 I'm a little confused about my preparation because my mind is still in the previous wonderful lecture about physics, so if I mix up manifolds in physics, you have to forgive me. I'm also giving this lecture on somewhat false pretenses because this seminar is called basic notions and one is supposed to tell some basic notion that all mathematicians or all physicists or all everybody should know and but that is kind of very well known to some experts but actually what I'll do is different. I will talk about something very basic that everyone should know but everyone does know what the question that I'll talk about means and what I want to explain is that rather surprisingly the answer is not as stupid as the question sounds. So there seems to be no chalk at all that's more than an eighth of an inch long. There are hundreds of very small pieces but I'll do my best. So I should say by the way right to be getting before I forget that this is joint work with Matthias Kreck in Bonn but actually it's his question and he told me about a year ago we were waiting for a train. I was accompanying him to the station, to the platform and the train was late so we had 15 minutes and I said you're a topologist, we're old friends, tell me something you do that might involve number theory. So he told me that 25 years ago he invented this question, what are the betting numbers of a manifold? He assumed that it was very well known but he looked in the literature maybe in that day not that Google but now he's tried Google at Wikipedia and there was nothing really known about it and then he told many colleagues what a nice question, everybody was completely bored and it's a pretty stupid question, why are you wasting our time? So he dropped it for many years and then a few years ago one or two people did get interested, one of such apologists called Jim Davis who gave it as a thesis program, a very special case to a student who solved it. Partially there's a lot of overlap between her results and ours, she's called Sue, maybe I'll write it later but I just want to say that this question has been around for a couple of decades but somehow nobody could see what the point was. So I want to tell you that there is some point and that it's actually quite an amusing story. So first I start as if there was an honest basic numbers seminar and I'll first answer the question what are the Betty numbers of the manifold, meaning what is the definition but then I'll come to the theme of the talk which is what are the Betty numbers of the manifold, meaning what is the answer, it's very different to define the question and to give the answer. So this title is intentionally a little provocative it sounds like everybody knows what that is but what everybody knows is the definition of the question not necessarily the answer. So first of all what is a manifold? Well a manifold has a dimension n which I hope I'll remember to always call n and not keep changing, n-dimensional manifold and in this talk all manifolds will be smooth so no boundaries so smooth means c infinity come back to no boundary and also connected because if you have a non-connected manifold like for instance two spheres it might be good for physics but in mathematics at least for the topology it's not very interesting you just add the Betty numbers and there's nothing really to be said. Now here this smooth you might think well what's the difference smooth topological pl some kind of manifold it turns out to make a big difference and if I ask to this in the topological category then the answer would be we don't know at all it's okay so what is first I remind you what a smooth manifold is you start with Euclidean n-space which is just the n-fold product of orbit itself and you have a topological space such that locally every point has a little neighborhood which you have identified in a specific way with a little neighborhood of a point in Rn but it's not that specific you give yourself some freedom you can have many such local charts and the requirement of smoothness is that you've done it in such a way that if you take two different ways of identifying your little piece of your smooth manifold with a little piece of Euclidean space if you take two different ways of doing that then on the overlap of those reasons if there is one the passage or let's say I've done it in different Euclidean spaces but the identification of a little bit of one Rn with a little bit of the other is smooth in the usual sense it's given by an n-tuple of c-infinity functions but I mean I'm not going to give a mini course in topology I actually assume that everyone knows what the manifold is it's something that looks like Euclidean space but locally of course it may not be oriented because it looks like Euclidean space locally but when you go around a path you may come back and find your Euclidean space is turned on his head a typical example would be the Mobius Strip but the Mobius Strip wouldn't do this because it's got a boundary it's a strip with with a twist so that's not a manifold with with no boundary but for instance the famous I'm going to have to move this so it doesn't keep crashing with my hand uh if you take the Klein bottle that would be an example so if n equals zero there's only one example because it's connected it's a point if n equals one there's also only one example it's the circle now the circle is orientable and then you can say you could give two different orientations so in this case the right morphic but I won't worry about orienting my manifolds but I will distinguish orientable and non-orientable in a moment for n equals two so this is orientable and here's non-orientable well there aren't any non-orientable zero or for that matter one manifolds so that's the classification here here you have of course the Riemann sphere or there's simply the sphere the two sphere then you have the torus then you have the double torus and so on you have the surface with g holes so with genus g where g is zero for the sphere one for the torus two for the double torus etc so you can and that's the complete classification that's been known for I don't know 150 years at least and in the non-symmetric world you start with rp2 which is s2 divided by identifying antipodal points then you have the Klein bottle I can't draw anything non-orientable actually I can't even draw anything orientable but you somehow make a bottle and then the end of the bottle is thirsty and goes inside the bottle and joins up with the bottom and in doing that it flips around anyway I'm sure you've all seen pictures of the Klein bottle and here too you can go on so this one is no handles one handles you have again a kind of a genus sometimes also called g which can take on the value zero one two and so on in dimension three there's already a great deal of choice and you get a million dollar prices for proving things but one thing you also have always have in every dimension is the end sphere which is what I think everybody has seen the end sphere it's the either the vectors of length one in Euclidean n plus one space or topology you can take an n-dimensional plane and add one point at infinity and kind of fold it together in a way that I think everybody has seen the pictures so those are what the manifolds are both oriental and non-oriental and then to each manifold m and n means it's an n manifold we associate the betty numbers so bi is the i-th betty number and the formal definition I'll mostly work over q so it's the i-th homology of m with coefficient q but it's also the dimension of the i-th homology of m with coefficients in q at least in the oriental case I guess always these things are those two spaces are dual so they have the same dimension if I work with coefficients zz they wouldn't be quite dual but anyway I want the dimension if you work over z the rank but intuitively and again I think everybody has seen this and knows it very well bi is the number of linearly independent so this means over z and therefore also over q because if you had something say that's torsion then well three times of d zero that would be a linear relation so when I say linearly independent over z it's really over q it's the number of i-dimensional and if you're very careful cycles and I won't try here because I mean that would be a basic notion but I think too elementary to explain exactly the definition of singular homology and singular comology you can think of i of bi of these cycles as being represented by i-dimensional submanifolds and if the i-dimensional submanifold happens to be the boundary of something i plus one dimension it's zero that's correct as far as it goes an embedded i-dimensional manifold does have a homology class and if it's the boundary of i plus one dimensional embedded manifold the boundary is zero in homology but not every homology class can be represented