The inverse Galois problem and the Monster - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:00:51Z http://mathoverflow.net/feeds/question/13851 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13851/the-inverse-galois-problem-and-the-monster The inverse Galois problem and the Monster aorq 2010-02-02T19:02:20Z 2010-02-12T16:34:27Z <p>I have a slight interest in both the inverse Galois problem and in the Monster group. I learned some time ago that all of the sporadic simple groups, with the exception of the Mathieu group $M_{23}$, have been proven to be Galois groups over $\mathbb{Q}$.</p> <p>In particular, the Monster group has been proven to be a Galois group over $\mathbb{Q}$. What techniques are used to prove such an assertion?</p> <p>Is proving that $M_{23}$ is also Galois over $\mathbb{Q}$ within reach? I assume that the same techniques do not apply, for it is a much more manageable group than the Monster.</p> http://mathoverflow.net/questions/13851/the-inverse-galois-problem-and-the-monster/13861#13861 Answer by Wanderer for The inverse Galois problem and the Monster Wanderer 2010-02-02T19:56:06Z 2010-02-03T15:02:28Z <p>Hi Charles,</p> <p>the monster is a nice example of how the so-called <strong>rigidity method</strong> for the inverse Galois problem works. There is a lot of beautiful mathematics behind this, I will sketch the different steps.</p> <p>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 <em>some</em> 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 $K = \mathbb{Q}$, but also other base fields can be considered - for example, for $K = \mathbb{C}(t)$ 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: $\mathbb{C}(t)$, $\overline{\mathbb{Q}}(t)$, $\mathbb{Q}(t)$, and of course $\mathbb{Q}$.</p> <p>(1) Start with the fact that each finite group $G$ can be realized as a Galois group over $\mathbb{C}(t)$. This follows from the theory of coverings of Riemann-surfaces; if $G$ can be generated by $n - 1$ elements, then we can realize $G$ as a quotient of the fundamental group of the punctured Riemann sphere $\pi_1^{\text{top}}(\mathbb{P}^1(\mathbb{C}) \setminus \{P_1,P_2,\,\cdots,P_n\})$, where we choose the points $P_1,P_2,\,\cdots,P_n$ to be rational.</p> <p>(2) We use the theory of the étale fundamental group to get an isomorphism $\pi_1(\mathbb{P}^1(\mathbb{C}) \setminus \{P_1,P_2,\,\cdots,P_n\}) = \pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\})$, which allows us to realize $G$ as a Galois group over $\overline{\mathbb{Q}}(t)$. [$\pi_1$ is the étale fundamental group - here the profinite completion of the topological version.]</p> <p>[Of course, this is already advanced material; see the <a href="http://en.wikipedia.org/wiki/%C3%89tale%5Ffundamental%5Fgroup" rel="nofollow">Wikipedia article</a> 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).] </p> <p>(3) There exists an exact sequence </p> <p>$1 \to \pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\}) \to \pi_1(\mathbb{P}^1(\mathbb{Q}) \setminus \{P_1,P_2,\,\cdots,P_n\}) \to \text{Gal}(\overline{\mathbb{Q}}|\mathbb{Q}) \to 1$</p> <p>(this is a very fundamental result; see again the books I mentioned) and basically we now want to extend a surjective homomorphism from $\pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\})$ to $G$ to a surjective homomorphism from $\pi_1(\mathbb{P}^1(\mathbb{Q}) \setminus \{P_1,P_2,\,\cdots,P_n\})$ 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 <em>it works for finite groups with a trivial centre, and a rigid system of rational conjugacy classes</em> - these are quite technical conditions, of course, and I will just state the definitions. An $n$-tuple of conjugacy classes $C_1,C_2,\,\cdots,C_n$ of $G$ is rigid if there exists $(g_1,g_2,\,\cdots,g_n) \in G^n$ such that the $g_i$ generate $G$, $g_1g_2\cdots g_n = 1$ and $g_i \in C_i$, and if moreover $G$ acts transitively on the set of all such $n$-tuples $(g_1,g_2,\,\cdots,g_n)$. A conjugacy class $C$ of $G$ is rational if $g \in C$ implies $g^m \in C$ 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...</p> <p>[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" :)]</p> <p>(4) The previous step allows us to descend from $\overline{\mathbb{Q}}(t)$ to $\mathbb{Q}(t)$, i.e. to realize $G$ as a Galois group of a <em>regular</em> extension - another technical notion which I won't explain, but it is not unimportant - of $\mathbb{Q}(t)$. To descend from $\mathbb{Q}(t)$ to $\mathbb{Q}$, there is <a href="http://en.wikipedia.org/wiki/Hilbert%27s%5Firreducibility%5Ftheorem" rel="nofollow">Hilbert's irreducibility theorem</a>, or some slight generalization (I don't remember exactly).</p> <p>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 :))</p> <p>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. </p> http://mathoverflow.net/questions/13851/the-inverse-galois-problem-and-the-monster/13986#13986 Answer by JSE for The inverse Galois problem and the Monster JSE 2010-02-03T15:48:43Z 2010-02-03T15:48:43Z <p>I'll add a brief commment to Arne Semeets's thorough and useful answer. If I fix three rational conjugacy classes c_0, c_1, c_infty in a finite group G, then there are finitely many isomorphism classes of unramified G-coverings</p> <p>X -> P^1 - 0,1,infty /Qbar</p> <p>with the property that the image of tame inertia at 0 (resp 1,infty) lies in the class c_0 (resp c_1,c_infty) of G.</p> <p>Call the set of such covers H. What is |H|? One can check (by comparison with the complex case) that the number of such covers is the number of conjugacy classes of triples (g_0,g_1,g_infty) with g_i in c_i and g_0 g_1 g_infty = 1. To say that (c_0,c_1,c_infty) is rigid is just to say that there is precisely ONE such triple.</p> <p>In that case, H consists of just one cover. But H is evidently preserved by Galois conjugacy. So this unique cover is defined over Q. (One has to be slightly more careful when G has nontrivial center, in which case what I've really proved is something more like "there's a cover whose isomorphism class is defined over Q," not quite the same in general as "there's a cover defined over Q."</p>