So last week, I introduced the idea that non-commutative geometry can be taken as a basis to unify all fundamental interactions.

And so what I did is that I defined what we mean by non-commutative space. I gave the definitions. And so certain properties. And just as a reminder, just set the spectral triple.

And I defined these five quantities and the relation among them. And then the question, of course, is if non-commutative space is the basis to unify the interactions, then the question is,

what kind of non-commutative space should we take? And as a starter, I made an assumption. And the assumption I made is that space-time R

is a non-commutative space which is formed as a product of continuous four-dimensional space

times some finite space.

Now, four-dimensional spaces are well understood. And then I went to see essentially what one should take for the finite space. What kind of space should one take? And so for that, started by classifying

spaces which fit certain physical criteria. And the physical criteria that we need are the following. That the Hilbert space, which would be space of fermions,

essentially, for us, would include all the fermions. And in this case, actually, the fermions should satisfy certain properties. And so essentially, we said we have fermions.

But because actually of the presence of what we call the reality operator, what really requires, essentially, what we would like to require, essentially, that the fermions are real, which in physical language means Majorana.

And in this respect, of course, actually, the presence also of the gamma operator, which is a current operator, would allow us always to impose this condition. So what we really have, we know, we say we have current spinners.

Now then, of course, actually, if you act on J on psi, you are really getting to get something like the conjugate spinner. And the question is that, is this an independent set of fields?

If it's an independent set of fields, then you really immediately face the problem that all your particles will be doubled. You are really going to get, for every particle we know, a mirror particle. And sometimes it's called the mirror fermion problem, or the mirror doubling problem.

And usually it's not easy to solve this problem. Now, we are really working in Euclidean space. And the issue, of course, is that if you really go to Euclidean space, and you said, I would like, sorry. But the physical space, of course, is Minkowski. And if you would like to impose both conditions that Jip psi is

psi, then this actually would impose certain conditions on the properties of the space. And as I have shown last time, that would impose a condition what we know as the KO dimension.

So in Minkowski space, you really need to impose both conditions, that Jip psi is psi, and gamma psi is psi. And this would imply, actually, that the KO dimension of this space is equal to 0, is equal to 0.

Which implies, actually, that J squared, which is epsilon, should be 1. Jd is, I think, epsilon prime, dJ.

And I need one more sign. J gamma. J gamma is epsilon double prime gamma J. And d gamma is minus gamma J. So these are the properties that we need.

And essentially, it really forces. Now, we know, actually, that in Minkowski, you have four-dimensional space with signatures, say, minus, plus, plus, plus, or the opposite. And these would somehow cancel each other. And then you are left with KO dimension 2,

which means that the internal space, or the finite dimensional space, should be dimension 6, KO dimension 6. So that's actually forced the finite space to have certain properties and KO dimension 6.

Then you need certain sets of epsilon. For example, epsilon is really minus 1. I forgot, actually, I think epsilon prime is equal plus 1 and epsilon prime is minus 1. Something like this. So then, actually, one goes ahead and tries to understand these finite spaces.

And what one discovers, and I will go through, not through the proof, but through the realization of the proof, is that one discovers that the algebras of the finite space,

or essentially what one finds, is that the complexification of the algebra. So essentially here, one assumes that the finite space is characterized by AF, HF, DF, gamma F, and JF. Everything is for finite. And one discovers that the complex extension

of this algebra, if you take the center of it, there are only two possibilities, which are either C or C plus C. And this, actually, is inconsistent with KO dimension 6,

which leaves us only one possibility, that AC is C plus C. And really, it corresponds to an algebra, which is a variant of, you know, let me call it M N of C plus M N of C.

In addition, when we imposed certain isometry on the first algebra, and this is a point, actually, we don't really understand, we cannot give it

a mathematical characterization why it should be like that, but it is like that. And the other possibilities are not relevant. And this really reduces this. Instead of matrices on the complex numbers,

you really take matrices over the quatorions for the first, and in which, actually, N is 2A in this case. So N is even. And also, because one would like to impose the chirality operator on this algebra, and elements of the algebra,

for example, gamma A equal A gamma. So in this case, what happened is that this 2A of H would split into MA of H plus MA of H,

for example, plus M N of C, or 2A of C. So the upshot of this is that really, the first non-trivial example you can do

is the following, that A should be 1 in this case. Sorry, A is, yeah, MA of H. This should be 4A or something. N is, let me see, M is, N is 2A, yeah.

So essentially, yeah, so essentially this, yeah. So essentially, really, the first non-trivial example is M1 of H, or H plus H. OK, let me write it like this. And this actually is like 4A. And in this case, the first example would be H plus H for the first algebra,

plus M4 of C for the second. Of course, other algebras are possible. What I wanted to say here is that you have the gamma, which is a diagonal matrix. On the first, right. For M2 of H. Right. You take the even part. It's the even part of M2 of H, which is H plus H.

Yeah, OK. So I have to say, OK, this I would say M2AH even. And then it really, at least for the first non-trivial case, it belongs to H plus H. Now, it's up to this point, you know, I went last time.

And the only thing where, you know, the assumptions I made are there. So at one point, one did impose on isometry. If one didn't impose on isometry, then you are really going to get, instead of H plus H, you're going to get M2C plus M2C. And one really can show that these cases would not

be physical. But one cannot rule them immediately, unfortunately. I was not imposing the isometry. What it is is that if you are here as a block, then you have three cases. Yeah. The symplectic case. Right. Yeah. The orthotic case and the orthogonal case. Exactly. There are three cases. OK, we choose, you know, the symplectic unit of case.

So this is a choice. This is something which is not imposed by the- But you could have chosen one. We could have chosen, yeah. Symplectic, symplectic, and- We could have chosen this section. This is the, really, the others actually are not relevant, because they would really give you real algebra. For instead of H plus H, you can take M2 of C plus M2 of C.

