 Okay, let's get started. Welcome to sadly the last of Tim's lectures on inverse Galois theory. Thank you, Tim. Yeah, thank you, Akshay. So to remind you where we stand, I started to talk a little bit about the group SL2 of three. The reason for that is, I mean, there are a couple of reasons it's the smallest group which is not semi-Abelian. And that makes it interesting because we haven't so far got any constructions out to get its families. And so this is an example of a group which is in fact an extension of C3 by Q8. But in this case, you see this group is non-Abelian here. And so our theorem doesn't apply. So what we're going to prove that in this case, actually the embedding problem is soluble and you can extend every C3 field, putting it inside into an SL2 of three field. So I finished last lecture by just writing down the list of subgroups of SL2 of three which are relevant. So it misses two subgroups that we don't care about but it has a whole tower for Q8 and then going up to SL2 of three. And by Galois correspondence, it gives us an indication of what we're going to try to do. We'll start with any cubic extension depending on what this field is, think of it as a Q. So cubic extension of Q or maybe Q of T. So then it's a family then. And then I would like to compute a quaternion extension on top of that, on which C3 acts like it acts on Q8 naturally as automorphisms by permuting i, j and k cyclically. So this is what we're going to try to do. And for that, we're not quite going, I mean, in theory, we know everything about Q8 extensions now we have a theorem due to Witt. But there is a slightly different characterization of quaternion extensions. A lot of people that were interested in quaternion extensions, which even goes earlier than Witt. It's a theorem of Bucht from 1910, which is essentially equivalent to what Witt has done in this case. And what he proves is the following, that if you have a bipoderative extension of some field, so let me call it capital K because that will be like in my diagram. And then I would like to compute a quaternion extension on top of it. So I start with a bi-quadratic and I ask, can I embed it into a quaternion extension? Bicoderatics are easy, take two elements and join the square roots, but quaternion extensions are harder. So, and here, because this problem is abstracted, if you remember, and in this case, what he showed is that such a bi-quadratic is embeddable in a quaternion extension, if and only if this U and V are up to squares in K, so squares in K obviously don't change which quadratic you get, but up to squares in K, U and V have the following sort of funny form. There exists alpha, beta and gamma in K, such that U is one plus alpha squared plus alpha squared beta squared times the same thing but shifted where you commute cyclically alpha, beta, gamma, similarly for V, when you shift this whole construction a bit further, and if you multiply these two together, you'll also see that one of these terms, maybe this one, becomes a square, so you can absorb it into square in K, so the expression for U, V, it's also very symmetric in this context. And now, so if you stare at this, well, first of all, I call this bi-quadratic square root of U and square root of V, but actually it has three quadratic subfields in it which are quite symmetric and they're given by square root U, square root V and square root U, V, so these are the three numbers whose square roots I'm joining, and you see his construction is very symmetric in U, V and W in a sense that if it respects the C3 symmetry, if I take this alpha, beta and gamma in such a way that they're invariant that C3 permutes them cyclically, it'll permute these three guys cyclically, and that's exactly what we want because we want not just any quaternary extension of the top of a C3, it'll give us a large risk product, but the one which respects the C3 action. And this, now with this, it's very easy both sort of in practice and in theory now to construct a sort of three extensions, so this is how we go about it. We take now, as I said, where essentially solving an embedded problem, so take any C3 extension, let's say of Q, and then you wait to put it in the family, so take any cubic Galois extension of the rational, which is going through the bottom of our diagram. So maybe let me go back for a second, so recall that again, in SL2 is three, quaternions are an index two subgroup, and then inside there are three subgroups of index two permuted by C3, and then there's a center and then this whole thing, and then identity, and the corresponding tower of fields is that we started with the cubic extension of little k, and then we are going to construct three quadratic extensions, which are permuted by C3. So what we are going to do, we take alpha to be, well, basically anything you want, so take it to be a general enough element of k, I'll comment to this in a second, sort of like when we constructed cyclic families and so on, when we said, well, take a general element of Q zeta n, and then apply its construction. So let's take alpha to be a general element in this cubic extension of Q, and now everything is fixed, because what I would like to do, I would like to have these three quadratics permuted by gamma, so I would like alpha beta at gamma to be permuted by gamma, so I'm going to let the gamma group of C3, so sorry, the gamma group of k over Q, the cyclic group of order three, let's pick a generator called G, I'm going to hit alpha with G, I get beta in k, hit it again by G, you get gamma in k, so alpha beta gamma would be three elements in this field, which are permuted by this C3 downstairs. Okay, this gives us alpha beta gamma, then we divide, define UV and W by this formula, and by his criterion now, this extension is first of all embedded into a quaternion extension, and secondly, everything is so symmetric that it's not hard at the quaternion extensions, if you recall from Witt's theorem, I essentially unique that it forces the whole thing to be Galois over Q with Galois group SL2 and 3. In fact, because Witt is very explicit, it tells you what is the extension that you get, you can unravel it, and then you find that the SL2 and 3 extension that we're after is given by this, very nice, very symmetric, explicit formula. It's K, I join root U, root V, so that's a bipodratic, and then after that, to get the quaternion thing on top, we're joined square root of this expression, which is, you see again, symmetrical UV and W. So the only thing is that if you do this theoretically, we haven't quite solved an embedding problem yet, because it could be, you know, that something goes terribly wrong and there are no such thing as general enough elements of K for this to work. So for example, for some reason, you always get smaller extensions, that all these elements, for example, for some reason, if you take them in less three extension, could be, I don't know, a square. So this style on top that we've constructed is going to be trivial. So