to the last week here. So let me share my screen and you'll see that the title of today's lecture is quadratic forms over the integers. So we're gonna be starting a new topic today. But first, we have to finish up our discussion from last time. So we're gonna finish the proof of Hossam and Kowski and then we'll wipe the slate clean and kind of start afresh with a new sort of theme quadratic forms over Z. Well, so let me recall what we were in the middle of doing. So the statement of Hossam and Kowski is that if F is a quadratic form over Q such that F new is isotropic, crawl places new of Q, then F is isotropic. So if you've got a non-trivial zero in the p-addicts for all primes p and in the real numbers, then you've got a non-trivial zero in the rational numbers. And so we just saw that this was true for n, the number of variables less than or equal to three. And today, we're gonna do n equals four. And then finally, we'll have enough raw starting power to be able to do an induction allowing us to establish the case n greater than or equal to five. Okay. And I just, before we get going on this, I just wanna remind you of a small demo we had from last time that we didn't actually use last time because it was actually only supposed to be used in the n equals four case. So if you have A and B and Q for any new, any place new, if you have A, B and Q new cross, and also C and Q new cross, but playing a different role, then the form AX squared plus BY squared represents C, if and only if there's a certain equality of Hilbert symbols. So if and only if AB new is the same thing as C comma minus AB new. And the precise form of it doesn't matter. The important thing is just that it's some relation between Hilbert symbol evaluated on C and that on A and B. So we proved that last time. I won't go over it again. And now let me explain the case n equals four. So the proof of the hospital has to be in Kowski for n equals four. So we're gonna, as usual, write our quadratic form in diagonal form.  Now, in the previous lecture, we said we can do a number of reductions. We can make A whatever we want, and then we can make BC and D square free integers, but none of that is gonna be important for this argument. So you could just have all of these be non-zero rational numbers. The further assumptions on them will not be relevant. So I won't use them. Okay. Now, what is the idea? The idea is we're gonna split it up into two. So actually, I'd rather change signs for the purposes of the argument. So I'm actually gonna write it as, I'm gonna change what C and D are. I'm actually gonna write it as a difference of two quadratic forms in two variables instead. And that's just, it's just gonna make the notation a little simpler. So for all new, we have this F new represents zero. So we can solve for this expression being equal to zero, which means we can solve for AX squared plus BY squared equal to CZ squared plus DW squared. And that tells us that we can find a scalar T new in Q new such that AX squared plus BY squared represents T new and so does CX squared plus DY squared. Well, I guess maybe I should use the same variables because the CZ squared plus DW squared also represents T new. So what this hypothesis that our four variable quadratic form represents gives us is that when we split it up into two, then the first half, we have a number which is represented by both the first half and the second half, so to speak. Now, let me make a small remark. We can assume that TV is non-zero. So TV is in Q new cross. Why is that? Well, suppose it were zero, then this form would represent zero. Therefore, it would be isotropic. Therefore, it would represent everything. There was this lemma we've had and used several times. So in fact, it would be the hyperbolic plane and the hyperbolic plane represents everything. And similarly, this one would represent everything. So then we could just choose any random non-zero scalar and it'll be represented by both. So we could just replace T new by that random scalar. So there would be no problem. So we can indeed assume that it's non-zero. So I could ask a question. So F sub new, that's just the original F when you take it. But why are you now saying that, so it's AX squared plus BY squared minus CZ squared minus DW squared that represents F new. So there's a bit of an abuse here. So when I say F new, I mean F, the same quadratic function, right? But viewed now as a function on Q new instead of just as a function on Q. So it's the same F you could say, but the statement F new represents zero means something different. Now it means that there are elements X, Y, Z and W and Q new such that the value is equal to zero. So, right, but you are omitting some of the summons in F new when you say that CX squared plus BY squared represents F new or A. Well, this, representing zero is the same thing as saying AX squared plus BY squared equals CZ squared plus DW squared. But that's not the FW, that's not FV. No. That's not F. No. So F is that four polynomial with four terms, right? Yes, this one right here. So F new also has four terms, but now you're just cutting F new and in half some. Yeah. So I haven't given names to these forms. Oh, are you calling the T? Is it some of these? No, no, T is a scalar, yeah? So T is some scalar that we know exists just because we can take the value on AX squared plus BY squared, call that T new.  Okay, sorry. I thought that T new is F new. Ah, yes. Yes, yes, yes. So there's a reflection that takes one to the other on the piece of paper, but they are in the way I write them, but they are different, yeah. Could you make it clear please for somebody who walks in later? Ah, I see what you mean. It's not that I'm looking at it myself. Yeah, it does not look clear at all. Okay, you know what? Let me see what I can do with making my form a capital F instead. And I hope I remember, if I don't remember, please remind me. So our form is gonna be capital F. Thanks for pointing that out right now. Yeah, I didn't realize that looked so confusing. Yes. Right. So we can assume T new is in Q new cross. And now by the lemma that we just recalled here, this computation with Hilbert symbols, well, this is equivalent to saying that, you know, some of the quality of Hilbert symbols, T new minus AB equals AB, and this is all at new. And similarly here, T new minus CD equals C comma B new. Okay. Now, we're going to use weak approximation. We want to turn, want to replace the T new by the same T in Q cross same. So all these T new user depend on you a priori, right? We want to say that we can replace them by something which doesn't depend on you. And in fact, you know, it comes from Q and we'll do that using weak approximation in a certain clever way. So the statements from your problem sets about, that's exactly a way of going from a bunch of local data to some global data, some local scalars to some global scalars. But to apply weak approximation, you have to have a finite set of primes at which you're operating. So you somehow have to be able to control everything that's happening outside of that finite set of primes. So what is our finite set of primes going to be? It's going to be the least obvious. I mean, the primes for which the behavior, at which the behavior of the quadratic form is hardest to control. So let S be the set of places which is infinity and two, those are always tricky, right? Those are kind of exceptional cases in the theory of quadratic forms. The real number is in Q two. So we'll always stick those in. And then we union the set of primes P for which P divides either A or B or C or D. So we just look at all the primes that are going to cause us trouble, potentially. And we make that our finite set. All right, yes, other areas. And I mean, can we say that for all the other primes that we have it included in S, this four dimensional form that we started with is going to be isotropic in Q new, right? Yes. Yeah, that's not exactly, we're going to use that, we're going to use that observation, that kind of trick more in the n greater than or equal to five case. That's not actually going to help us with this precise, this precise argument here. We've transferred everything to being represented by quadratic forms in two variables. Yeah, but no, we'll actually, we'll use a form of that pretty soon. Okay, so now let's apply weak approximation from your problem set. And that tells you there exists a T in Q as close as you want to T new for all new in S. So meaning there's a rational number T and you can choose the rational number such that the distance between T and T new in the new absolute value is as small as you like. Or this is equivalent to saying that the distance between T over T new and one is as small as you like. Well, you have to fix how small you want it to be and then you can make it that small. And this means that you can guarantee that T over T new is a square because the first problem on the problem set is that if you're close enough to one in any of these local fields, then you're a square. So, and that's all we're going to take out of this is a T in Q cross such that at all the exceptional places T differs from T new multiplicatively by a square. And what does this tell us? Well, the Hilbert symbol recall is bimultiplicative and it takes values in plus or minus one. So if you