 Okay, well, after my first number theory website seminar, I swore I would never give another one. This is it. Sorry, that's only a joke. That's a spiteful joke. I'm very pleased to, to give the second one, the technical preparations were somewhat tedious, but the intellectual preparations were great fun. So there's my title, some new, then you see a misprint, elliptic, and that's, that's my favorite misprint. I was getting my PhD ready for submission to the University of Cambridge in 1974, and it was all typed up and I checked it 140 pages or all mathematically correct and the title page then had the word elliptic, and it was all typed. So I had to get out a little, a little bottle of white liquid and a little brush, and type out and wipe out the P and the T, and then write them in again, imitating TypeScript and here is the now, yes, here is the, here is the, here is the thesis and the title page, you can't see it now, but it's a little bit tough. It's, I got through, I got my PhD. Okay. So, it's an old title, a strange title. It's almost like some new positive integers. Some new elliptic intervals have been a new positive integer, positive integers, but there are number two, please. Philip. Here is a new positive integer. Six, six billion four thousand and 49. And I first met this number in 1979 in Sydney. I was invited by Vanderporten. And I went into the library I was the University of Sydney I went into the library and found this book. Find this pic this find by inspection a factor of this number. Six billion and 4049 and I thought how can anyone find solve this problem inspection means you can't even write anything down. And it contains all kinds of all kinds of things like this. Here, here you have to factorize these huge large numbers. You have to factorize as far as possible, but these numbers you have to factorize completely. And the book was written in 1906 so you didn't have any, you had no electronic calculators or computers or anything like that to help you knew only these funny calculating machines. Okay. Now, number three, please, Philip. That's. Yes. 1979. The first thing new number. That's a prime number. And if you add these six digits here you get primes. For the first few terms, like this is a prime I can do color you see, and then, and then all these numbers are primes. And showed me that probably in 1979. And it's an exercise to show that there's no longer example that this is quite useful to know this kind of thing if you have something you want to check for large primes you just write this number down and try only the partial sums kind of thing. Okay, number four. Yes. So much for positive integers. And these arise from the concept of elementary integrability. So, one over X is integrable integrates to log X. One over X log X integrates to log log X. But if you try and integrate one of the log X, then it's not elementary in any sense, in the sense that I'll soon describe. Oh, I know two arrows going on. It has a name. It's essentially the logarithmic integral li X in prime number theory. But that doesn't qualify for elementary. What is the key property that is that the integral of FX DX can be written down using X and FX, and then the log function and it by symmetry the exponential function, and you throw in algebraic functions. So, for example, sheet five. This could be an answer to the integral X of square root log log X plus log one plus X FX. So you're allowed all kinds of combinations additions and iterations and so on. You can do a formal definition and the best formal definition I think is by differential algebra. I'll do it here, but the differential algebra. Completely sweeps under the carpet all problems of domains of definition, or even differentiability and this kind of thing. It doesn't, it doesn't even sweep under the carpet eliminates them completely. I just skip this phrase now and and I another reason I don't give this is that there's a simplification that's about around 1840. Newville showed that exponential is not really needed in the definition and even log log and algebraic functions are not needed either. At this point I would skip back to the previous one today. That's to tedious I think in the previous one we did see a log log, but it wasn't really a log log it was log of log and log was given to you in the integrand so. And the idea of his theorem which I don't state precisely FX, the integral FX DX is elementary if and only if it is a linear form in logarithms. And what does that mean. Please. F zero plus C one log F one plus up to CM log FM. This is the formula labeled you will. C one up to CMR constants in C, and F zero F one up to FN are in this large looking field here that you get by joining all the derivatives of F, and we don't have to say that F is in many different if we often differentiable. This is implicit in the in the differential algebra business. So, if you know that the integral has to have this special form. There's no exponentials and there's no log logs and the. There's no iterations and there's essentially only additions, then it becomes a relatively easy exercise to check that this DX over log X is not not elementary. The hint here is that this big, this big field that I introduced, I have my arrow here. Yeah, this one here is actually rather simple just generated by X and log X. And X and log X are