 Okay, welcome everybody. So the goal of my lecture series is to give a very whirlwind introduction to the subject of number field counting and class group distributions with the particular aim that there are many many connections of these fields to computational number theory. Both a lot of these questions have been inspired by computational number theory. Progress in in these areas maybe can be applied to computational number theory and computational number theory can can provide us a lot of evidence for for various conjectures and new conjectures as well. So that's that's the goal. There are notes for these lectures that are available on the Park City webpage or you can also find them on my webpage. This is my webpage and if you go down to conferences you can see here the notes for my course and it'll just open up a webpage if you don't have a Dropbox app and you can get the PDF of them there and the notes have many more details and a lot more references than I'll give in the talks and so you can can refer to those for a lot of the things I'll quickly brush over in the talks. There are also lots of exercises of a wide range of difficulty so if you look at some of them and you think oh that's trivial if you're you know been doing this a lot that's okay you can you can move on and but some of them really I even named exercise slash project because they're more of an in-depth investigation so they really range from very simple exercises to things that are really more projects than exercises. So I hope you enjoy those. The notes probably have a lot of typos we've caught some of them but I'm happy to hear any more corrections typos questions about the notes please just let me know or email me. All right so let me get started with my talk. So today I'm going to talk about number field counting. So number fields are finite extensions of the rational numbers so these are my number fields K that I want to count and so I want to count I want to ask this question here how many number fields are there and of course if I don't put anything in that blank we know the answer is that they're implementing and so what's interesting here is what you what exactly you want to count to try to understand in the zoo of number fields in the universe of number fields what do you see when you look out there in it so one of the kind of starting restrictions you might make is putting in this blank something like of degree D and say discriminant up to X I'm going to be talking about the absolute discriminant so much I'm going to write it as DK the absolute value of the discriminant of K and maybe you want this to be at most X for some large X and then in that case you are often interested not in the value for some particular acts like X equals a thousand but how this grows asymptotically you know we'll see all right so that's that this is a very fine question for which we I know the answer for only a few D and as even for the the first few D as you go beyond D equals two you start realizing well when you're talking about degree three fields there really two kinds of degree three fields there are the gawa cyclic cubic fields and the non gawa fields and they're pretty different and then when you get to degree four you see that also that gawa structure matters so it's often very convenient in this question and the answers come more naturally if we fix that gawa structure now what I mean by a gawa structure be in the degree three case there's the ones that are gawa and then the ones that aren't gawa and the best way to fix the what I want to think of is gawa structure is looking at the gawa group of the gawa closure so if I have a number field k over q all right k tilde for its gawa closure and now that just tells me about the gawa closure but if I want to know about how k sits inside of its gawa closure one way to do that is by looking at the action of the gawa group of k tilde over q as it acts on all of the embeddings the homomorphisms from k into the gawa closure and so that makes this a permutation group not just a group but a group with an action on a finite set and so when I have a field not necessarily gawa the sort of gawa structure that I will fix is the gawa closure of its the sorry the gawa group of its gawa closure as a permutation group in this way and then really the driving question in this area of number field counting is what are the asymptotics of this function well I'll call in g of x where g should be a permutation group the number of k over q where this gawa group is g and I mean here as a permutation group with the discriminant bounded all right so the the first first case of this is when the gawa group is c2 say acting on two things that and this is just counting quadratic fields all right so to count quadratic fields one one way to start is what we just know all of the quadratic fields right when we take algebraic number theory that's a thing we learn and let's just remember because it's a little bit relevant as we think about the higher degrees how do we know all the quadratic fields how was it that we we got a list of them well they each had to be say generated by some alpha and some algebraic number and we could always take that by clearing denominators to be an algebraic integer so some alpha that satisfies a polynomial like this alpha squared plus a alpha plus b equals zero and we could change by subtracting off an integer or a half integer from alpha and clearing denominators again we could say that the field was generated by say some alpha that satisfied an equation like this you know after we eliminate the a coefficient say alpha squared minus d equals zero so we're joining the square root of d and then also by clearing or unclearing denominators we could assume that d is square free