# Metric in $\mathbb{P}_2$

I have to prove that $\mathbb{P}_2$ with the function $\delta(P,Q)$ defined by "Sine of the angle between two vector in $\mathbb{R}^3$ such that they correspond respectively to P and Q" is effectively a distance. I am not able to prove the triangular inequality. I am trying with the brute approach, supposing that the three points are $$(1:0:0), (\cos(a):0:\sin(a)), (\cos(b):\sin(b)\sin(c):\sin(b)\cos(c)),$$ I get this one $$\sin^2(b)\cos^2(c)+(\cos(a)\sin(b)\sin(c)-\sin(a)\cos(b))^2\leq(\sin(a)+\sin(b))^2.$$ Note that the weird point refer to the vector $(\cos(b),0,\sin(b))$ rotated by the angle $-c$ with the axis of rotation corresponding to the z-axis.

Are you able to solve the inequality? Or have you got a different approach?

• F**k this problem! I have literally spent all the day on it! I am gonna do another excersize and leave this unsolved! – Nisba Jun 13 '15 at 13:22
• Here's an idea (not a hint since I don't know whether it will work). Use unit vectors on the sphere to represent your projective points. There great circle distance is clearly a metric. Perhaps spherical trigonometry will help now (see the wikipedia page en.wikipedia.org/wiki/Spherical_trigonometry#Sine_rules. – Ethan Bolker Jun 13 '15 at 13:34
• @Ethan: I suspect that the "clearly a metric" claim may be the sort of thing that isn't immediately allowable. – John Hughes Jun 13 '15 at 13:37

Here's a geometric proof.

Let $O$ denote the origin. If $P$ and $Q$ are two points on the unit sphere, then $\sin(\angle POQ)$ is twice the area of triangle $\triangle POQ$. Thus, it suffices to prove that $$\text{area}(\triangle POQ) \,+\, \text{area}(\triangle QOR) \;\geq\; \text{area}(\triangle POR)$$ for any three points $P$, $Q$, and $R$ on the unit sphere centered at $O$.

Note first that, by reflecting $P$ or $R$ around $O$ if necessary, we may assume that $\angle POQ$ and $\angle QOR$ are acute. We may also assume that $\angle POQ$ and $\angle QOR$ are both less than $\angle POR$, since otherwise the inequality is trivially true.

Now, rotate $\triangle POQ$ along $\overline{PO}$ so that it partially covers $\triangle POR$, and similarly rotate $\triangle QOR$ along $\overline{OR}$ so that it partially covers $\triangle POR$. Then both of these rotated triangles contain the orthogonal projection $Q'$ of $Q$ onto the plane of $\triangle POR$, and hence they overlap. It follows that they completely cover $\triangle POR$, which proves the inequality.

• You are a genius! – Nisba Jun 13 '15 at 19:27
• Only a imprecision, $\sin(\ldots)$ is twice the are of the triangle $\Delta\ldots$. The proof follows anyway. – Nisba Jun 13 '15 at 19:29
• @LucaMarconato Good point. Fixed. – Jim Belk Jun 13 '15 at 19:48

A probably-irrelevant remark: if you define the distance to be the sine of half the angle between the points (required to be in the same hemisphere!), then the triangle inequality is straightforward, because $2 \sin \frac{\theta}{2}$ is exactly the length of the chord from $P$ to $Q$, so the thing you're trying to prove becomes...the Euclidean triangle inequality for the three points $P,Q,R$.

On the other hand, a quick matlab test of a thousand triples of points on the sphere suggests that the original statement, using sine, is in fact correct, since that test didn't produce any counterexamples. But I don't know a quick and easy proof, alas.

• Thank you! Your are right, they need to be in the same emisphere, I have rewritten the text in a wrong way, the original problem "told it". I like the metric you propose, which is very elegant. But I (think) I have proved my problem in a geometrical way. Now I will write it. – Nisba Jun 13 '15 at 14:19
• I was wrong, I haven't solved. I will assume that the problem is stated correctly with \theta/2. Thank you again! – Nisba Jun 13 '15 at 15:06
• @LucaMarconato It can't be a typo in the problem, because $\sin(\theta/2)$ doesn't give a well-defined function on $\mathbb{P}^2$. (Try negating one of the vectors.) – Jim Belk Jun 13 '15 at 17:22
• Apologies, Luca! -- 2/3 of what I wrote earlier was nonsense. It's the COSINE that's negative for the points I suggested -- the sine is just fine. And it has the advantage that you don't need to put the points in the same hemisphere. It's true that if you DO put the points in the same hemisphere, then $\sin \theta/2$ is half the chord-length, so things work out nicely...but that's about the only solid thing in what I wrote. – John Hughes Jun 13 '15 at 17:37
• I mean \$\sqrt{\frac{1-\frac{<v,w>}{\Vert v\Vert\Vert w\Vert}{2}} – Nisba Jun 13 '15 at 17:38