you have to do a little bit of work to prove that this is not the case and these things exist, no matter what a little capital K or a little K is, but it's not terribly hard. And there is a theorem, which I think Fadeev was first to prove in 1945. So he proved that this, what I called picking in general enough element of Ks for this construction to work is always possible. So he showed that over Q or Q of T, or in fact, over any hill version field, this construction works and every C three extension of little K can be embedded into an SL two or three extension of little K. In particular, these things exist and there are regular families of SL two or three extensions. So again, this construction is kind of interesting is that it does solve an embedding problem in the case where the kernel is a non-ability group and it works well. And of course, you can ask yourself, well, like when does this work? Can you generalize it to other groups and semi-direct Q where N is non-ability? In other words, suppose I have a family which has a Galois group Q or some field Q or Q of T or some regression field, whatever, and I want to extend it to a family. I want to put it inside a larger group which has the following shape and semi-direct Q where N is an arbitrary group for which N and Q you can do this. If you could do this, at least soluble groups I think you could get very far with this process. This is not known. Well, in fact, this is not known for which class of groups it is possible but there's a big conjecture by the Bayer-Deschamps that say that it's always possible. So the conjecture says that every what we've now considered the obvious name split embedding problem. So lifting a Q extension to an N semi-direct Q extension for any N and any Q it's always possible over any hill version field. Now, of course, it's a difficult conjecture because it's much stronger than the inverse Galois problem. You can take N can be arbitrary and Q could be trivial. So for example, it says that you can embed any C1 extension into an extension with an arbitrary group, a Galois group. So this is a very strong statement but at least it gives you an indication of what we expect to be true and maybe hopeful to find some techniques which work and there are some very interesting groups that you could try to play with and see if you can generalize this construction in some way. So the difficulty here really is that what happens here secretly is that, remember again, the Quaternion group, it has an auto-automorphism of all the three and here we've got a family or construction whatever of Quaternion extensions which respects this auto-automorphism action. And this was what make, and this is now what makes it possible to push it further to the two or three extensions. And normally when you construct a family for some group G it is not going to be invariant on the auto-automorphism of G in any way. Well, unless maybe you try hard or I don't know. So it really raised an obvious question is how far you can push this. In chat, if there are any questions, if no. So I think that's all I want to say about this whole embedding problem and to finish by talking a little bit about two methods. So we've talked about semi-Abelian groups and split embedding problems. This is a very nice channel method and we know when it applies more or less except there are conjunctures that say that it applies much, much generally. But let us look at two other methods. One which is called descent to subgroups and one which called rigidity, which again, I think have a potential to be pushed further and they would be sort of very interesting I think to study as research problems. So one obvious question I talked a lot about with products, direct products, quotient groups, but well, except for an answer to one of the questions and one of the first lectures, I didn't talk much about subgroups and there is an obvious question. Suppose you constructed a family which has Galois group U, let's say, and then you have a subgroup G U and the question is can you find a subfamily which has this Galois group G? So for example, what Hilbert's construction of AN extension essentially does, it says let's start with an SN family such as X to the N plus AX plus B and then find some specialization. So try to find let's say A and B to be rational functions on some variable T that the discriminant of this thing becomes a square and the family drops from SN to AM. And you can ask yourself, when is this possible? And again, that's a big question. If this were always possible, you could solve the inverse Galois problem because every group is a subgroup of SN and we know how to construct SN families. So if we could always descend to a subgroup, then we could construct an arbitrary group G. We don't expect this really. As asked like this for an arbitrary family, this is certainly not true and you can not always do that. So let us look at example where you can construct C3 families starting from an S3 family like this by starting this descent problem. And already this case shows very nicely what sort of problems you encounter along the way. So again, what I would like to do, well, essentially this is Hilbert's case because this is A3 inside S3. I would like to construct families of extensions of cubic Galois extensions, let's say a cube starting from the fact that I know a lot of S3 extensions because just take an arbitrary cubic and then, well, you've got most likely it's going to be an S3 extension when it's going to be small. So let's start with the S3 family which is built in the package for no particular reason. It's just has small coefficients. It's an obvious family where you take a polynomial X cubed plus X plus A. Right, so again, what I would like to do, I would like to see if I think of this as a family of S3 extensions, so I take A to be one, two, three, any rational number, it'll give me an extension. And a question is, does it sometimes become dropped? Does the group sometimes drop to C3 and can you put these exceptions into a family? Well, how do you know whether it drops to C3? You have to compute the discriminant and see if the discriminant is a square. So you compute the discriminant of this cubic up to squares, it's minus 27 A squared minus four. And what I'd like to see is that if there are specific rational values for A or ideally a whole rational function, let's say A of T such that minus 27 A of T squared minus four is equal to a square. And if yes, then by plugging in instead of A, this sub family A of T will give me a family now over P1 with a variable T, a larger degree possibly, but now it will have the Galois group contained in C3. Now, unfortunately for this particular family, this is clearly impossible because you can't make this into a square. Minus 27 A squared minus four is very negative. So if you want this to be equal to B squared, this gives you a two to zero curve over the rationals which doesn't have any rational points because it doesn't have any real points. So there