that way and even if it can it may be zero in homology even if it doesn't balance something smooth so that's why you need what's called singular meaning not necessarily smooth homology and it's done carefully and actually fairly tediously in every course of beginning algebraic topology and I won't do it but you can believe me certainly that there's a definition so from this definition one sees immediately that bi is zero if i is bigger than n because there's no room for an i plus more than n-dimensional cycle is something n-dimensional we also see that b zero is one because it's connected so if you have your manifold then you can represent a zero-dimensional cycle as a point and any zero-dimensional cycle is just a number of points maybe taken positively or negatively so the group is z but you can move any point to any other point because it's connected so the group is just z so the dimension is one and finally bn the last one is one if it's oriented because the only n-dimensional thing you have room for an n-dimensional manifold is the manifold itself or maybe twice the manifold or three times or minus three times but since it's oriented cycles you only get that if it's if it's oriented manifold so the top petty number you know it's if not so we already have a tiny little bit of information and so the actual question is more precisely given some numbers b zero up to bn i don't have to specify more since we know that the higher ones are zero and actually i can start writing some of the conditions this has to be one and so on is there an n-manifold realizing these numbers in other words having exactly these petty numbers okay so that's the question and it still sounds certainly very innocuous certainly if so far i've only said very very standard and easy things that everybody has seen in any kind of course on topology so let me first write down some sufficient conditions well two basic ones and they both come from the Euler characteristic which is the simplest invariant of a manifold so the Euler characteristic i of the manifold is known to be the alternating sum of s j if you triangulate the manifold so for instance i have my two manifold and i triangulate it doesn't have to be trying because it could be something else it doesn't really matter so i divide it up into little polytopes uh in some way and then s j but let's tell you okay i could actually do it that way s j is the number of j dimensional simplices maybe i do want to triangulate then for this but that's the elementary definition but not very useful because it's quite hard to see that's what Euler proved for the two sphere but you have to do it for every manifold a manifold has lots of different ways of triangulating if there's no visible reason by this is always the same since these numbers s j depend on how you triangulate of some triangulation triangulation means you've divided up into polytopes so this is sort of the bad definition and the way that we know that this is an invariance you show that that's the same as the sum minus one to the i b i and b i are the many numbers and these are individually invariance each one whereas the s j are not individually invariance all of their alternating sum so the Euler characteristic which is the oldest top lots invariant by uh 150 years or something since Euler discovered it that's an invariant that you define in terms of non-invarying things and you have to work and then this formula writes it as a weak a weakening of giving the entire information about all the Betty numbers so now the that's one important fact the second important fact is Poincare duality and the third the most important for this lecture i'll come to bit later that'll be the Hertzberg signature theorem but Poincare duality tells you that if if m is oriented then you can intersect an i-dimensional cycle and an n-dimensional cycle and you get an integer which the number of intersection points even physically if you move them around so that they intersect nicely and transversely and that intersection number is a pairing so it tells you that h i and h n minus i have a non-degenerate pairing into q these are vector spaces over q this is non-degenerate and so that identifies h i with h n minus i and that implies that b i is equal to b n minus i in the oriented case and therefore the sequence of numbers is symmetric but it implies one more thing which is also very easy but not quite obvious can anybody tell me what else follows about just about the Betty numbers from Poincare duality come on come on let me give an example question a mini question is there an oriented i'm doing oriented here an oriented two manifold called the surface with Betty numbers maybe from now on when it's convenient just write the vector of Betty numbers i don't keep writing b zero equals b one equals b two equals so one one one is there such a manifold no certainly not if you look at the little table that i raced we had you know zero holes one hole two holes but the middle Betty number was twice the number balls a torus has two cycles one that way and one that way in fact the answer is no and so the other condition is that if n is congruent to two mode four which means it's twice an odd number then the middle Betty number is even so if n is odd there is no Betty middle Betty number the numbers are just symmetric but if n is even there's a middle Betty number it can be even it can be odd but it can only be odd if n itself is a multiple of four and the reason is very simple if n over two if n is two mode four then n over two is an odd number but the this is the if i think in comology which again everybody has seen then this is the cup product it goes into h n which is canonically as more than q and that's a skew symmetry pairing because it's an odd dimension and comology classes odd dimensions and to commute with each other and a skew symmetric non-degenerate pairing that's called a symplectic structure on a manifold can only exist at the dimension of the vector space is even and that's why in the two dimensional case but also in the six or ten or fourteen dimensional case the middle Betty number has to be even now about the non-orientable case so we already have b zero equals b n equals one so we know it starts with one it ends with one there's something in the middle but the something in the middle is at least symmetric and even in the middle if n is two times an odd number now what if n is non-oriented so these are sort of the triple conditions so I called it zero well here I already told you the zero condition it's still connected so b zero is one it's now not oriented to be one is one this condition the one that you all would have thought of that it's symmetric is missing because it's not oriented there's no duality between n and minus i but there is a condition one which is not quite obvious can somebody guess or tell me what the condition is for instance is there a non-oriented three-fold well any three-fold but it'll have been non-orientable with Betty numbers one three five zero so it's a three-fold so it has to start with one and end in zero if it's non-oriented can that happen can I have one three five zero answer no namely the second condition is that if n is odd then the alternating sum of the Betty numbers has to be zero and here one minus three plus five is not zero why is that well you see in the oriented case I don't have to write that as an extra condition but here you merely see if n is odd then chi is the sum by definition minus one to the i b i but for a practicality that's minus one to the i b n minus i which by rephrase sending i to n minus i is the sum minus one to the n minus i b i so it's minus one to the n times chi and so in the orientable case you see immediately that if the dimension of the manifold is odd like a three three manifold or five manifold then the order characteristic is always odd simply because the Betty numbers come in pairs and they cancel so I didn't have to write that as an extra condition it's a consequence of one here it does not follow from from that one because that one isn't true but it's still turned the reason is very simple I told you that the Euler characteristic is more elementary as an invariant than the Betty numbers but it's also much more robust you can do anything to topology and you never harm the Euler characteristic it's it's invincible you can mess up you know Betty numbers blow all kinds of things but Betty numbers like you know vibrates with things degenerate and so on the Euler characteristic survives all bad treatment so one of the bad treatments you can do is you can say let me not work over q let me work over z or z and now let me reduce bulk two now I'm back to coefficients this is slightly technical just for a second in z modulo two which is a field and over z modulo two unlike z modulo p for any odd prime p your manifold is oriented because