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(x^{μ}) 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} = X^{λ}∂_{λ}Y^{μ}
 Y^{λ}∂_{λ}X^{μ}

(2.23)

4) Transformation as a vector field:
Perform explicitly a coordinate transformation on the
expression (2.23) to verify that all potentially nontensorial 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: that Y(x^{λ} )=Y^{λ} and removing f 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!
Ex 2.04 Commutator of two vector fields.pdf
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