What would it really give essentially? It gives you, you know, it gives you an extra U1, actually, you know, which is anomalous. So one really can work it out. It gives, it's a little variation. Everything almost the same, except this little variation, it gives you an extra vector. An extra vector which is anomalous, you know.

Who knows, you know? Maybe there is an extra vector lurking in the spectrum. But, you know, people are talking, you know. Recently, there was some papers in which they said they observed something which deviates from the standard model. And, you know, some people said maybe it's an extra vector. So who knows? Anyway, but this is actually the most symmetric possibility.

The only assumption I made also that left action, this is the left action. And then this is, say, the right action. And then we said this is JB star J inverse.

And we said, and this actually is defined to be B opposite. So this is one of the axioms that left action and the right action commute with each other. So this actually has been used, this property really has been used up to this point. Now, from here, actually, one needs further assumption.

And, you know, I will discuss both possibilities. In that, and this is called order one condition.

And now, actually, we understand this condition. It's really a condition on the linearity of the connection. If one drops this condition, then what happens is that the connection would really get quadratic terms. And one really can work out the possibility when this is not so, when this is not satisfied.

But this I would leave to the last lecture to allow. Because actually, in non-commutative geometric, there are examples where this condition is not satisfied. And, you know, this is what the quantum spheres, OK? Anyway, so here, it will have really big advantage

that connections are linear. Everything is very nice. Everything, you know, falls into place, OK? What happens if it is not satisfied? You know, this leads us to grand unified theories, and patticellar models, and things like that. But this would come much later. OK, so what are the consequences

of this condition not satisfied? OK, so I would assume, actually, that this is satisfied. And on top of it, actually, now there's one more condition. Is the question that if we take the commutator of the Dirac

operator with the center of the algebra, is it 0 or not 0, actually? This is the question. Well, what we have shown is that there are two possibilities. If it is 0, then this symmetry is really intact. And if this symmetry is intact,

one can show, actually, that you are really going to get one work out of the model, which I'm really going to do a little bit later, is that you really get SU4 color, which really corresponds to this M4C, the symmetry of the quarks. So essentially, in this case, you really get a higher color symmetry, which

is not observed in nature. And in top, actually, you will find out that the neutrinos are all massless. They are exactly massless, which now we know that it's not true. So from the physics point of view, we know actually that the three must be different than 0.

Because if it is 0, then you are already in contradiction with the experiment that neutrinos would become all massless, which would have been OK until 1995 or something like that, because people didn't know or not sure, actually, whether neutrinos are massive or not. Because they have really this tiny mass that only limits

on the masses of the neutrinos. Now we know, actually, that they are really massive, but the mass is really extremely small. But that they are massive is well established. OK. Now, what happens, actually, if this is different than 0? Now, the beauty of it is that if this is different than 0, and for this condition to be satisfied,

the Dirac operator must have the following form. So in this case, actually, instead of telling you in description what really happens, I would start being more concrete. I will tell you exactly what are the representations, what

goes on, so that you have an idea that this actually, as I will say, field is not simply talk. You really can go and compute everything unambiguously, not much ambiguity, and with all the details.

So all right. So essentially, let me examine this in a minute. But I really can tell you that the fact that this is non-zero really gives a unique possibility that you really can give the neutrinos a Majorana mass.

And it will explain, actually, why the masses of the neutrinos are very small. It's an example of what was known as the SISO mechanism. So here, it really comes out naturally. And it solves the problem in a really very nice way. So let's see how things would look

like if I wanted to go ahead and do some computations. What should I do? So this is really my starting point. I'm really going to take H, the algebra A, to be H plus H plus R forced.

And this, remember, almost uniquely we were led to this. It didn't come out of the sky, out of classification. This was the first non-trivial result that we have.

So essentially, if I would like actually to represent an element of A, how would I write that? Then I'm going to write it in matrix form. I'm going to show you that everything that we do is really matrices, essentially. We really can use tensorial notation

if one would like to simplify his life. And recently, everything that we do can be put on the computer. So in the end, one doesn't do much calculation. You just write it on Mathematica, and press a button, and it gives you the answer. So I'm really going to use index notation.

So essentially here, what I'm really going to say, so in order to tell you my notation, I really have to tell you what are my spinners. So the spinner, actually, as you see, it belongs to this guy. So what I'm really going to write, I'm going to write as psi alpha i.

Where this alpha would see the h plus h, and the i would see the 4. So obviously, i is 1 to 4, and the alpha is also 1 to 4, but it's a different form. It's the h plus h. Now, and then, of course, actually, I'm going to write this as a dot a, where this has two

indices, like 1, 2, and this is 1 dot 2 dot. So this actually tells me that it's doublet under the quaternions. You know, doublet left, doublet right. And what we write, actually, we write as h right plus h left,

because we have already graded the algebra. And in this respect, my spinners would be psi a dot i and psi i. OK? So how many spinners I have? I have 16 spinners.

And of course, we have also the conjugates. But the conjugates, we're going to see that they're not really independent spinners, because the Fermion doubling problem is solved through the Majorana condition. We are not really going to say jep psi equal to psi. The problem in Euclidean space is really solved in a very clever way by the path integral formalism,

that only chiral formulas get integrated, and they give square root of determinant instead of determinant. So it's really solved in a really, very subtle way. You know, this is a nice discovery. You get the fulfillment. Anyway, so you see, actually, now, you know, without really doing any work, I have already obtained

the correct classification. First, actually, we already predicted that you have 16 Fermions, which is exactly, per family, of course, which is exactly what we have. We will also actually find, let me, you know, let me write this one and this. The I. The I eventually, you know,

when this condition, the d with z of a is different than 0, will also be broken into 1 and I, where I is 1, 2, 3. So that actually will be the color index, and this is the lepton index. So this is lepton, and this is the quark.

