OK. So what I'm going to explain is, in fact, I think quite in the line of this conference.

Because somehow what I am going to present is a succession of matrix models, essentially. And I mean, if we follow it quite closely, eventually we shall get to something which will display, if you want, in Euclidean version,

will display space-time. But with gravity coupled to the standard model. In fact, a small enhancement of the standard model. And this, everything will be coming from one very simple mathematical question. And the simple mathematical question would be, how can we encode four-dimensional compact spin

geometries? How can we encode this?

So this is a very simple mathematical question. And what I want, I want to have an answer which encodes them in the spirit of quantum mechanics. So this would be exactly in the spirit of what we were discussing yesterday, for instance, in the talk of Maxim and so on, of quantization. And I mean, to set the stage, I

will start by very, very simple considerations, very simple examples. So the first equation that I will write down will be the following. It will be the following equation between operators. u star du is equal to d plus 1, where d is self-adjoint,

not bounded, and u is unitary. So let's look at this equation. And what we look at, in fact, are irreducible representations of this equation.

So I mean, if we look at this equation with u unitary and d self-adjoint, what do we see? We see that if I have an element lambda in the spectrum of d, then this implies that lambda plus 1 is also in the spectrum of d. I mean, for the obvious reason that d is equivalent to d

plus 1 by unitary equivalence. So the spectrum has to be invariant under addition of 1, and of course, under subtraction of 1 for the same reason. And what this means, this means that if I look at an irreducible representation, automatically, in this irreducible representation, the spectrum will consist of an arithmetic progression,

inside the real line, because the operator is self-adjoint. So I mean, the spectrum of d will necessarily be in an irreducible representation of the form n plus alpha, where n belongs to z. So if you want z plus alpha in simple terms,

where alpha is some number, modulo 1, which I don't care about. And once you have this, what do you know? You know that if it's irreducible, also these eigenvalues will be simple. And if I take an eigenvector, so if I write the equation d of some eigenvector en for the eigenvalue n

is equal to n en, then what happens? Is that if I apply u to en, so if I take u of en, and if I apply d to that, and if I use this equation, I will immediately find that d of u en,

because I can permute u and d by adding 1, would be in fact equal to n plus 1 times u en. So what this means is that u is shifting the eigenvectors by 1. And when the representation is irreducible, it's very easy to show that, in fact, you

will generate the whole representation just by taking the vector e0 and then shifting it backwards and shifting it forward. So this will tell you that, in fact, there is only one sort of irreducible representation, except for the alpha, which is the following, which is that the Hilbert space H is L2 of z.

So it's square integrable series. The operator d is such that, with the obvious basis en, d of en is equal to n plus alpha. I should have put the alpha, but I forget it. n plus alpha en. And what about the u? So u is a shift, so u en is equal to en plus 1 for n.

Now, once you have this, you are happy. You have found the irreducible representations and so on. But what does it mean geometrically? What does it mean geometrically? I mean, it means the following. It means, by passing in Fourier, that, in fact, H

is also equal to the space of L2 functions on S1, with the measure d theta, the usual R measure, that the operator d is given up to the alpha by 1 over i times d by d theta plus alpha.

And that the action of the operator u, but more generally of the functions of u. So if I take a function, f of theta, f of u, if you want, this function will act by multiplying en, sorry,

by multiplying any vector, actually, L2 vector psi. And if I evaluate it at theta, this will be just f of exponential i theta times psi of theta. So if you want, the algebra is acting by multiplication operators. And the operator d itself is differentiation.

Now, you fall into something which is very familiar in non-competitive geometry, which is what is called a spectral triple. Namely, you have an algebra a, which is the algebra of functions of this u. You have an inverse space H, which is here. And you have an operator d. This operator d is telling you what is the metric.

Why? Because how do you compute distances between points? First of all, the points are given by the spectrum. So here, it was a spectrum of u. But in general, it will be the spectrum of the algebra as an algebra of operators. So the points are given by the spectrum. And the distance function, the distance

between two points, theta 1 and theta 2, it will be given by the supremum of the evaluation of a function at theta 1 minus the evaluation at theta 2 with the condition that the function doesn't vary too fast.

Namely, that its commutator with d is in norm less than 1. So what you can see in this example, in this very simple example, is that what one recovers completely from this operator equation is the full geometry of the circle of length 2 pi.

And there is only one. So what you recover is a circle of length 2 pi with its complete understanding of its geometry just from this operator theoretic data. Now, the circle is a very simple thing. And of course, you want to go further. Now, the first next example is the following.

First next example is an example where what will become obvious is that you will gain something by actually using matrices. Namely, if you want, instead of here obtaining the algebra of functions of u, which was commutative algebra, I will

from the start put in my algebra the 2 by 2 matrices. So now I am taking the following. I am taking the algebra M2 of C. And let me make the remark that if the algebra M2 of C is contained in any other algebra, then what you can do,

you can write any element of A. This is a very simple exercise. You can write any element of A, let me call it little a, as a11 times e11 plus a12 e12 plus a21 e21 plus a22 e22.