if you go around the path and come back you flip the orientation so something has gone to minus itself but in characteristic two minus x equals x so your manifold becomes oriental and suddenly panker a dwelling which is false over q false over z and false over z mod p for p odd it's true in characteristics z two and therefore the the new Betty numbers which are the dimensions of the z two topology has vector spaces over that lousy little field of two elements but still they're finite dimensions they're they now satisfy panker a duality they're alternating some seems to have nothing to do with the original one because that was the sum of integers which are much smaller in general but no it's the same that's the robustness the Euler characteristic is independent the field you calculated in so because the Euler characteristic over q is the same as over z which is the same as over z two where it is oriented there you have panker a duality and the results that this is true so therefore if we look up to dimension four here the possibilities so possibilities possible b vectors b which is b zero to b n for n less than or equal to four so I should define the word low-dimensional topology for most apologists low-dimensional topology means up to n equals four that's because around the time I was a graduate student Smale and other people proved the h-covidism theorem the hope from woodland every big conjecture of topology dimensions bigger than or equal to five and so topologists feel it's small if it's four or less that's why I took that but for this talk I'm a number theorist I don't play around with things like four small for me means less than or equal to one million so this will be a lecture about low-dimensional topology but in this somewhat extended sense but for the moment n only goes up to four so here's n and here's the oriental case and here's the non-oriental case well already told the dimension one the only possibility is the point and here there's nothing here the only possibility is the circle the betting numbers are one and one indeed the other characteristic is zeros it should be an odd dimension and it's nice and symmetric here there's nothing here already told you you have a number 2g but I'll call it 2a where here a and in the second b c are simply non-zero integers so it has to have the form one something one but the symmetry and the a is greater than equals zero and 2a has to be an even number and here it could be one anything but it can't be zero so I'll call it a plus one and then it's zero because it's non-oriental okay so a equals zero would be the rp2 equals one would be the Klein bottle then here again by symmetry you have one a a one now there's no restriction on a the alternating sum is automatically zero and here it's one a plus one a and zero because the order has to be zero this number is zero that means one and this the alternating sum is zero so it has to be one a plus one a and zero and finally here a priori you might have one a b a one there's no restriction that b is even because four is not twice not numbers twice an even number and here you could have one a b c zero and there's no restriction at all because the other characteristic doesn't have to be zero because it's all dimensional so what I've told you up to now says these are the only possibilities and up to here we've solved the problem they all occur I can state immediately the following theorem but as far as we know it's news in the joint paper which is half written up it's written i mean matthias proved it it's written it's a new theorem it seems not be in the literature theorem in the non-oriental case well the theorem I could just write in three letters if and only if in other words b occurs if and only the conditions that I told you are all satisfied so they're all non-negative b and is zero the first one is zero and chi is zero if and is odd that's not a complete the trivial theorem you use some surgery so there is some topology going into it but it's relatively elementary in any way the the result is kind of disappointing though the necessary the obviously necessary conditions are sufficient so you don't get any surprising new conditions now let me show you why these things occur before I show you that let me make a preliminary thing so this is actually I said there are four basic things I'll use Euler characteristic prank rate duality and the Hertzberg signature theorem but there's one other which is very easy so I'll put it now and that's what's called connected sum and then as I say I'll come still to the Hertzberg signature theorem a little later and that's the thing that makes it work it makes it interesting makes it non-trivial so first here I'll first just draw a picture let's have an n manifold m1 and another n manifold with the same n so here for instance both of them are two manifolds because I can't really draw anything else and then what you do if you're well but physicists actually to publish this usually do is they make a little hole here and make a little hole here and connect them by tube but actually it's easier but I can't move on the board to make a little hole here and then not make a little tube just move them together and glue them along the boundary so you cut out a little neighborhood of a point which by definition is an n-dimensional disc because locally near any point these things look like n-space and then those two discs are the same actually you have to do it in oriental reversing ways so you can cancel the two and you just glue together and it's very easy to see that if you do this with another initial point you get exactly the same thing this so well-defined operation on the types and when you do that then let me call it m and m prime double prime so m is it's actually called the connected sum and then bi is simply the sum of the bedding numbers except that of course b zero sorry b zero is one and which is equal to b zero prime which is equal to b zero double prime so because it's the connected sum you've taken out that point and a little neighborhood so you took out two points and put together just one or one little disc and that's why the zero bedding number is not added of course it can't because it has to be one but all the others just add so if you think about it now if I take sj I didn't say that but the smallest thing you might have is the n-sphere and the n-sphere is oriental for every n and it is of course it's euclidean space plus one point so if you move that point and therefore if you have any cycle of dimension less than n it can't hit every point let's say it doesn't hit infinity then it's sitting in euclidean space which is contractible so there's no homology in intermediate dimensions so you get one zero zero one but now if I take sk and sn minus k then I have to multiply one so it's easier to do with generating functions if I multiply one plus x to the k times one plus x to the n minus k then I get one plus x to the k plus x to the n minus k plus x to the n so assuming for the sake of argument that k is the smaller of these two numbers I'll have a zero in position a one in position one k n minus k and n and so I'll get a new manifold where I've added one to uh to those two positions but now more generally if m had any b zero b k uh b n minus k and b n then if I take the connected sum of m with this product as I told you I just add sorry this was also uh b n of course by point of radiality because you've taken out an n disc so except for the top and bottom they're just wrong but all the intermediate ones you just add so now I've I've got a one in two symmetric positions and now I've just added one to those two numbers and of course in the case when k equals n minus k that can only occur if n is even then I'll have b zero b n over two and I've added two to it because of course the product of the n over two sphere with itself has betting numbers one zero zero two in the middle zero zero zero one so I can always go up so now you immediately see so we can we can send m to m connected sum with sk across sn minus k and this will send bi goes to bi plus one if i is equal to k or n minus k this is of k let's say smaller than n over two and otherwise the same bi else and in the case when as I already said if k is equal to n over two then you send the middle of the betting number up by two and you don't change it otherwise so if we didn't have the two if we just had a one my problem would be rather trivial because I start with the sphere and I can add even I can add two ones in complementary position then I can make that number have any value a and the same value a by crank rate you have to have the same value so I can get everything but in the middle I can only add even numbers now if n is an odd number there is no middle like that doesn't bother me and if n is two times an odd number so two mod four it also doesn't bother