Covariant derivative of the metric

[vitaniarain]

hello!
just a quick question, does the covariant derivative of the metric give zero even when the indices(one of the indices) of the metric are(is) raised?
also another question not entirely related, does the covariant deriv. of exp(2 phi) where phi is the field, also give zero or not necessarily? thanks

 

[haushofer]

Act with the covariant derivative on a kronecker delta. You know that

$$g_{ab}g^{bc} = \delta_a^c$$

The covariant derivative of your exponential of scalar field is given by a partial derivative per definition; this will only be zero for constant phi.

 

[vitaniarain]

if u act on a kronecker delta with the covariant deriv will that give zero? :S

 

[haushofer]

Well, you know how a covariant derivative acts on a (1,1)-type tensor, right? Just write it down explicitly and check :) I would say that you'll find that it indeed is zero, provided the connection is symmetric.