Whereas the eij's are the matrix units. So for instance, e12 would be the matrix 0 0 0 1. That would be e12, and you can guess the other ones easily. And where the aij's now are in the commutant, so they belong to the commutant of M2 of C.

So this is a general fact. You can do it as a bigger size. And now what we shall do, we shall try to find the simplest possible algebra which contains M2 of C but non-trivially. And for that, what we shall do, we shall add to M2 of C a punctuation symbol.

So I will add, Jenny must already recognize what I am after. So I will add the punctuation symbol. And this punctuation symbol, I will call it y. And it will satisfy the rules that y is equal to y star

and y star is equal to 1, which means that there is no gain in repeating y. That's why I call it a punctuation symbol. I can put a comma only once. So now we write words. We write words. We write words like ayb plus yc plus.

We write words like this. We write words like this. And then once you are just working like this with words and so on, you can define an algebra. Using this rule here, we can define

the trace of an element as being a11 plus a22. Equivalently, you can define the trace of an element by averaging it over 2 by 2 matrices. And what we shall require, actually, we shall require one more condition, which is that the average of y is equal to 0.

In other words, when I write y in this algebra, which contains M2 of C, I require that y11 plus y22 is equal to 0. Now, theorem, which is quite meaningful, is that if you take this algebra, which

is if you want M2 of C together with y, with this condition, and this algebra is naturally endowed with a norm. Why is that true? Because you can take all of its representations

as operators in the inverse space and take the max norms that you get from representations. So for instance, if I take this element here, I will define its norm as being the maximum, the supremum, of all norms it has in representations. I mean, this is clearly bounded, because when you take 2 by 2 matrices, their norm is, of course, the same norm as in 2 by 2 matrices.

And when you take the other elements, because y squared is 1, they are also bounded. So it's bounded. And the theorem is that if you take this algebra with its norm, you get continuous functions

when you complete, of course. This is nothing. Continuous functions on the 2-sphere with values in M2 of C. The algebra generates 2 values in 2-sphere. So this abstract algebra, when you complete it, what you get is continuous functions from the 2-sphere with values in M2 of C. And the smooth elements

are obtained by only taking the smooth representatives, which is a fairly obvious thing to do. You take the smooth function of calculus. So what does this tell you? This tells you that if we had looked for the sphere, we would have to say, OK, we take this and that and so on. But here, what it is saying is

that it's simpler to actually take matrices of functions on the 2-sphere than on the 2-sphere itself. And of course, if you know a little bit of the principles of non-commutative geometry, you will know that the presence of the non-commutativity in the algebra of matrices is not at all a problem for other reasons,

because it gives you the gauge fields for free. So in other words, this algebra has inner automorphisms, but it also has, matrically speaking, what are called inner fluctuations of the metric. And this gives you the gauge fields. So how do you get the matrix? So how do we extend this relation over there

when we are talking about the sphere? OK, so what you do, so I should say that this type of work has been done with Jenny and with Michelle. I mean, we have gone to the quantum sphere and so on. But I mean, OK, so I just explained it like this. Now, if you take the metric problem,

how will you get the metric? Well, it turns out that you get the metric in a very, very easy way. I mean, in the simplest possible way. You generalize the condition which was there by, you know, first of all, rethink about the condition which was written there as being u star d commutator u is equal to 1.

Of course, it's the same equation. Now, once you think a little more, you'll find out that in this case, you can add another entry in the theorem. And the other entry in the theorem is that if now you consider the following equation, which is y d commutator y,

but now I have to square it. And the reason why I have to square it is that I am in dimension two. And now I no longer write that it's equal to one. I write that it's equal to gamma. What is gamma? Gamma is a chirality on the spinor representation.

Because now what you get as irreducible representation is a spinor representation. There is a chirality operator because we are in even dimension. And now the equation, which in the odd case was just equal to one, will always in the even case be equal to gamma. Where gamma is a chirality. So what is it, mathematically speaking?

It's something which satisfy gamma square is equal to one. Gamma is equal to gamma star, of course, as usual. But gamma d is equal to minus d gamma. And gamma commutes with the algebra. So gamma a is equal to a gamma for any element of the algebra a. And what is the incarnation of y?

What is the incarnation of y? It is the instanton. I mean, it is the, if you want, y is related to a projection. Because y square is one. So there is this relation, you see the y with y square is one, is very much the same thing. You can write y equals two e minus one.

And then e is a projection. And the corresponding projection on the two sphere is a well known formula for the instanton. So now what you have? You have a translation, okay? And it turns out, so as I said, that which are the metrics that are solutions?

So it's not only the round metric which is solution, which is beautiful, because in the case of the circle, we had only one circle. But in the case of the sphere, we get all possible spheres. The only condition on their metric is that the volume form of the metric is fixed. It is the volume form of the round metric, okay? This is the only condition. So here, if you want, we get all metrics.

All metrics. With the fixed volume form. Okay, so this is really the starter.

