# Why do we require manifolds to be a topological space?

Roughly speaking, we define a manifold $M$ to be covered by a set of charts $\{(U_i , \varphi_i)\}$ such that locally the $n$-dimensional manifolds looks like $\mathbb{R}^n$. One of the conditions is that all the $U_i$ are open sets of the topology of the manifold.

Why do we require the manifold to be a topological space? And why do we want $U_i$ to be open sets? What are the implications of these requirements on physics? (It appears to me that without these conditions the manifold still looks locally like $\mathbb{R}^n$.)

Edit to make my question more concrete: are there any physical theories that use manifolds that are not topological spaces? For instance, what would happen to general relativity if the spaces are diffeomorphic to each other, instead of homeomorphic (see answer below by Robin Ekman)?

• ...because it's meaningless to discuss continuous functions without topology, and calculus requires at least having continuity... Mar 30, 2014 at 23:59
• If you didn't require the manifold $M$ to be a topological space so that you have the notion of homeomorphism (topological equivalence) then in what sense would the set you're talking about still locally "look like" $\mathbb R^n$? Also, note that the notion you're referring to is that of "topological manifold." The concept of a manifold is rather more general in mathematics as a whole: en.wikipedia.org/wiki/… Mar 31, 2014 at 0:03
• @joshphysics I wasn't aware of that. Thanks for the link. Mar 31, 2014 at 0:10
• Just two comments: a) Doesn't a metric induce a topology anyway? If you want to have conventional spacetime, there will necessarily be some topology. b) There are doing physics in topoi, e.g. this, in which the notion of openness is a little broader (but not for manifolds, it doesn't start out with spacetime.) Mar 31, 2014 at 8:55
• What is $U$? Anyway, the Wikipedia article Metric space says in the fourth section how every metric space induces a topology. / The work I liked to is a little long. The main idea is to use topoi, which are frameworks which encompass the set paradigm, but generally don't need to play the same game/follow the same logical rules. This is related. And just for curiosity, lets me mention that the discussion in this question reminded me of Exotic R4. Mar 31, 2014 at 12:48

You have many comments to the effect that "topology is needed to describe continuity, calculus concepts, the notion of "looks like", homeomorphism and so forth". And these are all altogether right, but I'm getting that your question is about the global picture. Also, the following is mainly about a toplological or differentiable manifold; Joshphysics's link shows that there are many other concepts of manifold.

We begin with the notion of "locally looks like $\mathbb{R}^N$"; but you can have a set $\mathbb{M}$ of any kind of weird creatures whose subsets you can put into one-to-one, surjective correspondence with some open (more about this below) subset of $\mathbb{R}$ (wontedly a simply connected neighbourhood of the origin). For one of these subsets $\mathcal{N}$ you have a "labeller" map $\lambda:\mathcal{N}\to\mathbb{R}^N$. Then you notions of open, neighbourhood and all the rest of it arise by definition: a subset $\mathcal{O}\subset\mathcal{N}$ is open iff $\lambda(\mathcal{O})$ is open in $\mathbb{R}^N$. Likewise a "path" $\sigma:\mathbb{R}\to\mathcal{N}$ is $C^0,\,C^1\, C^\omega$ or whatever iff $\lambda\circ \sigma:\mathbb{R}\to\mathbb{R}^N$ has the same property. All topology, neighbourhood, calculus, differentiability and so forth concepts are then defined by "fiat", and the need for the concepts is why we want our zoo of weird creatures to "locally look like $\mathbb{R}^N$" in the first place so this is all highly intuitive and obvious.

So I'm guessing (also by reading your other probing questions on this site) that you already understand all this. So the crucial question is then that of transistion maps and how we glue all our local copies of $\mathbb{R}^N$ together. Going back to our subset $\mathcal{N}\subset\mathbb{M}$:there are other "local copies" of $\mathbb{R}^N$ that bestow our topological / calculus and so forth concepts on subsets of $\mathbb{M}$ other than $\mathcal{N}$. But these subsets must overlap, because, when we're doing calculus or toplogy or dynamics or whatever, we don't want suddenly to run into a "co-ordinate wall" and have to jump suddenly from one co-ordinate system to another. As an example, suppose we have a spacecraft in an Einstein manifold (universe that is a vacuum solution to the EFEs). For calculus, measurement and other mathematical concepts, we need always to be able to define the manifold in a neighbourhood around the spaceship: so, as the spaceship nears the boundary of one co-ordinate system, it must also be describable by another co-ordinate system wherein we can carve out a "neighbourhood": we couldn't do this if our co-ordinate systems didn't overlap but instead partitioned the manifold $\mathbb{M}$. Otherwise put, in relativity, the boundary between co-ordinate systems is an artifact of our particular mathematical description of the physics, it does not belong to the physics. Another, dramatic, example is the phenomenon of gimbal lock in Euler angle charts for the unit sphere that very nighly cost the Apollo 11 astronauts their lives, cost many pilots their lives in the years before then and is the reason why the software processing signals from fibre ring Sagnac gyros that keep you safe in a commercial jetliner either manipulate the aeroplane's calculated orientation in two overlapping charts covering $SO(3)$ or, more recently, model the aeroplane's orientation by unit quaternions in $SO(3)$'s double cover $SU(2)$.