and so this is this is how we knew that we could find all of the quadratic fields sort of among adjoining the square roots of of square free integers now and then we we check maybe you know in in our algebraic number three course we check that these are different now that might seem intuitively obvious but there is something to check there how do you know that when you would join the square root of three that you don't somehow in that field get a square root of seven I mean some things are surprising perhaps how various primes split when you were joining the square root of three might not have been a priori obvious but in any case for in many different ways you know such as checking the behavior of different primes we can see that that these in fact give us all different fields and moreover since I was so interested in counting these fields up to discriminate acts importantly we can in each case compute the discriminant and know what it is and in which case I mean as um Hendrik spoke about earlier finding the ring of integers in general is a challenging a challenging problem and if you're in some completely general setup so just because we know the field doesn't mean we know the ring of integers or the discriminant but here of course everything can be worked out because it's such a small case and so then roughly now to count these asymptotically we need to count something like square free integers in the notes I carefully give a version of this argument where you would actually count fundamental discriminants or discriminants of quadratic fields but then you have to be very fussy at two because these aren't quite square free integers they sometimes have these factors of two and it depends on other things all right so for simplicity in in this talk I'm going to count square free integers to sort of convince you in practice that we could count these discriminants of quadratic fields and then you can you can refer to the notes for for the real version so let's let n of x be the number of of square free integers from one to x and one way to think about it is well I have all these integers and then I want to get rid of the ones that have square factors so I'm going to let n sub n of x to be the integers that have an n squared factor in them so the d that can be written as n squared times d that are also in this set from one to x all right and if I want to count square free integers like I said I could start with all this in one of x that's just all the integers in the interval from one to x so this is actually counting square free positive integers here I guess I should say and then I should throw out all of the integers that are divisible by p squared for each prime p so this here is a sum over over p prime of n p of x this gets rid of everything that's divisible by p squared except the problem is if I have something that's divisible by p squared and q squared I've thrown it out twice and so I continue in the usual inclusion exclusion way and I have to add those back in and now I have to correct for the things that are divisible by by three factors etc etc all right so we write that all together what the inclusion exclusion does and that says that the number of square free integers is the sum over n of this Mobius function that is the sign that tells you whether you need to be throwing them in or out or times times this in n sub n and I'm going to use this then to estimate the count of these square free integers okay so firms are even only appearing right that we have that for n sub n of x to be non-zero for there to even be a term here we need that n is less or equal to the square root of x or we're not going to have any numbers that are divisible by n squared in this interval from 1 to x so that says that I can take this sum instead of overall in I can just take it up to the square root of x I've got my Mobius function here my sign and then this is actually something that's pretty easy to count this is just integers from 1 to x that are divisible by n squared so that's roughly x over n squared and in fact I really have very good control over how much I could be off by right it's the floor of n over x over n squared which is which is at most at most one off from from from x over n squared all right so now if I collect together these terms these x over n squared terms I put them all together with the Mobius coefficients now I'm going to sum overall in because this sum is much nicer overall in so I have to subtract off the terms from n that are greater than x squared which I do here and then these oh once this error term that could have been as big as one I got that for each of square root of x terms so I have some square root of x error here coming from these oh ones and now this summing overall in the x pulls out it is nothing to do with the sum and this sum of mu of n over n squared has this has the beautiful factorization of the product over primes of 1 minus p to the minus 2 and then this error goes into here but also this if you look at this sum again the x would pull out and whatever these signs are you could bound them by the absolute value and this is you know the sum of 1 over n squared starting at the square root of x so here's the integral comparison and get that that is again this big O of x over the square root of x and that just goes into the same error term all right so I have then this this is a just a number it's a particular number it's zeta 2 zeta 2 inverse right so but it doesn't particularly matter that's just the sort of convenient way to write that number times x and I wrote here little O of x because frankly a lot of the time we're not going to get nearly as good of error terms as this square root of x though here here we happen to do to do much better when I ask what