Exercise [14.25]

robin · 10 May 2010, 04:31

Although Penrose asked to show this result in the torsion-free case, it is a simple matter to keep track of what happens if we leave it in.

We shall use the following expressions that I derived in Exercise 14.8
$(\nabla_a\nabla_b-\nabla_b\nabla_a) \xi^d = R_{abc}^{\ \ \ d}\xi^c + <br />(\nabla_c \xi^d)\tau^{\ \ c}_{ab}$
and Exercise 14.19
$(\pounds_\xi \eta)^b = \xi^a \nabla_a \eta^b - \eta^a \nabla_a \xi^b + \xi^a\eta^c \tau_{ac}^{\ \ b}$

We have:
$( \nabla_L\nabla_M N )^d = L^a \nabla_a (M^b \nabla_b N^d) = L^a M^b\nabla_a\nabla_b N^d + L^a(\nabla_a M^b)(\nabla_b N^d)$
and by definition of the Lie derivative:
$( \nabla_{[L,M]} N )^d = (\pounds_L M)^b (\nabla_b N^d)$

Hence
$((\nabla_L\nabla_M-\nabla_N\nabla_M-\nabla_{[L,M]}) N)^d$
$= L^a M^b (\nabla_a\nabla_b - \nabla_b\nabla_a)N^d + (L^a\nabla_a M^b - M^a\nabla_a L^b)(\nabla_b N^d) - (\pounds_L M)^b(\nabla_b N^d)$
$= L^a M^b (N^c R_{abc}^{\ \ \ d} + (\nabla_c N^d)\tau^{\ \ c}_{ab}) + ((\pounds_L M)^b-L^aM^c\tau^{\ \ b}_{ac})(\nabla_b N^d) - (\pounds_L M)^b(\nabla_b N^d)$
$= R(L,M,N)^d + L^a M^b (\nabla_c N^d)\tau^{\ \ c}_{ab} - (L^aM^c\tau^{\ \ b}_{ac})(\nabla_b N^d)$

We see that the torsion contributions cancel out (which is what Penrose is talking about in the comment about torsion being automatically allowed for at the expense of the extra commutator term).