Classification of finite groups of isometries - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:47:35Z http://mathoverflow.net/feeds/question/37136 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries Classification of finite groups of isometries Mathieu Dutour Sikiric 2010-08-30T09:33:42Z 2013-02-01T00:54:02Z <p>Consider the problem of classifying the finite groups of isometries of R^n. --For n=2 it is cyclic and dihedral groups. --For n=3 they are well known, probably from Kepler and are related to ade-classification. --For n=4 we can get them by taking the universal cover of SO(4) which is isomorphic to SU2 x SU2, though I do not know where the classification is available.</p> <p>But my main question is for dimension n=5 and above. Does anybody knows the state of the art? A reference would be most helpful. Note that the finite subgroups of GLn(Z) are classified for n&lt;=10.</p> <p>Mathieu</p> http://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries/37152#37152 Answer by Roland Bacher for Classification of finite groups of isometries Roland Bacher 2010-08-30T12:39:52Z 2010-08-30T17:13:25Z <p>There are a few papers by Gabriele Nebe and Wilhelm Plesken on this topic, eg:</p> <p>Nebe, Gabriele Finite subgroups of ${\rm GL}_{24}(\mathbb Q)$. Experiment. Math. 5 (1996), no. 3, 163--195.</p> <p>Nebe, Gabriele Finite subgroups of ${\rm GL}_n(\mathbb Q)$ for $25\leq n\leq 31$. Comm. Algebra 24 (1996), no. 7, 2341--2397.</p> <p>Nebe, G.; Plesken, W. Finite rational matrix groups. Mem. Amer. Math. Soc. 116 (1995), no. 556, viii+144 pp.</p> http://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries/37176#37176 Answer by Richard Borcherds for Classification of finite groups of isometries Richard Borcherds 2010-08-30T16:19:17Z 2010-08-30T17:19:04Z <p>This is one of the problems that just gets hopelessly messy beyond a few small dimensions. The reason is that asking for all finite subgroups of isometries of Euclidean space is essentially the same as asking for all orthogonal representations of all finite groups, and since irreducible representations have dimension at most the square root of the order of the group, you have to use all groups of order up to at least n<sup>2</sup> to find groups of isometries of R<sup>n</sup>. A major problem in doing this is that there are huge numbers of nilpotent groups of order p<sup>n</sup> once n is larger than about 5; for example there are several hundred groups of order 64, all of whose irreducible representations have dimension at most 8. So my guess would be that classifying all groups of isometries in dimensions greater than about 10 will require a lot of obstinacy and a big computer. </p> <p>(Added later) On checking the literature, I find that people classifying such subgroups usually make some simplifying assumptions, by only looking for ones that are irreducible, maximal, and that act on an integral lattice. With these extra simplifications one can get a bit further: the state of the art seems to be around 30 dimensions. </p> http://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries/62734#62734 Answer by Geoff Robinson for Classification of finite groups of isometries Geoff Robinson 2011-04-23T11:41:39Z 2011-04-24T10:58:00Z <p>There is a vast literature on the classification of finite linear groups over various fields. Over the complex or real fields, all finite linear groups are conjugate to subgroups of the respective unitary or orthogonal group, so as remarked in one of the comments above, studying finite groups of isometries in this context is the same as studying all the finite subgroups of ${\rm GL}(n,\mathbb{C})$ or ${\rm GL}(n,\mathbb{R}).$ As Richard Borcherds remarked, this soon becomes a complicated problem. But strategies have evolved since the birth of representation theory to tackle the problem (for general fields) difficult as it is, in a systematic way. I'll discuss the real and complex cases. Generally speaking, we want to concentrate attention on linear groups which can't be described in some "obvious" way in terms of linear groups in smaller dimensions. The first reduction, then, is to concentrate on irreducible groups, those which leave no proper non-zero subspace invariant. Maschke's Theorem tells us that no information is lost in the reduction. Another question, for real representations, is what changes if we extend scalars to the complex field, where life is generally easier. An irreducible real linear group may become reducible when the scalars are extended to the complex numbers (this only happens when its character has squared-norm $2$ or $4$). In each case, the real finite linear group is isomorphic to a finite complex linear group in half the original dimension. So now I only speak of finite complex linear groups. As remarked in someone's earlier comment, the next natural reduction is to the case of primitive linear groups, those which (up to equivalence) be induced from linear groups of smaller dimension. There are strong restrictions on normal subgroups of finite primitive linear groups. In particular, the structure of primitive solvable finite linear groups is very tight, and is well-understood. Having reduced to the primitive case (back to the general finite group), the next question is