The coordinate transformation law (from coordinates x to coordinates y) for the Christoffel symbol is: $$\Gamma^i_{kl}(y)=\frac{\partial y^i}{\partial x^m} \frac{\partial x^n}{\partial y^k} \frac{\partial x^p}{\partial y^l} \Gamma^m_{np}(x) +\frac{\partial ^2x^m}{\partial y^k \partial y^l} \frac{\partial y^i}{\partial x^m}$$ Therefore the transformation law for the contracted Christoffel symbol would be: $$\Gamma^i_{il}(y)=\frac{\partial y^i}{\partial x^m} \frac{\partial x^n}{\partial y^i} \frac{\partial x^p}{\partial y^l} \Gamma^m_{np}(x) +\frac{\partial ^2x^m}{\partial y^i \partial y^l} \frac{\partial y^i}{\partial x^m}$$

$$\Gamma^i_{il}(y)= \delta^n_m\frac{\partial x^p}{\partial y^l} \Gamma^m_{np}(x) + \frac{\partial }{\partial y^i }(\frac{\partial x^m}{\partial y^l}) \frac{\partial y^i}{\partial x^m}$$

$$\Gamma^i_{il}(y)= \frac{\partial x^p}{\partial y^l} \Gamma^n_{np}(x) + \frac{\partial }{\partial x^m }(\frac{\partial x^m}{\partial y^l}) $$ My question is, does the term $\frac{\partial }{\partial x^m }(\frac{\partial x^m}{\partial y^l})$ equal zero? My reasoning is that since partial derivatives commute, the order of differentiation can be reversed: $$\frac{\partial }{\partial x^m }(\frac{\partial x^m}{\partial y^l})=\frac{\partial }{\partial y^l }(\frac{\partial x^m}{\partial x^m})=\frac{\partial }{\partial y^l }(\delta^m_m)=0$$ That would mean that the contracted Christoffel symbol transforms like a tensor, specifically a vector, and therefore is a tensor. However, I am unsure if it's correct to commute partial derivatives of different coordinates.

Any clarification would be greatly appreciated!

  • 1
    $\begingroup$ As a counter example to the commution consider $x^m=(x,y)$ and $y^l=(r,\theta)$ as the usual polar coordinates. If you calculate $\partial_y(\frac{\partial x}{\partial\theta})$ you get $-1$ which is not zero. You can't use commutation of partial derivatives because you are calculating something of the form $x(r(x,y),\theta(x,y))$ and the commutation breaks down (I can't explain exactly why but it should feel reasonable). $\endgroup$ Commented Nov 16, 2022 at 16:36

2 Answers 2


The commutation of partial derivatives doesn't work here. When you write $$\frac{\partial}{\partial y^i} \frac{\partial y^j}{\partial x^k}$$ you are utilizing shorthand in a dangerous way - in $\partial/\partial y^i$ you are using the $y$'s as variables with respect to which you may differentiate functions, but in $\partial y^j/\partial x^k$ you are using the same symbols to refer to functions of the $x$ variables.

Formally, a partial derivative is defined as $$\frac{\partial f}{\partial x^k} = \lim_{\epsilon\rightarrow 0}\frac{f(x_1,\ldots,x_k+\epsilon,\ldots) - f(x_1,\ldots,x_k,\ldots)}{\epsilon}$$

When we write $\partial y^j/\partial x^k$, we are taking $y^i$ to be a function of the $x$'s and then applying a difference quotient with respect to the $x^k$ entry. Simple enough. But now think about what it would mean to differentiate the result with respect to $y^i$.

The result of the initial derivative would be some function of the $x$'s. We would then have to write the $x$'s as functions of the $y$'s in order to perform the second differentiation - a process which is doable, but not taken into account when deriving the usual commutation of partial derivatives.

To avoid problems like this, we must not try to differentiate with respect to different sets of coordinates at the same time (or if we do, we certainly can't assume that such derivatives commute). We can undo your work by re-express the second derivative operator as $$\frac{\partial}{\partial y^i} = \frac{\partial x^\ell}{\partial y^i} \frac{\partial}{\partial x^\ell}$$ (which encapsulates the procedure I mention in the previous paragraph, in which we express the $x$'s as functions of the $y$'s) at which point we obtain $$\frac{\partial}{\partial y^i} \frac{\partial y^j}{\partial x^k} = \frac{\partial x^\ell}{\partial y^i} \frac{\partial ^2 y^j}{\partial x^\ell \partial x^k}$$ which is generically non-zero.


One easy way to see that this can't be right is to note the well known identity ($g$ is the determinant of the metric, and $\epsilon$ is the Levi-Civita symbol):

$\partial_{a}\ln\sqrt{g} = \Gamma_{ba}{}^{b}$

which you can prove by remembering that $g = \frac{1}{n!}\epsilon^{abc...}\epsilon^{ABC...}g_{aA}g_{bB}g_{cC}...$

Then, since it should be relatively obvious that the determinant of the metric does not transform like a scalar (compare polar coordinates and cartesian coordinates on the plane), then neither can the other side of this equation.


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