 So last week I introduced the idea that non-committal geometry can be taken as a basis to unify all fundamental interactions and so what I did is that I defined what we mean by non-committal space. I gave the definitions and so with certain properties and just as I remind you, just that there's a spectral triple and I defined these five quantities and the relation among them. And then the question of course is if a non-committal space is the basis to unify the interactions then the question is what kind of non-committal space we should be taking. And as a starter I made an assumption and the assumption I made is that spacetime or is a non-committal space which is formed as a product of continuous four-dimensional space times some finite space. Now the four dimensional spaces are well understood and then I went to see essentially what one should take for the finite space, you know what kind of space should one take. And so for that started by classifying spaces which fit certain physical criteria and the physical criteria that we need and the following that you know the Hilbert space which would be space of you know fermions essentially for us would include all the fermions and in this case actually the fermions would satisfy certain properties. So essentially we said we have fermions but because actually of the presence of what we call the reality operator one really requires essentially one would like to require essentially that the fermions are real which in physical language means Majorana and in this respect of course actually the presence also of the dam operator which is a current operator would allow us always to impose this condition. So what we really have we know we say we have current spinners. Now then of course actually if you act on J on psi you are really getting to get something like the conjugate spinner and the question is that is this an independent set of fields. If it's an independent set of fields then you really immediately face the problem that all your particles will be doubled. You are really going to get for every particle we know a mirror particle and you know something sometimes it's called the mirror fermions problem or the mirror doubling problem and usually it's not easy to solve this problem. Now you know we are really working in Euclidean space and the issue of course is that if you really go to Euclidean space and you said I would like sorry then but the physical space of course is Minkowski and if you would like to impose both condition that J psi is psi then this actually would impose certain condition on the properties of the space and you know as I have shown last time that would impose a condition what we know as the k-o dimension. So in Minkowski space you really need to impose both conditions that J psi is psi and gamma psi is psi and this would imply actually that the k-o dimension of this space is equal to zero is equal to zero which implies actually the J squared which is epsilon should be one J d is I think epsilon prime d J and I need one more sign J gamma is epsilon double prime gamma J and d gamma is minus gamma J. So these are the properties that we need and essentially it really forces now we know actually that the in Minkowski you have you know four dimensional space with signatures say minus plus plus plus or the opposite and you know this would somehow cancel each other and then you are left with the k-o dimension two which means that the internal space or the finite dimensional space should be dimension six k-o dimension six. So that's actually forced the finite space to have you know certain properties and k-o dimension six then you need certain sets of epsilon for example epsilon is minus one I forgot actually I think epsilon prime is equal plus one and epsilon prime is minus one something like that. So then actually one goes ahead and try to understand these finite spaces and what one discovers and I would go through not through the proof but you know through the realization of the proof is that one discovers that the algebras of the finite space are essentially what one finds and that the the complexity complexification of the algebra. So essentially here one assumes that the finite space is characterized by af hf df gamma f and gf everything is for finite and one discover that the complex extension of this algebra if you take the center of it there are only two possibilities which are either c or c plus c and this actually is inconsistent with k-o dimension six which leaves us only one possibility that ac is c plus c and really it corresponds to an algebra which is a variant of you know let me call it m n of c plus m n of c. In addition when we imposed certain isometry on the first algebra and this is a point actually we don't really you know understand you know we cannot give it a mathematical characterization why it should be like that but it is like that and the other possibilities are not relevant and this really reduces this instead of matrices on the complex numbers you really take matrices over the chtorions for the first and in which actually n is 2a in this case. So n is even and also because one would like to impose the chirality operator on this algebra and elements of the algebra for example gamma a equal a gamma so in this case what happened is that the this m 2a of h would split into m a of h plus m a of h for example plus m n of c or 2a of c. So the upshot of this is that really the first non-trivial example you can do is the following that a should be one in this case sorry a is yeah this is a matrix m a of h this is should be 4a or something n is let me see m is n is 2a yeah so essentially yeah so essentially this yeah so essentially really the first non-trivial example is m1 of h or h plus h okay let me write it like this and this actually is like 4a and in this case the first example would be h plus h for the first algebra plus m4 of c for the second of course other algebras are possible what is important to say here is that you have a gamma which is a diagonal matrix on the first right for m2 of h right you take the even part it is an even part