 Thank you very much for the introduction and for the invitation. So since the title already got was read, let me just kind of jump into the talk. Simply got to get on the right screen. There we go. So the title of the talk is about field of modular and fields of definition. And I'm going to start with fields of definition. So this is just a very, very general concept. So start with a field K and let K bar be a separable closure. I tend to just work with characteristics zero fields on algebraic closure. And let X be any of your favorite kind of object. A curve, a surface, an abelian variety, some other kind of variety, a morphism between varieties, where X is defined over K bar. So it's defined over some extension field. But within the K bar isomorphism class of this object, we can consider the fields L such that there's an object in the isomorphism class so that that object is isomorphic to your original X if you allow isomorphisms over K bar. But the object X prime itself is actually defined over L. So for example, if you talk about varieties, the variety is defined over L if its equations have coefficients in L or a map is defined over L if it's defined by polynomials with coefficients in L. So if you're given this X, which you want to think of as a K bar isomorphism class, we'll say that a field L is a field of definition for X. If there is an X prime, that's K bar isomorphic to X, but X prime itself is defined over L. And of course, I mean, if L is a field of definition, any extension of L will be a field of definition. So there are lots of them. And I'll write FOD of X or fields of definition of X to be the set of all L's such that L is one of these fields of definition. And one might expect if you haven't played with these things very much that there's a minimal field of definition. The natural thing to do is just take all the fields of definition and take their intersection. And that will be contained in all the fields of definition. And you'd expect that might well be a field of definition, but it turns out it often isn't. And that's really what I want to talk about today. But I'm going to start with an example that's probably the most familiar to people, where in fact, there is a smallest field of definition, which is the theory of elliptic curves. So let me let E be an elliptic curve, say defined over some number field. Then we know that the isomorphism class, the Q bar isomorphism class is uniquely determined by the J invariant. On the other hand, if I take a Bierstrauss model for E, so Y squared it was X, Q plus AX plus B where A and B are algebraic numbers, then the field I get by adjoining A and B is clearly a field of definition because I wrote down an equation whose coefficients are A and B, which live in Q of A, B. And of course, the J invariant is also in the field generated by A and B because the J invariant is just a rational function of A and B. However, it turns out, okay, so this means that every field of definition will contain the J invariant of the elliptic curve, but it turns out that in fact, E already has a model defined over J of E. So Q adjoined J of E is not only, is actually a field of definition. And explicitly, I wrote down a Bierstrauss equation, it's a little messy, but this E prime has the same J invariant as E does. So E and E prime are Q bar isomorphic to each other. E prime is visibly defined over the field where you get adjoining J of E. Okay. So the question is, is there a smallest field of definition? And if so, how would we find it? Well, as I mentioned, the natural thing to do is to take the intersection of all of the fields of definition, but in terms of analyzing what's going on, it's actually helpful to take an approach more using Galois theory. So let me let L be a field of definition for this object X, say as a variety. What that means is that there's some other object X prime that's defined over this field L and a K bar isomorphism from X to the X prime. That's the definition of being a field of definition. So now suppose I take an arbitrary element of the Galois group of K bar over L, so an element of Galois that fixes L, then this, can you see my pointer here? Yes? Okay. So then down here, if I apply sigma to X, I'll get a new variety or object, whatever it is. And since I had this isomorphism from X to X prime to find over K bar, if I apply sigma to everything inside, I get an isomorphism from X sigma to X prime sigma. But X prime sigma is X prime because X prime is defined over L. So it has equations with coefficients in L. And then I can apply the inverse to get back to X. So the moral here is if L is a field of definition and sigma fixes L, then applying sigma to any equation for X gets me an object that's isomorphic to X, although the isomorphism will be defined over K bar. So with that in mind, L being a field of definition means that there's a typo. There's supposed to be sigmas in Galois group of K bar over L. X will be isomorphic to X sigma, so this leads to the definition of the field of moduli. So the field of moduli of X is we look at the set of sigmas in the full Galois group such