 All right. Well, welcome everybody to this week's number three web seminar on behalf of Alina who's here and Philip who's unfortunately not joining us today. He's on vacation in the days of COVID. I don't quite know what that means, but I'd like to welcome you again. The usual disclaimer. So if you have any questions, please remain muted, but raise your hand or ask the question in the chat. Maybe some kind person will answer your question quickly in the chat. Otherwise, I will take it upon myself to interrupt Bjorn and ask away for you. Otherwise, please, please enjoy what's coming up without further ado. It gives me great pleasure to introduce Bjorn Poonin who will tell us about tetrahedra with rational dihedral angles. Bjorn. Okay. Thank you very much. So let me first mention Hilbert's third problem. So Hilbert around 1900 he asked whether you could take a regular tetrahedron and slice it into with with finally many planes and then reassemble the pieces into a queue. And it was discovered by Dane almost actually almost immediately as soon as he asked as soon as Hilbert asked the question that that this was not possible. And the way he did it was by introducing this invariant of a well it applies not just a tetrahedra but any polyhedra. So if you have a polyhedron inside inside three dimensional space. You can define the state invariant which which takes values in this tensor product group. And what you do is you for each edge of the polyhedron, you, you take the length, and you take the dihedral angle, which is the angle formed by the two, the two faces that meet along that edge. And if you take that tensor and you add them up, then you can it's pretty easy to check that that's that's invariant. It's it doesn't change if you if you if you if you subdivide your polyhedron to pieces. And so it must be preserved just like just as volume is preserved. If you slice something in pieces, you don't change the volume. And, and what so again and then, and then conversely and for the conversely, I said they're approved, is that you can tell whether to, whether to polyhedra can be cut up and reassembled into the other one. And the, and the, that'll, that'll be true if and only if they have the same volume, and they have the same data. Okay, so now one thing I want to mention in particular, because this angle is being taken as an element of our mod, the rational multiples of pi. If the, if, if you have a polyhedron for which all the dihedral angles are rational multiples of pi, and these angles will be zero in this group, and therefore these tensors will be zero. And so then the whole data invariant will be zero. So that would mean that that if you have such a polyhedron that means you can, you can slice it and reassemble it into a cube, because a cube is another, another, another polyhedron whose data invariant is zero. Okay, so, so now let me talk, not only specialize the case of tetrahedra. And so I'm going to talk about several, several properties that a tetrahedron could have. So one is, well, if it, as I was just saying, if you're in this bubble, if you have, if you have, if it has rational dihedral angles, then it must have gain invariant zero and therefore you can cut it up and make it equivalent to a cube. Another class of such, of such things are the ones, are the tetrahedra that can, that can tile three dimensional space. So if you have, if you have a tetrahedron and you can take instantly many other tetrahedra congruent to that one, and, and then, you know, cover our cover the cover three dimensional space where the tetrahedra meet only along the phases. Then, then, yeah, then, then, then it's a, it's a theorem proved by de Brunner. Well, in general, it is proved by de Brunner only in 1980, that these, those tetrahedra also have gain invariant zero. So they're also decomposable this. Okay, so the question is, can you figure out what, what tetrahedra have these properties. So this, this problem had a long history. Well, starting with Aristotle, who's who believed that the regular tetrahedron could tile RQ. And this is actually wrong. So, although it took, it took apparently almost 1800 years for for this error to be discovered. I don't know why it took so long but anyway, I guess it's not maybe it's not so easy to check. Well anyway, it's anyway so it's just, it was. Yeah, so this. So, Johannes Muller. So it's also known as Reggio Montanus. Anyway, he figured out that this is wrong. And yeah, and then, I guess, maybe nothing much happened until in 1895, he'll constructed a bunch of a bunch of while he constructed some isolated, some isolated tetrahedra, but also a couple, three one parameter families of tetrahedra. And he proved, well, actually he didn't prove that they tiled RQ, but he proved that they are, they have gain invariant zero that they're, and he proved that they could be reassembled into a Q. Okay, so actually, yeah, this, this, maybe this diagram is not completely precise because it doesn't show like what, I mean he, yeah, he proved that as I said, he didn't quite, it was only later that these were proved to also tile RQ. Okay, so, and then, okay, so then, then invented this gain invariant, and so now we know it's regular polyhedron action, the regular tetrahedron actually belongs out here. All right, and then, some of them found another, another, another tetrahedron that improved that this one tiled, tiled space, actually proved that all five of these, these dots actually tiled. And then nothing had much happened for another couple of decades, but then around the 1950s, 1960s, 1970s, people started finding, so these are in Goldberg, they started constructing lots of examples of other tetrahedra that had rational dihedral angles. Or that, or that otherwise were had gain invariant zero. Yeah, and by the way, these dots are not just, I mean, I just didn't, I mean, these dots actually are the correct number. This is actually how many they discovered, so. Okay, so, okay, so then, a subset of, oh, I should, maybe I should mention this is, I guess it was shown on my, my title slide, but this is all joint work with Geron Kedlaya, Alexander Kolplikov, and Michael Rubinstein. So, okay, so, so we started working on this problem, or the some of us started working on this problem back