replace anything in the Hilbert symbol by something which differs from it by a square then it doesn't change the value. So that means that T comma AB new is equal to T new comma AB, oh minus AB I should put new, which by hypothesis is equal to AB new. And that tells us that T is represented by, by AX, oops, plus BY squared. And similarly, also by CX squared plus BY squared. Okay, and this should be, sorry. And I should now I should be really careful because everything here is a rational number. So I mean in Q and U. So it's represented in Q and U by AX squared plus BY squared. Meaning you can find X and Y in Q and U such that T equals AX squared plus BY squared. So we've guaranteed that our number T is represented by AX squared plus BY squared in Q and U for these exceptional places in U, but what about all the other ones? We have to be able to control those as well. And for this, we actually need an extra hypothesis on T. So we're called a variant of weak approximation that says that we can also, this was also on the problem set. So we were allowed to guarantee that the new attic absolute value, well the P attic absolute value of T is less than or equal to one for all P that are not in S with possibly one exception, which we'll call P not. So you can make it P integral at all P except for one where you will have to, you have to have some kind of denominators. Okay, then for P not in S union P not, we have that T comma minus AB P is also equal to one because we're at an odd prime. All of our except, we put two as one of our exceptional primes. So we're at an odd prime and everything here is relatively prime to P. Oh, sorry, I shouldn't. Yeah, no, sorry, I should have actually said, I should have actually said, maybe I stated it wrong in the problem set. I wanna guarantee that it's a P attic unit, not just absolute value. I should have said that you can, I might have stated it wrong in the problem set. You can prove something stronger. Yeah, all right. You wanna actually guarantee that the P attic absolute value is equal to one so it's a P attic unit. But then the Hilbert symbol is equal to one, but that's also the same thing as the Hilbert symbol of AB P. So all of these are P attic units as well. So we have the same conclusion anyway. So T is also represented by AX squared plus BY squared in these QPs. So the only place left for which we don't know that T is represented by our guys is P not. But that's only one place. And by Hilbert's product formula, we must also have that T minus AB P not is equal to AB P not. Because Hilbert's product formula says that if you know, if you're taking a Hilbert symbols where the values are in Q and you know that, you know them for all but one place, then you know them for the remaining place because it's determined because it has to be equal to the product of the rest. And that's true for this side of the equation and that's also true for this side of the equation. So here's where we're using Hilbert's product formula or quadratic versicrosti in a kind of surprising way. So we see that also at the place P not, we have the same conclusion. So our grand conclusion is that T is represented by AX squared plus BY squared and CZ squared plus DW squared for all places new. So in representing Q new. But being represented by a two variable quadratic form is the same thing as saying that the three variable quadratic form you get by, you know, so AB minus T new is isotropic for all new and so is CD minus T new. And then by sort of doing a little bit of an induction here, so by the three variable case, we deduce that AB minus T is isotropic and so is CD minus T. Sorry, I'm fitting that in a very small corner there. Let me instead move on to the next page. So this implies that by the three variable case that AB minus T and CD minus T are isotropic over Q and that again tells us that you have X, Y, Z, W such that AX squared plus DY squared equals T and CX squared plus Z squared equals T. So that means these two things are equal to each other and you bring it over to the other side and you've solved your equation. So that's the four variable case. Okay, so now let's move on to the five or more variable case. So I'll just write it in five variables just for concreteness, but you'll see that the argument is inductive and it will work for any n greater than or equal to five. So let's just say n equals five. Then we have F equals, no, we'll use the same trick. Well, we'll start in a similar way. Actually, let me do A1 and B1 and C1. So we have AX squared plus DY squared minus A1 Z squared minus B1 W squared minus C1 V squared. And we're gonna do a slightly different argument here. So yeah, it's not gonna be obvious why I'm making this my finite set right now, but it'll come through in the argument. So we're gonna do another weak approximation, but it'll suffice to use the naive form of weak approximation in this case. It's supposed to, everything's supposed to become easier when you add more variables, right? That's this principle of the new, the U invariant that Achille talked about that, you know, that eventually over most fields, if you add enough variables then you're supposed to always have a zero. So, and in particular, I just wanna make a remark here. We saw that over a periodic field, as soon as you have five variables, you're guaranteed to have a zero. So our hypothesis is almost vacuous. Our hypothesis in Hasselman-Kauski that the form is isotropic at all places is almost vacuous. I say almost because at the real numbers, it's still saying something. It's saying that you're isotropic over the real numbers, but that's sort of all we should need to use. Now, in fact, in the argument, we're gonna use more than that, but this is just to show you that things are supposed, the argument is a little easier than the N equals four case. It's supposed to be a little bit easier to find zeros the more variables you have, and that makes sense. Okay, so we're gonna let S be the set of places where this A1, B1, C1 is not isotropic. And this is finite because for large enough primes, you know, for primes P not dividing two A1, B1, C1, you have a non-trivial zero in Fp by the mu invariant. Well, that was this counting argument that Achille gave that a quadratic form in three variables over Fp always has a non-trivial zero. And then you get one even in Zp by what's it called the Hensel's lemma. And therefore also in Qp. And this is where we're using the hypothesis that N is greater than or equal to five, by the way. It's that when you split it up into two variables and then the remainder of the variables that you have at least three variables left over so that you can make this, exactly this argument here that the set of places that you're looking at is finite. Okay, so now, so for mu, so to make a weak approximation argument, we need a scalar. So, yeah, just a sec, let me get my notes up. I'm gonna get a little lost otherwise in my argument here. Right, so we're again just gonna, we're gonna choose a random value, T need of AX squared plus BY squared. So let's say AX mu squared plus BY mu squared equals T mu for mu and S. And all of these, you know, and X mu and Y mu and T mu are all gonna be in Q mu. And again, we can assume, yeah, we can assume T mu is in, but we can assume they're all in Q and U cross actually. I don't wanna give the little argument for Y, but, oh no, wait, no, maybe I don't need that because no, I'm not using the, I'm using the naive form of weak approximation. No, maybe I do need it. Oh, sorry, guys, I'm a little confused right now. Let's just put it in, right. So now what we're gonna do is we're gonna apply weak approximation, not to T mu directly, but to the X news and the Y news, which is actually stronger because if we're close to X mu and Y mu, then we're also close to T mu just because of this algebraic expression and continuity of the algebraic operations.  And this implies, as I said, by continuity that T is also in, well, we can choose it also close enough so that by continuity, we're implied that T is close enough to T mu and in particular, it's a square off from T mu. Okay. So now we already have that T is represented over, so that in Q and U, T is represented by rational numbers X and Y. So now we just have to control the other half of the expression. We also want T to be represented by the other variables that we left out. So, right, so, but so yeah, what am I trying to say? I didn't make it precise enough, I'm sorry. I don't wanna choose a random value of A X squared plus B Y squared. I wanna choose, I wanna choose it just like in the previous one so that this is equal to A1 Z mu squared plus B1 W mu squared plus C1 V mu squared. So I wanna choose rather a zero of my quadratic form over Q mu and let T mu be the thing that's equal to, both of the halves, since I put a minus sign on. So we're doing it just like in the previous one. I'm sorry for getting confused about this. So then the fact that T differs from T mu by a square implies that T is, so T mu is represented by the other half of my quadratic form. And since T differs by T mu by a square, then T is also represented. Or in other words, T comma A1 B1 C1 is isotropic for all mu and S. But it's also isotropic and here's where the way we chose S comes in. Also isotropic for all and mu not in S because well, even if