The trivial answer is $$g_{ik}(x) = G(e_i,e_k)$$ where $G$ is the abstract bilinear symmetric map form tangent vectors at point $x$ to the the number field, defining the local geometry, and $e_i = \partial_{x^i}$ is the local basis of tangent vectors, defined by translation of a point with all coordinates constant except $x^i$.

The triviality is best seen by the mixed metric tensor, that pairs coordinate differentials a md tangent vectors $$g^i_k = G(dx^i,e_i) = \delta_ik = \partial_{x^1} dx^k$$

So the diagonal elements are the euclidean lengths squared, the off-diagonal elements are what is called the cos-theorem in euclidean geometry $$g_{ik} = e_i . e_k = |e_i||e_k| \quad \cos (e_i,e_k)$$ In space time $g_{00}$ measures the proper time squared along a curve, positive for a physical worldline and negative for spatial distances. The off-diagonal space-time elements $g_{0k}$ measure Lorentz boost angle between time and spatial directions in the sense of the cos-theorem with imaginary angles.

The trivial answer is $$g_{ik}(x) = G(e_i,e_k)$$ where $G$ is the abstract bilinear symmetric map form tangent vectors at point $x$ to the the number field, defining the local geometry, and $e_i = \partial_{x^i}$ is the local basis of tangent vectors, defined by translation of a point with all coordinates constant except $x^i$.

The triviality is best seen by the mixed metric tensor, that pairs coordinate differentials a md tangent vectors $$g^i_k = G(dx^i,e_i) = \delta_{ik} = \partial_{x^1}\ dx^k$$ So the diagonal elements are the euclidean lengths squared, the off-diagonal elements are what is called the cos-theorem in euclidean geometry $$g_{ik} = e_i . e_k = |e_i||e_k| \quad \cos (e_i,e_k)$$ In space time $g_{00}$ measures the proper time squared along a curve, positive for a physical worldline and negative for spatial distances. The off-diagonal space-time elements $g_{0k}$ measure Lorentz boost angle between time and spatial directions in the sense of the cos-theorem with imaginary angles.