Well, actually, I should have really started my first lecture with this. But since every time I say next time, but in the end,

I keep forgetting. So this time, I said I will do it in the beginning. But I'm sure in the end, I will forget it. And it's a, yeah, in verse 12, I cannot listen. I cannot pull 12, you know? So maybe I will mention it when I reach the point.

So actually, most of the material I talked about can be found, actually, in this reference, especially in the first and the last. And for the classification of finite spaces, it's this paper, Why the Standard Model, which is mathematical. There is a physical review version of it

which is not mathematical, where things are summarized. It's called conceptual explanation. I think physical review letters force us to change the title. I think it was called The Rest for the Beggar. I didn't like this terminology.

OK. So today, the last of my lectures, I will start quickly by summarizing what we have done, since it was a break of two weeks. Our basic assumption is that space time

is the product of a continuous space times, or a semi-direct product, times a fixed finite space. And this should be taken as a basis for unification. In other words, this has to be considered as geometric space.

And all geometrical entities built on this space should be used in constructing the action. And since it's a non-commutative geometry, it means there is a lot of information, actually, in both algebra and, well, three, algebra, Hilbert space, and Dirac operator.

And the Dirac operator has to study the spectrum. And the eigenvalues of the spectrum are the geometric invariants. So we really have applied this idea to the product space of the four-dimensional continuous manifold times a finite space. Now, which finite space we used?

We used a finite space. We classifiers, actually, we have classified finite spaces, subject to what we call weak physical requirements. And we ended up that the space is c plus h, let me see,

left, plus m3 of c. And I remind you, actually, that originally, this was really, due to the neutrino mixing, was really, these three come together in one algebra, and this is from the other.

And what really breaks the symmetry is the mixing between fermions and their conjugates.

And this implies the existence of a Majorana mass

to only one particle, which has to be neutral. And it's no coincidence that this happened to be the right-handed neutrino. What we have shown also is that the basic building

blocks, the basic building operator, is not only the d, but the d that includes fluctuation, which we labeled by da. And this da is equal to d, where d is, you know, it could be a Dirac operator, say, in a previous Dirac operator that

makes anybody in flat space, plus flat, no committed space, I would say, plus some connection supplemented by this term, where actually a is a connection,

and there is a universal formula for it. And the ai and the bi are elements of the algebra. So this actually gives all connections.

And then we have shown that the connection a contains, let me say, the following, you know?

So let me give you a drawing just to, so let me assume this is h right, h left, and then this here, m4 of c. And this, we said, essentially is broken into same terms

of the form lambda lambda bar. And then here, let me say it is quaternion q. And here, I have, you know, same lambda. And here, I have 3 by 3 meters. So these are the elements of the algebra. A tree takes this form.

And then what we have seen is that the connection a, in this case, according to the same terminology, would really have this form, which I'm really going to. So for this part, it will have a 0. And then it will have b mu.

And here, it will have some b mu with some power, and some SU2, let me call it w. And in the lower part, what it will have, it will have b, w, and v for quarks.

It will have everybody. And here, actually, there's a b, because it's broken. Now, what happened, actually, is along the off-diagonal elements, we really get the Higgs field. And here, what we really get, something that connects, actually, here, the H right with the H left.

And this happens to be the Higgs and H dual, whereas this is defined to be sigma to H star. In other words, actually, we really have one Higgs field, which would give mass to both the, say, electron and the neutrino,

and the up and the down quark. And it's known, actually, in the standard model, the Higgs field is not, the Higgs field that gives mass to the up and down quarks is the same Higgs field. There are no different Higgs fields. And this, actually, you know, yeah, but the question is that there is no, here,

the two things that are really independent of each other, actually. They are dependent on each other. They could have been independent of each other, but they're independent on each other, which shows, actually, that only, this is the consequence, that only one Higgs field is present. Now, for the Dirac operator, we

have seen, actually, that there is here, there is zeros here, zeros everywhere, actually. There is only a singlet entry. Sigma is a singlet that gives Majorana mass

to a right-handed neutrino. So this is really, actually, are all consequences of the construction, that I would say predictions of the construction, that a singlet would exist, and the singlet is extremely necessary. It plays a fundamental role in giving the Majorana

mass to the neutrino, Dirac-handed neutrinos. In addition, so it really can establish the SISO mechanism. In addition, yeah. Yeah, it's this, actually. 2010, so it was published already before Higgs was discovered.

And the reason we mentioned that it really has a big consequence on the stability of the Higgs doublet or the Higgs field in the potential, it shows that you really can extend, one can extend the standard model all the way,

one can extend all the way up to very high energies

beyond the 10 to the 11 GeV, where the Higgs coupling lambda, quartic coupling, becomes negative.

Without this field, it would become negative. But it stays positive because of this field. So that's a very, very important point. In the sense, you know that many people think that our model is ruled out because of the low Higgs mass, which is well over 25 GeV. But in fact, because we had this field, if we had computed the renormalization group

equations and so on, we would have found that it was bad. One can show, actually, from the stability argument that if you don't have the sigma, then there's a prediction that the mass of the Higgs is somewhere between 160 and 180 GeV. And this was the first thing ruled out. This is ruled out.

It was ruled out first. Before, the Higgs was discovered at 126. However, once we have this, there is no problem because it makes everything stable. And this argument of stability, which was used in getting this mass, not only, actually, in our case, not only stability, we have some mass relation which forces this prediction

to be in this range. Anyway, so now everything is OK because the sigma is necessary and it's really necessary because one really has to break the m4 of c into c plus m3 of c, that this is really necessary and the symmetry is broken.

