Math of Relativity: Topology Needed?

[davidge]

How much of topology one needs to know to have a great knowledge of the math of Special and General Relativity?

I'm asking this because I'm interested in really look at the theory of Relativity with the eyes of a mathematician.

I suppose that just knowing what a manifold is or even what a topological space is, isn't enough. But also, I suppose that Relativity doesn't require one having a "phD" in topology.

 

[PeterDonis]

I would check out Sean Carroll's lecture notes on GR:

https://arxiv.org/abs/gr-qc/9712019

Chapter 2 discusses what you need to know about manifolds (including topology) for GR.

 

[davidge]

Thanks Peter.
I have read these notes some time ago, after recommendation of a PF member.

What I don't like is that the text has no precision in some points, like when it says that a manifold is a topological space which locally looks like the Euclidean Space. What is the mathematical definition of locally?

 

[robphy]

What aspects do you want to look at?
Are you looking to prove theorems in (say) causal structure, singularities, or in the space of solutions?

 

[martinbn]

Thanks Peter.
I have read these notes some time ago, after recommendation of a PF member.

What I don't like is that the text has no precision in some points, like when it says that a manifold is a topological space which locally looks like the Euclidean Space. What is the mathematical definition of locally?

He actually gives a precise definition of a manifold, in the section called "Manifold". But if you find the lectures not mathematical enough, then take any introductory book on differential geometry that covers manifolds and you'll have what you're looking for.

 

[PeterDonis]

What is the mathematical definition of locally?

Within some open neighborhood of a chosen point. In other words, given any chosen point in the manifold, there will exist an open neighborhood of that point that has the topology of n-dimensional Euclidean space.

 

[davidge]

What aspects do you want to look at?
Are you looking to prove theorems in (say) causal structure, singularities, or in the space of solutions?

I'm looking for a mathematical explanation of why the space-time have to be open and compact, for example.

take any introductory book on differential geometry that covers manifolds and you'll have what you're looking for

The problem I noticed is that each one of the material I found so far (most of them pdf on web; 1 was a book that is available at the university library) appears to give a different explanation on the subject. So they don't appear to agree with each other when they are explaining, for instance, what a manifold is. I think this problem arises because most of (pdf's on web) them are most like a summary and don't go deep on the topics in question.

Within some open neighborhood of a chosen point. In other words, given any chosen point in the manifold, there will exist an open neighborhood of that point that has the topology of n-dimensional Euclidean space.

Ah, ok. Thanks.

 

[PeterDonis]

I'm looking for a mathematical explanation of why the space-time have to be open and compact

Why do you think spacetime as a whole has to be an open and compact manifold?

 

[davidge]

Why do you think spacetime as a whole has to be an open and compact manifold?

Taking a physical point of view, I would say it has to be open because then it could have singularities, but still we would not need to care about them, because they are not part of the manifold.

Now, I just don't know what compactness mean.

 

[PeterDonis]

I would say it has to be open because then it could have singularities, but still we would not need to care about them, because they are not part of the manifold.

I don't understand. Are you saying a spacetime has to contain singularities? If not, how does what you are saying rule out a closed manifold?

I just don't know what compactness mean.

Heuristically, a compact manifold is a manifold that has a "finite size". The more technical definition is that a manifold is compact if and only if every open cover (i.e., every set of open sets whose union is the entire manifold) has a finite subcover (i.e., you can pick a finite number of the open sets in the cover and their union will still be the entire manifold). A simple example of a compact manifold is a closed interval on the real line; and a simple example of a non-compact manifold is the corresponding open interval. It's a good exercise to satisfy yourself that it is impossible to find an open cover of the closed interval that doesn't have a finite subcover, whereas it is possible to find an open cover of the open interval that doesn't have a finite subcover.

 

[George Jones]

I'm looking for a mathematical explanation of why the space-time have to be open and compact, for example.

? Any compact spacetime necessarily contains closed timelike curves.

 

[davidge]

I don't understand. Are you saying a spacetime has to contain singularities? If not, how does what you are saying rule out a closed manifold?

No, I'm saying that maybe if an open manifold has a singularity, that singularity isn't a problem for us, because the singularity isn't part of the manifold.

The more technical definition is that a manifold is compact if and only if every open cover (i.e., every set of open sets whose union is the entire manifold) has a finite subcover (i.e., you can pick a finite number of the open sets in the cover and their union will still be the entire manifold). A simple example of a compact manifold is a closed interval on the real line

So, would this mean that e.g.

$$[10,20] = [10,12] \cup [12,14] \cup [14,16] \cup [16,18] \cup [18,20] $$ but $[10,12] = [10,11] \cup [11,12]$ and similarly for the other intervals, and the union of these intervals is equal to $[10,20]$.

Any compact spacetime necessarily contains closed timelike curves

Does this mean that given two different values for the parameter, we have the same point on the curve?

