## Friday, 28 September 2018

### Commentary on Appendix A: Mapping S2 and R3

Sometimes I almost despair reading this book. I cannot understand part of it or find a solution. When I do, the feeling of joy is superb. There is much air punching. This is a case in point.

## Why am I here?

I'm still thinking about commutators and wanted to understand Appendix B on Diffeomorphisms and Lie Derivatives. Lie (pronounced lee) brackets are the same as commutators. Appendix B starts by saying "we continue the explorations of the previous Appendix", so I was thrown back. Appendix A is fairly comprehensible and there is a good example at the end. I'll restate the first part here:

## The problem part 1

Consider the two-sphere embedded in 3, thought of as the locus of points a unit distance from the origin. If we put coordinates θ, ф on S2 (the sphere) and x,y,z on 3, the map  is given by

ψ (θ, ф) = (sinθ cosф, sinθ sinф, cosθ)          (1)

In the book this equation is written as
ф(θ, ф) = (sinθ cosф, sinθ sinф, cosθ)            (A.11)
ф is used in the equation to mean a map and a coordinate. I will avoid this.

Sticking the sphere into  in this way induces a metric on , which is just the pull-back of the flat-space metric. The simple-minded way to find this is to start with the metric

ds² = dx² + dy² + dz²                               (1a)

and substitute (1) into this expression yielding a metric dθ² + sin²θ dф² on .

This was not so simple and I struggled with the substitution for days. (1) is easy to show algebraically or geometrically. Getting to the metric in polar coordinates involves various differentiation rules: (trig functions and the product rule), but then they need to be applied to infinitesimals (dx, dθ etc) rather than proper derivatives (dx/dy etc). That aspect was very novel to me.

Along the way I made an interesting observation about writing the matrix for xyα. It is the transpose of the matrix for xyα. I will add this to my up coming opus on Tensors, Matrices and Indexes.

Five pages. You can read it here: Commentary App A Mapping S2 and R3.pdf

## It contains a very useful corollary

It contains a very useful corollary on how to get the metric of a non-Cartesian system.
If we can transform non-Cartesian components j into Cartesiani with equations

i = f i (p j)                                                (20)
i = 1..m,   j = 1..n

Then we can construct the (n x m) pullback operator

Multiplying this by its transpose gives the metric of the non-Cartesian system.

Of necessity this is a (n x m) x (m x n) = (n x n) matrix, as it must be.

## Wednesday, 26 September 2018

### 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.

## Thursday, 6 September 2018

### 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
I then searched for Commutator and found this
Commutator of two vector fields: