Metric Tensor Antisymmetry

Question

The metric tensor on a Riemannian manifold is given as a symmetric $n \times n$ symmetric matrix (so $g_{ij} = g_{ji}$). Is there an intrinsic reason for this symmetry? Why can't it be antisymmetric (so $g_{ij} = -g_{ji}$), and what would be the physical meaning of the antisymmetry?

Answer 1

Let's be very general. Assume we are taking the metric tensor of the given point $g = g_{ij}(x)$. Assume everything in arbitrary coordinate system. Assume $M$ to be a arbitrary differentiable manifold, for an arbitrarily given metric tensor $g$.

For a given vector x, y: $$ x\cdot y = g_{ij}(x)x^i y^j $$

For a commutative inner product of vectors implies a symmetric metric tensor: $$ x\cdot y = g_{ij}x^i y^j = g_{ij}x^j y^i = g_{ji}x^i y^j = g_{ji}x^j y^i = y\cdot x $$

This implies: $ g_{ij} = g_{ji}$. Also very important: $$ |x|^2 = g_{ij}x^i x^j = g_{ij}x^j x^i = g_{ji}x^i x^j = g_{ji}x^j x^i, \quad\forall x\in M $$

The symmetric property is so important that some times the metric tensor is defined to be this way.