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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.2012-12-03
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    @QiaochuYuan does changing it to "intrinsic diameter" change anything?2012-12-03
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    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 (http://en.wikipedia.org/wiki/Intrinsic_metric) but this is still not well-defined until you choose an intrinsic metric.2012-12-03
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    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.2012-12-03
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    @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.2012-12-03
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    @QiaochuYuan, it was a question unattached to a textbook, just "What are the intrinsic diameters of SO(2n) and SO(2n+1)?" The textbook we have been using is Frank Morgan's "Riemannian Geometry". My professor probably assumed that the choice of metric would be immediately obvious, but it is not, at least to me.2012-12-03

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