me because I know that the middle betting number has to be even anyway it's on the list so something now an easy fact is in the orientable case which we've done the non-orientable case one the betty the order the the obvious conditions so the punk radiality conditions obvious conditions are sufficient except n is zero mod four and the specified betting number remember I've given you some betting numbers that I'm asking for is odd so if n is not a multiple of four you can always do it and it then is a multiple of four but the numbers I offered you the middle one is even the others come in pairs then just by taking the sphere and adding a connected sum of product spheres you can realize anything so the only interesting case is the case when n is 4k and I want well I'm not saying we the others are trivial but in the middle dimension which is 2k well sorry these are just the dimensions but the betting numbers that you've specified are one some odd number and one and then here you've put something that you like then this case is not at all trivial it turns out so at least now we've produced what sounded like a somewhat structuralist problem to a very very specific one that I mentioned the manifolds multiple of four and we also so another fact so the other fact is if there exists and I'll call the letters rpp so let me define that definition a rational projective plane of dimension n is an n manifold n will have to be even in fact it'll have to be multiple of four so you'll see in a minute with betting numbers b zero equals b n over two equals b n equals one and else zero so by what I've already told you that can only happen if it's oriental because this is a one it can only happen if n is a multiple of four because otherwise the middle number had to be even and so n will certainly have to be a multiple of four why do we call it a projective plane so this remember these are the rational homology so it means that the dimension of h of the manifold because it's in q so it's over q it's one if i is zero n over two or n and it's zero else why do we call the rational projective plane but you have rp2 that I already mentioned that's non-oriental but the Euler characteristics over z2 are in that case one one one so it's exactly like this except now n is an odd number I mean it's one one one but if we take on the oriental manifold you have the complex projective plane which is extremely important in algebraic geometry it's the simplest surface the rational surface or p2 of c whatever you like to call it and p2 of c is betting numbers one zero one zero one so it's four-dimensional and similarly this is the complex projective plane but then as well as the complex numbers you have the quaternions and so you have p2 of the quaternions or qp2 it's unless called quaternionic one and that is betting numbers one zero zero zero one zero zero zero one so here n is eight and there's one more because we also have the Cayley numbers which are not even this is associated but not commuted Cayley numbers is nothing but it's still an eight-dimensional thing a division out of some sort so you have it also is a projective plane and then that one is one one two three four five six seven eight hold bunch of zeros and one so if n is four eight or sixteen there are actual projective planes which are specific manifolds associated to specific vector spaces dimension two four well complex dimension no real dimension four eight or sixteen you've divided anyway they're what they are and these are called projective planes and so these are called rational projective planes so in particular if n is four or eight we can realize the middle number with the simplest possible number which is one or one in the middle since we can also realize a zero in the middle with the sphere if there's a rational projective plane of dimension n so that's this rpp if it exists then we have both the sphere with betting numbers zero in the middle one of course at the ends and otherwise zero or the projective plane with the one in the middle and now if I connected sums with spheres we can raise the other numbers by one at a time symmetrically and the middle one by two at a time but starting at zero or one so we can produce everything so therefore I've now solved my problem that's why this table was complete for n equals four and also for n equals eight and also for n equals sixteen so if n is not zero multiple of four then the problem is solved it's always solved that I told you the unorientable case it's simply always true if the necessary condition is satisfied and in the orientable case there's only problem if n is zero mode four and if n happens to be eight sixteen or thirty two sorry four eight or sixteen then the problem is also solved so now we're still up with infinitely many numbers which are the multiples of n which are not four eight or sixteen so let me tell you first the answer and then explain a little where it's coming from so question question what about other n with right by what I said we can assume now it's 4k well the answer is very short I can tell you one sentence we don't know nobody's a clue in fact I can strengthen that statement I'm willing to bet anything but we can't settle the bet so it's worthless that we will never know humanity will never know if the theorem that I'm going to tell which has kind of a conjecture in it we will never really know okay that's an unusual statement because after all somebody might be you know cleverer than we are it's happened before and it'll probably happen again so how do I well not how do I know but why do I even think that so let me tell you the answer to the extent that I do know it so here's a theorem well there's one little theorem that I can tell you theorem if n is 4k and if it's not eight we already saw or already told you without really giving any details that n sorry if it's not four I mean then I don't need any details then you've cp2 and since it's multiple four bigger than four then there is no rational projective plane that's very easy and it's a special case of things I'll tell you later so actually you can eliminate the odd multiples of four right away except for four itself so in particular we already had we've done it for all non-multiple four so the only interesting numbers so far were four eight twelve sixteen twenty and twenty four four eight and sixteen occur and now I've told you that twelve and twenty don't so the first question mark is already dimension twenty four it is a very nice dimension you know twenty four we saw in the previous talk and whenever you see twenty four it's always related to z of two and z of minus one except today it's just the first one on the list which where the problem is not trivial okay so that's the first point we so actually from now on I won't even call n is 4k I'll call it 8k so it's not the same k I apologize but anyway now I want so theorem here's a nice theorem so this is crack and myself but but there were previous things proving that's the same sort of thing with weaker and weaker number theoretical conclusion by sue and two co-workers whose names suddenly escaped me and I don't have so this was sue inner thesis kennard and another one that I think fowler so I'll also put kennard and maybe there's also paper with your supervisor is jim davis so I don't want to be too careful about this none of these are terribly deep theorems anyway once you've seen how it works and what tools to use it's straightforward application of known theorems but the fun part is that the answer is is non-trivial so here's the theorem well you'll say what do you mean theorem you're just told us you don't know and now you're telling so I don't know exactly but I know the following so first of all the first statement is there exists an n-dimensional rational project of plane for the values we'll already told you four eight and sixteen twenty four in fact does not occur that sue already showed her thesis 32 dels she also showed that strange enough 64 dozen but 128 and 256 both occur so one two three four five that gives you six examples and for no other n less actually less than reequal but let's not quibble less than a million so I told the low dimension to policy for me today is n less than or equal to a million no other except possibly and now comes the list of exceptions five thousand five hundred and forty four is the first possible exception the next one is four thousand one hundred and sixty then eight two two four I hope you see the pattern if you do you're a better man or woman than we are because we certainly don't and I don't think there is a pattern we're actually the pattern these are the numbers we haven't yet succeeded with our computer in solving and every so often after weeks go by suddenly the computer announces that it's cracked another one and then one of these is removed from the list so the original