Double covariant derivative of tensor

Question

Consider the covariant derivative of a type $(0,2)$ tensor given in terms of the connection: $$ h_{ab;c} \equiv \partial_c h_{ab} - \Gamma^d_{ca} h_{db} - \Gamma^d_{cb} h_{ad} $$

What would the term $$ h_{ab;ce} = \nabla_e \nabla_c h_{ab} $$ look like? First term would be $$ h_{ab;cd} = \partial_e \left( \partial_c h_{ab} - \Gamma^d_{ca} h_{db} - \Gamma^d_{cb} h_{ad} \right) + (\ldots) + (\ldots) $$ What about the second one? Not sure how to juggle with the indices here.

$$ h_{ab;cd} = (\ldots) + (\Gamma^f_{ea} h_{fb;c} ) + (\ldots) $$

Does that look right?

Answer 1

So you have to compute $$ \nabla_a \nabla_b h_{cd} $$ First you have to think of $\nabla_b h_{cd}$ as a $(0,3)$ tensor so that $$ \nabla_a \nabla_b h_{cd} = \partial_a ( \nabla_b h_{cd} ) - \Gamma^e_{ab} \nabla_e h_{cd} - \Gamma^e_{ac} \nabla_b h_{ed} - \Gamma^e_{ad} \nabla_b h_{ce} $$ Finally, you expand out the $\nabla h$ so that \begin{align} \nabla_a \nabla_b h_{cd} &= \partial_a ( \partial_b h_{cd} - \Gamma^f_{bc} h_{fd} - \Gamma^f_{bd} h_{cf} ) \\ &\qquad \qquad - \Gamma^e_{ab} ( \partial_e h_{cd} - \Gamma^f_{ec} h_{fd} - \Gamma^f_{ed} h_{cf} ) \\ &\qquad \qquad - \Gamma^e_{ac} ( \partial_b h_{ed} - \Gamma^f_{be} h_{fd} - \Gamma^f_{bd} h_{ef} ) \\ &\qquad \qquad - \Gamma^e_{ad} ( \partial_b h_{ce} - \Gamma^f_{bc} h_{fe} - \Gamma^f_{be} h_{cf} ) \end{align}

Answer 2

For a vector field, $X^\mu{}_{;\nu} = \partial_\nu X^\mu + \Gamma^\mu{}_{\kappa\nu}X^\kappa.$

Because the covariant derivative should coincide with partial derivatives on scalars, and should satisfy the Leibniz rule, $$\nabla_\mu (X^\nu X_\nu) = (\nabla_\mu X^\nu) X_\nu + X^\nu (\nabla_\mu X_\nu) = \partial_\mu (X^\nu X_\nu) = (\partial_\mu X^\nu) X_\nu + X^\nu (\partial_\mu X_\nu)$$ we can conclude that for 1-forms, $$X_{\mu;\nu} = \partial_\nu X_\mu - \Gamma^{\kappa}{}_{\mu\nu}X_\kappa.$$

Now, because the covariant derivative should satisfy the Leibniz rule, for a simple tensor $X^\mu Y^\nu$, $$\nabla_\rho (X^\mu Y^\nu) = (\nabla_\rho X^\mu) Y^\nu + X^\mu (\nabla_\rho Y^\nu) = (\partial_\rho X^\mu) Y^\nu + X^\mu (\partial_\rho Y^\nu) + \Gamma^{\mu}{}_{\kappa\rho}X^\kappa Y^\nu + \Gamma^{\nu}{}_{\kappa\rho}X^\mu \ Y^\kappa $$ $$ = \partial_\rho (X^\mu Y^\nu) + (\Gamma^{\mu}{}_{\kappa\rho}\delta^\nu_\lambda + \delta^\mu_\kappa\Gamma^\nu{}_{\lambda\rho} )X^\kappa Y^\lambda. $$ Because any tensor is a linear combination of simple tensors, it holds for arbitrary (2,0)-tensors that $$\nabla_\rho T^{\mu\nu} = \partial_\rho T^{\mu\nu} + (\Gamma^{\mu}{}_{\kappa\rho}\delta^\nu_\lambda + \delta^\mu_\kappa\Gamma^\nu{}_{\lambda\rho} )T^{\kappa\lambda} = \partial_\rho T^{\mu\nu} + \Gamma^\mu{}_{\kappa\rho} T^{\kappa \nu} + \Gamma^\nu{}_{\kappa\rho}T^{\mu\kappa}. \tag 1$$

The generalization to tensors of arbitrary type is now clear, and can be summarized in the following rule:

For every up index, add a Christoffel symbol contracted with the tensor on that index. For every down index, subtract such a term instead.

