Intuition on Banach-Mazur distance

Question

Suppose two n-dimensional normed vector spaces $X,Y$ are isomorphic, and we define the Banach-Mazur distance between $X,Y$ as $$ d(X,Y)=\inf \{ \|T\|\|T^{-1}\|:T\in GL(X,Y) \} ,$$ where $GL(X,Y)$ is the space of all linear isomorphisms.

Also for $X,Y,Z$ normed spaces:

a) $d(X,Y)\geq 1 $
b) $d(X,Y)=d(Y,X)$
c) $d(X,Y) \geq d(X,Z)d(Z,Y)$

I have seen that the Banach Mazur distance can be interpreted as the infimum of the numbers $r\geq 1$ such that

$$ U_{Y} \subset T(U_{X}) \subset r U_{Y}, $$ where $U_{X}$ and $U_{Y}$ are, respectively, the closed unit balls of $X$ and $Y$.
My intuition:
This distance is somehow a "measure" for how isometrically isomorphic to normed spaces are.
That is if $X,Y$ are almost isometrically isomorphic then the distance $d(X,Y)$ will be closer to 1.

See here:
Are these two Banach spaces isometrically isomorphic?

https://mathoverflow.net/questions/300108/c-0-is-not-isometrically-isomorphic-to-c#comment746858_300108

that $d(c_0,c)=3$ so does this mean that $c,c_0$ are almost isometrically isomorphic because $d(c,c_0) \simeq 1$ .

Answer 1

Just a couple of comments. First, $\log d(X,Y)$ is literally a metric (say on the collection of normed spaces with underlying vector space ${\mathbb R}^n$).

Secondly, a reasonable sense of scale for this (not-exactly-a) metric space is that $d(\ell_2^n,X) \le \sqrt{n}$ for all $X$ (proved by Fritz John); that $\ell_1^n$ and $\ell_\infty^n$ are "about" $\sqrt{n}$ apart, but that suitably chosen random Banach spaces have average Banach-Mazur distance $\ge cn$ for some positive constant $c$, so the (multiplicative) diameter is essentially as large as it can be given the Fritz John result.

In answer to your explicit question about $c$ and $c_0$, I would say that these spaces are very close to being isometric based on their Banach-Mazur distance.