And then you can see that the lepton is the fourth color. It comes out of the fourth color. And in this respect, you are going to get psi 1 dot I, psi 2 dot, sorry, psi 1 dot 1, psi 2 dot 1, psi A, sorry, 1 dot 2 dot, and then I

have psi I, which I'm going to leave it as psi A1. Then I have actually psi 1 dot I, psi 2 dot I, and then I have psi I dot 1, psi 2 dot 1. What is this? You know, remember, this is A dot means,

when you see dot, dot means right. So this would be what I'm going to call new neutrino, right? You know, why I give it the name, because it comes out like that. The quantum numbers tell me that this will be a neutral guy. But this comes out. You know, I name it according to the usual naming, and it would agree, OK?

And then, of course, this is E, right? And this, of course, is upright. This is downright. Similarly, psi A1 is the doublet, which is a new A, a new E. It's a doublet.

You know, A is 1, 2, left. And this is up, down, left. OK? So you see, actually, without doing much work, we obtain the correct representations of the particle spectrum of all the fermions.

We know which representations they are. So one really can write that the 16 in this case is a 4, 4, 4. And this 4 is decomposed into 2 plus 2. And this 4 is decomposed into 1 plus 3. And then this is 2 right plus 2 left. And everything that we see belong to that representation.

Everything falls into place. So now, so these are my 16 spinners. But remember, actually, you know, I have the j psi. And so my spinners will be psi alpha i and then psi alpha i conjugate, which I'm really going to go psi alpha prime i prime.

They are not independent, but they are related through the j. OK? What does it mean, actually? It means if I would like now to write any element of the algebra, you know, I can write it as follows. I can write it as this one, you know.

Remember, this belongs to the second guy, term 4 of c. So it is something like this. And this is delta alpha beta. And this is like yij.

So you see, actually, how an element of the algebra is satisfied. And then you can form the b opposite, which is jb j star or j inverse. And you discover, actually, what happened is that here, you are really going to get the y times 1. So you get yij essentially transposed into 1.

And here, you are going to get the x alpha beta into delta ij. And if you look at this element and this element, obviously, they commute with each other. Why? Because then they just don't talk to each other. This is in the first algebra. Well, this is now in the second. And they don't talk to each other.

So essentially, gb star. gb star, yeah, absolutely right. And this is actually, I wrote it as transposed. But, you know, a star is, OK. So essentially, we can write it in this form. So the first a, b opposite is satisfied automatically.

Now, where do I go from here? Then I have to satisfy this condition. So this actually is really a condition. It's a condition on the d and the algebra, both, you know. Because what happened in the following,

you try to satisfy this condition. You work out the matrix representation. And you obtain, actually, and you need the commutation with the center, that is what. This is equal to that. So you are right here. No, this section, you are right. Yeah, it's exactly, yeah.

Different zero. Different zero. This is different zero. OK, so this actually, we have to assume it's different than zero. And then actually, this 3D would require the form of the Dirac operator to be as follows. One can show that you are going to get something like this.

Let me call it, I don't know, let me call it DAB or something. And here, we are going to get the bar. Here, we are going to get something. And then, one can show through a little algebra. Actually, I ran this through abstract proofs, which, you know, beyond me. But through a little algebra, one can show actually that the only operator that

satisfy is the following. That here, you can have only one non-zero entry. Only one. Anything else would not satisfy this condition. Of course, actually, this, together with that,

we have also to make sure that dA would be opposite is equal to zero, putting the two first, actually. First, we know that here, we must have a non-zero element. The question is that the whole thing could be non-zero. But because of this first condition, you discover actually that this is possible, if and only if you have only one non-zero entry. What does this mean, this one non-zero entry?

Let me write, actually, the expression for you. And let me see how does it look like. You know, I have, remember, when I write Dirac equation, what do I write? I write psi star deep psi. Or I can write it this way, you know, psi deep psi, OK? In a product. So this is my, say, Fermi action.

And this Fermi action, if I expand, and then I just look at this step, what does it mean? This, actually, if I look at that classification, it really tells me that you really have an element which is this one. If I have more, it would mean that, for example, the up,

the down, everybody will start getting Majorana mass. But you know, getting Majorana mass would break charge, would break the charge conservation. Because in this case, they would not have zero charge. So the only term that you, in physics-wise, that you can write is a term which is really neutral.

It must be neutral, otherwise, you destroy charge conservation. And it's not an accident, actually, then, that you can have only one non-zero term. And this one non-zero term would really correspond to the mass, Majorana mass of the right-handed neutrino. Now I'll explain later, actually, that why this is really extremely important in order

to solve, to explain, essentially, why the neutrino masses are so small. So this actually is the result. It tells us that in order to agree with the observation, the Dirac operator should take this form. It has only one non-zero entry, and the rest are zero.

This d and this d bar are correlated. They are one in the complex conjugation of each other. And the only thing I have to find is that what's the form of the Dirac operator in the finite space. So what shape does it take?

The whole idea that things really are really constrained. Because, in a way, we have to explain why the standard model. This is actually why the standard model. So if the hypothesis that everything

comes from logarithmic geometry, we must really arrive at this answer in an almost unique way. Hopefully, one day, one would have finally say, OK, here it comes. This is without any ambiguity. But there's always the possibility that you have to put some physics. Because if you don't put any physics, in principle, you can get anything. Because if I don't want to explain nature,

then I can start with the smaller space, or bigger space. You can't start with too small, because if you already assume that it's four-dimensional continuum times finite space, and the finite space has killed dimension six, and then you make a very minor hypothesis on that space,

then you are already led to the 16th. So it cannot be too small. This is the first example. But then you cannot rule out why it's not bigger. Oh, no. Of course, you cannot solve it. The question is that, can I say uniquely, this is the only thing I would ever get. But this is the first possibility that one gets.