list was longer I don't know if you really want the whole list maybe I'll leave out a couple I'm sure you all believe that I can copy numbers from a piece of paper so there are 19 unsolved cases up to a million but infinitely many unsolved cases up to infinity namely practically everything that we haven't yet got you'll explain where this comes from excuse me thank you very much k odd yeah I even said it I said it's the odd multiples we want can we can restrict to 8k but I forgot to write thank you very much yeah otherwise the the thing would be that is a doornail to say nothing of being completely false that all of these exams would contradict that theorem so you don't get any odd multiples of four except four itself so it has to be a multiple of eight that's so easy I want even go into the proof with the things I'm saying that comes out immediately so the first exceptional case is 544 which is eight times so you really have to write n is 8k and I found very surprising that is smallest a number of 68 using big computers we let we're doing this with the help of our computer expert in bond he's running at imperil very very fast mainframes in munich in the max monk society not even in our institute they ran for three months using the most sophisticated algorithms and it's you know many of them died they hit the bit the dust but 68 but it's a pretty small number is still unsolved now when I said I think humanity will never know the answer I'm perfectly willing to bet that humanity will know the answer even in the next five years maybe if not the next 5,000 years depends a little on computer progress for 68 so we will eventually know whether 544 occurs or not but the problem you have to solve I can already reveal something it's one about primality testing whether certain numbers which are certainly not prime they have no reason we they have lots of prime factors if their prime factors have a certain property which they almost certainly have if you get big enough so I can give give a guess the number of rational projective planes is probably almost certainly I'd say very probably exactly six maybe seven but it would surprise me that if this 544 succeeds and almost for sure less than or equal to eight I mean I have no specific reason except analyzing the probabilities if you work out the total number of probability that any of these 19 exceptions or any of the infinitely many numbers bigger than a million that we haven't yet tested that any of them fail the tests that were the very stringent tests or rather pass the very stringent test that they have to pass it's extremely unlikely and it becomes extremely much more unlikely as the number grows so if there isn't a further count example on this list there's almost certainly none beyond that and already this one it's very unlikely that it'll succeed but it has a small chance that any of the others do is extremely unlikely so I would say probably there exactly these six maybe this one is well then conceivably one more almost certainly on that short list but it's going to stop but there's no way to prove that and you'll see when I say the number theory that the actual number theoretical statement that then leads to something you check on the computer and this is the answer that's something that we have no reason to think can be solved you're essentially asking about something that has a quite random behavior if I say I can prove some property of manifolds if the nth digit of pi is a seven and then you know it's very unlikely that we'll ever be able to prove that there are infinitely many sevens in the expansion of pi maybe you can because that's you know proved that it's well distributed but if you're something even more esoteric something completely run like the like for instance between the nth and the two n digits of pi the digits are equidistributable that's surely true but you know there's no particular or take something very unlikely let's say you want some huge collection of digits of pi with no sevens at all then you could saftow while that's very unlikely to ever happen if the sequence gets longer and longer say the sequence from n to n n squared to n plus one squared it's extremely unlikely that those are all sevens but it's also I think very unlikely that that's a provable theorem not because humanity is too dumb but because it's not a theorem it's just a random experimental fact it may be true but not all three statements have to have a proof and there's absolutely no reason to think that there's any approved meaning a finite sequence of arguments that that shows that it's always true and here I think it's the same so this very harmless sounding problem led first of all to quite surprisingly hard results including that there are several tricky cases but but not to complete answer so another theorem maybe I won't write it down but I said that the full question is not just is there a rational project of plane remember that I said if there's a rational project of plane in dimension and then you get everything because you have one all zeros and a zero in the middle and you've won all zeros and a one in the middle but if there isn't a rational project of plane which is almost always the case presumably for every n except those eight then the only hard case I explained is one and an odd number in the middle then we can give a complete answer when you can have just an odd number in the middle but then we don't know that the other numbers are zero maybe you achieve that number with other numbers so we don't know but there is a theorem so theorem the problem is decidable about a specific algorithm we've done many cases for any fixed vector b of any number so fixed vector is a fixed length so if you give me a potential thing like one three two five two three one which satisfies the duality and to make it interesting is odd in the middle then in a finite amount of time I can tell you if there's a manifold with the appropriate dimension with those bedding numbers that's algorithmic but you have to do some verification of various number theoretical and algebraic criteria and we have no idea if they're true in general or when they're true so we have a couple of general statements so it is a decidable question it's reduced completely in the paper that took quite a lot of work mostly by Matthias it's completely reduced to questions of algebra and number theory but we don't know the answers to those but in any given case since there are finite numbers you can just check everything so now I want to tell you a little more where this comes from I should also keep an eye on the time I've but I started at 4 30 right so I've actually a little beyond so I've sort of 10 or 15 minutes left so I mentioned several times the Hertzberg signature theorem as the key thing just out of curiosity how many of you know I don't mean you could write down with every dotting all the i's but how many of you know the Hertzberg signature theorem well some of you certainly do and aren't saying it some here you certainly know sherry are you certainly know it I think we've got some overly modest people but I saw several hands don't not going up that probably belong to people who were quite correctly not raised in the hand because they don't know the Hertzberg signature theorem it's a wonderful theorem because it's true since my supervisor so all of his theorems are wonderful but actually most of his theorems were wonderful and this one really is so if you have an n manifold and it's only going to be interested what I'm going to say if n is divisible by four so it's called it 4k as I said my n might be 8k later it's a 4k manifold then with the middle dimension but since the middle dimension is even then you have this pairing which is now symmetric pairing remember if the middle dimension was odd it had to be an anti-symmetric pairing but non-degenerative that can only happen simple active pairing if the better number was even now the better number could be anything but we know that over the real numbers and even over the rational numbers not over z any symmetric quadratic form I'm any quadratic form so symmetric matrix can be diagonalized so you can always find a basis over q such that your manifold your thing is diagonal and then there are various invariance over q but if I just work over r then since every positive number is a square and I can rescale by squares I can assume that every star is either one or minus one so the basic thing you have here r plus signs and s minus signs so r plus s is this middle middle betting number and r minus s is called the signature so in the previous lecture on quantum physics we saw for instance signature one comma three according for a certain quadratic form with one minus sign and three plus well three one in my notation three plus signs and a minus sign so the signature theorem gives as its name said so we