Saturday, 11 February 2012

gr.group theory - The inverse Galois problem and the Monster

Hi Charles,



the monster is a nice example of how the so-called rigidity method for the inverse Galois problem works. There is a lot of beautiful mathematics behind this, I will sketch the different steps.



A general remark: it is known that every profinite group, i.e. every group which could be a Galois group of some field extension, is indeed the Galois group of some Galois extension. This is still quite elementary (a result of Leptin, also proved by Waterhouse). To make the inverse Galois question more interesting, we should therefore consider a fixed base field K. We can't hope that any profinite group will still be a Galois group over K - every Galois group over K is a quotient of the absolute Galois group of K, and therefore the cardinality of a Galois group over K is bounded from above, whereas it is easy to see that there are profinite groups which are "strictly bigger" in cardinality. So a very reasonable question is indeed to ask whether every finite group is a Galois group over some fixed base field K. The most natural case is to ask the question for K=mathbbQ, but also other base fields can be considered - for example, for K=mathbbC(t) the inverse Galois conjecture is true. In fact, the inverse Galois problems for different base fields K are sometimes closely linked; the method which I will sketch below is a perfect illustration for this, since we will have the consider four different base fields: mathbbC(t), overlinemathbbQ(t), mathbbQ(t), and of course mathbbQ.



(1) Start with the fact that each finite group G can be realized as a Galois group over mathbbC(t). This follows from the theory of coverings of Riemann-surfaces; if G can be generated by n1 elements, then we can realize G as a quotient of the fundamental group of the punctured Riemann sphere pitexttop1(mathbbP1(mathbbC)setminusP1,P2,,cdots,Pn), where we choose the points P1,P2,,cdots,Pn to be rational.



(2) We use the theory of the étale fundamental group to get an isomorphism pi1(mathbbP1(mathbbC)setminusP1,P2,,cdots,Pn)=pi1(mathbbP1(overlinemathbbQ)setminusP1,P2,,cdots,Pn), which allows us to realize G as a Galois group over overlinemathbbQ(t). [pi1 is the étale fundamental group - here the profinite completion of the topological version.]



[Of course, this is already advanced material; see the Wikipedia article for background. For a proper introduction to the theory, there is SGA 1 by Grothendieck; and the recent book "Galois groups and fundamental groups" by Tamas Szamuely is a very gentle introduction (and does all this in detail).]



(3) There exists an exact sequence



1topi1(mathbbP1(overlinemathbbQ)setminusP1,P2,,cdots,Pn)topi1(mathbbP1(mathbbQ)setminusP1,P2,,cdots,Pn)totextGal(overlinemathbbQ|mathbbQ)to1



(this is a very fundamental result; see again the books I mentioned) and basically we now want to extend a surjective homomorphism from π1(mathbbP1(overlinemathbbQ)setminusP1,P2,,cdots,Pn) to G to a surjective homomorphism from pi1(mathbbP1(mathbbQ)setminusP1,P2,,cdots,Pn) to G. Of course this depends heavily on the structure of the group G. This works for many finite simple groups; the construction is quite general, but for particular groups there is always some technical work to do to show that the method applies. In particular it works for finite groups with a trivial centre, and a rigid system of rational conjugacy classes - these are quite technical conditions, of course, and I will just state the definitions. An n-tuple of conjugacy classes C1,C2,,cdots,Cn of G is rigid if there exists (g1,g2,,cdots,gn)inGn such that the gi generate G, g1g2cdotsgn=1 and giinCi, and if moreover G acts transitively on the set of all such n-tuples (g1,g2,,cdots,gn). A conjugacy class C of G is rational if ginC implies gminC for all m coprime to the order of G. I won't explain why precisely these conditions give you what you want, since it is really technical. Szamuely explains this very clearly. The conditions can be generalized, but that doesn't make it more readable...



[References: section 4.8 in Szamuely's book I mentioned above, and also Serre's wonderful book "Topics in Galois theory", which should maybe be called "Topics in inverse Galois theory" :)]



(4) The previous step allows us to descend from overlinemathbbQ(t) to mathbbQ(t), i.e. to realize G as a Galois group of a regular extension - another technical notion which I won't explain, but it is not unimportant - of mathbbQ(t). To descend from mathbbQ(t) to mathbbQ, there is Hilbert's irreducibility theorem, or some slight generalization (I don't remember exactly).



According to Thompson, the Monster has a rigid system of three rational conjugacy classes of orders 2, 3 and 29. So the method will apply; of course, I guess that it will be very hard to construct these conjugacy classes, and it is clear that the classification of finite simple groups has played a very big role in these developments. (But I am not a group theorist, so anyone who knows how this works is welcome to give additional information about this construction :))



So I hope this gives you an idea; I wrote this up in a hurry, so suggestions to make this clearer or more coherent (or of course corrections of details which I got wrong) are always welcome.

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