Subspaces of $\ell_\infty^3$

Question

Let $a,b\in\mathbb C$ be suc that $\max\{|a+b|,|a-b|\}\leq 1$ but $|a|+|b|>1.$ According to this paper by Arias, Figiel, Johnson and Schechtman https://www.jstor.org/stable/2155206?origin=crossref#metadata_info_tab_contents, the two dimensional complex subspace spanned by $(1,0,a)$ and $(0,1,b)$ in complex $\ell_\infty^3$ is not isometric to complex $\ell_\infty^2$. Can anyone prove that?

Answer 1

A funny isometry invariant to distinguish these normed spaces is: The space of spheres of radius $2$ in the unit sphere of $(X,\|\cdot\|_X)$ which are maximal by inclusion, as described below. It turns out that for $\ell_\infty^2(\mathbb C)$ it is a torus $\mathbb S^1\times \mathbb S^1$, and for the space $Y:=\text{span}\big((1,0,a),(0,1,b)\big)\subset \mathbb C^3$ as defined above it is a disjoint union of two circles, $\mathbb S^1\times \mathbb S^0$.
Another one is: The length of the poset of all spheres of radius $2$ in the unit sphere of $(X,\|\cdot\|_X)$ ordered by inclusion. One gets respectively $2$ and $3$.

Consider the unit sphere $S_X:=\{x\in X:\|x\|_X=1\}$ of the normed space $(X,\|\cdot\|_X)$ as a metric space, and the set $\Sigma_2(S_X)$ of all spheres of radius equal to the diameter $2$ of $S_X$. This is both a metric space with the Hausdorff distance, and a POS with the inclusion relation, and these structures are clearly an isometry invariant of $X$. We can further consider the subspace $\Sigma^*_2(S_X)$ of all maximal elements of $\Sigma_2(S_X)$ w.r.to inclusion. For instance, in the case of $X:=\ell^2_\infty(\mathbb C)$, denoting by $\Delta$ the closed unit disk of $\mathbb C$, the sphere of radius $2$ and center $(x,y)\in S_X$ is either $\{-x\}\times\Delta$, or $\Delta\times\{-y\}$, or $\big(\{-x\}\times\Delta\big)\cup\big(\Delta\times\{-y\}\big)$ according whether $|y|<|x|=1$, or $|x|<|y|=1$, or respectively $|x|=|y|=1$. In particular, maximal spheres of radius $2$ are exactly those of the third type, that is with center $(x,y)\in\partial\Delta\times\partial\Delta$. Since these spheres are homeomorphically parametrised by their centers, we conclude that in this case the space $\Sigma^*_2(S_X)$ is topologically a torus $\mathbb S^1\times\mathbb S^1$. Note that this easily generalises for all $\ell_\infty^d(\mathbb C)$, in particular for $d=3$: now the maximal spheres of radius $2$ in $S_{\ell_\infty^3(\mathbb C)}$ are exactly those of center $(x,y,z)\in\partial\Delta\times\partial\Delta\times\partial\Delta$.

With some more computations, yet by elementary arguments, it is also easy to compute this object for the given complex space $Y:=\text{span}\big((1,0,a),(0,1,b)\big)\subset \mathbb C^3$ as defined above. We just have to look at the traces on $Y$ of the inclusion of spheres of radius $2$ in the unit sphere of $S_{\ell_\infty^d(\mathbb C)}$: it follows from the assumptions on $a$ and $b$ that the maximal spheres of radius $2$ of $S_Y$ are exactly those whose center $(x,y,z)\in Y$ has $|x|=|y|=|z|=1$. Moreover, for any $x\in\partial\Delta$ there are exactly $2$ distinct values of $y$ for which $(x,y,ax+by)\in Y$, which are of the form $y=\theta_1 x$ and $y=\theta_2 x$, for some complex numbers $\theta_1\neq\theta_2$ of unit modulus (reason: by the assumptions one has $|a+\theta b|\le 1$ for $\theta=1$ and $|a+\theta b|=|a|+|b|> 1$ for $\theta=\text{sgn}b /\text{sgn}a$, so the circle of center $a$ and radius $|b|$ has exactly $2$ intersections with the unit circle $\partial \Delta$, namely $|a+\theta_1b|=|a+\theta_2b|=1$ for $\theta_1|=|\theta_2|=1$, so that also $|ax+(\theta_1x)b|=|a+(\theta_2x)b|=1$ for any $|x|=1$). By similar arguments one checks that a strict inclusion of spheres of radius $2$ in $S_{\ell_\infty^3(\mathbb C)}$ gives a strict inclusion on the trace on the space $Y$, so that there are no other maximal spheres in $S_Y$.

So the set of the maximal spheres now is a space homeomorphic to the union of the graphs of $\partial\Delta\ni x\mapsto \big(\theta_j x, (a +b\theta_j)x\big)$, ($j=1,2$),thus to $\mathbb S^1\times \mathbb S^0$.

Rmk As a variant, we may consider the length of $\Sigma_2(S_X)$ as a partially ordered set (the maximum cardinality of its chains). Then, by the above computations, one gets length $d$ for $\ell^d_\infty(\mathbb C)$, and length $3$ for the space $Y$.

Answer 2

Somehow I missed this post two years ago. In the referenced paper, we proved something stronger than the statement that the two dimensional subspaces of $\ell_\infty^3$ mentioned by the OP are not isometrically isomorphic to $\ell_\infty^2$. It is easier to prove the weaker fact desired by the OP.

First, if a two dimensional space $X$ is isometrically isomorphic to $\ell_\infty^2$, then there are norm one linear functionals $f$ and $g$ in $X^*$ that are isometrically equivalent to the unit vector basis of $\ell_1^2$, and for every $x$ in $X$, $\|x\|= |\langle f,x\rangle|\vee |\langle g,x\rangle|$. Let $u:= (1,0,𝑎)$ and $v:=(0,1,𝑏)$ be as in the OP's post and let $X$ be the span of $u$ and $v$ in $\ell_\infty^3$. Take $f$ and $g$ as above in $X^*$ and observe that norm one extensions of $f$ and $g$ to elements in $(\ell_\infty^3)^* = \ell_1^3$ are also isometrically equivalent to the unit vector basis for $\ell_1^2$, so we can identify $f$ and $g$ with vectors in $\ell_1^3.$ But in any $L_1$ space, vectors that are isometrically equivalent to the unit vector basis for $\ell_1^2$ must be disjointly supported. So either $f$ or $g$ is supported in one coordinate. By the symmetry of the situation WLOG $f$ is $(1,0,0)$ or $(0,0,1)$. Suppose that $f$ is $(1,0,0)$. Then $g$ is of the form $(0,c,d)$ with $|c|+|d|=1$. Since $|\langle f,v\rangle|=0$ we have $1=|\langle g,v\rangle| = |c+ db|$, which forces $d=0$ and $|c|=1$. Now it is obvious that in order for $\|x\|= |\langle f,x\rangle|\vee|\langle g,x\rangle|$ to hold for all $x$ in $X$, we must have $|a|+|b|\le 1$.

The remaining case is that $f=(0,0,1)$, in which case $g$ is of the form $(c,d,0)$ with $|c|+|d|=1.$ The conditions on $a$ and $b$ imply that $|a|=|\langle f,u \rangle|$ and $|b|=|\langle f,v \rangle|$ are strictly positive and strictly less than one. Thus $|c|=|\langle g,u\rangle|=1=|\langle g,v\rangle|=|d|$, which contradicts $|c|+|d|=1.$