you throw out the T part, it's still isotropic. That was the definition of S. So for mu not in S, even this part is isotropic so you have a non-trivial zero here. So you can just make the first variable equal zero and you find that, oh sorry, minus T. You find that this one is isotropic too. Okay. And now by induction, so or the N equals four case in this case, we get that minus T A1 B1 C1 is isotropic over Q. So that tells you that T is also represented by A1 Z squared plus B1 W squared plus C1 V squared. And since we argued already with weak approximation that T was represented by the first half of the quadratic form, then we get that, well, so we get with our X and Ys that AX squared plus BY squared can be made equal to A1 Z squared plus B1 W squared plus C1 V squared. And that gives us our global solution, our solution in the rational numbers. Sorry, what's the last part of the previous page? The last part of the previous page? Yeah, you just moved it, yeah. Yes. Before A1 B1 C1, what? Because A1 B1 C1 is isotropic. So since this is isotropic, there's a zero, a non-trivial zero here, well, then you just add a var, you've added a variable to that. You can just set that variable equal to zero and you get a non-trivial zero there. So it's- Is that over Q new or over Q new? Yeah, yes, yes, over Q new. Okay. Yep, I didn't write that, thank you, yeah. The, and it'll be too cramped if I write it wrong. Okay, all right. So that was the Hasselman-Kauski theorem. And now let's, again, wipe the slate clean. So, and start, I don't wanna say start the course over, but let's start a new topic, which is integral quadratic forms or quadratic forms over Z. So let me start by recalling the definition of a quadratic space over a field. So a quadratic space was always, was a pair of V comma F. So where this is a finite dimensional vector space and F was a function from the, so let's call the field K was a function from V to K such that first property is that it should be homogeneous of degree two. So if you take F of lambda V, the same as you can, you pull out a lambda squared. And the second property was that if you define the pairing of V and W to be F of V plus W minus F of V minus F of W, this should be linear each variable. That's kind of abstract, but it's equivalent to saying that if you write, if you choose a basis of the, and then write F in terms of the coefficients, let's call them x one dot the dot x n, then oops, I was, should I call it capital F? Maybe it doesn't matter anymore than F is a homogeneous polynomial of degree two. So this quadratic space thing, it's just kind of a coordinate free way of talking about homogeneous polynomials of degree two. And I'm gonna slightly prefer this. Well, we'll use both perspectives, but for concreteness, I recommend you just think of a homogeneous polynomial of degree two. So for example, you know, F two, well, if you're in two variables, then you just get something like this, A x squared plus B x y plus C y squared.  But the fact that we started with an abstract description tells us that we're gonna be mostly interested in studying quadratic forms up to change of variables. So we would consider this quadratic form to be equivalent to or isomorphic to any other quadratic form we get by a linear change of variables and invertible linear change of variables in the x and y. So this is the context we've been working with since the very start of the Kiehl's lectures. Now, what about over Z? Well, we can actually just do this same definition. The only real question is what the analog of a finite dimensional vector space is. So, and it's, so you know that every, every, you know, yeah, well, the correct analog of a finite dimensional vector space over Z is just a finite free Z module. So it's a pair M comma F where M is a Z module or an appealing group which has a basis. So which is isomorphic to Z to the Z direct sum N for some N. So we just explicitly require that our getting group is a free of getting group of finite rank. And then we think it's analogous to a finite dimensional vector space or the closest we can possibly come to a finite dimensional vector space. And F is as before. And again, there was nothing in the equivalence between these two perspectives which really used that it was a field. So it's the same thing as a homogeneous degree two polynomial with integer coefficients. So before it was coefficients ABC in the field. Now it's coefficients line the green of the integers. So so far it doesn't seem that different. But well, I'm gonna make a series of remarks which will lead to a picture that actually there are quite a few differences between the theory over Z and the theory over field. So first remark is that Achille had this standing convention that a quadratic space is always non-degenerate. So before we said our quadratic spaces or quadratic forms should be non-degenerate. Or in other words, this associated pairing so V cross V to the field or Z or whatever should be a perfect pairing. So it increases an isomorphism of, oh, sorry, maybe I'm switching my M's and yeah. We're gonna drop this assumption now. Why? Well, our favorite examples are degenerate. It's not satisfied in our favorite examples. There are just too few for our purposes. Yeah, non-degenerate quadratic forms. So for example, one of the most famous and well-studied quadratic forms is also one of the simplest to write down. f of x, y is x squared plus y squared. And there are a lot of, you know, there's some cool theorems in number theory about this. So for example, then these are the kinds of theorems we're gonna be talking about. An odd prime P is represented x squared plus y squared if and only if P is convergent to one mod four. So that's a very, very fun fact in number theory. And do you wanna be able, I mean, these are the kinds of theorems we're gonna be talking about right now. Questions about representing numbers and in particular prime numbers by quadratic forms. And to get interesting examples, you have to allow degenerate. They're not really degenerate because they're non-degenerate over the rational numbers. So they're really just, yeah. They're not perfect. I mean, the word for it, the number theorists use in this context is unimodular, I suppose. So the associated lattice is unimodular. It just doesn't happen often enough. Right. So that's one difference. We're dropping this non-degeneracy assumption. By the way, how, so we had this, so how do you check non-degeneracy? So recall over a field, non-degenerate was the same thing as saying that the determinant of the matrix or that the determinant of the bilinear form is invertible, is in K cross. So you know, in terms of a basis, you can write out, in terms of a basis for your finite dimensional vector space, you get a square matrix by just plugging in the different possible values of your bilinear form on the basis vectors. And if the determinant, if that's an invertible matrix, then your perfect pairing is non-degenerate and that's actually an if and only if. This is also true over Z. But so in other words, it's if and only if the determinant of the matrix you get. Now you don't wanna say it's non-zero. What you wanna say is that it's a unit. So, but now the difference between the integers in a field is that the integers, well, one of the differences between the integers in a field is that there are very few units in the ring of the integers. So this is just plus or minus one, right? So to get a non-degenerate one, you have to have this property that the determinant of the matrix you write down from the quadratic form actually had a determinant of plus or minus one, not non-zero, but plus or minus one. So for example, if you look at the matrix for fxy equals x squared plus y squared, you actually get I think 2200, no, no, sorry. Yeah, yeah, that's right, 2002. And that means the determinant is equal to four. So yeah, but this, the fact that the units are so small also gives us something nice. So recall also, so that this determinant depends on the basis of the vector space, but only up to squares, only up to squares of the units. So recall that when you do a change of basis on your quadratic form and isomorphism on your quadratic form, if M is the matrix for the quadratic form and A is the change of basis, then the matrix for the new guy is AM, A transpose. So when you take determinant, you see the square of the determinant of A multiplied by the determinant of M. So again, all of this just works in fact over an arbitrary commutative ring and in particular over the integers. So now, but now we learned that over Z, the determinant of our quadratic form is independent of the basis. And this is because if you take the square of any unit, you get one. So nothing's really changing. So that's nice. So we don't have to remember that the determinant is a coset and more it's really a determinant or discriminant. I don't know. There's some sign, right? But yeah, it's well-defined on the nose. So it's just an integer. Okay. So now, right. I'm going to now specialize to, there's a lot to say about the general theory of integer quadratic forms, but I think it's a nice idea to just specialize to the first non-trivial case. Cause you see, well, there's some special phenomena which help you, which is, which make the discussion easier, but you see also a lot of the general phenomena that appear in the general period. So we'll specialize to the first non-trivial case. So this is two variables, also known as binary quadratic forms, which is a weird word, but it must be some old fashioned thing that stuck. Binary, I guess means stands for two variables. I don't know. So, so right. And so again, that just means that in coordinates, we have FXY is equal to AX squared plus BXY plus CY squared and A and B and C are integers. And if you calculate the matrix for the associated bilinear pairing, I think I'm not going to go through this, but you get two A, two C, B, B. So you're always going to get two times on the diagonals and then it's always going to be a symmetric matrix. And in fact, you can go backwards. If you have a symmetric matrix where the diagonal entries are even, you can write down an associated quadratic form from which it comes. But anyway, I just want to remark right now that the determinant is equal to, well, some familiar quantity. Well, it's all right, minus the determinant is equal to B squared minus four AC. You're all familiar discriminant from high school, right? So let's call this the discriminant.  