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For something like $g^{ij} n_i h_{kj}$, how do I know which one should the metric operate on? $n_i$ or $h_{kj}$? The results could be $n^j h_{kj}$ or $n_i h^i_{k}$, which are different.

The question also appears when encountering partial derivatives too, like $g_{ij}\partial_i h^i$

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  • $\begingroup$ The results could be $n^j h_{kj}$ or $n_i h^i_{k}$, which are different. Why do you think those are different? $\endgroup$ – G. Smith Apr 11 at 17:39
  • $\begingroup$ What I was caring is the notation. Do you mean both give one-form as results so they have the same physical meaning? $\endgroup$ – Simon219 Apr 11 at 17:42
  • $\begingroup$ When you studied Special Relativity, did you understand that $a^\mu b_\mu$ and $a_\mu b^\mu$ are the same contraction $a\cdot b$? $\endgroup$ – G. Smith Apr 11 at 17:47
  • $\begingroup$ I thought that case is different from this one cuz $g_{\mu \nu} a^\mu b^\nu$, no matter how I operate, it still gives me a vector and an one-form. While this case gives vector with (0 2) tensor and one-form with (1 1) tensor. Now I see summing up gives same h_k, thanks! $\endgroup$ – Simon219 Apr 11 at 17:59
  • $\begingroup$ But what about the partial derivative case? For example, consider $\partial_i \partial_j h^i$, I could think of $\Box h^i$ and $\partial_i \partial_j h_j$. It seems to me that they are not the same. $\endgroup$ – Simon219 Apr 11 at 18:04
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Those two expressions are equivalent. When contracting tensor indices, it doesn’t matter which one is up and which is down. So it doesn’t matter which one you contact with the (inverse) metric first.

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