Invariance of determinant of metric tensor Given any 2-tensor on a Riemannian manifold $M$ equipped with metric $g,$ we have a coordinate-free definition of its trace: 
$$\operatorname{trace}(T)=g^{ij}T_{ij}= T_i^i.$$ In particular, we have $$\operatorname{trace}(g)= g^{ij}g_{ij}= \dim(M).$$
On the other hand, the determinant as I have seen it defined is simply the usual determinant of the matrix $(g)_{ij}$ written in some coordinate system. In this definition, the determinant depends on coordinates. (Example: the riemannian volume form is $\mu= \sqrt{\det(g_{ij})} \,dx^1 \wedge \dots \wedge dx^n$)). 
My (possibly naive) aquestion: is there a coordinate-invariant notion of a determinant, which makes the determinant of a metric $g$ a smooth function on the manifold? 
 A: The determinant of the metric tensor does indeed depend on the coordinates. This already happens in the basic example of the Euclidean space $\mathbb R^3$. In the standard coordinates $(x_1, x_2, x_3)$ the metric tensor is just $$
\mathbf x\cdot \mathbf x= x_1^2+x_2^2+x_3^2.$$
The associated matrix is the identity and its determinant is 1. Changing coordinates, letting
$$
x_1'=\frac12 x_1, \quad x_2'=x_2, \quad x_3'=x_3, $$
we have that
$$
\mathbf x\cdot \mathbf x= 4(x_1')^2+(x_2')^2+(x_3')^2.$$
The associated matrix now is
$$
\begin{bmatrix} 
4 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1 
\end{bmatrix}
$$
and it has determinant $4$.
What does this little example tell us? That the determinant of the metric tensor will never be a coordinate-invariant quantity, because it measures a property of the coordinate system. In this Euclidean example, the determinant of the metric tensor measures the squared volume of the parallelepiped spanned by the triads $(1,0,0), (0, 1,0), (0,0,1)$ (for the coordinates $(x_1, x_2, x_3)$) and $(2,0,0), (0,1,0), (0,0,1)$ (for the coordinates $(x_1', x_2'. x_3')$).
Remark. In the Schwarzschild model of general relativity, the metric tensor apparently degenerates on a certain sphere; choosing the "naïve" coordinate system the determinant vanishes. Changing coordinates, the determinant does not vanish anymore, thus showing that that sphere was only a coordinate singularity, not an actual singularity. I am not writing more details, but I wanted to record this instance in which the dependence on the determinant on coordinates is actually a useful fact, not a nuisance.
