# Why is a real manifold (used in GR) defined in the way it is?

In this question one can read, in defining a real manifold $$M$$, the following condition must hold:

$$(1)$$ Each $$p\in M$$ lies in at least one $$O_\alpha$$, i.e. $$\{O_\alpha\}$$ cover $$M$$.

$$(2)$$ For each $$\alpha$$, there is a one-to-one, onto map $$\psi_\alpha: O_\alpha\to U_\alpha$$, where $$U_\alpha$$ is an open subset in $$\mathbb{R}^n$$.

($$3$$)If any two sets $$O_\alpha$$ and $$O_\beta$$ overlap, $$O_\alpha\bigcap O_\beta\neq \emptyset$$, we can consider the map $$\psi_\beta\circ\psi_\alpha^{-1}$$ which takes points in $$\psi_\alpha[O_\alpha\bigcap O_\beta]\subset U_\alpha$$ in $$\mathbb{R}^n$$ to points in $$\psi_\beta[O_\alpha\bigcap O_\beta]\subset U_\beta$$ in $$\mathbb{R}^n$$. We require these subsets to be open and this map to be $$C^\infty$$.

The sets $$\{O_\alpha\}$$ are open subsets of a continuous collection of (non-equidistant) points which are represented by a column (or row) of $$n$$ real numbers. The intersection of two open subsets, $$O_\alpha\bigcap O_\beta\neq \emptyset$$, is open too.

Condition $$(3)$$ implies that different $$\psi_\alpha$$'s acting on the same open subset transform this subset into different subsets of $$\mathbb{R}^n$$. That's clear. It is required that these different subsets of $$\mathbb{R}^n$$ are open. The images of open sets can be open or closed tough.

Why is it so important, in stating the conditions for a collection of real points to form a real manifold, that all subsets, {$$O_\alpha$$}, {$$U_\alpha$$}, {$$\psi_\alpha[O_\alpha\bigcap O_\beta]\subset U_\alpha$$}, and {$$\psi_\beta[O_\alpha\bigcap O_\beta]\subset U_\beta$$}, are open? (Doesn't, by the way, condition $$(2)$$ imply condition $$(3)$$?)

General relativity requires the existence of real (curved) manifolds. What's so important about the condition that the different subsets should be open? If you use a manifold without this condition (so instead of using open subsets in the definition of the real manifold above you use closed subsets), will this give another general relativity? Isn't the same structure of a real manifold obtained if the subsets are closed? Does the metric (associated with the $$\psi_\alpha$$'s) become discontinuous in that case? Or what?

• Where in what you quote does it say anything about closed sets? Commented Apr 30, 2021 at 12:57
• @jacob1729 Doesn't [...] mean that the set is closed? Commented Apr 30, 2021 at 13:04
• @DescheleSchilder The [...] here looks like it's just a grouping symbol. In the original question, (2) implies that the sets $O_\alpha$ are open. The intersection of open sets is also open. The condition (3) is about the images of those open sets. For arbitrary functions the image wouldn't have to be open, but this condition (3) places a restriction on the functions. Commented Apr 30, 2021 at 13:45
• @Brick I replaced the closed subsets by open subsets. Commented Apr 30, 2021 at 14:22
• @DescheleSchilder it turns out that open sets in some ways have vastly more structure than closed sets: in particular every point of an open set has an open neighborhood, but there isn't always a similarly graceful way to characterize closed sets.
– TLDR
Commented Apr 30, 2021 at 15:09

The images of open sets can be open or closed tough.

Not under homeomorphisms. The image $$\psi(O)$$ of an open set $$O$$ under a homeomorphism $$\psi$$ is also open. One can see this immediately by noting that $$\psi(O)$$ is the preimage of $$O$$ under the map $$\psi^{-1}$$, and since $$\psi^{-1}$$ is continuous, $$O$$ must be open.

Of course, open and closed are not mutually exclusive options, so the $$\psi(O)$$ might be closed as well, provided that it is also open.

If you use a manifold without this condition (so instead of using open subsets in the definition of the real manifold above you use closed subsets), will this give another general relativity?

No, it would give a nonsensical definition of a manifold - or at least a definition which doesn't capture our motivating desire to construct spaces which look like patches of $$\mathbb R^n$$ if you zoom in closely enough.

Consider $$\mathbb R^3$$ as a manifold with the standard topology of open sets defined in terms of open balls. Consider the origin $$p=(0,0,0)$$.

• The singleton set $$\{p\}$$ is a closed set containing $$p$$ which is homeomorphic to $$\mathbb R^0$$
• The line segment $$U_1 = \{ (x,0,0) \ | \ x\in [-1,1]\}$$ is a closed set containing $$p$$ which is homeomorphic to a subset of $$\mathbb R^1$$.
• The closed disk $$U_2 = \{ (x,y,0) \ | \ x^2+y^2\leq 1\}$$ is a closed set containing $$p$$ which is homeomorphic to a subset of $$\mathbb R^2$$.
• The closed ball $$U_3 = \{ (x,y,z) \ | \ x^2+y^2+z^2 \leq 1\}$$ is a closed set containing $$p$$ which is homeomorphic to a subset of $$\mathbb R^3$$.

So you see, talking about closed sets doesn't give us what we want. If we require an open neighborhood to be homeomorphic to some $$\mathbb R^n$$, then it's not hard to see that the only possibility is $$n=3$$.

Doesn't, by the way, condition (2) imply condition (3)?

No. You can have two different charts such that the chart transition map isn't smooth. Consider the manifold $$\mathcal M=\mathbb R$$ and the charts $$\psi_1:x\in \mathcal M \mapsto x$$ and $$\psi_2: x\in \mathcal M \mapsto x^3$$. Both of those are perfectly viable charts, but the transition map $$\psi_1 \circ \psi_2^{-1} : x \mapsto x^{1/3}$$ is not differentiable at $$x=0$$.

• Very enlightening! Thanks. Especially, "No. It would give a nonsensical definition of a manifold" (which was somehow on the back of my mind). But also the rest of the answer is great!. Commented May 1, 2021 at 10:21