Problem
Let $M = (\mathbb{T}^2, d)$ where $$ d(u, v) = 3 - [\cos(u_1 - v_1) + \cos(u_2 - v_2) + \cos((u_1 - v_1) - (u_2 - v_2))] $$
Prove that $M$ is a metric space.
Progress
So I know that axioms for a metric space are the following:
- $d(u, v) = 0 \iff u = v$
- $d(u, v) = d(v, u)$
- $d(u, w) \le d(u, v) + d(v, w)$
I am able to prove axiom 1 and 2, however, the triangle inequality (axiom 3) is the one giving me the problem. I do not even know how to begin proving this. I am also not educated in real or complex analysis so I do not have the knowledge of those theories at my disposal. If a proof is given in those terms, however, I would be able to follow and understand. I can't off the top of my head figure out a proof using them though (if that makes any sense).
If it helps, the above metric was simplified (with a few changes) from the following:
- Let $h(s, t) = (\cos(s) - \cos(t))^2 + (\sin(s) - \sin(t))^2$ where $s,t \in \mathbb{T}$.
Then the metric is,
$$ d(u, v) = \sqrt{h(u_1, v_1) + h(u_2, v_2) + h(u_1 + v_1, u_2 + v_2)} $$
Any and all help would be much appreciated. Thanks!!
Map $\Bbb T^2=\Bbb S^1\times \Bbb S^1\to\Bbb S^1\times \Bbb S^1\times \Bbb S^1$, $(u,v)\mapsto (u,v,u+v)$ and then each of the three $\Bbb S^1\to \Bbb R^2$, $u\mapsto (\cos u,\sin u)$ (during this, we identify $\Bbb S^1$ with $\Bbb R/2\pi\Bbb Z$). So in total you have a continuous injective map $\Bbb T^2\to \Bbb R^6$. Then $d$ is just the Euclidean metric on the image.