of m2 of h okay so I have to say okay this I would say m 2a h even and then it really at least for the first non-trivial case it belongs to h plus h now it's up to this point you know I went last time and the only thing where you know the assumptions I made are there so at one point one did impose on isometry if one didn't impose on isometry then you are you going to get instead of you know h plus h you're going to get m2c plus m2c and one really can show that these cases would not be physical but one cannot rule them immediately unfortunately yeah exactly there are three okay we choose you know the simple case so this is a choice this is something which is not imposed by the we could have chosen here we could have chosen this actually this is the really the other is actually an element because they would really give you real algebra for instead of h plus h you can take m2 of c plus m2 of c what would it really give essentially it gives you know it gives you an extra u1 actually you know which is anomalous so I really can can work it out it gives it's a little variation everything almost the same except this little variation it gives you an extra vector an extra vector which is anomalous you know who knows you know maybe there is an extra vector lurking in the spectrum but you know people are talking you know recently there was some papers in which they said they observed something which deviates from the standard model and you know some people said maybe it's an extra vector so who knows you know anyway but this is actually the most symmetric possibility uh the only the only assumption I made also that left action this is the left action and then this is say the right action and then we said this is j b star j inverse and we said and this actually is defined to be the opposite so this is the one of the axioms in that left action and right action commute with each other so this actually has been used this property really has been used up to this point now from here actually one needs further assumption and you know I will discuss both possibilities in that and this is called order one condition and now actually we understand this condition it's really a condition on the linearity of the connection if one drops this condition then what happened is that the connection would really get quadratic terms and what really can work out the possibility when this is not so when this is not satisfied but this I would leave to the last lecture to allow because actually in non-competitive geometry there are examples where this condition is not satisfied and you know this is what the the quantum spheres okay uh anyway so here it will have really big advantage that connections are linear everything is very nice everything you know falls into place okay what happens if it is not satisfied you know this would lead us to grand unified theories and pati-selam models and things like that but this would come much later okay so what are the consequences of this condition not satisfied okay so I will assume actually that this is satisfied and on top of it actually now there is one more condition is the question that if we take the commutator of the Dirac operator with center of the algebra is it zero or not zero actually this is a question well what we have shown is that there are two possibilities if it is zero then this symmetry is really intact and if this symmetry is intact one can show actually that you are really going to get when one work how the model which I'm really going to do a little bit later that you really get SU4 color which really corresponds to M4C as a symmetry of the quarks so essentially in this case you really get a higher color symmetry which is not observed in nature and in top actually you'll find out that the neutrinos are all massless exactly massless which now we know that it's not true so from the physics point of view we know actually that this really must be different than zero because if it is zero then you are already in contradiction with experiment that neutrinos will become all massless okay which would have been okay you know until 1995 or something like that because people didn't know we're not sure actually whether neutrinos are massive or not okay because every this tiny mass that only limits on the masses of the neutrino now we know actually that they are really massive but the mass is really extremely small but that they are massive is well established okay now what happened actually if this difference zero now the beauty of it is that if this is different than zero and for this condition to be satisfied the dirac operator must have the following form and so in this case actually I instead of telling you you know in description what really happens I would start you know being more concrete I will tell you exactly what are the representations what goes on so that you have an idea that this actually as I will say field is not simply talk you know you really can go and compute everything unambiguously not much ambiguity and with all the details so all right so essentially let me examine this in a minute but I really can tell you that the fact that this is not zero really gives a unique possibility that you really can give the neutrinos a Majorana mass and it will explain actually why the masses of the neutrinos are very small it really it's an example of what was known as the CISO mechanism so here it really comes out naturally and it solves the problem in a really very nice way so let's let's see how things would look like if I wanted to go ahead and do some computations you know what should I do so now so this is really my starting point I'm really going to take h the algebra a to be h plus h plus m4 and this remember almost uniquely we were led to this is that didn't come out of the sky you know out of classification this was the first non-trivial result that we have okay so if essentially if I would like actually to represent an element of a how would I write that then I am going to write it in matrix form you know I'm going to show you that everything that we do is really matrices essentially we really