# SO$(4)$ (& SO$(n)$) characterization?

I believe it is the case that any finite subgroup of SO$(3)$ (the $3 \times 3$ orthogonal matrices of determinant $1$) is either a cyclic group $C_n$, or a dihedral group $D_n$, or one of the groups for one of the five Platonic solids.

Q, What is the generalization of this characterization to SO$(n)$, $n>3$, and in particular (my specific interest), to SO$(4)$?

This is certainly well-known, so this is just a reference request. Thanks!

(Image from John O'Connor: S3(Cube): red:face(cents); green: vertices; blue:edge(midpts) .)

• Some trivial comments. First, in order to embed a finite group $G$ into $O(n)$, it is enough to embed it into $GL(n,\mathbb{R})$: if you have the latter embedding, start with any scalar product on $\mathbb{R}^n$, and use averaging over $G$ to turn it into an invariant scalar product, and you have an embedding into $O(n)\subseteq SO(n+1)$. Second, $G$ always embeds into $S_{|G|}$ by Cayley's theorem, and hence into $SO(|G|+1)$. These observations suggest to me that a general answer to your question is probably too much to hope for... – Tobias Fritz Jul 31 '15 at 0:46
• All representations of finite groups admit invariant inner products. Conversely, any finite subgroup has a faithful dim n representation by restriction. Therefore, your question is the same as asking for faithful representations of finite groups in dim n (or 4). Thus, to answer the question for general n, I believe you would have to classify all representations for all finite groups. – Henrik Winther Jul 31 '15 at 0:48
• For the case of SO(4), see this MO discussion, of which the current question actually seems to be a duplicate: mathoverflow.net/questions/37136/… – Tobias Fritz Jul 31 '15 at 1:39
• @TobiasFritz While I agree that this question is similar to the one in your link. This question seems to focus on $SO(4)$ while the other question focuses on $SO(n)$, $n>4$. – Neil Hoffman Jul 31 '15 at 2:37

As mentioned in the comments, for general $n$ this is pretty hopeless. For $n = 4$ we can take advantage of the fact that $SO(4)$ is double covered by $Spin(4) \cong SU(2) \times SU(2)$, which more or less reduces the problem to classifying finite subgroups of $SU(2) \times SU(2)$. This in turn more or less reduces to classifying pairs of finite subgroups of $SU(2)$ (via Goursat's lemma), and since $SU(2)$ double covers $SO(3)$ you more or less already know what the answer looks like. I believe this is spelled out explicitly in Conway and Smith's On Quaternions and Octonions.

• It is indeed. In Tables 4.1 and 4.2 in Conway & Smith you can find a list of the subgroups. There is one small correction to that table, though: the penultimate entry in Table 4.1, corresponding to the haploid subgroup $+\frac12 [D_{2n} \times C_{2m}]$ is missing the condition that both $m$ and $n$ be odd, which is in any case implied by their choice of generators. – José Figueroa-O'Farrill Jul 31 '15 at 12:39

For finite irreducible linear groups in low dimension (at most 4) , the answer was already known and discussed in books by Blichfeldt, and Miller, Blichfeldt and Dickson in the early 20th century. However, Blichfeldt did miss at least one case. Dealing with real representations complicates the issue somewhat, but if the real representation splits into two complex conjugate representations when the field of scalars is extended to $\mathbb{C}$, we are looking at finite subgroups of ${\rm GL}(2,\mathbb{C})$. It is also possible that the real irreducible representation could split as the sum of two equivalent irreducible complex representations, but the $2$-dimensional question for complex representations is easy.

For general $n$, the classification of irreducible finite subgroups of ${\rm GL}(n,\mathbb{C})$ can be refined in various ways. First, consider only primitive linear groups (where the representation is not equivalent to one induced from a proper subgroup). Next, consider representations which (even after taking central extensions) can't be decomposed as the tensor product of two representations of smaller dimension.

Next, consider representations which can't be tensor induced from a representation of a proper subgroup (even after taking central extensions).

This leaves two residual configurations for $G$: in one case, there is a quasisimple irreducible normal subgroup of $G$. In the other case, there is an "almost extraspecial" irreducible normal $p$-subgroup $U$ for some prime $p$, where $n = p^{k}$, and $G/UZ(G)$ is isomorphic to a subgroup of ${\rm Sp}(2k,p).$

Prior to the classification of finite simple groups, the situation was fairly explicitly understood in dimension up to $11$. Using the classification of finite simple groups, B. Weisfeiler got a close to optimal bound for Jordan's Theorem for $n > 70$ or so.

Recently, M.J. Collins has extended Weisfeiler's work to determine the maximal possible index of an Abelian normal subgroup of a finite subgrouup of ${\rm GL}(n,\mathbb{C})$. For $n > 72$ or so, the bound is $(n+1)!$, which is attained by $S_{n+1}$in its irreducible $n$-dimensional representation.

(Later edit: Perhaps I should have made clear that passing between complex representations of finite groups and orthogonal real representations is straightforward, as has already been touched upon in comments: a complex irreducible representation of degree $n$ of a finite group $G$ is equivalent to a unitary representation. A unitary irreducible complex representation of degree $n$ of $G$ may be replced by a real orthogonal representation of degree $n$ in standard fashion, replacing the complex entry $a+bi$ by the real $2 \times 2$ matrix $\left(\begin{array}{clcr} a& b \\-b &a \end{array}\right)$. If the character of the original representation was not real, the resulting representation is irreducible as a real representation. If the original character was real, the resulting representation is equivalent to twice an irreducible if the original character had Frobenius-Schur indicator $+1$, and is irreducible as a real representation if the Frobenius-Schur indicator was $-1$).

(Later remark: In general, for finite subgroups of ${\rm GL}(n,\mathbb{C})$, we can take scalar multiples of the elements to ensure that all elements have determinant $1$ without affecting the structure of $G/Z(G)$. This trick may not be available for real representations.)

• Added line breaks where your paragraphs were. Hope you don't mind. – Qiaochu Yuan Jul 31 '15 at 21:27
• @QiaochuYuan : No problem. In fact, thanks! – Geoff Robinson Aug 1 '15 at 6:37

A natural method (at least for me) of classifying finite subgroups of $SO(3)$ is to classify the orientable two dimensional orbifolds covered by $S^2$. This is done in several places, in terms of this discussion I would point to Thurston's notes Chapter 13.

A similar method can be employed to classify subgroups of $SO(4)$, namely the classification of elliptic 3-dimensional manifolds and orbifolds, which was accomplished over several works. The first is a classification of elliptic 3-manifolds which can be found in several places including chapter 4 of Thurston's book. This covers subgroups of $SO(4)$ that act freely on $S^3$. The remaining cases are those groups that act non-freely which were classified by William Dunbar over two papers Geometric Orbifolds and Nonfibering Spherical Orbifolds.

This classification is geometric and slightly more refined that the question asks for. For example there are elliptic 3-manifolds and elliptic 3-orbifolds with isomorphic groups. For example, lens spaces have finite cyclic fundamental group as do the so-call orbi-lens spaces. A good treatment of this can be found in the background of Boileau, Boyer, Cebanu, and Walsh's Knot commensurability and the Berge conjecture. This is pointed out to stress the fact that the list of fundamental groups above contains duplicates.

See the rather lucid account given by Bruno Zimmermann in his notes here.

• These notes assert that you can find a list of the finite subgroups of $SO(4)$ in Homographies, Quaternions and Rotations by P. Du Val. I don't have a copy of this reference, so can't confirm it. It would be nice to have an explicit list as an answer to the OP's question! – Nick Gill Jul 31 '15 at 9:26