-2
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Metric tensor multiplied by its inverse. I always see this with different indices.

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2
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Since $g_{ab}g^{cb}=\delta_a^c$ is the identity matrix, taking the trace gives $g_{ab}g^{ab}=D$ in a $D$-dimensional spacetime.

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  • $\begingroup$ And spacetimes are defined as $D \geq 2$, so it won't be 1 $\endgroup$ – Slereah Jan 5 '18 at 6:38
  • $\begingroup$ Why the downvote? $\endgroup$ – J.G. Jan 8 '18 at 12:44

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