I'm studying General Relativity by Hobson and got stuck in understanding the derivation of the transformation law of Christoffel symbols (p. 64). Let $x^a$ and $x'^a$ be two coordinate systems on a manifold. In this book, the Christoffel symbol (or the affine connection) is defined by $$\frac{\partial \vec{e}_a}{\partial x^c} = \Gamma^b_{\,ac} \vec{e}_b \tag{3.12}$$ where $\vec{e}_a$ denote the basis vectors corresponding to the $x_a$ coordinates. After some calculations, the book reaches the transformation law below which can be found in many other sources, e.g. equation (10) here: $$\Gamma'^a_{\,\,bc} = \vec{e}'^a \cdot \frac{\partial \vec{e}'_b}{\partial x'^c} = \cdots = \frac{\partial x'^a}{\partial x^d}\frac{\partial x^f}{\partial x'^b}\frac{\partial x^g}{\partial x'^c}\Gamma^d_{\,fg} + \frac{\partial x'^a}{\partial x^d}\frac{\partial^2 x^d}{\partial x'^c \partial x'^b} \,. \tag{3.16}$$

This is all very well, but I have a trouble with the next step. The book says that by swapping derivatives with respect to $x$ and $x'$ in the last term on the right-hand side of (3.16), we can arrive at an alternative but equivalent expression: $$ \Gamma'^a_{\,\,bc} = \frac{\partial x'^a}{\partial x^d}\frac{\partial x^f}{\partial x'^b}\frac{\partial x^g}{\partial x'^c}\Gamma^d_{\,fg} - \frac{\partial x^d}{\partial x'^b}\frac{\partial x^f}{\partial x'^c}\frac{\partial^2 x'^a}{\partial x^d \partial x^f} \,. \tag{3.17}$$

My question is:

  1. I cannot figure out how the last term in (3.16) can be converted to that in (3.17). Where does the minus sign come from?
  2. I am skeptical about the last term in (3.17) because I reached the same conclusion with this post that this term vanishes. The author actually uses equation (3.17) in p.68 of the textbook, so I'm really confused.

I appreciate any help.

  • 1
    $\begingroup$ The math stack exchange question you linked is incorrect. The primed coords are functions of the original coords $x'^{\mu}(x)$, so his second line is incorrect. Hint: expand the partial derivative of the kronecker delta (which is zero) using the chain rule to get two terms. Using them you quickly see the last terms of (3.16) and (3.17) are equivalent. $\endgroup$
    – Eletie
    Commented Mar 31, 2021 at 17:16
  • $\begingroup$ @Eletie Thank you. Your answer clarified what I was missing. I also proved the equivalence of the last terms thanks to your hint. By the way, is there any other way to see this equivalence without differentiating Kronecker delta? This method doesn't come intuitive. $\endgroup$
    – asdf
    Commented Mar 31, 2021 at 17:40
  • $\begingroup$ Glad the hint helped! And I'm not too sure tbh, but I agree it doesn't look very intuitive. $\endgroup$
    – Eletie
    Commented Mar 31, 2021 at 17:55
  • 1
    $\begingroup$ I think another way of understanding how (3.17) follows from (3.16) is to write an equation similar to (3.16) but for $\Gamma$ in terms of $\Gamma'$, schematically : $\Gamma = M^{-1}MM\Gamma' + M^{-1}\dfrac{d^2x'}{(dx)^2}$ (where $M = \dfrac{dx'}{dx}$). If you want to express $\Gamma'$ from this equation, you bring the non-homogeneous term to the other side (that's how it acquires a minus sign) and multiply both sides by $M M^{-1} M^{-1}$ and as a result you have the non-homogeneous term multiplied by $M^{-1} M^{-1}$, as we see in (3.17). $\endgroup$
    – Viking
    Commented Apr 2, 2021 at 19:24

1 Answer 1


I was also blocking for a time on this point in the book, I came accross your question which was helpfull for me.

In fact, I think the intuitive way is using the derivative of the product of functions:

In 3.16, let assume $f$ to be the first term $ \frac{\partial x^{'a}}{\partial x^{d}} $ and $g'$ to be the second $ \frac{{\partial}^2 x^{d}}{\partial x^{'c}\partial x^{'b}} $ ):

Then considering that $(fg)'=f'g+fg' \rightarrow fg'=(fg)'-f'g$ (this is where the minus sign comes from.

So you then have: $ g=\frac{\partial x^{d}}{\partial x^{'b}} $, $ f'= \frac{{\partial}^2 x^{'a}}{\partial x^{d}\partial x^{'c}}$

which yields that $fg = \delta^a_b$ which derivative is $0$: $(fg)'=0$

and $ -f'g=-\frac{{\partial}^2 x^{'a}}{\partial x^{d}\partial x^{'c}} \frac{\partial x^{d}}{\partial x^{'b}} = -\frac{{\partial}^2 x^{'a}}{\partial x^{d}\partial x^{f}} \frac{\partial x^{d}}{\partial x^{'b}} \frac{\partial x^{f}}{\partial x^{'c}} $, the second term in 3.17

Is all of this correct?

  • $\begingroup$ Yes, it's correct. Your way is exactly what Eletie gave as a hint in the comment. Thank you for your answer! $\endgroup$
    – asdf
    Commented Apr 21, 2021 at 9:54

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