0
$\begingroup$

It's very standard to view rotations about the origin in $\mathbb{R}^2$ as a group $\mathbb{SO}(2)$, with the zero rotation as an identity and composition of rotations as addition. This can also be viewed as the interval $[0,2\pi)$ with $0$ as the additive identity and addition defined mod $2\pi$.

We can lift this to a Rng structure by defining multiplication $\hat\times:[0,2\pi)\times[0,2\pi)\rightarrow[0,2\pi)$ as $$a\hat\times b=\frac{ab}{2\pi},$$ so each rotation is treated as a fraction of a complete rotation of $2\pi$. Since $a,b<2\pi$ we have that $\frac{ab}{2\pi}\leq\min\{a,b\},$ with equality holding iff $a=0$ or $b=0$. This is a rng and not a ring because the multiplicative identity we would like, $2\pi$, cannot be allowed in the set on consequence of non-uniqueness for the additive identity.

Is this a well known construction, or obviously trivial as a geometric structure for some reason I'm missing?

$\endgroup$
  • 2
    $\begingroup$ It seems multiplication does not distribute over addition: $\pi\hat{\times}(\pi+\pi)=\pi\hat{\times} 0=0$, but $\pi\hat{\times}\pi+\pi\hat{\times}\pi=\pi/2+\pi/2=\pi$. $\endgroup$ – Julian Rosen Dec 3 '17 at 22:57
  • $\begingroup$ @JulianRosen Ah, thank you. This may be what I was missing. On further consideration, this lack of well-behaved interplay will likely make this a useless construction. If you'd like to post your comment as an answer I'll gladly accept. $\endgroup$ – Alec Rhea Dec 3 '17 at 22:58
  • 3
    $\begingroup$ Can we please stop using all those stupid "rng", "rig" and similar puns? How can it be good mathematical notation when you never know if the author made a mistype or is discussing some special structure? $\endgroup$ – Anton Fetisov Dec 4 '17 at 0:48
  • $\begingroup$ How would you call them then? $\endgroup$ – Fosco Dec 5 '17 at 8:50
  • $\begingroup$ @Fosco Loregian: I call them ring, and I call a ring with one a ring with one. Different areas of mathematics have different generic examples, for some having a multiplicative identity is always satisfied, for some this is a strong condition. So every talk dealing with rings should start with "A ring is a commutative ring with one" or whatever conventions you prefer. $\endgroup$ – Jan-Christoph Schlage-Puchta Dec 6 '17 at 11:31
7
$\begingroup$

The multiplication does not give a Rng structure because it does not distribute over addition: $$ \pi\hat{\times}\big(\pi+\pi)=\pi \hat{\times} 0=0\neq \pi = \frac{\pi}{2}+\frac{\pi}{2}=\pi\hat{\times}\pi +\pi\hat{\times}\pi. $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.