is a local obstruction in this case. And therefore, you can forget about this family. It doesn't have any C3 specializations. You might as well sort of throw it away and start over with you. So let's do that. Let's take another S3 family which is X cubed plus AX plus one. So now, I let this last coefficient to be one and the linear term to vary with A. The discriminant of this is minus four A cubed minus 27, oops, sorry about that. This B squared shouldn't be there because B is one. So it should be just minus four A cubed minus 27, like here. And again, you can ask yourself, can this expression be a square sometimes? So in other words, what happens if I equate this expression to B squared? What's the set of rational solutions to this equation look like? Well, if you look at the equation B squared equals minus four A cubed minus 27, this is what's called an elliptic curve. It's a curve of genus one. It may have rational points. So it is quite possible that you can find values for A and B such that this expression here on the right is a square. I didn't check this to be honest. But in any case, what we know for sure is that because it's a high genus curve, it certainly doesn't admit any maps of P1 into it. So you cannot map rationally P1 into an elliptic curve. And therefore there is no rational function like this A of t such that minus four A of t cubed minus 27 is always a square because this is a non-singular curve of genus one. So in other words, using this family here, you may be able to construct C three extensions over Q. So for example, to solve inverse go problem for C three over Q, by just saying, well, let's take A to be this number, then this thing just happens to draw to C three, the Galois group, but you cannot solve the inverse Galois or remove Q of t with it. And what you find out if you play with this in practice, a lot of families that are built in, for example, in the package, if you do this, so you take a family for a specific group and let you say, for instance, what happens if I try to go to a, try to descend to a quadratic subfield. So this quadratic subfield will be like here, given by an equation B squared or X squared, if you want equals some polynomial in A. And sometimes this will be a genus zero curve, which doesn't have rational points. Sometimes it will be a high genus curve like here. Sometimes you're lucky and it will be a genus zero curve with rational points, but that's not going to happen very often. So there will be very few situations where this descent is going to work for one-dimensional families. Yeah, so can you explain again why the discriminant being a square means that there is C3 extension? So generally, when you have an S3 extension, so remember what an S3 extension looks like. So S3 is a normal subgroup, which is C3, and that corresponds by Galois theory to a quadratic extension of Q contained inside the specific field. So when you have, for example, this S3 extension, the quadratic inside it, it's given by the square root of the discriminant of this polynomial. So inside this field here, there is a Q a joint square root of this. But of course, if this expression is a square, then Q a joint square root of this becomes a trivial extension. And in that case, your group, the Galois group is not S3 anymore. It has to be a smaller group. In fact, it's has to be contained in the alternating group A3. So in general, when you have a family, let's say given by a polynomial of degree N, its Galois group is some, let's say transitive subgroup of SN, transitive if the polynomial is useful. But if the discriminant of that polynomial is square, then it is in fact contained in the alternating group A. So if you can force a discriminant to be square, you can force your group to drop into the alternating group. And that's what I'm trying to do here because the alternating group here is just C3. So I'm really forcing the discriminant of the cubic to be a square. Okay, and now, but what happens here is that if you forget for the moment, like I spoke from most of my lectures so far, about one dimensional families, about families with just one parameter eight. We said we're interesting in an inverse Galois problem of Q of t or Q of a. If you allow families with larger number of variables, then suddenly you can push this method much further. And let me illustrate it in this case, let's take the generic S3 family, XQ plus AX plus B where A and B are two parameters. So this now in fact covers all S3 extensions of Q. And it's discriminant again, it's given by this formula and minus four AQ minus 27 B squared. And again, I'm asking myself the same question, can I make this expression of square? Can I find some substitutions for A and B, some rational functions, which satisfy the property that minus four times, this rational function Q minus this rational function squared is always a square. So in other words, I'm interested now in equation C squared equals minus four AQ minus 27 B squared. And this equation is not a curve anymore, it's a surface because it's now given by one equation, three variables. And because this equation has the largest degree, you see here is three is what's called the cubic surface, it's a surface of degree three. And such surfaces, if you're lucky and they have rational points at all, then there was called rational. So they are birational to the projective plane. So there is a map, which is defined as an open subset of P2 from a projective space with two variables, let's say capital A, capital B, or three variables you make homogeneous to our surface S here. And I left a similar exercise, I just put it up on the web for slightly more interesting groups that S3, but various computer algebra systems including magma have a machinery to do that. So you can give it a surface and you ask, is the surface rational? And if he asks, can you give me, can you give me a rational parameterization please? And in this case, what it says, it says yet surely the surface is rational and it can be parameterized as follows. If you take A to be this expression here minus a quarter of 27 A squared plus B squared and B equals this and C equals this, then you take this polynomials and capital A and capital B, you plug them in here, you find out that this equation is universally satisfied. So this gives a rational parameterization P2 to S. So this map is more or less an isomorphism in the sense that there are rational functions which give you that the other way expressing A and B in terms of these things which is kind of quite easy from these formulas. And yes, that's probably true. Yes, thank you, Bjorn. Yes, in this case, you could easily see that this surface is rational and just so for T. Yeah, so you don't need much for this case here. Yeah, I think you're right. So but in any case, either way, once you found a parameterization like this, then you see that you can substitute this A, B and C into the family of mine, into this family here. Yeah, so C I don't really care about. It's some sort of, it's