 Kaj načo je prijavno, da vzupajte površenje lajkšne interakcije? Vzupajte je, da učimo vzupajte genetik, če se vzupajte na taj unifikacij. In zelo, da se metri, ki je vsak sem etikov, in zelo, da je zelo jazda modelov, nekaj pozdravil na neko semetri, kaj je skončil s geometrikih starča. Vse, nekaj zelo, da se vse zelo zelo vizivamo framework. in we fix the geometry by specifying what we call the geometrical data, which consists of an algebra, Hilbert space, drug operator, reality operator and the chirality operator. to je fizika ljubljana. In, kaj so vse proporti, kako se komutatori vs. vse prišli, tudi, kako, nekaj nekaj komut, nekaj nekaj komut. In, kaj smo prišli, da smo prišli izgledati kako smo našli na helberi vse, in smo zvonili vse začet, in tudi zvonili jasno jasno, jasno jasno, jasno jasno, zvonili jasno jasno, in smo zvonili našli vse začet, našli vse začet, in vse začet. Zato vidim se, da je vse bila vseh vseh. Vseh je izgleda, da je izgleda, da je za vseh, da za vseh pravne opravljene vseh, vseh je vseh pravne opravljene in vseh je vseh, ali vseh je vseh vseh, da je vseh, da je vseh, ta retiča pa izgleda. Zato bodo se da zaostavila poničnje korektivno. Ordoč se zelo naša idejavče prioritoupnega geometrija izgleda. To je da nekaj hidrati. Tkaj biti se prav, da se ono neko komitivnički spas je na odmah, naže... vse malo, da jim se napravimo, kroz phi njič. In sem zelo, na zašli h m kroz h phi njič, g m kroz g phi njič. Vse zelo, da se predaj prejdi ojevno vse zelo, da jim se kroz m kroz k phi njič. In najbolj, in da d je je dm cross 1 plus gamma 5 tensor d phi nite. In druga zvonja je, če je tudi tudi tudi tudi tudi tudi tudi tudi tudi dimensično in non-committive spasje. Zato tudi tudi tudi tudi tudi tudi tudi tudi non-committive spasje. Se vam češ smo mogli te čas v denetosti. Tukaj tudi nolisilo, na zemljenju doloženja tudi tudi tudi tudi nisem. Staja, na svečenju doloženju tudi nema metrini, kar se je z nimi njal. A čas se je v pade, pa je v počkori, da je njega, kar je počkori, počkori, ki je zelo pade, na našem tudi in vzelo. Pa je vsez vzelo. Čas je to tudi vzelo, nekaj je vzelo tudi vzelo. Tudi, tudi tudi, In potem smo zelo vzouto vzouto algebri, kaj je tudi c, zelo c plus c, in v zelo klasifikaciju, zelo to je neprostavno, ko je ko 6, b je to 6, in to je tudi tudi tudi poslustvanja, In tukaj smo prišli, da algebra v finačnih spesih, AF, je vsešelj, da je MNC plus MNC, in variacije, kaj je zelo prišli kompleks in prišli kompleks v Katerniu. Zelo tukaj so pošli. To, če bi se izgleda na izometriu na več Balj. To se nismo izgleda. Bezgleda je to, da je očinitga sa tegoalanju keternjev na radiobadi. Pake se to, da je to po povaču, vsak je m2h vzip? To je m2h vzip, podjeli da je, da je očinitga krenjev, In je tudi, da je tudi nekaj. Kaj je m4 h, plus m8 c. En tako. Tu je to taj tuk, ki je potrebno algebras. Če je tudi simplez. Tudi simplez. Če je tudi simplez. Če je kaj je tudi kaj je m2 h. Premajdej, če je m2 vse, če je tukaj tukaj tukaj matriče, kaj je elementi katerničnih. Tukaj je tukaj tukaj, če je tukaj q1, q2, q3, q4. Če je tukaj katernični? Katerničnjič nekaj prezentuje tukaj tukaj matriče v formu alpha, beta, minus beta bar, alpha bar. Bar je kompleks konjugacija in tukaj notacija. Kaj je matrič vse formu, če je tukaj tukaj prezentuje vse katerničnih v tom symbolizacijen formu. Zdaj je tukaj pomekaj tukaj. Kaj katerničnjič nekaj prezentuje tukaj vse m2 vse h, če je tukaj elementi, če je tukaj katerničnih, če je tukaj katerničnih, če je tukaj vse komučaj v stojku, ki izgledajteanzu, ki je Smoke ismi tukaj minus 1.docudija, kajo je tukaj z tezmi metrički, a tak je je si to na pohelo in ubrali h minus h. Na koncluzion in tak izgleda, algebra do finaj tukaj stav je zeprič f h doma plus h do Assistant. Zostar jo. Enče poz.) Zazasam tukaj algebra do finaj tukaj stav, da mora ovoju s vsem. na prvoj izgled. In je to dobro vse. Tem, ko je vse z naprej zelo za 16-lamenčno. Zato sem da se nekaj nekaj nekaj. Kaj je tko z 2 lefci in je je 4. Tko je 3, 4 x 4, in nekaj se početila, kot jaz si 4 colo. in te da. 2 r. 2 r. 2 r. plus 2 r. OK? So.. So asrefajs espeeno index. Hier actually is re, not a spinner index, it's acted on by the katerium. I it's a doublet action, is a doublet on the sg2, right and sg2, left. So as you see it. Now further, what did we have? We also said In vήmanje, ki bomo priježili, vse spinrje, ovo pa niste Return to 16, da znam 16 in vozbago si se 16o konjugirite vse. Kaj 16 o konjugirite ne gv lense jo. In bomo ima na tem spesih, 16-16 konjugirite. In to je drzok operator. Lej ste pa zdaj 16 ben arm in 16 konjugit in star. Aspečno. Wneč ne bo v tem drzok operator, The allocation of this expect is alright. Here we are going to get a 16x16 matrix. This is another 16x16. So here we really have 2 possibilities in that that the derivative operator, whether je da jaz v kratu amparti, nekako. Kaj so jaz v kratu amparti v kratu, izglede je izgleda, da bolj jaz z te spinas in konjigrana spinas ne wovali za vsih. Ča ima, da se ta metri m14 ne je obrostila, njih nič izgleda, in da proč se odnjužijo. In ne, da bomom izgleda zamordov na stvarom zestanje. is to have only one non-zero entry. In this, you know, as you see, is a 16 by 16 matrix, but you really can have only one non-zero entry. Otherwise, the, what we call, the properties a be opposite is equal to zero and da be opposite is equal to zero, which actually this is called order one condition. I will come later, actually, to this, I will explain this in great detail in the next lecture, is what's the significance of the order one condition, but I can't tell you the significance is that the connection is really linear. You are not really going to get quadratic terms in the connection. If you don't take this condition, then it means that you are really going to have, you know, the connection, which is d into d plus a plus j, a star j inverse, you are going to get an extra piece, which is really quadratic in the filter. So, this actually is a new phenomena. However, you know, we are really going to require that connections are linear, and if connections are linear, it means that this property is satisfied, and if this property is satisfied, one can prove unambiguously that you only can have only one nonzero entry in this matrix. What is the significance actually of this? This really has great significance physically, and the significance it will have in the following. It will turn out, okay. So, if you have a nonzero entry, what would happen, actually, that this algebra does break? You know, what's the algebra here? You know, you have like two by two matrices, and what I'm telling you that this entry somehow becomes distinguishable from the other entries, and the h right, which is a doublet, which I have written as two, would really break down into c plus the hl plus this m4 of c actually also breaks into c plus m3 of c, but the c's really become identified with each other because of this condition, and this symmetry really breaks down uniquely into this symmetry. This is the left entry. So, this actually would be the final algebra that satisfies all the conditions of the non-computative geometric space. In addition, I told you actually that the fermions were in this representation, two right plus two left, and four. So, this now breaks in as follows. This is would be one right plus one prime right. Two left does not broken, and four is broken into one plus three. This one is the lepton color, and this actually are the three colors, red, the yellow, blue, or whatever. Now, if you look at these representations, then you really can write this as one right, two left one, plus one prime right, two left one, plus one right, two left three, plus one right, two left three. Now, we can give them names. It's a state. This turned out to be the right-handed neutrino. This turned out to be the right-handed electron. As you see, it's one, it means lepton. So, these are the two leptons. And here, this would be the, sorry, it's not, I made a mistake, I couldn't, one right. Yeah, yeah. No, there's no, you