So essentially, I really can summarize for you, actually, what are the positive or the predictions of the model. Before going on, so let me repeat, actually,

what did we assume and what did we get? What we put in and what did we get out? What we put in is that this is assumptions and that space-time is a product

of a continuous four-dimensional space times a finite space. So this is an assumption.

The second assumption, while we were doing the classification we did, is that one of the algebra's m4c is subject to a symplectic symmetry, reducing it to m2

of h. So that actually limits two quaternions.

Three, we did assume that d of za is different than 0. And four, we did assume at some point that the Newton algebra u of a for the whole thing

is restricted to sq of a. Actually, this condition somehow implies

anomeric cancellation. But as you see, one has to assume, one has to put some input, and this is the input we put. One can argue, why you have to put this input or their way out mathematically, but it's not easy. We did try. And it was not really easy to reduce this set. For the first condition, yes, it's possible.

For the first condition, it just comes from the fact that in m4 of c, you have a weak inner automorphism, so it works out this thing. For the first condition. For the whole thing? No, no, for the whole thing, exactly. For the whole thing. And then when you reduce, you get this condition. So now this is really understood.

So maybe this can be weakened, actually. This can be weakened, yeah. OK, predictions. First, the number of fundamental fermions per family is 16.

The algebra is c plus h plus m c of c. This comes out, actually, as a consequence of the assumptions. And we obtain the correct representations

of the fermions, with the 16, with respect to SU3 cross SU2 cross 2. This is not trivial, actually, to get exactly, especially the young hypercharges, really very difficult to get.

But really, you get the 3, 2, 1. For each one of them, really comes out. The fourth is that we predict the existence of a Higgs doublet. The phenomena of spontaneous symmetry breaking.

And that actually really comes from the consequence of having the correct hypercharges.

It's a consequence, actually, especially with the negative mass term for the Higgs, which is minus mu squared h bar h. So it comes with that.

A prediction, more or less, actually, of the top quark mass of top quark, compatible with the experiment, we'll get it within, say, 4% or something.

I would say, actually, why we're not able to pinpoint everything to the last digit. We also predicted the see-saw mechanism to give very light left-handed nutrients.

And I will add, actually, something I mentioned last time at the end of the lecture, is that you get the correct Gibbons-Hawking term

in addition to the Hilbert-Einstein action, which

is that when people talk about the minimum kappa with how it is, they take the sum of the two terms. But we do get a cost term, which is, in fact, also

predicted by Feynman, which is the term in rh squared. So I mean, it's not really the sum of the gravity plus the sum. Conformer couplings of scalars to gravity.

What terms do you do like h? rh squared. So you get terms like root g d for h, rh bar h term. Also, r sigma squared terms. These terms are really present here. They are there with a conformer value, you mean?

Yeah. This is a thing which is like a theorem, actually. In the spectral action, you get, actually, first this cosmological term, which is lambda. Then you get an Einstein term, cosmological.

Einstein-Hilbert term, which is root gr. And also, you get, actually, h bar h term and sigma squared term. Then the next term in this expansion, in this asymptotic expansion, you get really what I call a conformally invariant action,

which is root gr. And then you get, actually, all the d mu h squared. You get f mu nu squared. And you get r h bar h. And it's not now h bar h. No, no. This is a conformal, yeah.

C mu nu sigma squared, yeah. Plus, and it comes with a Weyl tensor squared. And there is, of course, actually, another term, which is Gauss-Bonnet is also there. So this is reconformant. So what does it mean, actually, if I'm really

assumes that this is an effective action from the Wilsonian point of view, it means your starting point is conformally invariant, at least for all terms which have dimensionless coupling. All these terms are known. They have dimensionless coupling.

And so, of course, the coefficient is logarithmic. And that's why it comes. It's exactly actually conformal, every single term. So these, actually, I can say, these are the main. They are more subtle points. But I will not go through the more subtle points.

I can explain, actually, why the thing, OK, what are the drawbacks? So maybe, actually, one has also to say few things which, no, no, only one term in the expansion is conformal. Yeah, the whole block, actually, is conformal.

Or quartic, let me see, order four, are reconformant. It comes exactly conformal. Yeah, yeah, they are reconformant. Yeah, but the meaning of the conformal RH squared is only with respect to the first line. Conformal means I change. Yeah, yeah. No, this, actually, of course, these two terms destroy conformal invariance. The cosmological and the R.

If I change R here, G, the metric, I'm something proportional to 1 plus h squared, OK? Yeah. I will change the coefficient of RH squared. So what is the meaning if I have something conformal? You said the coefficient was 1, 6, which is the conformal.

Right, yeah, yeah. Is it 1, 6? 1, 6, yeah. 1, 6 means that when I change the metric by a conformal thing, I can absorb completely this RH squared in the metric, so I can make this appear.

No, well, actually, OK. This, no, no. No, no, I say only one part in this expansion happens to be conformally invariant. One can prove, actually, why this term in the asymptotic heat kernel expansion is conformal, actually. The term is conformal. If you change the coefficient of RH squared, it becomes conformal.

Let me say something, actually. Last time, I did say that the Gibbons-Hockin term comes exactly with the right sign coefficient. However, I also added that if I use not the Dirac operator but Laplacian, you get the wrong coefficient. In the same story, actually, suppose that I didn't use d squared in my expression.

I say f of Laplacian or something. Then, of course, this term, this part of the expansion, will not be conformal. It has to do with the Dirac operator. Yes, the Dirac operator has more conformal properties than Laplacian. Yes? Maybe for some reason, the reaction is really forward. No, actually, I would come to the question of scales,

actually. What does it mean to have this conformal invariance? I'm coming to it. But that's the reason, actually, I didn't want to stress so much that I have this RH bar because you're right. Either the whole action is conformally invariant or it's not conformally invariant. I would say the whole action is not conformally invariant because when I go even to even higher orders in my expansion,

this conformal invariance breaks down. The part which has no lambda is conformally invariant. The coefficient is logarithmic. There is no scale there, OK? Yeah, I think in this explanation. Yeah, OK, well, actually, not completely.

