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Let $p,q\in [1,\infty)$. Note that $p,q\neq\infty$. Let $m\geq 2$ be a natural number.

The paper Isometries of Finite-Dimensional Normed Spaces by Felix and Jesus asserts that if $(\mathbb{R}^m,\|\cdot\|_p)$ is isometric to $(\mathbb{R}^m, \|\cdot\|_q)$, then $p =q$.

I am interested in the case when they have different dimensions. More precisely,

Let $m,n\geq 2$ be natural numbers such that $m\leq n$ and $T:(\mathbb{R}^m,\|\cdot\|_p)\to (\mathbb{R}^n, \|\cdot\|_q)$ be a linear operator (Note that the dimension of domain and codomain are different). If $T$ is an isometry (not necessarily onto), does $p = q$?

By the paper above, if $m=n$, then we have $p=q$. However, if $m<n$, I am not sure whether the same result holds.

If there is a reference that cites this result, it would be good if someone can provide it.

Idonknow
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    It is known that $\ell_2^2$ does not embed isometrically into any $\ell_p^n$ if $p$ is not an even integer. It is also known that $\ell_2^m$ embeds into $\ell_4^n$ for some cases of $(m,n)$. Embedding $\ell_2^m$ into general $\ell_{2s}^n$ seems largely open. You can check any survey paper or book on the isometric theory of Banach spaces (for example, "Handbook of the Geometry of Banach Spaces, Vol 1", Chapter 21) – user58955 Jul 26 '20 at 15:19
  • I want to add that I meant the exact value of smallest $n$ is known for some cases of $\ell_2^m\hookrightarrow\ell_4^n$. And it's known that for $\ell_2^2\hookrightarrow \ell_{2s}^{n}$ for $n=s+1$, and this is the smallest $n$. – user58955 Jul 29 '20 at 05:41

1 Answers1

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In the linked paper is written that Lyubich and Vasertein [9] showed that if there exists an isometric embedding of $(\mathbb{K}^n,\|\cdot\|_p)$ to $(\mathbb{K}^m,\|\cdot\|_q)$ where $\Bbb K=\Bbb R$ or $\Bbb K=\Bbb C$ and $1 < p, q <\infty$ then $p =2$, $q$ is an even integer, and $m$ [I guess such minimal $m$. AR.] satisfies the inequality $${n+q/2-1\choose n-1}\le m\le {n+q-1\choose n-1}.$$

For instance, it can be checked that a linear map $T:(\mathbb{R}^2,\|\cdot\|_2)\to (\mathbb{R}^3, \|\cdot\|_4)$ such that $T(0,1)=(a,b,c)$ and $T(1,0)=(b,a,-c)$, where $b=\sqrt{\frac{2\sqrt{2}-\sqrt{6}}{6}}$, $a=(2+\sqrt{3})b$, and $c=\frac{\sqrt[4]{2}}{\sqrt{3}}$, is an isometry, because $(a,b,c)$ is a solution of a system of equations $$\cases{ a^4+b^4+c^4=1\\ a^3b+b^3a-c^4=0\\ 12a^2b^2+6c^4=2}.$$

Alex Ravsky
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