in 1995. And we found a, just by, by, by computer search, we found a lot of, a lot of new examples of rational dihedral angles, and this is how many we found. And we also found one new one parameter family of such things. Okay, and then, but we kept searching, we didn't find any more. So we were wondering whether these were all of them. And this is, this is what we finally proved. And while we also constructed 10 more family, one parameter families of things that are gain invariant zero. And yeah, so, so, and so this is what, so this is, it's this theorem I want to explain, it's the complete classification of all the, all the tetrahedra, who's such that all six dihedral angles are rational, are rational multiples of pi. Okay, so, all right, so let's get going. So here, okay, here, here's the, here's the final theorem. So I don't know if you're, well, if you're really good at counting dots, you will notice that there are 59 dots in this, in the circle. And so they're, so they're, they're, they're those two one parameter families, one discovered by Hill and then one discovered by us, and then, and these are the six dihedral angles. And this is depending on this parameter and so and so is this one. And then outside of these two, there are 59 sporadic ones. And so here's some examples of sporadic tetrahedra given by their dihedral angles. So this one involving multiple pi over seven and pi over three and multiple pi over 60. Okay, so, so how do you, okay, so, and oh, here's the complete, here's the complete list of all 59 tetrahedra. You don't, you don't want to really read this. I've, we've written them in a way so that I'm just going to fit on the slide, you're supposed to take each of these sequence of integers and multiply it by pi divided by this number. So, so this, this first one, for example, they're all multiples of pi over 12. And, okay, and so on. Okay, so. All right, so how do you prove something like this. So, well, first of all, let me say, I mean, why is this an interesting problem to begin with I mean, why can't you just specify six angles, and just find a tetrahedra with those six angles. And if there's sort of an obvious reason why you can't do that is that they're not just not enough parameters. If you think about how many parameters you need to describe a tetrahedron. They're unique. One way to describe tetrahedron is to up to congruence is to give the six edge lengths. They're subject to some inequalities like triangle inequality stuff like that. But yeah, so, so, so you might think there's a six dimensional space of tetrahedra, and also a six dimensional space of diagonal angles. But I mean, if you have, if you if you take, if you have similar tetrahedra, if you just scale all the edge lengths by, by a number, then you, you get the same angle. So, so actually, so actually if you're thinking about tetrahedra up to similarity, there's only only a five dimensional space of those. And so they they satisfy. Yeah, inside, you can think of it and project the six tuple of edge lengths in here, what I mean, then, and then there's some inequality, so it's an open subset there. And similarly, the, the space of possible angle tuples. I mean, the space of all angle tuples will be six dimensional, but because it, the ones that are actually realizable by tetrahedra can only be at most five dimensional because it's the image of this five So that means that if it, if the image is only five dimensional, that means there has to be some, and this some analytic map, then the, the, there has to be some I mean, there has to be some relation that these that six angles have to satisfy in order to come from an actual tetrahedron. And so, so our first task is to actually figure out what is this equation, what is the, what is the equation involving the six diagonal angles that has to be satisfied in order for it to actually come from a tetrahedron. All right, so, so here's how you can get started with this. So if you have a tetrahedron, you can consider the four outward unit normals. So you build an outward, the outward unit normal on each face. You can get four vectors. And instead of considering the diagonal angles you could consider the angles between formed by these, these outward unit normals. And if you do a little bit of geometry, you can easily see that those angles, but like the angle between V1 and V2 here is going to be Pi minus the dihedral angle along this edge. So yeah, so this is the dihedral angle and these are these angles between these vectors. Now you can, so using that you can restate the problem or you, well it's almost a restatement that if you're trying to classify these dihedral, instead of doing that. By the way, there's somebody in the waiting room. I don't know if you want to let them in. Yeah, this is for the host. Okay, anyway. So yeah, so the tetrahedra, these classifying these tetrahedra dihedral angles is essentially the same problem as classifying four tuples of vectors in R3. And for it to be non-degenerate, you want to make sure that no three of them are dependent. And then, and those angles should form, should have the rational multiples of Pi. Now going backwards, it's not, we can't quite go backwards. There's an extra condition that you need on those four vectors for them to come to a tetrahedron. If they have to be, if you have four-hour unit normals, the zero has to be a positive, a non-negative combination or a positive combination of those four vectors. So zero has to be sort of in the convex hull of those four vectors. But if you have four vectors that are, that no three are dependent and they don't satisfy that property, well, there's going to be some linear relation that these have satisfied because they're four vectors in a three-dimensional space. And so whatever that linear relation is, all you have to do is negate some of the Vs, and then you can make it so that the relation they satisfy will then have positive coefficients, and then they will come from a tetrahedra. So there's really almost no difference between these two problems. And so you can, so for now, and I think I'll talk about this, the second problem of classifying these four tuples, or four element subsets of non-zero vectors in R3, such that no three are in a two-dimensional subspace. And