Trouble understanding ##g^{jk}\Gamma^{i}{}_{jk}##

shooride
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Hi friends,

I'm learning Riemannian geometry. I'm in trouble with understanding the meaning of ##g^{jk}\Gamma^{i}{}_{jk}##. I know it is a contracting relation on the Christoffel symbols and one can show that ##g^{jk}\Gamma^i{}_{jk}=\frac{-1}{\sqrt{g}}\partial_j(\sqrt{g}g^{ij})## using the Levi-Civita connection, etc-- rhs is similar to the divergence of an antisymmetric tensor field (!)--. But, how should one interpret and call ##g^{jk}\Gamma^i{}_{jk}##?! I'm interested in this term since it appears in the Laplacian of the function ##f## (Laplace–Beltrami operator). Especially, ##\Delta f=div ~\nabla f= \partial^2 f-g\Gamma\partial f## (not using the Einstein notation). The interpretation of first term in the rhs is clear; Is there any clear/simple interpretation for second term?!
 
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I guess I should name ##g^{jk}\Gamma^i{}_{jk}##, ##a^i##:biggrin:.
 
First, do you understand why the covariant derivative of a vector has 2 terms? (I.e., a partial derivative and a term involving ##\Gamma##) ?
 
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strangerep said:
First, do you understand why the covariant derivative of a vector has 2 terms? (I.e., a partial derivative and a term involving ##\Gamma##) ?
Yeah, since the covariant derivative is a covariant :biggrin:. I think I was a bit confused when I asked this question :oops:. Anyway, do you know whether there is a particular name for the identity ##g^{jk}\Gamma^i{}_{jk}=\frac{-1}{\sqrt{g}}\partial_j(\sqrt{g}g^{ij})##?!
 
shooride said:
[...] do you know whether there is a particular name for the identity ##g^{jk}\Gamma^i{}_{jk}=\frac{-1}{\sqrt{g}}\partial_j(\sqrt{g}g^{ij})##?!
No, I don't know a name.

BTW, for Riemannian geometry stuff involving indices, (which mathematicians usually hate), you can sometimes get more answers by asking in the relativity forum.
 

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