OK. So now, actually, where do we go from here? And what we do is the following. So we would like, actually, to see what type of Dirac operator can we have, this DAB. Now, remember, actually, the D satisfy many properties.

I've written them up. So you really have to satisfy, to show, for example, you have to verify that what the R is, that gj and jd. Still, I'm talking about the finite space, remember, is equal to j, and gamma j equal minus j gamma,

and d gamma equal minus gamma j. So these properties, I have to satisfy. And so if you do that, you'll immediately discover, actually, that the Dirac operator of the finite space really takes simple form.

Remember, let me write it in terms of new right, e right, because I don't want to write one dot and two dot, because then it's more physical. And you write e right, and then I have, say, new left, e left is a doublet. And then up right, down right.

And up left, down left. OK? This is really a doublet, but these are really singlets. OK? Why it's a singlet? Because, actually, this sigma really breaks algebra

from, you know, for this condition and the other condition to be satisfied. And then you put the sigma. What you discover in the following, that this algebra, h plus h plus m4 of c, is really broken.

Obviously, it must be broken, because you put the sigma. Remember, this is like 1, and this is the i. And this really breaks the color index. It breaks the color index into 1 and 3. And in this respect, the m4 of c is really broken into c plus m3 of c.

However, because that condition dA be opposite is equal to 0, it really, what happened, and then it breaks, actually, this is also broken. You know, because this is one index, we'll talk this one index. And then, actually, essentially, this is a c plus c. It breaks. And the three are not independent.

What one discover, actually, that the element of the algebra A takes this form, in this case. So you are going to get, you know, x, x bar. This is my first h. These are my first quaternion. In other words, actually, remember, a quaternion takes the form like alpha, beta, alpha bar, minus beta bar.

It tells me that in the first h, it must be of this form. Now here, what do I have? I have a quaternion. So this is my first four. And this is my bottom four. And in the bottom four, what do we really have? We discover, actually, that this is x.

And of course, this is 0. And this is m, where m is an element of m3 of c. In other words, actually, what we discovered is that the three c, the c here, the c here, and the n, are not independent. And they are essentially one c. And the finite algebra of the finite space really breaks down to c plus h plus m3 of c,

which is really a symmetry of a central model. It really comes out of the requirement that there is a Majorana mass for the neutrino. No, this really has drastic consequences. OK, so with this, actually, an element of the algebra,

I go with the d. And then we can really classify the d, satisfying all these properties. And, you know, I want to cut the story short. What happened is that the only non-vanishing elements of the d, there is one more requirement that we have to put in order to really completely single out the d.

And that the d commutes with, no, actually, yeah, d commutes with the element which is lambda lambda bar lambda lambda bar in here. If you do that, then you discover, actually,

that only the non-Newcarver coupling of the Dirac operator are non-zero. And we have the Dirac operator completely singled out. If I want to write it down, I can write it down. Let me write it down, actually. So how does it look like? Many zeros.

I think here you are going to get k new. And here you are going to get ke, which is actually the Yukawa coupling of the neutrino and Yukawa coupling of the e. And similarly here, you are going to get,

this is new left, e left. And this is k new right, ke right, like that. And similarly for the quarks and the only off-diagonal elements you are going to get.

And yeah, here also. So this is k up, k down in diagonal form. So in other words, actually, after some matrix algebra, one would know exactly what the Dirac operator is. So what do we do with this now?

What we do with this is the following. Having determined that... By taking into account this competition... Exactly. So you'll get many, the essential point is that you get many zeros. By, you know, gamma d minus d gamma, I'll tell you all diagonal elements are zero, you only get non-diagonal elements, and so on. So it's a little matrix algebra, nothing much to it.

And however, once one is finished, one can proceed as follows. Notice that we have not really talked about dynamics or anything, we just talked about classifications and representations and satisfying the properties of the non-commutative space.

And you are led almost uniquely, actually, to what do I call the death of the standard model. Now, remember now, I said that the Dirac operator of the full non-commutative space is equal Dirac operator of the manifold cross one plus gamma five cross dF.

We have been, what I was talking about up to now is the dF. We know now exactly how the dF looks like. dM, of course, we know because it is the Dirac operator of the usual, say, remaining manifold.

And similarly, we can say A equal A of the manifold cross AF and J. Everything is a product, actually. Don't have to write it down. All right? Now, remember, actually, here,

one always starts with non-fluctuating d. So we start with a non-fluctuating d. Say, let's start with a d, which corresponds to, say, some flat space. Starting from a d from a flat space, for example, you can really start with, you know, gamma mu d mu, like that, actually, for the d.

And we know, actually, that you really can generate the metric of the dM out of fluctuations. These are the outer fluctuations. And this is easy, actually. So what you have to do is that you say this is e mu A gamma A. You make this as a function of x. You discover that you have to add a spin connection

in order to make the thing, in order to make the thing covariant. And so that actually is easy to do. Similarly here, we have seen last time that the d always go into dA, where dA is a d plus A plus J.

But that's good for the global. Yeah? What you have written there is for the global. Here is the global. Here is the inner automorphism, no? No, no, but I mean, if you want to do the fluctuations, you do that for everybody, of course.

I do for everybody, exactly. We get both inner, but I'm trying to say that if you'd like to get the gravity, you have to do the outer, and you want to get the other, the A. Yeah, but these are inner fluctuations by definition, So the A are one forms in this language, and it is ADB. This looks actually very simple, but remember actually,

the A I've written is what? The A we have written was a 32 by 32 matrix. The d was a 32 by 32 matrix, tensor do the Clifford algebra. So here we are talking about. Now in addition, actually, we said everything we know is really for one family, and they have written here,

for example, K, and K I've written as a number, but in reality, this K is a 3 by 3 matrix in generation space. So you mean it's a 96 by 36? It's a 396, so 32 times 32, and tensor do the Clifford algebra,

so it's 384 by 384 matrix. This is actually why it's much easier, instead of writing matrices because they look huge, is to put in tensor notation, because in tensor notation you just put an index

and you don't care how much trans, because you know where does it sum. So it becomes really efficient to do it. Of course, we didn't do it like that in the beginning, because for years you have to struggle with all these matrices to see what acts, and it took some time actually. This was progressing developments.

