Now according to the rules of multiplication, of the fore-going paragraph, the magnitudes
dξ_{σ} = g_{μσ} dx_{μ}
forms a co-variant four-vector, and in fact (on account of the arbitrary choice of dx_{μ}) any arbitrary four-vector. If we introduce it in our expression, we get
ds^2 = g^{στ} dξ_{σ} dξ_{τ}.
For any choice of the vectors dξ_{σ} dξ upside down?]_{τ} this is scalar, and g^{στ}, according to its defintion is a symmetrical thing in σ and τ, so it follows from the above results, that g^{στ} is a contravariant tensor. Out of (16) it also follows that δ^ν_{μ} is a tensor which we may call the mixed fundamental tensor. Determinant of the fundamental tensor.
According to the law of multiplication of determinants,
we have
| g_{μα} g^{αν} | = | g_{μα} | | g^{αν} |
On the other hand we have
| g_{μα} g^{αν} | = | δ^ν_{μ} | = 1
So that it follows (17) that | g_{μν} | | g^{μν} | = 1.