yeah, and the key condition is that every pair of them, the six angles formed are all rational multiples of Pi. Okay. All right, so now let's get the equation. So the way to get the equation is by considering the gram matrix. So the gram matrix is the four by four matrix of all the dot products between these vectors. And because they're unit vectors, these dot products are just going to be the cosines of the angles. So you get this, you get this kind of matrix. But these are, as I said before, these are four vectors in R cubed. So they're linearly dependent. And whatever that linear dependence is, it'll also be a dependence between the columns of this matrix. So that tells you that this matrix, if the columns are dependent, that means the determinant must be zero. And that's an, if you expand that out in terms of these cosines, that's an equation in all, in the six angles. And that's the equation that the six angles must satisfy in order to come from, from these four vectors or to come from tetrahedron. Yeah, and there are also, I mean, there are also some inequalities, but the main point is this equation. Okay, so all we have to do is solve this equation, set this determinant equals zero and find all the solutions where all six angles are rational multiples of Pi. Yeah, okay. Now as, okay, so the first thing, first step to do that in doing this is, well, everybody knows that if, if, at least in pure math, if you ever see a cosine or trig functions, things become better if you express it in terms of complex exponentials. But let's first do that. So let's take each of these cosine betas and use, use this identity for cosine data and to express it in terms of E the i theta. And the nice thing about that is that if, if theta is a rational multiple of Pi, then this will be a root of unity. And so you can re express this equation then in terms of all these roots of unity. Okay, I guess some obnoxious person is probably going to complain that I'm reusing i, using i to mean two different things here in the same, in the same equation, but okay, we can deal with that. All right, anyway, so, okay, so when you do this, you expand it, you, you plug, you plug this in each cosine becomes like a z plus the inverse. Well, over two, and then you expand it out and you maybe multiply by 16 to get rid of the twos, and you get this nice equation here. Okay, it's not as nice as it looks because if you, I mean, each of these terms is really a sum, where you sum not only over all the possible sine choices, independent sine choices for the x-men, but you also apply, you take the orbit of the S4, this, you permute the four coordinates in all possible ways. So for example, this one, they're actually, this, this, this sum here actually has 48 terms in it because there are three possible, the orbit of this under S4 is three but then you have three possible sine choices. This is actually 48 terms. And then it's one, yeah, and so on. And so all together this, this term has 105 monomials, and, and, and six variables. Yeah, okay, so now there's a long history of people solving root, solving polynomial equations in root community. And yeah, so, and yes, in fact, doing, using this to solve geometry problems where you have rational multiples of pi. So let me, let me now try to say what is known about this kind of problem. And so this, so let me say more in some more algebraic geometry terms. So you can, you can take the algebraic group whose, whose group of points is, is, is C star to the N. Yeah, so GM by itself without the N is the stands for multiplicative group. That's a multiplicative group of complex numbers, just viewed as a variety as an algebraic group. And then this is just the N power of that. So that's what's called an algebraic torus. And then if you have an algebraic subgroup in there. Like, like, for example, in GM squared, you could take the set of x comma y that satisfy x cube equals y to the fifth. And that the set of pairs that satisfy that would be a subgroup under pair y under coordinate wise multiplication. So that would be an example of a sub torus. And you could also, and such sub torus are also going to have lots of solutions. They're going to have lots of those kind of equations like, like, like y cube equals x to the fifth that's going to have lots of solutions and roots of unity. And it was the same will be true if you take a coset of one of these subgroups, take a coset as long as you take a coset where you multiply each coordinate by another by a root of unity. And the other called torsion cosets because you're you're multiplying the subgroup by an element of fine order in the in the group. And so those torsion cosets will also have a dense is risky dense set of points whose coordinates of roots of unity. And the and the big theorem is that whenever if you have any sub variety. And if you think of a given by some polynomial equation inside here, then if you want all the points on that variety that have coordinates that are roots of unity, then they'll all be explained by these torsion cosets. They will all all those solutions will be contained in a finite union of torsion torsion cosets that are each of which is contained in X. And so concretely in terms of equations with this tells you is that whenever you have a polynomial equation to be solved and roots of unity. Then the solution set is going to look like it's going to have a nice fine description it'll have. There'll be maybe some isolated sporadic examples but and then there might be some, some one parameter families and a couple of two parameter families, but they'll only be finally made families. And so, in principle, it's not a surprise that what I showed you on that in that Venn diagram, what that or that in a theorem after that that that they're going to be that there'll be finally many parameterized families and then a couple of isolated examples. We know in principle that it's going to look like that of course the problem is actually figure out what those what those families are and how do you know they're only 59 sporadic ones. And then actually a lot of people have worked on this. I know I'm admitting some names here so but well, I mean there's so many people have worked on it. Yeah, anyway, I, I, I, I'm there are probably