Ellipsoids and their defining inner product

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

• 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. – user131781 Nov 4 '19 at 2:40
• 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. – Pavel Nov 4 '19 at 9:55
• 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... – Pavel Nov 4 '19 at 9:59
• 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 $g_2(T.,T.)=g_1(.,.)$. – Pavel Nov 4 '19 at 22:32
• And using the same computation, the inner product g such that E=E(g) is unique. – Pavel Nov 4 '19 at 22:39

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.

• 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. – Pavel Apr 15 at 13:12

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.