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For $x,y\in \mathbb R^n$ define $$d(x,y)={\|x-y\| \over \sqrt{1+\|x\|^2} \sqrt{1+\|y\|^2}}$$ Here $\|x\|$ is the euclidean norm of a vector. How to prove that $d$ (the spherical metric) is indeed a metric?

Progress so far:

  1. $d(x,y)\ge 0$ is obvious.
  2. $d(x,y) =0 \iff \|x-y\|=0 \iff x=y$, so the positivity holds.
  3. $d(x,y) = d(y,x)$ is clear from the formula, so symmetry holds.

But I am having difficulties with the triangle inequality. Writing it out in coordinates leads to a complicated inequality with square roots all over the place in denominators. Is there a better way?

Bozo Vulicevic
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  • Cauchy Schwartz inequality ? – Maman Jan 03 '15 at 23:35
  • Haven't tried that, how would it look?does it work? – Bozo Vulicevic Jan 03 '15 at 23:42
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    @Suzu Hirose I apologize for not being honest, i have tried that to no avail, like many other techniques as well. None have worked and i apologize again for not displaying my efforts for vividly for the rest to see. What does not work is not worth mentioning i believe. A lack of effort on my part is not present. – Bozo Vulicevic Jan 04 '15 at 00:44
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    I would use stereographic projection: introduce 3D coordinates $u,v,w$ related to $x,y$ by the projection formulas; check that the distance function you have corresponds to the usual Euclidean distance in 3D space; conclude. –  Jan 04 '15 at 01:01
  • Very impressive way of seeing this. I'll definitely give it a try. – Bozo Vulicevic Jan 04 '15 at 01:13

1 Answers1

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The quickest way I know is to "cheat"${}^*$ with Stereographic projection. Introduce the map $F:\mathbb R^n\to\mathbb R^{n+1}$ defined by $F( x)=( z,t)\in\mathbb R^n\times \mathbb R$ with $$ z=\frac{x}{1+\|x\|^2},\quad t= \frac{\|x\|^2}{1+\|x\|^2}$$ (This is a projection onto the sphere $\|z\|^2+(t-1/2)^2=1/4$, but this fact isn't needed.) Direct computation shows that $$ \|F(x)-F(y)\|^2 = \frac{\|x\|^2}{(1+\|x\|^2)^2}+\frac{\|y\|^2}{(1+\|y\|^2)^2} - \frac{2x\cdot y}{(1+\|x\|^2)(1+\|y\|^2)} + \frac{1}{(1+\|x\|^2)^2}+\frac{1}{(1+\|y\|^2)^2} - \frac{2}{(1+\|x\|^2)(1+\|y\|^2)}$$ which simplifies to $$ \frac{1}{1+\|x\|^2}+\frac{1}{1+\|y\|^2} - \frac{2x\cdot y}{(1+\|x\|^2)(1+\|y\|^2)} - \frac{2}{(1+\|x\|^2)(1+\|y\|^2)} $$ and subsequently to $$ \frac{2+\|x\|^2+\|y\|^2- 2x\cdot y -2}{(1+\|x\|^2)(1+\|y\|^2)} = {\|x-y\|^2 \over ( 1+\|x\|^2)\,(1+\|y\|^2)} $$ Thus, $\|F(x)-F(y)\|=d(x,y)$, and the triangle inequality for $d$ follows from the triangle inequality for the Euclidean distance in $\mathbb R^{n+1}$.

$(*)$ I think this is not much cheating, because what use is this metric to us without knowing its relation to the sphere?

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    The proof uses inner products. It is interesting to ask if the result hods any normed linear space. – Kavi Rama Murthy Jan 10 '18 at 06:36
  • @KaviRamaMurthy: That's an interesting question. Why don't you ask it in another thread? – Giuseppe Negro Jan 16 '19 at 10:13
  • I've now stared at this calculation for longer than I'd like to admit, but I do think that the $t$-coordinate needs to be defined as $t = \frac{1}{1 + |x|^2}$ instead. Otherwise it is not clear to me why you'd have terms with enumerators independent of $x$ and $y$ in the calculation of $|F(x) - F(y)|^2$. However, that would mean this map isn't directly related to stereographic projections anymore, I believe. Rather, we'd have the equation $|z|^2 + |t|^2 = \frac{1}{1 + |x|^2}$... – Lukas Miaskiwskyi Nov 12 '20 at 18:49