algebraically independent functions so you can call y equals log X and then you have a problem in a function field. You can call it transence degree to and that gives you plenty of room to work in. But if you are now. Look at integrals of algebra functions square root of x cube minus x. Let me just, let me just get rid of the. This, this integral also has a name is two times P to the minus one X, where P is the viastras function corresponding to a certain elliptic curve or G to G three, and the the verbal name is an elliptic logarithm. And then the transcendence degree of this apparently large field is just one and that means you haven't got so much space, and the thing is not quite so elementary. One has to consider zeros and poles on this elliptic curve defined by y squared it was x cube minus x and I multiply by four, because I will soon be in in viastras mode. This is an easy exercise if you know a bit of theory here. And then, once you've done that, it's just a small step to show something a bit more complicated. But if you take now this integral so DX over y is just the same as this one more or less because of the y. And then you put this rash simple rational function in one over X minus tau. Then this being elementary implies that a certain point is torsion on the. This the point with ordinate with abscissa tau, and then the, the ordinate is to root tau cube minus tau. This is a key formula as well it shows how torsion comes into the whole thing. And with torsion comes number theory and so on. Okay. So six extended footnote. Please. Yes, so that's a proof of this implication I think I'll skip this on grounds of time. There's not much to it a bit of elliptic. Oh, yeah, there's a bit of elliptic theory but I call it Whitaker Watson section 2013 and 2014. These are the old this is what I first learned elliptic functions from good old, good old. Where am I. Oh, I know. Here is here is Whitaker Watson almost falling apart at the seams. Then, again, and then there's not much to it. If you know a little bit about elliptic curves. Unfortunately, you can't reserve reverse the argument. And that's the whole. Okay, so now pay page seven, please. So in fact, I say that the, the opposite implications force a great pity as you'll see. So we finally arrived at elliptic in the title. The difficulties here can be illustrated by the fact probably known to Arbel and Chebyshev that if you take now DX over why, and put this rational function in. Before we put one over X minus town now I put this explicit rational function in. Then this is actually elementary becomes elementary again, even though without this thing it's not elementary. And that has M equals one in Louisville. So in Louisville, you had the logarithms and M was the number of logarithms. So that's elementary. With a single logarithm, but if you, it's no easy feat to work it out if you give this to maple it's completely baffled. Now to find the Porton and Tran in the year 2000 that gives many more examples. My favorite example is oilers in 1780. And now you put this rational function in much more complicated. This was in disguise. I've transformed the Arbel and Chebyshev were working with square roots of quartic polynomials and Pells equation, but I don't want to introduce that here. And so I had to transform it and I transformed oilers to this one. So this is heavy disguise. This one is then number eight, please. It's elementary with M equals two. So there's two logarithms now and you can't combine the two logarithms into a single logarithm. And just a kind of vague remark in the classical literature, which is rather copious and fascinating. We never found any natural examples involving three logarithms. I may make a later remark on this. And then we'll see later exactly how the difficulties arise. They were clarified and summarized in a short 1970 paper of Robert Rich. So now we come to the main theme of the talk in 1981 in his PhD thesis, James Davenport. So now the picture which I thought there's a picture of sheep, the sheep photograph. Philip, can you find that asset here is here is the man James Davenport. He's being declared a freeman of the city of London. And he has to swear these oaths he will be obedient to the mayor of London. And, but he will warn the mayor. If he finds out that there are plots against him. And there are other various things he had to pledge. And the fake sheep here symbolizes right to drive sheep over London Bridge. So he has the theoretical right to drive real sheep over London Bridge. But he says on this is all on his website he says that the Metropolitan Police of London take a very dim view of anybody trying to exercise this on their own initiative. So he has the theoretical right and I think there's a there's a particular day where these people can all join and drive their sheep over. And now the photograph, the other photograph which you thought Philip was of seagull. Yeah, this is not seagull. This is his father Harold Davenport. So, and this is because the confusion can arise. This is Harold Davenport either either clapping or washing his hands I don't quite know which, but it whatever. And that is the ubiquitous cigarette he was never without a cigarette. So that's how I hope that. And now page nine please. So asserted this is James Davenport who asserted that if you take now fxt