know, it's, this is plus two. Plus, there is plus everywhere. God, made a mistake. I let me erase actually otherwise. Yeah, so plus is not the triplet. See, there is a U1, but that actually comes out. So, this actually would be one right one, one prime one, plus two left one, two left one. Okay, I think I better do it this way, plus one right three, one prime left three, and then I have two left one and two left three. Okay, this happens to be the right-handed neutrino. This happened to be the right-handed electron. This happens to be the upright, sorry, not upright, one right, and this is down right, and this happens to be the electron neutrino doublet, and this happens to be up-down doublet, which are left-handed. So, you really get all the representations automatically without any, you know, maneuvering, just in addition. The idea of the right-handed neutrino. No, you have right, and this is left, actually. This is up-down left. You know, this is the e-new left, up-down left. So, you really get everybody. So, the 16 representation, and one in addition. What's the significance of this sigma? The significance of this sigma, as we are really going to show that the hex field is there, and now we really can, you know, figure, somehow anticipate, we will get, you know, in this picture, as you see, everybody, you have only chiral thermos, which means masses can only be obtained after the symmetry breaking, and after the symmetry breaking, it means that the hex field gets an expectation value, and in particular, for, you know, everything will get, you know, for example, a new e-left, when you have 8-star, and then you have, for example, you know, e-right, and then you can take, yeah. So, all masses are really obtained, and similarly, you know, you are going to get a new right, 8-star, and then there is sigma 2, and a new e-left, and so on. So, actually, all masses are obtained only after the hex field gets its expectation value, which is, actually, is a standard teacher in all this unification picture. In addition, actually, the most important part is that now, because of the sigma that I talked about, you really get an additional Majorana mass for the right-handed neutrino. So, the right-handed neutrino, you are really going to get sigma r with the c, where c is a charge conjugation matrix, we are transposed, we are coupled to the sigma. And this, actually, I will show you later that this really gives an explanation why the masses of the neutrinos is extremely small. It's of the order, you know, the upper limit, I think is that something like 10 to the minus 2 eV, the present upper limit. So, this, actually, would explain why the neutrinos have small mass, and what really can work out, for example, is really clear that the left-handed neutrino would get a mass, which is, let me say, as follows. So, here you are really going to get the sigma for the right-handed, and then you are going to get some v, v, and zero, and it means if you work out the two eigenvalues of the right and the left, you really get mixed states, and the mixed states, so, you know, this sigma, let me call it w, you find, actually, one of the eigenvalues is, you know, approximately w and the other is v squared of rw, and v, say, of order 10 to the 2, and w is order 10 to the, you know, 11 or 12, gEv, which is really the masses of where the right-handed neutrinos would sit, and then if you really work out what is the mass of the left-handed neutrino mass of the nuL, it will approximately equal v squared of rw, and which is really, you know, it's 10 to the 4 over 10, 10 to the minus 8 or 10 to the minus 9 gEv, you know. So, it... A few eVs. A few eVs, yeah. Actually, you know, one really can work it out, it's really, you can easily make it 10 to the minus 2 eV and without any problem, and so this would really explain, and this is known, actually, it's called the seesaw mechanism. The advantage that you really get it automatically, we don't really plant the seesaw mechanism, the seesaw mechanism comes out, is extremely necessary, and it's necessary because you need to break the color group, and then you need to put something, and then the nice thing that, you know, you can only have one singlet, which physically means only the right-handed neutrino can have Majorana mass, which makes sense because Majorana mass, it means that the particle should not have any charge, it should be neutral, and then it's only the new right. You can ask why the new left actually doesn't have a Majorana mass, of course it cannot have because the new left sits with the electron in the same doublet, and you know, without breaking the SU2 left symmetry, you really cannot do it, you know, so. But you are an ER, do not make it a doublet. The new, and the right-handed, and they are an electron, do not make it a doublet. Yeah, not in this picture, actually. You know, in the other picture. And the left sector makes a doublet. Exactly, yeah. Left sector makes a doublet. If you really go to unification scale the picture may change, but you know, this is a different story, that has to do with this order one condition, if you really remove it, and then you really can have, you'll go to unification picture, but that's a different story, you know. It means, you know, what really holds at unification, whether we have a desert or we don't have a desert. If you assume that this market, DA with B up, V up is zero. This, exactly. This is assumption. The question, what happens if you remove this assumption. But in your model you assume that. Now, you know, it seems that this is the right picture is to assume actually this property. In the range. You know, if you believe that there is nothing up to, up to all things. But if you believe that at unification scale you have a higher symmetry which is, you know, h right plus h left and you have more fields, then you can remove that condition. The picture becomes much more complicated because then, of course, many more new hicks fields, which are necessary to break the higher symmetry. By the way, so you have both an h and a sigma. I have both h, yeah. I have h, actually, okay. So now actually I can talk about the spectrum of the particle. Sigma is a mean value. Sigma is a mean value of h. No, no, no, no. Sigma is not. Sigma is a new particle which will be heavy actually whose mass is something like 10 to 12 j. It's a new part. It's a new field. And this field is extremely for the stability of the standard model as we are going to show actually very soon. This field is what is the scalar? It's a scalar field, yeah. No, real. And the h field is complex? H field is complex, yeah. Where is the doublet also? H, okay. Now I'll talk about the spectrum of the fields. Where this h actually lives. So, you know, we said actually usually what you do, you can start from arbitrary Dirac operator and then you can generate actually the connection through fluctuation with inner automorphisms of the algebra. So, in this case, d goes into da, which is d plus a plus j a star j inverse. And what's a? a is the one form in this language which is adb. You know, this actually is a strange way of saying that, you know, you can always write a vector field a mu as summation of scalar functions, yeah. d mu of bi. You know, this is representation. As you see, you can have many, you know, you can really make a requirement that a i bi is one without any loss in generality. This is, you know, this is a arbitrary set of ai bi, an element of infinity of m, for example. So, these are functions and you can always write the gauge fields in this language. This is actually what it means, okay, for the usual Yang-Mills fields. So, these are algebras that you generate, you know, you generate a non-Abelian gauge field. But this is the picture to a is adb. Now, however, you know, we have to remember that this d that I've talked about in this huge matrix according to this picture is what? It's 32 by 32 matrix. It's a 32 by 32 matrix. And in addition we have to assume that we have more than one generation. You know, this is an assumption that we make because