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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.

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<=10.

Mathieu

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You mean linear isometries, right? - Otherwise, considering all the affine isometries, in dimension 2 you have the 17 crystallographic groups. –  Qfwfq Aug 30 '10 at 11:19
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The finite subgroups of SO(4) are listed in the book by Conway and Smith on quaternions and octonions. I have recently checked that it is correct and recovered it from the classification of finite subgroups of Spin(4), which we needed for a separate project: see the paper arxiv.org/abs/1007.4761 . –  José Figueroa-O'Farrill Aug 30 '10 at 11:29
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You may also look at the references in this answer to a previous MO question: mathoverflow.net/questions/17072/the-finite-subgroups-of-sun/… –  José Figueroa-O'Farrill Aug 30 '10 at 12:46
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@unknown: If $G \subset GL(n,\mathbb{R})$ is finite, then we may change coordinates so that the origin is at the centre of mass of some orbit $Gx$ (for an arbitrary $x\in \mathbb{R}^n$). Then the origin is fixed by every element of $G$, so linearity follows from finiteness. On the other hand, if you want to consider discrete subgroups, then you do need to allow for the possibility that some of your isometries may be affine transformations. –  Vaughn Climenhaga Aug 30 '10 at 19:35
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It's also worth pointing out that if G is any finite subgroup of GL(n,R), then one can introduce an inner product with respect to which G acts by isometries; in particular, there is an inner automorphism of GL(n,R) that maps G to a finite subgroup of Isom(n,R). Thus classifying finite groups of isometries is equivalent to classifying finite groups of linear transformations. –  Vaughn Climenhaga Aug 30 '10 at 19:38

4 Answers 4

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 n2 to find groups of isometries of Rn. A major problem in doing this is that there are huge numbers of nilpotent groups of order pn 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.

(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.

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Another simplifying assumption is that of primitivity, which means that the representation is irreducible and not induced from some proper subgroup. That excludes in one fell swoop all nilpotent groups. There are still a lot of them left however. –  Torsten Ekedahl Aug 30 '10 at 17:54
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I recommend using nilpotent groups as an excuse to give up. Otherwise you shortly afterwards run into solvable groups, which are much worse. –  Richard Borcherds Aug 31 '10 at 4:31

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.
In both cases b) and c), the respective subgroups $E$ and $S$ are minimal subject to being normal, but not central.

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@Richard and Geoff: Why should we consider "irreducible representations" of groups? In language of representation theory, the question will be simply "To find faithfull representations of finite groups over $\mathbb{R}$"; not necessarily irreducible. –  joseph Sep 10 '11 at 7:44
    
At least over the complex numbers, and for finite groups, all finite dimensional representations are equivealent to unitary representations, and are hence completely reducible. Hence trhey are direct sums of irreducible representations. So if we can understand irreducible representations, we can undersatnd al representations. –  Geoff Robinson Sep 10 '11 at 14:37

There are a few papers by Gabriele Nebe and Wilhelm Plesken on this topic, eg:

Nebe, Gabriele Finite subgroups of ${\rm GL}_{24}(\mathbb Q)$. Experiment. Math. 5 (1996), no. 3, 163--195.

Nebe, Gabriele Finite subgroups of ${\rm GL}_n(\mathbb Q)$ for $25\leq n\leq 31$. Comm. Algebra 24 (1996), no. 7, 2341--2397.

Nebe, G.; Plesken, W. Finite rational matrix groups. Mem. Amer. Math. Soc. 116 (1995), no. 556, viii+144 pp.

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  1. Surprisingly, I found explicit lists of discrete subgroups of the orthogonal group O(n) for up to n=8 dimensions on the wikipedia page for point groups, 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.]
  2. There is an old sequence of two long papers by Threlfall and Seifert, part I Mathematische Annalen 1931, Volume 104, Issue 1, pp. 1-70, part II 1933, Volume 107, Issue 1, pp. 543-586, 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.]
  3. I have a rather wild conjecture (true up to three dimensions).

    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.

    [UPDATE: Norman Johnson pointed out counterexamples: 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?]

    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.

    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).

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