What is a commutator?

Wikipedia says: "Conceptually, the Lie bracket [X, Y] is the derivative of Y along the flow generated by X."

A Lie bracket is called a commutator in the book.

Question: Reductio ad absurdum on commutators?

I am studying Spacetime and Geometry : An Introduction to General Relativity by Sean M Carroll and have a question about commutators of vector fields. A vector field on a manifold can be thought of as differential operator which transforms smooth functions to smooth functions on the manifold. For a vector field X and a function f(xi) we write
X(f) = g, where g is another function. We then define the commutator of two fields X and Y as

[X,Y](f) = X(Y(f)) - Y(X(f)

In the exercise I am working on (2.05), we are asked to find two vector fields whose commutator does not vanish. An important step is to show that if the commutator vanishes for one function f, it vanishes for all functions. This is implied by the question but not proven.

I proved it this way using 'Reductio ad absurdum'. Is this correct?

Our starting point is f  ≠ 0 and [X,Y](f) = 0. We have another function g ≠ 0 and [X,Y](g) ≠ 0.
We already know that commutators are linear (from the previous exercise), so

[X,Y](f + g) = [X,Y](f) + [X,Y](g)

or

[X,Y](f + g) = [X,Y](g)

Therefore f = 0, which breaks our starting assumption, with which there must be some error. The only non trivial possibility is that [X,Y](g) = 0. QED?
The question is at
https://www.physicsforums.com/threads/spacetime-and-geometry-vanishing-commutators.954767/
I then searched for Commutator and found this
Commutator of two vector fields:
https://www.physicsforums.com/threads/commutator-of-two-vector-fields.950661/
One answer contained
"[X,Y]  describes how far the endpoints of a rectangle vary if you go along X followed by Y or the other way around. Commuting vector fields mean the two path end at the same point;"

Perhaps I should have put my question in  the physics / Special and General Relativity forum.

Conformal Cyclic Cosmology (CCC)

An article in New Scientist (here) attracted my attention to Conformal Cyclic Cosmology (CCC) which sounds very interesting. The article presents some evidence for CCC. In the magazine the title was "A glimpse of a previous universe". I have a copy of it.

The 10 Commandments of Index Expressions and Tensor Calculus

This article looks extremely useful. It's at
https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

Exercise 2.03 2D Torus is a manifold

Question

Show that the two-dimensional torus T² is a manifold, by explicitly constructing an appropriate atlas. (Not a maximal one, obviously).

Answer

Fig 1: A Torus, with help from Wikipedia

We can define a torus as points in R³ that lie on the surface created by rotating a small circle of radius r round a larger circle of radius R. The larger circle is in the X-Y plane. Any point on the torus can be given by (θ , κ) as shown in Fig 1. There are no limits on θ , κ.

I realised that I could map to infinite cylinders touching the inner and outer edges of the torus and from them to annuli as in Exercise 2.01. No doubt our cunning author had set this trap. Having been on that wild goose chase, I came to a linear map from T² to the annuli and realised that we can also map from T² to two open rectangles much as one can map from S¹ to R¹. Fig 2 shows a section at some θ through our torus. We almost have the two charts U_O (Outer) and U_I (Inner) we just need to cut them and unroll them to make open rectangles. We cut the outer U_O at θ = 0 and the inner U_I at θ = -π. If we cut them both at the same θ it would lead to disaster.

Fig 2: Section through torus and rectangular maps.

This is where it gets complicated and we end up with a few maps, so;

Fig 4: Showing the transformations. T² is exactly U_I  ∩ U_O. 

We now want to check that the charts obey the strict conditions, as given in the book before Fig 2.13 at the bottom of page 29, to make an atlas from T² to R². I have adapted them for the exercise.

1. "The union of the charts U_I  , U_O is equal to T²; that is, U_I  , U_O cover T²."

We have two charts U_I  , U_O; their union manifestly covers T² by construction. ✓

We could check that there are no cracks in the union by combining the limits on ϕ_I  , ϕ_O. We have  -π< θ<2π and -π<κ<2π. That covers T². ✓

2. "The charts are smoothly sewn together. More precisely, if two charts overlap, U_I ∩ U_O ≠∅, then the map (ϕ_I ∘ ϕ_O^-1) takes points in ϕ_O (U_I ∩ U_O  ) ⊂  R² onto* an open set ϕ_I (U_I ∩ U_O ) ⊂ R², and all these maps must be C^∞. The reverse for  (ϕ_O ∘ ϕ_I^-1) also applies.

