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Let $X$ be a set and $(M, d)$ be a metric space. Let $f:X\to M$ be an injective map. Then we define $f∗d := d \circ (f \times f): X \times X →(0,\infty)$, i.e. $f*d(x, y) = d(f(x), f(y))$ for $x, y \in X$. Let $d^{k}$ denote the standard euclidean metric on $\mathbb{R}^k$. Is there an injective map $f: \mathbb{R}^{2}\to \mathbb{R}$ such that $d^2 = f∗d^1$? Hint: Look at a triangle in $\mathbb{R}^2$.

Usually I would put up an attempt at a solution first but I've tried a few that were utterly fruitless to the point that I'm starting to think this must be a trick question and that there is no such $f$. Any help would be very much appreciated.

Vinícius Novelli
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excalibirr
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  • is there anyone with whom you can discuss this in person? A study group, perhaps? – Will Jagy Oct 17 '17 at 01:08
  • unfortunately not as theres a hurricane where i live right now, so colleges are closed until wednesday. my thinkin on the question so far though is that im supposed to be looking for an isometry between d^1 and d^2 but i dont see how its possible to have an isometry between dimensions... – excalibirr Oct 17 '17 at 01:12

1 Answers1

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Suppose such a function existed. Then, it would satisfy for $(x_1,y_1),(x_2,y_2) \in \mathbb{R}^2$, $$ \sqrt{(x_1-x_2)^{2}+(y_1 - y_2)^{2}} = \left|f(x_1,y_1)-f(x_2,y_2)\right| $$

Choosing one of these points to be the origin, you get $|f(x_i,y_i)-f(0,0)| = \sqrt{x_i^2+y_i^2}$, for $i=1,2$, which means that $$ f(x_i,y_i) = f(0,0)\pm \sqrt{x_i^2+y_i^2}. $$

Plugging this into the original equation, we get $$ \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}} = \left|\sqrt{x_1^2+y_1^2}\pm\sqrt{x_2^2+y_2^2}\right|. $$

Choosing $(x_1,y_1) = (1,0)$ and $(x_2,y_2)=(0,1)$ you check that neither of these choices are possible. Therefore there is no such $f$.

Vinícius Novelli
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