1
$\begingroup$

Context:

In Lecture 1 of the Central Lecture Course, Professor Schuller begins with the so-called "physical key definition" of space-time:

Spacetime is a four-dimensional topological manifold with a smooth atlas carrying a torsion-free connection compatible with the Lorentzian metric and a time orientation satisfying the Einstein equations.

At the coarsest level, "spacetime" is a "set". Not enough to talk about continuity of maps. Why do we as physicists want to talk about continuity of maps? Well, in classical physics we have the idea that curves do not "jump".

Weakest structure that can be established on a set which allows a definition of continuity of maps is called a topology.

Not all the first terms in the first paragraph are clear to me yet. However, this question thread is solely based on what I understood (and didn't understand) from the first lecture.


As far as I understand:

  1. If $M$ is a set, a topology $\mathcal{O}$ is a subset $\mathcal{O}\subseteq \mathcal{P}(M)$ satisfying

    (i) $\phi\in \mathcal{O}$

    (ii) $u,v\in\mathcal{O}\implies u\cap v \in \mathcal{O}$

    (iii) $u_{\alpha}\in \mathcal{O} \implies (\bigcup_{\alpha\in A}u_{\alpha}) \in \mathcal{O}$ where $\alpha$ comes from an arbitrary (not necessarily countable) index set.

  2. We can define "continuous maps" between two sets $M$ (equipped with a topology $\mathcal{O}_M$) and $N$ (equipped with a topology $\mathcal{O}_N$).

    A map $f:(M,\mathcal{O}_M)\to (N,\mathcal{O}_N)$ is called continuous if

$$\forall \ V \in \mathcal{O}_N: \text{preim}_f(V) \in \mathcal{O}_M$$

Now, the problem is that I'm not being able to relate "there cannot be jumps in classical physics" to "spacetime is a set with a certain topology".


Questions:

  1. What exactly is meant by "there cannot be jumps in classical physics"? Are they simply referring to the fact that particles in classical mechanics follow "continuous trajectories"? Or do are they referring to something more fundamental to classical mechanics? By the way, I'm also not completely sure that "trajectories are continuous" is an absolute necessity in classical mechanics.

  2. To define a continuous map (in this context), we will need two sets equipped with their own topologies. If spacetime is considered to be a set with its own topology, what is the other set (to which continuous maps can be defined)? (As of now) it sort of seems meaningless to me when they say that spacetime is continuous in general relativity without mentioning what the other set (equipped with a certain topology) is.

  3. If the answer to point (1) is "yes", then how exactly is the fact that "trajectories in classical mechanics are continuous" related to the fact that "spacetime is continuous", and in turn how is that related to the fact that "continuous maps can be defined between two sets equipped with their own topologies". Basically, I'm not being able to connect these three notions. At least, not mathematically.

  4. How are we certain that the weakest structure on a set, which allows for continuity of maps is a topology? Is it a mathematical fact? If yes, could someone point me out a proof and/or a reference which discusses this?

$\endgroup$
  • $\begingroup$ Complicated stuff. Fortunately it is a purely mathematical subject. $\endgroup$ – my2cts Jun 25 '18 at 19:30
  • 1
    $\begingroup$ Sorry, unhelpful comment. I am just flashing on the "torsion free" part. When I was a PhD student in the physics department in which John Moffat was teaching, we got a copy of MACSYMA to play with. And at the point where you enter the torsion part of the metric, if it was non-zero, it said "You must be working on a new theory of gravity... Which is not supported by this package." And then it exited. $\endgroup$ – user93146 Jun 25 '18 at 20:16
  • $\begingroup$ Basically, there are two ways of viewing such things. There is the math way, which basically says you lay down some postulates and then only claim what you can prove. Then there is the physics way, which basically says, assume the math you need, unless there is some good reason that the math is wrong. So the "weakest structure" might be the weakest structure from either point of view. Maybe it can be proved from some postulates. Or, maybe it means the weakest structure that gets the physics answered desired. Not clear which applies. $\endgroup$ – user93146 Jun 25 '18 at 20:19
  • $\begingroup$ I don't currently feel 100% confident in answers to the first three points, but for question (4), a topology is how continuity is defined. Without a topology, you have no notion of continuity, because you have no notion of open sets, and thus no notion of neighborhoods of a point. If nothing else, the fact that topology is the study of continuous functions should be a tip off that it's the weakest structure allowing continuity, because otherwise there would be some other structure people would study. $\endgroup$ – Jared Dziurgot Jun 25 '18 at 20:23
  • $\begingroup$ IMO the restriction to classical physics is not relevant at all with respect to continuity. Quantum mechanics is continuous as well. The time evolution of the system is given by a unitary transformation that is a continuous function of time. What is probably more relevant about classical vs quantum here is that your professor is starting with a depiction of spacetime as a 3+1-dimensional manifold, whereas in quantum mechanics time is a parameter, and is not treated in the same way as the spatial coordinates. $\endgroup$ – Ben Crowell Jun 26 '18 at 1:33
3
$\begingroup$