that X sigma is K bar isomorphic to X. This gives me a subgroup of the Galois group. And then the field of moduli is the fixed field of this Galois subgroup, if you like. And that's a particular field. So X has lots of fields of definition, but it only has one field of moduli. It is the fixed field of this of this group. And I'll write FOM, so field of moduli of X for the field of moduli of X. And fairly easy to see, well, actually the boxed formulas in the previous slide essentially prove A that if L is a field of definition, then it will contain the fixed field of this group up here. So it will contain the field of moduli. And it's not so hard to show that this definition is the same as the intuitive definition that this field of moduli is the intersection of all the fields of definition. Okay. So a key, well, two key questions. The first one is the field of moduli, a field of definition. Keep in mind from here, oops, sorry. The field of moduli is the smallest possible field of definition. So now I'm asking, is it actually a field of definition? Can I find a model for X that's defined over this minimal possible field of definition? And, well, given away, the answer is not always. So if not, then one might ask how close can one get to it? So I'm going to write D of X for the degree of the minimal extension, well, the smallest degree of a field of definition over the field of moduli. Okay. So just to say it in slightly different words, you're given X, you can find this field of moduli that's the minimum possible field of definition. But if it isn't, how big an extension do I need to make to get to a field of definition? Okay. So hopefully it's clear what the questions are. And an example for an elliptic curve, the field of moduli is a field of definition. That's because it's the field generated by the J invariant. So the defect in this case is one. Elliptic curves have defect one. However, there are curves of higher genus and there are a billion varieties of higher dimension where the field of moduli is not a field of definition. And in fact, this is the first place that I ran across this problem was in Chamorra's arithmetic theory of monomorphic forms where he has a fairly detailed discussion about fields of moduli of a billion varieties. So why do we care about this? Well, in the first place, when you're working with something in arithmetic geometry, it's nice to be working over the smallest field that's necessary. But for explicit computations, you actually kind of need to write down an explicit model for your object. If it's a variety, you'd want equations for the variety. If it's a map, you'd want equations for the map. So you can get a model defined over any field of definition. So if you know some sort of bound on how big a field of definition has to be over this minimal field of moduli, it gives you some information there. So it will get you a model to work with. It's also useful for proving uniform badness results because a k-rational point on a moduli space essentially represents a k-bar isomorphism class of the objects. So the abstract, I hope people won't be too upset that I cheated a little bit. I said I was going to talk about the classical case of curves and a billion varieties a little bit, but I realized I won't get to the main thing I want to talk about if I spend time on that. So I'm going to jump right into the actual objects that I'm interested in, which are maps of projective space to itself. So the objects that I want to look at, which I've denoted N and D, the endomorphisms of projective N space of degree D, and I want to look at the the maps that are morphisms. Okay, and we'll again, we'll want to look at k-bar isomorphism classes of those and then ask, you know, what's the minimal field where we can find a map in the k-bar isomorphism class. We'll get to that in a minute. So for the moment, this is the space of objects that I want to study. And I'll note that I can give it the structure of a quasi-projective variety, indeed, an aft line variety. Actually, so that was a good question someone asked in the chat, which some of the answers. Someone asked if the minimal field of definition might have, or a minimal field of definition might have infinite degree. The answer is basically no, at least in the settings I'm talking about, because every object, you can write down some model for it. That model has finitely many coefficients and they generate a finite extension. Okay, good. So we're looking at this space of degree D morphisms of projective space, which is an aft line variety. So I'm going to write it down explicitly for maps of P1. So what does a map of P1 look like? It's just given, I mean, you should be able to look at the second line here. It's given by two homogeneous polynomials of degree D. And each of those polynomials has D plus one coefficients. So a map from P1 to itself, a degree D map of P1 to itself, is given by a 2D plus 2 tuple of numbers of elements of your field. But of course, if you multiply all the coefficients by the same quantity, you'll get the same map because everything's homogeneous. So that 2D