Ellipsoids and their defining inner product

Question

$E\subset\mathbb{R}^n$ is an ellipsoid if $E = E(g):= \{x\in \mathbb{R}^n \mid x^t g x \le 1\}$ for some inner product $g$ on $\mathbb{R}^n$. Given an ellipsoid $E\subset\mathbb{R}^n$, how unique is $g$ such that $E=E(g)$? Is there a formula for $g$ such that $E=E(g)$ (see the note below for what kind of formula I envisage)? If $T\in SL(n,\mathbb{R})$ satisfies $E_1 = T(E_2)$ for two ellipsoids $E_1$ and $E_2$ of the same volume, does it follow that $T^t g_1 T = g_2$? What if $T$ is orthogonal or symplectic? Thanks for giving a hint or a reference.

Note: Given an ellipsoid $E$, then, I guess, $A_{ij} = \int_E (r^2 \delta_{ij} - r_i r_j) dV$ and $B_{ij} = \int_E r_i r_j d V$ are inner products; they don't give $g$, though, but are related to it. It is clear from these formulas that if $E_1 = T(E_2)$, then $T^t A_1 T = A_2$ provided $T\in SO(n)$, and $T^t B_1 T = B_2$ provided $T\in SL(n)$.

Answer 1

The Binet-Legendre metric of an ellipsoid $E\subset \mathbb{R}^n$ is defined as $g_F$, the metric dual to $$ g_F^*(\xi,\eta)=\frac{n+2}{\operatorname{Vol}(E)}\int_E \xi(x)\eta(x) dx. $$ where the volume and integral are computed using a translation invariant Lebesgue measure. Note that rescaling the choice of measure has no effect. Vladimir S. Matveev, Marc Troyanov, The Binet-Legendre Metric in Finsler Geometry, arXiv:1104.1647 prove that $E=\{x\in \mathbb{R}^n|g_F(x,x)\le 1\}$ for any ellipsoid $E$. This is not quite an explicit integral formula, because you still need to invert the symmetric matrix $g_F^*$, in any linear coordinate system, to get $g_F$, but matrix inversion is an explicit algebraic map.

Answer 2

Let $E\subset\mathbb{R}^n$ be an ellipsoid, and let $g$ be an inner product on $\mathbb{R}^n$ such that $E = E(g)$. Let $\|\cdot\|$ denote the corresponding norm. If $\|\cdot\|'$ is another norm such that $E = \{x\in \mathbb{R}^n \mid \|x\|'\le 1\}$, then it holds $$ \|x\|' = \bigl\| \frac{x}{\|x\|}\bigr\|'\|x\| = \|x\|\quad\text{for all }x\in \mathbb{R}^n\backslash\{0\}, $$ where the second equality holds because $\|\cdot\|$ and $\|\cdot\|'$ are continuous, and thus $\{x\in \mathbb{R}^n \mid \|x\| = 1\} = \partial E = \{x\in \mathbb{R}^n \mid \|x\|' = 1\}$. Therefore, an inner product $g$ such that $E=E(g)$ is unique.

If $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a linear map such that $T(E_1) = E_2$ for two ellipsoids $E_1$ and $E_2$ with inner products $g_1$ and $g_2$, respectively, then $$ \|T(x)\|_2 = \bigl\|T\bigl(\frac{x}{\|x\|_1}\bigr)\bigr\|_2\|x\|_1 = \|x\|_1\quad\text{for all }x\in \mathbb{R}^n\backslash\{0\}. $$ Because $$ g(x_1,x_2) = \frac{1}{4}\bigl(\|x_1 + x_2\|^2 - \|x_1 - x_2\|^2), $$ it follows that $g_2(T\cdot,T\cdot)=g_1(\cdot,\cdot)$.

Given an ellipsoid $E\subset \mathbb{R}^n$, the unique inner product $g$ such that $E=E(g)$ can be recovered from the formula above using the Minkowski functional $$ \| x \| = \inf \{r>0\mid rx\in E\}. $$ Thanks @user131781 for pointing this out.

This was a partial answer. I still wonder if there is any relation of $g$ to $A$ or $B$ or an integral formula (see my question). One has to perhaps compute $A$ and $B$ and see what one gets.