have this invariant the signature and you notice that the signature is an integer and because it's the the sum of the betting numbers is b n over two and this is the difference it's certainly at most the middle betting number and it's uh and it's at least minus the middle betting number and it's also congruent to the middle betting number modular too because r minus s is congruent to r plus s month two so there are that already pins it down but there's an exact formula and the exact form is extremely beautiful you take what's called the Hertzbruch L-class that's a homology class on the manifold and then you take its n-dimensional piece which topologists write as the pairing with the fundamental class or if you think of uh of homology classes given by differential forms or if you're a physicist you might think of integrating the corresponding form over the whole manifold and it's a specific class to write down its form in a second so what you do is the manifold because it's a manifold it has a tangent bundle which is a bundle it's an r n bundle over the thing and this thing uh an oriental bundle has a Pontriagin classes so it has Pontriagin classes p0 of tm which one just writes as p0 of m is zero-dimensional it's always one but then you have p1 p2 and so on and pi is a four i-dimensional class with coefficients in z and z and of course it stops uh well it continues for all i but of course if i is bigger than k then they're they're zero because the homology group is zero so you have these classes and they're given by the theory of characteristic classes they're computable if you know enough about the manifold and now here's through gifts a formula so these pi's belong to p and what you do is you take the total turn class the total Pontriagin classes called the Pontriagin classes they're very spellings depending how you transcribe Pontriagin into English uh p of m you write as one plus p1 sometimes people put an x there to remind you of the dimension you don't really have to i'll drop it if it helps you i can put it back in i don't have to put it it's a generating series but since p1 is in dimension four p2 and dimension eight they're separated all by themselves but if you like you can put p1 x p2 x squared and so on maybe it is helpful just remind you that x to the i means that you're in dimension four i but the pi tells you and now you write that formally as a product of some numbers alpha new where alpha new is four dimensional but actually it turns out that the formulas will be easier if i call it alpha new squared this is just formally now you can't actually factor this thing in the homology just like you can't factor a polynomial over q until then your factors but you can pretend you can factor you can imagine the formal roots and call them even the square roots of the formal roots called them alpha new squared alpha new the square roots so if i do this then we know from the theory of symmetry polynomials that any polynomial in the alpha news which is even and therefore a polynomial in the alpha new squared and which happens to be symmetric can always be written as a polynomial in the elementary symmetry polynomials of the alpha new squared but those are exactly the pi because you see p1 is the sum of the alpha new squared p2 is the sum new less than u prime of alpha new squared of new prime squared etc but these are the elementary symmetric polynomials and we know from elementary algebra that any polynomial with coefficients in z even, which is symmetric in the alpha i is nu squared, can be re-expressed as a polynomial in these elementary symmetric things, and that makes sense. So if I do anything symmetric with these alpha nu's, the final answer makes sense even though the alpha nu don't exist. And now what you do, that's the famous Heusbruch formula, is you take the product alpha nu divided by hyperbolic tangent of alpha nu. I think it's hyperbolic, sometimes it's ordinary tangent, depending if I put alpha nu squared or minus alpha nu squared, and I can never remember which sign you have to take to get the correct class. Actually it doesn't matter for anything I'm doing, it's, I had it written down and it just disappeared, it's hyperbolic tangent, the way I wrote it. So that reminds you that alpha over tangent alpha tangent, hyperbolic tangent of alpha, which is the quotient of sinh over cosh, is an odd power series. So therefore alpha over that is an even power series, so it's a power series in alpha squared and it starts with plus alpha squared over 3 minus alpha to the fourth plus 2 alpha cubed over 945 plus, and the denominators of these, these are essentially what Euler found, it was mentioned in the previous lecture, what Euler found in 1834, he found that for instance, zeta of 6 is pi to the sixth divided by 945, which is half of this number and similarly zeta of 4 is pi to the fourth over 90, which is half of this number and so on. So Euler found in 1734, a long time ago, almost 300 years ago, almost 300 years ago, that the values of the zeta function are proportional to the coefficients of tangent and these numbers up to a scaling factor called Bernoulli numbers, I'll write it down in a second. So that's there through a signature theorem, so you do this very odd procedure, you take the Pontragon classes, which are things you know or can calculate for any given manifold, they're integral classes, but now we work over cubed because this thing is full of denominators, we write P formally as the product 1 plus alpha nu squared and then we take this product and so if you do this, then you see that this is the product by what I just told you of 1 plus alpha nu squared over 3 minus alpha nu to the fourth over 45 and so if you multiply that out, you see it's 1 plus the sum of the alpha nu squared but that's P1 over 3 and the next one I don't remember but the part is 7 P2 minus P1 squared over 45 and the next one is 62 times P3 minus 13 times P1 times P2, so in general you get all partitions of i for the i-class, so that's the beginning of the expansion, so the L series is a sum, it's called a piece in dimension in degree 0, 4, 8, 16 and for this, this is simply this is equal to the 4k dimensional piece because that's the dimension of this manifold piece of the L-class, so although it looks very abstract now to break at the end of the day, if you have a 12 manifold, 12 is 4 times 3 so I have to look at the third L-class, I look at this combination of Pound-Reyen class evaluated on the manifold, that's an integer because they're integer classes and then by the Hertzberg signature there, that integer will be divisible by 945 and if I divide it, the number I get will be the signature of my manifold, okay, so it's kind of an amazing thing, so now I was going to actually give a couple of mini proofs that the number theory part of the time is running out so I'll skip it but the things I say are extremely elementary I mean to actually prove and actually the next thing I say is going to be in is already in Hertzberg's habitation, the famous topological methods at the foundations of algebraic geometry or inalgebraic geometry, so let me just tell you briefly, I hope everybody knows at least the existence, that I think I once did give a basic notions talk on I think, but at least the word and maybe roughly the definition of the famous Bernoulli numbers, maybe the most important thing to know about Bernoulli numbers is they discovered plus or minus a couple of years in 1742 but not by Bernoulli but actually by Seiki, independently Seiki Takakauzu is a very important Japanese mathematician that even many Japanese mathematicians now have forgotten because somehow when Western mathematics came to Japan at the end of the Edo period then they purposely the government kind of decreed that all previous Japanese mathematician should be forgotten and not taught in the schools because they hadn't helped them to make as good candidates as the Americans had and so they stopped doing Japanese mathematics was and it's a big pity and now many young Japanese mathematicians don't even know their own heroes. Seiki was somebody discovered many many things that were discovered sometimes a little earlier quite often a bit later at the same time in the west also student Takebi both were absolutely great mathematicians for instance he invented determinants independently of Leibniz at the same time possibly a year too earlier and he invented Bernoulli numbers at the same time roughly as Bernoulli for the same reason anyway whoever invented them here they are well they're they're all zero for even insist except b1 which is no interest so I can put down a few and it starts one one six this is related to this age of minus one that we saw in in a teacher's lecture on quantum field theory or not with quantum physics for mathematicians then it's minus a third if they alternating sign 691 