Suppose I use Laplacian. Then I can show you, actually, you don't get y squared. You get something else. So it's not only that, actually, there's some hidden aspect. OK, one really can say also things which are not perfect.

And the thing which is not perfect, actually, that here there is a prediction on the unification of the coupling, of gauge coupling. What is this?

It tells me that 5 over 3 g1 squared is equal g2 squared equals g3 squared. And if you run the RG equations and you look at if you define alpha i to be gi squared over 4 pi, i is 1, 2, 3, you find that alpha 1 inverse, alpha 2

inverse, and alpha 3 inverse, they almost meet, but not quite. Depends, actually. One, of course, has to run. If you do to 0th order, just turn on first order. If you run that, you discover they don't. But in reality, you really have to take loop corrections

into account. Now, the loop corrections have been worked out even up to three loops, actually. Up to three loops, things have been worked out. And it's known, actually, that the running does change. And especially for the alpha 3 inverse, actually, does change is like 16% correction or something.

So however, the problem that nobody has worked out the equations in the presence of the singlet, because we know the singlet does change the Higgs, and the Higgs itself changes the other coupling. And nobody has done, actually, even two loops. They have done only two-one loop. We know two-one loop things do change, but still, that thing don't really completely meet.

We are not really sure, actually, of the story, because we are waiting until someday, somebody will compute the two-loop order corrections. And the guess is that things would really improve, actually. However, this is subject to discussion, actually, whether is there something beyond the standard model

or not. This is a topic that I will discuss later, whether is there any physics beyond the standard model, besides the fields that I have written. Because if you look at this point of view with this classification, it seems that we only predict the standard model plus the sigma

field, which is new. Later, actually, there is a dilaton, which I have not talked about yet. That's also a possibility. You can add, actually, an anti-symmetric field, which is an axion, through torsion terms, by assuming the connection, actually, in the manifold part here.

In the four-dimensional manifold part, you can assume that the connection has a completely anti-symmetric portion, in which the B field would appear only through its field strength. This you can do. And in this case, you'll obtain the usual axion, actually, as a contribution. So that's a possibility.

But apart from this, this is it, actually. Then you say, OK, there is nothing else. So this actually is. But of course, there is a question mark. We cannot really make a definite statement now. They don't meet. But something may happen, because nobody has computed the two-loop re-normalization group equations.

This will come at some point, I'm sure. The next, of course, actually, many things we cannot explain, like why there are three generations. The structure of the Yukawa coupling, this Dirac operator in the finite space, has many entries. And every entry happens to be a Yukawa coupling for one

of the observed fermions. So if one can predict, actually, the values of each, of course, you have solved the eternal problem. But that seems to be really tough. And the question is, what determines the Yukawa coupling section? This we have not answered. What determines?

Or the entries, let me say. What determines the Dirac operator in the finite space completely? This is a big question.

But if one can do that, you have really a full-scale prediction of everything that is observed. There are a number of generations, also. We have nothing to say about that. If you want to multiply this by three generations, it means the algebra is the Dirac sub-node.

No, no, no, no, the algebra doesn't change. The inverse space changes. Just you repeat three times the inverse space. But it's important to say, yeah, if you want that, the difference with the usual problem for Yukawa coupling is that now that it's geometric meaning. And this means that if one could

make a guess of what is the corresponding geometric non-commutative space, then you could make a guess for the Yukawa coupling. Yeah, I think now the question of the Yukawa couplings of the firm has been changed into a geometric question. Is that what determines the Dirac operator of the finite space? We have not really assumed anything about this operator.

And this is the reason why, you know. Let me make, for instance, a wrong guess, but just to give some idea. It's known, for instance, if you look at spectra and so on, you'll find that there is something which is corresponding to the usual music, and which has no incarnation for finite spaces,

but which, if you think about non-commutative space, you will obtain what is called a quantum sphere. Now, the quantum sphere has a spectrum which is powers of Q. And it's difficult to see that when you look at the structure of the quarks and all that, there is a little bit of this geometric structure, in the sense that the masses are more like powers,

like rather than an arithmetic progression. So what one would need to do is a bright guess of what is the underlying geometric space. But I think the point that is made is that now the question of Yukawa coupling has been changed into a geometric question. What's the nature of a geometric space remains to be seen. We need to have a huge basket of geometries,

non-commutative geometries, including quantum spheres and everything, and look at this basket for what would correspond to that. This is the idea. But according to my information, there are some examples, but not that many, actually. Are there many examples? Yes, there are plenty of examples. The thing that you need to understand is what would make examples to be finite.

Every year, there are plenty of infinite examples, with infinite spectra. But of course, we know it's a finite space. So a finite space, you have to somehow cut this quantum sphere or something, do something like that. OK, deal at all.

When I introduced the spectral action, I said that the spectral action is trace of f

of d squared over lambda squared. And we put this by hand, this scale. However, the idea of a scale which is inserted by hand, not very natural. And then it's more natural, actually, to say I'm really going to, I can replace d squared

by e minus phi d squared e minus phi. Why, actually? Because here, if you let phi goes into phi, say, plus little lambda, it really gives me that d squared would go into d squared over lambda squared. OK, so let me call it this d phi.