plus 2 tuple is really a point in projective 2D plus one space. So each map gives you a point in projective space. These point in projective space gives you a map of degree D, almost. There's one slight hiccup, which is I want to look at maps that are actually of degree D. So if these two polynomials have a common factor, then that would cancel out. Because we're using homogeneous coordinates, you get a lower degree map. So what I really want is all the a comma b pairs such that the f sub a and f sub b polynomials, these two homogeneous polynomials have no common root. And that's given by throwing out the a comma b pairs where the resultant is zero. So the endomorphisms of degree D on P1 is the set of pairs a, b where the resultant of f sub a and f sub b is not zero. That resultant is just some messy polynomial in the coefficients of a and b. And in fact, the same thing is true for maps in projective n space. In that case, you'll have n plus one polynomials of degree D in n plus one variables. You get a huge number of coefficients. You have to throw away the points where those polynomials have a common root. But that's the zero locus of the Macaulay resultant. So this is something like, there's a higher dimensional resultant there. So the net result is that the space of maps at the top of the slide here is simply the set of points in some high dimensional projective space throwing away complicated hyperservice. Okay. So I don't want to actually consider sort of one map. I want to consider isomorphism classes of maps, just as for elliptic curves, it's really the isomorphism class of the elliptic curve. So when are two maps isomorphic on PM? Well, basically they're isomorphic if you can get from one to the other just by changing coordinates, right? Because change of coordinates really doesn't affect much. But there are many different ways to change coordinates. So I could just change coordinates in the domain or in the range when both of them, but independently, or in both of them, but not independently. And there are actually situations where each of these answers sort of is the right answer for the problem that one's studying. For example, just change, well, if you replace Pn by the first Pn by a curve and take the other Pn just to be P1, you might want to just change coordinates in the base P1. This would be sort of, if you're studying Desson-Gonfond, for example. But that's not what I want to study here. What I'm interested in, I'm interested in in these maps as being dynamics. So I'm interested in iteration. Okay. So I won't actually be talking about dynamics, but I need to mention this because this is where the problem comes from. So if we think of a map from Pn to Pn as generating a dynamical system, I'm going to iterate app and look at orbits and stuff like that. But I really only care about f up to isomorphism. So when are two maps dynamically isomorphic? Well, that means up to changing coordinates in a manner that respects iteration. And the correct thing there is that you want to conjugate the map by an automorphism of Pn. Okay. So we'll say that two maps are dynamically isomorphic if they differ by conjugation by an element of the automorphism group of Pn, which is just n plus one by n plus one matrices, but up to scaling. And the reason that's good for dynamics is because if you iterate, well, I actually wrote this explicitly as an action of Pgl on the space of maps by sending a map to write an f superscript C to be f conjugated by C. And then iteration commutes with this action of Pgl. Because when you iterate, the feet the inverses cancel out in the middle. Okay. So you get an isomorphism class of app. The dynamical isomorphism class is the set of all conjugates of that, where we're conjugating by elements of Pgl. Okay. And now we're going to take f to have coefficients in some number field. But I want to try to conjugate it to get myself down to a smaller number field. So now we get to the main problem I want to discuss, which is field of moduli versus fields of definition for dynamical systems on projective n space. And for the rest of the talk, I'll fix K to be a number field, K bar is an algebraic closure, and N and D will be integers and is the dimension of our projective space. So I want that to be at least one and D will be the degree of the map. And I want that to be at least two. It is interesting to study maps of degree one, but things are very different. So I'm going to restrict to the case of maps of degree two or bigger. Okay. And the two key questions is if you have an F whose, so this is a map given by polynomials, if the coefficients are in K bar, I can find a field of definition for F. That's the smallest possible, I'm sorry, field of moduli. That's the smallest possible field of definition. And is it actually a field of definition? And if not, how big an extension do I need to go to? So here's an example. I'm going to do two examples. Here's the first one. Here's a map on P1, which I've de-homogenized. So it's a polynomial, but you can re-homogenize it if you want. So x squared plus 2ix minus 4 minus i. So q adjoin i