Non-orthogonal transformations of the inertia tensor

Question

The inertia tensor for a point is given by 
$$ I_{ij} = m(delta_{ij} ||x||^2-x_i x_j)$$
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:

$$A^k_i A^l_j I'_{kl} = A^k_i A^l_j m(delta_{kl} ||y||^2-y_k y_l) = m(delta_{kl}A^k_i A^l_j ||y||^2-x_i x_j)$$

and now I don't see why for a non-orthogonal transformation we should have

$$delta_{kl}A^k_i A^l_j ||y||^2 = delta_{ij} ||x||^2$$

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

$$I'_{kl} =  m(delta_{kl} ||y||^2-y_k y_l)$$

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.

Phoenix87 · Answer

With $g$ denoting a metric in the Euclidean space $mathbb R^3$, The tensor of inertia of a point mass can be expressed as

$$I = m(g(mathbf r,mathbf r)operatorname{Id} - mathbf r^flatotimesmathbf r^flat),$$

where $operatorname{Id}$ denotes the identity operator on $mathbb R^3$, and $flat$ is the musical isomorphism. The "vector" $mathbf r$ of $mathbb R^3$ are coordinates, and coordinates need not satisfy tensor transformation laws. However, when restricting to the orthogonal group, $mathbf r$ behaves like a (1,0)-tensor, so that the expression of $I$ is a (0,2)-tensor under $O(3)$.

Gary Godfrey · Answer

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.

Non-orthogonal transformations of the inertia tensor

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