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Let $X = \{v_1,\ldots,v_n\}$ be a set of vectors non-zero vectors $v_i \ge 0$ and such that the vectors are pairwise linear independent. Define a function on this set $X$:

$$d(v,w) = 1-\frac{2 \langle v, w \rangle ^3}{|v|^2|w|^2(|v|^2+|w|^2)}$$

My question is, if this function is embedabble in Euclidean space. I am searching for a mapping $\phi : X \rightarrow \mathbb{R}^N$ for some $N$ such that, for each $i,j$ we have:

$$|\phi(v_i)-\phi(v_j)| = d(v_i,v_j)$$

where $|.|$ denotes the Euclidean norm.

Is this possible?

(This would answer the specific question about the embeddability of the abc-metric $d(a,b) = 1-\frac{2 \gcd(a,b)^3}{ab(a+b)}$ in Euclidean space in a positive way and would prove that $d$ as above is a metric for $X$.)

Thanks for your help!

Edit: As stated above, the question has a negative answer. There exist examples of vectors which violate Schoenbergs criterion of positive definiteness of the Gram matrix $\frac{1}{2}(d(x_0,x_i)^2+d(x_0,x_j)^2-d(x_i,x_j)^2)$. Hence I am hoping that the construction below will work somehow:

Here is an example for the application I have in mind: Let $X_a = \{ a/k | 1 \le k \le a\}$ for a natural number $a$. Then $X_a \cap X_b = X_{\gcd(a,b)}$, $|X_a| = |a|$.

Given natural numbers $a_1,\ldots,a_n$, we let $m := \operatorname{lcm}(a_1,\ldots,a_n)$ and $Z := X_m = \{ z_1,\ldots,z_m\}$, where the elements of $Z$ are ordered in some way, which does not change.

To each $a_i$ we assign the vector $\phi(a_i)\in \mathbb{R}^m$ whose $j$ entry is equal to $1$ if $z_j$ is an element of $X_{a_i}$, $0$ otherwise.

In this way one has the following:

$$\gcd(a,b) = \langle \phi(a),\phi(b) \rangle$$ $$ a = |\phi(a)|^2$$

which leads to :

$$d_{ABC}(a,b) = d(\phi(a),\phi(b))$$

For example, for the numbers $1,2,3,6$ we have $m=6$ and the we assign the following vectors to each number correspondigly:

$$v_1=\phi(1) = \left(1,\,0,\,0,\,0,\,0,\,0\right)$$ $$v_2=\phi(2) = \left(1,\,0,\,0,\,1,\,0,\,0\right)$$ $$v_3=\phi(3) = \left(1,\,0,\,1,\,0,\,1,\,0\right)$$ $$v_6=\phi(6) = \left(1,\,1,\,1,\,1,\,1,\,1\right)$$

For instance $X_3 = \{3/3,3/2,3/1\}, Z = \{6/6,6/5,6/4,6/3,6/2,6/1\}$ and comparing the elements we find that $$\phi(3) = \left(1,\,0,\,1,\,0,\,1,\,0\right)$$

So you can assume, that all vectors $v_i$ above are generated this way.

  • I understand the question as whether it is true that for every $n$ and $n$-tuple of vectors, the given function can be embedded into a Euclidean space as described. I this the correct interpretation (I'm confused by "Is this possible?") – YCor Oct 20 '19 at 18:32
  • yes. thanks for your comment. thats how i meant it –  Oct 20 '19 at 18:34
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    It sure seems weird to describe this as embedding the function. Surely the right thing to do is to say that you are embedding the metric space $X$, where the metric is as given? (I guess, per your other questions, that it is a metric.) – LSpice Oct 23 '19 at 13:36
  • Also, what does it mean to say of a vector $v$ that $v \ge 0$? Do you mean, as in your example, to speak of rational numbers rather than abstract vectors? – LSpice Oct 23 '19 at 13:37
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    Whether a finite metric space embeds into Euclidean space has a sharp characterization by Schoenberg (see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X13003841 ). I am not sure whether it is at all feasible to use that to check your case though. // I am also not sure why you have to disallow collinear vectors: it seems that as long as $v \neq w$, the function $d(v,w) > 0$. – Willie Wong Oct 23 '19 at 13:46
  • @LSpice, $v \ge 0$ means all entries of $v$ are $\ge 0$. I need this for $\langle v, w \rangle \ge 0$. –  Oct 23 '19 at 14:56
  • @WillieWong: I know of Schoenbergs characterization. Thanks for the link. I will have a look at the other criterion! –  Oct 23 '19 at 14:58
  • @williewong you could have one vector being the multiple of the other resulting in the distance being zero –  Oct 23 '19 at 15:16
  • @WillieWong: I have tried the criterion of Schoenberg, but I could not prove in general for $d_{ABC}$ that its Gramian Matrix is positive definite. –  Oct 23 '19 at 15:40
  • If $w = \lambda v$, you have $d(v,w) = 1 - \frac{2\lambda}{1 + \lambda^2}$ which vanishes iff $\lambda = 1$. – Willie Wong Oct 23 '19 at 20:54

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