people in this audience I think who have worked on this to that I, that I, I apologize to all of you in advance but so the lot of people have worked on this. But I want to explain to you now why none of the, as far as I can tell, none of these methods that have been developed will apply directly to the equation and we want to do. So here's, here's why so Oh, here, let me, let me tell you about two of these methods for solving these equations by. Let me explain this first method, which goes back to serentate and was developed more by these, these authors. So, as a warm up let's consider just to solve solving an equation two variables, where, in fact, let's not look at all roots of any solutions but just where the roots of any or both of odd order. So if you have an equation like that where F is a polynomial with rational coefficients, what you can do is you can apply a Gala automorphism to your solution. And there's going to be some Gala automorphism that squares your odd, your roots of unity that sends each root of unity or odd order to its square. And so if you apply that, then, then from the fact that f of x y equals zero, you'll get that f of x squared comma y squared is also zero. Yeah, then it's, it's, it's not hard to figure out what x and y have to be because these are not too independent. Well, maybe they're independent, but most of the time they will be independent. You have two, two polynomial equations and two variables. And so you can just solve them using resultants or whatever or elimination theory. And so you can just solve them and you'll get finally many intersection points and that'll tell you, and then the solutions must be among those. And so that's an easy way to get these solutions. At least when x and y are roots of unity of odd order. And there's a variant of this, it also works if one of the roots of unity has even order. And so that actually works very well in practice. And in principle, you can do this in higher dimensions too by doing some kind of inductive argument on the dimensions and number of variables. So, and this, as I said, this works very well when you have two variables. It works. It kind of works when there are three variables. And for four variables, it seems that it's usually impractical. It just, the computer won't, often won't stop. And we need to do it for six variables. So, the other approach is to, it depends on the number of monomials. So what the other approach is to classify all the ways you can have. Because each monomial in the polynomial represents a root of unity, because it's a product of powers of roots of unity. And so you can try to classify all the ways that m roots of unity can sum to zero. And so, and this classification was completed by various authors doing more and more with larger and larger values now. I mean, it gets, it gets more and more painful as you go up and 12 is already painful enough that the only reason that we did this is that we had a nice application in mind. And, well, we just, we just need to go increase this from 12 to 105. I mean, this, this, I mean, this, I mean, this, this approach, I mean, the complexity of this grows, I think probably grows faster than exponentially. And I think same, same for this one. So, yeah, so it's not looking good. So, and also that it's not also our problem is not just a sum of 105. They're also coefficients involved. But anyway, minor, minor problem. So just to give you a sense of what's involved here, here are all the list of ways that 12 roots of unity. I'm not going to explain everything on this slide. But in fact, these are not all the ways that 12 roots of unity can sum up to zero. You can you can break some of the rays of the 12 roots of unity. Some to zero, you can decompose into, you can break them into subsets that separately some to zero. So what I'm showing here are the indie composable ones. And, and the number of most of these ways are actually they're not this, they're a lot, they're all together they're about 107. I should know there are 107 ways. There are 107 of these relations, just going just up to 12 roots of unity, something to zero. And so it's going to get worse as this increases. In fact, yeah, okay, so obviously we didn't do that. Alright, so I'm just to show you why we did this. I mean this is sort of a side issue but why, why did we want to know this. This came up in another geometry problem, which was trying to classify, try to count the number of intersection points formed by the diagonals of a regular end gone as a function and so this is this is the picture for n equals 30. So yeah, you're just trying to count. So there's one intersection here, there's another one here. And the problem is to get a formula for the intersection point as tricky because I mean you might think oh you just take each pair of any four points, any four vertices. We'll determine a unique pair of intersecting diagonals that that's true, but the problem is that those entries for intersection points, some of them coincide like a lot of them coincide here. And, but there are other points where they coincide so you can see that here there's their points were more than two diagonals intersect. And so that some of the intersection points are like I'm getting this is this is not the problem I talked about. But that's what we did in 1995, but that was only involving a sum of 12 roots of unity. And so, going back to our problem, we have this sum of well integer combination of 105 roots of unity, and we want to solve it. So, well, can you guess what we do. So there's something special about this polynomial that we use. Nobody wants to guess. Okay, well I'll tell you that. So the good the trick is to reduce mod two. The point is that most of these coefficients are even. And so if you reduce this equation mod two, you get an equation that has only 12 roots of unity in it. You get an equation and it's just, you just get a sum of 12 roots really well it's not equal to zero anymore, but it's congruent to zero mod two. And when I say my two, I really mean modulo the ideal generated by two in this in the big cyclotonic ring. So this is the ring generated by all roots of unity. And so we're working module to so I mean to is not a prime ideal in this ring because this is ramified at two. But