General Relativity - Double Covariant Derivative

[Tangent87]

I know that for a scalar \nabla^2\phi=\nabla_a\nabla^a\phi=\nabla^a\nabla_a\phi. However what is \nabla^2 for a tensor? For example, is \nabla^2T_a=\nabla_b\nabla^bT_a or is it \nabla^2T_a=\nabla^b\nabla_bT_a? Because I don't think they're the same thing.

Thanks.

 

[dextercioby]

The metric is a constant wrt to the covariant derivative. So the 'wave'/D'Alembert operator is defined by

\Box =: g^{ab} \nabla_a \nabla_b = \nabla^{b} \nabla_b = \nabla_a \nabla^{a}

which is to be applied on any tensor object.

The contravariant derivative is

\nabla^{b} = g^{ba} \nabla_a

so

\nabla^{b} T_{ac} = g^{bd} \nabla_d T_{ac}

The last part you can compute using the normal rules for covariant differentiation.