## Tuesday 14 September 2021

### Question

Take the mathematician's view that tangent vectors and directional derivatives are the same thing, $u\equiv\partial_u$. Let $u,v$ be two vector fields and define their commutator in the manner familiar from quantum mechanics [not familiar to me!]$$\left[u,v\right]\equiv\left[\partial_u,\partial_v\right]\equiv\partial_u\partial_v-\partial_v\partial_u$$(a) Derive the following expression for $\left[u,v\right]$ valid in any coordinate basis$$\left[u,v\right]=\left(u^\beta v_{\ \ ,\beta}^\alpha-v^\beta u_{\ \ ,\beta}^\alpha\right)e_\alpha$$Thus despite that it looks like a second-order differential operator, $\left[u,v\right]$ is actually of first order - i.e. it is a tangent vector.
(b) For any basis $\left\{e_\alpha\right\}$ one defines the "commutation coefficients" $c_{\beta\gamma}^{\ \ \ \ \ \alpha}$ and $c_{\beta\gamma\alpha}$ by $$\left[e_\beta,e_\gamma\right]\equiv c_{\beta\gamma}^{\ \ \ \ \ \alpha}e_\alpha;\ \ c_{\beta\gamma\alpha}=g_{\alpha\mu}c_{\beta\gamma}^{\ \ \ \ \ \mu}$$Show that $c_{\beta\gamma}^{\ \ \ \ \ \alpha}=c_{\beta\gamma\alpha}=0$ got any coordinate basis.
(c) Calculate $c_{\hat{\beta}\hat{\gamma}}^{\ \ \ \ \ \hat{\alpha}}$ for the spherical noncoordinate basis of exercise 8.1.