How to show that $\mathbb R^n$ with the $1$-norm is not isometric to $\mathbb R^n$ with the infinity norm for $n>2$? Could you please give me a hint to prove that
$\mathbb{R}^n$ with the 1-norm $\lvert x\rvert_1=\lvert x_1\rvert+\cdots+\lvert x_n\rvert$ is not isometric to $\mathbb{R}^n$ with the infinity-norm $\lvert x\rvert_\infty = \max_{i=1,\ldots,n}(\lvert x_i\rvert)$ for $n\gt2$.
I do not see how the critical case $n=2$ enters the picture.
Thank you!
 A: But one can do without the Mazur-Ulam theorem just as easily. I find it convenient to put the exponent $p$ in subscript, freeing the superscript for dimension: $\ell_\infty^n$ and $\ell_1^n$ 
Suppose $F:\ell_\infty^n\to \ell_1^n$ is an isometry (which I do not assume to be surjective or linear). By translation we can achieve $F(0)=0$. Let $V\subset \ell_\infty^n$ be the set of vertices of the unit cube, i.e., the set of all vectors with entries $\pm1$. This set has three interesting properties:  


*

*all points are at distance $1$ from the origin

*all points are at distance $2$ from one another 

*the cardinality of the set is $2^n$. 


Let $W=F(V)$. The set $W\subset \ell_1^n$ also has the properties 1,2,3 listed above. For each vector $v\in W$ consider its positive and negative supports $\mathrm{supp}^+v=\{1\le j\le n : v_j> 0\}$ and  $\mathrm{supp}^-v=\{1\le j\le n : v_j< 0\}$. If $n>2$, then $2^n>2n$. Therefore,  we can assume that at least $n+1$ vectors in $W$ have nonempty positive supports. But then at least two of them, say $v'$ and $v''$, have overlapping positive supports. This creates cancellation when we compute distance between them: $\|v'-v''\|_1<\|v'\|_1+\|v''\|_1=2$ , a contradiction. QED
The proof actually shows that $\ell_{\infty}^n$ does not admit an isometric embedding (linear or not) into $\ell_1^N$ for $N<2^{n-1}$.
A: To wrap the comments up (assuming $n \geq 2$ since the cases $n=0,1$ are uninteresting):
From the Mazur–Ulam theorem we know that a surjective distance-preserving map between normed spaces is necessarily affine. After composing with a translation we may assume that our isometry preserves $0$, so we may assume that the isometry is linear right from the start.
A surjective linear isometry must map the extreme points of the unit ball of the domain onto the extreme points of the unit ball of the range. A moment's thought shows that the unit ball in the $\ell^1$-norm has $2n$ extreme points, namely the $n$ coordinate vectors and their $n$ negatives; and that the unit ball in the $\ell^\infty$-norm has $2^n$ extreme points: the vectors all of whose entries are $\pm 1$. Therefore isometry is only possible if $n =2$, where there is an isometry, given by $e_1 \mapsto e_1+e_2$ and $e_2 \mapsto -e_1 + e_2$.
