Exercise 2.04

Verify the claims made about the commutator of two vector fields at the end of Section 2.3 (linearity, Leibniz, component formula, transformation as a vector field).

Linearity 

$$ \begin{align} [X,Y](af+bg)&=X(Y(af+bg))-Y(X(af+bg)) \\ &=X^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(af+bg))-Y^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(af+bg))\\ &=aX^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(f))+bX^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(g)) \\ &\qquad -aY^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(f))-bY^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(g)) \\ &=a\Big(X^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(f))-Y^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(f))\Big) \\ &\qquad +b\Big(X^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(g))-Y^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(g))\Big) \\ &=a\Big(X(Y(f))-Y(X(f))\Big)+b\Big(X(Y(g))-Y(X(g))\Big) \\ &=a[X,Y](f)+b[X,Y](g) \end{align} $$

Leibniz rule

$$ \begin{align} [X,Y](fg)&=X(Y(fg))-Y(X(fg)) \\ &=X^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(fg))-Y^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(fg))\\ &=X^{\mu}\partial_{\mu}(fY^{\nu}\partial_{\nu}(g)+gY^{\nu}\partial_{\nu}(f))-Y^{\nu}\partial_{\nu}(fX^{\mu}\partial_{\mu}(g)+gX^{\mu}\partial_{\mu}(f)) \\ &=fX^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(g))+\cancel{ X^{\mu}\partial_{\mu}(f)Y^{\nu}\partial_{\nu}(g) }+gX^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(f))+\cancel{ X^{\mu}\partial_{\mu}(g)Y^{\nu}\partial_{\nu}(f) } \\ &\qquad -fY^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(g))-\cancel{ Y^{\nu}\partial_{\nu}(f)X^{\mu}\partial_{\mu}(g) }-gY^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(f))-\cancel{ Y^{\nu}\partial_{\nu}(g)X^{\mu}\partial_{\mu}(f) } \\ &=f\Big(X^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(g))-Y^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(g))\Big)+g\Big(X^{\mu}\partial_{\mu}(Y^{\nu}\partial_{\nu}(f))-Y^{\nu}\partial_{\nu}(X^{\mu}\partial_{\mu}(f))\Big) \\ &=f\Big(X(Y(g))-Y(X(g))\Big)+g\Big(X(Y(f))-Y(X(f))\Big) \\ &=f[X,Y](g)+g[X,Y](f) \\ \end{align} $$

Component formula 

$$ \begin{align} [X,Y]&=[X,Y]^{\mu}\partial_{\mu} \\ &=X^{\lambda}\partial_{\lambda}Y^{\mu}\partial_{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu}\partial_{\mu} \\ &=(X^{\lambda}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu})\partial_{\mu} \\ &\hspace{1pt}\Downarrow \\ [X,Y]^{\mu}&=X^{\lambda}\partial_{\lambda}Y^{\mu}-Y^{\lambda}\partial_{\lambda}X^{\mu} \end{align} $$

Transformation as a vector field 

$$ \begin{align} [X,Y]^{\mu'}&=X^{\lambda}\partial_{\lambda}Y^{\mu'}-Y^{\lambda}\partial_{\lambda}X^{\mu'} \\ &=X^{\lambda}\partial_{\lambda}\left( \frac{ \partial x^{\mu'} }{ \partial x^{\mu} }Y^{\mu} \right)-Y^{\lambda}\partial_{\lambda}\left( \frac{ \partial x^{\mu'} }{ \partial x^{\mu} }X^{\mu} \right) \\ &=X^{\lambda}Y^{\mu}\partial_{\lambda}\left( \frac{ \partial x^{\mu'} }{ \partial x^{\mu} } \right)+\frac{ \partial x^{\mu'} }{ \partial x^{\mu} }X^{\lambda}\partial_{\lambda}Y^{\mu} \\ &\qquad -Y^{\lambda}X^{\mu}\partial_{\lambda}\left( \frac{ \partial x^{\mu'} }{ \partial x^{\mu} }\right) - \frac{ \partial x^{\mu'} }{ \partial x^{\mu} }Y^{\lambda}\partial_{\lambda}X^{\mu} \\ &=(X^{\lambda}Y^{\mu} -Y^{\lambda}X^{\mu})\frac{ \partial x^{\mu} }{ \partial x^{\lambda} }\partial_{\mu} \left( \frac{ \partial x^{\mu'} }{ \partial x^{\mu} }\right) + \frac{ \partial x^{\mu'} }{ \partial x^{\mu} }\Big( X^{\lambda}\partial_{\lambda} Y^{\mu}- Y^{\lambda}\partial_{\lambda}X^{\mu} \Big) \\ &=\cancel{ (X^{\lambda}Y^{\mu} -Y^{\lambda}X^{\mu}) \delta ^{\mu}_{\lambda}}\frac{ \partial^{2} x^{\mu'} }{ \partial^{2} x^{\mu} } + \frac{ \partial x^{\mu'} }{ \partial x^{\mu} }[X,Y]^{\mu} \\ &= \frac{ \partial x^{\mu'} }{ \partial x^{\mu} }[X,Y]^{\mu} \end{align} $$