with a parameter in. And it's an algebraic function of x and t. Now to algebraic functions. And this means that f is in the algebraic closure of see a joint x and t, such that the integral fxt the x is not elementary, identically in t. Then there are most finitely many complex numbers tau, such that the specialized integral when you specialize the T from the tower is itself elementary. Identically in T can be explained by an example. The integral of one over x plus T is log x plus T. So that's elementary. And then here you can simply sub specialized T to tau and you get elementary there. So when you have this identically in T business, then the thing become stays elementary under all towels, except perhaps for a finite number. So as long as this doesn't hold, then you get this finalness. This is a rather bold assertion and somewhat original. If you take classical specialization theorem of number theory Hilbert's irreducibility theorem, then this doesn't hold in such a strong form. So x squared minus T is irreducible. But if you put T equals n squared, then x squared minus n squared is not. So you don't have fineness in this case, but he asserted that you have fineness in this case, in our elementary integrativity case. Page 10, please. But in fact, it was too good to be true. And on the 9th of February 2016, I think it was about quarter past three in the afternoon. After about three years work on better as any and I found two count examples. The first was this one so we take again the old DX over why and we put another rational function in, which is slightly more complicated now degree two. And this is not identically elementary. Not quite so easy to see it, but it becomes elementary at T equals tau if and only if this point that we've seen before is torsion of order at least three on the curve on this curve. That looks like this formula torsion, but that has only only if so here is if and only if and here we can go backwards. This point is torsion for infinitely many. This gives infinitely many towels you can just choose your order as you like. And so you get infinitely many towels and so this is a counter example to his assertion. And if you think about it this has some sort of strange flavor. If we take order 1979, which we happen to know is prime. Then the integral becomes has m equals to it's a linear for me logarithms it's a C1 log F1 plus a C2 log F2 as a new bill. But F1 F2 there now in the function fields of the elliptic curve CXY they must have degrees in X at least 989. Whereas the original thing had degree only two. So if you take an arithmetically complicated towel. Then the integral gets a geometrically complicated number. Number 11, please. So arithmetic complexity implies geometrically complexity in some sense. This may remind people of Belly's theorem where you have a similar kind of odd behavior. So this. This count example is not one of the new integrals of the title. This is an old new integral. This elliptic curve has complex multiplication in this quadratic field Q root minus one. And for example, the action of minus one on X Y is minus X times minus X and then root minus one Y. We had a second count example and that had a similar phenomenon in Q root minus three. We all know these are the only imaginary quadratic fields with non-true units. And it seemed to me that our proofs used these units. Footnote here we already knew that every elliptic count example must involve complex multiplication. But so but in this in our paper of 2020 we only had the root minus one root minus three. Number 12, please. And in our paper we we had to we had to restrict to algebraic coefficients. I said that F was algebraic over C XT here we have to have algebraic coefficients in. And then we showed that such count examples are very rare. It was very difficult to find so we called them elusive. That was about the time that the Higgs boson was found. And also lucid. But there was a puzzle and it's mentioned in our paper. Some 10 years ago 1010 years before previously Daniel Betron found count examples to a different conjecture. And these were in every Q root minus D. Why do our elusive seem to stop at root minus three. And what about root minus two in the middle. Finally, relatively recently 2021 we could after all construct elusives with respect to every Q root minus D. And I just will now I didn't write them down. They're not so difficult to write down. But the method is analytic. So we're back to Whitaker and Watson and bias stress. So take your square root Q take your D a positive integer. And then you have the this field and there's an elliptic curve certainly with this complex multiplication. We can put it in bias just form with G2 and G3. And then you can always do this without your break G2 and G3. There are page 13 please associated bias just functions PZ which we saw already, and also Zeta Z. Then you pick any delta zero non zero in the endomorphism ring. There is a functional equation involving values of the Zeta function Zeta of delta Z is delta bar times Zeta of Z delta bar is the complex conjugate. Plus a couple of other terms one is a linear term alpha Z. And the other one is essentially elliptic function. And it's an old elliptic function. So it's the derivative of the P function times a polynomial in the P function. This alpha then depends on Delta. And it's