I define it this way. You know, it gives you. So it's natural to replace the scale. And you hope that the field phi is dynamical. And then it will get some expectation value,

you know, since the only scale is. I just want to make one remark. It's about at this point, because imagine that f is just a cutoff function. Then when you take f of d squared over lambda squared, you look at the eigenvalues of phi, which are smaller than lambda. That's the same thing. But now when you do the data, what you do is you look at where it's smaller than it's going to shut down.

So it's longer at the global, but it's localized somehow. So this is what it means, this data. Well, actually, see, in physics now, it's not that the best thing that if you really want a number, the best thing is to put a field and let it get an expectation value. And this expectation value would really, OK. So now the problem then reduces, actually,

to this computation, trace of this d squared phi like that. The question, what's the answer? So one does the calculation. And one discovers, actually, the following, that this becomes. You know, it's really easy to see here what happens that if you have something like g mu, g mu.

OK, so what's the leading operator, the leading term? The leading term would be e minus 2 phi g mu d mu plus. It's known, actually, that this elliptic operator starts with g mu nu d mu d nu. Now you discover that there is e minus 2 phi.

So you say, OK, let me define a new metric, which is e minus 2 phi g mu nu. And with this metric, actually, this would be equivalent to going to the Einstein frame. This gives me the capital G. All this bad notation, actually, is to write capital G, because in general, it gives you capital G refers to the Einstein metric.

But let me take Einstein frame. And then one observes the following, that in this Einstein frame, in which the metric in this guy, in which it has been rescaled with the g, what happened is that the spectral action really becomes like this. This phi, this appears.

And then the next term will be root g R of g. And then, for example, even for the mass term, you'll get, actually, let me say this is important. You'll get h prime bar h with some mu squared or something. Well, actually, there's h bar prime h prime.

And there is sigma and everything. And where we have defined h prime to be e minus phi h. And in addition, there is a term, which is g mu nu d mu phi d mu phi.

And then we go on, and then we get g with c mu nu rho squared, which is conformal invariant. In other words, actually, the action in terms of the new frame, which has been rescaled with the electron field, is exactly the same action as without, except for one term,

one and only one term. And this term is the kinetic term for the dilaton. It gets a kinetic term, so it's really a physical field. In addition, everything is rescaled. For example, phi prime will become e minus 3 over 2 phi. h prime is e to the minus phi h, and so on.

So in other words, the physical fields you observe have been really scaled with the dilaton. What is now actually in reality, one really can go and compare this, the model you obtained, with what was known as the Randall syndrome model. You really find exactly the same model. In that picture, what happens is that you go to a five-dimensional theory.

In this five-dimensional theory, you have in the fifth direction, you assume, first of all, actually, you assume that you have periodic coordinates. And so, well, actually, they actually assume that you have two brains, and this would correspond

to the x5 is 0. This corresponds to the x5 is pi. And what happens is that some fields get rescaled on one copy and not on the other. However, in our case, it's a similar situation. The terms are identical, actually, except the fact now we know that h prime is e minus phi h.

Perturbatively, one really has to compute the action to all orders, which we cannot. But it's really clear that at the perturbative level, the phi doesn't really get the potential term. The higher terms will be derivative terms.

And this situation is very similar to what happened in string theory, in which the field phi doesn't really get the potential. And so one assumes that at the non-perturbative level, phi gets some expectation value.

You have the scale somewhere. I mean, the Higgs potential of the scale minus mu squared. So exponential minus. Here, well, actually. Well, actually, here, there is a scale, OK, lambda squared. Or it's really m prime squared in all this. OK. You are saying there is no algebraic dependence on phi,

exponential phi, anywhere? They were the mass squared, one squared. No, this is the kinetic term, but the mass, like the Higgs minus mu squared, h squared. Because h is the scale. They're not exponential minus 2 phi. OK, let me see. You are really going to get through g, I don't know.

In the Higgs potential, there was a scale. Yeah, OK. H is the scale, by the scale. H squared. But if it's like that, phi equal to 0 is a solution, you are saying. It's really decoupled, then. Except from the kinetic term, yeah.

Yeah, but it's decoupled. I mean, if phi equal constant, then this is an exact solution. It does not couple to physics. It doesn't couple to except through derivative terms, yeah. So one has to assume that the non-perturbative level it may get the potential. This is not easy to see, but there's a possibility.

No, but even if it does, this is an extra scalar field. It's not a dilapant anymore. I mean, a dilapant changes something. Here, it's called a scalar field, which is on the side. Yeah, you will not see, but it has a consequence. For example, that if the field H. After the scaling, it disappears. The only term that is the kinetic term,

it's called a master scalar field. Is that what you call conformal in Venice? No, that's what I think, because it's not conformally conformal. But it's not conformal, because this term destroys conformal in Venice. I think the problem that now, it's a question of interpretation. Because now you can say, actually, the field H prime,

if the field H would get an expectation value, we have seen that. Now we have to look at the symmetry breaking with H prime now. We forget about H. If what you are saying is correct, let me forget about the H and all that. And I have only five somewhere on the side, and then I have H prime. So H prime, we get the value 1, let's say.

The five independent value of H. But before, the value of H prime was like a mu, which was lambda.

And therefore, as you have modified lambda exponential, the value of H prime should be a exponential 5. The square term cannot be correct. It's only the conformal piece which is in I, but not this term. Yes. So this term cannot be the same. Yeah, this term cannot be the same, because the volume is

scaled at the first power. This term is scaled at the second power. So there is an exponential in front of this column. Then it is a de-lactamic field. Sure. Sure, sure. Then you see it changes. This term is not correct, really. Because the volume is scaled at the first power, not from this term.

It cannot be correct, this term. I have to check it just a second. One needs to see the ratio between H4, because there is an H. No, H4 is correct, but not this term. This term has to be the same. The scale on the square.