is clearly a field of definition, because I wrote down an equation for F of x whose coefficients are in Q of i. But in fact, I can conjugate F by the map x plus i. So Fv means, what is it? It's phi composed with F composed with phi inverse. Or maybe it was the other way around. Anyway, one or the other. And if you do that conjugation, you'll see that the new equation becomes x cubed minus three. So the coefficients are in Q. So Q is a field of definition for F because I can change coordinates to get to a model for it that has coefficients in Q. And then the defect is one because, well, Q is the absolute minimal possible field. You can have this field of modulator field of definition. So this one isn't that. How did a quadratic become a cubic under conjugation? Thank you. That is called a typo. Sorry. This three here is supposed to be a two. Okay. Here is a more interesting example. Again, Q of i is a field of definition, because you can see the coefficients live in Q of i. So it's this cubic rational function, which again I've de-homogenized. There's a map from P1 to P1. And this time, thank you. I do mean x cubed, not x squared. And it turns out if you conjugate by the rational function minus one over x, I mean, I did the computation here in gory detail. You don't have to read it. But just read the first thing and the last thing. If you conjugate by minus one over x, you'll end up not with the original function f of x, but minus f of x. On the other hand, if I apply the non-trivial element of the Galois group of Q of i, namely i goes to minus i, that also changes f of x to minus f of x. So f conjugated by this rational, by this element of PGL, namely f conjugated by phi, is the same as applying the element of the Galois group to the coefficients of f. So the field of modular, so that means that sigma is in that Galois group whose fixed field is the field of moduli. So this f of x, even though it has a coefficient, this complex coefficient, Q is the field of moduli. f of x is PGL 2 conjugate to every Galois conjugate. It's another way to say it. However, I won't give the proof, but one can prove that Q is not a field of definition for f. And in fact, there is no way to conjugate f of x with a linear fractional transformation with coefficients in C to get to a rational function that even has real coefficients, much less rational coefficients. So the field of moduli or field of definition defect of this f is 2. Its field of moduli is Q. There is no, Q is not a field of definition, but there is a field of definition that's a quadratic field, for example, Q of i. Okay, so hopefully this explains what the problem is and shows at least that there are non-trivial examples. So there's something worth looking at here. Now for maps of P1, at this point we have a fairly good understanding of fields of moduli versus fields of definition that's encompassed in sort of three results that I've amalgamated into one theorem. First, if you take a rational function on P1, so or morphism of P1, and it has even degree, then the field of moduli is always a field of definition. That's why in this example I have a map of degree 3. I would have done a map of degree 2, except there aren't any. Second, if you take a polynomial map of P1, the field of moduli is always a field of definition. Again, that's why I used a rational function here rather than a polynomial, which would have been easier, but there aren't any. And then Hildago a few years, well it's almost a decade ago now, showed that for maps of P1, if the field of moduli is not a field of definition, you never have to go to more than a quadratic extension to get to a field of definition. And that's not really not at all obvious because if you took a rational function of very high degree, you might expect to have to take bigger and bigger extensions to get fields of definition. Okay, so this gives it sort of a uniform bound or how big an extension you need to go to, but it's only for maps of P1. And the proof of actually all of these results are to a certain extent a case by case proof that look at all the finite subgroups of PGL2, of which there aren't that many. Well, there's two infinite families and then a few exceptional ones. And you actually have to do them individually. Okay, but since there is this nice uniform bound, let me look more generally for maps and projective n space. And each particular map will have its own defect, namely how big an extension do I have to go to over the field of moduli to get a field of definition. And for a fixed project, fixed n, so projective n space, fixed d, so degree d morphisms, I can look at the maximum possible degree that I need. Okay, so example is really the theorem on the previous slide that on P1, for maps of even degree, you never need defect more than one. And for maps of odd degree, you never need more than two, but you do sometimes need to take an extension of degree two. And the theorem that I want to talk about for remaining about 20 minutes is a theorem that John Doyle and I proved a few years ago, which partially generalizes the results on P1, which says that