also algebraic and Q also depends on Delta. It's a rational function of this new variable X formerly known as Twitter. And it also has algebraic coefficients. This function equation looks completely classical and I still don't know. I still don't know how to find it in the classical literature. As far as I know, it seems to be in written down only around 1975 independently by Dale Brownwell with Ken Kubota and by me and that was in my PhD thesis with a misprint. So if you differentiate this and then use the Zeta dash is minus P, then you get another functional equation. This time involving just P, and it looks much nicer because this differentiates to alpha. You get simply P of Delta Z is R of PZ where R depends on Delta as well. So this rational function. Now we're nearly at the different nearly at the count examples. They come in page 14. Now take Delta, not in Z. So a real a real complex modification. But first we had a differential we called it the usual suspect because it was always was somehow the obvious one to be to have this property. So it's omega now I talk about differentials I don't use integral signs anymore. So the usual suspect is only go usual suspect, depending on the and also the Delta. And you have the usual DX over why, and then you have these, these terms here this is also a rational function of X a simple rational function of X, but involves these rational functions of T as well. So you have X minus RT minus Delta, Delta bar over X minus T. And if you take the situation of the first two elusives, then it really does more or less lead to what we found for a while we thought it works for root minus two. And if you work at this out for root minus two, it's this thing five T squared plus 40 T plus 62 times X plus T Q plus 80 squared plus 70 T plus 144 over X minus T, then times two T plus eight times X plus T squared plus 40 plus 18. And the why has changed now. This is the, this is a convenient form of elliptical with complex multiplication by square root of minus two. So we have coefficients 1356. But this is definitely not elusive. And if you put T tower, it almost never becomes elementary. I think it becomes elementary just three values, which are easy to describe. It doesn't give you infinitely many values at all. So that doesn't work and the, and the final step was to make a final adjustment. Omega is the usual suspect minus a multiple a constant multiple of DX over why this is a constant this is a constant. With regard to X just involves the T. And the Q is the rational function in the fun now I would flip back, but I can't. Q is the rational function in the functional equation there. Thank you. This is this Q here. Thank you. 14. R is R is also as in the previous one. So you have these two related rational functions R and Q. So we make this function co-operation then we get to know this. Page 15. Please. We omit the proof. A footnote in Whitigan Watson section 2053 plays a minor but non trivial role. So, as luck would have it the, the modification to the, to the thing on the previous slide which turned out not to be elusive is just to add one to the rational function. And then after some simplification you get this, which looks a bit shorter than what I wrote down on the previous page but the degree of X has gone up to two. X squared plus two T plus 10 X plus two T plus 18 of X same denominator X minus T to T plus X to T plus X to T plus eight X plus T squared plus 40 plus 18 DX over Y. We could then show it's the simplest one. And essentially unique. There's no unique one because you can, in fact, you can always replace T by any other rational function of T. And this, this is essentially the simpler, simplest one. And in view of the enormous amount of time required to find this, we called it super elusive. As a color corollary all elliptic elusive can be written down now and they all have two logarithms. And this is this problem of three logarithms, and it's still conceivable that you can find somewhere three logarithms for something sporadic, which doesn't come from a specialization of a family. But this is a somewhat vague assertion. These calculations were very tedious involving sign endless sign problems, which is why I like to work in characteristic to where it doesn't the signs don't matter. Here they matter a lot. And we calculated the above integral at this value of tower so two plus three root six, and that corresponds to a point of order three. It is m equals two so it's C one log F one plus C two log F two where page 16 C one. There we are, they're written down. I've taken some trouble always to read out formula to slow the expedition down but I'm not going to read these out. You can see that here you can see that the C one and C two, their ratio is this root minus two. So it's irrational. And that means you can't combine the two logarithms. How does one actually do this integration maple as I said is is out of depth with much simpler things and and here you your. This looks like a number field of a degree eight here with I in it and square root of two and square root of three not dimensions, square root of a square root here. So it's a field degree 80 or eight or 16 and it's just out to can't handle these things at least I don't know how to make it. There's a, there's a, there's a computational machinery called Frick us, and that