Exercise [14.04]

robin · 04 May 2010, 04:28

In short, once we have defined the covariant derivative on scalar fields and vector fields, we can work out the action on more general tensor fields

by induction using the identity: $(\nabla T)\cdot U = \nabla (T\cdot U) - T\cdot\nabla U$ , where

is either an arbitrary vector field or an arbitrary covector field, the choice being such that the convariant derivative of the contracted quantity $T\cdot U$ is known. Since

is arbirtrary we can then recover $\nabla T$ .

Because it will be useful for later questions, I will work this out explicitly.

For each coordinate vector field $\partial_c := \partial/\partial x^c$ , define the coefficients: $\Gamma^a_{cb} := \left[\nabla {\partial_c}}\right]^a_b$
(Note that $\Gamma^a_{cb}$ is tensorial in the

and

indices by construction, but in general it will not be tensorial in the

index).

Now for an arbitrary vector field $\boldsymbol{\xi}=\xi^c \partial_c}$ we use the Leibniz rule and the rule for the action of a covariant derivative on a scalar field:
$\nabla_b \xi^a = \left[\nabla\boldsymbol{\xi}\right]^a_b = \left[\nabla\xi^c \partial_c\right]^a_b\\ = \left[(\nabla\xi^c)\otimes\partial_c\right]^a_b + \left[\xi^c\nabla\partial_c\right]^a_b\\ = \left[(d\xi^c)\otimes\partial_c\right]^a_b + \xi^c\Gamma^a_{cb}\\ = \left[{\partial\xi^c\over{\partial x^d}}dx^d\otimes\partial_c\right]^a_b + \xi^c\Gamma^a_{cb}\\ = {\partial\xi^a\over{\partial x^b}} + \xi^c\Gamma^a_{cb}$

We now calculate the covariant derivative of a covector field $\boldsymbol{\alpha}=\alpha_c dx^c}$ . The Leibniz rule gives:
$(\nabla\boldsymbol{\alpha})\cdot\boldsymbol{\xi} = \nabla(\boldsymbol{\alpha}\cdot\boldsymbol{\xi}) - \boldsymbol{\alpha}\cdot \nabla\boldsymbol{\xi}$
In coordinates this becomes:
$(\nabla_b \alpha_a)\xi^a = \nabla_b(\alpha_a\xi^a) - \alpha_a \nabla_b\xi^a\\ = {\partial (\alpha_a\xi^a)\over{\partial x^b}} - \alpha_a\left( {\partial\xi^a\over{\partial x^b}} +\xi^c\Gamma^a_{cb}\right)\\ = \left( {\partial\alpha_a\over{\partial x^b}} - \alpha_a\Gamma^a_{cb}\right)\xi^a$
(n.b. I changed dummy indices on the $\xi$ term in the last step)
Since this is true for any vector field $\boldsymbol{\xi}$ we have:
$\nabla_b \alpha_a = {\partial\alpha_a\over{\partial x^b}} - \alpha_a\Gamma^a_{cb}$

Using induction, it is then straightforward to show that for an arbitrary tensor field

we get the following:
$\nabla_b T^{i_1\cdots i_p}_{j_1\cdots j_q} = {\partial T^{i_1\cdots i_p}_{j_1\cdots j_q}\over{\partial x^b}} + T^{c i_2\cdots i_p}_{j_1\cdots j_q}\Gamma^{i_1}_{cb} + \cdots + T^{i_1\cdots i_{p-1}c}_{j_1\cdots j_q}\Gamma^{i_p}_{cb}\\ \text{\hspace{4cm} } - T^{i_1\cdots i_p}_{a j_2\cdots j_q}\Gamma^a_{j_1 b} - \cdots - T^{i_1\cdots i_p}_{j_1\cdots j_{q-1}a}\Gamma^a_{j_q b}$

I'll prove the inductive step in the case of an arbitrary covariant tensor. The general case is similar (except that you need to contract contravariant indices with a covector field instead of a vector field).
Suppose we have proved the identity for covariant tensors with up to

indices. The case for a covariant tensor with q+1

indices is handled by contracting with a vector field $\boldsymbol{\xi}$
$(\nabla_b T_{j_1\cdots j_q k})\xi^k = \nabla_b (T_{j_1\cdots j_q k}\xi^k) - T_{j_1\cdots j_q k}\nabla_b \xi^k\\ = {\partial (T_{j_1\cdots j_q k}\xi^k)\over{\partial x^b}} - T_{a j_2\cdots j_q k}\xi^k\Gamma^a_{j_1 b} - \cdots - T_{j_1\cdots j_{q-1}a k}\xi^k\Gamma^a_{j_q b} - T_{j_1\cdots j_q k}\left({\partial\xi^k\over{\partial x^b}} +\xi^c\Gamma^k_{cb}\right)$
$=\bigg( {\partial T_{j_1\cdots j_{q+1}}\over{\partial x^b}}-T_{a j_2\cdots j_{q+1}}\Gamma^a_{j_1 b} - - \cdots - T_{j_1\cdots j_q a}\Gamma^a_{j_{q+1}b} \bigg) \xi^{j_{q+1}}$
(where we have renamed dummy indices in the final step). This is true for any vector field and so the induction step is proven.