### 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.

### Answers

This was straight forward. But I learned that Hermann Schwarz outshone Euler, Lagrange, Cauchy who were all very clever. It was Schwarz who proved that partial derivatives commute. The others didn't get it quite right apparently. So I repeated his proof and now I know when partial derivatives commute and when they don't. Hurrah.