Non-orthogonal transformations of the inertia tensor The inertia tensor for a point is given by
$$ I_{ij} = m(\delta_{ij} \Vert x\Vert^2-x_i x_j)$$
where $m$ is the mass of the point and $x_{i}$ are the (covariant) coordinates of the given point in an orthonormal coordinate system. I fail to see how this formula fulfills the transformation law for a $(0,2)$-tensor for transformations others than orthogonal ones.
I strarted like this:
Let $A \in GL_3(\mathbb{R})$ be a transformation such that the new coordinates $(y^i)$ and $(y_i)$ behave in the following way
$$A^k_i x^i = y^k$$
$$A^i_k y_i = x_k$$
Then if $I'_{kl}$ is the inertia tensor in $y$-coordinates, we should have
$$A^k_i A^l_j I'_{kl} = I_{ij}$$
The computation then goes as follows:
$$A^k_i A^l_j I'_{kl} = A^k_i A^l_j m(\delta_{kl} \Vert y\Vert^2-y_k y_l) = m(\delta_{kl}A^k_i A^l_j \Vert y\Vert^2-x_i x_j)$$
and now I don't see why for a non-orthogonal transformation we should have
$$\delta_{kl}A^k_i A^l_j \Vert y\Vert^2 = \delta_{ij} \Vert x\Vert^2$$
which would finish the computation.
My only explanation is that maybe the inertia tensor is in fact not given by the same formula for a non-orthonormal coordinate system, so it is not true that
$$I'_{kl} =  m(\delta_{kl} \Vert y\Vert^2-y_k y_l)$$
Do we then simply define
$$I'_{kl} := (A^{-1})^i_k (A^{-1})^j_l I_{ij}~? $$
I would be very thankful for any clarification.
 A: With $g$ denoting a metric in the Euclidean space $\mathbb R^3$, the tensor of inertia of a point mass $m$ can be expressed as
$$I = m\ \left[g(\mathbf r,\mathbf r)\operatorname{Id} - \mathbf r^\flat\otimes\mathbf r^\flat\right],$$
where $\operatorname{Id}$ denotes the identity operator on $\mathbb R^3$, and $\flat$ is the musical isomorphism. The "vector" $\mathbf r$ of $\mathbb R^3$ represent the coordinates of the point mass, and as such needs not satisfy tensor transformation laws. However, when restricting the attention to the orthogonal group, $\mathbf r$ behaves like a (1,0)-tensor, so that the expression of $I$ is a (0,2)-tensor under $O(3)$.
A: I think the confusion is that for GL the metric $g_{ij}$ with 2 lower indices is not the delta function $\delta_{ij}$.  You also have to specify which column the indices are in. Then your equations should be
$$ A^k_{\quad i } \ x^i = y^k $$
$$  x_i (A^{-1})^i_{\quad k} = y_k  \quad  same \ as \quad  A^k_{\quad j} \ y_k= x_j $$
$$ I_{ij}=m(g_{ij}||x||^2-x_ix_j) $$
$$ A^k_{\quad i} \ A^l_{\quad j} \ I'_{kl} = A^k_{\quad i} \ A^l_{\quad j} \ m(g'_{kl} ||y||^2-y_k y_l)= m(g_{ij} ||x||^2-x_i x_j) = I_{ij}  $$
$$ g'_{kl} A^k_{\quad i} \ A^l_{\quad j}  = g_{ij} $$
$$ ||y||^2 = y_k y^k = x_i(A^{-1})^i_{\quad k} A^k_{\quad j } \ x^j = x_i \ \delta^i_{\quad j} \ x^j = x_i x^i = ||x||^2 $$
$$ I'_{kl} =  m(g'_{kl} ||y||^2-y_k y_l) $$
The orthogonal subgroup of GL leaves the diagonal $\delta_{ij}$ function invariant, but non-orthogonal (=strains} of GL do not.