can use tensorial notation if one would like to simplify his life and you know recently you know everything that we do can be put on the computer so in the end one doesn't do much calculation you just write it on Mathematica and press a button and it gives you the answer so you know it's all right so the I'm really going to use you know index notation so essentially here what I'm really going to say all right so in order to tell you what my notation I really have to tell you what are my spinners okay so the spinner actually as you see it belongs to this guy so what I'm really going to write I'm going to write as psi alpha i where this alpha is would see the h plus h and the i would see the force so obviously i is one to four and the alpha is also one to four but it's a different form it's the h plus h now and then of course actually I'm going to write this as you know a dot a where this is has two indices like one two and this is one dot two dot so this actually tells me that it's a doublet under the catorians you know doublet left doublet right and what we write actually we write as h right plus h left because we have already graded the algebra and in this respect my spinners would be psi a dot i and psi i okay so how many spinners I have I have 16 spinners and of course they we have also the conjugates but the conjugates we're going to see that they're not really independent spinners because the fermion doublet problem is solved through you know the Majorana condition we are not really going to say jeb psi equal to psi the problem in euclidean space is really solved in in a very clever way by you know in the path integral formalism that only chiral fermions get integrated and they give square root of determinant instead of determinant so it's really solved in really very subtle way you know this is as discovered you get the fulfillment you know anyway so you see actually now you know without redoing any work I have already I've already obtained the correct specific first actually we already predicted that you have 16 fermions which is exactly per family of course which is exactly what we have and we would also actually find let me you know let me write this what is this the i the i eventually you know when this condition the d with z of a different zero will also be broken into one and the i where i is 1 2 3 so that's actually be the color index and this is the lepton index so this is lepton and this is the quark and then you can see that the lepton is the fourth color it comes out of the fourth color and in this respect you're already going to get psi 1 dot i psi 2 dot sorry psi 1 dot 1 psi 2 dot 1 psi a sorry 1 dot 2 dot and then I have psi i which I'm going to leave it as psi a 1 or then I have a psi 1 dot i psi 2 dot i and then I have psi a dot 1 psi what's in this you know remember this is a dot means when you see dot dot means right so this would be what I'm going to call new neutrino right you know why I give it the name because it comes out like that the quantum numbers tell me that this will be a neutral guy but this comes out you know I name it according to the usual naming and it would agree okay and then of course this is a right and this of course is upright this is not right similarly psi a 1 is the doublet which is a new a a new e it's a doublet you know a is one two left and this is up down left okay so you see actually without doing much work we obtain the correct representations of the particle spectrum of all the fermions we know which representations they are so one really can write that the 16 in this case is a fork for four and this four is decomposed into two plus two and this four is decomposed into one plus three and then this is to right plus two left and this everything that we see belong to that representation everything falls into place okay so now so these are my 16 spinners but remember actually you know I have the j psi and so my spinners will be psi alpha i and then psi alpha i conjugate which I'm really going to go psi alpha prime by prime they're not independent but they are related through the j okay what does it mean actually it means if I would like now to write the any element of the algebra you know I can write it as follows I can write it as this one you know remember this is belongs to the second guy to m4 of c so it is something like this and this is delta alpha beta and this is like yh so you see actually how an element of the algebra is satisfied and then you can form the b opposite which is jb star or j inverse and you discover actually what happened in that here you are going to get the y times one so you get yij essentially transposed into one and here you are going to get the x alpha beta into delta ij and if you look at you know this element and this element obviously they commute with each other why because then this is don't talk to each other this is in the first algebra well this is now in the second and they don't talk to each other so essentially jb star absolutely right and this is actually I wrote it as transposed but you know a star is okay so essentially we can write it in this form so the first a b opposite is satisfied automatically now where do I go from here then I have to satisfy this condition so it is actually a condition it's a condition on on the d and the algebra both you know because what happened in the following you try to satisfy this condition you work out the matrix representation and you obtain actually I saw you are right here no this section you are right yeah it's exactly different zero this is different zero on okay so this actually we have to assume it's different than zero and then actually this really would require the form of the rack operator to be as following actually one can show that you are going to get something like this let me call it I don't know let me call it d a b or something and here we are going to get the bar here we are not going to get something and then one can show through a little algebra actually I landed through abstract proofs which you know beyond me but through little algebra one can show actually that the only