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It is well known that the alternating group $A_n$ is simple unless $n=4$. It is likewise well known that the special orthogonal group $SO(n)$ is essentially simple unless $n=4$ (specifically, the group $SO(n)$ is simple for odd $n$ and the group $SO(n)/\{\pm I\}$ is simple for even $n\neq 4$).

My question is: are these two facts equivalent? The non-simplicity of $SO(4)$ can be proved by observing that the double-cover of $SO(4)$ is $SU(2)\times SU(2)$ which, being a direct product, is very much not simple. This double-cover is closely related to properties of the quaternions (see Stillwell's Naive Lie Theory). Is there an analogous proof of the non-simplicity of $A_4$ based on a geometric structure related to the quaternions?

P.S. This relationship is an example of the "field of one element" heuristic. Can that be formalized?

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Lie algebra is simple iff Weyl group modulo center does not split as a direct product. Is it what you are after? –  Misha Oct 16 '13 at 20:32
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$SO(3)$ is simple but $A_3$ is not simple. –  YCor Oct 17 '13 at 11:54
    
@Misha: what is the Weyl group of an arbitrary Lie algebra? or do you mean semisimple (finite dimensional in characteristic zero [over algebraically closed field?]) –  YCor Oct 17 '13 at 14:47
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I added the f-1 tag. I wasn't sure how specifically the question "Can that be formalized?" was intended, but browsing through the other f-1 questions may be of interest. –  Hugh Thomas Oct 17 '13 at 20:53
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$A_{3}$ is simple; it's just an abelian simple group (i.e., a cyclic group of prime order). –  DavidLHarden Oct 23 '13 at 19:46

3 Answers 3

First, the non-simplicity of $A_4$ has a very beautiful proof, which I heard summarized by Gromov as : $2+2=4$, or rather $4=2+2$.

Or rather, there are 3 ways to pair 4 objects 2 by 2. The action of $S_4$ on the 4 objects therefore induces an action on the 3 pairings, hence a non-trivial morphism $S_4\to S_3$, whose kernel intersects $A_4$ in a non-trivial simple subgroup.

The sequel is a bit rough, but when looking at $SO(4)$, you can try to reinterpret the same proof, using the elements of a basis instead of the permuted objects. $SO(4)$ naturally acts on the set of direct orthonormal bases of $\mathbb{R}^4$; with each basis comes 3 decompositions of $\mathbb{R}^4$ into pairs of orthogonal planes (which should correspond in some sense to 3 complex structures satisfying the quaternionic relations, probably using that the planes are endowed with particular bases). So $SO(4)$ acts on such triples of complex structures, which if I remember well identifies with $SO(3)$, the unit tangent bundle $S^2$ (choosing the first complex structure $I$ is picking a point on the unit sphere in purely quaternionic numbers, then you have left to choose an orthogonal pure unit quaternion). Considering dimension you get a relatively large non-simple subgroup.

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Here is an argument using the finite quaternion group.

Let $Q_8$ be the quaternion group of order $8$, namely $$Q_8= \{{\pm 1, \, \pm i, \, \pm j, \, \pm k\}}.$$ It is well known that $\textrm{Aut}(Q_8)=S_4$, and this is usually proven by constructing an explicit isomorphism between $\textrm{Aut}(Q_8)$ and the symmetry group of the cube, see for instance here.

On the other hand, since $\textrm{Z}(Q_8)=\{\pm 1\}$, one also has $$\textrm{Inn}(Q_8)=Q_8/\textrm{Z}(Q_8)=V_4,$$ where $V_4$ denotes the Klein group of order $4$, which is isomorphic to $C_2 \times C_2$.

Finally, the inner automorphism group is always a normal subgroup of the full automorphism group, so the argument above shows that $S_4$ contains a normal subgroup isomorphic to $V_4$. Using again the identification of $\textrm{Aut}(Q_8)$ with the symmetry group of the cube, it is no difficult to show that such a normal $V_4$ consists of even permutations, in other words it is contained in $A_4$.

This shows that $A_4$ is not simple.

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Here are some general remarks which hold for arbitrary semisimple Lie algebras ${\mathfrak g}$ relating its algebraic properties to that of its Weyl group $W$.

  1. ${\mathfrak g}$ is simple if and only if the standard linear action of its Weyl group $W$ is irreducible. However, $W$ itself might still split nontrivially as a direct product (this happens in few cases); let's call such $W$ "reducible". This was analyzed by Luis Paris in http://arxiv.org/pdf/math/0412214v2.pdf. The main focus of his paper was on infinite Coxeter groups, but he also classified reducible finite Coxeter groups whose root system is irreducible (section 7). In all cases, the factor of $W$ by its center is irreducible, i.e., does not split nontrivially as a direct product. Therefore, the statement is: ${\mathfrak g}$ is simple iff $W/Z(W)$ is irreducible.

  2. As for simplicity of $W$ itself, in the case of "classical" root systems, $W$ always has the form of semidirect product of a permutation group and a finite abelian group. Thus, if one is willing to divide by the abelian normal subgroup and then pass to the alternating group, then in the classical case one does obtains that simplicity of ${\mathfrak g}$ is equivalent to simplicity of a certain "subquotient" of $W$. I do not know enough about exceptional Coxeter groups to make a similar conclusion in general.

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I guess you work over an algebraically closed field of characteristic zero? Or how do you define the Weyl group? –  YCor Oct 17 '13 at 23:12
    
@Yves: Yes, I do. I did not think through the argument to conclude what happens over the reals. However, one can define root system and Weyl groups for real semisimple Lie groups. –  Misha Oct 17 '13 at 23:43

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