Do you remember? Oh, I think, yes, yes, right. Cause I want this, I mean, B squared minus four AC, that's gotta be the discriminant, right? And you don't want it to be minus the discriminant. That would be confusing. Yeah, so yes. Now, there are two, there's a fundamental dichotomy. Whoops. So, well, we're gonna throw out discriminant equals to zero. That's gonna be too degenerate for us. So, there's D positive and D negative. Oh, yes, right. D positive and D negative, which can be read off. I mean, the discriminant, you can, the discriminant is defined for an integer quadratic form. But if you, when you have an integer quadratic form, you get a quadratic form over an arbitrary field, right? Because any field you like, you can do this, these are integer coefficients. So they make sense in any commutative ring, hence in any field. So you could just view it there. So if you have any invariance for any quadratic field, quadratic form over any field, you get an invariant for quadratic forms over Z, right? And the discriminant, so in particular, and the discriminant doesn't change because it's just the same matrix. So we can always, well, we can look over the real numbers and we have a two variable quadratic form and we have the Sylvester's law of inertia, which says that it's isomorphism type is recorded by just which signs you have on the diagonal, right? So pluses or minuses. And the determinant greater than zero, that corresponds to the case where you have a plus and a minus, which means that this is indefinite. That's just the definition of indefinite, over the reals, you have pluses and minuses, both pluses and minuses. But it means that, well, yeah. F takes both positive and negative values. And if the discriminant is negative, this means this is the same thing as saying that you're, over the reals, you're definite. And that's the same thing as saying that F's values, I mean non-zero values are always either positive zero or greater than zero. Greater than zero, less than zero. So, and we can choose, in the second case, in this case here, we can arrange all values to be positive by changing F by a sign. So it just replaced F by minus F and we switched between these two cases. Now those F and minus F are not isomorphic, right? They can't be because they take different values, but it's clear that the study of one will easily reduce to the study of the other. So, so if we're gonna, so what I'm trying to say is we're gonna focus on this case. This is the simplest case. For a reason I'll explain in just a second. And we're also going to just change the sign of F to make it so-called positive definite instead of negative definite. So why is, so I'll give, I wanna give you a hint of the arithmetic significance of this dichotomy. We're not gonna prove this, but it'll be in the background for this distinction. So you can also read this off. So D is greater than zero. It turns out your orthogonal group of our quadratic form is a finite group. Whereas if D is less than zero, it's infinite. I guess I'm saying the same thing twice because I put the reverse implications too. Well, actually you'll see, you'll prove this on your problem set. In fact, a generalization of this to quadratic arbitrary number with an arbitrary number of variables. This is a little more subtle to prove, but you'll do it in one case at least. I want you to study a particular indefinite quadratic form and find its orthogonal group. It's a good exercise of the number theory. And this is basically the reason why we're ignoring the indefinite case because, oh shoot, did I do it backwards? Ah, sorry, sorry, sorry, sorry. Yes, in the definite case, it's finite. In the indefinite case, it's infinite. I'm sorry. Yeah, we're sticking to the definite case because it's much simpler to have a finite group of automorphisms. In fact, it's gonna be very, very small. You can eat, yeah. It won't be very big at all. It's not finite. It's also easy to determine. It's almost always just of order two or order four, I guess, I don't know. And yeah, right. Having an infinite group of automorphisms makes things a little more complicated. Okay, right. Ah, there's one more thing I want to mention, which is another change in conventions that takes place when you move to the integers. So this is a little obscure at first sight, but. I'm sorry. Can you define order, please? This is order? Or what's that O sub F? This is the orthogonal group of the quadratic form. I think this notation was used by Achille. I hope that, but yeah, what it means is it's the automorphisms of the quadratic form. So it's those linear change of variables which leave the form invariant, so which don't change the form. So for example, if we look at X squared plus Y squared, if we replace X by minus X, it doesn't change the quadratic form. So that would be in the orthogonal group that linear change of variables. If we switch X and Y, it doesn't change the quadratic form. But if we replace X comma Y by X plus Y comma X, that's also an invertible change of variables. That will change the quadratic form. That will not be in the orthogonal group. That will be an isomorphism between this quadratic form and another quadratic form. Thank you. You're welcome. So when we work over Z, it's a good idea to use a finer form of equivalence than isomorphism. So it's called strict isomorphism or strict equivalence. This will mean that they can be, that F and G are related by an invertible linear change of variables. Excuse me. An invertible linear change of variables. So far I've just said the definition of isomorphism, right? But now I'm gonna add another condition with determinant one. So you're kind of throwing away half the possible changes of variables. You want to require them to be determinant one. The reason for this is not obvious at first sight. Well, it certainly gives us more information. So a priori, it's a good idea to do it. Quadratic forms up to this equivalence is finer. So you're potentially seeing more, right? But it turns out that many of the results are much nicer to state when you use this finer notion of equivalence than if you use the coarse one. It comes, same as become much more uniform. And you can understand how to go from the fine one to the coarse one. And you can see, once we developed the theory, you can see exactly why what's causing the fact that you get more uniform results with this notion of equivalence versus the other notion of equivalence. Right. So I think, so this was just setting the stage and I think maybe I'll stop for now and we'll actually get to some results, some non-trivial theorems and results in the next lecture. Yeah, but maybe I should say, it's not, how would you do that? We had this nice abstract notion of a quadratic space. So this is a concrete thing with polynomials being related by changes of variables. You can make it into an abstract thing by saying that a quadratic space, by talking about an oriented quadratic space. So like a finite free Z module with a quadratic form on it and an orientation. So an orientation, you could define or an orientation on the real vector space you get from that free Z module of rank two anyway. So there is an abstract perspective on this too. I'm just gonna stick to the concrete one. Yes. So any questions? Maybe I should actually say one more thing because I didn't get quite as far as I thought I would. So there are a couple of problems on the problem set that, well, I guess you can still do them. Yeah, you can still do them. You just don't have the context for the discussion. So you doing them will then make you better appreciate the context which is coming in the next lecture. Let's put it that way. Yeah, so problems three and four, I thought I was gonna be giving a little bit of context for, but it doesn't make it any easier to solve it to have the context, but yeah. Okay, so fine. Yes, Sundar, exactly. A basis, a matrix belonging to GLNZ. Yeah, so an invertible matrix with integer coefficients. Invertible in the sense of matrices, you can find out. That's the same thing as having determinant plus or minus one. And now I've just excluded determinant minus one for the strict equivalence. Yes, Kinta. Quick question. Is there an explicit example of things that are two forms that are isomorphic but not strict or isomorphic but not strictly isomorphic? Yes, it's not so easy to say right off the top of my head because you do need to go to pretty high discriminant. I think the lowest discriminant for which it happens is minus 23. So we'll have better, it's a good question. It's not obvious. We'll have a better context for it after the next few lectures and then I'll be able to give an example. Thank you. You're welcome. Yes, Latharius. So over Z, if we consider the quadratic form of X square plus Y square, this binary quadratic form if we write the matrix and it's two times the identity if I'm not mistaken, which would mean that the determinant of that would be equal to four, which is not in Z star, which is not in plus minus one. So would that mean that this form is degenerate over Z? Well, I would hesitate to use the word degenerate but it means that it's not, but the world that people use when working over Z is unimodular. So it's not unimodular. I mean, it's not real, I wouldn't really call it degenerate because you're never gonna have like, because over the rational numbers, it's not degenerate. So it doesn't really feel degenerate.  