No, it's clear. You don't have to look. This term is clearly changed. You can look at the previous thing. When phi is constant, it just changes what was lambda squared before.

And lambda squared changes. And then it gradients phi somewhere. Yeah, yeah, sure. And everywhere there was a lambda squared, it becomes exponential. But probably, this would cancel with root G. No, no. We cannot do two truss ratios at the same time. Root G is a determinant, so it scales as a first power, not as a second power.

OK. So I think the important point is that if this is H, then the mass has to be H prime, which is e to the minus 5 H. So essentially, if you can think that H can get an expectation value of the order of the Planck, and the phi is of order 40,

then H prime would have a normal mass. You know, this is the term. Which is your first time, sine phi or phi prime? Here. Ah, this is for the Fermi's, you know. Because the Fermi's are also rescaled. Everybody is rescaled, actually, in this language, because you have to.

The phi doesn't appear. You see, in the Fermi's, it disappears also. No, it's 3 half of phi in the exponent. Yes, yes, yes. And you just want 3 half. That's my question. Yes, 3 half of phi. To be disappeared from phi. It gives you the dimension of analysis that you should know.

Psi squared is the cube of the mass. Anyway, so the important thing, actually, that there is a scalar field, phi, a dilaton, which is in the game. And this, somehow, the only also, we discovered that the only other particles, other fields, which are present beyond the standard model

is the sigma, the phi. And maybe you can have a torsion term through a B mu nu that only appears through its field strength. So the phi, you didn't analyze the phi. You didn't look at the dilaton. And are you sure the phi does not come in front of F mu mu squared?

No, of course, because everything is conformally invented, yes. Can you say that the dilaton comes in front of it? It's not the dilaton of string theory. This, I can say. It's different, yeah. All right, one more thing maybe I should mention is that the question of parity violation,

which is written on the theta problem in QCD. See, we have written this spectral action trace of F

of d squared over lambda squared, say. We can also write, actually, another trace in which we insert the gamma. Remember, actually, this non-commodary geometric data includes the gamma, the chirality operator.

And in reality, you can compute this guy. And for trivial models, of course, if you do it naively, you obtain terms like this. Mu nu rho sigma R mu nu AB R rho sigma AB.

You get such terms. You get also terms like epsilon mu rho sigma F mu nu rho sigma with a trace on the internal space. This is known for QCD. It gives you what's called the theta parameter. This will be present.

You can compute, and you'll get, OK? So this is the Euler terms, actually, which destroy parity because it has only one epsilon.

I think in generativity, what they call it, barbaro immersi or something, terms. It has some name in the generativity. Now, so if we do that calculation for the spectral action here, for this, what do I really get? This, as I said, is a huge trace,

where this is an arbitrary function, but the trace is 384 by 384 matrix. And in reality, you re-obtain this term, but the term have a coefficient 0. 3 comes to 24 minus 24 equal to 0. So for spectral action, this term is really absent, this parity-violating term.

And in addition, for QCD, I don't know. We call it g mu nu or g mu nu or sigma 4 SU3 is also absent. It's also absent. It's not absent for everybody. For example, you get a term like epsilon mu nu sigma

B mu nu B rho sigma. But this is a total derivative, actually. Doesn't count. However, there's a term, which did bother me, actually, at some point. And this has to do exactly the same as the theta term of QCD, but for the weak interaction.

But then I went, and I discovered that people actually, this term is observed, actually. And it's taken into account. You see? It's not that it's not present. It is present, and it's there, actually. It's a parity-violating term for the SU2, because it's known that SU2 does actually break parity.

But you know, I need a high combination to the case of 1 plus gamma over 2 times f of mu. Well, actually, I'm really general. I take this, and I take that. This we have already computed. This actually is not parity-violating. This term is parity-violating. This term is an index popular term.

Yeah. So if you compute, you find, actually, that in reality, it's Euler term, but it's absent. And similarly, it's absent because of some anemone cancellation. It's a numeric section, and everything is rigged so that's. There are sectors that are less quite than that. Exactly. I think this is the good answer.

This is the good answer. It's left to right symmetry. Make sure that this term is not there, actually. Anyway, how about actually, of course, at the tree level, we have the theta at the tree is 0, yeah? But it's not enough to guarantee if it's also true at the loop,

actually. At the loop, one can show that, actually, at one loop, you have to make an arrangement to cancel. But if you manage to arrange it to cancel at one loop, it will cancel to all orders.

However, there is a condition which we don't know what it means here. The determinant of the up and the down quark coupling should be real. If you can arrange that, one can show, then, that the theta would be absent to all loop orders. This is the condition that it's eliminated at one loop order.

But if you arrange that, it will be eliminated at all loop orders, actually. So again, this is a question which is related to the structure of Dirac operator in the finite space. Why should this condition be there? But at least, actually, at the tree level, it's not there. And there's no fine tuning. It's natural, and it's automatic. So I think that's one of the positive things from the next.

Any questions before I move on? Yeah, I just want to say one thing about this, when you say the connection not to be symmetric, when you have this anti-symmetric terms.

Because, I mean, of course, it's very important that one characterizes abstract into Dirac operator, so on a deception. Yeah, usually, you have to make sure that psi, deep psi. Well, there is a J, but OK. Let me say deep psi, psi.

I can do it this way. It has to be Hermitian. And this Hermiticity, it means you really have to integrate by parts. Integrating by parts, in generativity, it means it's a condition on the spin connection. And this condition on the spin connection,

it's really, I think you can call it like this. Omega mu, mu B is equal to 0. You know, it means the trace of the connection should be equal to 0. And this, of course, is satisfied by connections where the torsion is completely anti-symmetric. But what I would say is that if you want to characterize algebraically

the Dirac operator, or the manifold, the condition that you obtain very naturally for the cognitive geometry, they still have a little bit of figure. And it corresponds to this figure. So this field might be interesting to investigate. Of course, we didn't look at it. Yeah, the axiom is most irrelevant to dark matter, CP violation, and things like that.