what solves what makes, it's a square root of this expression. But the A and B I do care about. So if I plug A to be this and B to be this, then I find a family like this of cubic equations which is now, they have square discriminant and the discriminant is C squared where C is this expression here. So I find now a family of C3 extensions of QAB and in fact, this family is generic in a sense that it covers every single cubic extension, Galois cubic extension of Q, which is easy to see because every cubic extension of Q can be given by an equation like this. So it does correspond to a rational point on the surface and because we found that by rational isomorphism like this, we covered all rational points on the surface possibly with some exceptions, but I don't think in this case they matter. So this family here is what's called generic. It covers every single C3 extension of Q or in fact, over any extension of Q. So this is sort of a nice baby example. And it turns out that this method works in examples which are much more interesting. So let me give you one such example and look a little bit at the smallest groups for which the inverse Galois problem is unknown over Q of T because we've talked a lot about methods which allow you to go this far. So if you've looked at the list of semi-herbillian groups of order less than 64, you would have found that there are, I forgot I mentioned last time, I think six exceptions, one of them is A5 which we know how to do over Q of T by Hilbert and the other all relate to SL2F3. And it turns out that once you can construct SL2F3 by again fiddling with split extensions which you can now do for that family, you can construct the other groups there as well. So the smallest groups which are not known to be Galois groups with regularizations over Q of T are groups of what is 64. So again, if you've ever generated this list of non semi-herbillian groups, you'd find that there are 10 groups of what is 64 which are not semi-herbillian. So in a small group database, there's 250, I think groups of what is 64, it's a long list but there is 10 of them with numbers 8, 9, 10, 11, 12, 13, 14, 41, 42, 43 that turn out to be non semi-herbillian. To answer the question, is there any indication when a family will be generic, not as far as I know. So I mean, there are quite a few babies where people prove that certain families are generic, but they always like in my case essentially kind of generic by construction rather than you first construct a family and then you try to figure out whether it's generic or not generic. So I do not know, I don't know what answer to this. We're just looking at a family to decide that it covers every single extension of with that Galois group. I don't know if any such results to be honest. Okay, so there are 10 groups of what is 64 which are not semi-herbillian, in other words, which we don't know how to construct over QFT. And now groups, they all need to be generic. Groups, they all need potent and the new potent groups are classified by what's called the new potency class, the length of the central series. If you take the center of a group and divide by the center, you take the center of a group again and so on until you exhaust the whole group which you can always do in a new potent group. I mentioned very briefly there is a result by Thompson that every group from the potency class two is semi-herbillian and of these 10, seven are have new potency class three and three have new potency class four. And there's a paper by Leila Schnepz where she proves by extending this semi-herbillian, no, sorry, by looking, by studying the obstructions in this case of various lifting problems, she proves that the new potency class three groups of these seven, they in fact have realizations over QFT. So the three groups, which as far as I know are not known to be realizable over QFT are these three, the small group of 64, 41, 42, and 43. Let me just call them G1, G2, and G3. And it turns out that for at least one of them, this descent method works even though it's reasonably convoluted because you don't need to have quite a few steps in it. And you can construct a family for the first one of these three groups using this method. So if you look at G1, small group of 64, 41, then it's a transitive group of 60 points, number 156. It's a group of over 64. And it says as a maximum subgroup of some group of 128, which sits in a maximum subgroup, specific group of 256, 512, 1024, of which the last one, this particular group of 1024 still acting on 16 points, turns out to be a nice semi-direct product, which you can construct very easily. You can construct lots of families for this group using just resultant. It's essentially a with product kind of group. And then you can do this descent. So you can construct a family for this group and then you can ask yourself, can you find, can you descend to a subgroup of index two? Turns out that you can. Can you descend again? Turns out that you can. Can you descend again? Turns out that you can. And finally to the group G1. Along the way, you get some rational curves or rational surfaces. And in all cases, they have rational privatizations. There are non-local obstructions and you can construct a family for this group G1. G2 is very similar, except that I haven't been able yet to construct, to find a family for this group here, the top group on top, which I would be able to do this set all the way. I can only get up to the last step up to here, but not up to this specific group. And the third group G3 only acts on 32 points rather than 16, which makes the whole thing just a little bit more cumbersome. So in any case, what I just want to say about this descent method is that it does seem to be very powerful in practice. So it has been very helpful to construct a lot of groups and especially nice families for a lot of groups that haven't been known before. But I find it's very difficult to analyze theoretically. That is to say, if I start with a group and I start with a subgroup or even a maximum subgroup, somehow predicting in advance, that this descent is or is not going to be possible, whether you will end up running into local obstructions or not, I do not know both methods for doing that. So it's a little bit hard to, as far as I'm aware, to prove theoretical results with this method. That would be, I think, very interesting if one could do something like this rather than just do it group by group, family by family. Well, you can do this, but it's hard to do it on some sort of general basis. So to predict at least some situations when it works, I think that would be quite amazing. Okay, so that's, I think, all I want to say about this descent thing and except that, as I said, it does work very nicely in practice. And I gave, again, one exercise, how to construct, in fact, a generic family for fraternian groups using this method here, where you need to do a descent through a delpets of surface of degree four and parameterizing. It's a very nice