So I think one should come up with a new word. I mean, they unimodular, but then that only really works over Z. So I don't know. I don't think it's a good idea to call it degenerate, but what it means is that, it's not, you don't get nice morphism from the finite free Z module to its dual. You get something which differs from nice morphism. It's off by some finite amount in some sense. It's an injection with a finite co-kernel, you could say. I see, thank you. You're welcome. I had a question. Yes, please. Yeah. So is there a relation between the orthogonal group of the quadratic form over the integers and the unit group of the corresponding quadratic field? Because we see that there is a similar phenomenon happens with the units that, in the real quadratic fields, for instance, the units are generally infinite. Yes, indeed, there is. Imaginary quadratic fields, the units are finite. Indeed, it's, we're going to see, we're going to say that these co-ks are indeed very closely related. Yes. So just a spoiler alert. We're going to see that a source of quadratic forms is ideals in rings of integers in quadratic fields. So for every ideal in a ring of integers, we're going to introduce all these concepts. Yeah, but for every ideal in a ring of integers in a real quadratic field, you can write down a quadratic form. And if you have an isomorphism, and then if you have any unit, you can act by it and you get an automorphism. Yeah. Well, sorry, I mean, you have to exclude, in the real quadratic case, it's not as simple, which is another reason for excluding that. But in the imaginary quadratic case, yes, every unit gives an isomorphism of the corresponding quadratic space. And so you at least get an injection from the units in the ring of integers of the quadratic field to the orthogonal group of the quadratic space. Yeah. Well, no, maybe there is no subtlety in the real quadratic case. We'll see when we get there. Yeah. Excellent question, Zindara. Zindara asks, for quadratic forms over Q, we have a theorem like Hasselman-Kowski. Over Z, do we have such an analog? The answer is no. So, and in fact, one of the problems on your problem set is going to be, I'll give you two quadratic integer quadratic forms, and I'm gonna ask you to prove their isomorphic over ZP for all P and over R, but they're not isomorphic over Z. And this is, so we'll talk a bit more about this phenomenon in the next lecture. It's not an easy problem, I don't think. I'm curious to see how you guys attack it. Yes, so there it is. So just looking back at the Hasselman-Kowski theorem, wouldn't it follow from this theorem that if I have, because I think I've read this theorem somewhere and I just wanna see if I remember it correctly, that if I have a quadratic form over the rational numbers that has dimension at least five and is indefinite, then that would be isotropic, right? Yes, that's right. Because the indefinite gives me the isotropy in R by Sylvester, and the dimension five and above gives me the isotropy in QP by the UN variant. Correct. So then I go to global by Hasselman-Kowski. That's a nice corollary of the Hasselman-Kowski theorem. Thank you. You're welcome. Okay, sorry. What was the same amount of the corollary? All right. Well, Eleftherios, would you mind repeating yourself? Yeah, that's fine. So the corollary of the Hasselman-Kowski theorem would be this. So suppose I have a quadratic form over Q and I know that it is indefinite in the sense that, I mean, basically in the sense that it represents both positive and negative values. So similar to the usual sense in the real numbers. So I know it is indefinite, and its dimension is at least equal to five. It's five or above. And then it would follow from Hasselman-Kowski that this form is actually isotropic because the indefiniteness makes it isotropic over the reals and the dimension being five or above makes it isotropic over QP for every P by this U invariant equal to four. Therefore by Hasselman-Kowski, I go from local to global and it's isotropic over the rationals. Oh, okay. Thank you. Oh, you're welcome. This observation in fact proves, I think, Lagrangian standard theorem in every number. So we just take this positive number on the other side and we get using this observation. So Lagrangian is still the every positive number. Oh, really? Well, I didn't know that. Because, so let us take this four dimensional form, one, one, one, one, and then we take this N minus N. Yeah. So using He-Lefterio's argument, we get a rational solution for this. Yeah. But then will that prove Lagrangian's? It proves the form of Lagrangian's theorem, at least for where you allow rational numbers. But I don't know a priori, I think a priori, it's a different question. Representing N by sums of squares of rational numbers or by sums of squares of integers. It's probably worth thinking about, but yeah. At the moment, I don't see an argument for going from one to the other, but that's an interesting line of thought. Yes, He-Lefterio? In the beginning of the course, I remember that Professor Matthew started his first lecture presenting some theorems on sums of squares, I mean, mainly for integers, like there's, I mean, for like sums of two, three, and eventually four squares. Are we gonna be ending up proving some or all of those? Well, I don't, actually at the moment, I don't remember which theorems like he'll stated, but- Like for example, Lagrange's theorem for like every natural number, some of four squares. For now, we're gonna be sticking with two variables. So we won't be talking about Lagrange's theorem. Yeah, that's fair. I see, thanks. Okay, I might say we're gonna give us, we're gonna do study x squared plus y squared, right? At the very least, and then remind me at that point, and I'll tell you how to modify it to handle x squared plus y squared plus z squared plus w squared. I'm sorry, I guess he thought that Sarah, I mean, in the course in arithmetic book, he does deduce like Lagrange's theorem and these various theorems for some of the four squares and also three squares, but I guess, yeah, I'm forgetting a little bit. Yeah, there is some extra argument needed to go from a rational solution to an integral solution. But the argument exists, say? The argument exists. Interesting. So there's so much stuff in this Sarah book.  Yeah, here we go. Oh, there's a lemma of Davenporting castles. Okay, all right. I see, okay, that's interesting. Well, people always say Sarah's a really good writer. I don't know. I always find it really difficult to read this book, but it's also jam-packed, you know? And it feels, some of the arguments feel kind of conflicted with their many references. But so, yeah, but there is this section in this book, so I just found it, so it's on paid, in the edition I have at least, it's on page 46. There's a lemma B attributed to Davenport castles that allows you to go from Q to Z in the case at hand. So I will refer you to the book with my sympathies for how difficult reading it actually is. Yeah, but I guess it looks like, yeah, it's something special to, there is this condition that you have to impose on the quadratic form. I mean, this condition that he puts in parentheses H and then under that condition, which applies to sums of three squares and I guess sums of four squares, or at least sums of three squares. It's, yeah, you can go from an integral, the rational solution to an integral solution. Oh, yeah, so it's still from, yeah. So for four squares, he reduces to three squares by some more, by some elementary, not eight considerations, well, anyway, okay. So yeah, there's some tricks, yeah. And then Asingar, your proposal does work out after some tricks, yeah. And again, everybody, please do feel free to correct me if I'm mispronouncing your name. I, a lot of, yeah, a lot of your names are unfamiliar to me, but I'm happy to learn. Any more questions? Okay, well then I'll see you in a couple hours or however, yeah, see you when the time comes. Good luck with the problem set and have fun the rest of the day.


