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Let $\Gamma$ be a discrete cocompact subgroup of the euclidean motion group $$ G={\mathbb R}^d\rtimes O(d). $$ Let $\phi:G\to O(d)$ the projection homomorphism. Is it true that $\phi(\Gamma)$ is finite?

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Aakumadula, why not post this as an answer? It seems sufficiently precise and comprehensive –  Yemon Choi Mar 6 '13 at 16:13
    
Yemon Choi, I thought this was just a reference. But I see your point, that it will look like an unanswered query. –  Aakumadula Mar 6 '13 at 17:04
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up vote 6 down vote accepted

The answer is yes. This is a theorem of Bieberbach (see Corollary (8.26) of the book "Discrete Subgroups of Lie Groups" by M.S.Raghunathan (Springer Ergebnisse Tract).

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Interesting (to me!) that this is an established result, since without the co-compactness assumption the conclusion fails: an irrational winding in SO(2) combined with the one-dimensional discrete subgroup of translations immediately gives a non-discrete image in SO(2). –  paul garrett Mar 7 '13 at 0:47
    
Professor Garrett, I do not understand your remark. If I have an element of the form $(v,r)$ in the semi-direct product of ${\mathbb R}^2$ with $SO(2)$, where $r$ is an irrational rotation, there is no guarantee that itw powers (vector component) is discrete: the $n$-th power vector component is of the form $(1+r+\cdots +r^{n-1})v$, which can well converge. –  Aakumadula Mar 7 '13 at 1:16
    
Of course, you can take $v\in {\mathbb R}^n$ $n\geq 3$ which is fixed by the rotation $r$, in which case you get a discrete group. When you assume "co-compactness", the vector parts are sufficiently many that the rotation parts do not always fix the vector part. –  Aakumadula Mar 7 '13 at 1:24
    
@Aakumadula: indeed, this is a funny situation. It is late for me here in my time-zone, but (in case you'd not already in your own time thought through the facts) I'll add comments tomorrow which I hope will clarify the thing that'd been in my mind. Irrelevant to the question as posed. yes. –  paul garrett Mar 7 '13 at 2:00
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