are the asymptotics of you know the number of quadratic fields of discriminant x this is the sort of first kind of thing that would be an answer what I mean by an asymptotic count is that you got some main term and you know that the difference between the actual value and the main term is little O of the main term meaning you know when you divide by the main term in the limit it goes to zero so this is this is the asymptotics and so of course I was just counting square free positive integers here in the notes this same argument now it has the more funky numbers in it but it counts quadratic fields and gives you actually the same answer so there's all these funny things happening at two and they all somehow miraculously cancel each other out and you get the you get the same answer you and you also have negative discriminants and positive discriminants and all the funniest of it too and it and it works out oh sorry yes thank you I don't need the inverse because I put in the denominator I'm just so used to writing zeta 2 inverse it just like I never write zeta 2 like that doesn't that number doesn't come up in my life so okay thank you um all right are there any questions about that yes yes you should multiply by two and you would need to take into account the negative discriminants but then you need to do all sorts of other things at two because you have you know these discriminants like eight and uh right and and twelve that aren't square free because they have these factors of two and so you have to also multiply by a bunch of other factors to accommodate that and then they all cancel out you can see that in the notes that it works out yes you would think shouldn't it be twice as much but then there's this phenomena that we we push off and make some of the discriminants bigger by multiplying them by four and that that all magically magically cancels good question good question okay great done with quadratic fields so let's go to higher degree all right we can start the same way um uh we can say okay look a cubic field it's got to be generated by some algebraic number satisfy some polynomial again I can you know clear denominators and I can get rid of the the alpha squared term so now I'm just basically looking at say you know p q integers here that parameterize some polynomials that algebraic numbers of degree three could satisfy but now I have some questions that become much harder than the quadratic case when did if I have you know give you two p pairs of p q when to a p q and a p prime q prime give the same field um I mean among other things we know for for um cubic fields the discriminants no longer discriminate the fields there can be fields with the the same discriminant and and and how can we is there any easy way to see I mean the real answer is that there's not any easy way to generally see for this kind of counting um win two p and q give the same field and then we also then need to know even if we have a p and q that give us some field what is the discriminant of that field um so we can compute the discriminant of the polynomial quite easily and we know that the discriminant of the field divides the discriminant of the polynomial and it's just off by some square factors but getting those you know getting those right um is not something that one can just easily write down for all cubic fields all at once like we did in the quadratic case these become much much thornier questions um so of course in any individual case you can answer you know maybe the numbers are too large or something you can answer these uh these questions but in nothing like the systematic way we answer it for all quadratic fields and where we can just say look here's a list these are all the quadratic fields um so and not systematically enough to count um maybe uh to count let me add a word here to count easily all right so the moral just my point here is that you know often when people are new to number field counting the first question is but you know number fields basically just correspond to polynomials so aren't they isn't it just about polynomials and um the isomorphism classes of number fields are not the same thing as polynomials because in particular it's hard i mean for these two reasons it's hard to to know systematically when two polynomials will give the same field um and it's hard to know what the discriminant of the field is given just a polynomial um and we'll see more and and one sees more in the example of the notes that that this actually makes some of the asymptotic and the statistical and the counting phenomena quite different if you say take a random number field from looking at all number fields up to discriminant x versus if you take a random polynomial and you see what number field you get it's a it's a different kind of thing um nonetheless uh even though uh there is is this gap between between sort of pinning down number fields and just writing down polynomials in general um that among among other things so actually looking for algebraic numbers via this general direction of approach still gives us the best general approach to say finding what the degree d fields are but um not in some way that you can can write down a simple formula but just in terms of computation doing a tabulation of fields by which i mean making tables listing each field of bounded discriminant um at most x ones and so in um you know in arbitrary degree that's that's uh the best way that we have to to to start finding fields and in fact um with enough with heuristics that account for these two features how much over counting you're getting from counting the same field from multiple algebraic integers and how much your discriminants are off so for with accounting for those two things and some some heuristics