we cannot predict it. We know actually that we need more than one generation. The question, what's the number? We know that you need at least three if you'd like to obtain the CKM matrix and things like that, so at least there are three generations. So, the number of generations have to be imposed by hand and this Dirac operator, it means actually that every element has to be in generation space and in this case actually it is 32 cross 32 cross 3 cross 3, you know. So, in this case is really 96 by 96 matrix. In addition this has to be tensored with the Clifford Algebra. Why? Because remember d is like gamma it starts with gamma mu d mu. So, you see that the gamma and you know, we said that d it's dm cross 1 plus gamma 5 cross df. So, you are really tensoring with the Dirac Algebra. So, in other words actually, all these are Clifford Algebra valued Dirac operators and in reality you know, you still have to 4 times 6 is 3 384 by 384 meter. Yeah, it's okay. They look huge, but however you know, we invented some notation where everything can be written in indices and you know, one really can do it on the computer with really little effort. You know, you just put it in Mathematica and gives you all the expressions. Feet in terms of feets, you know. I mean, d is linear in gamma mu and the formula you write for a is still linear in gamma mu. Yeah, it's actually it's just like that. It's gamma mu tenser d mu. Yeah, but it's at the level of the gamma mu, not the algebra. Yeah, yeah. No, I mean, I said you are right. What I said that the Dirac operator, if I want to present them as matrices, they would be that of that dimension. Okay? Yeah. It's, you know, tenser. You tenser the products. So, you know, the calculation is it looks complicated, but it's really not because in the end it's nothing but matrix multiplications. That's what I'm saying. All right, so let's see actually how this A would look like. So you say, all right, this A. Okay, so now actually I can tell you how the elements of the algebra would look like and the elements of the algebra because we said this h right plus h left. It really looks like, you know, I don't know, maybe I should use it. It's like x, tenser 1 and 1, tenser y where the x would act on this h plus h and the y would act on m4 of c. So, in reality, you know, we really can represent it simply in terms of matrices. And if you do that, what would happen? If you do that, then you discover that this A field or even, you know, I can write immediately what the Dirac field is. So Dirac field has, as I said, it has, you know, 32 by 32 matrix, but, you know, I can simply concentrate on 16 by 16 of them. How does it look like this 16 by 16? 16 by 16, and then you start actually enumerating, you have an element of the first left and the first right and you discover actually that this d is essentially like gamma mu d mu and then you are going to get b mu in it, where b mu would be the u1 charge and the u1 charge has to do with this c and then you are going to get, I don't know, so usually you are at i g and then i g prime w alpha sigma alpha. So then this really would come into matrices which has to do with h left. And again actually, when I add this j A star j inverse, it really brings things from down to up and it really gives me if for quarks and at least for the quarks, it gives me also the g3 v mu i, I don't know what I'm doing. What I would like, what I'm saying is that you really get exactly the correct, when you write psi d psi, you get all the correct vector interactions in addition because because of the tensoring here gamma cross df, you also get hexfeeds. What are the hexfeeds? The hexfeeds are nothing but the connections along discrete directions. Because it's really if you do it easily, I give you an example maybe with 2x2 matrices. If I have just like A, B, C, D, this is an algebra of this form and actually here actually it's and if I do the calculation with gamma 5 so essentially I can take d to be gamma mu d mu and here I can take gamma 5 some scale gamma 5 comes scale and then I have gamma mu d mu. You'll discover actually that when you compute d I think here I'm going to have to take diagonal or something dA you'll discover actually that there are cross diagonal terms and you are really going to get a gay shield A mu and here some gay shield B mu and here you're going to get h and h star. In other words because this is with a gamma mu and this is with a gamma 5 the off diagonal elements in the operator d will be nothing but scalar fields and depending on the representation for example if you take sq2 cross u1 here in this simplexic picture so this would be 2 by 2 and this would be 1 by 1 you'll discover this h will become a doublet it belongs to the 2-1 representation so you really get a doublet. Now the nice thing about this is that you only get 1x doublet you don't get 2x doublets one a priori one expected that for the right handed electron the neutrino you should really get independent but they are really related to each other because the algebra is fixed. So the upshot of all of this is that we really get only 1x doublet and we get all the correct representations all the right couplings of both the vectors and the hicks. So in this respect the hicks field is nothing but the gauge connections along discrete directions or along finite directions. All right, so what do I do with this? Now what I need actually is really to get dynamics. Which by the way is the limit of the Kaluza client case where yeah but not completely because actually if you do Kaluza client yeah but there is difference if you do Kaluza client you know that you always get only adjointer presentations. Why? Because after all the gauge feeds suppose I work in 5 dimensions ok? In 5 dimensions then I take, you know, suppose I take a Yang mystery in 5 dimensions so I write it f mu nu. Let me write like fmn squared, yeah where m is like mu and then the 5. So obviously this becomes f mu nu squared and then you are going to get d mu I don't know h squared, yeah actually in this case ok, you are really going to get a scalar field but it's really a commutator if it's done abelian if it's done abelian then the phi and the a mu are both in the adjointer presentation and I remind you actually that you know a because am is am i ti where ti usually is in the adjointer presentation and you are really forced to take all scalar fields and not be able to get the doublet actually so here you really get the doublet and in addition you really have no dependence on the extra coordinates because here actually the problem is that in this picture a priori phi is a function of x mu and x5 and then what you do is that you take a Fourier series of x mu I don't know e to the i whatever x5 with it to pi or something and you really get all this oscillators and you say ok I only keep the lightest of them which is massless all the other states will have mass m squared and m squared will have to do with that is related to the radius of your circle so you get an infinite tower of states which is ok in string theory is not ok in field theory but in string theory is ok ok because then you really can control the infinite tower of states but in field theory it's a 5 dimensional theory it's not that complicated no, no, but I mean for quantum the question that are you going to throw all these or not going to throw 5 dimensional quantum theory so people actually use Kaluza Klein they throw all this infinite tower of states in string theory don't throw anyway Kaluza Klein would not really give you this picture it doesn't give you that the fermions all will live in one representation things like that the nearest thing to this of course is the azotan alright, so let me see continue and see actually what we can do with that so the next thing is that we have to describe the dynamics now in this picture actually the fermionic action is given by have gpsi and one really can show that all fermi fermi which are kinetic plus fermi fermi vector plus fermi fermi Higgs are all included you generate them from simply writing psi