Everybody Matt here for late in sports cards rip in 2023 tops finest baseball hobby a box full case break number 59 pick your team We have Alex M. Cardinals Twins rays Alfonso a Dodgers RER Guardians Brian M. Braves KDNC pirates Chad T Royals Chris F. Astros Tigers Marlins DW Blue Jays H2 Giants James E Orioles Jeffrey S. White Sox Joe T Mariners Joshua H. Brewers Kareem a angels Kenny K Rangers Rockies Corey K Nationals Kyle C. Phillies Padres Matt O A's Michael B. Red Sox Philip T. Metz Yankees pink T D-backs Roger R. Reds Timothy P Cubs That was it. All right, here we go Two cases of finest This one pick your team style the next one random division style. Yeah, let's find you some Philip This is sky blue rookie to 300 DL Hall This one came out a few weeks ago Yeah, like three weeks ago Jordan Walker rookie Aronato and JT real mutile refractor But we know hoppy rookie rookie design variation Vinny Pasquantino Adley Ruchman rookie Headliners Bryce Harper and Brewer's rookie green wave auto to 99 Ethan small But I sang a rookie not a refractor Masatake Yoshida rookie world's finest Paul Goldschmidt Jonathan Aranda rookie green lava to 99 rays Summon Woods Richardson and Jordan Walker rookie refractor Tovar Headliners Adrian Beltray Reds rookie Otto Graham Ashcraft Otani and Riley green rookie purple refractor to 250 Christopher Morel Max Meyer rookie refractor Headliners Nolan Ryan Francis go Alvarez rookie box number two They're the they're the best team from the West in the NFC Alex Pregman That's green speckled 125 and behind it Jazz Chisholm Jr. Orange refractor to 25 Nice and Marlins. There's Riley green rookie. I'll to Vade to Grom Blue Bob and jazz Chisholm blue 150 Marlins color match Josh Young I feel like the Marlins have like four four different colors. It could be color matches. Yeah We have five to go 2223 Donner's team all basketball 20 box full case number two random teams that is all we have to fill today for the daytime breaks nice Rangers rookie gold refractor Otto to 50 Josh young congrats Texas That is Kenny K. Pull a nice Texas card right after talking trash about a Texas football team Matthew we were tore rookie refractor With a headliners each row Michael Harris rookie and Adley Ruchman. That is rookie aqua refractor to 199 Orioles Brett Bady rookie Byron Buxton Here at Mitchell Headliners are in Otto and Nolan Gorman rookie Volpe rookie Garrett Cole blue Aqua vapor 250 and Dodgers rookie Otto Miguel Vargas Garrett Cole Gunner Henderson rookie they were tore Randy Johnson Hey, that's nice Matthew liberator rookie design variation and Gunner Henderson rookie liberator Ronald Acunha refractor Garrett Cole and Rose gold Shohei Otani to 75 angels Nice one LA Astros finest moments Otto you're on Alvarez And you're not for Walker rookie Headliners Dave Winfield, you know hoppy say a Suzuki refractor Adley rookie and Oscar Gonzalez rookie purple refractor to 250 could I sang a rookie beer by us and mark McGuire rivalries die-cut Marino Yoshida rookie Joy vato refractor and tigers rookie Otto terrain Palacios Machado Devers headliners Simeon Woods Richardson and Nets to 50 Francisco Indoor gold refractor rookie refractor out Thomas refractor headliners Tatees and Waldechuck Cardinals traded the best left-handed starter on the market Another major picture For three minor players I mean, I know a couple of their names Oh Wally many Machado lot of refractor to 100 Corbyn Carroll rookie Logan Ohopie your rookie refractor Alonso Tom art a sky blue 300 and Rangers rookie blue refractor Otto to 150 Ezekiel Doran Riley green rookie Cartoon bull All-to-bay Shohei Otani headliners Just in that picture yeah Tatees refractor Josh Young rookie Steer DL Hall Vinnie Pasquantino rookie Aqua shimmer to 175 Byron Buxton blue Aqua vapor to 250 JJ Bladet Osvaldo Cabrera rookie refractor Michael Harris rookie Kunya headliners and Rockies rookie Otto Wynton Bernard Colorado on the board Raleigh and Logan Ohopie rookie Aqua refractor to 199 Jordan refractor and Nolan Gorman Anthony Volpe Headliners cow Ripken Contreras and Paraza five to go on Donner's team all basketball Fred Beatty and Matthew Libertura Blue Aqua vapor rookie to 250 Anthony Volpe Vlad refractor Nolan Gorman long grissom rookie Lou Bob sky blue Libertura headliners Vlad Gunner Henderson your Cole and Masataki Yoshida rookie refractor Jordan Walker rookie A's rookie Otto Dermis Garcia Kunya and Alvarez Logan Ohopie Vlad junior and brewers. I'm sorry. That was Aqua shimmer to 175 and brewers rookie Otto Ethan small Adley Ruchman rookie Jonathan Aranda rookie refractor by his headliners Ken Griffey junior Kodai Senga rookie Masataki Yoshida Michael Harris rookie design variation And Moreno Ronda cow Mitchell rookie refractor headliners mookie Mike trout one second spots rolling in for tonight's breaks. We're down to four now Four to go on Donner's team all Mack Scherzer Orioles rookie Otto Kyle Stowers Michael Harris rookie show heo Tony refractor fellow terrain red Beatty and Cubs rookie blue refractor to 150 Christopher Morel Go nice one Chicago Anthony Ope Headliners Nolan Aranado Nolan Gorman rookie Rissom Shailanga Lears rookie refractor Libertura Masataki Yoshida that is rookie Green Lava to 99 red socks Nice one Boston. This is nationals rookie blue refractor Otto to 150 Jeter Downs Wander and Gunner David Ortiz rivalries die-cut Aaron judge That's lava to 100 Rara very wise There's a Garrett Mitchell rookie refractor and rookie design variation Kinsu Kroy black gold Nolan Gorman nice one Cardinals you go st. Louis Yeah, let's get you something Jordan Walker rookie Randy Johnson Logan O'Hoppe Austin Riley purple refractor Lee rookie Jayrod Refractor Kodai Senga rookie Yankees auto playhomes Christopher Morel and Vlad green refractor to 99 Chisholm Adley-Ruchman rookie refractor and Guardians rookie design variation green refractor auto to 99 Oscar Gonzalez Ray Kevally and Bo Nailer rookie green speckle to 125 Francis Galvarez rookie Juan Soto headliners Corbin Carroll rookie Lonzo Shane Bieber and Mickey blue refractor to 150 Riley green rookie world's finest mookie bets steer by us and rookie aqua refractor to 199 Max Meyer Josh Young oh headliners redemption coming. There's Oscar Gonzalez and Finest rookies design variation or design autograph variation gold refractor parallel to 50 Metz rookie Francisco Alvarez congrats New York Phillip Grats on the Alvarez So that's the rookie design the finest rookies design variation auto so we like this but a gold refractor auto Francisco Alvarez Heck yeah it is Harper Scherzer Benny Pascuantino rookie refractor Michael Harris rookie probably headliners Yellich Mbello Last box do a recap at the end Yeah, that's cool. Appreciate that Tuglia hunter green purple refractor to 250 And Dodgers rookie auto James Altman There you go. Nice one Dodgers Tovar onion sosa simian woods Richardson rookie refractor Christopher Morrell and Oscar Coloss rookie sky blue refractor to 300 headliners Derek Jeter Wally Mookie Betts refractor Francisco Alvarez rookie Alec Minoa refractor and Corbin Carroll rookie Trey Turner world's finest and Giant skilled refractor to 50 Brandon Crawford This is nationals rookie green refractor auto to 99 Joey Manessus Hunter green Michael Toglia rookie refractor Riley green rookie Oscar Coloss headliners Ichiro Tel Marte and Jacob de Graham blue aqua vapor to 250 Josh young rookie Spencer steer lenient sosa rookie refractor Last pack Miguel Vargas headliners Albert Pujols And Bryce Harper on the back. All right Let's do a recap. Yeah, please be sure to hit the like button We very much appreciate that if you haven't already subscribed or followed. Please do so. All right We have for to go on 22 23 Donneress basketball team all addition Geez, that's a lot of letters Try to type two to five letters Like I just went on the keyboard Yeah, for to go. There's your link. All right. We had rookie design variation kinsu kroy black gold null and gorman cardinals Orioles rookie aqua refractor to 199 Adley ruchman Cubs rookie blue refractor to 150 Christopher Morel Red Sox rookie green lava to 99 Mastaka Yoshida Guardians rookie green speckle to 99 bow nailer Angels rose gold to 75 shohei otani Marlins orange refractor to 25 jazz chism jr Astros finest moments Otto your non avarice Yankees Otto clay holmes Tigers rookie Otto germain palacios Brewers rookie Otto Ethan small Orioles rookie Otto Kyle Stowers Reds rookie Otto Graham Ashcraft A's rookie Otto Dermis Garcia Rockies rookie Otto Winton Bernard Dodgers rookie Otto James Altman Dodgers rookie Otto Miguel Vargas Nationals rookie blue refractor Otto to 150 Jeter Downs Rangers rookie blue refractor Otto to 150 Ezekiel Durand Brewers rookie green wave Otto to 99 Ethan small Guardians rookie design variation Otto green refractor to 99 Oscar Gonzalez Nationals rookie green refractor Otto to 99 Joey Manessis Mets redemption that is finest rookies design autograph variation gold refractor parallel Francisco Alvarez And rangers rookie gold refractor Otto to 50 Josh Young That will do it for the break. Thank you everybody. We have case 60 up next


me is always seeking union that's literally the purpose of creation the very purpose of creation is to undergo this dreamed evolutionary ascent where the self-awareness goes from a place of being veiled by its forgetfulness of its true nature and going through a seeking process to pierce that veil in to union with its true nature so part of intelligent infinity uses the mechanism of this electromagnetic field architected universe that appears as a space-time that appears as planets orbiting stars that have evolved life from single cell to civilization and then those self-aware individuated mind-body-spirit complexes going through a process of seeking externally for peace and happiness and union with material objects or relationships or substances or experiences then they get hit by the drill sergeant of awakening which is suffering with a big sledgehammer which catalyzes that turn inward like in the parable of the prodigal son and then they find like Rumi said that diamond necklace that was already there the whole time and so they taste that union of that infinite conscious nature the absolute unbounded mysterious nature the unified field nature and then they go through a process of purifying those last bits of conditioning and then that is how this expression of intelligent infinity is architected and there are infinite other possibilities for expressions of intelligent infinity to undergo this dreamed ascensionary process of piercing through a veil of forgetting and awakening uniting with true nature


