The Banach-Mazur distance between two centrally symmetric convex bodies $K,L\in\mathbb{R}^n$ can be defined as
$$ d(K,L) = \inf \{ r : \exists T\colon \mathbb{R}^n \to \mathbb{R}^n \text{ linear such that } T K \subset L \subset r T K \} .$$
If $B^1=\mathrm{conv}\{\pm e\_1, \ldots, \pm e\_n\}$ and $B^\infty=[-1,1]^n$ denote the standard cross-polytope and cube, and if $M$ is a [Hadamard matrix](https://en.wikipedia.org/wiki/Hadamard_matrix) (i.e. a Matrix with only $\pm 1$ coefficients and whose rows are mutually orthogonal), one has
$$MB^1 \subset B^\infty \subset \sqrt{n} B^2 = M B^2 \subset \sqrt{n} M B^1 ,$$
where the first inclusion follows from the fact that the vertices of $M B^1$ form a subset of the vertices of the cube $B^\infty$, and where $B^2$ is the unit ball for the Euclidean norm.
Thus if the dimension $n$ is such that there exists a Hadamard matrix then $d(B^1,B^\infty)\leq \sqrt{n}$.
I am under the impression that equality should hold but I can't find an argument.


**Is it known, conjectured or disproved that $d(B^1,B^\infty) = \sqrt{n}$ ?** (assuming that $n$ is such that there exists a Hadamard matrix)



Banach Mazur distance between the cube and the cross-polytope in the dimensions for which a Hadamard matrix exists
As pointed in comment by Bill Johnson, this equality is (trivially) **disproved** for $n=2$, since in that case $M B^1=B^\infty$ and thus the distance is $1$.


Less trivially, it also fails for $n=8$, in which case the distance is $2.5 < 2.82... = \sqrt{8} $, see [Fei Xue (2017, arXiv)](https://arxiv.org/pdf/1705.01353.pdf) where $d(B^1,B^\infty)$ is explicitelly computed up to dimension $8$.


Note that it holds for $n=1$ (in which case $B^1=B^\infty$) and $n=4$ (again, see [Fei Xue](https://arxiv.org/pdf/1705.01353.pdf)).