anyway, you can still work module to and. And so, but so what we did is we, we were able to take what we did back in 1995 and not not just classify the way the 12 roots really can sum to zero exactly, but the ways that 12 roots really can sum to zero my two and turn actually turns out the classification is basically the same. And so, and then once you do that, then that gives that with those ways of that those consumption zero that gives you constraints on these six roots of unity. And when you, and when you compose those constraints, you no longer have six variables, you just get lots of families where each family is parametrized by three variables at most. And so now you get a whole bunch of families, actually, a rather large number of families, each of which has only three variables in it. And then you take those parameterizations you plug them back into the original equation. Now you have these complicated polynomials, but they're only but now they're just complicated polynomials and three variables. And so now you can go back and use the seratate Booker Smith's alia method, and, and you can, you can use that inductive method to solve each of those. And then so that's what we did. And so it was, yeah, okay, it was still a lot of work, but it was sort of barely barely doable with with well with with enough computers. Okay. So, um, yeah, so again, so, so the final result is this one again, that we get those these, these two one parameter families, and 59 spread catch media. And as I was saying for this is a quote, this, this classification is equivalent to classifying these four vector configurations inside three dimensional space, where no three of them are dependent. And, and every and all the pairs of angles formed by these are rational multiples of pi. Okay. Yeah, don't, don't be hypnotized by this thing, but it's just, I know, I was having fun. So, okay, so, um, alright, so let, let me know. Um, consider a more general problem. I mean, suppose you get rid of the condition that's the three or that this condition about them being in general linear position, you can try to classify more generally for vectors and are three. Again, but in the key condition still is that the angles formed are in rational pies of rational multiple pie. So now they're a lot more, they're a lot more possibilities. And you might, by the way, you might as well consider the vectors. I mean, it doesn't matter if you take a scale of multiple one, because that's not going to change the angles. I mean, even, I mean, if you take, even if you take a negative multiple of a vector. I mean that will change the angles, but it'll only change the angle, some of the angles will change to pi minus the original angle. So it's not a big deal. So let's consider vectors up to scale in multiple, up to real scale in multiple. And then, yeah, and there's no point in having two vectors be the same because obviously, anyway, it doesn't matter. So that's not interesting. So if you, if you, so up to those, if you, if you, so if you consider these conditions, then you can classify these, and you get the same kind of prime tradition, except now there's, well, there's some higher dimensional families, there's three primary family. Actually, can you guess what that three primary family is? It's a really stupid family. You just take four vectors in a plane. And then if you, if you make the, if you make the three angles, rational multiple of pi, then all, all of them will be rational multiple of pi. So, so, and then up to, up to, up to, up to congruence, they're all there to three primary families because you basically get to specify these three, these three angles. Yeah, anyway, so there are a couple of other two primary families and then you get a whole lot, a lot more sporadic solutions, but, but you can, but you can classify all those as well. All right, so now what I want to do is ask, well, this was just for four vectors in space. Suppose you have five vectors in space. Can you, and suppose you have, and such that every pair of those forms a rational multiple of pi at an angle. Can you classify those as well? So you can, you can try to take the, all the angles. I mean, if you have five vectors, now you have, I guess, 10 angles need, and so it's actually quite, it's quite a strong constraint now because you have 10 angles and all 10 of the angles need to be rational multiples of pi. And you can, you can understand these by, in terms of the previous classification because of this, there's a, there's a nice fact here that it's not, not too hard to prove that if you haven't, if you, if you try to specify those 10 angles. So you can put them in if you want into this symmetric matrix. Then that will be realizable by some configuration vectors in space, if and only if every four vectors, every four elements subset of those is realizable. So for example, I mean here, here you can, you can impose a condition that these four vectors have to form rational angles and these four vectors have to form rational angles, and these four vectors rank rational angles. Yeah, so, so you can imagine now how you might solve this problem because you, you would just start with a four vector configuration, where they, where they form rational angles, and then see if you can stick a fifth vector in there somewhere. And the constraint that that fifth vector has to satisfy is that it together with any three of the others has to be something that's in your, that's in your, that's in your classification. And so, so it's a, it's a big combinatorial problem. I mean it's a little bit tricky because these, you have to, I mean, you could, there are a lot of, I mean, not when I can sit when I say that for example at 425 sporadic ones. That's really up to congruence, and that's ignoring the order, that's not, that's not, that's not fixing the order of the vertices. So if you actually want to have fixed, if you want to think about sort of marked tetrahedra or marked four vector configuration or ordered ones, then you have to really multiply that number of this 425 by four factorial. Anyway, so it's a very large number of, and you also have to deal with the fact that when you're adding a new vector, it might, it's not just there, it's not just there finally many possibilities