# Is this posible in GR $g_{ab}g^{ab}=1$? [duplicate]

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

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.
• And spacetimes are defined as $D \geq 2$, so it won't be 1 – Slereah Jan 5 '18 at 6:38