there's a beautiful paper of uh rural shaker and jacob zimmerman where they give a conjecture for the number of um the asymptotic count of s d number fields so degree d number fields whose gavel closure is group uh gavel group s d and they actually are able to unconditionally prove enough of the heuristics to use this to count non-gawa cubic fields asymptotically i'm going to talk a little bit about an older approach to doing that that's a little simpler but even though their their approach is more more complicated than what we're going to talk about it it it it does it is exciting because it says that you know with enough work there is something you can do at least in one case to overcome um these these two challenges of the sort of over counting and the lack of control of of of the discriminate um and actually asymptotically count fields um so okay and still this sort of approach of looking for algebraic numbers instead of fields also gives us what the in cases where we can't count fields asymptotically gives us the best currently known upper and lower bounds um on those on those numbers um but i should say these were all i was saying these are these asymptotics for s d fields by which i mean degree d fields whose gavel group uh of the gavel closure is s d so as big as possible but in general this approach gives much less access to understanding um g extensions when g isn't this full gavel group of s d because um we know from Hilbert's irreducibility theorem that a generic degree d polynomial in an inappropriate sense but also a sense that can be made quantitative uh is it has gavel group full s d and so if you another problem with this approach is if you're interested in g extensions that aren't quite so generic then you may be looking a long time at algebraic numbers before you find one before you find one that generates uh the g that you want but i think i don't know if anyone is in here i know there's a group this week like trying to find a uh field with a particular gavel group so they have a sense of how how hard that is all right so um however there are some different approaches uh beyond uh just looking for algebraic numbers of the appropriate degree that can be used to count asymptotically number fields of with this uh gavel structure given by g um that have i mean these are these are three approaches that have yielded many results there are some there are other approaches as well but as a brief taste i'm going to you know very quickly give you a sample of what you know three three different other ways besides looking for algebraic integers are um to count to count number fields all right and so the first approach um the first approach is to use class field theory and this of course applies uh quite well when g is abelian uh and because in that case class field theory tells us about um the uh the abelian extensions of a number field now i set up for simplicity this talk over q where everything works in terms of q but of course it's not hard to imagine and one does ask these questions over a general number field uh or even function fields and um so even though over q one can work with class field theory quite explicitly um using the cyclotomic fields i'm going to do it in a more general way which may feel like overkill if you're just working over q but the reason that i'm going to do this is that it works very well over general global fields and so it's so so this is meant to be um a taste of of of the a method that can do much more than what we're going to we're going to see today and what we're going to see today is the count of cyclic cubic fields so where the gaol group is the cyclic group uh on of order three acting on three elements and this counting was done um via this kind of approach by um by cone many years ago so what does class field theory tell us it tells us that these c three extensions and another way to talk about c three extensions is to think of them in terms of homomorphisms from the gaol group of q bar over q the absolute gaol group to c three now these aren't quite c three extensions because you know there's the trivial homomorphism and there are two homomorphisms that correspond to each field but okay these are roughly the the these c three fields and they correspond to homomorphisms from the edel class group of q uh to the cyclic group of order three and uh this is it is gone through in the notes in a little more detail but for this for this purpose the um the edel class group it's isomorphic to the product over p of uh of z p star times the positive reels um and so the the homomorphisms from this absolute gaol group to c three or the c three fields are simply given by homomorphisms of this into c three continue everything need to be continuous homomorphisms and there's a restricted some restricted products here um which means that only in these homomorphisms only finitely many of the homomorphisms can be uh non-trivial and so are the positive reels don't have any continuous homomorphisms to to c three since it's it's discrete and so this says to pick a cyclic cubic field i need to pick at each p a homomorphism from z p star to c three all right so that is a way in which class field theory makes very concrete kind of what are all um what are all the fields in a way that's very good for counting now it's not necessarily uh giving you a you know giving you those those polynomials those algebraic integers so it's quite different it's not uh that that takes a lot more more work but it's very good at just saying like here you can understand the set of the c three fields all right because homomorphisms from z p star into c three aren't so complicated i'm going to ignore p equals three um uh though again i think that is done carefully in the notes and so say if