example. Okay, and finally, the last method that I would like to talk about, it's called rigidity. It's a very powerful method which is on the opposite spectrum of what we've looked so far. We've mostly looked at soluble groups and this method applies mostly to simple groups or groups which are very close to simple. So let me, there are many variants of it. People did try to push it quite far, even though I think the work there is certainly not exhaustive, I'm sure you can do more with it. But let me at least state some of the basic versions of the rigidity method. So first of all, let us start with something that's called the Riemann existence theorem, which, as I, again, briefly mentioned before, it settles the inverse Galois problem over C of T and are analogs by Harbater and Pop. And there is other people who replace C here by a complete field, but over C, this is a very classical and very strong result due to Riemann. Since the following, suppose you have a finite group and you pick any number, any generators of G that you want, which multiply to one. So you pick elements G1 up to GL and G, completely arbitrary with the only condition here that G1 times G2 times and so on GLs just in this order is equal to one. And then you take arbitrary L, so it's the same L as this number of element, points on a projective line, so distinct points P1 up to PL. And then the Riemann existence theorem tells you that there exists a Galois cover of P1. So in other words, this is the compact Riemann surface which maps to P1, which has G, this group here, acting when it is an automorphism group, such that if you take the quotient of this group by of this Riemann surface by G, you get P1. So in other words, you get a map next to P1, such that G acts faithfully, so most points, they have exactly the order of G pre-images and such points are called already-fied, except, finally, many points where this is not true and this points, over these points, you say that the map phi is Riemann-fied. So, and the Riemann-ification locus is exactly these points P1 up to PL here, at least assuming the G is our identity. And so in other words, this Galois cover here is un-Riemann-fied outside this fine set of points S. And associated to every Riemann-fied point is what's called the Riemann-fied group. It's exactly the same as certification groups you see number theory. If you think of a cover of curves like this, in terms of their fields of rational functions, it gives you a Galois extension of fields of rational functions, where it goes the other way. So the field of rational functions here is just Q of t and you embed it into a larger field which has Galois group G over Q of t, sorry, C of t. We're in the complex numbers, I have to remember that. So C of t here and a fine extension is Galois group G here. And points of P1 you think of as places of the field of rational functions and the ramifications, the usual notion of ramification and the ramification group is the usual notion of our ramification group, except that here the residue field complex numbers algebraically closed. So this group here is always cyclic. And so what's in existence says is that you will discover to be un-ramified outside S and exactly have prescribed ramification groups. So ramification group over PI is generated by GI. But to be precise, ramification group is only defined up to conjugacy because if you take a different point over PI, you get a conjugate group. But one of these conjugates is exactly the group generated by GI. So it's an extremely powerful statement. And in particular, it tells you that the inverse Galois problem is very much true over C of t because so let's just prove this. In other words, every finite group has a regular realization over C of t. So you take any group G, you take any generators which multiply to one. So for example, you take any generators of G call them G1 up to GL minus one. And then you define GL to be the inverse of G1 GL minus one is supposed to be inverse. So the product is equal to one. And well, you apply Riemann existence here. So you take completely arbitrary points P1 up to PL. You can take any choice you wish. And then by Riemann existence, you find a cover like this phi with Galois group G which again by Galois theory corresponds to a field extension where the bottom field is C of t the field of rational functions of P1 and the top field, the field of rational functions of this curve X has Galois group G over it. And of course, it's automatically irregular because the field of constants here is algebraically closed. So there are no algebraic elements in this field over C except C itself. So that's it. So over C of t, you can construct covers like this and there's a lot of choice involved. And the problem is descending from C to Q because in general, you can write down some sort of variety which is called Hurwitz schema when it's one-dimensional Hurwitz curve which parameterizes things like this which parameterizes covers with a specific group and specific verification data. But this variety is difficult to compute, difficult to understand and so on. And they don't necessarily have rational points. So the problem really is to somehow get from here to the rational maps and to do the same sort of thing but over Q of t. And the trick is there is a general principle sort of a number theory and Galois theory that you can construct any sort of object over the complex numbers which is unique then just uniqueness forces it to be defined over the rationales. Basically because if it wasn't defined over rationales you could hit it by Galois and find another object like this. But now if it's unique, then it's not possible. Of course, this is a more philosophy than a theorem but it very often works and it works here in the same way. So generally the covers which are given by this theorem are not unique. There's a lot of choices involved and choices of these points PI, choices of these GIs and so on. And but you can what's called rigidified by putting enough conditions on this conjugacy classes. So on this GIs and selecting PI is sort of fixing them to be certain rational points in such a way that discover can be proved them to be unique and then it's can also be proved to be defined over Q. So that's what this rigidity method does. So this is how it works. So suppose G is a fine group and you pick L, it's the same L as before as I had this elements G1 up to GL. You pick L conjugacy classes. So let's say you fix them, let's call them C1 up to CL. So conjugacy classes of G and then you look at, well, basically what we're looking up there, the set of L tuples of elements G1 up to GL where GI each lies in the corresponding conjugacy class CI and their product is one. So G1, G2 up to GL is equal to one and they generate the whole group. So the group generated by G1 up to GL is equal to G. Now, if you just pick group at random and conjugacy class at random, this set may be empty because for example, you could just say, I want all my CI to be identity. Well, sure