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I was reading Schutz, A First Course in General Relativity. On page 9, he argued that the metric tensor is symmetric:

$$ ds^2~=~\sum_{\alpha,\beta}\eta_{\alpha\beta} ~dx^{\alpha}~dx^{\beta} $$ $\text{Note that we can suppose}$ $\eta_{\alpha\beta}=\eta_{\beta\alpha}$ $\text{for all}$ $\alpha$ $\text{and}$ $\beta$ $\text{since only the sum}$ $\eta_{\alpha\beta}+\eta_{\beta\alpha}$ $\text{ever appears in the above equation when}$ $\alpha\neq\beta$.

I don't understand his argument. If someone can explain why, I would really appreciate it.

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marked as duplicate by innisfree, Kyle Oman, ACuriousMind, John Rennie general-relativity Jun 16 '15 at 5:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ That question has some good answers already... $\endgroup$ – innisfree Jun 15 '15 at 18:02
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    $\begingroup$ I don't think this is a duplicate. The proposed dupe is asking if there are antisymmetric metrics; this question is asking how the quoted argument means that the metric is symmetric. $\endgroup$ – Kyle Kanos Jun 15 '15 at 18:37
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Assume $\eta_{\alpha\beta} \neq \eta_{\beta\alpha}$. Because it's irrelevant what letter we use for our indices, $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}.$$ Then $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2}(\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}$$ so only the symmetric part of $\eta_{\alpha\beta}$ would survive the sum. As such we may as well take $\eta_{\alpha\beta}$ to be symmetric in its definition.

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    $\begingroup$ what about the anti-symmetric contribution to e.g. $\eta_{ab} p^a q^b$ such that $p^a q^b \neq p^b q^a$ $\endgroup$ – innisfree Jun 15 '15 at 18:00
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    $\begingroup$ If you assume $\eta_{\alpha\beta}\ne\eta_{\beta\alpha}$ then you cannot claim the next relation in your answer. Because of the commutation between $dx^\alpha$ and $dx^\beta$, you have essentially said "$\eta_{\alpha\beta}$ is symmetric, therefore we can assume it is symmetric". Where have you established that the metric is equal to its symmetric components? $\endgroup$ – Jim Jun 15 '15 at 18:09
  • $\begingroup$ @Jims it says that only the symmetric part of $\eta_{ab}$ contributes in a contraction with a symmetric tensor $S^{ab}$ such as $S^{ab} = dx^a dx^b$. $\endgroup$ – innisfree Jun 15 '15 at 18:18
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    $\begingroup$ huh? i would have said, we aren't Math.SE so stuff like that can be taken for granted! $\endgroup$ – innisfree Jun 15 '15 at 18:28
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    $\begingroup$ What identity? I obviously just relabeled the indices. $\endgroup$ – FenderLesPaul Jun 15 '15 at 18:47
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The metric tensor is created from the spacetime interval equation. On top of that, $[dx^\alpha,dx^\beta]=0$. Suppose we have a 1+1 dimensional spacetime, if you are given an interval equation resembling:

$$ds^2=-a\,dx_0^2+b\,dx_1^2+c\,dx_0dx_1$$

Obviously, $\eta_{0\,0}=-a$ and $\eta_{1\,1}=b$. Now, you can assume that, say, $\eta_{0\,1}=c/3$ and $\eta_{1\,0}=2c/3$ such that you still get the $c\,dx_0dx_1$ term, but why would you? You know $c=\eta_{1\,0}+\eta_{0\,1}$ and that's the only requirement you have. Since $dx_0$ and $dx_1$ commute, it is easier for everybody if you just say the tensor is symmetric and set $\eta_{1\,0}=\eta_{0\,1}=c/2$. Nothing changes except that you have saved yourself some trouble later on.

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