 

[George Jones]

Does this mean that given two different values for the parameter, we have the same point on the curve?

Yes. In any compact spacetime, there exist observer worldlines, with the same observer having different wristwatch times at the same event.

I'm looking for a mathematical explanation of why the space-time have to be open.

Every spacetime is a topological space (plus other structure), and every topological space is both open and closed. This is a defining feature of any topology.

 

[PeterDonis]

I'm saying that maybe if an open manifold has a singularity, that singularity isn't a problem for us, because the singularity isn't part of the manifold.

You do realize that, as you state it, this is self-contradictory?

Putting that aside, this still does not require any spacetime manifold to be open, and in the post of yours I was originally responding to in this subthread, you said you thought a spacetime manifold had to be open. Do you still think that? If so, why?

would this mean that e.g.

No, because you are describing unions of closed sets, not open sets.

Here is more detail about what I was describing. Consider two intervals: the closed interval $[0, 1]$ and the open interval $(0, 1)$.

Here is an open cover of the open interval $(0, 1)$ that does not contain any finite subcover: the infinite set of open sets: ${ (1/4, 1), (1/8, 1/2), (1/16, 1/4), ... }$. The union of this infinite set of open sets is the open interval $(0, 1)$, but no finite subset of them has that property.

However, there is no way to construct an open cover of the closed interval $[0, 1]$ that does not have a finite subcover. One way to see this is to imagine trying to alter the open cover I described above to cover the closed interval, by adding two open sets: $(- \epsilon, \epsilon)$ and $(1 - \epsilon, 1 + \epsilon)$. This makes the cover an open cover of the closed interval $[0, 1]$ (note that in this case, the union of the open sets is not exactly the closed interval, but the closed interval is a subset of it--that is actually the fully correct definition of an "open cover", since obviously no union of open sets can ever exactly give you a closed interval). But it also means that we can pick out a finite subcover, by leaving out all open intervals in the infinite set that extend less than $\epsilon$ up from zero, i.e., the last open interval in the finite subcover will be $(\epsilon / 2, 2 \epsilon)$, which is enough to make the union of that finite set of open sets contain the closed interval.

 

[davidge]

Every spacetime is a topological space (plus other structure), and every topological space is both open and closed

Oh, I see.

You do realize that, as you state it, this is self-contradictory?

Putting that aside, this still does not require any spacetime manifold to be open, and in the post of yours I was originally responding to in this subthread, you said you thought a spacetime manifold had to be open. Do you still think that? If so, why?

That's because I thought a topological space had to be open. Now that George Jones pointed out that it's in fact open and closed, I see that is hasn't to be open.

Here is more detail about what I was describing. Consider two intervals: the closed interval $[0, 1]$ and the open interval $(0, 1)$.

Thanks for working on this example. Now I understand what is meant by open cover and subcovers.

Here is my current understanding of a topological manifold

A manifold is an collection of open sets that locally (as you explained what locally means in post #6) looks like the Euclidean plane $\mathbb{R}^n$. So, for example, in the case $n=1$, suppose $T$ is the manifold consisting of all points $t$, such that $d(t, x_o) < r$ for some $x_o \in T$ and $r > 0$, and that $f: T \rightarrow \mathbb{R}$ through $t \mapsto tan(t)$.
- If we choose $t \in (0, \pi / 2)$, the image of the map would be $(0, \infty)$ $(1)$.
- If we choose $t \in [0, \pi / 2]$, the image still would be $(0, \infty)$ $(2)$.

In case $(1)$ there are a homeomorphism between $T$ and $\mathbb{R}$, because $f^{-1}( \mathbb{R})$ is open in $T$ each time that $f(T)$ is open in $\mathbb{R}$.
But in case $(2)$ it's not a homeomorphism, because the domain of $f$ is a closed interval.

So case $(1)$ is the correct one for GR, because it has one of the requirements for the space-time. The another requirement would be the map to be one-to-one.

Now I'm trying to translate this to the case of the four-dimensional spacetime.

Is this correct?

 

[PeterDonis]

A manifold is an collection of open sets

More precisely, a topological space (a manifold is a particular kind of topological space) is a set $X$ together with a collection $\tau$ of subsets of $X$ that satisfies three axioms: (1) the empty set and $X$ itself belong to the collection $\tau$; (2) any union (finite or infinite) of members of $\tau$ also belongs to $\tau$; (3) any finite intersection of members of $\tau$ also belongs to $\tau$. The subsets in $\tau$ are then called the "open sets" of the topology.

(A closed set is then any set whose complement in the set $X$ is open. You can see from this definition that the empty set and $X$ are both closed and open. Note also that the terms "open" and "closed" here are different from the terms "open" and "closed" as I was using them to describe intervals on the real line! Terminology here can quickly get confusing since the same terms can have several different meanings depending on the context.)