OK, next. The question is, is there anything

beyond the standard model? In other words, you know, it's now known that experimenters start looking for new physics, most in the form of supersymmetry, but in the form of anything, actually.

Everything is open for possibilities. And it's an open question, is there any physics beyond the standard model? We have seen that in our case, these are the only possibilities. And they have consequences, something at 10 to the 11 GEV with the sigma and with the neutrino masses and things like that.

But the question, is there anything observable at low energy? In other words, if tomorrow, something at sun, they announce that they discovered in the data new signals, what would it correspond to? So the question for us is that,

is there any room for generalization? Have we been exhaustive in our analysis? Well, there is a root point, obviously, in our analysis, which is the breaking from HR position and so forth. So obviously, we remember that at one point,

we said there is a non-trivial mixing, and the sigma does exist. And this broke into C plus H left plus M3 of C. And this we did it by assuming that this condition holds.

Now, I mentioned before that this condition does imply that the connection is linear. What does it mean, actually? It means if I define dA, which is d plus A plus absolute JA J inverse, then you know that

under, so for example, if you let psi go opposite,

then dA would go into, well, at least in this case, it would be U dA U star, where U is an element of A. And U is unitary, provided this transformation is OK,

provided this condition is satisfied. Do you have psi on both sides? No, because it's not U dA U star.

It's U J U J star. The U, no, it's not U dA U star. It's U J U star. Here. No, no, psi is U psi U star, yes, and the action on the right is given by J. U psi U star, OK. Which is U psi, OK, here, we have to take, actually,

it comes to the right as J U J U J times psi. It's U J U J times U J times psi. So J makes a bi-module. So this, actually, let me note, is U U hat into psi,

where we define U hat to be J U star J. I don't know, if you write it like this, put G U J. U J. So this is quite an important point, because what happens is that because of the J,

the hyper space is a bi-module, and the fermions are in the adjoint of an operation. This is very, very important. Moreover, when you write dA, it's not U dA U star. It's U U hat times dA times U U star U. Now, we can see, actually, that in this case, actually, in reality, if we want to define the A, one has to define the A as something like A A hat.

What you have written in the graph is not correct. It's U U hat. dA goes to U U hat. Capital U. U U hat? No, no. I mean, it's not the U in the algebra that you have here. It's U U U hat.

The H transformation, you have two Us. I have to try to have the A transformed, actually.

I wonder how dA holds this. Yeah, but dA is not U dU star. It's U U hat dU star. U U hat. Yeah, U U hat.

Essentially, if I define U to be U U, then dA would go to U dA U star. Well, actually, remember, I was talking about the case where I didn't assume this condition. The thing is that if you assume this condition, then,

of course, this reverts to the other, because all the hats will drop out. No, they don't drop out. No, they don't drop out. They have a U term. They don't drop out at all. You see? I thought, actually, that we have A. One second, A1 here.

But I thought that A1, if you assume this condition, would go into U A1 U star. And then you also have plus U hat. Oh, plus the other term, yeah. Yeah, OK. So it's 3D. Similarly for the other terms, OK.

So in other words, actually, in reality, one has to introduce such transformations and see what is the invariant operator or what's the invariant quantity you can produce. And in this respect, I want to discover, actually,

that the A has to be written as something like Ai Ag hat into D Bi Bj hat.

You know, there is a tensor operator between them, but I will not write tensor operators, obviously. So here we can make without any loss in generality

the summation of Ai Bi is 1. Why? Because you can always increase your set by 1, in which whatever you have, you add one more element. And so that this is satisfied. And here, summation of i and j. Now, remember, actually, Ai Bj hat is equal to 0.

This was the 0 order condition, or order 0 condition, which implies that left and right actions commute with each other. Now, if we try to compute this object,

We discover the following, that you have ai aj hat. And then I'm going to get dbi bj hat plus bi dvj hat, summation on i and j.

Let me look at the last term first. So the second term can be written as ai summation i on j, aj dvj hat. Why? Because this, the bi and the aj hat is 0.

It goes through. It commutes. And of course, actually, this term, because ai bi is 1, these terms become aj hat dvj hat. So obviously, I obtain my next term, which is this way. Now what about the first term?

The first term, this one, I have to commute, actually, this guy with this guy. So obviously, I get the term, which is aj hat dbi into bj hat plus another term, which

is ai dbi aj bj. But this is 1, summation on the j. And this, of course, is a1. Now we call it 1. But we are left with this term.

Now this term, remember, we did assume the order 1 condition, which tells me that the a hat commutes with the db. And this term drops out. Now suppose that I don't assume this. What does it mean? It means that the a connection will have a new term. And this new term can be written as, because this commutes with this in reality,

I can take it in. And this term is nothing but it can be written as a commutator of a1 with aj hat aj hat a1.

Note that it depends on bimodule terms. In other words, it depends on the elements in the a and the elements in the a opposite.

And it's mixed up. That's why it's quadratic in the dependence. It depends on a and adb, but from the other side. So these are quadratic terms. But once we admit, actually, that this term is there, we introduce this connection, then in reality, you can have more general connections than the one we considered.

And these connections would have certain properties. So this is the most general connection you can consider without assuming the order one condition. I think it's very important to say two things here, because at the conceptual level, it's extremely important.

So the first thing that we discovered is that when you drop this order one condition, which was making things very simple, you find out that the inner fluctuation of the venting is added connection. They form a surmised loop. And how did we find that? We found that because you have this quadratic term, so you can get very scared.

