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)

  • 1
    $\begingroup$ mathoverflow.net/questions/237567/… $\endgroup$ Mar 12, 2021 at 16:29
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 12, 2021 at 16:32

1 Answer 1


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).


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.