dpsi ok so the fermionic action is ok the question of course we need this da or the gauge fields to be dynamical if I have to really find in order to get a good basis at least at the classical level for my action I have to find a good action that represents the dynamics of the vector and the scalar fields so what is that so this actually so we hypothesized at one point the bosonic sector so how to represent the bosonic sector one thing actually has to do with the observation that if you look at the specter of the eigen values of the Dirac operator then the eigen values are known to be geometrical entities they really depend on the geometry and depend on the curvature of the of the space and the curvature of the connections so so what we said in that the spectral action which can be found uniquely which satisfies the property that if you really have two disjoint sections like two disjoint operators like d1 and d2 then you need the action to be additive so that you describe these two disjoint sets then one can show that this property and this action is invariant under automorphisms and would satisfy that if you have d equal d1 plus d2 then trace if they are disjoint then the action becomes the sum of the two pieces and we take actually f to be positive function one can argue whether it should be positive or not but at least in the Euclidean version we take it to be positive it would really make good what we call passive integral in that but this actually looks very arbitrary so I say trace of fd2 you tell me what is f and the answer is that for our purposes it doesn't really matter what choice of f we make we simply do a asymptotic expansion of this function and we keep so here actually you would immediately object because you tell me look this f of d2 because d has a scale remember d is gamma mu d over dx mu and whatever so it already has dimension 1 so in order to you have to make it dimensionless and you have to divide by a scale later I will make actually this scale to become a scalar field so that we can control the scale dynamically also because you can write this it is dilaton so you can put the dilaton and then you can see how far you can violate the scale and it has interesting consequences alright so the question how can we handle that and the way we handle it is that simply you write first f of d2 you know I would insert lambda squared it doesn't matter you first write it as fsd squared minus s and then use the so you put it on some kind of of Laurent series and then use the property that d2 let me see it's what 1 of gamma of s 0 to infinity e minus d let me see I forgot exactly that last time here it is ok so what we have suppose that I can use this question trace of p minus s p would be d2 here is 1 over gamma s t s minus 1 so I have trace of p minus s trace of e minus dp and for this I use I seem to take expansion which is the cd expresses this trace in terms of cd do it coefficients so we write that summation and larger equal to 0 n minus 4 4 actually in this case the dimension of the manifold over 2 so it's n over 2 and it's 2 so it's like this in this case and a n of x p p view of x and the nice thing about it that actually these objects only depend on the operator d squared in this case now for many force without boundary we know that a even sorry a odd is equal to 0 for many force without boundary we have a very nice situation because we can predict something that occurred in quantum gravity and we can predict something called the Gibbons Hawk in time that's needed for the consistent quantization for the consistent for Hamiltonian formulation of general relativity and x p is the operator p is the operator p is d squared in this case in reality one really can write d, p is anyway so essentially the nice thing about this actually t expansion because you did not explain why we are allowed to do a t expansion for the moment there is a function f as you see however because of the presence of this lambda because of the presence of this lambda it would look like as follows you are really going to get lambda 4 a0 ok and then let me say it is f4 usually actually we keep changing the rotation so f2 lambda squared a2 plus f0 a4 plus f2 over lambda squared a6 ok, it looks like that now if lambda is the unification scale and we are really going to show that lambda must be of the unification of the order of the unification scale we are really talking about this is suppressed actually see as long as you are really below say plank all this would be suppressed because you are dividing by the lambda squared these terms would be suppressed and the question is that I think this I have to call f4 anyway see I think what is f0 it is uf of u du between 0 and infinity so if you know the function it is a number and similarly f2 is the integral of f u du between 0 and infinity and f4 is equal to f of 0 alright now I think even if you use this function which is simply a cutoff function it means that the function if u is less than 1 remember u is d squared over lambda in this picture lambda squared so if u is less than 1 you take it to be 1 and otherwise you take it to be 0 you really get you know we are going to show that you really get an excellent answer with this cutoff function of course one can improve you know here you have somehow arbitrary coefficients with this so a0 what's a0? the cosmological constant is simply there is a number so all these have been at least the general formula of how to extract the a's out of the form of the d squared have been given by gilki so all what we have to do is to use his prescription and extract as I said we have 384 by 384 but this is a mere technicality because after all there is simply 10 sub product of matrices and you can simply trace and take products so this is really mechanical in that so we get that a0 is essentially is a number times root g to the 4x and what's a2? a2 is another number into with a negative sign into r plus h bar h you know some coefficient here and so obviously we start to get the Einstein term and we really get the mass term for the Higgs field with a minus sign ok that's the level of lambda squared and the level here as you see this is lambda squared it tells you know how can you get the mass of the Higgs which is of the order of the of the problem this problem can be solved by putting the dilaton as I show you in principle one has to take the dilaton into account and but I will talk about the stator the problem of fine tuning could be reduced because you get something which is known as the Randall syndrome model actually you know so you get it but this was done before Randall syndrome so we won't use the name but it's the effect so yeah so you are saying the first prediction is that the mass of the Higgs is equal to the plant mass remember it's a mass term it's not the mass of the Higgs the mass of the Higgs of course as you know even in the standard model is plagued by quadratic divergences so even if you set it to be you know 100 or whatever becomes infinity anyway if you don't keep tuning it at every order so usually the mass of the Higgs is really controlled by the lambda coupling and more than the mu squared mu squared this is a fine tuning problem of the Higgs field which is a different story but again actually I would say that the physical Higgs would have a mass if you put the dilaton into account would really have the physical field it's not the h but it will be e minus phi h so the dilaton will come in and if the dilaton acquires a potential and acquires an expectation value then it's enough that this phi to be of order 40 or something in order to really have a tiny mass so this is possible actually but still you are talking about geometric unification we have Einstein we are very happy to have Einstein with the plant mass we are saying H is like a geometric partner the Higgs is the beauty and at bare value the Higgs has a master equal to the plant mass bare master for the gravity it does not it does not change the plant mass remains in the low energy physics of the gravity because this has to do with