And then you can say, OK, imagine that I do an inner fluctuation. I get this new Dirac operator. Now imagine I start from this new Dirac operator, and I do again an inner fluctuation. I should get the quadratic term, because I got this quadratic term. No. There are cancellations. You still get the quadratic term. And so what does it mean? It means that, in fact, underlying this whole story,

there is a surmised loop. And this surmised loop only depends on the other one. And it's an extension of the gauge group. And it governs all the inner fluctuations. So this is a big change with respect to the only non-linear picture. But it means that linear fluctuations are, in fact, the action of the semi-group.

And when you perform by an element of the semi-group, you control the elements of the semi-group. So this is what Alain was trying to say, that these inner fluctuations that you considered

with the element u, capital, and that they form a semi-group, and what does mean that if you fluctuate with respect to the new generated inner fluctuations, you fluctuate again, it still gives you terms of the same form.

So inner fluctuations are still, again, inner fluctuations, which are the semi-groups. And one can describe the semi-group. So the semi-group is simply you take elements of the algebra which satisfy sigma of ai di is equal to 1. And you compose them by left and right multiplication. And they still satisfy the same thing.

We call it the step of perturbations of a. So this is the property, opposite summation on j, which is an element of a across a opposite and such

that aj bj is 1. Actually, as I said, it's not really strong condition because you can always make it, and such that aj bj opposite is equal.

It's the star of itself. It's bg star aj opposite. I'm not writing tensile products. There are tensile products actually here. So the definition of the semi-group. Anyway, now I'm not going to because the computation are

quite involved. Show spacetime point of view, this new term in the connection, a2, what does it look like? Yeah, I'm going to present it now here. It's only if the Dirac operator was not first order. But for instance, the reason why we are also there at s for space time is that we know that the Dirac propagator is

dressed by the quantum corrections. So this means that the two Dirac operators will not be order one because it will have a form factor. And then it does not satisfy the order one condition. So this perturbation will be necessary for this case, even for the ordinary case because of the quantum corrections.

So let me try to answer to your question what happens actually if we allow for this case, practically, how those things would look like. So now, no more breaking because there's no order one condition to break this. This has to come dynamically, in other words. The breaking must be attained dynamically.

And you simply say, OK, now let me compute the Dirac operator. And you discover that how that Dirac operator looks like. First of all, you have H right, H left. It would imply that your gauge groups is SU2 right and SU2 left.

And in the bottom, you have SU4 color, actually. So diagonal elements in this would be right, left, and color. So obviously, you know that the gauge group in this case is SU2 right plus SU2 left.

The dimension of the Hilbert space is still 16. So the prediction doesn't change. But now your 16 is written as 2 right plus 2 left and a 4. So 4 times 4 is 16.

And this is the way it's decomposed. And of course, this is known as the Pati-Salam model in which it has lepton comes out as the fourth color.

Now, what about actually how the Dirac operator looks? We know we have learned, actually, that whatever happens in the off diagonal elements would be Higgs fields. So you are really going to get new Higgs fields.

The new Higgs fields would look like as follows. Let me give you an example. So this would be, let me say, actually, I'll call it sigma a dot b.

Sigma a dot b, a phi a dot b, something like this. So it's 2 right, 2 left. So you really get this type of new Higgs fields. And along the off diagonal elements, you are going to get, I don't know, sigma a dot i bj. So you really get higher representations with respect

to the SU2 right cross SU2 left cross SU4. So you get the new Higgs fields. So there is no more one bubblet, which is necessary because in the end, you need more Higgs fields to break this symmetry

into the lower symmetries. Now, there are technical details because if you really start from Dirac operators that satisfy the order one condition because of the semi-group structure, you would obtain actually a new phenomenon

where the Higgs fields in this are composite of Higgs fields that appear here. So you don't really get independent fields. You get fields that depend on only a very small set of Higgs fields, which is the 2 right, 2 left. And you get another one, which is delta a dot i.

This is 2 right, 2 left. And this is 2 right and a 4. And yeah. So now, actually, you may ask, what is this old sigma? So if you look, the question, which configuration

would generate for me the standard model as in its set? And it is the following. Yeah, this one. So a dot i, if we take it to be,

you know, if we take this index to be 1, one of them, it means it's OK. This is the expectation value. And if you take delta i to be 1, it means the lepton color would get a wev. And it's root sigma.

Then that, and similarly, actually, this guy, you take it phi a dot b. You take it to be delta a dot 1 epsilon b c h c, like that. Then you can show everything reduces to standard model.

So you can obtain the standard model as one of the vacuum expectations of the full theory. But of course, actually, there are many vacua. The question, which vacua to consider depend? Of course, I can say the following.

The Higgs structure is completely fixed. It's not that I unlike gauge theories. In gauge theory, what killed grand unified theories, like SU 5, is the following. In grand unified theories, first of all, the Higgs sector is completely arbitrary, first.

And the second, you have the proton decay problem. Yes?

I mean, once you have Higgs in some representation, you have two or three scalars invariant. So you start completely arbitrary. Well, look, I give you how arbitrary. Let me take, say, the SO 10 model. Then what are the possibilities? You can have 45, you can have 210, you can have 120,

you can have 126, and all these actually you can put in. And if you look at the potential, you will discover that it can have a principle 20 times or so, all with arbitrary coefficients. So group theory cannot help you in this respect, because there are many possibilities. Once you start taking project representations, then many things are possible.

And in reality, I always give an example of the SO 10 grand unified group. Why, actually, in the end, things really become very complicated for grand unified theories. Because one can say, OK, why don't you consider SO 10? Because an SO 10 is like us. The fermions all live in the 16 dimensional representation.