you can, but then they are never going to generate the whole group. So these conditions here may be sort of usually exclusive, but whatever this set is, let's say assume for the moment it's non-empty, G acts on it by conjugation because you can take an element little G and G and then conjugate all the GI by G and then you'll see that these conditions are going to be preserved. Conjugacy classes of course are preserved by conjugation. This product will still be one if you put here G and G in versus everywhere and they will still generate the whole group G because essentially they generate a conjugate of G inside G. So in terms of freedom of surfaces, what this conjugation does, it actually produces an isomorphic group of surface. So, okay, and normally this group G will act freely on it by conjugation. So in other words, all elements except identity are not going to have any fixed points. So let's look at this condition a little bit. So if G, little G has a fixed point on sigma, so in other words, suppose you have such a tuple G1 and 2GL that you conjugate by G and you find exactly the same tuple. So again, G acts on the tuples by conjugation. So it means that G conjugates G1 to G1, G2 to G2 and so on. So in other words, G commutes with all the GI's but because GI's generate the whole group, G commutes with the whole group and that means that G is in the center of G. So this condition is just so having fixed points equivalent to being in the center. So from now on, when we're going to talk about this rigidity method, we're going to assume that the center of G is trivial. As I mentioned before in one of the first lectures, this rigidity is a powerful method. It applies to a lot of interesting groups like the monster groups, but it certainly is restrictive. And for example, it never applies to groups such as milpotent groups because of this condition, the center has to be trivial. So then the action G on sigma is free. So it's just a union of orbits which are just regular orbits. Each one is sort of like a little copy of G. And we say that this tuples C1 up to CL is a rigid L tuple of conjugacy classes. If the size of this set, well, it's not empty, but otherwise in that it's as small as possible. So the size of this set sigma is exactly the order of G. Equivalently, you have only one orbit. The action of G on sigma is transitable. So every such tuple can be obtained from every other such tuple by conjugation by G. So this is the set up. So this is the most important definition here. So think about like this, you have a group, you have a bunch of conjugacy classes, you compute the order of this set, and if it turns out to be order of G, then you're like, okay, great, this is a tuple to which rigidity method, at least some form is probably going to apply. And a second condition, which you have to impose in some form or other to make the resulting family defined over Q, is what's called rationality for conjugacy classes. So it's also even easier to define. So if you've got a conjugacy class in G, it's called rational. If it has the following property, you take an element in that conjugacy class and you raise it to a power, which is co-prime to the order of G. So for example, you take the conjugacy class of elements of order three, then you take an element there and then you square it or you raise it to the power of four or power five anything co-prime to three because you don't want to get identity or something which is in a smaller conjugacy class but something co-prime to the order of G. And then the condition is conjugacy class is rational is that you always end up with the same class. So for any element G in the conjugacy class and any power which is co-prime to the order of G, this power here lies again in the same conjugacy class. So there are some examples of groups which are called rational groups. In fact, where every conjugacy class is rational and the most famous of them is the symmetrical SN. So if you take G to be SN, then it's conjugacy class to be correspond to cycle types. So for instance, in S4 you have a cycle type, four cycles, three cycles, double transpositions and so on. And our cycle type is unchanged under this operation. If you take, I don't know, a four cycle and you raise it to the power of three anything co-prime to four, then it stays a four cycle. So you end up in exactly the same conjugacy class. And that means that every conjugacy class in SN is rational. So SN is one of these groups to which this rigidity applies quite nicely, except we don't particularly care because constructing SN extensions is not very exciting. You just take an arbitrary polynomial, you will get an SN extension more or less. So this is not a very interesting group from our perspective, but nevertheless it illustrates at least its definition of rational conjugacy classes. And to give an example of a non-rational conjugacy class and what sort of more typically happens, let's take the smallest group which is not a SN really, cyclic group of what is three. So it has three elements, one G and G squared. And because the group is a billion, every element is its own conjugacy class. So this group has three conjugacy classes, identity, G and G squared. And of course you see already here that when you take an element and you square it, and that's an operation which is allowed, it's called prime to threes to the order of the element, then you end up in a different class. So you've got three classes, one G and G squared, and this operation of squaring an element, it has a property that this identity classes definitely always going to be preserved, but these classes G and G squared are going to be swapped. So this is an example of a group where not every conjugacy class is rationed. And the reason here I draw a character table for this group is that there is a very nice little lemma which says that a conjugacy class is rational, if and only if from point of character theory, every irreducible character takes a rational value on it. So you can always spot them in the character table. If you look at the character table of a specific group, you just look at the table where all entries are integers because algebraic integers, they're rational, they're automatically integers. And if you find such column, then you've got to solve the rational conjugacy class. And if not, then it's not. And like here, the operation of squaring or cubic or whatever an element will swap it with some other class and you can see which ones they are again by looking at the character table. So in this case, for example, if you look at this character table, complete conjugation swaps zeta three and zeta three squared and so it swaps these two columns and it tells you that these two classes are not rational. And more generally, this operation if you want by taking a conjugacy class associating to it the corresponding column in the character table and hitting it by Galois defines a Galois action on the set of conjugacy classes which factors to some a billion