Hi. How are we, Emma? How are you all done? You're still all alive and here and everything. It's brilliant. That was a slightly terrifying introduction because the one thing that I'm not really doing today is presenting any new illusions, anyone who is a little bit of an auditory illusion nerd. Sorry. There are a couple of things before I get started, like housekeeping things. I'd like to get out of the way. The first one is I just want to talk about this title because I think I might actually be able to lay claim to have the worst title for any talk and the entire thing. So it's kind of like I was going for a pun, but I don't know what it was a pun on. Anyway, the second one is I've made a big mistake since I got here, which is having ideas. So the talk that I'm just about to do is not exactly the talk that I had on Friday. So the new bits, yeah, hopefully they'll work. Anyway, with those caveats out the way, I'll stop apologising before I've done anything.  Overall, I'll be talking about three different auditory illusions. They're basically just my favourites. I think they all give us a little bit of an insight into the mechanisms behind our perception of the world. But I just like them because they're cool, basically. So, sine wave speech. It's easiest with this one to start with an example. I've also put both audio and video in a PowerPoint presentation, so let's see how this goes. Sine wave speech sounds like this. It doesn't sound like anything. Have I got? Try again. So, hopefully, out of interest, could anyone kind of get any decipherable speech out of that sound that you just heard? No, no one. It's a clanger. Yeah, that's it. I've actually got a box of wild clangers behind the stage. If it's anyone's birthday, you come and find me at the end and you can have one. So that is an example of sine wave speech. So that is, it's generated from a recording of audio, and that audio is human speech. So, when I got into the green room before, I just asked one of the organisers to record me, you know, like a little sample that I could use for this demo. So the original sample sounds like this. No, it just moved it. This is why you auto-play your samples. That was the wrong sample. Excuse me a second. Does the laugh mean that people could understand that? Or you just like the weird sound? Booking Ben was a mistake. Actually, now that's almost prophetic. So for anyone that didn't hear that, that was... Booking Ben was a mistake. So like the interesting thing now, if I go back to the original sample, is hopefully hearing this. So could people now kind of hear human speech like sounds in... Yeah, I can see a few people nodding. So it worked eventually. So what this means is that I can make other recordings like these. Sorry guys, mouse issues. So hopefully still no one can really understand that. Or if I play this, then probably the same again. So it's good that probably no one can understand those. Because actually both of those recordings are massive spoilers for Game of Thrones. So if anyone gets out of line during the presentation, then I'm afraid we're going back to that. And anyone who's not up to date is a bit screwed. So I just quickly want to kind of get into this a little bit and tell you guys about how it works essentially. So it's time for everybody's favourite part of everything, which is audience participation. You guys feel like audience participating? Okay, that's enough of you feel like audience participating. That's not just going to be me. So what I'm going to do is show you pictures of things and see if anyone in the audience can tell me what they are. So does anyone know who that is? Sorry, at the back. So I think I heard someone say Niels Bohr. If they did, yeah. So this is Niels Bohr. So from now on, whenever you see this picture appear on the screen, I want all of you to just yell Bohr at me. Just Bohr. No Niels. So the next one is this. Can anyone tell me what this is a picture of? Ah, amazing. See, this is what I was expecting with Bohr. Apparently he's not a big enough celebrity. The last one I thought was far too easy. So I've kind of obfuscated it a little bit. Does anyone know who this is? Okay, so we've got Bohr, Sand and Cox. So let's just give this a quick go. That's just for my own entertainment, that last one. Okay, so the next one. What I want you to do is do the same thing again, but try and say all of the words without closing your mouth. Without, yes, without closing your mouth. Get it the right way round. So let's try this. Any minute? Okay, so a little bit more difficult. Next one, what I want you to do is... Hold up guys, you're ahead of me.  If anyone is likely to get short of breath, please don't, well, breathe. If you need to, then breathe. But it's not worth passing out for the sake of science. So with each picture, take a big breath now and then still try and say them just without breathing out. We've gone back. Keep holding your breath. All right, here we go again. Go on. There we go. Okay, so... I have a feeling that not everyone was holding their breath there because if so, you guys are really, really good at annunciating because what you should actually get from this is you can only kind of get up or a... and say, can you kind of see what I was doing there? Just trying to get a mass crowd beatboxing event going on. Just basically because I wanted to demonstrate that but I didn't want to do it myself so I thought I'd get you guys to do it. So what we've actually done there is we've made the voiced and the unvoiced parts of speech. So the voiced is when you try to talk with your mouth open without closing it and the unvoiced parts are these kind of type sounds. So, if I get, for example, the original recording of a Game of Thrones spoiler and play it like this... Everybody dies. Also, there are boobs. So... and the other one for good measure. Hey, isn't she from Hollyoaks? I wrote this talk a long time ago so I know that's like proper season one, but anyway. Incidentally, I'm not up to speed on Game of Thrones so no one come and talk to me about it afterwards. I'm like two seasons behind. So what you hear in those original recordings is both the voiced and the unvoiced parts of speech at the same time. With sine wave speech, what we've actually done is created a sound which follows the voiced part of the speech without the sort of the pts sounds. But it's not actually a recording of that sound so you can kind of tell when you listen to it it's got that clangory type vibe to it. And so the first time you heard it you would have just heard that whistling sound. But the second time you hear it you can make it out, right? So... I've got one of my next slides. I've rearranged my slides. This is going to happen every now and again. We're just going to have to deal with it. Ah, yeah. So what we've got in these sine wave speech recordings is not like we've got the Game of Thrones spoiler kind of obfuscator. I haven't put something in the way of it. I also haven't kind of blurred it out or anything. What I've done is reconstructed something which has some of the essential parts of the original recording that are not all the bits we need to be able to understand it like this. So your brain, with the knowledge of what that speech should sound like is able to reinsert all of these important bits of speech that we need to be able to understand it. So that's basically sine wave speech. And I think it's a really interesting example of how perception is incredibly subjective because it's something that we actively create all the time, basically. We're constantly doing this. We're bombarded with so much information and so much ambiguous information that we have to be able to use our knowledge about the world to try and figure out exactly what's going on. If I figure with this sort of crowd, it's worth mentioning. If anyone's interested in the sine wave speech demos, there is a piece of software and a script that you can use to generate them. It's really easy. I'll tweet the link later. How do I do it for time? Not too badly. For now, I just want to go on to the second one, which is... Oh, no, I'm not actually. Again, rearrange the slides. Can anyone tell me what this is a picture of before I carry on? So the people who are saying frog, have you seen it before or can you just see it? Is there anyone that can't see a frog? A few people. Yeah, okay, quite a lot of people. It's all right. You're not supposed to be able to. We've just got some kind of mad perceptual effect going on where people are just too good. So this photo is generated from this photo. For those of you that couldn't see it in the first place, this will probably make a lot more sense. This is just a visual homolog of the same effect that you're using your existing knowledge about the world to kind of make better sense of your perceptions. So I think that really is the last slide. It is. Okay, the next one is yes. You're just really not a fan of sine wave speech. Okay. Well, you may regret saying that because for this next bit I need a volunteer. Do you want a volunteer? Right, come up onto the stage.  