This is really the starter. I mean, this is what the situation was. I mean, until the beginning of 2012 or something like that, when my collaborator, Ali Shamseh-Din, looked again at these papers of the 2000,

and he came with an idea. And I mean, of course, this idea existed also, I mean, implicitly, also in the work with Michel. But he came up with the following idea, to extend it to higher dimensions. The idea is that, you know, when I was talking to you about M2 of C,

the algebra of two by two matrices, and I was imposing this strange condition, Y equals zero. In fact, there is a more conceptual way to think about it. And the more conceptual way to think about it, is that in fact, I am secretly writing Y is equal to Y mu times gamma mu,

where the gamma mus are the three Pauli matrices. Okay, and you see, of course, all of these matrices have trace zero. And so I get all two by two matrices, but with this condition. So when you write that, Y equals Y mu gamma mu, and of course, when I write that,

I am in the same domain as when I was writing the element A as combination of elementary matrices. So the Y mu are assumed to commute with the gamma mus. So if you want the Y mus belong to the commutant of this Clifford algebra, generated by the gamma mus, okay.

And then what you will find out, okay. In fact, it will give you a proof of the case of the sphere, because when you write down the conditions that Y is equal to Y star, well, this gives you that the Y mus are self adjoints. Okay, I call to the adjoints, I mean, this is trivial. But when you write down the conditions

that Y square is equal to one, okay, what you find out is that, you know, if you write it in terms of gamma matrices, you would have cross terms. You would have gamma mu, gamma nu, gamma nu, gamma mu. Now they anti-commute. So what this tells you is that the Y mus commute with each other, okay. So this tells you automatically that Y mu, Y mu

is equal to zero, okay. So this is what you get. And of course now the fact that Y square is equal to one also tells you another thing. It tells you that sigma of Y mu square is equal to one.

So what do you have? You have commuting operators, which square adapt to one, okay. So of course, you have a mapping to the sphere, okay. You have an obvious mapping to the sphere. Okay, so I mean, what we started by investigating with Alisham Sedding and Slava Mukhanov,

we started investigating this equation in higher dimension and also this equation, okay. So of course, I mean, then we were also looking at the equation Y times commutator DY to the power 2M. So I am only in the even case. And I have average over the Clifford algebra

and this is equal to gamma. So this was a basic equation that we started to investigate, okay. And I mean, this equation is quite interesting. Let me tell you why. Because when you start about to talk about spectral triples in general, AHT, okay,

the operator D is a Dirac operator, I mean, as defined by Attia and Singer. It is them who really defined the Dirac operator for Riemannian spin manifolds. I mean, of course, it was known to physicists, it was even known to Hamilton. But the point though is the following. The point is that when you write down this equation here,

well, the operator D is like the momentum. It's a slash of the various components of the momentum. It assembles all these components into a single object, which is very economical.

Now, this was missing on the side of the components because on the side of the components, what we were doing, even when you write the algebra, you don't assemble them into a single thing. But now the Y actually assembles them into a single thing because it is this Y equals Y mu gamma mu. So now the coordinates themselves are assembled into a single thing, okay.

And I will come to the mathematical meaning very soon. So we investigated this condition, and what did we find? What we found with this equation was disappointing because what did we find? We looked at this equation, we looked very closely at this equation,

and what we were able to do, of course, is to do it for the sphere. So if you take the sphere as 2m of even dimension, then exactly as in the case of dimension two, you will find that for any metric which has the same volume form as a round sphere, you have a solution, okay.

This is not difficult at all. But yes, yeah? This expectation is into the commutant? No, into the commutant of the gamma mus, okay. So commutant of the gamma mus, okay. You see, the gamma mus, they generate a key for the algebra C on the commutant of C. So what you can do, you can average if you want on C,

on the unitaries of C. I mean, this is the best way. It's a conditional expectation. So that's the condition. Okay. It's a gain of dimension two or higher? No, m is arbitrary here. Oh, yeah. M is arbitrary, okay. So of course, you get the sphere as a solution. This is not a problem, and you get all metrics

which have the right volume form in the sphere. I mean, this is not difficult at all to see. But now, what we tried to do, we tried to find a solution to this equation when we are taking a manifold, m, and when we were taking d to be the Dirac operator on this manifold, and so on. And what did we find?

We found something which was quite disappointing, because we found that if there is a solution, y to this equation as a function, you know, and so on, on m, then actually y defines a map from m to the sphere, s to m, okay, because of the components. Obviously, it defines a map from m to the sphere.

And then what we found is that the Jacobian of this map never vanishes. Why it never vanishes? Because actually, when you write down this equation, it's an equation on the Jacobian of the map y, and what it tells you really, I mean, in other words, this equation is really exactly telling you

that when you pull back the volume form of the whole sphere, you get the volume form of the remaining manifold, okay. So of course, it implies that the Jacobian cannot vanish. But if the Jacobian cannot vanish, what it means is that you have a covering. You have a smooth covering of the sphere. Now the sphere, unfortunately, is simply connected.