over I've forgotten this one is 76 this is a minus sign it's I forget the denominator but any of the denominator is not the interest it's a product I think it's 3072 or so it's some stupid number that I've written down right here but the interesting the thing is the numerator Seiki gave a little table of the Bernoulli numbers when he discovered them and he stopped before this 691 he gave them up to the one before he even gave all the zeros so if you put them all in in Seiki's works you you find even the zeros but he stopped at 11 and I think it's not a coincidence well his page was he was kind of at the end of the page there wouldn't have been room for another row but actually I believe because he was doing it using Pascal's triangle but I believe that he calculated b12 found the 691 and since all the others if you divide by k which is more natural have no numerator he assumed he had made a mistake and didn't put it in his book so he didn't actually tell us the 691 but I don't think it's a coincidence that he stopped just short of the first interesting one anyway these are the famous Bernoulli numbers and they're definable in many different ways one of which is at the same time the theorem that anyway I need which that alpha over tangent of alpha is the sum well it's only the even ones because I said the opens are zero anyway except b1 which we're not interested in okay so that's the relation to Bernoulli numbers and there's also a relation the log of this thing if you take the log of this thing then the coefficients is a very easy form there are also multiples of the Bernoulli number now involving not just a power of two but also power of two times a power of two minus one so it's slightly different and using that you can easily prove what I'm now going to tell you which is that if you take l of m remember in dimension in every dimension it's a polynomial in the p's where the total degree is the degree we're looking at like here degree three it's actually degree 12 because pi is degree four i so this is the 12 dimensional piece so you can ask what's the linear part so I can ask how about when does pk I mean every pk will occur because every partition like the partitions three one plus two and one plus one three they all occur with some coefficient so I can start with the triple partition just k gets partitioned as k so let's call that coefficient sk it's some rational number and similarly another one I'll be interested in that's called the tk is pk squared but of course there are many others there's pk times pl and there's pk cubed and so on but then we just define right give notations for those two numbers okay so let me call them sk and tk tk and the formula is for them which you get easily from the formula that I didn't show you for the log so this is an exercise is sk is 2 to the 2k times 2 to the 2k minus 1 minus 1 I told you there's a power of 2 minus 1 over 2k factorial times b2k actually the absolute value but as I told you they alternate in science I also put minus 1 to the k plus 1 it would be the same that's sk and tk therefore it's a more complicated number but the formula is simpler is it's sk squared minus s2k over 2 so you can write it down shortly briefly but then you have to put in the formulas for the s's so that's very easy and the first formula at least is given here it's broken both are very easy to prove oops I'm not allowed to do that because there's a microphone in it won't work so we have these numbers so now let's look 4k manifold and then I think I'll more or less stop and not tell you although it's just when it's beginning to get fun so we look at the park race series of our manifold let m be my rational project of plane so we already know I told that the dimension has to be well 4k but very quickly it'll be k will have to be even that's the theorem I already mentioned so in dimension 2k the comology well it's one dimensional that's what I'm assuming and it's actually it's ignoring torsion if I think there is even no torsion in this case it's just generated by some class x maybe plus torsion then h4k is generated by a multiple of x squared and it's really x squared it's not 17 x squared or x squared over 5 that's why punk radiology x paired with itself has some multiple of x paired with some multiple has to give the unique generator in the only ways it's x and x squared okay and all the other comologies zero well h0 of course just generated by one so therefore the pantheon class is equal to simply one if k is odd then the only dimension that's a multiple of four is pk so it's simply sk times pk but if k is even let me call it k is 2l so n is 8l then p will be 1 plus sl times pl which I don't really care about and then it will be s2l times p2l plus tl times pl squared because there aren't any other pontriagin numbers sorry excuse me I'm talking nonsense I'm talking nonsense I'm talking complete nonsense p in the first case is 1 plus an integer times x because there isn't any times x squared there is no other comology class but in the second case since the middle dimension is maybe I should call this n since the middle dimension exists it is this form where n and n are integers and now if I work out the Headsbrook l class then by what I told you here in the first case it'll be 1 plus s maybe I'll do it the other way n is 4l in the other case n is 8k because I'm more used to the k and I'll make mistakes so in this case we have just pk and by what I just told you this will be sl times this number n times x squared but in the second case you'll get 1 plus sk times nx plus but now I'll have two contributions first of all I'll have s2k times the 2kth pontriagin class which is n but I'll also have I hope I didn't erase it as a tk times m squared times x squared and that's all there is because there isn't any other comology so the Headsbrook signature theorem will tell me that the signature is equal to in the first case sl times some integer and in the second case s2k times some integer plus tk times some integer squared but what is the signature well it's pretty easy the middle dimension is one-dimensional a one-dimensional quadratic form it's either plus or negative the signage is plus or minus one you can even make it one by reversing the orientation so it's one so we get this very strong theorem well you can already see that the first one is never going to work because sl you know it has to be in a sub-integer multiple of one and the s k's go very rapidly to infinity on top of their even this it means that l is one and k is four that's the thing I already told you in the case of an odd multiple of four you only can get dimension four that just drops out but in the second case you get this and so we have to be able to solve this and there are no obvious further restrictions on m and m it turns out there is a further re-restriction also coming from Headsbrook's work but it's something else he did it was the invention of topological k theory together with Artya and there's a topological version of k theoretical version of these theorems and that gives a certain restriction also involving the signature class but involving k theory and it involves instead of the e to the alpha you take e to the alpha minus 2 plus e to the minus alpha that's also an even power series in you and products of this and you do something I'm not going to go into it it gives you a slight extra restriction and it comes from work of Sullivan and of Hattoria and Ston what the exact restriction is but let's just stick to this main one which comes straight from the signature theorem and let me now just tell you in a couple of words what this leads to well the first point is this the Bernoulli number is always an odd number divided by two times an odd number actually there's a complete formula very famous the Von Schleil-Klausen formula for the entire denominator of pk it's a product of certain primes to the of bk over k even or no bk it's a product of certain primes but for the prime two it just occurs once so here the power of two this is a one two and the denominator this is no twos at all here the number of twos is a famous four here the number of twos is 2k and here the number of twos is a famous formula which is roughly 2k or it's exactly 2k minus the number of ones in the binary expansion of k so here i've one two too many but here i've one fewer so i've a two but there's a two in the denominator so this thing is odd if and only if k has only one power in its binary expansion which means it's power of two and it's two month four if and only if k is a power a sum of two different powers of two and otherwise it's zero month four so in other words just very trivially these sk's are always divisible by uh by two except if k is the power of two and so now if you look at this then you see that to make this one that both numbers can't be odd i mean if s2k and tk