for the, for the how it interacts with the others ones. It could be a member of, of these, these parametrize families. So you have to, you have to keep track of all these parameters that come along as well. Hi, Bjorn, could I just interrupt you for a second? We have a question from Andrew Granville Andrew, could you unmute and ask your question, please? Yeah, when, when you're talking about these parameters and rephrase the problem in terms of vanishing sums of roots of unity, where are the parameters occurring? So the parameters, I mean, for the original problem of like 12 or 3 summing to zero, the, the roots of unity, there are the parameters themselves. These parameters sort of is a root of unity. So, I mean, but if, if, so, if you look at the solutions though, I mean, you can't, those aren't independent. So you won't have, I mean, you can't, it won't be a 12 dimensional family because, yeah, it will have to do with the vanishing subsums. And each time you have a vanishing subsum, you can rotate it relative to the other one. So for example, if you had a, if you had a, if you had a, if you had a, if you had a, there's a relation of weight of with eight roots of unity, where you have the five cube, you have the, no, not the five two roots of unity, you have the three cube roots of unity and the five fifth roots of unity. And you can sum those and that'll give you a, a relation with eight roots of unity, but you could also take the three cube roots of unity and multiply them by a seventh root of unity to shift it relative to the fifth root of unity. And so you'll get it like a one primary family that way. And so it's, it's by, it's by independently rotating the, the indie composable subsums. So it's, it's, it's what you, it's what they all come from. Is this where parameters come from? That's where the, at least up to 12, I think that's, yeah, that's where that, I'm not sure if that persists forever, but at least up to 12, that's true. Okay, thanks. Okay, any other questions. All right, so, okay, so I saw, I was describing this, this algorithm that we had in mind for, for, for passing from the classification of four vector configurations to try to get the five vector configurations and six vector configurations. And you would think that each time you stick another vector in there, it becomes harder and harder to get another vector in there because each time, I mean, I mean, for example, if you're up to 10 vectors, you've got 10 vectors in space. And you're trying to put 11th one in there. It's going to have to form rational angles with all 10 of those. That's pretty serious. Those are pretty serious constraints. So you, so we were expecting that this thing would sort of, as you started adding vectors of this thing would peer out pretty quickly. And we probably end up with, we'd have a complete classification. But actually what happened is that it just sort of kept going for a while. Well, there's one stupid reason why it keeps going because they could all be in a plane. You could have any number. You can have as many as you want in a plane and have them be at rational, rational angles. But even excluding those, we found we were, we were getting a lot of 10 vector configurations and they extend to 11 vector, 12 vector, 13 vector, and it kept going until we got, we found a 15 vector configuration and, and finally stop. And so, so the final result was, was, was this. I'm showing you, so I'm not, so we, we ended up classifying all the ways, all possible configurations of vectors in, in our cube of any number of vectors that that have such that all pairs of them are rational angles. And this is the result. So there's, I'm only showing you the maximum ones, the one, I mean, some of these, I mean, as I said, there are lots of 11 vector configurations, but they all extend they're all part of a larger configuration. So I'm just showing the, the, the max ones are maximal in terms of number of vectors in terms of not being able to extend. Okay, so all right, so this is the result, how many of them and there are a couple of, there are a couple of parameterized families and they're, these sort of isolated examples and they're so now the question is, what, what do you think these things are? And what are the, I mean, there's, there's obviously something special going on here. I mean, what, what is this 15 vector configuration? Well, and what's this, if not vector configured, well, you can probably guess with this one. Well, actually, maybe it's not, it's almost clear what this is, because somebody gets what this is. How can you fit alphanaut vectors in, in, in, in space, and such that every pair of them form a rational and multiple tie can happen. Okay, so you can put them all on a plane, but that's not actually quite maximal. If you put all of them in a plane, you can take, yeah, you take all the rational angles on a plane, but a rational multiple of a plane, but actually, and we're only considering up to, up to scale multiple, but if there's actually one more you can put, after putting all these things in a plane. I think we have one of you guessed it in the chat. Yeah, you can put one, yeah, you put one sticking out of the plane. It's perpendicular to all of them. So that, that's, that's the stupid configuration here. Yeah, that, that's, and that's maximal. And how about, how about this one? How can you still make guess what this one is? So I'm considering vectors up to scale multiple. So I don't, if I have a vector, of course you can put a negative in there too. So, but I'm not considering, I'm not counting that as an extra one. So, so anybody want to guess what this 15 vector configuration is? And we didn't actually guess this until we found that there was, it was popping out. We just saw, we just saw from, we just started drawing, I mean, we knew what the angles are, we just started drawing it on the sphere, see what we're getting. And also the nine vector vector configuration also is kind of special. And I'll talk about that one too. So if somebody's guessing a platonic solid, that's actually very close. You might think, oh, maybe we'll just take the, like the vertices of an icosahedron or something. But if you take the vertices of one of the platonic solids, it turns out that the only platonic