What follows is a very rough sketch which I hope may help. See the end for some pointers.

The trick here is that, as you say, to define continuous maps you need two sets with topologies. And we then choose one of these sets to be a set we understand very well with a topology we understand very well: the set $\mathbb{R}^n$ with the usual topology.

The usual topology on $\mathbb{R}^n$

The easiest way to define the usual topology on $\mathbb{R}^n$ is to do it backwards:

  1. for any $x = (x_1,\ldots,x_n), y = (y_1,\ldots,y_n)\in\mathbb{R}^n$ define $d(x,y) = \sqrt{(x_1 - y_1)^2 + \cdots + (x_n - y_n)^2}$ -- this is the normal euclidean distance function;
  2. for any point $p\in\mathbb{R}^n$, define an open neighbourhood of radius $r > 0$ from $p$ as $N_r(p) = \left\{q\in\mathbb{R}^n: d(p,q) < r\right\}$
  3. now define the open sets of the topology as $\emptyset$, all the $N_r(p)$ for all $p$, their unions & finite intersections and $\mathbb{R}^n$ itself.

It's pretty easy to show that this:

  • is a topology;
  • is the one you'd expect for $\mathbb{R}$ ($d$-neighbourhoods are just open intervals);
  • is not very sensitive to the details of $d$ -- the topology is not dependent on the particular distance function and many distance functions will give the same topology.

It's also possible to show that for this topology the topological definition of continuity is exactly the same as the traditional $\epsilon$-$\delta$ one. So the continuous mappings using this topology between, say $\mathbb{R}$ and $\mathbb{R}$ are just what we would hope they would be, and all our intuitions of what it means for functions on the reals to be continuous work.

Topological manifolds

So, then the trick is that we define a topological manifold to be a set $M$ with a topology, and the the requirement that each point of $M$ has an open neighbourhood which has a 1-1, continuous, map onto an open neighbourhood of $\mathbb{R}^n$.

So long as we use the usual topology on $\mathbb{R}^n$ this really means that $M$ is 'locally like' $\mathbb{R}^n$ with the usual topology. And this in turn means that our intuitions about continuous functions on the reals also work in $M$, at least locally.

Note that there is no requirement for there to be a single, global, 1-1, continuous map between $M$ and $\mathbb{R}^n$, and there often is not (for instance the 2-sphere has no such single map). But thanks to the nature of open sets in the usual topology the maps will end up with overlaps (or it's possible to define them so that there are always overlaps by shifting them around in obvious ways).

An example: the unit circle

This is a simple example where you can easily visualise what's going on. The unit circle, which I'll call $S$, is $S = \{(x,y)\in\mathbb{R}^2: \sqrt{x^2 + y^2} = 1\}$. This is a one-dimensional manifold, so it wants to be mapped onto (open sets of) $\mathbb{R}$.

The first obvious trick is to say that $(x, y) = (\cos\theta, \sin\theta), \theta\in\mathbb{R}$. This makes it easy to talk about the topology we want on $S$: it's just the usual topology on $\mathbb{R}$: for some $\theta_0$ the neighbourhoods of it are $(\theta_0 - \epsilon, \theta_0 + \epsilon), \epsilon > 0, \epsilon < 2\pi$. Note I am now using $(a,b)$ to mean $\{x: x > a, x < b\}$ & similarly for $[a,b]$, $(a, b]$ & so on, rather than to mean a tuple, and note also the $\epsilon < 2\pi$ condition which avoids us looping around $S$.

