Monday, 13 August 2018

Exercise 2.04 Commutator of two vector fields

Question

Verify the claims made about the commutator of two vector fields at the end of section 2.3 (linearity, Leibnitz, component formula, transformation as a vector field).
The commutator of two vector fields X and Y is defined by its action on a function f(xμ) as
[X,Y](f) ≡ X(Y(f))) - Y(X(f)))
(2.20)
The claims made were:

1) Linearity: if f and g are functions and a, b real numbers, the commutator is linear:
[X,Y](af + bg) = a[X,Y](f) + b[X,Y](g)

(2.21)
2) It obeys the Leibnitz rule:
[X,Y](fg) = f[X,Y](g) + g[X,Y](f)

(2.22)
3) The components of the vector field [X,Y]μ are given by:
[X,Y]μ = XλλYμ - YλλXμ


(2.23)
4) Transformation as a vector field:
Perform explicitly a coordinate transformation on the expression (2.23) to verify that all potentially non-tensorial pieces cancel and the result transforms like a vector field.

Answers

The first two were pretty easy. They just involved using the definition, using the fact that X and Y must be linear and / or obey the Leibnitz rule and then cancelling and / or using the definition again. My answers are almost line for line in agreement with the UCSB solution.

The third was a bit more difficult. I used the fact that the vector field is also a derivative operator. There were two slightly dubious steps in my proof: thaY(xλ )=Yλ and removing  entirely to get equation 0.5. But I think they are OK especially as they give the right answer!

I eventually understood the fourth question and knew where to start and where it should end. The four fairly hairy steps in between eluded me. I have followed the steps in detail in my pdf.

The UCSB solutions also proved the Jacobi identity for vector fields. It is

[ [X, Y ], Z ] + [ [Y, Z], X ] + [ [Z, X], Y ] = 0.
The proof was easy!

Read my solutions and comments here. Ex 2.04 Commutator of two vector fields.pdf

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