However, there it's not the only representation allowed. You can have, actually, many other representations, like, as I say, this 126 is a spin representation. So you can have the 126 spin representation. For us, it is the only representation allowed. 16 is the only thing which you can have.

In addition, when you compute the spectral action, all the Higgs interactions are fixed. You compute, you get an answer, and this is the end of the story. There is no room for maneuver. The nature of the Higgs shields, it's more, actually, we really find two cases, let me say. The two cases, one case in which

you start from a d, which does not satisfy order one condition. Unarbitrary, you know, the thing that before you fluctuate, you allow a d which looks random. Then in this case, you are really going to generate Higgs fields, which are listed,

which you can list, actually. I can write them. It's what, two right, two left, and 15, and you know. So I write them.

Yeah, I written them, actually. Two right, two left, 1 plus 15, 3 right, 1 left, 10, 1 right, 1 left, 6, and so on.

So this actually you can have. Or if you start from a d, that does satisfy the order one condition. Because of the semi-group structure, you will go there and you will stay there. You cannot get out. And in that case, the Higgs fields that you can have

is only these two guys that I have written here, actually, which happens to be two right, two left. And this happens to be two right, 4. These are the only Higgs fields you can have. And then this potential does become simplified. The problem, actually, in this case, of course, is that the connection becomes quadratic in these fields.

Quadratic in these fields, then one has to understand the phenomena, whether the semi-group structure would help you with the normalization scheme or not. This we have not investigated. We don't know the answer. It may help or it may not help. But the worst possible case is that you can have all of this. Now, all of this can have this break-in mechanism.

And in this break-in mechanism, you would recover the standard model. And in this case, if you ask me, what would happen if this was the case? Then, of course, you are really going to generate this SU2 right, for example. And SU2 right has extra gauge fields.

And these extra gauge fields, of course, it depends which scale you are going to do the breaking. If you do the breaking at the intermediate scale or you do the scaling at the unification scale, you are really going to get different physics. So it's too early to tell, actually. There could be consequences in which one really

can go beyond the standard model. And this thing would have consequences. But this thing has to be studied seriously. And we have not studied this thing seriously because nothing has been observed yet. We're not going to go to jump and say, OK. Because as we say, the possibilities are plenty. The model is fixed. The model is fixed.

So in principle, with enough expertise, one can analyze this model. One thing about the proton decay problem is known, actually, that SU2, the Pati-Sala model, SU2 right cross SU2 left cross SU4 does not have the proton decay problem. Proton does not decay with this symmetry.

So you avoid it. And in reality, actually, what really killed the SU5 model and the SO10 bosonic model, not the supersymmetric one. Supersymmetric one is not dead, actually. And the reason is that the unification scale in supersymmetry is raised to 10 to the 16 GEV. And there are some cancellations.

So there is proton decay which is invisible. You cannot see it. You need at least four orders of magnitude in order to become observable. So supersymmetric unification models are still OK. But here, we don't have, actually, the problem. So this problem, although it's bosonic, but it has no problem of proton decay.

It has the advantages of unification. It can give you physics beyond the standard model. But the physics beyond the standard model is somewhere at unification scale, which is something like 10 to 16 GEV. And it could be even higher, actually, the symmetry,

depending on where the coupling constants meet and whether you want to say the gravity. Because there's three orders of magnitude, it depends on some numbers that come in which unifies gravity with other interactions. And it is possible that gravity does unify.

All the gravity is at the plank. It does unify with 16 because you have relations with numbers coming in. And it's easy to explain 1,000 with all the numbers coming in. So the conclusion is the following in that if we allow the most general possibility in which we

drop one of the conditions on the non-commutative geometric properties, which is what we call the order one condition. Order one condition means connections are not linear. They could be quadratic. They could have quadratic dependence.

Then you can go to the Pati-Sella model. But again, this is unique. You really can do nothing else apart from that. Another issue which I did not discuss is that what happens if you go and consider, we stopped at the first case where the dimension of the Hilbert space is 16. But we know that the next dimension of the Hilbert space,

this was 4 squared. The next dimension will be 6 squared or 8 squared, depending actually whether we want to consider. Let me see. You have to take M6 of c. And maybe I'll take M3 of h or something. But if we insist that this to be even again,

this would really take us into M4 of h plus M8 of c. This, it means you really have 64 fermions. It means you'll have a new fermion. It's too big.

It's too big, except actually some people are saying, OK, maybe. But I'm not really going to go into really something very speculative. Remember, we took the product of space-time times. So there is a spinor index. And then you say, OK, maybe this M4 of h

has also the Lorentz group inside. One can say such things, but one has to prove it. And it's a realistic picture. Some people did assume that, like Litzy and company.

But it's suspect, I would say. It's a suspect. They have not shown that it works. And so there are possibilities. But it really would need too many new fermions. Until they find something, I don't think it's worth venturing into.

In fact, there was a prediction where from the top mass, it wouldn't change. So I don't believe this. Yeah. Remember, actually, something which probably I said, I said it in words, but I didn't say, is that we have this relation. This would never change, that the summation of the square of the weighted, I would say, M squared, in this case,

E plus M squared, U plus whatever. It's not a square, you call it that. Yeah, but there's weights, actually, with colors. This will always be there. It's independent of what you do, actually.

It has to do with the scaling of the Higgs field. Once you scale the Higgs field to have canonical kinetic energy, then you are stuck with this relation. Is it usually as a regulation? So obviously, it's not really possible to go way beyond what we have. And it seems that the picture is extremely tight,

I would say, extremely tight. My time is up. I will stop here. If you have any questions, I would be happy to answer.