So, we now need some 1-1, continuous mappings between $S$ & open subsets of $\mathbb{R}$. It's tempting to just pick a single one, $m(\theta) = \theta$. This is continuous but it's not 1-1, because $\theta + 2n\pi, n\in\mathbb{N}$ is the same point as $\theta$. Well, we could limit the range of $\theta$ to be $(0, 2\pi)$ but this interval doesn't include $0$ and so does not map the whole of $S$. We can map the whole of $S$ by having $\theta\in[0, 2\pi)$, but this isn't an open set of $\mathbb{R}$ in the usual topology.

So, it turns out we need at least two mappings. There's a lot of freedom in choosing them, but for instance

  • $m_1(\theta) = \theta, \theta\in(0, 3\pi/2)$
  • $m_2(\theta) = \theta, \theta\in(\pi, 5\pi/2)$

Note that these overlap: in $(pi, 3\pi/2)$, $m_1(\theta) = m_2(\theta)$, and in $(0, \pi/2)$, $m_1(\theta) = m_2(\theta - 2\pi)$. And between them they map all the points of $S$ in a 1-1, continuous way, to open subsets of $\mathbb{R}$.

It should be obvious that we need more than one mapping: although $S$ locally looks like $\mathbb{R}$, it doesn't look globally like $\mathbb{R}$ at all. And that's part of the point of the whole topological manifold thing: we want to be able to talk about objects which locally are like $\mathbb{R}^n$ but which globally may not be like it at all.

A note on overlaps. Above I said that 'open sets in the usual topology always end up with overlaps'. What this means is easy to see by considering the intersection of two open intervals of $\mathbb{R}$: $(a,b) \cap (c,d)$. This set is either empty, if $c \ge b$, or it contains more than one point, if $c < b$. It never contains just a single point. This is a property of the usual topology on $\mathbb{R}$: it's not true for all topologies, but it is true for the usual topology. As an example it is not true for the discrete topology where all sets are open, because then $\{x\} \cap \{x\}$ is a single point, and $\{x\}$ is open in this topology: that's why the discrete topology is useless. I don't know the name of this condition: it might be the Hausdorff condition or an implication of it, but I'm not sure.

So long as we stick to a topology where this is true then we can be sure that if we have maps for all of a manifold, then those maps will have overlaps, and this means that we can work our way between them using the overlaps. And that means that there are no 'bad places' in the manifold where continuity breaks down: if there are going to be such bad places then we need to cut them out of the manifold somehow (such bad places are singularities, of course).

The questions

  1. What is meant by 'there cannot be jumps' is that, for instance, trajectories of particles in the manifold are continuous, which translates directly to trajectories in the images of the maps between the manifold and $\mathbb{R}^n$ being continuous in the way we expect from basic analysis. There are, in fact, no jumps.
  2. You don't in fact need two sets, or rather the two sets can be copies of each other. However the trick sketched above of defining a manifold by requiring there to be well-behaved mappings between it and $\mathbb{R}^n$ means that you always do, in fact, have two sets, and you can bootstrap things like continuity from elementary notions of continuity on the reals.
  3. The manifold is continuous because it allows these continous mappings. If it was not continous then these mappings would not exist at all.
  4. I think this is backwards: defining a topology on a set defines what 'continous' means: if you define a different topology (there are various more-or-less pathological ones such as the discrete topology (all sets are open) and the trivial topology (only the whole set and the empty set are open)) you will get a different notion of continuity.

Pointers

  • I have always liked Geometrical methods of mathematical physics by Schutz, which gives a much better version of the sketch I've given here. It was the first book I read which made me understand what topology actually is and why it's useful.
  • Analysis, manifolds and physics by Choquet-Bruhat, deWitt-Morette with Dillard-Bleick is a much more serious approach which starts with a fairly fierce overview of topology and analysis. I used to use this as a reference.

There are certainly more modern references and certainly other equally old and perhaps better references: these are just the books I used before about 1990.

$\endgroup$
  • $\begingroup$ Thanks for the answer. "But thanks to the nature of open sets the maps will end up with overlaps" <--- what do you mean by overlaps, in this context? It would be nice if you could add some illustrations to convey your point regarding the 2-sphere. $\endgroup$ – Blue Jun 26 '18 at 5:58
  • $\begingroup$ @Blue: I am working on an example which I hope should make it clearer. I'd like to add pictures but I don't have any suitable drawing thing (and you really don't want to look at my hand-drawings!) $\endgroup$ – tfb Jun 26 '18 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.