All right, so this actually here. You say, all right, I would like to compute this guy. How does it look like? Okay, remember, so I will give you an example. The reason I want to show you that things come out, that it's not that you say, ah, the gauge fields come out, the Higgs field come out,

and people think that you are cheating, because we know the answer, we put it in, and then it's not true. Everything comes out in a very well-defined way. One really cannot cheat in any way whatsoever. So I can say, okay, AB, and then you can put indices like AA prime,

and then you can do DB like AC, CB, and so on. So then I can look actually at all the matrix elements AB. And how do they look like? So I say, okay, how do they look like? The following. I go back to my representation, and then I'm really going to get an element here,

like, you know, one dot, one, one dot, one. Remember, you know, I said A alpha I beta J, and this alpha could be one dot or two dot, and the I could be one and the I.

And you see, now I have to take all possibilities into account. If I think about this sort of this, this would be the gauge field acting on the right-handed neutrino, because we define the right-handed neutrino to be on this corner, okay? And in this respect, you know,

you'll discover actually that this guy would be zero, because the neutrino, the right-handed neutrino, is massless guy. Sorry, is a neutral guy. Is a neutral guy. So, you know, then of course actually the next guy would be A two dot one,

two dot one, and so on. And this actually we discover is a vector, and then, you know, it is given by minus I over two G, I don't know, B mu. So you know it's a vector, and with a gamma mu. Everything is tensed with the Clifford algebra, so you get that this guy is a vector.

In addition, you start actually to get something like, you know, a guy like A one dot one, one dot A, you know? And then, you know, it's a doublet index. A, remember, is an S G two index, and you discover actually, and then you put A is one two, and what do you discover? That this is really proportional to gamma five A H A.

And this is where the X enters. So the X enters as inner fluctuation, and now I say what is this I know? Why I went this way? This is an off-diagonal element. An off-diagonal element, it means that, you know, it's really trying to connect two elements of the left and right algebras with each other.

I mean, in fact, it's very easy, when you write D equals D N times one, plus gamma five times D F. I mean, you know, so D N times one will generate the gauge both of them. Right, yeah. And the other one will generate the X. Oh, yeah, yeah. One really can see it. But of course, you know, you really cannot see. For example, here it's really easy if you see that it's a doublet.

What's not so easy, actually, is to show, for example, the other guy. The A two dot two dot, for example, one A. In this case, you say it's gamma five. And what happens, actually, it is A B H bar B. You really cannot get ahead of time

that you are only going to get one Higgs and not two Higgs doublets. The fact that you get only one Higgs is really related to the fact that these guys are related. Remember, I told you you have C plus C. And then, because we are talking about the same C, it tells you that, okay, you know, you really cannot have different Higgses. You are really going to get the same Higgs.

So the upshot of this is that one really can do this calculation. It's a matrix calculation, no complications. And after some algebra, one finds the Dirac operator, including fluctuations.

What does it mean? It gives me the connections. It gives me the connections on the space. And the connections are nothing but the gauge fields and the Higgs fields. Nothing else. So up to here, you know, we have, we can achieve that. And before doing anything,

I can tell you that the Dirac action for the fermions, okay, okay, you know, one really has to write the gpsi. I'll tell you actually now, maybe I can tell you why we can write the gpsi. The reason we write the gpsi, because in Euclidean space, I really cannot say gpsi equal to psi.

Yeah. So in order to get, once you write gpsi, what really happens is that when you integrate on the gpsi and the path integral, then you really, this is an anti-symmetric product and you don't get determinant of dA, you get the path integral of dA and the path integral of the square root,

which means that it's really a chiral integration and you have not doubled the degrees of freedom. Now, anyway, if I expand this, what do I really get? I really get all ferminic kinetic terms,

for example, nu bar d nu plus e bar d e plus left and right and everybody, plus vector Fermi interactions,

plus Higgs Fermi. You get exactly the correct hypercharges and everything, actually. Why one would get the correct hypercharges? Exactly because of this, actually.

If you really work out the J A star J inverse, they really come and they bring, actually, the u1s in a really completely funny way, and in order to have the matching, and here, actually, one actually makes an assumption that one takes SU of the algebra, not the algebra. In other words, one has to make a requirement

that the trace of the A should be equal to zero. This is another condition that we put in. Okay, so here, actually, it's a first check. Everything ferminic-wise works and it works perfectly.

So what's next, actually? What's next is that you need to make... So the fermines are dynamical, but the bosons are not. So how do they get the dynamics? After all, we say that we have a curved space, a curved space should have dynamics, and we should be able to describe the full interactions of the system.

See, up to now, I have been speaking classically. Everything here is classical. So here, actually, comes the idea of the spectral action.

And the basic idea is that the Dirac operator, which was really the building block of the non-committed space, in principle, one can study the spectrum of this Dirac operator. So you study the eigenvalue problem, and you look at the eigenvalues,

and we know, actually, from studying, say, remaining geometry or the Dirac operators on remaining manifolds, that the eigenvalues are geometric invariants. In other words, all the geometric invariants are included, actually,

or, you know, lambda is a function of all the geometric invariants. So now the question is that how can we extract this dynamic? How can we extract the information about the lambda?

Of course, it's not easy to simply work out lambda of every space, so we need something more general. And so the basic idea is that, since the spectrum knows about the geometry, we took, as the dynamics of our system,

is conjectured to be given by the trace of the function of a d, where f is a positive function.