can actually do it. But it is not guaranteed to turn up an answer. Again one looks at the poles of the differential with T equals tau. They're simple. And if you remember the denominator is could we have 15 back yes. The denominator gives you the poles here. And so X equals tau is, is gives you one, actually two to a one pair of poles. And this other value of X gives you a rational function of T, and they're written down in the sheet 16 and now 17 I think is needed. That's 16 that's tau pi one, and now she's 17 is pi two and that's this this abscissa here. And in fact pi two is square root of two times pi one. And then you can find logarithmic differentials, the F one over F one df two over F two with the same poles. If you have this, if you take tau suitably to correspond to a torsion point, that would be pi one is a torsion point and therefore pi two is automatically torsion point. We then find C one C two such that this linear combination has the same residues. So now you've got the same poles and the same residues all the poles are simple. And so if you take the difference between the original differential and this linear combination, you've got no poles at all. Now the only differentials with no poles are constant multiples of DX over Y, constantly with respect to X. And one just works out C to 100 decimal places if, if one's desperate and the one gets C equals zero. So this C is related this final adjustment coefficient to delta Qt in the differential but we hit zero here because we kind of took it over to the other side and made it zero. We calculated some other super elusive and we went up to Q root minus 163 for fun. This is the last one where you can stay over the rationals because of class number one. The degree in T has gone up to 162. The degree in X is still two. And some of the coefficients have 535 digits. I could show it on the on my screen here while I can't. In fact, but I have it on maple but it takes about half an hour to scroll through. So page 18. And now I will change the theme a little bit. And I can link between the two themes is the Weistar sigma function, sigma dashed over sigma zeta. And there is a function equation for this as well of the same sort, sigma delta z here is sigma delta z and here is sigma z. And this is to a power delta delta bar that we've seen before and the whole thing is squared. And then you have this exponential polynomial. Alpha is as before and P now is also depends on delta but is a polynomial. So you've got gone from rational functions to polynomials and this is this is rather handy for calculations. If you differentiate logarithmically this one you get the functional equation for zeta, which then differentiated gives the functional equation P. So it's kind of a hierarchy of things and this, this one for sigma is somehow the fundamental one. And it also supplies an indirect connection between the elusive integrals and the rebate curves. So there's no clue how much time I have now. Could someone give me an estimate. About 10 minutes you can go because it started later. So it's up to you. Okay, I'll have to stop and really see. I'll go ahead with this. This. I wanted to talk about the rebate curves and the connection is Madeline Mumford, the well known in diaphragm and geometry circles about torsion points. The simplest case here is to solve completely x1 plus x2 equals one in roots of unity. Here it can be done with a simple picture page 19. It does not contain the simple picture. After that one can treat a general algebraic curve in GM cross GM with the multiplicative group GM of nonzero complex numbers. So you've got a general algebraic curve replacing this line after that one can work with a general inside a general commutative algebraic group G. But then you can have now an obedient variety. But just as in the Davenport assertion one must allow a parameter. And this puts you in the so-called relative Manin Mumford situation. And our own work in 2020 involved mostly a billion varieties then parameterized by a single parameter. But we also had to contend with so-called additive extensions of elliptic curves G and these sit inside an exact sequence or an elliptic curve E here. And GA is now a long hand for the for the additive group of complex numbers. And now you allow parameters. You can't allow parameters in there. You can see involves no parameters, but you could allow parameters in E. And most of the work it's the this is this is the genre. It's the genre of curve page 20 please. Defined by y squared it was xx minus one x minus t for example. So that was the situation for elusive as regards. And by contrast the rebate curves sitting multiplicative extensions G. And now it looks much the same. You have G and you have an elliptic curve and you have GM here. So that's that's now multiplicative. So I put the one instead of the zero. So it doesn't look like a big change. In fact now the theory also forces complex multiplication on you. Which forces E to be constant say again let's put it in by a stress form. As we did before. And, and then we have G. Now the parameter seems to have vanished again again there's no parameter in C star. And now there's no parameter in E because it's constant. So where's the parameter gone. And here it comes in a more subtle form. These multiplicative extensions are far from unique. And