if you take a number field, then this defect, but the sort of uniform defect for all degree d maps on Pn is bounded, that there is a finite bound that depends only on the dimension of the projective space and the degree of the map. So in words, if you take any degree d morphism of Pn to find over a number field and you take its field of moduli, you can find an extension field where that's a field of definition where the degree of the extension field doesn't depend on F very much. It only depends on the dimension of the space and the degree of the map. It would be great if we could find a uniform bound that didn't depend on the degree either. That's still an open question, as I've written here. So for those who don't want to listen to the rest of the talk, they can work on proving this and put the proof into the chat sometime in the next 30 minutes. Anyway, even for n equals 2, by the way, I think this would be very interesting. And it might be feasible to do in that case. If you couldn't find a good sort of theoretical way to do it and wanted to do a case by case, you'd have to look at all of the finite subgroups of Pgl3, which again, there is a classification. Okay. So the proof, actually, it also works for piatic fields for finite extensions of Qp. And the proof is a fairly intricate galachomology computation. And it turns out the most difficult maps to analyze are the ones that have non-trivial automorphisms. So I want to discuss that a little bit. First, I just want to tell you what the automorphism group of a map is. Namely, if I have a map from Pn to Pn, it's possible that there's some change of coordinates on both the domain and range, of course, that I get the same map back. It doesn't change the map at all. And that's the automorphism group of that, which I'll denote by this sort of script a sub f. It's the set of automorphisms of projective space so that if I conjugate the map f by that automorphism of projective space, I get the same map back. And this is clearly a subgroup of Pgln plus 1. So here's an example. f of x equals ax over bx squared plus c. This is a map of P1. And phi of x equals minus x is the automorphism because if I conjugate, think about what that means. First, I substitute x equals minus x, but then I multiply by minus 1 because I'm conjugating the other side. So the minus 1s cancel out. Okay. It's not that hard to prove that the automorphism group is a finite subgroup of Pgl. Although I'll mention it's not entirely trivial because if you look at maps of Pn to Pn that are dominant, but are rational maps, not morphisms, then in fact, they can have infinite automorphism groups. Okay. So somehow somewhere you need to use the fact you have a morphism. But much deeper and a fact that I'm going to need later is a theorem of Alan Levy's that says that the automorphism groups of degree D maps of Pn have size that's bounded independently of the map. Bounded just depends on the dimension of Pn and the degree of the map. Okay. And this is not so obvious. But I'll mention also that the bound in this theorem definitely has to depend on the degree D because for example, the map A, this is a generalization for instance, the map AX over BD plus C has an automorphism group that includes the D3 roots of unity. CD here is zeta Ds or D3 to unity. So you can get a, even on P1, you can get automorphism groups of arbitrarily large size. You can get cyclic groups of order D for any D. You can also get the hydral groups of order 2D in fact for any D. And eventually the theorem that I stated that John Doyle and I proved, the upper bound in fact is going to depend on the size, the automorphism group of F. So we're going to have to quote Alan's theorem. And unfortunately that bound definitely has to depend on D. Okay. So how does one get group homology coming in? Well, got a homology coming in. Let's take an F that's defined over K bar. Let me let K be its field of moduli. How can I tell if K is a field of definition? What's the obstruction? Well, the definition of the field of moduli says that for every sigma in the Galois group, and I'm going to write the Galois group as G sub K going forward because, just for notational convenience, for every element of the Galois group, if I apply sigma to the map F, I get an isomorphic F, namely there's a there's a V sigma in PGL that will be isomorphic to F sigma. So I get a map given an element. I mean, if I start with the map F, then given an element of the Galois group, I get an element of PGL N plus one, namely this V sigma. Almost. The thing is the problem is this equation doesn't define V sigma uniquely, because I can always change V sigma by an element of the automorphism group of F. So what I really get is a map from the Galois group to PGL N plus one, but cosets of the automorphism group. And for people who have played with this kind of thing, I mean, one expects this map to be a co boundary, to be a co cycle, and if it's a co boundary, then things are nice and so on and so on. And that kind of works. But there's an issue that is set here. First, a sub F tends not to be a normal sub. Well, first, PGL