So when we found that, we were quite disappointed, because we said, you know, I mean, of course, this sphere will be of Planck size, when you think about physically. I mean, I will come back to that. So the sphere will be very tiny sphere. So I remember when we were discussing with Ali and Salava,

I was talking, you know, about these little bubbles that you do when you play when you are a child, you know. So I mean, what I was telling them is that at the moment, we are only able to find Euclidean space times, which look like a big collection of bubbles. And I mean, this is not very satisfactory from the point of view of physics, okay.

So now it turns out that there is a beautiful solution to this problem. We shall find all spin for manifolds with quantized volume, and so arbitrarily large volume. But in order to explain to you the solution, I have to make an intermezzo through mathematics, okay.

So it turns out if you want that to find the solution after this example, this concrete example, it turned out that one had to think very carefully about what was really going on behind the scene when I was writing these equations, okay. And in fact, when I was writing this equation,

as I said, I was using the Dirac operator, okay. The Dirac operator, I mean, but what is the meaning of the Dirac operator in topology it is what is called the, it's a cycle in what is called keromology.

And I mean, it's a great contribution of Michael Attia, actually. You know that Michael Attia with Ersebrock, they invented K-theory after Grothendieck, of course. They transported the ideas of Grothendieck of K-theory to topology. But in topology, there is something which is very nice,

which is that whenever you have a keromology theory like K-theory, there is automatically a dual theory, which here is called keromology. Now, the great discovery of Attia, and also Singer, but I mean, it's Attia, I think, who wrote first the paper, is that the keromology cycles are actually given by Fredholm representations of an algebra.

So in particular, the Dirac operator, the way you should think of it in these topological terms is as defining a keromology class. Moreover, the y, because it satisfies y equals y star and y equals 1, of course, it's of the form y equals 2e minus

1. So y is a K-theory class. It's an element of K-theory. It's a K-theory class. And somehow, the equation which we have written there is an equation which tells you that you have a non-trivial pairing between keromology and K-theory.

It's a kind of roof of the theory, telling you that the two pairs non-trivially. Now, this is too naive. And this is why we didn't find the solution there. And why is it too naive? It's too naive. Why?

Why is it too naive? Well, it's too naive because of the work of Sullivan in the 70s, and many other people, but especially of Sullivan's thesis. So what did Sullivan prove? You see, when you do geometry, you have various classes.

You have smooth manifolds, which we think we understand, at least for the definition. Then you have what are called PL manifolds, piecewise linear. So we have PL manifolds.

Then you have all sorts of refinements and so on. And then what you have is Poincare duality. I won't write down all the intermediate steps. But what you have is Poincare duality in homology, Poincare duality spaces, if you want. And this is an ordinary homology.

So if you want to, when you take a space which is a closed, compact manifold, oriented, it has a distinctive property in homology, which is that it has Poincare duality. And let me assume that pi 1 is trivial. Otherwise, you have to take the fundamental group

into account, but I won't bother about that. I assume pi 1 is trivial. So what you have, when you have Poincare duality in homology, there is a theorem of Spivak, which tells you that you can take your space, you can put it in a Euclidean space, and you can put it in Euclidean space in such a way

that it will have a kind of normal bundle. So it will have something like a tubular neighborhood, if you want. It will have a normal bundle. Now, this normal bundle, in general, it will not be a vector bundle.

It will be what is called, I mean, it's something which is a bundle in homotopy theory. So it's like a sphere bundle. I mean, the unit sphere bundle. So it's something which is not linear at all. Micro bundle, in the sense of mean, null, and so on. So I mean, why is this a buzzer? It's a big buzzer, because if it were a true bundle,

then René Tombe would tell you that you apply a little bit of transversality, and you have a manifold. You know, you cut it through, and you have a manifold. So then, if it were a vector bundle, the rest thing would be fine. So the whole issue of, if you want,

somehow improving the structure from that side to up to that side. But I mean, the real note is the PL, is the problem of transforming the micro bundle, the normal micro bundle, into, hopefully, vector bundle. But if it's a PL bundle, it's good enough. Then you're getting the right.

Because, for instance, in dimension 4, PL and smooth is the same. Okay, so I mean, what was the great discovery of Denis Sullivan? The great discovery of Denis Sullivan was that the condition that you can actually lift this micro bundle to a true PL bundle

is exactly that you have K-O homology, that you have a K-O homology class. Okay, so the obstruction is in K-O homology. And what does it mean if you can lift it as a K-O homology class? Well, you pass to the orthogonal, which is a tangent bundle.

And then what do you get? You get two things. You get Poincare duality, but no longer in homology. You get Poincare duality in K-O homology. Now you might think that these are nuances, fancy nuances for topologies. Not at all. Why? Because you see, in topology, there is a chain character, which goes from a K-O homology to ordinary homology.

Okay? And when you compute the chain character of the fundamental class in K-O homology, you get the Pontryagin classes of the manifold. And the Pontryagin classes of the manifold are not at all homotopy invariants. They are not given homotopically.

They can be extremely different for homotopic manifolds. So they tell you the whole structure of the manifold, essentially, I mean, at least rationally, okay? So in fact, it's very, very important for the manifold structure that you are able to lift the structure from ordinary homology to K-O homology.

