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George Keeling

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- Should I try to prove this with non-diagonal metric?

M. Blennow's book has problem 2.18:

Show that the contracted Christoffel symbols ##\Gamma_{ab}^b## can be written in terms of a partial derivative of the logarithm of the square root of the metric tensor $$\Gamma_{ab}^b=\partial_a\ln{\sqrt g}$$I think that means square root of the determinant of the metric tensor (and it does in the next question). I don't know how to take a square root of a tensor.

I start with $$\partial_a\ln{\sqrt g}=\frac{1}{2}\partial_a\ln{g}=\frac{1}{2}\frac{1}{g}\partial_ag=\frac{1}{2}g^{-1}\partial_ag$$and$$\Gamma_{ab}^b=\frac{1}{2}g^{bc}\left(\partial_ag_{cb}+\partial_bg_{ac}-\partial_cg_{ab}\right)$$If the metric is diagonal it is pretty easy to show in ##n##-dimensions that those are both the same as$$\frac{1}{2}\sum_{i=1}^{i=n}{g^{ii}\partial_ag_{ii}}$$Should I be trying to prove ##\Gamma_{ab}^b=\partial_a\ln{\sqrt g}\ ## for a non-diagonal metric too?

(Carroll has a similar exercise which is restricted to a diagonal metric. Perhaps he is just not so cruel!)

Show that the contracted Christoffel symbols ##\Gamma_{ab}^b## can be written in terms of a partial derivative of the logarithm of the square root of the metric tensor $$\Gamma_{ab}^b=\partial_a\ln{\sqrt g}$$I think that means square root of the determinant of the metric tensor (and it does in the next question). I don't know how to take a square root of a tensor.

I start with $$\partial_a\ln{\sqrt g}=\frac{1}{2}\partial_a\ln{g}=\frac{1}{2}\frac{1}{g}\partial_ag=\frac{1}{2}g^{-1}\partial_ag$$and$$\Gamma_{ab}^b=\frac{1}{2}g^{bc}\left(\partial_ag_{cb}+\partial_bg_{ac}-\partial_cg_{ab}\right)$$If the metric is diagonal it is pretty easy to show in ##n##-dimensions that those are both the same as$$\frac{1}{2}\sum_{i=1}^{i=n}{g^{ii}\partial_ag_{ii}}$$Should I be trying to prove ##\Gamma_{ab}^b=\partial_a\ln{\sqrt g}\ ## for a non-diagonal metric too?

(Carroll has a similar exercise which is restricted to a diagonal metric. Perhaps he is just not so cruel!)