So, our overlap is very much needed, so many, if not all, regions in the manifold can be described by more than one local copy of $\mathbb{R}^N$ with more than one labeller. So, suppose we have two regions $\mathcal{N}_1,\,\mathcal{N}_2$ with labellers $\lambda_1:\mathcal{N}_1\to\mathbb{R}^N$, $\lambda_2:\mathcal{N}_1\to\mathbb{R}^N$: we must make sure that these labellers yield consistent notions of opennes, neighbourhood, differentiability and all the rest of it in a region $\mathcal{N}_1\cap\mathcal{N}_2$. So, a set $\mathcal{O}\subseteq\mathcal{N}_1\cap\mathcal{N}_2$ must be open as reckonned by labeller $\lambda_1$ and $\lambda_2$ and so $\lambda_1\circ\lambda_2^{-1}$ and $\lambda_2\circ\lambda_1^{-1}$, the "transition maps" between charts, must be local homeomorphisms, analytic, diffeomorphisms, or whatever the relevant notion is for the kind of manifold in question. Likewise for all other calculus and topological concepts we wish to speak of. This is most readily achieved if the charts (ranges of the labellers $\lambda_j$) are open, and their intersections are open as reckonned by all local copies of $\mathbb{R}^N$ that are applicable to the overlap. So we have two axioms for manifolds further to the obvious one that every point in the manifold must belong to the preimage of at least one labeller:

1. An intersection between two "patches" (domains of labellers) must be open in the topology as reckonned by each of the two labellers for the overlapping charts;

2. The transition maps must be local homeomorphisms, diffeomorphisms, ....

3. Some authors also add the axiom that the manifold should be Hausdorff ($T_2$) in each chart but in many fields, notably Lie groups, $T_2$ is enforced by other structure (the group laws) so this axiom is redundant here.

The easiest way to do this is to kit the manifold globally with a topology whose base is the open sets as reckonned by their images under the labellers, or, written backwards, the base for the topology is the collection of all preimages of sets open in $\mathbb{R}^N$ under the labellers.

Hopefully you can see that the notion of consistency as reckoned by overlapping charts, and thus the notion of the global manifold topology, is very much bound up in the physical concept of covariance and the Copernican notion that Nature's behaviour cannot depend on our mere description of it.

• Thanks! This is exactly the answer I was looking for. I needed to make a connection between the maths and the physics, and what you've written makes perfect sense. Mar 31, 2014 at 2:10
• Hello Mr Vance, I was wondering whether you could help me with this question: physics.stackexchange.com/q/215764 whenever time allows. Cheers Oct 31, 2015 at 18:18

Strictly speaking, if there can be defined charts covering a set (an atlas), you can give that set the topology induced by defining the charts to be bicontinous. That is, a set is open iff it's the domain of a chart in the maximal atlas.

If your set already has a topology, the topology induced by the atlas will agree with that one under some conditions. (I think it's Haussdorff and second countable but I have to check this.)

Why would you want a manifold to have a topology? Well, you want to say that on a set small enough to be covered by a chart the manifold looks like Euclidean space. But then you have to define what "looks like" should mean. Being homeomorphic is one possible choice. Being diffeomorphic is another.

• Ok, thanks, that makes sense. In physics we (always?) assume that the manifolds we are working with have a homeomorphism from the $U_i$ to $\mathbb{R}^n$. Are there also physical theories that don't have this condition? What would happen to, say, general relativity if the spaces are diffeomorphic to each other, instead of homeomorphic? Mar 31, 2014 at 0:13
• Actually in physics we usually require diffeomorphism, because we want to talk about derivatives. So we require that the change-of-coordinate functions are smooth. Then we define $f : M\to \mathbb R$ to be smooth if $f \circ \varphi^{-1}$ is smooth. Here $\varphi : M \to \mathbb R^n$ is a coordinate function. A diffeomorphism is a homeomorpism that is smooth. With this definition the manifold becomes locally diffeomorphic to Euclidean space. Note that we had to define what it means to be differentiable on a manifold using the coordinate functions. Mar 31, 2014 at 0:30