Now, why were we too naive before? We were too naive before because we had forgotten the O. We are talking about K-O homology, not K-O homology. O refers to orthogonal?

Yeah, O refers to orthogonal. O refers to the nuance, if you want, that there is. And how is this nuance translated when you look at Attia's version of K-O homology by means of Fredholm representations? This nuance arises, I mean, this is explained in the book of Attia on K theory.

So the nuance between K and K-O, nuance between K and K-O, okay, what is it? It's an additional structure. It's an additional structure on the Hilbert space,

on the Hilbert space H. And what is this additional structure? It's beautiful. It has three meanings. Concretely, it looks very trivial.

Concretely, it's a real structure, real structure. By what I mean, I mean an operator J, which is anti-linear, anti-linear from H to H, okay.

It satisfies that its square is plus or minus one. I will denote this by epsilon. It satisfies some commutation with D I don't care about, but it satisfies some commutation with gamma, which I care about, okay. And which is also by plus or minus one. Now, I am in the even case,

and this epsilon and epsilon double prime, they actually distinguish between four different dimensions. These dimensions will be valid modulo eight. They are valid modulo eight. They are related to the Clifford algebra classification. And I mean, there is a table. Don't ask me much about this table.

I don't know if I will remember it. So there is epsilon, there is epsilon double prime. There is a dimension zero, dimension two, dimension four, and dimension six, and then it repeats. Dimension zero, of course, epsilon is one, epsilon double prime is one. Dimension two is very interesting

because it has minus one and minus one. Dimension four has minus one and one, and dimension six has one and minus one, okay. So that's what you get, that's what you get. Now, what are the three meanings of J? So this is extremely important. So what are the three meanings of J?

Three meanings of J, what are they? So the first meaning of J is that unlike K-theory, which has just an even and a odd incarnation, the KO-theory has eight incarnations, okay.

So I'm only talking about the even ones, there are only four ones, and they are governed by this rule of science, okay. So this is the first meaning, this is the first meaning. Now, so what is the second meaning?

The second meaning is well known to physicists. The second meaning is that J is known to physicists as a charge conjugation operator, okay, in Euclidean. I am working in Euclidean. So for physicists, J is a charge conjugation.

But okay, this is not very surprising so far, okay. I mean you are sort of in well paved waters and so on, I mean, you know. But there is a third meaning, and this third meaning is a key, it's a big key, okay. And what is this third meaning? It turns out that this J, exactly with that notation, is a core of Tomita's theory,

and what Tomita has found, he has found a marvelous theorem, if you want, which is that if you take an algebra in inverse space, and that algebra is sort of wild enough, it's just most general, what does it mean? Well, you have to assume that the algebra A acts in H,

okay, and you don't want it to be all operators in H. You don't want it to be too small either. So what you ask is that it has a cyclic and separating vector, okay. If you don't know what this means, it doesn't matter.

Okay, so you ask this condition, that it has a cyclic and separating vector, and then Tomita tells you that there is an operator J that does a kind of hat trick. It sends the algebra to its commutant, okay. So I mean, what Tomita tells you is that you have a J now which is an operator satisfying the same rules as here,

and which would be such that J A, J inverse, will commute with B for any A and B in the algebra. So this means, if you want, in some sense, that this operator J is restoring commutativity. It's a resource, it's a resource which is there,

and tells you that even though the algebra is not commutative, you have this operator which is a substitute for commutativity, okay, which restores some commutativity. Sorry, sorry, sorry, sorry, sorry. Yeah, sure, zero, zero, zero, of course. This is always zero, okay. All right, so now, if you want, now we understand that we were too naive

when we were writing this equation. This equation here, we were too naive because we were not really telling that the space we are considering, which is generated by, say, the components of the y's,

I mean, the spectrum, is a Poincare duality manifold. We are not telling it in K-O homology, which was a pity because we had the same language as K-O homology. So now we have to include in our picture, and in the equation as well, we have to include this operator J.

Okay, we need to include this operator J. Okay, and this is what we did with Alisham Seddin and Slava Mukhanov. And then what happens, what happens is incredibly surprising. I mean, we're really incredibly surprised

because you imitate what is done here, you put in the J, okay? You imitate what I have done here, you put in the J. And what do you find? Well, you find that because you put in the J, okay, because you put in the J,

then somehow, if you want, what is going on is that, well, and you know, what we want is dimension four. We want dimension equals four, okay?

But it turns out that this dimension four doesn't allow you to write down the correct approximation when you look at the action for fermions. And we want to correct this dimension four into dimension two. Okay, so we would like four to become two.

Everybody in string theory knows what to do. I mean, you add six, plus six is equal to two two. Module eight, of course, okay? I mean, you all know that. Okay, and this dictates the structure you are looking for as far as J is concerned. Namely, if you want to have that,

then what you need to have, if you want, is you need to have a situation like this where you're in the space, if you want, it will have two pieces. I mean, it will have a piece where, well, so, I mean, if you want a Y, we'll satisfy now Y four is equal to one,

not Y square is equal to one. There will be a piece in which Y square is equal to one, and there will be a piece in which Y square is equal to minus one, okay? And these pieces will be intertwined by J. So the J will flip the two pieces, and the J will also satisfy the Tomita condition

of the commutativity. So, I mean, from this, you find a structure which emerges, which is completely canonical, and this structure actually gives you that, if you want, the type of Clifford algebra which I was using before, that one, with the gamma mu's, which was allowing me to put a slash. Now, if you want, there will be two types of gamma mu's.

