Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From wikipedia the definition of diameter is the supremum of the distance function of the set. But what if there is no obvious distance function, say for the set $SO(n)$. Also how does this work when distance is just some function, what if I replace distance $d$ with $D(p_1,p_2)=2d(p_1,p_2)$, then the supremum of their differences would be bigger, but this doesn't change the diameter?

I do know that $SO(2)$ is isomorphic to $S^1$, does that mean I can say the diameter is 2?

share|cite|improve this question
"The diameter of $\text{SO}(n)$" is not well-defined until you choose a metric. – Qiaochu Yuan Dec 3 '12 at 2:52
@QiaochuYuan does changing it to "intrinsic diameter" change anything? – Mike Flynn Dec 3 '12 at 2:59
It sounds like you're working from a specific textbook or paper and it would be much easier to just say what textbook or paper that is. "Intrinsic diameter" sounds like it refers to an intrinsic metric ( but this is still not well-defined until you choose an intrinsic metric. – Qiaochu Yuan Dec 3 '12 at 3:06
One candidate for a canonical intrinsic metric on $\text{SO}(n)$ is as follows. $\text{SO}(n)$ admits a bi-invariant Riemannian metric which is (I think) unique up to scale. There is a unique choice of scaling such that the volume of $\text{SO}(n)$ is $1$. The intrinsic metric induced by this Riemannian metric might be the desired metric. But I have no way of knowing because I don't know what source you're working from. – Qiaochu Yuan Dec 3 '12 at 3:07
@Qiaochu: There is unique (up to scale) bi-invariant metric on $SO(n)$ for $n\neq 4$, but it's not unique for $SO(4)$ since $SO(4)$ isn't simple. Working on $SO(4)$'s double cover $S^3\times S^3$, all biinvariant metrics are products of round spheres, but you can change the scale on each $S^3$ individually. – Jason DeVito Dec 3 '12 at 3:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.