In Spivak's book Calculus on Manifolds, he describes the definition of a coordinate system as being, given some $k$-manifold $M$, a 1-1 differentiable function $f: W \to \mathbb{R}^n$ which is such that $f(W) = M \cap U$, $f'(u)$ has rank $k$ for each $y \in W$ and where $f^{-1}:f(W) \to W$ is continuous.
Now, there are several quantities that he is able to show are coordinate independent, such as the tangent space of $M$ at $x$, and also the unit normal vector $n(x)$.
I am stuck on an exercise that extends these concepts: Showing that $\det((g^{ij})(h_{ij}))$ is independent of the choice of coordinate, where it is defined that $(g^{ij}) = (g_{ij})^{-1} = (D_if \cdot D_jf)^{-1}$ and $(h_{ij}) = D_if \cdot D_jn$. Here, these $D_i$ are partial derivatives, $f$ is a coordinate system and $n$ is the unit normal vector to an oriented manifold $M$.
The unit normal $n(x)$ is defined as being the unique vector, given $f(0)=x$, which is equal to $f_*(v_0)$ for some $v_0 \in \mathbb{R}^k$ with $v^k < 0$.
I am not sure how to show such coordinate independence. I tried first expanding the determinant using rules of determinants of inverses, and when doing this I get stuck with a fraction involving determinant of the $h$ in the numerator and determinant of $g$ in the denominator. I am not sure how to proceed though, and further I'm not sure I see how this formula is coordinate independent. In particular, I don't see how to incorporate the unit normal vector into the proof.
Any help would be greatly appreciated! Thanks!
Say we have two coordinate systems $y$ and $x = x(y)$. In $x$ coordinates we have the function $\phi(x) = \det(G(x)^{-1}Df(x)^TDn(x))$. In $y$ coordinates we set $\tilde{\phi}(y) = \det(\tilde{G}(y)^{-1}D\tilde{f}(y)^TD\tilde{n}(y))$. We are asked to show that $\phi(x) = \tilde{\phi}(y)$. Using formulas like $\tilde{f}(y) = f(x)$, $D\tilde{f}(y) = Df(x)Dx(y)$, it is straightforward to show that $$\tilde{G}(y)^{-1}D\tilde{f}(y)^TD\tilde{n}(y) = Dx(y)^{-1}G(x)^{-1}Df(x)^TDn(x)Dx(y).$$ Using the product rule of the determinant, we obtain the result. Note that the above shows even that the linear transformation $G(x)^{-1}Df(x)^TDn(x)$ is coordinate independent when interpreted as a linear map from $T_{f(x)}M$ to $T_{f(x)}M$.