# Inferring properties of spacetime manifolds from those of chart maps

This question is related to topological manifolds, as discussed in one of Frederic Schuller's lectures on gravity : https://www.youtube.com/watch?v=93f-ayezCqE.

A bit of background: Briefly, to study spacetime, we assign it the structure of a topological manifold $M$ because one can then map its open subsets to open subsets of $\mathbb{R}^d$, via chart maps. One can then infer the continuity of the "real" curve in spacetime (say, $\gamma : \mathbb{R} \to M$) from the continuity of the curve (that it's mapped to) in $\mathbb{R}^d$. More generally, one can infer the properties of spacetime by looking at the properties of its chart maps (atlas).

The lecturer stresses that while the curve in the manifold is a "physically real thing", the curve in $\mathbb{R}^d$ isn't, since it depends on our choice of the chart map, which in turn can be arbitrary. Therefore, we are concerned with those properties of $\mathbb{R}^d$ curves that are independent of the choice of the chart map, and we can only talk about such properties being applicable to $\gamma$. Continuity happens to be a chart-independent property and so we can talk about "continuity of $\gamma$".

Question: Doesn't the notion of the continuity of spacetime curves itself depend on our choice of the topology on $M$? Does that not rule out continuity as a "physically real property" of a "physically real" curve, since it's not independent of the choice of topology? If that is indeed the case, why go through all the effort of inferring the continuity of $\gamma$, instead of being satisfied with the continuity of chart map ($\mathbb{R}^d$) curves?

Cartesian spaces are real spaces, so a curve in it is a real curve. Now, we understand cartesian spaces very well, so we define topological manifolds in terms of it. This is why we use charts. We are defining an unknown something in terms of something well-known.

First, going by the question you're confused by open sets. A curve in 2d space is not an open set, but an open disk is; however in 3d space, an open disk is not open, but an open 3d ball is. And so on.

A manifold is defined by its charts; these are maps from a manifold to cartesian space; thus a curve in the manifold can be mapped to many different curves in the charts. This is why we see a curve in a manifold as a 'real' thing and it's image in a chart as a representation of that curve - so less 'real'.

Now the topology of cartesian space is very well understood; and what we are doing by these charts and their compatibility conditions is to transport the topology of cartesian spaces to the manifold.

This is why a manifold is understood as a topological space that is locally like a cartesian space.

Question: doesn't the notion of the continuity of spacetime curves itself depend on our choice of a topology on the manifold

The situation is exactly the reverse. There is no topology on the manifold. It's just a set of points; and we use the charts to transport the topology on the charts to manifold making sure that they are compatible with each other.

• But the way it's defined in the video lecture (lecture 1 of the same series), continuity should depend on the topology on $M$. One can't talk about an open set of $M$ without specifying the topology. Do you mean to say that we don't explicitly specify the topology on $M$, and that it is implicitly specified by our choice of topology on $\mathbb{R}^d$? Even if we transport the topology on the charts to $M$, the notion of continuity for a curve on $M$ would still ultimately depend on the topology on the charts, which we specified to begin with. – Shirish Kulhari Dec 12 '17 at 6:40
• (cont'd from above) The question still stands: does that rule out continuity as a "physically real property"? – Shirish Kulhari Dec 12 '17 at 6:42
• @Shirish Kulhari: for some reason the sound doesn't play on my smart-phone so I can't check out the video lecture you've mentioned. Yes, the topology is not explicitly stated, it's implicitly stated by defining the charts - this is what I mean by transporting the topology to the manifold. If you think about it you can see this makes sense philosophically - we are defining something unknown in terms of something known. – Mozibur Ullah Dec 12 '17 at 6:50
• For sure, the topology ultimately depends on cartesian space; but once we've transported the topology to the manifold we can do a lot without having to go back to the charts. For example, we can show that the Cartesian product of two manifolds is a manifold - and to do this we have to go back to the charts; but once we've done this, we can say that an open line segment x circle gives us a topological cylinder, and we can say this without going back to the charts. – Mozibur Ullah Dec 12 '17 at 6:51
• Continuity is probably not a physically real property; it's likely at very small distance physical continuity fails; but this perspective comes from quantum mechanics. For GR it's best to take it that it stands. – Mozibur Ullah Dec 12 '17 at 6:55