I think you have to put a scale, otherwise it doesn't make sense. I was to go and talk about the scale in a minute, because, yeah, d has dimension on operator,

remember it is, say, gamma mu, say, d over dx mu, and this actually has dimension one, therefore you really have to divide it by something of dimension one. Now, of course, actually, you may ask what is lambda, and I will talk a little bit about, you know, dealing with the scale later, because this is a very sensitive issue,

what you should do with the scale, can you get rid of the scale, or what are we talking about? I want to add something, which is that this expression, trace of a function of d, is the only expression that you can write, which is additive when you take disjoint spaces. See, the action should have at least the properties that if you take two disjoint spaces,

it should die. Right. And this is the only expression you can write. So you want to say f, but d1, d2 should be disjoint in this case, or what? Yeah, sure. If you take d1 and d2 to be disjoint, so it takes the full direct sum,

f of d1 plus d2, okay, then it's certainly true that, you know, the trace of that will be the sum of it. Or thereof it. Or thereof the function. And the problem is that if you take the function, which is specified, and fulfills this additivity property, then it has to do all this work.

But what about the positivity? Positivity, because, you know, Positivity? No, no, no. But it's important for Euclidean actions, no? No, no, of course. I mean, if you z is required on top of that, it's Positivity. Yeah, because some people have been, you know, toying with the idea that maybe it's not positive and things like that.

Anyway, so now essentially, how can we deal with this space? Obviously, we really, for an arbitrary function, it's not that easy to work out, but what we can really work out is the asymptotic expansion. This way, one would know how to do. And the idea of the following, we use methods of heat kernel expansion.

You know, it's not the only methods, but here, actually, they, at least for remaining manifolds and for product almost commutative manifolds, they are really very, very effective. So what's the basic idea here? You say, okay, you know, suppose that I have a function of some operator P.

You make some, you know, I don't know, a normal expansion. You say, say, SPS, and then trace, and then you have something like trace of P minus S.

The new identity trace of P minus S is given by, you know, you use the Mellon transform formula, and with the Mellon transform formula, you obtain that, you know, must be the opposite of the definition of the gamma function, essentially.

TS minus one, trace E minus TP, where the real part of S is larger than or equal to zero. And then the next step would be to use formulas of trace E minus TP

using heat kernel expansions, and then, of course, actually, this is given by AN of the, sorry, T N minus, actually in this case, minus N over two minus two.

Okay, at least in four dimensions, and where N is larger than or equal to zero, and this is AN of XB over DV.

And these actually are called the series of coefficients. And they are defined in terms of connections of the Dirac operator.

So, yeah? It's an asymptotic series. It's an asymptotic series, yeah. Okay, I will say why it's useful in physics, because exactly at one point, this is, you know, the fact of D over lambda, and then, to be clear, actually, when I write the expansion, it will be an expansion of one over lambda,

so the first term will be one over lambda to the four, then you are going to get lambda four, sorry. Then lambda squared, then one, then one over lambda squared, and so on. Now, assuming actually the scale is extremely large, then these terms will become, you know, less relevant. Of course, actually, the expansion will break down

once you approach the scale. And, you know, where the scale, you know, you can take the scale to be, say, unification scale or Planck scale. In this respect, actually, the expansion will definitely break down. And in that case, one has to learn different ways

of dealing with such expansions. But, you know, we did some calculations of this expansion for, you know, Robertson-Walker metrics and things like that, and really it holds almost up to the Planck scale now. Up to the Planck scale, it's only when you, you are orders of, you know,

like twice or three times the Planck scale, then you start to see that it breaks down, but not before, you know, which is really amazing that this action is so accurate that it goes all the way up almost, you know, to the Planck scale, but not beyond, you know. It's just there. All right, so what do we do with this expansion?

It doesn't give us anything sensible because, obviously, this looks far-fetched, I would say. That you take an arbitrary function, you say, okay, this is an arbitrary function. This is, the fabric part is straightforward. You say psi, deep psi, okay? And you get, you know, the usual expression. You can say it's by construction.

But for the botanic part, it's not really clear that this idea should work. You know, that it should give you anything accurate, essentially. After all, you are using some expansion. All right, so how to proceed, actually? Here, you know, it's known, as I said,

here we assume that the continuous manifold is four-dimensional, and in this respect, in this respect, the expansion, yeah. So, what happens is the following,

that only a n equal to zero, if n is odd, for many-folds without boundary, okay?

What happens for many-folds with boundary? We have already addressed this question. It's more, it's actually very interesting, but I will talk about it later. But for many-folds without boundary, a n equals zero and is odd, what it means, actually, the only non-zero ones are the a zero.

the a2, a4, and things like that. And these formulas were worked out long ago by Gilkey. He has some, you know, standard formulas. And because of that, actually, what really becomes relevant in this is the d squared over d.

d sub a squared. d sub a squared. Yeah. And the trick of the following, you write this as g mu d mu plus a remainder. And this d mu will be d mu plus omega mu, which is a connection.

And this omega mu, OK. This also is given by minus g mu d mu d mu plus a mu d mu plus b. So what Gilkey did, actually, you know, he gave an expression.

This actually is, there are two important terms in this expression. The e, which is an invariant, and the connection, the omega mu, and you know, curvatures which are constructed out of this connection.

And this omega mu, one reads from the following. You write this expression into this form, into this form. And then we even know, actually, that a mu is given by, you know, some expression. It is half g mu a mu plus, I don't know, plus comma mu.

This is a Christoffel, contraction of a Christoffel. In other words, actually, one really can compute all these expressions in a unique way. And remember, we have already computed our d. So we have this d, huge matrix. You take it, you square it, you get another huge matrix.

And you write it in this form. And then you try to find the curvature of the space. Remember, this is a generalized curvature now. And you also get the e invert of the space. Another huge matrix, you know, 384 by 384. Okay?