extension of q. It's a reasonably simple kind of action but it's not always trivial. Okay. So now with this notion of rational conjugacy classes we can state the first rigidity theorem which is usually referred to as basic rigidity theorem and combined work of the variety of people to Billy Fried, Matzatschik and Thompson. So following, suppose you've got a rigid l-tuple of rational conjugacy classes. So here are two conditions and I've got a group g trivial center, so three conditions. First condition is trivial center. Secondly, I take a tuple of conjugacy classes which are all rational. And thirdly, I want to be rigid in the sense that I described. So if you do this column elements, little g i in here which multiplied to one, the number you get is known zero but otherwise smallest possible equal to the order of g. Take p1, pl to be arbitrary but now rational points on p1. Then there exists and in fact, it exists unique which is its uniqueness which forces it to be a query. There exists a unique regular g covering from x to p1 over q which is like in the Riemann existence theorem except now this is the current defined over q. This covering here is defined over q. It's still unregified outside p1, pl and the inertia group at p i is generated by an element of this conjugacy class c i. So this is a nice theorem. It doesn't often apply because, well, the two conditions, rigidity and rationality. Rigidity is kind of, well, you have to have it otherwise there is no hope but rationality that we asked for is a little too strong for applications. There are just not many groups to which this applies. So there's a variant which I think is due to salary in this combination, in this formulation that says almost the same but it weakens rationality a little bit in an obvious way. It says, well, suppose you have a rigid tuple again of conjugacy classes c1 and cl but we don't want every class to be rational. We have an action of gawa, q bar over q on this set and we just want it to be gawa stable. So in other words, like in this example, if I took this conjugacy class and this conjugacy class, then although these two conjugacy classes are not rational, gawa permutes them sort of like with divisors, a rational divisor consists of points such that the whole thing is gawa stable but the points are not necessarily defined over q. It's exactly the same thing here. If you take these two classes, then they are not rational but they form a gawa stable set. And then you take the points pi in a projective line which are now not rational but which correspond to this action and that's always easy and possible to do. So you choose points p1 up to pl, now points in p1 q bar which are anti-isomorphic as a gawa set to this set c1 up to cl. And then there exists exactly as before a regular g covering which you can make unique once you sort of say carefully, state carefully what happens with inertia groups. Again, covering defined over q under a modified outside p1 pl and which has inertia at qi generated by an element of ci just a slightly restricting condition of rationality. And this now applies to a few interesting groups though not so many still and a lot of work by people like Batsat, Thompson who pushed it to sporadic groups and so on they looked at various variants and pushed this even further to write down specific variants or specific groups which work in their context and again, guarantee the existence of such a covering. But this method sort of look a bit at Hock which you can do if you're dealing with 25 sporadic groups because only 25 of them but I don't know of like a very clean general result which is more general. Okay, so in the last, what is it, seven minutes let me just finish with an example of an interesting group to which this rigidity applies. So the group here is a second simple group that there is we've already dealt with a five that's a smaller simple group non-Abelian simple group for the 60 the next non-Abelian group has order 168 and it's PSL2F7 or GS3F2 same group. And here is its conjugacy classes and a character table. So it has six conjugacy classes and six irreducible characters, elements of order one, two, three, four and two conjugacy classes of size seven, sorry, of elements of order seven and I wrote the sizes here of conjugacy classes and I wrote the six irreducible characters and the character table. And you see here immediately from this table that these four classes are rational and these two classes are not they're commuted by Galois as you can see by these numbers here which are commuted by Galois which are not rational numbers. Okay, so now what we can do we can find a rigid tuple of Quanget's classes and what, I mean, this is such a small group that you can just write a code, you know which goes through elements in these classes and just finds tuples which multiplies to one it's not very hard to do and the process will terminate very quickly in large groups it's harder to do but there's a very nice neat formula that allows to count in any group L tuples of elements such as GI lies in CI where CI given conjugacy classes and which multiplied to one exactly the thing that you want except the condition that GI generate the whole group but the formula is very, very nice it's given in terms of characters it says that in any group G and conjugacy classes C1 up to CL if you are interested in a number of elements so L tuples GI in that class CI whose product is equal to one then the number is equal to one of the order of G times the product of the size of these conjugacy classes and then there is a sum of irreducible characters of the values of that character like here are the values of irreducible characters of various classes on the class C1 up to CL multiplied together and then divided by the dimension of chi to the power L minus two I mean it looks maybe a little convoluted but because character tables are easy to compute normally for even large simple groups this gives a very efficient way to find these things so in this group PSL 27 there are three classes which I sort of try to mark with bold which is these guys element of order three and the two classes of order seven so again this is a collection of so C3, C5 and C6 that's going to be my triple of conjugacy classes and these two are not rational but the whole collection is defined over Q so the Gallo group over Q permutes these two so this is a rational in the sense of serves Gallo stable triple of conjugacy classes and what we're going to check we're going to check that this is a rigid triple so first let's apply this formula here and count the number of triples of elements element in this class element for three in this group element for seven in this group another element for seven in this group such which is in different conjugacy class which all multiply to one and this number is given by this formula here so you can just compute it you don't even need computer algebra system for this so it's one over the order of group which is 168 times the product