1
$\begingroup$

$E\subset\mathbb{R}^n$ 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 $T\in SL(n)$.

$\endgroup$
  • 5
    $\begingroup$ If $E$ is an ellipsoid, then its Minkowski functional is a norm on the underlying space. Furthermore it satisfies the parallelogram law and so there is a standard way to use it to rediscover the inner product. $\endgroup$ – user131781 Nov 4 '19 at 2:40
  • $\begingroup$ Great, thanks! And by expressing the Minkowski functional as $|x|_{E}=\frac{|x|}{|E\cap\langle x\rangle^+|}$, where $|.|$ is the Euclidean metric and $\langle.\rangle^+$ the positive span, it is easy to see that an orthogonal map $T$ with $T(E_1)=E_2$ preserves the Minkowski functionals, and hence $g_2(T.,T.)=g_1(.,.)$ for the associated inner products. $\endgroup$ – Pavel Nov 4 '19 at 9:55
  • $\begingroup$ I would still be interested in whether one can relax the orthogonality of $T$ to being symplectic or just volume-preserving. I suspect that the first case would work and the second not... $\endgroup$ – Pavel Nov 4 '19 at 9:59
  • $\begingroup$ Ah, I am sorry, one can just write $|T(x)|_2=|T(\frac{x}{|x|_1})|_2 |x|_1=|x|_1$ and see that any linear $T$ with $T(E_1)=E_2$ satisfies $g_2(T.,T.)=g_1(.,.)$. $\endgroup$ – Pavel Nov 4 '19 at 22:32
  • $\begingroup$ And using the same computation, the inner product g such that E=E(g) is unique. $\endgroup$ – Pavel Nov 4 '19 at 22:39
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much. This is an interesting fact I did not know. I will have a closer look at the Binet and Legendre ellipsoids to see what role they play in mechanics of rigid bodies. $\endgroup$ – Pavel Apr 15 at 13:12
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.