How to show that this expression with tensors reduces to zero?

In summary, the problem reduces to proving that the following expression reduces to zero:$$\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x'^d}{\partial x^h}g^{eh}g_{hg}\left(\frac{\partial^2 x^h}{\partial x'^c\,\partial x'^d}\frac{\partial x^g}{\partial x'^b}-\frac{\partial x^h}{\partial x'^b}\frac{\partial^2 x^g}{\partial x'^c\,\partial x'^d} \right )$$However, I couldn't
  • #1
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Homework Statement
Following a book "Introducing Einstein's Relativity" by Ray d'Inverno, I came across this problem (6.16):
Show directly that $$\Gamma^a_{bc}=\frac{1}{2}g^{ad}(\partial_b g_{dc}+\partial_c g_{db}-\partial_d g_{bc})$$ transforms like a connection.
Relevant Equations
The definition of ##\Gamma^a_{bc}## given above, and the usual rules for transforming tensors, i. e. $$g'^{ab}=\frac{\partial x'^a}{\partial x^c}\frac{\partial x'^b}{\partial x^d}g^{cd}$$

A useful result is also
$$\partial'_b g'_{dc}=\frac{\partial x^e}{\partial x'^b} \frac{\partial x^f}{\partial x'^c} \frac{\partial x^g}{\partial x'^d}\partial_e g_{gf}+\frac{\partial x^e}{\partial x'^b}\frac{\partial}{\partial x^e}\left (\frac{\partial x^g}{\partial x'^d}\frac{\partial x^f}{\partial x'^c}\right ) g_{gf}$$
I simply just wrote down the definition of ##\Gamma'^a_{bc}##, and inserted the transformations of ##g'^{ad}##, ##g'_{dc,b}##, and the like terms. After some rearranging and cancelling out,

$$\Gamma'^a_{bc}=\frac{\partial x'^a}{\partial x^e}\frac{\partial x^f}{\partial x'^b}\frac{\partial x^g}{\partial x'^c}\Gamma^e_{fg}+\frac{\partial x'^a}{\partial x^e}\frac{\partial^2 x^e}{\partial x'^b\,\partial x'^c}+\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x'^d}{\partial x^h}g^{eh}g_{hg}\left(\frac{\partial^2 x^h}{\partial x'^c\,\partial x'^d}\frac{\partial x^g}{\partial x'^b}-\frac{\partial x^h}{\partial x'^b}\frac{\partial^2 x^g}{\partial x'^c\,\partial x'^d} \right )$$

The first two terms correspond to the usual transformation law for connections, hence the third term should vanish. If my calculations until this point are correct, the original problem reduces to proving that the following expression reduces to zero:
$$\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x'^d}{\partial x^h}g^{eh}g_{hg}\left(\frac{\partial^2 x^h}{\partial x'^c\,\partial x'^d}\frac{\partial x^g}{\partial x'^b}-\frac{\partial x^h}{\partial x'^b}\frac{\partial^2 x^g}{\partial x'^c\,\partial x'^d} \right )$$

However, I couldn't show this, despite making lots of attempts. One of more promising attempts is swapping the dummy indices ##g## and ##h## in the second term, and arriving at
$$\frac{1}{2}\frac{\partial x'^a}{\partial x^e}\frac{\partial x^g}{\partial x'^b}\frac{\partial^2 x^h}{\partial x'^c\, \partial x'^d}\left( \frac{\partial x'^d}{\partial x^h}g^{eh}-\frac{\partial x'^d}{\partial x^g}g^{eg}\right )$$
which would have to be proved to be equal to zero.

Unfortunately, I only know the basics of tensors, and don't know how such equation could be proved. As such, I am kindly asking you to post some insight on how such a problem could be tackled. Of course, I have almost certainly already made some errors, and I would be happy if you'd pointed that out as well.
 
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  • #2
Use the fact that the metric is symmetric.
 
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  • #3
I used this fact to derive the last expression in the post, but I haven't been able to go much further. Could you be more specific of where to apply this fact?
 
  • #4
In the last expression, note that ##g## and ##h## are dummy indices, so the two terms in the parentheses are equal.

Your last bit of work wasn't necessary though. In the earlier line, you had this factor
$$\left(\frac{\partial x^g}{\partial x'^b}\frac{\partial^2 x^h}{\partial x'^c\,\partial x'^d}-\frac{\partial x^h}{\partial x'^b}\frac{\partial^2 x^g}{\partial x'^c\,\partial x'^d} \right ),$$ which I rearranged slightly to make its antisymmetry with respect to the indices ##g## and ##h## obvious. Since your multiplying it by ##g_{hg}##, which is symmetric, the product will vanish.
 
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  • #5
vela said:
Since your multiplying it by ghg, which is symmetric, the product will vanish.
Oh, thank you!

vela said:
In the last expression, note that g and h are dummy indices, so the two terms in the parentheses are equal.
I really wanted to use this fact, but I was worried because ##g## and ##h## also show up in the factor before the difference as well. Is this not a problem when cancelling out the terms in the difference?
 
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  • #7

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