p is not three then um z p star so the the the the piatic units they have a map to z mod um p z star and the kernel here is a propi group so if i'm looking for a map to c three it's not going to care at all it's not going to see any of this kernel it's just going to factor through z mod p z star and we know very well what z mod p z star is it's isomorphic as a group to z mod p minus one z so i have a cyclic group of order p minus one and i want to know about its group homomorphisms into c three all right so when p is one mod three i get two non-trivial maps exactly and when p is two mod three i don't get any non-trivial maps and of course i always have the trivial map so um so understanding what these homomorphisms are from the absolute gal group to c three simply is saying okay ignore three um a sweeping three under the rug otherwise is at each prime is one mod three i um get to choose do i want a trivial map or do i want one of the non-trivial maps or do i want the other non-trivial map also another useful feature of the the homomorphism of class field theory is that it tells us in terms of this data what is the discriminant of the corresponding field and we're just going to say the field was unramified at at three say we took the trivial map at at three uh though again in the notes the full case with three has worked out um and here the discriminant is simply the product over the p's where we chose a non-trivial map all squared all right so these are these z p's are the are the inertia groups in uh in this a gal group and so a prime is ramified if it only if the z p maps non-trivially to the to c three and you know away from three you um uh know that in the case of a cubic acyclic cubic field a ramified prime will defy the discriminant exactly twice and so not only do we have a sort of easy way to see in data like what are all the c three fields but we have an easy formula for their discriminants and so we're now um to uh we're now to a question that's quite to the question we had before with quadratic fields where we wanted to count square free uh integers we now want to count these integers well you know up to the fact that they're all squared we're roughly we want to count square free integers um that are only products of primes that are one mod three but since we have these two choices at each of the ramified primes we need to put in factors for that and that is is um gets sufficiently complicated that it's worth introducing a more general tool to count these things and also this general tool will allow allow us to count a billion g extensions for any g okay so we're going to bring in a tool from analytic number theory that says if i build a Dirichlet series so i've got some a sub n coefficients on into the minus s and i think about it as a function in a complex variable s the asymptotics of the um oops there's a term here that shouldn't be here um the asymptotics of a sub n up to x so the asymptotics of the the the partial sums of the coefficients up to x can be understood in terms of the rightmost poles of this function so that's a really beautiful and important thing when you want to count something if when you put the things you want to count as coefficients on a Dirichlet series you can say something about the analytic behavior of that function now i think we know from many things in number theory just because you put some coefficients on a Dirichlet series doesn't mean that you understand the analytic behavior but but when you do it's it's it's very useful so in this case what does this look like so this Dirichlet series where we put the things that we want to count as the coefficients here like how many how many of our homomorphisms to c3 above had you know discriminant in so what does that look like it looks like the product over primes that are 1 mod 3 times 1 plus 2 times p to the minus 2s right because at each p that we were going to pick um that we want to let ramify we had sort of two choices for that that map and if we choose one of those two non-trivial choices then p is going to contribute to the discriminant and otherwise it won't so this is the Dirichlet series of things that we were wanting to count above um and indeed this is a Dirichlet series whose analytic behavior um we can understand something about at least up to its its rightmost pole um okay so i may uh here so i'm going to continue to ignore everything happening at p equals 3 it doesn't really change it's just one factor it's not going to change any big analytic results so the way that we can under oops let me erase the way that we can understand the asymptotic behavior of this Dirichlet series is by comparing it to some other functions that we know uh sorry the analytic behavior of this Dirichlet series we can compare it to some other functions that we know um the analytic behavior of and given this form it's natural to look at the the following product of functions so the zeta function we're going to evaluate it at 2s because we see this is all kind of happening in 2s times the Dirichlet l function for the Dirichlet character mod three again evaluated at 2s and so each of these functions can be written with an Euler product and at p that are one mod three um they each have a one minus p to the minus 2s to the minus one factor so you collect that same factor when p is one mod three now when p is two mod three from the l function the Dirichlet character you get one of these minuses becomes a plus and those two things multiply to one minus p to the minus four s so um now when you first look at this uh that oops i mean except that they're both Euler products you might say okay that looks like quite different than this this doesn't have any p equals two terms and the terms are all different um but in