I know that the Levi-civita connection preserves the metric tensor. Is the divergence of the inverse of metric tensor zero, too?! I'm not so familiar with the divergence of the second ranked tensor. However, I think one can write $$\nabla_j g^{ij}=\partial_jg^{ij}+\Gamma^i_{kj}g^{kj}+\Gamma^j_{kj}g^{ki}$$ using the identity $g^{jk}\Gamma^i{}_{jk}=\frac{-1}{\sqrt{g}}\partial_j(\sqrt{g}g^{ij})$ and $\Gamma^j_{kj}=\partial_k~log~\sqrt g$, so $$\nabla_j g^{ij}=\partial_jg^{ij}-\frac{1}{\sqrt g }\partial_j(\sqrt g g^{ij})+\partial^i~log~\sqrt g$$ therefore $$\nabla_j g^{ij}=\partial_jg^{ij}-\partial_jg^{ij}-g^{ij}\partial_j~log~\sqrt g+\partial^i~log~\sqrt g=0$$ which is weird to me. Where am I doing anything wrong?
$\begingroup$
$\endgroup$
6
-
2$\begingroup$ $0=\nabla_k \delta^i_j=\nabla_k(g^{mi}g_{mj})=g_{mj}\nabla_k g^{mi}$ $\endgroup$– AccidentalFourierTransformCommented Feb 1, 2016 at 20:10
-
$\begingroup$ @AccidentalFourierTransform what I can't understand is the divergence of metric..BTW, why didn't you write the second term?!... $\endgroup$– AFZQCommented Feb 1, 2016 at 20:18
-
$\begingroup$ which second term? $\nabla_k g_{mj}=0$ by definition of $\nabla$... $\endgroup$– AccidentalFourierTransformCommented Feb 1, 2016 at 20:19
-
$\begingroup$ @AccidentalFourierTransform ..you mean Levi-civita connection?! I've asked the divergance..BTW, I do not think one can write sth like $\nabla_k (g^{ki}g_{kj})$..it's an abuse of notation..or I don't get your point.. $\endgroup$– AFZQCommented Feb 1, 2016 at 20:28
-
$\begingroup$ 1) yes, we are both talking about Levi-Civita $\nabla$ 2) note that once you know what $\nabla_k g^{mi}$ is, you can easily calculate the divergence by contracting the $ki$ indices 3) ... yes, that notation makes no sense, but I never wrote that... $\endgroup$– AccidentalFourierTransformCommented Feb 1, 2016 at 20:45
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
0
We know that the Levi-Civita connection satisfies $\nabla_a g_{bc} = 0$ and the product rule. The definition of the inverse metric $g^{ab}$ is $g^{ab}g_{bc} = \delta^a_c$. Therefore, we have:
$$\begin{align} 0 &= \nabla_a \delta^b_c \\ &= \nabla_a (g^{bd}g_{dc}) \\ &= (\nabla_a g^{bd}) g_{dc} + g^{bd} \nabla_a g_{dc} \\ &= (\nabla_a g^{bd}) g_{dc} \end{align}$$
Upon multiplying both sides by $g^{ce}$ and renaming indices we get that indeed $\nabla_a g^{bc} =0$. If you want the divergence $\nabla_a g^{ac}$ then it's just a matter of setting $b=a$; we still get zero.
