It is known that there are -up to conjugation- 5 classes of discrete subgroups of SU(2). One way to show this is by means of the McKay correspondence. My question is more regarding products of $SU(2)$, say with $n$ factors. Well, since I am definitely not familiar with the correspondence above, I would like to ask whether I have any hope to classify all finite subgroups of the product of 2 or 3 copies of $SU(2)$ by following the same line of thought. Or in case someone knows a more economical alternative, I would be very happy to listen to it.
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2$\begingroup$ I don't quite understand the term "classes" in the first sentence. Anyway it's unnecessary to mention the McKay correspondence, since the finite subgroups of SU(2) were known much earlier. By now the determination of those finite groups has made its way into elementary sources (such as M. Artin's textbook on algebra. $\endgroup$– Jim HumphreysCommented Aug 16, 2013 at 14:32
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$\begingroup$ Yes that is right, maybe the term "classes" is not appropriate. I meant just that such subgroups are conjugate to either a cyclic one or a binary polyhedral one. As for the Mckay correspondence, thanks for the comment. $\endgroup$– user38651Commented Aug 20, 2013 at 10:05
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2 Answers
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Subgroups of a direct product of groups can be classified in principle using the Goursat Lemma. To see how this is done for $SU(2) \times SU(2)$ you can look at a paper I cowrote with Paul de Medeiros: arXiv:1007.4761. The case of $SU(2) \times SU(2) \times SU(2)$ is in principle doable from the results of that paper, but I would not undertake that task lightly.
answered Aug 15, 2013 at 20:19
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$\begingroup$ Ok, thank's for the quick answer. I will try use the approach to classify all the subgroups SU(2)×SU(2)×SU(2). $\endgroup$ Commented Aug 16, 2013 at 12:29
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$\begingroup$ Sorry, but I'm not sure what you mean by table. It may be beneficial to read the introductory section "How to use this paper". There you will read that Tables 8, 9 and 10 contain the desired subgroups, written as twisted products, which is how they are described using Goursat lemma. $\endgroup$ Commented Aug 16, 2013 at 16:17
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$\begingroup$ Ok, so I think I have understood the general ideas in your paper. Theoretically it should be extendable for three factors and the main difficulty would be in discriminating all possible compatible pairs $(A,B)\subset S^3 \times (S^3)^2$. The first problem is more concerning the subgroups of a fiber product - In fact I don't see how you identified this twisted fiber products, for instance of cyclic gropus with the product of cyclic groups, even less in the case of bynary polyhedral groups - $\endgroup$ Commented Aug 20, 2013 at 13:43
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$\begingroup$ The second problem I see is that the amount of members in the list would increase vastly! - I wouldn't say douplet, but rather exponentially- . And after that I in fact have to do something else with those ... Do you think it is sort of sensible investing time on this ? $\endgroup$ Commented Aug 20, 2013 at 13:46
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$\begingroup$ - sorry for the typos, I wish I could edit comments - $\endgroup$ Commented Aug 20, 2013 at 15:53
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Incidentally, $SO(4)=SU(2)\times SU(2)/\{\pm (I\times I)\}$, so the classification of discrete subgroups of $SO(4)$ is (almost) equivalent to the classification of discrete subgroups of $SU(2)\times SU(2)$ (every discrete subgroup of $SU(2)\times SU(2)$ will correspond to a subgroup of $SO(4)$ containing $\pm I$). The discrete subgroups of $SO(4)$ correspond to orientable 3-dimensional spherical orbifolds, which have been explicitly worked out by Dunbar (see also here).
answered Aug 17, 2013 at 15:47
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$\begingroup$ Thanks, but due to the paper due to Prof Figueroa, I am left with the problem for three copies. $\endgroup$ Commented Aug 20, 2013 at 15:50