referencing an embedding?

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  • $\begingroup$ It's a little unclear what you are asking. Would it answer your question if I give an example of a manifold which is not defined by any embedding into any $\mathbb R^n$? One such class of examples are the projective spaces $\mathbb R P^n$. $\endgroup$
    – Lee Mosher
    Dec 12 '20 at 17:08
  • $\begingroup$ I can recommend an excellent introductory to the subject, Manfredo Carmo: Differential Geometry of Curves and Surfaces. In chapter 4. it introduces the intrinsic geometry of the surface and show how computations can be made without “leaving” the surface. In section 5.10, it defines the abstract surface (differentiable manifold of dimension 2) and the geometric surface (Riemannian manifold of dimension 2). $\endgroup$
    – Zsombor
    Dec 12 '20 at 17:14
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The issue here is that a circle is most naturally a subspace of $\mathbb{R}^n$, usually $\mathbb{R}^2$.

But we can choose other ways to think about a circle that don't describe it as an embedded submanifold of $\mathbb{R}^n$. For example, $[0,1]$ with the points $0$ and $1$ identified is a circle. The quotient $\mathbb{R}/\mathbb{Z}$ is also a circle. Each description of a circle has its uses, but we usually choose the embedding into $\mathbb{R}^2$ for visualization purposes.

The moral is that some manifolds in some contexts most naturally present as an embedded submanifold of $\mathbb{R}^n$, but they also exist as abstract manifolds (a topological space plus smoothly compatible charts).

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Perhaps an example might help.

Consider the definition of projective space $\mathbb R P^n$, which is the quotient space of $\mathbb R^{n+1} - \{0\}$ under the equivalence relation $p\sim q$ if there exists $r \ne 0$ such that $p=rq$. This space is not given to us by an embedding in any Euclidean space.

Nonetheless, we can work out coordinate charts for this space, and verify that the overlap maps are smooth, and so this is a manifold. These coordinate charts are as follows.

Let $[p] \in \mathbb RP^n$ be the equivalence class of $p = (p_0,...,p_n) \in \mathbb R^{n+1} - \{0\}$.

For $i=0,...,n$ let $U_i = \{[p] \mid p_i \ne 0\} \subset \mathbb R P^n$. Then the function $\phi_i : U_i \to \mathbb R^n$ defined by $\phi_i[p] = \frac{1}{|p_i|}(p_0,...,p_{i-1},p_{i+1},...,p_n)$ is well-defined, and it is a coordinate chart for $\mathbb R P^n$. Taking these all together we get an atlas of coordinate charts $\{\phi_i : U_i \to \mathbb R^n \mid 0 \le i \le n\}$.

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I think the problem here is that we are so used to visualizing the circle as a geometric object that we consider it (intuitively, and our intuition is correct) embedded in $\mathbb R^2.$ But, if we simply define $M=\{(x,y):x^2+y^2=1\}$ where we give $M$ the subspace topology of $\mathbb R^2$, then, with the charts as in your post, $M$ becomes a bonafide manifold in the abstract sense. The fact that we already know inuitively that $M$ is locally homeomorphic to $\mathbb R$ is something that the abstract definition of $M$ is designed to make rigorous. And the fact that $M$ is a $\textit{subset}$ of $\mathbb R^2$ is incidental. An example of an abstract manifold for which geometric intuition at least for me, does not work, wold be a Lie group, $\mathrm{GL}_n(\mathbb{R})$ for example.

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A smooth manifold $M$ of dimension $d$ is locally homeomorphic to and open subset of $\mathbb{R}^d$, if $\varphi$ is a homeomorphism of a connected open set $U\subset M$ onto an open subset of $\mathbb{R}^d$ then $x_i = r_i\circ \varphi$ is called a coordinate function, and together $(U,\varphi)$ a coordinate system (here $r_i(a) = a_i$ for $a\in\mathbb{R}^d$). A smooth manifold inherits a differentiable structure $F = \{ (U_\alpha, \varphi_\alpha): \alpha\in A\}$ which is a collection of coordinate systems satisfying \begin{align*} a)& \bigcup_{\alpha\in A}U_\alpha = M\\ b)& \varphi_\alpha \circ \varphi_{\beta}^{-1} \text{ is smooth for all }\alpha, \beta \in A\\ c)& F\text{ is maximal with respect to $b)$; if $(U,\varphi)$ is a coordinate system such that }\varphi \circ \varphi_{\alpha}^{-1}\text{ and }\\& \varphi_{\alpha} \circ \varphi^{-1} \text{ are smooth for all $\alpha\in A$ then }(U,\varphi)\in F. \end{align*} This is paraphrased from Frank Warner's Foundations of Differentiable Manifolds and Lie Groups ($F$ typically has some script font, but I couldn't find an appropriate one). Here I use smooth to indicate infinite differentiability. Essentially instead of embedding the manifold in some other space, just use it's local Euclidean properties.

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