the fact remember everything is bare and then you are going to get the physical action and you really have to run all the coupling constants and it's known actually that the Newton doesn't run very much this is what people know Newton doesn't run very much while the other does run a lot actually so here they are geometrically unified so you would say well actually if we look at a couple everybody runs differently we will discover that some couplings run a lot much actually and it's not the only one because if I now go to the A4 and then what do I have A4 A4 you have some constant and you are really going to get actually now 5 over 3 g1 squared b mu nu squared plus g2 squared w nu squared plus g3 squared b mu nu squared so this is v1 hyper charge this is su2 left is the weak Yangmills and this is the su3 angles the kinetic terms so you get actually and they have they come exactly in this form ok plus then you are going to get d mu h squared you get d mu sigma squared squared and then you get lambda h bar h squared actually this lambda really comes as a fixed value when I say lambda it means I'm going to get a number and this number is determined by the Yukawa coupling of the particles so here everything really is well defined it's not that you get something arbitrary and the ratios for example g3 squared is this ratio you get 5 over 3 1 1 which is known to be exactly what you really get if you have su5 unification model you mean really g1 squared or 1 over g1 squared how do you normalize the thing because usually it's 1 over g squared with a geometric normalization yes you know we have so essentially what we do your field they include the g or they don't include the g so here as you see so I get this term and then what do I require I have here some f0 so I require actually this number into f0 to be 1 over 4g squared and obviously if I would like to normalize thing simultaneously this really requires for me that g3 squared equal g2 squared is 5 over 3g1 squared and this is my g squared and then I can factor 1 over 4g squared and then you get actually without any g's in your expression why are you imposing this unification I have to normalize I have to normalize this comes on equal footing at this I have to normalize my kinetic in simultaneously because I can also redefine my fields so in the end of course in the end actually what do I really get I would get you know minus quarter b minu squared well this is 5 over 3 I think you have to plus w minu squared plus b minu squared because this is the normalized kinetic energy it should come out with minus quarter this is well defined the kinetic energy of vectors is always minus quarter so I have to normalize in order to normalize this I have to define this to be like 1 over 4g squared I have the right to define it to be given so I take it to be over 4g squared or the g squared cancel and in this respect this is my unification you are forced to have this unification because you can always redefine your fields right but you have to redefine them in such a way you get the canonical kinetic terms in order to get the canonical kinetic terms it forces this unification you know similarly here actually with all this for example the lambda coupling which is the quadratic coupling of the Higgs field is proportional to g squared which is a situation which is familiar in supersymmetry in supersymmetry the lambda and the g squared are proportional to each other so here you have also the same story that the lambda coupling of the Higgs is proportional to the gauge coupling and it means you know it would have similar argue equations anyway, what else do we have we really get actually plus here we get C minu rho sigma squared this is the conformal tensor now we don't really and then we get Gauss Bonet of course when I say RR star it means you know epsilon nu rho sigma epsilon A B C D R mu nu A B R rho sigma C D this is retropological term so you get this topological term and you know you have surface terms and things like that but the important thing is that you read this conformal squared and of course you know here you stop now here one may argue that if you really have the conformal tensor squared then your graviton propagator if you want to quantize the theory would really have a Gauss mode it's known that in graviton propagator you get something like K squared plus K to the 4 if you go to the and it's like this it's 1 over K squared plus K to the 4 with coefficients which you can write as 1 over K squared plus here maybe you have lambda squared or something you know and the other K squared and this you can write as the difference of two terms which really shows you that there is a Gauss term however actually this it makes it well behaved at infinity but then you have to say it's non-unitary if I go all the way to the Planck scale so you have actually problem with the uniterity at the Planck scale but anyway since we already declared that the Planck scale something bad happens this is not a serious problem so the theory actually you really can work with it and you can do normalization and you can do everything and you really can work with an effective theory the claim that this is a good effective theory so where do I go from here anyway so the essential point F0 contains the cosmological cosmological constant F2 contains the cavitation master and 4 contains kinetic term for all Gauss fees plus conformal squared plus kinetic term for the H kinetic term for the sigma so you really get everybody with fixed coefficients there is no potential for sigma yeah there is sigma I didn't try it but you know it comes with another with sigma to the 4 there is he has the same expression actually you get the factor of half with different the difference is that how they couple these coefficients really depend on for example for this it depends on the square of the Yukawa coupling while this would come on the quartic power of Yukawa coupling so everything is well defined the important thing is that everything is well defined third level contains all these kinetic term for the gauge fees exactly ok where do I go from here here at no mass term for the gauge fees no mass term for the gauge fees of course here if you break symmetry they generate mass term if you break symmetry they generate mass term so what happens is the following now we really can start talking about whether we can take this action seriously as you know the dynamical action of all the bosons that we know there are really two things that we can do first of all we can immediately give a prediction actually to the top quark mass this we can predict and the reason of the following remember I told you that all terms all masses actually are generated by coupling to the Higgs that's why in this popular thing they call it got particle because it's one responsible for masses for all and here actually you really have some k where k are you call a coupling it's all scale psi bar psi h but this is direct mass because it's psi bar and psi no where you are now the Majorana of course we can have k in u r whatever but you are right the Majorana is a different story so it doesn't really affect this one so here what do we do actually we notice that this term here there is some coefficient which depends on the yukawa coupling and the kinetic term of the Higgs field comes with this a and a depends on the square of the yukawa couplings then you say ok I really have to normalize to make a proper kinetic term I have to normalize the kinetic term so that in the end it should be only half d mu h bar square it should be just half without anything so in order to that we really have to normalize h into h over square root of a f0 where f0 is not every term now if we do that this term becomes k over square root of a psi bar psi h but this a then actually I discover that if I square this term k squared over a if I square this term I really get one with a summation because a is defined to be the sum of k squared so from here we deduce that at the three level we have a formula which tells me essentially mass of the electron