these Euler factors it's the sort of leading the leading terms that contribute uh to the to the asymptotics and this because this is a minus and there's a minus two here actually has the same leading term a two p to the minus 2s if you were to sort of write out the the geometric series that you get from taking this to the minus one or to the minus two okay and so let me just say how that happens so if we look at the ratio of the Dirichlet series of the things that we want to count to the product of these functions that we know very well the rayman zeta function this Dirichlet l function what we get and you know okay it's up to my arithmetic being poor for calculating these integers we get an Euler product where the leading term terms are at p to the minus 4s and so when you just think about when you when you multiply this out and what you get and what it looks like this um this converges for the real part of s greater than a fourth so this is analytic this ratio is analytic for the real part of s greater than a fourth so that says that in terms of looking for poles um that that this Dirichlet series of the things that we want to count and this Dirichlet series that we know have the same poles up to real part of s is a fourth um so uh that means that this Dirichlet series has its rightmost pole at s equals one half um like this function which we very well know has a pole at s equals one half uh from uh the zeta function evaluated at 2s there uh so so the comparison uh between the Euler products for the thing that we want to count which we got kind of a pretty Dirichlet series from from class field theory and some functions that we know and we know about their their poles let us uh use use this and it it tells us that the asymptotics of um uh the number of cyclic cubic fields up to discriminant x is like a constant times x to the one half plus some term that asymptotically is smaller the oops I should that's not a very useful result what I wrote right there we need to put a one half in here a term that grows smaller than than x to the one half all right um and so let me just say right so this this one half here and so in the notes I have to bear in theorem that you can apply and see in general how you use this rightmost pole to get an asymptotic count but just the sort of basic thing is that this one half becomes this one half right and this constant here comes from the residue um uh at that pole and actually you can compute that residue at the pole very explicitly because you know the residue of this function and then this uh you can just evaluate it at one to see how it changes the residue so you can actually compute oops you can compute that c um see very explicitly from this um it's a lot more work keeping track of all the things there are to keep track of this general approach uh can be used uh to count all abelian fields meaning for each abelian g to count the asymptotics of g extension so this was first done by um Machi and uh many others have have have worked on this problem and use this approach referenced in the notes like right and uh fray long run and newton um and so you can and the the general strategy has has come a long way but the rough outline of okay class field theory tells us the set of these things we can put the things as coefficients on a Dirichlet series uh we can use then analytic techniques um to count the things uh still still continues um and i'll just say that um even though uh class field theory does not over general base fields say give um easily give the explicit polynomials in some systematic way that you can just write down a formula for it is it it is reasonable as a strategy for for tabulation so certainly much better you don't if you want to be looking for say for some you know degree 12 uh abelian group the fields with that gawa group you don't want to just start looking at degree 12 polynomials you want uh to to use to use class field theory so any questions about that yes it works yes so the reason that i presented it this way is because it actually works um over over general um general global fields number fields or function fields now you know a few things are a little different this is not quite the adel class group but it's off by class group and it's off by units and you can write those down and and there you know there are more complications but this general strategy um works over over an arbitrary global field i don't know if anyone has actually done it when the characteristic of the global field is the same divides the order of the group but certainly over any number field this um this strategy has been been applied so good question yeah and so that's kind of why i presented it this way it's not maybe the most concrete thing you could do for c3 over q but it presents a strategy that works you know for for all billion g over all number fields with with significantly more work certainly um okay so um all right um the um approach i want to talk about uh is from parametrizations and geometry of numbers and i realized i was supposed to have a piece of paper up here with me so i should pull it out here okay all right so um another another way to think about okay what could these number fields be and we you know we say we're counting number fields but then we when we take the discriminant of course the discriminant really belongs to the ring of integers not to the number field uh and and because there's only one maximum of integers we kind of say you know but it's really about the ring of integers so maybe we should think about you know what are these maximal orders of um of cubic fields so i'm going to present a strategy that was used for example by davin porne habron to count cubic fields uh so it's going to turn out that they're more non-gawa cubic