whether the underlying module is a tensor product of two non-trivial modules of smaller dimension. At this point, it may be necessary to take (still finite) central extensions of the group you started with. If there is a non-trivial tensor factorization, then we are reduced to questions in smaller dimension. If there is no such factorization (even allowing for central extensions), then the structure of the residual groups is very restricted indeed. The given representation may be "tensor induced" from a representation (of smaller dimension) of a proper subgroup. Tensor induction was introduced by Serre. If it can't be tensor induced from a lower dimensional representation (again, even allowing for central extensions), then the only possibility that remains is subgroup of a central extension of the automorphism group of a finite simple group (containing all inner automorphisms). Many mathematicians, for example, Guralnick, Tiep, Zalesski, have calculated (relatively) low dimensional complex representations of (central extensions of) finite simple groups in recent years. My answer is therefore: yes, it is a difficult question, but one which can be addressed systematically in any given case, and for which much hard-won theory is available in the mathematical literature. Addendum: Just as it becomes impractical to list all groups of a given finite order relatively soon, and we have to content ourselves with understanding the "building blocks", that is, the finite simple groups, so it is with finite linear groups. There are three types of building blocks for finite complex linear groups: a) 1-dimensional cyclic linear groups. b) Finite complex linear groups $G$ of dimension $p^{n}$, for some prime $p$ and integer $n > 0$, which have an irreducible normal $p$-subgroup $E$ (extraspecial of order $p^{2n+1}$ and exponent $p$ when $p$ is odd; either extraspecial or the central product of an extraspecial group of order $p^{2n+1}$ with a cyclic group of order $4$ when $p = 2.$). In this case, $G/EZ(G)$ is isomorphic to an irreducible subgroup of the finite symplectic group ${\rm Sp}(2n,p)$. c) Finite complex linear groups $G$ of degree $m$ which have an irreducible quasisimple subgroup $S$ ( this means that $S = S^{\prime}$ and $S/Z(S)$ is a non-Abelian simple group). Then $G/SZ(G)$ is a subgroup of the outer automorphism group of $S/Z(S)$. The third type of building block naturally does not occur for solvable linear groups.<br> In both cases b) and c), the respective subgroups $E$ and $S$ are minimal subject to being normal, but not central.</p> http://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries/119420#119420 Answer by Günter Rote for Classification of finite groups of isometries Günter Rote 2013-01-20T19:55:53Z 2013-02-01T00:54:02Z <ol> <li>Surprisingly, I found explicit lists of discrete subgroups of the orthogonal group O(n) for up to n=8 dimensions on the <a href="http://en.wikipedia.org/wiki/Point_group" rel="nofollow">wikipedia page for point groups</a>, with rather unspecific references, however. Point groups is another name for discrete subgroups of O(n). [UPDATE+CORRECTION: For dimensions n=4 and larger, only the point groups which are generated by reflections (Coxeter groups) are listed. In particularly, subgroups of SO(n) (which include no matrix of determinant $-$1) are missing.]</li> <li>There is an old sequence of two long papers by Threlfall and Seifert, part I <a href="http://link.springer.com/article/10.1007/BF01457920" rel="nofollow">Mathematische Annalen 1931, Volume 104, Issue 1, pp. 1-70</a>, part II <a href="http://link.springer.com/article/10.1007/BF01448910" rel="nofollow">1933, Volume 107, Issue 1, pp. 543-586</a>, where they apparently do the classification of discrete subgroups of SO(4) by associating to each element of SO(4) a pair of rotations from SO(3). (Although my native language is German, I had a hard time reading (through) this, because I am not used to the terminology that was used at that time.) [Addition: These results are mentioned in the book by Conway and Smith on quaternions and octonions; Conway and Smith say that the list is complete, but contains duplicates.]</li> <li><p>I have a rather wild conjecture (true up to three dimensions).</p> <blockquote> <p><em>Every discrete point group in n dimensions is the symmetry group of an n-dimensional polytope which is the Cartesian product of regular polytopes, or a subgroup thereof.</em> </p> </blockquote> <p>[UPDATE: Norman Johnson pointed out <strong>counterexamples</strong>: The symmetries of the root lattices E6, E7, E8 in 6, 7, and 8 dimensions. (I could not yet fully convinced myself that they are indeed counterexamples.) So dimensions 4 and 5 remain open. If I extend my conjecture to include the polytopes which have those E6, E7, or E8 symmetries, in addition to the regular polytopes, in which dimension would the next counterexamples be?]</p> <p>For example, the symmetries of an $m$-gonal anti-prism in 3-space are contained in the symmetries of the $2m$-sided prism, which is the 1-simplex $\times$ the regular $2m$-gon.</p> <p><s>Since the regular polytopes are known in all dimensions, this would give an easy way to obtain all finite point groups. (at least in principle).</s></p></li> </ol>