So what Gilke tells us, look, you know, this a0, the first term in the expression is given by 1 over 16 pi root g trace 1. Trace 1, of course, is 384 because it's 384 by 384 matrix.

So this is actually nothing to compute. It's the part which is independent of lambda, obviously. Yeah, sure. This is, no, no, this is actually lambda 4. This is really the cosmological constant. And it's huge. You know, the cosmological constant is huge here. Why it is zero, that's a different story. Okay?

This is not quite high actually, because coefficient is not 1 in the sense that you have the things effect, you know, you have to talk about effective. You want to talk about it? I'm talking about in zeroth order cosmological constant. Of course there are contributions coming from the vacuum expectation value of the Higgs.

But this is supposed to be much smaller actually. I'm saying that when you compute this coefficient a0 and when you take into account the Majorana mass matrix you get another term. Yeah, okay. The negative sign. Yeah. That actually, but, but that comes from Higgs, no?

You mean the G squared term. Yeah, sigma squared. This is a sigma actually. Ah, sigma, yeah.

Yeah, but, you know, okay. This is actually after. Sigma would come later because I see sigma lives here now. Sigma lives here. Because a priori we don't know, okay. Anyway, so the interesting part actually is this guy.

So you get actually r over 6 is the Einstein term. And then you say trace e. Now you, let me actually give you an idea what trace e is in this case. Or what e is actually.

You know, e you have to read, you have to read by squaring the d. By squaring the d. And so obviously let me remind you that you have all this d slash part. You have here gamma 5h part, gamma 5h dagger. You have some d squared, okay. When you look at the d squared, yes,

the off-diagonal terms are really going to give you what? H-h dagger terms. Which is really nothing but the Higgs mass, okay. So you discover actually essentially and then you trace it and you say you only get trace h squared. So if you really work it out actually in this example, what do you really get? You're really going to get the following actually.

You get, this gives up to a factor. So you see, this is the guy that

would eventually give mass to the neutrinos. And this is the Higgs mass. These are essentially Yukawa coupling, you know, square of Yukawa couplings. Square of Yukawa coupling. So we look at this expression and then we notice the following. The curvature term,

you mean the whole minus sign? Over all minus sign, yeah. See I write minus, okay. So if you'd like, yeah.

Okay. Minus 2 over pi squared if you wanted this. F2 lambda squared. Yes? Gd for x.

There is a minus sign here. Well the notation that r is negative for Euclidean spaces. Anyway, so you see we get a mass term and with a minus sign this is really extremely important if you, as we'll see, for the standard model because the standard model is really

characterized with a potential V as you see lambda over 4 h h bar h squared minus m squared over minus mu squared over 2. And it is the minus sign that gives the potential the Mexican hat. It's essential that you get the minus sign. Now obviously you are really going to get the minus sign.

Why? Because if you look at it and then you are taking trace of e minus t p p has d squared, d squared has s squared. Obviously you are going to get the first term in the expansion, the minus sign the next term will be plus sign it alternates. Which is extremely nice because of course

but you really cannot guarantee that. In principle, before you start. We don't know. So this actually, important points we can point that the curvature which is the Einstein term and the mass term Re3 is equal footing so h bar h is like

curvature of the antenna space and ok, you know scale lambda squared. What's lambda squared? Of course actually one would like this lambda squared to be related to

8 pi g or 16 pi g or whatever ok? It should be related to the, so this actually is the Planck, Planck squared. So from here one would read that if you'd like to make sense of the scale lambda that we started with, the natural

interpretation of it is that it's almost of course you have numbers you know you have f2, it's a maline transform of the function, so you can get factors of 10 or things like that, you have the pi squared so it is of the order of the Planck mass. It's not exactly the Planck mass because you have some factors here

but at least in order we know that it's of the order of the Planck mass The next term the A4 and according to Gilkey what's the formula? The formula tells me that, what is it? Ok, you have

let me write here, ok I'm not

going to write the surface terms so what re-enters

is the curvature squared

everything is squared actually here e squared, curvature squared now you know this e squared, the omega mu squared, they really give me invariance in terms of curvatures square of curvatures what are the squares of curvature?

of course the r mu nu squared is there but here we are really going to get what? we are going to get f mu nu squared which is curvature squared and we are going to get kinetic energy of the x term and the sigma term and things like that so and exactly, Yang-Mills terms

is nothing but curvature squared terms you know, because what's f mu nu? you know, if you take d a squared, as you see d plus a squared you know, you are really going to get f the square of the curvature the square of the connection is the curvature and it really, it lives here, it lives here

and one really can compute it some long calculation but in the end is well defined calculation, the important thing is there is no ambiguity at any point, ok so, in this action we are going to get f zero over two pi squared

and y squared Gauss-Booney plus

plus conformal factors

kinetic terms for the h quartic terms for the potential and similarly everything I write here I have to write also

for the sigma field and so on

r sigma squared anyway, see the important thing is that first of all, this term there is no scale second we note that we have a Weyl squared term now this is really important for, if you are really doing say

normalization of gravity this term would make it would improve the propagator of the graviton it makes the theory of course non-unitary this we don't really worry about because in that case we are cutting you have to cut the expansion and if you would like to treat unitarity you should not cut the expansion you have to take the full-fledged theory

but at least this conformal tensor or the Weyl tensor squared is important for the propagator of the graviton and now here we see here actually that we really have a kinetic term for the E1 for the electromagnetism

a kinetic term for the SU2 weak and a kinetic term for the color SU3 exactly, and then we look here and then we immediately find a relation that tell me that G1 squared equals G2 squared is 5 over 3 sorry G3 squared G2 squared 5 over 3 G1 squared which is the famous unification

one obtained for SU5 or S over 10 grand unification theory we will have actually, we have to talk about this later in the next lecture because this is not exactly satisfied it's not exactly satisfied because as we are going to see next time

