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so I am quite confused regarding the spatial metric tensor $g_{ij}$. If I have $g_{ij}g^{ij}$ I essentially get the trace of the metric tensor $g$ right? Or, do I get $\delta^i_i = 3$ instead?

The second part to this is, does curvature matter? In a Minkowski metric, I understand that I'll get $3$ either way, but what if im considering spherical coordinates with non vanishing curvature constant? Is it still $3$ or is it something totally different? Thank you for your help

Qmechanic
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fp007
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    $g_{ij}g^{ij}=4$ not three for Minkowski metric. Also please refer https://physics.stackexchange.com/q/212421/330899 and https://math.stackexchange.com/q/4642539/1053268 – GedankenExperimentalist Mar 21 '23 at 01:01
  • @GedankenExperimentalist Referring specifically to the spatial part of the metric so $g_{ij}^{(3)}$ Does it change from $g_{ij}g^{ij} = 3$ if we include curvature and go to the spherical system or is it always 3 ? Thanks – fp007 Mar 21 '23 at 02:07
  • I don't think this question should have been closed - people seemed to miss the point about the three-metric. – Eletie Mar 21 '23 at 09:51

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The inverse metric is defined such that $g^{ij} g_{jk} = \delta^i {}_k$. This implies that $g^{ij} g_{ji} = \delta^i {}_i = D$, the dimensionality of the space you're in. This is true regardless of the coordinates you use or whether the metric is curved.

Michael Seifert
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