There will be the gamma mu plus here, and there will be the gamma mu minus here. Now, how many of them there will be? There will be five. Remember, for the case of the two-sphere, we had three gamma's, okay? So here there will be five gamma mu's, five of them.

There will be five of these gamma mu minus, five of them, okay, and they will satisfy some rules, the rules of the Clifford algebra. And then you have everything that you need to go, okay? You have everything that you need to go,

because you have the replacement for the gamma mu's. You will have the replacement easily for the y, okay? And you will write down the very similar equation which I will write down in short time. But our great surprise came here. Our really great surprise came here, because we looked in a table. You can look in a table.

What do we need? We need an irresistible representation of this gamma plus mu, and an irresistible representation of this gamma minus mu, okay? You take a table of Clifford algebras, okay? What do you find?

So you take, for instance, the book of Lausson, or any books of physics, and you look at the Clifford algebras. And we look for an irresistible representation of the gamma mu's, the gamma mu plus, and the gamma mu minus.

Sorry, what? Plus five plus five. Five plus five, yes. So you find an algebra which is C plus, plus there are some C minus. The C plus will correspond to an irreducible representation. Irreducible representation of the five pluses of Cliff, if you want, of plus, plus, plus, plus, plus.

And the C minus will correspond to an irreducible representation of Cliff of minus, minus, minus, minus, minus. Now, remember that our problem was a geometrical problem.

Remember that our initial problem was a geometrical problem. You look at tables, what do you find? You find that this algebra, an irreducible representation is two by two matrices over quaternions. Maybe I am flipping the two,

but no, I don't think so, actually. And this one is M4 of C. Now, at this point, we were totally mystified. We are totally mystified, because in my work with Shamsedin and Walter van Salekom, we had from bottom up,

we had found that an algebra that would be marvelous in order to encode the gauge transformations and so on of the standard model before symmetry breaking would be M2 of quaternions plus M4 of C. It is slightly different from the standard model in the sense that it's asymptotically free

and it is the smallest asymptotically free extension of the standard model actually that you can get, because the U1 is replaced by an SU2. Okay, so at that point, we were really totally mystified. Totally mystified, because what does this mean? This means that you ask a geometrical question.

You start from a mathematical geometrical question. And the answer is, look, it's much more beautiful if you don't take scalar-valued functions. If you take matrix-valued functions. And this will give you, of course, the gauge transformations and everything else. Okay, so now, what is the equation?

So we wrote down an analog of this equation, you know, of the equation, sorry, which is very simple to write. It's Z, again, DZ to the power of four is equal to gamma. This Z is a little bit more complicated, because you have to use the J. So remember that there was a relation

between the Y and the projection. Okay, the projection was obtained by taking Y was equal to two E minus one. So similarly, what will be the Z? The Z will be related not to E, but to E times GE, G in this. So if you want, the Z actually uses the J

to put the projection not only in the algebra, but also in the commutant, which is actually essential for Poincare duality. I mean, it's the correct thing to do for Poincare duality. So now we had a geometric problem, we had a beautiful geometric problem, because now, you see, it's no longer true as it was before, before we had this awful obstruction

that the Jacobian couldn't vanish. But when you write the new equation, equation which I wrote down here, you no longer find that the Jacobian of the map Y is equal to the volume form. What you find is something slightly more subtle. What you find is a different equation.

You find that the volume form is, again, can be written, but it can be written now as a sum of two Jacobians, because now you have two maps. You have the map if you want Y plus, which is coming from the plus. So you have a map Y plus to the sphere, S four, let us say, okay, from wherever you are, M.

Okay, and you also have a map Y minus, which goes to the sphere. And what is the condition, actually, when you write down the condition, you find that the condition is really that when you pull back from Y plus the volume form V of the round sphere, and you add the pullback by Y minus

of the volume form of the round sphere, you find the volume form square root of G d four X, if you want, of the remaining manifold. Okay, so this is the equation. And now, at this point, with my two collaborators, we had a problem.

Is it true that if I take a four manifold, I can find two maps to the sphere, so that the sum of the Jacobians of these two maps will never vanish, you know? I mean, if it never vanishes, it's good, because then it's by a theorem of yoga and mother, you can make it any volume form which has the same volume.

This is not difficult. This is a beautiful theorem of yoga and mother. But the problem is that it shouldn't vanish, okay? So, I mean, the issue is, does this vanish or not? I mean, is it possible to find two maps so that this doesn't vanish? This was a purely geometric problem.