operator that satisfy in the following that here you can have only one non-zero entry only one anything else would not satisfy this condition okay of course actually this together with that we have also to make sure the d a with b opposite is equal to zero putting the two first actually first we know that here we must have a non-zero element the question that the whole thing could be non-zero but because of this on the first condition you discover actually that this is possible if and only if you have only one non-zero entry what does this one non-zero entry let me write actually the expression for you and let me see how does it look like you know I have remember when I write the rack equation what do I write I write psi star deep side or I can write it this way you know psi deep side okay in a product so this is my say fermic action and this fermic action if I expand and then I just look at this term what does it mean this actually if I look at that classification it really tells me that you really have an element which is this one if I have more it would mean that for example the up the down everybody will start getting myurana mass but you know we're getting a myurana mass would break charge would break the charge conservation because in this case they would not have zero charge so the only time that you're in physics wise that you can write is a tab which is really neutral it must be neutral otherwise you destroy charge conservation and it's not an accident actually then that you can have only one non-zero term and this one non-zero term would really correspond to the mass myurana mass of the right handed neutrino now I'll explain later actually that why this is really extremely important in order to solve it to explain essentially why the neutrino masses are so small okay so this actually is the result it tells us that you know in order to agree with the observation the dirac operator should take this form it has only one non-zero entry and the rest are zero this d and this d bar are correlated they are you know one of the complex conjugates of each other and the only thing I have to find is that what's the form of the dirac operator in the finite space okay so what shape does it take you know the whole the whole idea the things really are really constrained you know because in a way we have to explain why the standard model this actually you know why the standard model so if the hypothesis that everything comes from not going to geometry we must really arrive at this answer in an almost unique way hopefully one day one would have fine and say okay here it comes this is without any ambiguity but you know there's always the possibility that you have to put some physics because if you don't put any physics in principle you can get anything right because you know if I don't want to explain nature then you know I can start with uh with smaller space or bigger space okay if you already assume that it's four dimensional continuum times finite space okay and the finite space has killed dimension six yeah and then you make a very minor hypothesis on that space then you are already led to the 16th so it cannot be too small yeah this is the first example but then you cannot throw out why it's not bigger oh no of course you cannot throw out you know the question is that can I say uniquely you know this is the only thing I would ever get you know but this is the first possibility that one gets okay so now actually what do we where do we go from here and uh what we do so we would like actually to see what type of Dirac operator can we have this dab now remember actually the d satisfy many properties I have written them up so you really have to satisfy to show folks you have to verify that uh what are in the g d with jd still I'm talking about the finite space remember is equal to j and gamma j equal minus j gamma and d gamma equal minus gamma j so these properties I have to satisfy and so if you do that you'll immediately discover actually that the Dirac operator of the finite space really takes simple form remember you know let me write it in terms of new right here right because I don't want to write one dot and two dot because then it's it's more physical and uh in your right here right and then I have say in your left e left is a doublet and then up right down right and up left down left okay this is really a doublet but these are really singless okay why it's a singlet because actually this sigma really breaks algebra from you know for this condition and the other condition to be satisfied and then you put the sigma what you discover in the following that this algebra h plus h plus m4 of c is really broken and obviously it must be broken because you put the sigma remember this is like one and this is the i and this really breaks the color index it breaks the color index into one and three and in this respect the m4 of c is really broken into c plus m3 of c however because that that condition d a b opposite is equal to zero it really what happened and then it breaks actually this is also broken you know because this is one index we'll talk this one index and then actually essentially this is a c plus c it breaks it breaks and the three are not independent what one discover actually the element of the algebra a takes this form in this case so you are going to get you know x x bar this is my first h these are my first cation in other words actually remember a cation takes the form like alpha beta alpha bar minus beta bar it tells me that in the first h it must be of this form now here what do i have i have a cation so this is my first four and this is my bottom four and in the bottom four what do we really have we discover actually that this is x and of course this is zero and this is m where m is an element of m3 of c in other words actually what what we discovered is that the three c the c here the c here are not independent and they are essentially one c and the final algebra of the finite space really breaks down to c plus h plus m3 of c which is really a symmetry of a central model it really comes out of the requirement that