fields so we could also sort of say to count non-gawa cubic fields so but any cubic field it has some maximal order and we know that that maximal order as a z module is just um is just isomorphic to to z cubed and so we can write down a basis with three elements and you can think it's exercise in the sheet so you can pick one of your elements to be one and so now if i have a z module um basis with three elements of a of a of a z algebra that is is free of rank three is a z module what do i need to know the ring that i've got what do i need to know i know how to add everything in terms of you know i have the this z module i need to know how to multiply things and what do i need to know how to multiply i need i already know how to multiply one by everything so i need to know how to multiply um omega by theta and how to square omega and how to square theta and if i know how to do those three things if i can fill in uh numbers integers here that tell me for each of these multiplications what these things are in terms of the z module basis one omega theta i will know the ring and i know how to do everything now i can add multiply use the distributed property um and find the ring so i can say okay well um you know you could you could sort of start by saying all right so there's just some sort of numbers here a one a two a three a four a five etc all right but now i want to simplify this a little bit and this is a little bit like the step in the quadratic fields where we cleared the the linear term and the polynomial i want to clear some terms so actually i can modify omega and theta by some integers i could replace omega with say omega plus k for some integer k and i can replace theta with theta plus l for say some integers k and l such that i can assume that these two numbers are zero all right because i just say omega theta gave me some number of omegas and then i subtract that off of theta so you can you can check that in the exercises that i by changing omega and theta by an integer in fact there are unique integers such that i can do this that's that's that's important such that i can assume that these two coefficients are zero and now i've got one two three four five six seven coefficients to write down and if um if i i write them down here i'll try to get the right letters oh no i already used l okay um uh there's r here okay so let's say i write them down i'm just using letters so these things are all just going to be some integers um all right so a b etc these are all integers now um so now every cubic field has some real integers and i have captured that ring of integers and it's isomorphism type in these seven integers um but now i could wonder like well how often do these seven integers actually come from a ring of integers in a cubic field and one of the the first things that you might want to check actually is that this multiplication that you have defined is associative because nothing nothing about this this multiplication table forced it to be associative and certainly the multiplication in okay is associative so it turns out um that uh if you write down the equations of associativity so the equations of associativity beautifully exactly tell you that these three values have to be certain things up to some sign which no matter how many times i do it i seem to always get wrong but i really hope these are the right signs um okay this minus a a b um and minus a c and minus b d so that's a sort of miracle that didn't wasn't a priori anything like this was going to happen and it tells us that in fact there are really only four parameters here um there are four four parameters a b c and d and then associativity forces um forces all of these values and so that's a lot better than um that is a lot better than uh that seven parameters and now you could start this say over a quartic for a quartic field or a quintic field or any degree field and you could always formally sort of write down parameters and then you could look at the equations of associativity and unfortunately for every degree higher than three you just get a bunch of equations you don't get this beautiful elimination of variables um and so so it's really that coincidence here that we get such a clear elimination of variables and in fact it's it's um maybe i should make this make this a if and only if when you put the coefficients like this with a b c d and these are as as specified that this does always define an associative algebra so you get the associativity exactly eliminates um three of of your seven variables with no remaining equations um and so that tells us that the rings of integers of cubic fields along with we had to choose a z basis a z basis of but not a z basis of okay because we picked one we we there was no choice there and this omega and theta we had some unique shift of each of them by an element of z to make these two numbers here zero so it turns out that that's equivalent to choosing a z basis of okay mod z all right so okay the ring of integers with a z basis that that data turns out to exactly correspond to some tuple of integers a b c d in in in z to the fourth um all right and so care just about these okay's we don't want a choice of basis we want to forget the choice of basis so a different basis of okay mod c um means that we've we've we've changed this by some element of gl to z since all bases of z squared are related by an element of g l to z so that means that there must be some gl to z action on these tuples of four integers such that um you know such that the um such that that the orbits correspond to the different choices of basis and so it turns out that you can work out explicitly um what the action is on z to the fourth so that's done in the notes and in exercises and also which a