is not a billion and second, a sub F is not a normal sub group. So this is just a set of cosets. Okay. However, if the automorphism group is trivial, then life actually works much better. In that case, this map sigma to fee sigma is well defined. It gives you a one co cycle with values in PGL N plus one. If you take equivalent, so in other words, maps that differ by some automorphism, then the one co cycles differ by a co boundary. And K will be a field of definition if and only if the fee you get, the signal goes to fee sigma is a one co boundary. And then you can start asking, well, what field trivializes the cosmology class? So that's good. And working along these lines, and this is under the assumption that the automorphism group is trivial, Ben Hudson, Michelle Mayness proved something actually a little stronger than what I wrote here, but what they proved didn't fit on one line. So I took something slightly weaker. But what this essentially is, it's a generalization of the statement that when on P1, maps of even degree have field of module, equal field of definition. This is a similar sort of congruence type. Well, similar sort of thing. And I couldn't find an explicitly in their paper that Michelle can correct me if I'm wrong. But basically following the argument in their paper, you can pull out the fact that if the automorphism group is one, then there's always a field of definition whose degree is at most N plus one over the field of module I. So you get a fantastic uniform bound here. And I suspect that may actually be best possible. In fact, anyway, the problem is that this is only under the assumption that the automorphism group is trivial. Okay, so I mentioned that the group uses the fact that you have a map from H1 of PGL, which is by the way, it's only a pointed set because PGL N plus one is non-Abelian. So this comology set is a set. But there is a sort of a long exact sequence it injects in the category of pointed sets into this comology group, the Bauer group. And the image of H1 is in the N plus one torsion. And that's what gives this N plus one here. Has this N plus one here. Anyway, so that's useful. The problem is this doesn't quite work if the automorphism group isn't one. And that's where life gets more interesting. So in this case, we only get a map from the Galois group to a set of cosets. Now there is still there's a still still a one co-cycle formula. Here it is written in turn, but it's for cosets. There's a one co-boundary formula. There's the fact that you do get a field of definition, if and only if the co-cycle is a co-boundary. But again, this is all with co-cycles. And the problem is since if A sub F is not a normal subgroup of PGL N plus one, which it normally won't be, then two two co-cycles differ differing by a co-boundary. That's not even an equivalence relation. So co-group homology is kind of a mess in this setting. So here's what we do for the proof. I have about seven, eight, nine more minutes. So I'm going to start sketching the proof of what one does when this automorphism group is non-trivial. So first let me let N sub F be the normalizer of the automorphism group in PGL N plus one. Remember that the automorphism group is finite, but this normalizer probably isn't, right? The most extreme case of the automorphism group is trivial than the normalized result of PGL. But the first thing we do is we prove a proposition that if every element of the of the automorphism group actually is given by matrices with coefficients in K, which we'll be able to do by going to a finite extension, then the image of this sort of co-cycle, actually the image lies in the normalizer. And we have the nice fact that a group like the automorphism group is a normal subgroup of its normalizer. That's essentially the definition of the normalizer. So now I actually get an honest to God one co-cycle in this cohomology group where N sub F mod A sub F is, oops, is at least a group. I mean, it's a non-Abelian group generally. So this H one is just a cohomology set, but at least it makes, it's a reasonable cohomology set because N sub F mod A sub F is a group. And we have the trivialization property that we want the K as a field of definition if and only if that cohomology class is trivial, if and only if the co-cycle is a co-boundary. Of course, it normally won't be a co-boundary, but then we can say, well, if we replace K by an extension field so that when we replace G sub K by G sub L for the extension field, we turned the co-cycle into a co-boundary. So this is the standard way to trivialize co-cycles. And doing that, what John I actually proved was that you can find a field of definition whose degree is bounded by a constant that depends well on the dimension of the projective space that's always got to be there. And not the degree of the map, but simply by the size of its automorphism group. So if you bound the automorphism group size, you get what you want. And Alon's theorem says that the automorphism group is bounded by a constant depending on the dimension of the degree. So you get the theorem I stated originally. Unfortunately, as