I am beginning to study the Riemannian geometry, and in particular the shape of the volume in a Riemannian collector $ (M, g) $. I was first introduced as a differential $ n $-to form $ dV $ for which $ dV (e_1, cdots, e_n) = 1 $ for any choice (positive) of orthonormal basis $ e_1, cdots, e_n $ in a tangent space $ T_p M $. It's easy to see that in this local framework, with a dual base $ omega ^ 1, cdots, omega ^ n $, we have:

$$ dV = omega ^ 1 wedge cdots wedge omega ^ n $$

That is well defined, as a change of basis to another positive orthonormal framework, say, $ tilde {e} _1, cdots, tilde {e} _n $ with double base $ tilde { omega} ^ 1, cdots, tilde { omega} ^ n $ yields:

$$ dV = det (A) tilde { omega} ^ 1 wedge cdots wedge tilde { omega} ^ n $$

Where $ A $ is the change of base matrix, which must satisfy $ det (A) = 1 $ because both bases are orthonormal and positive. However, I have now seen that in *local coordinates*, the shape of the volume looks like this:

$$ dV = sqrt {g_ {ij}} dx ^ 1 wedge cdots wedge dx ^ n $$

And I'm not sure how to move from the previous representation to this one. I can see that we will have:

$$ dV = det (B) dx ^ 1 wedge cdots wedge dx ^ n $$

Where $ B $ is the change of base matrix with components $ omega ^ i ( frac { partial} { partial x ^ j}) $but i'm not sure how to relate $ B $ in a sensible way to metric. I guess we can also put $ g $ in local coordinates such as:

$$ g = omega ^ i otimes omega ^ i $$

But this does not seem manageable.