TITLE: Ellipse naturally associated with a polygon
QUESTION [25 upvotes]: My colleagues and I have stumbled onto a way to associate an ellipse, or equivalently a positive definite symmetric matrix, to a polygon that is different from other better known ways. We want to know if anyone has ever seen this before. Before describing this, I do want to note that there are other well-known ways of associating an ellipse to a polygon, including the matrix of second moments of the uniform distribution supported on the interior of the polygon and the so-called John ellipses, which are the ellipse of largest volume inscribed in the polygon and the one of smallest volume circumscribing the polygon.
Here is the ellipse we found: Given a polygon $P$ that contains the origin in its interior, let $\ell_1, \dots, \ell_n$ denote the lines that contain the sides of $P$. For each $i$, let $n_i$ denote the unit vector orthogonal to $\ell_i$, $h_i$ be the distance from the origin to the line $\ell_i$, and $s_i$ be the length of the side lying in $\ell_i$. For each $v \in R^2$, define
$ q(v) = \sum_{i=1}^n (v\cdot n_i)^2\frac{s_i}{h_i}. $
Let $E_P = \{ q(v) \le 1\}.$ The ellipse $E_P$ does not behave particularly nicely under translations of $P$. It however, behaves nicely under linear transformations. In particular, if $A$ is an invertible linear transformation with determinant $1$, then $E_{AP} = AE_P$. It is scale-invariant in that $E_{tP} = E_P$ for any $t > 0$.
The ellipse $E_P$ can in fact be defined without using the inner product and is a linear invariant of the polygon $P$. Its definition can be extended to any body $P$ containing the origin with a sufficiently regular boundary. It is an example of a matrix-valued valuation on the set of all convex bodies that contain the origin in their interior. Its volume is maximized if and only if the body $P$ is itself an ellipsoid centered at the origin.
Does anyone recognize this association of an ellipse to a polygon? I would be grateful for any information or references.
EDIT: Vladimir's comments below, especially the first sentence, indicate clearly that I omitted something crucial. The definition of the ellipsoid depends on the choice of the volume form in the ambient vector space (and the dual volume form in the dual vector space). Changing the volume form will rescale the ellipsoid by a factor. However, if you fix the volume form, then the ellipsoid is invariant under rescaling of the convex body.
This observation, however, makes the existence of a scale-invariant ellipsoid associated with the polygon much less surprising.

REPLY [8 votes]: Maybe it would have been more convincing if you had given the definition of $q_K$ in a unimodularly equivariant way from the beginning.  That's not at all hard to do, and it's also easy to see how to create many more such (and how to create many more in all dimensions).
For example, suppose $K\subset V$ has vertices $v_0,v_1,\ldots,v_n=v_0$ (cyclically ordered counterclockwise, say).  Let $\alpha_i\in V^\ast$ be the unique element that satisfies $\alpha_i(v_i)=\alpha_i(v_{i+1})=1$.  Then 
$$
q_K = \sum_{i=0}^{n-1} \Omega(v_i,v_{i+1})\ {\alpha_i}^2
$$
(where $\Omega$ is the area form.)  This is clearly unimodularly equivariant with respect to $K$ and, since, for $K' = tK$ (with $t>0$), one has $v'_i = tv_i$ and  $\alpha_i' = (1/t)\alpha_i$, it follows that $q_{tK} = q_K$.  Of course, anything like this would have worked.  For example, you could have taken
$$
\tilde q_K = Area_\Omega(K)\left( \sum_{i=0}^{n-1}  {\alpha_i}^2\right),
$$
and this would also have had the same equivariance property.
In dimension $n$, I think that the right formula would be to define, for each face $F$ of $K$, the element $\alpha_F\in V^\ast$ to be the linear function that equals $1$ on $F$, let $\Omega(F)$ denote the volume of the cone with vertex $0\in V$ whose base is $F$, and then set
$$
q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2.
$$
If you want it to be invariant under scaling, you should take
$$
q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)^{2/n}\ {\alpha_F}^2.
$$
Perhaps, better, though, would be to take 
$$
q_K = Vol_\Omega(K)^{(2-n)/n}\ \left(\sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2\right),
$$
since this is also invariant under subdivision of the faces of $K$.