Why does the determinant of the metric tensor depend on the coordinates

Question

It is my understanding that the matric tensor field $g$ of a diff. manifold $M$ is a section of the tensor bundle $T^*M\otimes T^*M$. So that $g:p\mapsto g_p\in T_p^*M\otimes T_p^*M$ is the assignment of an inner product on the tangent space at $p\in M$ and it induces an isomophism $\hat{g}_p:T_pM\to T_p^*M$ given by $X\mapsto \bigr(Y\mapsto g_p(X,Y)\bigr)$ However, this assignment is defined independent on the charts, and thus it would seem like $\text{det } g_p$ is well defined independent on any coordinate choice. However, when dealing with, for example 'spherical coordinates' on $\mathbb R^3$, we can write the metric tensor field $g$ as
\begin{align}
g=dr\otimes dr+r^2d\theta\otimes d\theta+r^2\sin(\theta)d\varphi\otimes d\varphi.
\end{align}
Now, in the basis $\{\partial_r,\partial_\theta,\partial_\varphi\}$ and $\{dr,d\theta,d\varphi\}$ for $T_pM$ and $T_p^*M$, the linear map $\hat{g}_p$ takes the matrix form
\begin{align}
\hat{g}_p=
\begin{pmatrix}
1 & 0 & 0 \\
0 & r^2 & 0 \\
0 & 0 & r^2\sin(\theta)
\end{pmatrix}.
\end{align}
This matrix has determinant $r^4\sin^2(\theta)$, where as in cartesian coordinates $\det \hat g_p=1$. I do not know how to reconcile this with the fact that the determinant is well defined without any choice of basis. What is it that I am not understanding?

Answer 1

(Too long for a comment)

The isomorphism you stated clearly does not depend on the choice of basis.

In order to be more specific, let $V$ be a vector space and $\varphi:V\to V$ be a linear map. In this case you can define the determinant of $\varphi$ by taking the determinant of some particular matrix representation. This definition is independent of the choice of basis, since the matrix representation of $\varphi$ with respect to another basis and the original matrix representation are related by conjugation. (and the determinant is invariant under conjugation)

However, if you have two different vector spaces $V,W$ and a linear map $\psi:V\to W$, then there is no way to define the determinant of this map (in an invariant way). Therefore it doesn't really make sense to talk about the determinant of the map that you called $\hat{g}_p$ since it's not a map from $T_pM$ to itself.