Amazing. So you know how the McGurk effect works? Yes. Okay, brilliant. What I'm going to try and do is essentially perform it. Okay, cool. So what we're going to do here is give you conflicting visual and auditory information at the same time. Hopefully. By the way, quick caveat. This is one of the bits that I've added and I haven't tried this before and I don't mean like haven't tried it in a talk. I mean I haven't tried this before. So I literally have a browser window open in the background with a video that demonstrates it really well. So if this goes horribly wrong you'll still get the effect but we'll explain how it works. It's not the BBC one, it's another one. Okay, so I just need to do a quick bit of setup for this performance. So are you all right doing the spoken part? Yes. Okay, so that's the word. Okay. So all I want to do... Too close to the microphone. I'll tell you when to go. Okay, can we switch to the camera please? By the way, before we get onto the camera bear in mind I've been living in a field for three days and I'm going to have to get pretty close for this to work so don't judge me. Probably use the mic. What I want you guys to do is pay attention to what you see on the screen here, however horrific it is you need to be looking at the face for this effect to work. So on the red. Bass. Bass. Bass. Bass. Bass. Bass. Bass. Bass. Bass. Bass. Bass. Okay, that's probably enough. I just really hope that some people walk past behind then. Some kind of weird avant-garde theatre performance going on in here. Avant-garde minimal techno music, amazing. Sorry, what was your name by the way? Chemical Adam. Chemical Adam, thanks to Chemical Adam. Round of applause. So I was going to say did anything unusual happen to anyone there? Did anything like related to the topic unusual happen to anyone there? Did you get any unusual sort of perceptual experiences? So did some people experience it? It felt like you heard different words on different repeats. You got a few people. Okay, so it kind of worked. Can those people just put their hands up again? And then can the people who are just being nice put their hands down because I really want to know how well it works? Okay, so it did work for a few people. For the sake of everyone else, there is this lovely YouTube video which I'm just going to steal. So this is a much better example of the same effect. So again, the sound is not changing. Between the different faces here. So we've got like 15 minutes left. Auto play. So I can play that another 15 times or we could do something else. What do you reckon? Something else. Okay, so I mean this is, you can kind of maybe see how this is related to this effect in sine wave speech that you're getting conflicting information and you have to build kind of a subjective perception out of this kind of the facial movements that you're seeing and the sound that you're hearing. And as it turns out, the facial movements are strong enough stimulus, kind of salient enough stimulus for you to kind of incorporate that into what you decide the sound is. So this is, I was looking at some stuff about this recently. It's fascinating. Apparently this can work if you are just feeling the facial movements of a person. And it's reduced in tonal languages. So I thought that was interesting. Hopefully some of you do as well. But there's, you know, I thought this is kind of great. You can do this with lots of different types of conflicting information. So I can even, this effect is strong enough that I can do it with something that you're all probably familiar with. So I imagine quite a lot of you will recognise this as soon as it comes in. But hopefully I'll be able to kind of prod your brain into having a slightly different than usual sensory experience. So, um, yeah. So I mean you can do that with almost any song with words that sound vaguely kind of humorous. If you got my sense of humor. So, an interesting kind of side effects of this demo is, I've done this, I've used that video a few times. And the last time I used it, I was giving a talk in London somewhere, and they were filming them. And they put up not all of the set, not all like nine minutes of what I was saying, but just a clip of me stood next to that. So if you YouTube searched my name, there were like two videos of me. One of them is something to do with work, and the other one is just me stood by that.  So that's the second one. The third illusion that I wanted to go on to is speech to song illusion. Now, I have a slight issue with this, which I've had to fix at the last minute. Does anyone know who these guys are? Okay, not too many of you, that's not too bad. But has anyone been listening to Domestic Science on Radio 4? Right. So you will be aware of the fact that they did exactly what I'm about to do on Radio 4 about two weeks ago. I didn't know this until about three days ago when I was listening to it on Catch Up on Demand. Having already put this together, like... Oh. To be honest, I thought that more people would have listened to it. So shame on the rest of you, listen to it. It's great. Episode one, they do so much better of a job than I'm doing. But I thought, you know what? I'm not going to let myself be bullied by this massive broadcasting corporation. I'm going to do the demo anyway, but to extract a small petty revenge out of it, I'm going to demonstrate the speech to song illusion with a recording of Steve Mould from Festival of the Spoken Nerd describing the speech to song illusion just because I felt a bit meta. So this is a recording of Steve Mould from Festival of the Spoken Nerd explaining a little bit of the background of the speech to song illusion or what it's doing. Sorry, I'm terrible with track pads. Your brain has imbued that phrase with a musical quality it didn't have before. This effect was discovered by an amazing psychologist called Diana Deutsch, who presumably had a cassette player in too much time. I'm actually just stealing that joke at the end there as well, because why not? So what I'm going to do is, incidentally my main aim with this is just to have more weird noises and repetition than the Algarave stage. So I apologise beforehand for anyone who doesn't know what's about to happen. I'm going to take a small bit of what Steve was saying there and I'm going to repeat it until you all leave. Diana Deutsch, who presumably had... That didn't repeat. For any of the less observant people. Diana Deutsch, who presumably had... I was thinking of the fish in seconds. I thought he was actually looping. So did anyone have an odd experience there when they were listening to... In the last repetitions, do you experience anything different from the beginning? You lose a bit of the sense of the content, but specifically bearing in mind that the effect is called the speech to song illusion. Did anyone have the effect that I would want you to have had? Yeah, at the back. Sorry? Yeah, and you get to a point where it sounds like it's almost got a melody to it, right? No? Yeah, okay, thank God. Because otherwise the rest of this is just not going to work. So that is, like he said, all the spoken nerd guys on the first episode of Domestic Science do a great job with that. Loads of other really interesting auditory illusions, so do go and listen to that. But I was a little bit like, okay, they've done that. I need to do something else now. So I started reading around it a little bit and tried to figure out what else I can do with that illusion. And so as you probably picked up from that recording, that was discovered by Diana Deutsch, who presumably had quite a lot of interest in these types of illusions. And so she actually apparently discovered this because she released a lot of CDs, I think, of auditory illusions. And she was editing some of her own voice and had to loop one bit for ages and this happened. I can't remember what the original phrase was. But can you imagine anything more maddening than not just that, but that with your own voice? Apparently this wasn't maddening enough, so she did a bunch of research on it. They discovered, unfortunately I can't demonstrate this because we've already heard it. I've already basically broken you guys with that repeating. But if you change the pitch of the repetitions, it doesn't really work. If you rearrange the words, it doesn't really work. Not very surprising. But they did what I thought was a really, really nice little study because they were interested to know, that everybody hear the same melody in this illusion.  And then at the end, they asked these people to sing back to them what they heard as the melody in this sample after they'd heard it so many times. And they overlaid them all. And they found that people did seem to hear the same melody. So I thought maybe we could try the same thing. I mean you've been fairly gracious with your audience participation now so I'm just going to push you to like breaking point because I thought we could do a bit of auditory illusion karaoke. So I mean some of you might be thinking oh you know I just can't quite remember what it is. Diana Deutsch who presumably had... So that was Diana Deutsch who presumably had... I don't know why that's cutting off too late. Anyway, it doesn't matter.  So sorry. Here are the words if anyone wants to remind themselves before we do a sing-ong. But this will be the end of the talk. So it would be really, really nice if on the count of... on go, right, because this is always a mistake. Three, two, one, go. If people could sing along with one last repetition of this just to see if we all got the same melody. So three, two, one, go. Diana Deutsch who presumably had... Beautiful. I mean, yeah, I think I heard it. Did you guys hear it? Did that sound right to you? Okay, perfect. So thank you so much for actually getting involved in that because I was slightly terrified when I put that together that nobody would do it. So you guys are awesome. I've got a few minutes for questions at the end if anyone wants to know anything more. But that's it. Thanks.