there is a Majorana mass for the neutrino no this really has drastic consequences okay so with this action element of the algebra i go with the d and then we can really classify the d satisfying all these properties and you know i want to cut the story short what happened is that the only non vanishing elements of the d there is one more requirement that we have to put in order to really completely single out the d and that the d commutes with no actually um yeah d commutes with the element which is lambda lambda bar lambda bar in here if you do that then you discover actually that only the non yukawa coupling of the of the diac operator are non zero and really we have the diac operator completely singled out if i want to write it down i can write it down let me write it down actually so how does it look like many zeros i think here you are going to get k new and here you are going to get k e which is actually the yukawa coupling of the neutrino and yukawa coupling of the e and uh similarly here you are going to get uh this is new left e left and this is k new right k right like that and similarly for the quark and the only of diagonal elements you are going to get and um yeah here also so this is k up k down and diagonal form so in other words actually you know after some major accelerations one would know exactly what the diac operator is so what do we do with this now what we do with this is the following having determined exactly so you know many you'll get many essential point that you get many zeros by you know gamma d minus d gamma i tell you all diagonal elements are zero you only get non diagonal elements and so on so it's little matrix algebra nothing much to it and however one song is finished one can proceed as follows not that we have not really talked about dynamics or anything we just talked about classifications and representations and satisfying the properties of the non-computed space and you are led almost uniquely actually what do i call that of the standard mode now remember now i said that the diac operator of the full non-computed space is equal diac operator of the manifold cross 1 plus gamma 5 cross df we have been what i i was talking about up to now is the df we know now exactly how the df looks like dm of course we know because it is the um the diac operator of the usual say remaining manifold and uh similarly we can say a equal a of the manifold cross af and j everything is a product actually don't have to write it down all right now remember actually here one word when all starts with in what's non fluctuating d so we start with a non fluctuating d say let's start with a d which corresponds to say some flat space starting from a d from a flat space for example you can really start with you know gamma mu d mu like that actually for the d and we know actually that you really can generate the metric of the dm auto fluctuations these are the outer fluctuations and this is easy actually you also have to do that you say this is emu a gamma a you make this as a function of x you discover that you have to add a spin connection in order to make the thing uh in order to make the thing covariant and uh so that actually is easy to do uh similarly here we have seen last time that the d always go into d a where d a is a d plus a plus j but that's a for the global yeah what you have written there is for the global here is the global here is the inner automorphism no no but i mean if you want to do the fluctuations if you don't find nobody of course i do for exactly you know we get both inner but i'm trying to say that if you'd like to get the gravity has to do the outer and you want to get the other the a but these are enough fluctuations by the finishes exactly so they are one forms in this language and it is adb this looks actually very simple but remember actually the a i have written is what the a we have written was a 32 by 32 the d was the 32 by 32 matrix tensor do the clip for the algebra so you're talking about now in addition actually we said everything we know is really for one family and i have written here for example k and k i have written as a number but in reality this k is a three by three matrix in generation space it's a 396 so 32 times 32 and tensor do the clip for the algebra so it's 33384 by 384 matrix this is actually why it's it's it's it's much easier instead of writing you know matrices because they look huge is to put in tensor notation because in tensor notation you just put an index and you don't care how much it runs because you know you know how where does it some so it becomes really efficient to do it of course we didn't do it like that in the beginning because you know for years you have to struggle with all these matrices to see what acts and it took some time actually you know this was a progressing developments all right so this actually here you say all right you know i would like to compute this guy how does it look like okay remember so i will give you an example you know the reason i want to show you that things come out that it's not that you say oh the gauge fields come out the x-field come out you know and people think that we are cheating because we know the answer we put it in and then it's not true you know everything comes out in a very well-defined way one really cannot cheat in any way so you know i can say okay a b and then you can put indices like a a prime and then you can do db like ac cb and and so on so then i can look actually at all the matrix elements a b and how do they look like so i say okay how do they look like in the following i go back to my representation and then i'm really going to get an element here like you know one dot one one dot one remember you know i said a alpha i beta j and this alpha could be one dot or two dot on the i could be one and i and you see now i have to take all possibilities into account if i think about this what are this this would be the gauge field acting on the right-handed neutrino because we define the right-hand neutrino to be on this corner okay and in this respect you know you discover actually that this guy would