# Banach Mazur distance between the cube and the cross-polytope in the dimensions for which a Hadamard matrix exists

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 (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)

• mathoverflow.net/questions/237567/… Mar 12, 2021 at 16:29
• The lower bound of $K_G^{-1} \sqrt{n}$ follows from Grothendieck's inequality. I think it is known that there is not equality when $n$ is a power of $2$. When $n=2$ the distance is one, of course. Mar 12, 2021 at 16:32

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) 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).