solid for which those angles, the form by those vectors are a rational multiple of pi is the octahedron. And which is sort of, well, it's sort of obvious because those are like the coordinate axes, but the other ones don't, it turns out that you get things like cosine inverse of one third. And contrary to what many freshmen will tell you, there is, there's not a rational multiple of pi that is cosine is one third. So, yeah, so, ah, somebody's, so edges of icosahedron. Yeah, that, that's, that's, that's, that's right. So if you take the midpoints of the edges of icosahedron, there are 30 edges of icosahedron. And if you take their, the midpoints of those, and you, you can form, then you get this thing that's called the icosahedrodecahedron. That's formed out of, it's a polyhedron that's formed out of 20 triangles and 12, 12 pentagons and so, and so you get this thing. So if you take those vertices, the vertices of these things and, and take the vectors that, so let me blow this up by a factor of two and just show you the, and show you the, the vectors. So you can then you get these 30 vectors. And it's true that all 30, any pair of those 30 vectors forms an angle. That's a rational multiple pi. I mean, some of these are in planes like it turns out that they're inside this. There are six decagons inside this. I don't know, oh yeah, there we went. Okay, so you can, every once in a while you can see 10 of these vectors line up and, well, you can, the plane containing 10 of them lines up. But anyway, so yeah, so that's, so right, that's the 15 vector. So this is one of the, that thing is one of the Archimedean solves. Another thing to do is you can take, if you take each of these vectors and you put a perpendicular plane at the, at the endpoint of each of these. So you get, you get a, then those, those bound another polyhedron that has 30 sides. And that one is known as a rhombic tricont, wait, what is it called rhombic triconti-contihedron, I think. I mean, and if you play Dungeons and Dragons, you know it by a simpler name, which is a D30. Anyway, so here's, here's the nine vector configuration. It's, it's one of the root lattices. So if you've studied, if you study lead groups or algebraic groups, the algebras, you know that there's a classification of root lattices. And this is one of them. This is one of the three dimensional root lattices. So this is called the B3 root lattice. And this one you get by taking a cube and you take the, you take the midpoints of, sorry, you take the, it's like the center is of each of the six faces. And then you also take, but then you also take the centers of each of the edges, which gives you 12 more. So it's six plus 12. So it gives you 18 vectors, but we're only considering them up to, up to scale of multiple. So each pair of, it's really nine vectors that are sort of independent. Yeah. Okay, so that's the nine vector configuration. And so those were, yeah, so going back to this. So, so there's, so that, so that explains these first three. This eight vector configuration is, well, I mean, so they're, they're, they're five of them. Each of these is, I don't, I don't really understand what these are. I mean, they're obviously also somehow special because, well, I don't know, they're, they're the only ones of the type and there's nothing that's in seven dimensions. So, but these are kind of weird. These are each of these consists of seven vectors in a plane, and then one vector sticking out of the plane. That's not perpendicular to the other ones. So, and the seven vectors in the plane in each of these cases are at the vertices of a six, there's some of the vertices of a 60 gone. And beyond that, I don't really understand what these things, why these things exist. But anyway, maybe somebody can figure it out. Yeah, and then these are, okay, and then these are all these other sporadic ones and a couple of one primer families that aren't contained in the larger ones. Okay, so, alright, so that's, that's it. So, now what I want to do at this point is, is go back to the tetrahedra and try to explain those 59 sporadic solutions. So, the explanation is going to use something called rej symmetry, which is something. Okay, which is which is sort of, it's sort of a miraculous fact of, of, of geometry that I think people still don't really understand. It was discovered by two physicists who are studying the some, they're studying what I don't know what they're called six j symbols, whatever that is, and they are in some representation theory. But anyway, let me tell you what they discovered. So, if you take a tetrahedron. And you look at its edge lengths, let's call them X, Y, A, B, C, D. And I'm ordering these so that each pair is opposite. So if, so these, the edge length, the edge with length X is does not, is not touching the edge with length, length Y, they're opposite. And that's true for each pair. And similarly, and you, for each edge, you give name to the corresponding dihedral angle along the edge. And I tried to choose angle names that kind of looked like the edge length. I didn't quite succeed, but I tried, I tried my best. So this is my X, this is my Y, angle, A, B, C, D. Maybe that one should have been the Y. I don't know. Anyway, all right. Anyway, so, and yeah, you see, you have your initial tetrahedron. You have your edge lengths, some baggage angles and volume and it has its gain and variant. And then you, you make these numbers, these six numbers where you take the X and Y you leave alone. And for each of the other ones, you take sort of the A plus, you take this combination of these four divided by two and you take S minus A, S minus B. And then it turns out that there's a tetrahedron with those six edge lengths, which is not so surprising because the only condition on six edge lengths to actually come from tetrahedron are some inequalities. But that's true. But the surprising thing is that that tetrahedron not only do the edge lengths transform that way, but the dihedral angles transform exactly the same way. So the dihedral angles of this tetrahedron with those edge lengths also has the angles being transformed this way. And not only that, but it has the same volume as the original tetrahedron and it has, and there was also observed later about