Let $x_1,x_2,\cdots,x_n\in\mathbb{R} $ or $\mathbb{C}$. By the non-degeneracy of Vandermonde matrix

enter image description here

the maps

$$ f: \mathbb{R}\longrightarrow\mathbb{R}^n,$$ $$ x\longmapsto (1,x,x^2,\cdots,x^{n-1}) $$ and $$ f:\mathbb{R}^2=\mathbb{C}\longrightarrow \mathbb{R}\times\mathbb{C}^{n-1}=\mathbb{R}^{2(n-1)+1}, $$ $$ z\longmapsto (1,z,z^2,\cdots,z^{n-1}) $$ satisfy the condition:

(C) for any distinct $x_1,x_2,\cdots,x_n$, their images are linearly independent.

Question: how to construct maps satisfying (C)

$$ f: \mathbb{R}^m\longrightarrow \mathbb{R}^{m(n-1)+1} $$ for $m\geq 3$?

  • 1
    $\begingroup$ What is a motivation for requiring such a dimension? It does not seem to be sharp. Say, for $n=3$ dimension $n+2$ is enough, as a map $x\rightarrow (1,x,F(x))$ for a strictly convex $F$ shows. $\endgroup$ – Fedor Petrov Nov 16 '15 at 11:59

(This answers the previous version of the question.)

For example, you may choose large enough $N$, say $N=(m-1)n(n-1)/2+1$, $N$ linear functions $\ell_1,\dots,\ell_N$ in general position on $\mathbb{R}^m$ (any $m$ are linearly independent) and take the following map from $\mathbb{R}^m$ to $\mathbb{R}^{1+N(n-1)}$: $$ x\rightarrow (1,\ell_1(x),(\ell_1(x))^2,\dots,(\ell_1(x))^{n-1},\ell_2(x),\dots,(\ell_N(x))^{n-1}). $$ For any $n$ distinct points $x_1,\dots,x_n$ in $\mathbb{R}^m$ there exists index $i$ such that $\ell_i(x_1),\dots,\ell_i(x_n)$ are distinct reals. Indeed, for any pair $(x_p,x_q)$ there are at most $m-1$ indices $i$ for which $\ell_i(x_p)=\ell_i(x_q)$, totally at most $(m-1)n(n-1)/2$ not appropriate indices. Now linear independence of images follows from linear independence of $n$ Vandermonde vectors $(1,\ell_i(x_j),(\ell_i(x_j))^2,\dots,(\ell_i(x_j))^{n-1})$, $j=1,2,\dots,n$.


A map $f: M \to \mathbb{R}^n$ is said to be $k$-regular if whenever $x_1, \dots, x_k$ are distinct points of $M$, then $f(x_1), \dots, f(x_k)$ are independent. There is an abundance of literature on $k$-regular maps. Blagojević, Lück, and Ziegler - On highly regular embeddings gives obstructions and a nice history of the problem, as well as many references.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.