you plot the thing, okay, the logarithm of of the developments of the alphas, which is G squared over 4 pi alpha inverse of the alpha inverse, then you discover that they don't exactly meet, but you know this is a different story, you know, whether they

meet, they don't meet, they almost meet but in principle, you know, there's so much in between that we don't really know, we don't control and it's premature to say whether this is a bad point or not a bad point but these are always the same order you know, it's almost, you know

okay, within, let me say, within 10% it works within 10% but you have extrapolated 15 orders, 16 orders of magnitude actually, which is huge okay and you know, things we have discovered that things have not really been calculated, you know

to two loops in the model that we have because this field actually turned out to play an very important role this sigma, which priorly we ignored, at one time we even make a prediction that the Higgs mass is given by a certain number and this turned out not to be correct because we have ignored actually this

this field, however, if you take this field into account, then you know, you're completely in agreement with the experiment and in addition actually, the renormalization group equations taking the sigma into effect has only been carried to one loop order nobody has carried it to two loop order so we really don't know, so maybe it will change

a little bit actually, you know, my guess is that it will change, but I don't know whether it will change in the right amount, in the right direction because this is up in the air so so let me summarize again so we started from

a very general idea the general idea that space-time can be approximated by a product of a continuous-time discrete manifold, the real answer we don't know, we know it may be just completely discrete, but to a very good

start to say is, because we know that generativity works very well so we know the continuous part up to the Planck scale works very well, so we can up to that energy level, or a little bit below we know that the you know, the Riemannian manifold part works, so to assume that

we have an invisible extension which makes it almost commutative but in reality non-commutative that we can extend it by a finite space, we entertain this idea and we discovered that to

almost uniquely we can predict the structure of the model we know the Fermi-Nieck representations we know actually why there are 16 fermions, we know what are the representations of these fermions we know the gauge group one generation one generation this actually, I'm saying that what we can

the things what we cannot do is much more of course, because what we cannot do for example, predicts say the electron mass, and you know, all the masses because in principle you know, it's if there is more restrictions on one should be able for this

remember, the finite space contains all the Yukawa coupling, so it contains actually information about so if you know that for example that k is like 10 to the minus 2 or something then you know that the ratio of the electron mass top quark mass is like 10 to the 4 or something, so the minus 4, so in that case

this information we are putting in by hand we are not predicting, but if one really is completely successful you would, everything comes out, but you know this is really far-fetched for the time being because you need some extra structure which we really cannot put our hands on, not yet you know, so I think there is another aspect

the criteria and the properties that we verified do not tell us actually why there are three generations this we don't know, this must come out somewhere, you know, so this aspect, we have no idea no one has any explanation we need an explanation no one has any explanation for this

there is no I was even going to guess whether there should not be a fourth generation this actually almost rolled out now actually there are two things to say there is one, these two Japanese who got the Nobel prize had a reason why we needed more than two generations

and that's the capital of Kobayashi-Baskawa because it involves a complex number only if you have more than two, strictly so I mean C is the first number of generations for which you get a complex number, so for which you get a violation of charge communication but then, okay I think now experimentally it's known that there is no fourth generation

there are very strong tests which are good yeah, but so why three, actually it's not that we don't know, nobody knows it's up in the air why the masses of the firms are the way they are, you know, the standard model after all has what 17 or 18 parameters sure, but there you can say one thing

it's exactly like if somebody would come and tell you why the metric of space-time the way it is, okay, you could say well, Kepler are top guess, but you know you could say well, it's like this, it's because of circumstances so the issue is what is the geometry of the finite space is it due to circumstances

or does it and this we don't know this we don't know one would hope, actually, that if maybe well, 20 years ago okay, you know before, say, the advent of local geometry if somebody asked you why the standard model you see, you could have answered the same, but now actually

I think we are very close to knowing that it is, it must be like SU3 cross SU2, it must be like 16, so we must have a Higgs field, so this actually may have been answered successfully but we are far from, you know, the full answer and this is the there could be more than one Higgs or not not here, actually, here, okay

if later if, you know, we take off this condition and we really want to push everything to grand unification then of course there are many more Higgses actually not only one, we are going to get but that actually will be relevant at higher energies and we will not be able to see what really goes on at lower energy except for

the sixth doublet, you know it's not clear but that's another story, you know anyway I would stop here, I think, because next time I will continue by analyzing the quantum aspects of the model and maybe, you know, theorizing a little bit about

so what we have so far is the first quantization just to answer what Timo has last time it's the first quantization it's a variation of geometry and some upper space setup and so on but what is quite

one should stress one thing, you know, which is that there are seven or eight sides which have to be correct and they are all correct and moreover you know, it encapsulates this enormous complexity of the Lagrangian into something which is I think, you know, if this thing was done before, you know, this would have been like and then people come and verify it

it would have been like a revolution but the thing actually lost the luster because people didn't think that, okay, you know the answer you are fitting it and this is the you know, biggest thing that one has to fight and the misconception that there are billions of possibilities and simply we are given an interpretation of this of the one possibility which is correct

and there are many other possibilities which is not really correct because you know, according to this classification that I went through there are not really many possibilities, you know it's almost, it comes uniquely, you know but then one would have to explain why if you want the correction by the finite space improves the ordinary four-dimensional continuum and the reason why

it improves it, it changes the k-th dimension it changes the k-th dimension from four and that's a big improvement because then you can you need to be you need to be finite well, it's the most economical to be finite, I mean it could be that the splitting is actually more complex but I mean the simplest way to see it is

that you are a four-dimensional space and then you make it two-dimensional into a dimension without changing the metric dimension which is remaining but if you have, you know, if you start finite then you will have to fight all this Kaluza-Klein modes, as I would say you know, you have many extra degrees of freedom which we don't see

so there's a new dimension by something finite exactly, by something finite which doesn't have this infinite number of modes like in Kaluza-Klein I mean, that's the way very basic