b c d in z to the fourth actually correspond to okay for some cubic number field k uh and that is not done explicitly in the notes but many references are given and it's sort of talked about it's mainly sort of local conditions at at every prime um and one sort of global condition but the key to all of these conditions is that in some sense generic tuples of four integers a b c d actually correspond to okay so it's not like we're hunting for needles in a haystack it's most of these a b c d's actually came from the things that we want to count and they're just some of them some of them we need to throw out and so that is that is is helpful and what it what allows of what allows counting and how that counting uh is that um i said there's you know a gl to z action so we only want to count one element in each gl to z orbit over here to count each okay once and that comes from counting um a b c d in a fundamental domain for that action so there's some fundamental domain i do this picture not only because it's probably all of your favorite fundamental domains but literally like you can use some image of this fundamental domain to find the fundamental domain for this problem maybe not a big surprise because sl2 and gl2 are similar uh oops uh and so then you have some problem where you have a fundamental domain some some region and you want to count lattice points in that region and um the techniques from geometry of numbers can be used there's some difficulty the regions have some cusps uh that are very thin and you have to make some arguments so there's there's more to do but i wanted to sort of outline the the general strategy which is once you reduce it to a problem of counting lattice points in a region um okay and so not only does this give the count oh i think i didn't i think i didn't write it down i wrote it um write it write it in the in the in the notes but this gives the count um the asymptotic count of non-gaoa s3 cubic fields which um it's actually this um not only does it give give the count um this general strategy of then looking for these appropriate lattice points in a fundamental domain gives very fast tabulation of cubic fields um and green bail boss actually has some software that is still available online and still super useful if you want to find yourself some cubic fields um uh by by finding them in this way um and i just want to mention i talk about this more in the notes but i this so this strategy of parameterization and geometry of numbers has been used by barba to count um portic and quintic uh sd extensions and uh that in the way that this strategy led to fast tabulation of cubic fields i think has some potential to lead to fast tabulation of quartic and quintic fields so i think there's some really interesting directions there all right um so all right i am uh about at the end of my talk and we'll tell you about um uh quickly about the final uh strategy um which is one also sometimes can count extensions by counting extensions of extensions so for example d4 quartic extensions um so this means quartic number fields whose gaol group is d4 and so this is say my quartic number field and the gaol group is d4 and i made the gaola diagram over here i've got d4 which is say the subgroup of of automorphisms of a of a square and then um uh i've got uh k the quartic field corresponds to 24 and then inside of here there's another uh subgroup here which corresponds to some quadratic field so a d4 field is a quadratic extension of a quadratic extension and so if we're very good at counting quadratic extensions and i said for a billion extensions we can count them over a general base field then if i can count extensions and i can count extensions of extensions um i should be able to put that put that together so over a general base this say is the counting function of counting fields and if i count quadratic extensions of quadratic extensions they don't have to be d4 fields but the other possibilities are a billion and like i said we can already count all the a billion things and so i could say well i'm going to count something like d4 fields by summing overall quadratic extensions and now i'm going to count their quadratic extensions up to this discriminant and uh you know the discriminant of a composite field you have a formula for that so you put that in to see what you need to put it put in here um and let's say we could do that for each for each quadratic field we can count its uh its quadratic extensions with a similar formula the problem is that this is an infinite sum and that doesn't behave well with this little o of x or this limit like for any finitely many quadratic fields fine but if i have infinitely many quadratic fields these error terms it's they could possibly swamp out the main term um and so you need some additional information some kind of bound on how big these counts can be um absolutely uh and not just with like a little o or limit and so this is for example the kind of of tail bound that you could use and uh when when you have a bound like that you can then um uh you can then sort of use this use this to um sort of interchange not the sum and the little o because you don't always write the little o you know to interchange with the sum um and that again is done done uh carefully in the notes uh and is another you know another strategy that uh for many gala groups sort of allows us to to count extensions of extensions you have to have as you see in something like this tail bound a little more information than just an asymptotic count of extensions at each layer um but if you do have that you can sometimes put them put them together all right uh so that is it for today do we have any quick questions before lunch all right let's thank melanie again ash okay