they can't both be even then i'll never get one sk is essentially always even but sk squared is divisible by four so when you divide by two it's still even but s2k maybe just barely even i think i'm off by one so what you find is that this implies too adequately already that k is a power of two or a sum of two powers of two which actually includes this case if you love them to be the same power of two so that's an extremely strong restriction they're only about 120 numbers up to a million that are sums at most two powers of two so already my list that i reduced from a million to 19 i've already reduced from a million cases to about 120 and that's you know my computer only had to look at those 120 and is now eliminated five six at them roughly so that's very easy and by the way this you can give a complete theorem that i should say there exists a manifold with petty numbers one all zeros until the middle dimension n over two zero zero and now i don't say what it is i just put odd then the same argument will tell me i have to work a little well i have to work harder but if i get got it right this is true now if and only if n is maybe 4k and k is the sum of two powers at most two powers of two roughly so if you don't ask to have a rational projected plane with the one in the middle but just anything odd but also take care of the zeros you just having an odd number at all in the middle that will happen if i'm not sure enough i'm getting it right only if it's a sum of two powers of two so there are other cases than the rational project of planes so this the first condition but now there are two more conditions because there are two other things that could go wrong for a prime so the conditions on k and that's what makes it so restrictive and then i'll stop the conditions on k remember n is 8k and i'm assuming there is a rational project of plane so the first one already told you that's already extremely restrictive which is a k or n is a product of only two powers of two or one power of two the second one is not restrictive at all so far as i know it says that b2k and b4k or their numerators are co-prime because if some prime number some large all prime divided b2k it would be an sk so it would be here if it also divided b 4k would be an s2k and so we divide both of these and therefore p would divide both s2k and tk and then that's impossible because then every linear combination would be a multiple of b that can so as i said this is extremely restrictive 10 to the 6th goes down to about 120 or 150 this is not restrictive at all if i just looked at the 120 they don't they're also exactly 120 they don't change it never happens that these numbers are already of a very thin set but even if i look at all numbers i've gone up to about 500,000 it took days on the computer i've never found a case where a bernouy number and the bernouy number twice the index had any factor in common never found one however i conjecture that they're infinitely many and there's an easy reason which i won't tell unless anyone asks so question can you find any pair of bernouy numbers well it has to even index so 2k and 4k which are both divisible by the same prime does that ever happen it does not happen up to you know several hundred thousand but like i'm sure that it happens infinitely often but extremely rarely but so rarely that it intersects with this other very thin set of sums of two powers of two is undoubtedly zero so i'm absolutely convinced that there will be never a case where you pass test one and you need test two but still i have to add it to the list because we don't know but there's one other thing which is slightly it's a little hidden but you actually can see it is somebody in the room really good with number theory and can just see immediately what there's one last condition in order to be able to solve this equation so i've already said we have to be able to solve since the signature is one i have to be able i think i already wrote it i have to be able to solve the equation tk m squared plus s2kn equals one tk and s2k i told you what they were and m and n are just some integers so we don't know anything about what is the condition well this appear the condition this is just a congruence because since we don't know anything about n it just as the tk times the square is congruent to one multiple s2k right it has nothing to do with these numbers it's just a multiple so it's a congruence condition it's true about the chinese remainder theorem even only if it's true for every prime that divides s2k if p doesn't divide the numerator of s2k even worse at p's of the denominator there's no restriction at all one case is what if a prime divided both s2k and t2k then it couldn't work well that prime would either be two that's how i got the condition on the powers of two or it could be not prime that's why i got the co-primality but there's one other thing this tk is not a random number it is equal to sk squared minus s2k over two but since i've already taken care of the tuatic stuff to ask this periodically i can multiply it by two i haven't lost anything and so i want this model to s2k but now we see that t2k times two is minus a square except i want plus a square no it's plus a square not the s2k tk times two is sk squared so what i get is that there's a square which is two although it's 2k that is not possible for every number because this is possible for a given number s that two is a square not that number again but the chinese remainder theorem if and only if for all p dividing my number you can have a square which is congruent to two mod p but that's by gals is the same as saying that p is congruent to one or to seven mod 8 and therefore we have the last criterion which is that all p dividing s2k are congruent to plus or minus one mod p so now let me go back to the theorem with that long list which i think have erased and let me just show you a wave in the air my handwritten table of factorizations and the top factorization maybe you can see even there there are three lines of very long numbers that i caught by hand 50 digit numbers that's the factorization of the renewing number that i need for 32 remember that was one of the dimension the last dimensions eight times 32 256 that number is one two three four five six primes the last one is one one one one two nine one six one six seven nine eight i've read about a tenth of it it's an 80 digit prime but still it might be one or seven mod 8 the chance are 50 50 since this number have six primes the chance was one in 64 that it would pass all six tests but a small miracle happened one chance in 64 that particular number so the b of four times 32 i think it was the numerator is a product of six primes but they're all congruent just by chance to plus or minus one mod 8 so there's no problem and since that's the only necessary condition it works now when you get to 68 it doesn't sound rather harmless but it's b four times 68 it's a 300 digit number it is one small prime but it's it's not a useful one it works and then there's this huge number and the computer worked for weeks and weeks and weeks and it's using an algorithm based on the theory of elliptic curves that can almost guarantee that any prime small prime factors meaning up to 10 to the 25 would have been found but unfortunately this number is 300 digits and so if it's a product of two primes they might have 150 digits for product of three it might be 100 digits for product of four they might be 80 digits so they're much bigger and we haven't found any of them so if it is three primes well the number itself is one mod 8 or i wouldn't be testing it so if there are three primes there's one chance in four that it works if there are seven primes which is more likely then there's one chance in 100 if there are 20 primes that's impossible because then it would have to have some small prime factors we've already checked so that number has a few primes we don't know how many probably somewhere between three and maybe even two but even if it is only two the chance is only 50 50 that it'll work but the next number in my list already is a thousand digit number that's going to have dozens of primes and the chance that they all work is essentially zero that's why i said that almost certainly we'll never have success again so i'm finished but i showed you that this top logical problem led to three necessary conditions one of which is very restricted but very easy some of two primes of two one is amusing and probably not restrictive at all but leads to a non-trivial problem that i don't know if it's ever been studied can two Bernoulli numbers whose indices differ by a factor of two have a common factor and the third leads to this weird problem of factorization of Bernoulli numbers that is almost certainly unsolvable and to the classification up to dimension a million so thank you