A manifold is a topological space with the additional property that every point has an open neighborhood which is homeomorphic to $\mathbb{R}^n$, i.e., to Euclidean space of the appropriate dimension. Note that this does not mean the manifold as a whole must be homeomorphic to $\mathbb{R}^n$; the n-dimensional sphere, for example, is not.

suppose $T$ is the manifold consisting of all points $t$, such that $d(t, x_o) < r$ for some $x_o \in T$ and $r > 0$,

This is not sufficient as you state it, because you have not defined which subsets of the set $T$ are the open sets. To be strictly correct, you either need to do that explicitly or say something like "with the standard topology on the real numbers", which means that the open sets are just the open intervals. (Note that with this topology, the open interval $T$ itself is also a closed set in the topology, but it is not a closed interval--in other words, this is an illustration of the difference in meanings of terms that I referred to above.)

and that $f: T \rightarrow \mathbb{R}$ through $t \mapsto \tan(t)$.

Just to be clear, this is not part of the definition of the manifold.

If we choose $t \in (0, \pi / 2)$, the image of the map would be $(0, \infty)$

Yes.

If we choose $t \in [0, \pi / 2]$,

Then you are contradicting your definition of the manifold $T$, since that definition specified $d(t, x_o) < r$, and this closed interval only satisfies $d(t, x_o) \le r$.

the image still would be $(0, \infty)$

Even leaving aside the objection I just made, this is not correct. $0$ itself is included in the image, since $\tan t = 0$ if $t = 0$, and $t = 0$ is now included in the domain. And for $t = \pi / 2$, $\tan t$ is undefined, so $f$ is no longer defined on the entire domain, so it isn't a valid function on the closed interval.

The question it looks like you are trying to get at here is, can the closed interval, $t \in [0, \pi / 2]$, be considered a manifold? The answer is no, but not for the reason you give. The reason is that every point in the manifold must have an open neighborhood that is homeomorphic to $\mathbb{R}^n$ (which in this case is just $\mathbb{R}$). $f$ still defines a homeomorphism from $T$ to $\mathbb{R}$ on every open interval in $T$ if $T$ is the closed interval $[0, \pi / 2]$; that's not the problem. The problem is that the endpoints, $t = 0$ and $t = \pi / 2$, do not have an open neighborhood in $T$ at all! Any open set on the reals that contains $t = 0$ or $t = \pi / 2$ must contain points that are outside the set $[0, \pi / 2]$. That is why that set cannot be considered a manifold.

 

[davidge]

More precisely, a topological space (a manifold is a particular kind of topological space) is a set $X$ together with a collection $\tau$ of subsets

The subsets in $\tau$ are then called the "open sets" of the topology.

So are you saying that the subsets $\tau$ themselves have subsets that are the so called open sets?

Note that this does not mean the manifold as a whole must be homeomorphic to $\mathbb{R}^n$; the n-dimensional sphere, for example, is not.

It's good to know.

Then you are contradicting your definition of the manifold $T$, since that definition specified $d(t, x_o) < r$, and this closed interval only satisfies $(t, x_o) \le r$.

Even leaving aside the objection I just made, this is not correct. $0$ itself is included in the image

I see

$f$ still defines a homeomorphism from $T$ to $\mathbb{R}$ on every open interval in $T$ if $T$ is the closed interval $[0, \pi / 2]$; that's not the problem. The problem is that the endpoints, $t = 0$ and $t = \pi / 2$, do not have an open neighborhood in $T$ at all! Any open set on the reals that contains$t = 0$ or $t = \pi / 2$ must contain points that are outside the set $[0, \pi / 2]$. That is why that set cannot be considered a manifold.

Oh, I got it. Now could we take then an open interval on $[0, \pi / 2]$, say $(1 / 2, \pi / 4)$, to be the manifold?

 

[PeterDonis]

the subsets $\tau$ themselves

$\tau$ is a collection of subsets of $X$. Each member of $\tau$ is called an open set.

could we take then an open interval on $[0, \pi / 2]$, say $(1 / 2, \pi / 4)$ to be the manifold?

Any open interval on the real line can be considered a manifold. But if you pick, for example, $(1 / 2, \pi / 4)$ to be the manifold you are interested in (more precisely, the set $X$ is $(1 / 2, \pi / 4)$ and the collection $\tau$ of open sets is the standard one for an open interval of the real line), then the closed interval $[0, \pi / 2]$ does not appear at all.

 

[davidge]

$\tau$ is a collection of subsets of $X$. Each member of $\tau$ is called an open set.

Ok. I understand it now.

and the collection $\tau$ of open sets is the standard one for an open interval of the real line

What is this standard collection?

 

[PeterDonis]

What is this standard collection?

In the standard topology for an open interval of the real line, the set $X$ is that open interval, and the collection $\tau$ contains all open intervals that are subsets of $X$. In other words, the open sets in this topology are just the open intervals contained in the one we choose as the manifold.