General Relativity/Differentiable manifolds

<General Relativity

A smooth $$n$$-dimensional manifold $$\mathrm{M}^n$$ is a set together with a collection of subsets $$\{O_\alpha\}$$ with the following properties:


 * 1) Each $$p\in\mathrm{M}$$ lies in at least one $$O_\alpha$$, that is $$ \mathrm{M}=\cup_\alpha O_\alpha$$.
 * 2) For each $$\alpha$$, there is a bijection $$\psi_\alpha:O_\alpha\longrightarrow U_\alpha$$, where  $$U_\alpha$$ is an open subset of $$\mathbb{R}^n$$
 * 3) If $$O_\alpha\cap O_\beta$$ is non-empty, then the map $$\psi_\alpha\circ\psi_\beta^{-1}:\psi_\beta[O_\alpha\cap O_\beta]\longrightarrow\psi_\alpha[O_\alpha\cap O_\beta]$$ is smooth.

The bijections are called charts or coordinate systems. The collection of charts is called an atlas. The atlas induces a topology on M such that the charts are continuous. The domains $$O_\alpha$$ of the charts are called coordinate regions.

Examples

 * Euclidean space, $$\mathbb{R}^n$$ with a single chart ($$O=\mathbb{R}^n,\psi=$$ identity map) is a trivial example of a manifold.
 * 2-sphere $$ S^2 = \{ (x,y,z) \in \mathbb{R}^3 | x^2 + y^2 + z^2 = 1 \}$$.
 * Notice that $$S^2$$ is not an open subset of $$\mathbb{R}^3$$. The identity map on $$\mathbb{R}^3$$ restricted to $$S^2$$ does not satisfy the requirements of a chart since its range is not open in $$\mathbb{R}^3.$$
 * The usual spherical coordinates map $$S^2$$ to a region in $$\mathbb{R}^2$$, but again the range is not open in $$\mathbb{R}^2.$$ Instead, one can define two charts each defined on a subset of $$S^2$$ that omits a half-circle. If these two half-circles do not intersect, the union of the domains of the two charts is all of $$S^2$$. With these two charts, $$S^2$$ becomes a 2-dimensional manifold. It can be shown that no single chart can possibly cover all of $$S^2$$ if the topology of $$S^2$$ is to be the usual one.