In physics we have a particular tensor called inertia tensor. In abstract notation, this tensor can be written as:
$$ \textbf{I} = m [\langle \textbf{r},\textbf{r} \rangle \textbf{Id} - \textbf{r} \otimes \textbf{r}]$$
Where, $m$ is the mass, $\textbf{r}$ is the position (contravariant) vector and $\textbf{Id}$, is the identity tensor.
But now, consider this term
$$\langle \textbf{r},\textbf{r} \rangle \textbf{Id}$$
On the one hand, this is clearly a scalar $\langle \textbf{r},\textbf{r} \rangle$ times a map (identity tensor). On the other hand, if I say that the Manifold is a euclidean space $(\mathbb{R}^{3},g)$ then of course that the metric tensor is unitary, and then:
$$ \langle \textbf{r},\textbf{r} \rangle = \delta_{ij}r^ir^j = r^2$$
but in general case that the metric is not unitary, then:
$$ \langle \textbf{r},\textbf{r} \rangle = g_{ij}r^ir^j \neq r^2$$
So, my question is, in a general case, the following equation is valid?
$$\langle \textbf{r},\textbf{r} \rangle \textbf{Id} = \textbf{g} \otimes \textbf{Id}$$
and then (in general case):
$$ \textbf{I} = m [\textbf{g} \otimes \textbf{Id} - \textbf{r} \otimes \textbf{r}]$$