Banach–Mazur compactum

Remarks

The function $\delta$ is a metric on isometry classes. It is symmetric because $\norm{T^{-1}} \, \norm{T}$ is the same number for $T$ and $T^{-1}$. The triangle inequality comes from $\norm{ST} \le \norm{S} \, \norm{T}$, so $d(X, Z) \le d(X, Y) \, d(Y, Z)$. If $d(X, Y) = 1$, then a compactness argument gives an isomorphism with $\norm{T} \, \norm{T^{-1}} = 1$, and a multiple of it is an isometry.

By John's theorem, $d(X, \ell_2^n) \le \sqrt{n}$ for every $n$-dimensional normed space $X$, where $\ell_2^n$ is $\R^n$ with the Euclidean norm. So the diameter of $\mathcal{Q}_n$ in the multiplicative form is at most $n$. Gluskin proved that this order is right: there are $n$-dimensional spaces at distance at least $c n$ from each other, for a constant $c > 0$ that does not depend on $n$.