* We remind ourselves that onto means each point of the target has at least one point of the source mapped onto it.

The first part was easy as shown. The second part was fiddly and you can read the detail and wild goose chase in Ex 2.03 2D Torus is a manifold.pdf. It has helped me greatly to understand manifolds.

Links broken!

Sadly on 19th August I accidentally deleted all the files in my Space time and geometry. I then recovered them from the Microsoft bin. That was a mistake because it changed the URLs of all the files in the directory, so many links prior to this date are broken. If you need to find a file whose link is broken please look at the List of All File link in the LINKS section. Read more for how how to avoid this in future.

Exercise 2.01 One chart for infinite cylinder

Question

Just because a manifold is topologically non-trivial doesn't necessarily mean it can't be covered by a single chart. In contrast to the circle S¹, show that the infinite cylinder R × S¹ can be covered by just one chart, by explicitly constructing the map.

Answer

The first thing about the chart is that it must be open. That is its boundary is not included. The reason this is important is because the chart must be differentiable everywhere and if any boundary is included, it will not be differentiable there. It would be a sharp edge. This is why S¹ cannot be mapped on one chart.

Fig 1

Any point on the infinite cylinder can be given coordinates t, θ. t is the distance above or below some arbitrary point, with values from −∞ to +∞ . θ is the angle round the cylinder. 0 ≤ θ < 2π. We cannot simply unroll the cylinder because then we would get an infinitely long strip closed on the left an open on the right as shown in figure 1.

So I started by trying to map to a line which spiralled up the cylinder at some very small angle. Any point on the cylinder would map onto the line and vice-versa. This didn't seem to work.

I then asked Google and found two possible answers.

Physics forums solution

The first was on physics forums from July 2016 at

https://www.physicsforums.com/threads/infinite-cylinder-covered-by-a-single-chart.879193/

The first reply in the thread was from andrewkirk:

Annulus. Wikipedia

Map each line on the cylinder that runs parallel to the cylinder's axis of rotation, to a radial line in the annulus. Then set units along that radial line by parameterizing it in such a way that one approaches but never reaches the outer (inner) circumference of the annulus as t→∞  (t→−∞). The parameter t can represent signed distance along the line on the cylinder from an arbitrarily chosen starting point.

This solution might seem to suffer from the problem with the circle S¹: Two charts are needed to map it. However this may not the case as I showed in "Commentary 2.2 Covering by a single chart".

Following andrewkirk's instructions we could use the tangent function to parameterise the line as shown below

Fig 3. Section through infinite cylinder, showing tangent curve (blue),  and mappings of three points on the cylinder (green)

The tangent curve will rotate through 2π to form a very tall wine glass, without a base and a hole at the bottom. Our original cylinder has radius r and an outer cylinder has a radius R. It limits the outward extent of the curve. The equation of the curve is

t=tan((x- (R+r)/2) π / (R-r))    (1)

It is not trivial to derive the equation. Click Read more below, if you care.

Consider the point P at (t , θ) which maps to P' at (x , θ). From (1)

x = (R+r)/2 +(R-r) (tan^-1t) / π   (2)

Therefore, more conventionally, the mapping is 

θ' = θ

t' = (R+r)/2 +(R-r) (tan^-1t) / π

It is easy to show that this is C^∞. (see pdf for details.) 

The most interesting thing about this is that it shows that one can map an infinite area (the cylinder) onto a finite open area - the annulus. It is quite easy to imagine deforming the cylinder into the annulus. The 'mapping' from infinity to the open boundary is also interesting and adds to our intuitive understanding of why an open boundary is differentiable and a closed boundary (or sharp edge) is not.

University of California Santa Barbara solution 

The second solution was at University of California Santa Barbara saved here.

It also contains and answer to Exercise 1.04 on super luminal velocity and I discovered a small treasure trove of answers which I have catalogued in Answers to Exercises..

The answer for the infinite cylinder starts with a half cylinder and ends with a punctured plane. It is more complicated. It should be noted that a punctured plane has striking similarities to an open annulus.

For my pdf on this subject with prettier equations and further links look here
Ex 2.01 One chart for infinite cylinder.pdf