I mentioned, these automorphism groups can have size that depends on the degree of the map. And as the degree goes up in principle, their size can go up. In any case, this is the theorem that we actually want to prove. And here's a sketch of the proof. What I'm going to do is describe the steps leaving out a bunch of details, but just to give you an idea of the tools that are used. So the first thing we do is we have this automorphism group that's sitting in PGLN plus one. And I lift it to SLN plus one because SLN plus one surge axle into PGLN plus one and the kernels finite. And I have this finite subgroup. So let me let M be the exponent of that group. So I have a finite subgroup of SLN plus one such that every element has order dividing M. And first thing I'm going to do is I'm going to put a joint use of M to K. And then I'm going to use this nice theorem of Browers that says if you have any finite subgroup of SLN plus one and every element of the subgroup has order dividing M, then you can conjugate that subgroup, the entire subgroup, even the non-Abelian subgroup, so that all of the matrices have coefficients in the six atomic field QE join UM. This, by the way, is a very useful theorem in representation theory. But I'll use it here by simply replacing F by F conjugated by this matrix coming from Browers theorem. And now I can assume that every element of the automorphism group is in PGL of K. And that lets me apply the proposition that I mentioned earlier. So now I'm reduced to the case that I have a co-cycle in the normalizer subgroup modulo of the automorphism group. And I want to trivialize it. Okay. So we have this automorphism group A. I have this normalizer M. I'm also going to look at the centralizer, which, remember, is the set of elements that commute with everything in A, basic group theory. The centralizer, of course, is a subgroup of the normalizer, and I can map that to this quotient group. And that induces a mapic homology. And a short ish calculation shows that after replacing K by a field of bounded degree, the co-cycline interested in that a priori is taking values in the normalizer modulo, the automorphism group, is actually lives in the homology with values in the centralizer. So that helps. So replacing F by a conjugate, I can assume F is actually a one co-cycle taking values in the centralizer subgroup. Then we use constructive pairing on the centralizer across the automorphism group to K bar star, which I won't bother with the definition, and letting CF0 be the left kernel. Then some calculation and an appeal to an earlier lemma shows that, in fact, the co-cycle takes value in the left kernel of this map. So you see what I'm doing? I keep making my co-cycle is taking values in a smaller and smaller group, so it's easier to deal with. So we now lift everything to SLM plus one, and study the irreducible representations of this lifted automorphism group, using Broward's theorem so that we can get the irreducible representations based here defined over K. And we do some further calculations using Shura's lemma. You have to use Hilbert's theorem 90 for GLN. And using a map I mentioned a long time ago to the Broward group, and we use the fact that for number fields, at least, and for periodic fields, in the Broward group, the period equals the index. The period is the order of an element of the group. The index is the smallest field where the co-cycle trivializes. And that eventually gets us a field where our co-cycle trivializes, becomes a co-boundary. And that finishes the proof. More generally, you can do all of this, but the bound that you get, which I just want to mention, will depend on this index period problem for the multiplicative group. So I just wanted to mention there's this generalization. And I am going to go one more minute over by mentioning there's an alternative approach. For a billion varieties, instead of doing all this co-amology, you can simply add some level structures so that automorphism go away. And as I mentioned earlier, basically, if there are no automorphisms of your object, life's easier. And you can do that just to do the Abelian variety case. It's actually easier than the sort of proof that I just outlined. So John and I are in the middle of doing an alternative proof of the theorem I stated, where we will adjoin level structure, but in this case, the level structure's periodic points, which aren't as nice as torsion points. And there are a whole bunch of complications, which we're in the middle of working out, but we're hopeful that, fairly soon, we'll have a pre-print, give you an alternative proof that this defect is finite. But this alternative proof also, because of the fact we have to adjoin level structure, will not be uniform in the degree D. So this problem of getting a uniform bound that depends only on n and not on D is still very much an open problem, even for n equals 2 in dimension 2. Okay, so I will stop there. I thank everyone for holding on for a whole lecture. And I especially thank the three organizers for inviting me to speak. And I'll open the floor to questions.