of course I am talking about now three families mass of the neutrinos plus three masses of the d plus three masses of the op is 8mw squared because it has to do with the wave of the h and the wave of the h has to do the w mass we really get an identity and this identity in this one we really get it yeah we know you can write at the term of ukawa coupling it is the same because in the end everything is multiplied with v squared and plus 3m squared op would be m I think g squared or something you know maybe it is 4g squared or something I think something like this this really already tells me we know actually the mass of the top physically the mass of the top is dominant if you square it it is much larger than anything else remember it is of the 173gv where the rest is like the largest is like 5gv so if you square it 3 dominates everybody and it means what is the mw 890 something actually I can write it as 4g squared you are right it is 80 it is like 80gv anyway it tells you that the m squared of the top is like 8 over root 3mw squared I can take the square root so it is 2 root 2 over 3mw and however this is a 3 level and then of course you really have to run it down which I am going to write actually the organization group equations how they run down if you run it down actually you cannot really ignore this term because you would expect that the neutrino mass is small but it is not really small what we know is that it is only the siso mechanism that makes it small so there is no reason actually for k and u to be small however if we ignore it in the 0th approximation you really get something like 175gv immediately for the top cork mass so it is really one of the predictions then we discover actually later that you cannot ignore this and then you can predict it but not to decimal points you have some room you have actually one variable in the story so let me then continue around it is very close to the experiment value it is very close to the experiment value however now let's look at the hex sector because in the end there is a prediction of the hex sector and I thought last time we did make a fatal mistake and we predicted the wrong hex sector and the culprit in all of this is that in our analysis we dropped we assumed that the sigma is irrelevant we said look it is 10 to 12 or who knows it is very high it will not affect anything it does I am going to show it does tremendously actually it gives you a factor of I don't know 0.8 whatever the hex mass is because of this neutrino contribution so that actually we ignored and in a way we really paid a heavy price for it but why do you put 10 to sigma as a potential also because what is nice in unification is indeed as you said you make predictions so at some scales everything is equal to everything else many things there are other conclusions for the mass is the mass of the sigma equal to the plant mass equal to the mass of the hex at lambda for instance as you know the problem with this we cannot say we cannot say what is exactly the mass of the we only can say that in order to get good properties that we need to get neutrino masses of a certain order for which now this is another fine tune remember actually there is fine tuning you should say there exist a lambda where the mass of the sigma is equal to the mass of the hex is equal to the plant mass at this lambda and then you can as you know the problem in the following we really have quadratic divergences in the model in any and all these things if you don't have supersymmetry you will face immediately the problem of quadratic divergences for the scalers and this is really the main so essentially the lambda the quadratic couplings we can run and we can determine completely but what we really cannot control is the quadratic divergences and then of course these would affect actually exactly the mass of the hex the mass of the sigma so we can only say that look the vacuum expectation value that is something like 10 to 12 gv and then everything we can make predictions that comes also from another for I would say actually where does it come why should be 10 to 11 and not something else but still I mean the bare value of all the masses of sigma is equal to 10 to the 19 gv square that was a prediction now so you should not be afraid of quadratic divergences because you start from then they can be diminished if the sign is okay by quantum fluctuation when you go to lower but they are uncontrollable the problem is that the question because of quadratic divergences it means that if you do one loop you get infinity, you do two loops you get another infinity you have to do it at every order so this is the hierarchy problem you really cannot rely once you have quadratic resistance you say okay I cannot control you have a cutoff so you should take the logic over the values of the running at all because of quadratic divergences we don't know, nobody knows why not, I don't understand you have a differential equation dm squared over d lambda squared equal something, a number and you have a boundary value which is mplank squared so you can see if it goes up or down no but the problem is it's infinity it's not that you know it's not infinity you have a cutoff and what is happening to us I have an f of d over lambda so you must put a cutoff otherwise there is no meaning in this therefore you have a cutoff therefore you can start the ODE the renormalization group at this cutoff and see where they go why do you abandon if you have predictions here, use them I think the problem, I don't know the problem is that people when they treat because they are I think nobody deals with them I can tell you that nobody deals with the quadratic divergences once people have them but because they do in reverse they start from low energy and then they say ok I go up to infinity you are doing the reverse you start from mplank which is for u finite and you go down so this is like a well defined scheme ok, but continue anyway, the way I know that it's not even there is not defined so this actually this is a fine tuning that we do why it's just we don't know fine tuning and this fine tuning is also needed for as I said, the CISO mechanism because otherwise you don't really get the right order of masses for the neutrinos in addition the standard model will break down anyway as I am going to show you in a minute standard model breaks down because if you would like to get 125 GeV Higgs mass then you really cannot you cannot ok, if you run up it will become negative at 10 to 11 or 10 to 12 and it becomes unstable alright, so let me see actually what happened let me write for you actually what's the Higgs potential in this respect after I rescale because as I told you, we really have all these fees they come not scaled and then we have to rescale them to make them in the physical we have to scale the kinetic times and we really make we make an approximation that the top quark mass mtop is much larger than m of all fermings you know which is an assumption you make it simplifies the analysis you don't have to worry about many fermings you worry about one so we can write actually ok, so now actually h bar h essentially is set to be 0h after you expand around the vacuum expectation value we know the other particles will be absorbed so I am only going to keep the Higgs which should be observed in the laboratory so it is of this form and then you have lambda h sigma into h squared sigma squared plus lambda sigma sigma top form which is the usual stuff and then actually we have something like I don't know, minus 2g squared over pi squared f2 lambda squared and what's lambda h will be something like n squared plus 3 over m plus e squared 4g squared lambda h sigma and we have assumed that k nu is root nk up in other words we assume that the neutrino mass is not small the direct neutrino mass or the Dirac type in neutrino mass is not small is some number n actually this number n you know to fit the experimental values somewhere is between 2 and 3 actually or 1.6 and 2.4 or something so we really have this now what are these, these