be zero because the neutrino right-hand neutrino is massless guy sorry is a neutral guy is a neutral guy so you know then of course actually the next guy would be a two dot one two dot one and so on and this actually we discover it's a vector and then you know it is given by minus i over two g i don't know b mu so you know it's a vector and with a gamma mu everything is tensile to the cliff or the algebra so you get that this guy is a vector in addition you are you start actually to get something like you know a guy like a one dot one one dot a you know and then you know it's a doublet index a remember is an sq2 index and you discover actually and then you put a is one two and what do you discover that this is really proportional to gamma five h and this is where the x enter so the x enter as inner fluctuation and now i say what is this i know why i went this way this is an off diagonal element and off diagonal element it means that you know it's really trying to connect two elements of the left and right algebras with each other oh yeah one really can see it but of course you know you really cannot see for example here it's really easy if you see that it's a doublet what's not so easy actually is to show for example the other guy the a two dot two dot for example one a yeah in this case you say it's gamma five and what happened actually it is a bh barb you really cannot get guess ahead of time that you are only going to get one higgs and not two two higgs doublets the fact that you get only one higgs is really related to the fact that these guys are related remember i told you i have c plus c and then because we are talking about the same c it tells you that okay you know you really cannot have different higgs you are really going to get the same higgs so so the upshot of this is that one really can do this calculation is a matrix calculation no complications and after some algebra one finds the the Dirac operator including fluctuations what does it mean it gives me the connections it gives me the connections on the space and the connections are nothing but the gauge fields and the x fields nothing else so so up to here you know we have we can achieve that and before doing anything i can tell you that the Dirac action for the fermions okay okay you know one really has to write the gypsy i'll tell you actually now maybe i can tell you what that why we can write the gypsy the reason we write the gypsy because in in euclidean space i really cannot say gypsy equal to psi yeah so if in order to get once you write gypsy what really happened is that when you integrate on the gypsy and the path integral then you really this is an anti-symmetry product and you don't get determinant of dA you get the faffian of of dA and the faffian in the square root which means that it's really a chiral integration and you have not doubled the degrees of freedom okay no anyway if i expand this what do i really get i really get all fermionic kinetic terms for example nu bar d nu plus e bar d e plus left and right and everybody plus vector fermi fermi interactions plus x fermi fermi you get exactly the correct hypercharges and everything actually why one would get the correct hypercharges exactly because of this actually if you really work out the jA star j inverse they really come and they bring actually the u1's in really completely funny way and in order to have the matching and here actually one of us actually makes an assumption that one takes a su of the algebra not the algebra in other words one has to make a requirement that the trace of the a should be equal to zero this is another condition that we put in okay so so here actually it's the first check everything fermionic wise works and it works perfectly so what's next actually what's next is that you need to make so the fermions are dynamical but the bosons are not so how do they get the dynamics after all you know we say that we have a curved space the curved space should have dynamics and we should be able to describe the full interactions of the system see up to now i have been speaking classically everything here is classical now so here actually comes the idea of the spectrum action and the basic idea that the dirac operator which was really the building block of the uncommitted space in principle one can study the spectrum of this dirac operator so you study the eigenvalue problem and you look at the eigenvalues and we know actually from studying say remaining geometry or the dirac operators on remaining manifolds that the eigenvalues are geometric invariants in other words they all the geometric invariants are included actually or you know lambda is a function of all the geometric invariants so so now the question that how can we extract this dynamics how can we extract this the information about the lambda of course not easy to simply work out lambda of every space so we need something more general and so the basic idea that since the spectrum knows about the geometry we took as a dynamic the dynamics of our system is conjectured to be given by the trace of the function of a d where f is a positive function i think you have to put a scale as well why is it yeah yeah exactly i was to go and talk about the scale in a minute because yeah d has dimension on operator remember it is say it's gamma mu say d over dx mu and this actually has dimension one therefore you really have to divide it by something of dimension one now of course actually you may ask what is lambda and i will talk a little bit about you know dealing with the scale later because this is really a very sensitive issue what you should do with the scale can you get a little of the scale or what are we talking about of course i want to add something which is that this expression trace of a function of d is the only expression that you can write which is additive with disjoint spaces see the action should have at least the properties that if you take two disjoint spaces it should have right this is the only expression you can write so you want to say f but the one d2 should be disjoint in this case or what yeah sure if