These are my calculations $$x'^i_j x^j_k = \sum_{j=1}^n \frac{\partial x'^i}{\partial x^j}\frac{\partial x^j}{\partial x'^k} = \sum_{j=1}^n \frac{\partial x'^i}{\partial x'^k} =n \delta^i_k\ne \delta^i_k?$$

But in general relativity, they always just take $x'^i_kx^j_k=\delta^i_k$ in any dimension of manifold? Surely that only holds in a $1$-manifold?

Notation: $x'^i_j=\frac{\partial x'^i}{\partial x^j}$, $x^i_j=\frac{\partial x^i}{x'^j}$, where we have two coordinate systems $x'=(x'^1,\cdots,x'^n)$ and $x=(x^1,\cdots,x^n)$.

  • $\begingroup$ Related: physics.stackexchange.com/q/66394/2451 $\endgroup$
    – Qmechanic
    Commented Mar 15, 2017 at 7:41
  • 1
    $\begingroup$ What are $x^i$ and $x'^i$? $\endgroup$
    – Javier
    Commented Mar 15, 2017 at 12:35
  • $\begingroup$ @Javier See edit. Apologies for not writing this. $\endgroup$ Commented Mar 18, 2017 at 3:19

1 Answer 1


You've misapplied the chain rule; we have $\dfrac{\partial x'^i}{\partial x'^k}=\sum_j \dfrac{\partial x'^i}{\partial x^j}\dfrac{\partial x^j}{\partial x'^k}$, so $\sum_j x'^i_j x^j_k = \dfrac{\partial x'^i}{\partial x'^k}=\delta^i_k$.

  • $\begingroup$ Have you used Einstein notation in the last steps? I.e. that is a sum over $j=1,\cdots n$ and hence is actually $n\delta^i_k$? $\endgroup$ Commented Mar 15, 2017 at 9:50
  • $\begingroup$ I'll update the answer to make this issue clearer. $\endgroup$
    – J.G.
    Commented Mar 15, 2017 at 13:20
  • $\begingroup$ I still don't understand, it would appear that: $$\sum_j x'^i_jx^j_k= (\sum_{j=1}^n 1)\frac{\partial x'^i}{\partial x'^k}=(n)\frac{\partial x'^i}{\partial x'^k} = n\delta_k^i ?$$ $\endgroup$ Commented Mar 18, 2017 at 3:16
  • $\begingroup$ @DisplayName the chain rule already includes the sum over $j$, you can't factor it out. $\endgroup$
    – Javier
    Commented Mar 18, 2017 at 3:34

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