actually lambdas are governed by renozition group equations so we really can write the argument the Yukawa case are governed by renozition groups and they are logarithmically dependent so you have a prediction I mean you cannot impose like all the k's are equal no, no, but I this is my 3 level no, but remember actually all these conditions that you run but you have, we have actually a relation which holds at unification at this scale so this relation I can start from it and go down that's what we do but is it not saying that all the Yukawa's are equal? no, they are because the Yukawa's as you see now we ignored all the Yukawa's except for the top quark mass because the rest the rest are negligible compared to the top quark mass this actually is the dominant factor you can get, by including others you can get the correction but it's known to be less than 1% actually yeah, if one of them starts going up but here you want that two Yukawa's are comparable which one? ah, this one, for this actually and for this we have to run both actually yes, so I mean this is an ODE so I mean two directions have to go up and others go down but that you can check you should not make an assumption you have to show that there are directions where two Yukawa's get stay large or not in principle these are logically divergent they are not quadratically divergent so you cannot argue by hand waving you have to prove something in principle ok, what you are saying that the question whether this is actually we only do this at one point ok, we didn't do you have a point where everything is normalized yeah, yeah, no, this actually is ok, in order to simplify our analysis we said this is related by some number we don't know what this number is we don't say in the end what this number we cannot determine, we are fitting it to experiment but you should, you have a lot of information the beauty of geometric engineering is to make predictions we are doing phenomenology saying let me forget completely about the predictions and tune it to the data that's what you are saying we are not tuning, we only said that no, you don't face what are the predictions this actually all these relations hold at unification and this actually we are going to write the RG equations not taking this and this actually is only a boundary condition on the differential equations all these are at unification exactly, this only boundary condition is at unification because after all, RG equations are differential, first order differential equation so you need some boundary condition so these are all boundary conditions fix by the geometric unification so you cannot impose create of n no, no, but n actually is just a parametralizing number to get an idea what the n should be, without that we know no, but you know whether it's one or two-third that's what I'm saying if you just numberize thing at lambda everything is determined from the geometry yeah, but the problem that I agree with you in principle since we don't know what k nu is now because nobody has measured the direct Dirac in utrinuma so this actually is something that we cannot compare with the data so we have to assume there is some relation if we knew the value of k nu now then we would have actually looked at the equation but since we don't know the initial value or the present value sorry we don't know because nobody has measured what is the present neutrino mass because that's also related to this Majorana mass the CISO mechanism there is an unknown there if this unknown is one day measured of course we'll be able to make a prediction for that, okay but for a time being we don't know that so we assumed that it's this actually is parametralizing our ignorance probably you have less ignorance yeah, I would agree with you if we knew the value now because then we can run it up, it's true but since we don't know the value now I don't know, there is no point in running something you don't know up, this is what I'm saying anyway, look what happens so what happens is that I can write all this RG equations as Thibou said for example the beta of the GI is very simple is given by 1 over 16 pi squared GI cube and this actually are three numbers which are 41 over 6 minus 19 over 6 minus 7 and this simply tells me that if I define alpha i to be GI square over 4 pi and I plot alpha 1 inverse and alpha 2 inverse and alpha 3 inverse then I get something like this this is alpha 1 inverse alpha 3 inverse we plot it actually and one really can plot it from the present experimental values and then you can check actually they don't exactly meet there is some like 4% discrepancy between the meeting points however actually these equations could be improved because this is you know the first order the second order loop corrections unfortunately the second order loop corrections has not been worked out taking the they have been worked out to three loops but not taking the singlet sigma into account because we discovered that sigma makes a big difference so I'm asking people who do these calculations and seems that nobody has already started doing this anyway we know actually how the top quark mass so we get some relations minus 17 over 6 so people have really worked out these things so we are really going to get a system of coupled differential equation and the important thing of the following delta d lambda of h over dt will have terms which are proportional to the h and the h sigma coupling plus the usual stuff and simply the lambda h should move so they all actually really get coupled system of differential equations for all the h and the sigma coupling and this is the thing that we missed we said that they are really independent and we didn't put however if you do that what happened is the following because you forgot about the sigma we forgot why actually we have it in that we wrote the Lagrangian but at one point we said ok in our RG analysis we ignore the sigma because we assume they are the coupled because we have not looked at people have already done the calculation but we are not aware that people have done the calculation they did it like 2 years earlier but we didn't know we are not phenologists after all but of course it's not to be aware of something anyway so what happens is that they really will give you a plot of so the nice thing actually all this could be worked out using mathematics actually you simply give the equation and the mathematics gives you the solutions to the plot and everything so this is I think this is lambda sigma and this is H sigma and this is H and zero point I don't know 2 or something you know this starts with 2 which really becomes weak at low energies now what's the interesting thing is the following is that m squared obviously the H and sigma states are mixed so you really have to diagonalize your system and you really get actually 2 eigenvalues for the m plus squared a minus squared and this actually would be lambda H v squared into 1 minus lambda squared H sigma over lambda H lambda sigma and the important thing is the following this is the thing we predicted below the sigma and you can come something like 170 gv it's 3 between 160 and 175 or something this is the actual this is what we predicted some time ago ok now sigma as you see because if we assume it's 10 to 12 of course it will be 10 to 12 depending on the expectation value that you are going to assume for the expectation value of the sigma once that is done then of course the mass of the H is really given by this this is the usual as I said this is the 170 squared or something but then it is really reduced by this factor and this factor you have to measure at this point if you measure at this point you get that this factor is something like 0.8 so it reduces to 170 ok well of course here everything is squared by a factor of 0.8 so the essential point is that if you really work it out and you get a number which without doing anything is somewhere between 120 and 130 gv ok now this is really important because suppose that I didn't have the sigma at my disposal and then I worked out simply the lambda