# Metric Tensor times its inverse (non-zero curvature)

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

• $g_{ij}g^{ij}=4$ not three for Minkowski metric. Also please refer physics.stackexchange.com/q/212421/330899 and math.stackexchange.com/q/4642539/1053268 Mar 21 at 1:01
• @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 Mar 21 at 2:07
• I don't think this question should have been closed - people seemed to miss the point about the three-metric. Mar 21 at 9:51

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.