Now, when taking second covariant derivatives, it has to be remembered that the Christoffel symbols are not constants, so one has to take derivatives of them, also. And, the first covarian derivative adds an index, so for the second step, we need to use the formulas for tensors of the appropriate type.

E.g., for a vector, $X^{\mu}{}_{;\nu}$ is a $(1,1)$-tensor, so $$X^\mu{}_{;\nu\rho} = \nabla_\rho(\partial_\nu X^\mu + \Gamma^\mu{}_{\kappa\nu}X^\kappa).$$ Now, in the operand, neither term is tensorial on its own -- remember: partial derivatives and the Christoffel symbols are not tensors Thus, $\nabla_\rho \partial_\rho X^\mu$ doesn't strictly make sense -- the covariant derivative can only act on tensors. But we can proceed formally using the appropriate version of (1) for each term, and adding them up, we do get the a tensor. Anyway, that gives something like $$X^\mu{}_{;\nu\rho} = \partial_\rho \partial_\nu X^\mu + \Gamma^\mu{}_{\kappa\nu,\rho}X^\kappa + \Gamma^\mu{}_{\kappa\nu}\partial_\rho X^\kappa + \Gamma^\mu{}_{\kappa\rho}(\partial_\nu X^\kappa + \Gamma^\kappa{}_{\lambda\nu}X^\lambda) - \Gamma^{\kappa}{}_{\nu\rho}(\partial^\kappa X^\mu + \Gamma^{\mu\lambda}X^\lambda). $$

It is simple but extremely tedious and involves an awful profusion of indices to extend the calculation to the second covariant derivative of a $(2,0)$-tensor.

Answer 3

To answer this, you need to know what the covariant derivative of a rank $(3,0)$ tensor looks like. When in doubt, you can just use an example tensor $A_{\nu}B_{\rho}C_{\sigma}$ and use the product rule for the covariant derivative:

$$\nabla_{\mu}(A_{\nu}B_{\rho}C_{\sigma})=\left(\nabla_{\mu}A_{\nu}\right)B_{\rho}C_{\sigma}+A_{\nu}\left(\nabla_{\mu}B_{\rho}\right)C_{\rho}+A_{\nu}B_{\rho}\left(\nabla_{\mu}C_{\sigma}\right)\\ =\partial_{\mu}\left(T_{\nu\rho\sigma}\right)-\Gamma^{\alpha}_{\,\,\mu\nu}T_{\alpha\nu\rho}-\Gamma^{\alpha}_{\,\,\mu\rho}T_{\nu\alpha\sigma}-\Gamma^{\alpha}_{\,\,\mu\sigma}T_{\nu\rho\alpha},$$

where we have written $T_{\nu\rho\sigma}=A_{\nu}B_{\rho}C_{\sigma}$. Since a general rank $(3,0)$ tensor can be written as a sum of these types of "reducible" tensors, and the covariant derivative is linear, this rule holds for all rank $(3,0)$ tensors. This method can be used to find the covariant derivative of any tensor of arbitrary rank.

Apply this to your problem with $T_{\nu\rho\sigma}=\nabla_{\nu}h_{\rho\sigma}$ and you'll be off to the races!

I hope this helped!

Answer 4

When we talk about a covariant derivative $\nabla$ we must specify the vector field along which we take derivatives. Small example with vector fields $X,Y$ in terms of a chart \begin{eqnarray} \nabla_{X}Y = \nabla_{X^i\frac{\partial}{\partial x^i}}Y^j \frac{\partial}{\partial x^j}=X^i(\frac{\partial Y^q}{\partial x^i}+Y^j\Gamma^q_{ij})\frac{\partial}{\partial x^q} \end{eqnarray} The $\nabla$ is linear in the coefficient function, so it pulls to the front and the $\Gamma$ is a "connection coefficient" function, which expresses in the given chart how the basis is changing. This is what is usually meant by those indices. Try apply this formalism to your problem. Tell me how it goes.