H coupling so the h coupling starts from this very small number here and it really goes and if you don't have the sigma it really goes like this actually and it becomes 0 at 10 to 11 gv and it becomes negative afterwards this is actually why people were not very happy when 125 or 106 gv when the Higgs was discovered to be 126 because it really make the standard model if you would like to note it to higher energies higher than to higher energies assuming that nothing else is found it will break down it means it becomes unstable at 10 to 11 gv and lambda becomes negative not so if you put at 10 to the 2 10 to the 3 probably so anyway but this actually people assume that something would be found you don't worry about this problem because you say something else will happen but if you believe that the standard model is a good model up to higher energies then you are immediately in trouble with this at 10 to 11 and the only way out is that you would assume that there is some scalar that enters in this which would have similar coupling so people assumed actually all these properties and then you said you can save the model if you introduce a single now the advantage that we have is that we didn't have to assume anything the sigma was there and we have the other advantage that the initial values of these is fixed for us more or less fixed because we have this N which we still don't know because of our ignorance about the Dirac mass of the neutrino but more or less for a wide range of the N you always get the same pattern so essentially we can give a good value for the Higgs mass which now one thing I can mention what about the sign of mu square of the Higgs minus it comes with a minus sign it's easy to understand why it comes with a minus sign because after all remember we are really expanding this guy so you get minus Td squared of course you have to do integration but it tells you that you are going to T squared and this would have opposite signs this is the kinetic term and this is the mass step heuristic argument of course in the beginning I was completely surprised but it's nice actually that you really get the minus sign that you are really going to get lambda H to the 4 and this relative sign is really extremely important that otherwise you would not have any symmetry breaking so this is quite nice since my time is up actually I would like to mention one final thing about maybe one would doubt actually why the spectral action should be taken seriously I will give you I will give what we call supporting evidence that the spectral action is a good action even for gravity support that I was only doing gravity and I want simply I would ignore gauge fees I want to do generalitivity then I take this trace of Fd squared and as I told you get a cosmological constant and you get root gr now and you know I am really getting say a minus sign so I can you know tune my first coefficient to be minus 1,16 pi g but assume that my manifold has a boundary yeah then is known that you have to add this term to the Einstein action what do you have to add this term the reason you have to add this term in the form it depends on the geometry yeah this actually is called the second fundamental form and this is the curvature on the boundary and this is the metric on the boundary so and it is known actually that you have to give exactly the sign and the coefficient to be in certain conventions once you fix the conventions there is a factor of 2 what does this come from now it is known actually that R as you know now I am going to write it cryptically is d gamma plus gamma gamma where gamma is Christopher connection what is gamma again cryptically is g inverse dg you agree therefore the first term will give me dg inverse dg plus g inverse d squared g and the second term will give me g inverse dg squared ok so we see actually that the gamma gamma and the first derivative are of the same form g inverse dg dg inverse dg inverse this is the same guy however I have a term with g inverse d squared g now if you want to do Hamiltonian formulation of general relativity then you have to find the momentum which is dL over delta dL over delta so you have to take actually with respect to time derivative you have to find the momentum now this term d squared g would prevent you from doing that you have d squared g so what do you do you integrate by parts you integrate by parts but if you integrate by parts what do you really get you really cannot get rid of this term you really have to get rid of this term you have to introduce a boundary term ok to integrate by parts but you have to cancel the boundary term the manifold as a boundary and the thing that cancels the boundary term exactly is root of this hk that's why actually Hawkin and Gibbons they proposed that the Einstein theory in the Hamiltonian formalism would only make sense even actually in the Euclidean path integral would only make sense if you introduce this term there are many arguments and you have to have exactly time and coefficient this is a factor of 2 between here and here ok because of the integration by parts in order to kill this quadratic term in the derivatives so now let's see suppose you see this I don't know, I can't get root gr but I have no control on the dm so on the other hand actually this hk expansion has also been worked out for manifold with boundary and there is a formula for it so you go and you do actually some calculation not easy actually but well defined and what you really get you get exactly this combination of terms to assign without any ambiguity you get that so in other words one of course can explain it why it really gives you the right time because essentially you are really solving an eigenvalue problem and you have to tell me what's the problem with psi, deep psi must be hermitian this hermiticity would force you actually take a boundary condition of the psi on the boundary that it should be zero and if you do that once you take this condition you apply the prescription of computing the heat, the series with coefficients and I told you actually that a odd for example a1 equal to zero for manifold but for manifold with boundary it's not equal to zero you compute you get exactly this term with the factor of 2 so this shows actually that this spectral reaction knows there ah there is so the spectral reaction knows about you know quantization quantization of just saying it is a better functional because the simple argument is this is a functional of the metric differentiable only if you include the boundary terms so this is normal but the spectral action which has indeed assumed to be this contents it you know I would go along with you assume that I don't take d as my fundamental object I take it to be that a plastic and then I take f of delta you discover that you don't get the wrong answer ok you don't get the 2 you get a different number so you really have to take the Dirac in order for it to work and this is not a coincidence because it's really very precise I can take any other operator I always get the wrong answer except to the Dirac operator and you really get exactly the right answer and we did it exactly for the whole standard model and we really get exactly the same coefficient not for this simple art theory but for the whole theory action and you know and that was a rough calculation but you really get exactly the right answer so there is something something deep goes there which we don't fully understand I think you know so I think I will stop here however maybe I have to ask you because Tibo told me that next week there is some many events we know with Maxim so what about postponing for a week the last lecture would you? I am in favor I am not here next week ok so maybe I think I can announce what about you Tibo would it be ok that we don't do it next week we do it the week after so I think you know we will announce that it will be postponed one week so that people would like the Maxim celebration then they can do that alright? ok so there is also a strict meeting next week some people will be I don't know