# $x'^i_j x^j_k = n\delta^i_k$ rather than $1\delta^i_j$?

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 $x=(x^1,\cdots,x^n)$.

• Commented Mar 15, 2017 at 7:41
• What are $x^i$ and $x'^i$? Commented Mar 15, 2017 at 12:35
• @Javier See edit. Apologies for not writing this. Commented Mar 18, 2017 at 3:19

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