of the size of conjugacy classes which are 56, 24 and 24 easy to see if you compute so centralizes of various elements sorry I lost my page pressing something, there it is again and then you need this sum over the characters the values so you go through all the characters and then you look at these values on this class multiply by the value on this class multiply by the value on this class and now of course there's lots of zeros at my table so for these characters so two rows three, four and five the product value here times value here times value here is going to be zero so in actually it's only two characters that are relevant which is this one and that one where the values are known zero so what you find is 168 times 56, 24, 24 and then one times one times one with a trivial character divided by one dimension of chi and then for this last character here minus one times one times one divided by the dimension of chi to the power my L is three so to the power one so over eight so this is seven eighths I guess one minus one eighths so you multiply one over three times seven times eight by seven by eight by three by eight by three by eight and by seven eighths and if you do that you see that one of the sevens cancels one of the eights cancels and one of the threes cancels and you find that you get 168 which is the order of the group so in other words the number of triples which satisfy this condition there is exactly 168 of them and it turns out that every such trip will automatically generates chi I don't know a very quick way to see this but one way to see this for example is just to look at this group just find one triple of elements like this which does generate chi just find one example by hand and then once you know that there is one example we know that there is 168 of them because remember again the set of elements which generate chi they form a union of regular orbits and therefore you know that all of these 162 triples have this property and therefore this triple is rigid you could probably do something a little bit better by proving that two of these elements already generated group of order 21 at least element 43 and element 47 if you could just prove that this group has larger order then it will be enough to prove that it's whole group but I'm not sure I'm not entirely sure how to finish off this argument so in any case it's easy to find an example it's a small group and then you find that this triple is therefore rigid and by the rigidity theorem therefore this is a Galois group of q of t so this example is I mean by itself it's maybe not super interesting because I left as an exercise to prove that for some reason that I don't exactly know it's very easy to realize it's global q of t anyway you look at LMFDB you look at a bunch of fields there which have PSL2F7 group there's lots of them and by just basically interpolating them as I did there in an exercise you find lots and lots of one parameter families and you can in fact push it to two, three or even four parameter families that's the paper of Malle which does this so this group is actually quite easy to construct over q of t but rigidity applies to a lot of other groups where this is certainly not the case and generally just to give a last comment on PSL2F7 so Schich has proved a theorem using a different method which involves module forms and module curves but PSL2FP is a Galois group over q whenever two or three or seven is a quadratic non residue on UP and this particular in particular it covers PSL2F7 so what we just did is basically we proved one simple case of his result but there are many other methods and especially the other groups especially sporadic simple groups where it's only using rigidity that we know how to construct how to construct these groups so generally it's a very powerful method and it complements a lot of things that we've done so far which mostly applies to solute groups okay I think I exactly run out of time so this is a really good place to stop for me thank you very much okay let's thank Tim both for that lecture and for all his lectures we're a little past 10 but I think there's a break before the next problem session so I think we have time for questions are there modifications to the rigidity results if Z of G is known trivial that's a very good question so I think I've seen several so I've seen one paper where I think it's a book in Malematsat who say that it's you don't need a center to be trivial you need it to have a complement to the group so a subgroup which together with the center generates the whole group and which have a trivial intersection but as somebody else remarks in different paper that means that your group is a direct product by the center of something else a smaller group and in a smaller group you still have rigidity so in fact it doesn't add anything interesting so to answer your question I do not know personally of any modification which is interesting in a sense that it would allow you to do more groups than you could do otherwise with this rigidity method so it seems somehow a bit of a fundamental issue is this method because of the action of the group by conjugation and the fact that the center is just invisible Other questions? Could I ask a question? Please Yes, I heard there was a question Oh yeah so other questions so part of the problem is this variety which has a map to P1 there needs to be defined over Q so sort of for the most but is there any connection with Galatasent in this complex? Well certainly in a sense that what we are trying to do here is you know Galatasent we are trying to we start with a curve over C or more generally as I mentioned there is a theory of who with schemes and who with curves where you try to understand all colors which have this property and then to descend the corresponding scheme to Q or find and then after that find its rational points so this is I think a special case of Galatasent and in fact I think the proof in Serge's book Topics and Galatas theory of this basic rigidity theorem it's you know it's very much Galatasent like it uses arithmetic fundamental group and it's very much in that flavor and I think he even mentions connections to Galatasent theorems in there Okay, thank you and there was a question what is the status of PSO2FP? Yes, by now it's known for all primes so Sheikh proved it under these conditions of two, three and seven being quadratic non-residue and then it was extended I think by Matzat two and five as a quadratic non-residue and by now it is completely settled but I think it is apart from the alternating groups it's possibly the only family like complete family of simple groups for which it is known for which you know that these groups are known to be realized over Q of t There's also a work of Zhiyue Yun that did some exceptional groups I think G2, G2FP and E8FP but I guess those are yeah I think they're Okay simple Yeah, absolutely Thank you Okay, if there are no other questions let's thank Tim again for all of his nice lectures