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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?

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"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

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