they're parametrized, they come in families whereas the additive extensions are. The additive extensions have much more rigidity and they're practically unique. They are 100% unique in current number theory terminology. And they're all isomorphic to each other. Here the isomorphism classes are parametrized themselves. And they're parametrized not by a simple T but by points pi on E itself. It's not clear how to see this. One can imagine one can see it through the exponential maps of the of the algebraic groups. The exponential map on GM is is is the exponential map E to the W from C to GM. But on G in here this G. Now that's get that's two dimension two dimensional. So you'll have functions in C to of Z and W. And the key functions are. Page 21. X equals PZ. And also you have P dash said but that's, that's, that's kind of somehow superfluous. Parameter is U. And this is given by Sigma of little Z plus big Z over Sigma of little Z Sigma big Z times E to the W. So that's the E to the W that we saw before with this little modification. And so we have little Z and W and we also have a big Z floating around the big Z is more or less Pi. So if you write Pi parameterized Pi analytically it's PZ P dash P dash big Z P big Z P dash big Z. So with these two algebraic coordinates we can sort of see G in C to some kind of projection on it but good enough for practical purposes. And if we throw in T was PZ into here, then we get a much better view and that's in C three. And now you've got the parameterization into account even though you've still only got a group, an algebraic group in C two you've got an object in C three whose fibers of the various various genes. And now you have algebraic coordinates X you and T. And now in this C three, we would like to construct the rebate curve. We pick delta now in the endomorphism ring, even more special so it's purely imaginary delta bar is minus delta. So using this star thing, make a strange choice of big Z and W in terms of little Z and big Z is delta times Z, and W is Z times Z to delta Z minus a half alpha. And then I keep the dependence on delta here times Z squared. So, there's only one analytic independent analytic variable left that's little Z. We're going to get a curve of Z variables, various, but it looks, it looks like a horrible analytic curve, because this is horrible analytic this is even sort of essential singularity start up here. But in fact it's an algebraic curve. So, we can see this with a little bit of calculation and we can see one relation between these three coordinates. T is PZ and big Z and P big Z is P delta Z from here. And here we have the functional equation that's our delta of PZ and that's our delta of X. So T is a rational function of X. And so we have one, at least one algebraic relation between these two to three coordinates. Page 22, please. So an algebraic relation. Now, what about you? You now becomes so big Z was here but now it's delta little Z. And then we had some exponential parts and they simplify to to this. And here you see sigma delta Z. This suggests the functional equations for sigma. Here you see sigma of one plus delta Z. And so you're going to need the functional equation with delta plus one as well. And they involve squaring so you're going to have to square the whole thing. And then miraculously the exponential part goes away. And you get you squared is P delta plus one X over P delta X. And P was a polynomial. So this is a second algebraic relation, definitely different. And that gives you your curve. And I think here this slide is the first appearance of the of of ribbed curves in print, at least in this kind of two dimensional form for delta is minus root minus one it's the boring. U squared is minus two minus one X, but for Delta it was minus two the the source of all our troubles. It's this longer thing to use squared X plus four is one minus two minus root minus two X squared plus four plus 10 minus two X plus 14 plus X root two for delta equals root minus 163 it is page 23 about time to stop. So, do I have a, let me see. Do I have say three minutes. Yes, yes, please stick. Am I, I don't want to go over. No, no, please, please go ahead. Just perhaps one more slide. It reminds me of a story about that someone was in the next hotel room to the famous cellist Rostropovich, and they could hear him practicing a piece. But they knew it wasn't the piece for tomorrow's concerts and so they were puzzled. And when they got to the concert, they found that he was actually practicing his own course. So here are my own course which I've practiced and written down and this is just open problems. So complex coefficients to algebraic coefficients. There's no work on this at the moment, but it's not expected that you get any new elusive things. And then even over here, this doesn't, this doesn't automatically point to elliptic you have to allow curves. See say of higher genus, we've been dealing with genus one, a genus zero is easy there are no elusives. Already we had seen in our paper that if, if there is an elusive on C, then the Jacobian must contain an elliptic curve with complex modification. And this more recent work enables us to reverse this implication in other words, there's an elusive on C if and only if the Jacobian contains an elliptic curve on complex modification. I think I'll stop here.