The basis expansion of inertia tensor I want to express the inertia tensor in terms of the tensor products for cartesian coordinates in a Euclidean space, and for a general manifold
Consider the Euclidean space: $(\mathbb R^3, g)$ (where $g$ is the euclidean metric in cartesian coordinates) and the Angular momentum (in cartesian coordenates):
$$\textbf{L} = m\textbf{r}\times(\textbf{w}\times\textbf{r}) $$
where $\textbf{w}$ is the angular velocity vector.
So, we know that
$$\textbf{r}\times(\textbf{w}\times\textbf{r}) = \langle\textbf{r},\textbf{r}\rangle \textbf{w} -\langle\textbf{r},\textbf{w}\rangle\textbf{r}  $$
Then,using the summation convention in some parts,:

$$\textbf{L} = m[\langle\textbf{r},\textbf{r}\rangle \textbf{w} -\langle\textbf{r},\textbf{w}\rangle\textbf{r}]  \implies  L_i = m[ (\delta_{ij}r^ir^j)w^i- (r^i w^j) r^j] = m[r^ir_i (w_j\delta ^{ij})- (r^iw^j) r^j] = m[r^ir_i (w_j\delta ^{ij})- (r^ir^j) w^j] = m[r^ir_i (w_j\delta ^{ij})- (r^iw^j) r^j] = m{[r^ir_i\delta ^{ij}- (r^ir^j)]w_j} = \displaystyle \sum_{i} \sum_{j} m{[r^ir_i\delta ^{ij}- (r^ir^j)]w_j} = \sum_{j}I^{ij} w_j = \sum_{j}I_{ij} w_j $$

where: 
 $w_j\delta ^{ij} = w^i\\ r^iw^j = \langle\textbf{r},\textbf{w}\rangle \\ r^ir_i = \delta_{ij}r^ir^j = \langle\textbf{r},\textbf{r}\rangle$
and $I^{ij} = I_{ij}$ because the coordinate system is cartesian i.e. $g = \delta_{ij}\mathbb d\mathbb r\otimes\mathbb d\mathbb r$.
So how can I write the tensor in terms of tensor products?
My guess:

$$I_{ij} = m \sum_{i} \sum_{j} [r^ir_i\delta ^{ij}- (r^ir^j)] = m \sum_{i} \sum_{j} [r^ir_j - (r^ir^j)]\iff \\ \textbf{I} = g - \mathbb r \otimes \mathbb r = \sum_{i} \sum_{j} \delta_{ij} \mathbb d\mathbb r^i\otimes\mathbb d\mathbb r^j - r^ir^j\mathbb e_{i} \otimes \mathbb e_{j} $$

 A: Your index balance isn't good: for example, the expression $\delta_{ij}r^ir^j w^i$ is meaningless for Einstein's convention. When you wrote $L_i$ on the left-hand side, $i$ is not a dummy index anymore, so to write $\langle {\bf r},{\bf r}\rangle$ you must use some other indices, as in $\delta_{jk}r^jr^k$. Moreover, vectors have contravariant components, so we should go with $L^i$ instead of $L_i$ to avoid confusion. So: $${\bf L} = m(\langle {\bf r},{\bf r}\rangle {\bf w} - \langle {\bf r},{\bf w}\rangle{\bf r}) \implies L^i = m(\langle {\bf r},{\bf r}\rangle w^i - \langle{\bf r},{\bf w}\rangle r^i),$$and now we have $$L^i = m(\delta_{jk}r^jr^kw^i - \delta_{jk}r^jw^kr^i).$$Einstein's convention says we have a double sum $\sum_{j,k}$ there. I assume the definition of the inertia tensor (in coordinates) is $L^i = I^{ij}w_j$, where $w_j = \delta_{j\ell}w^\ell$. Note that we can actually keep the quantity $\langle{\bf r},{\bf r}\rangle$ as it is a number, to prevent indices from proliferating like a plague. Procceeding, we have $$L^i = m(\langle{\bf r},{\bf r}\rangle \delta^{ij}w_j - r^jr^iw_j) = m(\langle{\bf r},{\bf r}\rangle \delta^{ij}-r^ir^j)w_j \implies I^{ij} = m(\langle{\bf r},{\bf r}\rangle \delta^{ij}-r^ir^j).$$
In coordinate free notation, the last expression would be ${\bf I} = m(\langle{\bf r},{\bf r}\rangle {\rm Id} - {\bf r}\otimes {\bf r})$.
