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If I understand correctly, one way to get the components of a metric tensor (treating it like a matrix here) is to look at the ds interval. Isn't that interval always in terms of sums of dr squared plus d theta squared etc, meaning that the metric tensor will only have nonzero values for x^ix^j (using the caret for super scripts) when i=j? If this is not the case, can anyone give an example?

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  • $\begingroup$ The carrot is a caret, really. $\endgroup$ – Mariano Suárez-Álvarez Jun 24 '14 at 4:43
  • $\begingroup$ Oh haha oops, thanks. $\endgroup$ – Yadnarav3 Jun 25 '14 at 1:07
  • $\begingroup$ I would vote both for the answer but it seems I can't. $\endgroup$ – Yadnarav3 Jun 25 '14 at 1:13
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Being diagonal is a coordinate-dependent concept: the components of the matrix associated to the metric tensor depend on the coordinate system you use. Thus a very simple example of a non-diagonal metric is the standard Euclidean metric $\delta = dx^2 + dy^2$ on $\mathbb R^2$ in the coordinate system $(x,z) = (x, x+y)$, where it has the coordinate expression $$\delta = dx^2 + d(z-x)^2 = 2dx^2 + dz^2 - 2 dx dz.$$

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No, in fact, there's some very famous solutions that have non-diagonal metrics. Such as the Kerr metric for a rotating black hole in General relativity.

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