And, well, okay, then, it took some time to resolve this problem. Took some time to resolve this problem. I mean, I had the help of several geometers, in particular Simon Donaldson, and also several geometers in Lyon. And, first of all, so you ask,

I mean, is it always true that if I take, well, let's climb in dimensions, because it would be easier. So, you see, you can ask the same question in any dimension. So you can ask, if you are with a sphere Sn, and if you have a manifold of dimension N, is it possible to find two maps to the sphere Sn

in such a way that the sum of the two Jacobians never vanishes? Okay, you can ask this question in dimension N. I ask it in dimension N, according to the well-known principle in mathematics, that if you have a problem, you should first generalize it, and then specialize it to a simpler case.

Okay, now, in the case of S2, okay, I mean, M2 will be a compact Riemann surface oriented, and this is very easy, because it's always a ramified cover of the sphere, okay, I mean, by well-known theorems. And if it's a ramified cover of the sphere,

I mean, this is bad, of course, because you have ramification points. But what you can do, you can move the ramification point by a small diffeomorphism. You can move them away from where they are, and then, of course, the sum of the two Jacobians will never vanish, okay? So, I mean, dimension two is sort of obvious. In dimension three,

when you look in dimension three, S3, it's a little bit more complicated, because now, in dimension three, what happens? It's not at all, first of all, you see, in general, if you take a map from a manifold M of dimension three to the three sphere S3, okay, the place where this map will have a vanishing Jacobian

will be a co-dimension one. I mean, it's an equation, the determinant to be zero. But in fact, this is the beauty of complex analysis. Complex analysis tells you that you can always find the ramified covers where the ramification will be of co-dimension two, okay?

And this is exactly because of complex analysis, okay? Because of the map, if you want z goes to z to the power n. Yeah, it's dimension three, so it's a complex field. So of course, but it doesn't matter, because you imitate the complex case and then you get the same thing topologically, okay? So what this tells you is that you can first find a map to the three sphere

so that the place where the determinant of phi vanishes, okay, where it ramifies, is of co-dimension two, okay? In fact, it will be a knot, okay? Or link, you know, it will be something like a knot. Now again, what you can do, you can move this knot away from itself

by a diffeomorphism, and then you are back to problem, okay? But now in dimension four, this is no longer the case, because in dimension four now, what do you have? You will have a map from the manifold M to S4, okay? With a little bit of nice choice,

you will make this map to be ramified in co-dimension two, okay? This co-dimension two, what it will be? It will be a surface in M, okay? But now a surface in M, it has co-dimension two, okay? So, and what do you have? You have that N minus two plus N minus two, okay, is equal to N.

So it's the same thing as saying that N is equal to four. Okay, so N equals to four is really the critical case. Okay, it's really the critical case. Now if you begin to look a little more closely, of course this requires some knowledge in geometry, if you begin to look a little more closely, you'll find that it doesn't work for P2 of C.

So if you take P2 of C, okay, if you take the manifold P2 of C, M, it is not possible, I claim, to find two maps from P2 of C to the sphere S4, which have the properties that the sum of the Jacobian is never zero. And what is the proof of that? The proof is not so difficult, because,

so this is due to the geometries in Léon. What is the proof? Well, you see, if you could find such a pair, in fact you could find an open cover of your manifold by two open sets, and such that if I take this church covering,

if you want, the tangent bundle would be trivialized on each of them. Okay, of course, because the Jacobian was not vanishing on each of them. And then it implies that the square of the second Stiefel-Whitney class is actually vanishing. That's an exercise, okay. So if it exists, you would have that W2 squared is equal to zero.

But this is not true for P2 of C. When you look at the projective space, P2 of C, the square of the Stiefel-Whitney class doesn't vanish. So at that point, we were quite worried with Ali. We said, okay, maybe we don't cover. However, there is a beautiful theorem, I think it's due to this geometer in Switzerland.

Well, anyway, I mean, it's a fact, which is not so difficult to prove, that a manifold is spin if and only if W2 is equal to zero. If and only if the second Stiefel-Whitney class

is equal to zero. So then we said, oh, okay. Then the obstruction, at least, is not there, okay. So spin means W2 is equal to zero, okay. Now, so and eventually we found the proof

that if the manifold is spin, you can find two such maps. Not only you can find two such maps, but you can find two such maps so that the sum of the degrees of the map is as large as you want. And this is extremely important. I will give you a picture of these maps. Sorry, four, in dimension four.

So the theorem that we proved is that in dimension four, so in dimension four, okay, for any integer n greater or equal to five,

okay, there exist two maps, phi and psi, or y plus minus, if you want, y plus minus, from the manifold M to the sphere S4, and which are such that the sum of the degrees, the degree of y plus plus the degree of y minus is equal to n, and moreover, the Jacobian of y plus

plus the Jacobian of y minus never vanishes. It's always different from zero. Everywhere, everywhere, okay. And what was the proof? The proof is highly non-trivial, in fact.

First of all, because it chooses a beautiful result from Italian geometers that for any manifold of dimension four, you can write it as a ramified cover of the fourth sphere with at most five layers. So what you do is you take the fourth sphere, you write it as a ramified cover. You write the manifold as a ramified cover

with at most four layers. But in fact, you can easily add as many layers as you want, I will show you the picture, okay. But then you are worried, because then you get this surface inside, okay, where the Jacobian actually vanishes, which is of co-dimension two, okay, where the Jacobian vanishes. However, this is a beauty.