Joe Roberts that it also has the same gain and variant. And not only that, but okay, yeah, so it's, so it's a mystery. Okay, so yeah, not only that, but this theorem holds also for spherical tetrahedron. If you take tetrahedron on a sphere and spherical geometry, or if you take tetrahedron hyperbolic geometry, you have the same theorem holding. And so nobody really knows why this is true, I think, at least there's no simple geometric explanation. I mean, you might think, oh, maybe there's just some easy way of cutting up or, I don't know, you take the midpoints of the edges of the T or something and you make T prime. But there doesn't seem to be any simple construction. No, no, there's, they're the best explanation I've seen so far. Are they given these two very recent papers, one of which is still is still preprint, I think. So, but yeah, I feel as if there must be some simple explanation for this, but it's just too good to be true. I mean, why? But anyway, for our purposes, this is nice because this means that whenever you have a tetrahedron such that the six triangular angles are rational multiples of pi, you can make another one by doing this symmetry. And in fact, you can do this in more than one way because you can, you don't have, you don't, you can choose any pair to be your X and Y. You can do also the regis symmetry where you choose C and D and then operate on the other four edge lengths. So, you get this whole orbit under, there's some group that acts on the tetrahedron that preserves the property of that angle as being rational multiples of pi. So using this, sorry, there's some questions here. Should I, oh, let's see. So, um, well, okay, earlier slide, I found some sets of sprites and for the same size one inside a set and one outside. Does this symmetry account for that? I, I'm not sure, but maybe my next slide will answer this. So here I'm going to, these are, I'm going to try to describe, I'm trying to give a geometric explanation for the for the answer for the answers to the original problem. What are these 59 sprites tetrahedron? What are these two one parameter families? So here's how you can get the one parameter family. Start with three unit vectors in space. I'm going to get three, okay, here are three, three unit vectors in space. And suppose you make them so that their end points, so that they're, they're unit vectors and they're independent and they're, they form equal angles. So, so that's the end points of these three vectors formally equilateral triangle. And then if you take the, if you take the numbers of the form C1, V1, C2, V2, C3, V3, if you let all the variables range from zero one, then you would get the parallel of pipet that was spanned by those three vectors. But if instead of taking all as C1, C2 and C3 in the whole interval from zero one, if you impose the ordering of those, so one of the six possible orderings between those, those coefficients, then you'll get one sixth of the parallel pipet, and that gives you a tetrahedron. And, and that tetrahedron obviously tiles our cube because you can take six of these and tile the, and fit six of these into a parallel pipet and then just translate the parallel pipet around. And so, so that tiles our cube and therefore it must, this one must have diagonal angles. Well, as long as it, yeah, oh, I should, well, as long as it, well, sorry, it'll, it'll, it'll have diagonal angles in cube pi, as long as for the, for correct values of the parameter, I mean there's a parameter here, which is what the angle between the vectors you choose is. So you can like open, open and close up your, your tetrahedron, you can tetrahedron. And, but for suitable values, it will, of the parameter, it will all have diagonal angles. And so this gives you one of the parameter families. And then once you have that family, you can apply one of these regis symmetries. And it turns out by doing that you get, you get another family. And you can keep playing regis symmetries, but that's all you get as far as families go. And as for sporadic tetrahedron, well, you can get lots of them by taking these special, these special, these special 15 vector configuration and nine vector configuration and take all the four elements subsets that are in linear general position. And then those give you a lot of other tetrahedron. And then, and then you can apply regis symmetries to all of those to generate even more. And that gives you almost all of them. The only ones that haven't been explained are the three left that, that I guess I still don't really have a geometric explanation for is that these things involving the pi over seven. And, and the ones you can get, and there are two others you get by applying regis symmetries to these. And so, yeah, I mean, so, so we almost have geometric explanation for everything. Well, I guess, except that we don't really have a geometric explanation for the regis symmetry, at least not a nice one either. But, okay, but anyway, we sort of understand where all these things come from so at least, so at least you can say put some order into this, into this, into this zoo of, of sporadic tetrahedron. So, okay, so, all right, so I'm so this is so I'd like to end here so here's again just the, the summary of what, of what we've done now so we've at least started to treat your bow, we've classified all the ones with whose dihedral angles are rational multiple and then these other interesting questions about which ones tile are, are cubed and there really haven't been any other new ones discovered in the last couple of decades. So besides these three one parameter families and these isolated examples. So but I mean we haven't, it's actually pretty hard to test whether, I mean how do you, whether a given texture it does tell you because there are lots of ways you can put it together and I mean it's pretty hard because if something doesn't tile our cube. Well, I don't know if it's so actually some of these might actually tile our cube they might not be in position correctly, and same for some of these other examples of we're not really sure. But, okay, so let me end here so thank you for all the questions and I'll be happy to take more questions. Thanks very much everybody please take this opportunity to unmute and thank you.