What happens is that if you take a tubular neighborhood of the surface here, it turns out, if you look closely, that you can prove that this is parallelizable. This is coming from the spin structure. And because the tubular neighborhood has homotopical dimension two, and then you have lifted to the spin cover,

so everything is fine, because the spin group has no homotopy in dimension less than three, and then the classifying space in dimension less than four. So I mean, it's very easy, okay. What do you do then? Well, then you apply a great theorem of Valentin Po-Enerault. Valentin Po-Enerault proved the following theorem

by using Smale's theory. He proved that if you take a parallelizable manifold, called open manifold, open is crucial, of dimension n, and if it's parallelizable, you can immerse it in Rn, in the same dimension. You can immerse it in Rn with the same dimension. I'm stopping, okay.

So now, when you put things together, what you do, you use this process as a way to glue the bubbles together, if you want. And then what you have, you have this cover of the original sphere, okay, by these two maps, and you have the fact that the sum of the Jacobians

actually doesn't vanish, okay. So this is the main result. Now, so, might be all of this is a coincidence, but I mean, my very, very strong belief is that what we have here, what we have found, if you want, is a way to make geometry be born

out of very, very simple quantum mechanical equations, okay. Of course, our next step is to quantize this, and I mean, it's related to sigma models, I mean, because of the map to the sphere, and so on, and that's what we are working on. But somehow, if you want, the framework

is quite intriguing, because it's motivated by quite profound mathematics about understanding what is a manifold, and at the end of the day, when you use the spectral action and everything, I mean, you know, it recovers gravity coupled to matter. So, I mean, maybe it's, maybe there is someone who is laughing at us,

but okay, there is also the other possibility. Dimension four. Dimension four. Dimension four is crucial, of course. Okay, so I think that we stop here. I just, I didn't get this name of Ballantin.

Poinaro. Poinaro is a great geometor who lived in France, he's a Romanian, he migrated to France, and he worked for many years on the Poincare Conjecture. This means every non-compact parallelizable is immersed.

Immersable, if it's parallelizable, it's immersible in iron, in the same dimension. So every non-compact legal? Sure, sure, sure. Sure. If we want to dream a bit about quantization. Yes, sir. In the PPL class, if you start from pre-ambulation that you do.

You get a small fraction that are smooth manifolds. Many of them are quasi-manifolds, and they do have singularities. Sure, but here... You operate quasi-manifolds with singularities into your... Okay, well, you see, I mean, singularities, I don't know... What I know, I have not actually completely characterized the solutions.

What I know is that the manifolds, which are ordinary smooth manifolds, are solutions, but I have not classified all solutions. I have proved the theorem, which is called the reconstruction theorem, which tells you the extra conditions to get a smooth manifold. My question is almost the same as having a 3D boundary. Oh, yeah, okay. Why do we need boundaries? We are also thinking about it. Of course,

I mean, this is, you know, why do we need boundaries? Because the link to quantum gravity, of course, is not in four dimensions. It's taking two three-dimensional manifolds and taking all co-boundaries, co-boardisms between the two and making the functional integral over this Euclidean functional integral. For that, you need to know the

case with boundaries. So, but, you know, I mean, we are all a little bit lazy, so, I mean... Yes. A small detail. Yes. M2H is of dimension 16. Yes. And 4C of dimension 32, but why 16?

Yes, okay, but this is what, when we, with Ali, before we found that. We had these two algebras, M2 of H and then 4C, and we are saying it will be impossible to find a rationale behind that, because one of them is dimension 16, the other is dimension 32. What is the reason behind this? The reason is that Clif, where is it?

Yes, Clif of plus plus plus, it's this plus M2 of H. But you only need one irreducible representation, so you don't care. Maybe one more question. Yeah, at the beginning you started with this continuous mapping from S2 to M2C.

So is there any, because... Arbitrary continuous maps, I mean, this is algebra of continuous maps. S2 is some complex continuous space, if you introduce a complex structure, and, or maybe any Riemann surface, because Dirac operators are kind of concealment. Well, this is a specific example of a metric given by some spin structure, that's all.

So does generalize to this? Yeah, you can, you can, you get all possible examples of Dirac operators, in particular those that come from other structures, yes. But in general, complex structure doesn't give you a spin structure, it gives you a spin C structure. So this example, generalizes to this? Yeah, sure, sure. But you need the, you need the spin structure,

because of the J. I mean, the J operator is very, very important. All this structure is extremely tight, you see, and I mean, it owes a lot to Michael Attia. In fact, with a friend, we have written now, if you want, a long paper about Michael Attia,

his contributions to mathematics, and so on. In particular, I wrote a long section about non-commutative geometry, how his ideas played a crucial role in non-commutative geometry, his idea about k-theory, k-o-omology, and all that. People take it for granted, but you know, I mean, it's, it was a great, really great contribution that he made.

Any other questions or comments? So we thank, again, our students.