you take d1 and d2 to be disjoint take the fact the full direction after d1 not d2 okay but it certainly proves that you know the trace of that will be the sum of it okay i got it okay i got the function and but there is a provence to that it can be that it has a functional which is spectral and full field is its additivity properties and it has to do with this one but what about the positivity because you know but it's important for euclidean actions because some people have been you know toying with the idea that maybe it's not positive anyway so now essentially how how can we deal with this space obviously we really for an arbitrary function it's not that easy to work out but what we can really work out is the asymptotic expansion this way one would know how to do and the idea the following we use methods of heat cannon expansion you know it's not the only methods but they actually they at least for remaining many force and for product almost commuted very force they are really very very effective so what's the basic idea here you say okay you know suppose that i have a function of some operator p you make some i know i don't know a normal expansion you say it's a sps and then trace and then you are you have something like trace of p minus s the new's identity trace of p minus s is given by you know you use the melin transform formula and with the melin transform formula you obtain that you know it's the opposite of the definition of the gamma function essentially ts minus one trace heat minus tp with the real part of s is larger than equal to zero and then the next step would be to use formulas of trace e minus tp using heat cannon expansions and then of course actually this is given by an of the sorry tn minus actually in this case minus well n over two minutes okay at least in four dimensions and where n is larger than equal to zero and this is an of xb while dv and these actually are called the series of and they're defined in terms of connections of the dirac operator so it's an asymptotic series okay i would say why it's it's useful in physics because exactly of at a point that this is you know of the fact of d over lambda and then it would be clear actually when i write the expansion it would be an expansion of one over lambda so the first term will be one over lambda two four then you're going to get lambda four then lambda squared then one then one over lambda squared now assuming actually the scale is extremely large then these terms will become you know less relevant of course actually the expansion will break down once you approach the scale and you know what the scale you know you can take the scale to be say unification scale or plank scale in this respect actually the expansion would definitely break down and in that case one has to learn different ways of dealing with such expansions but you know we did some calculations of this expansion for you know robertson walker metrics and things like that and really it holds almost up to the plank scale no up to the plank scale it's only when you you are orders of you know like twice or three times the plank scale then you start to see that it breaks down but not before you know which is which is really amazing that this action is so accurate that it goes all the way up almost you know to the plank scale but not beyond you know it's just there all right so what do we do with this expansion doesn't give us anything sensible because obviously this looks far fetched i would say that you take an arbitrary function you say okay this arbitrary function this is the permanent part is straightforward upside upside okay and you get you know the usual expression you can say it's by construction but for the bosonic part it's not really clear that this idea should work you know that it should give you anything accurate essentially after all you're using some expansion all right so how to proceed actually here you know it's known as i said here we assume that the continuous manifold is four dimensional and in this respect and in this respect the the expansion what happens is the following that only an equal to zero if an result for manifolds without boundary okay what happens for manifolds with boundary we have already addressed this question is more is actually very interesting but i will talk about it later but for manifolds without boundary an equals zero and the thought it means actually the only non zero ones are the a zero the a2 a4 and things like that and these formulas were worked out long ago by by gilki he has some you know standard formulas and because of that actually what really becomes relevant and this is the d squared of the d because and what the trick of the following you write this as g mu d mu plus a remainder and this d mu will d mu plus omega mu which is a connection and this omega mu okay this also is given by minus g mu d mu d mu plus a mu d mu plus b so what gilki did actually you know he gave an expression this actually is there are two important terms in this expansion if the e which is an invariant and the connection the omega mu and you know curvatures which are constructed out of this connection and this omega mu he one reads from the following you write this expression into this form into this form and then you can know actually that a mu is given by you know some expression it is half g mu a mu plus i don't know plus comma mu this is a crystalline contraction of a crystalline in other words actually one really can compute all these expressions in a unique way and remember we have already computed our d so we have this d huge matrix you take it you square it you get another huge matrix and you write it in this form and then you try to find the curvature of the space remember it's v a generalized curvature now and you also get the e invariant of the space another huge matrix you know 384 by 384 okay so what gilki does us look you know this a zero the first time in the expansion is given by 1 over 16 pi is root g trace 1 trace 1 of course is 384 because the 384 by 384 matrix so this is actually nothing to compute this is no no this is actually lambda 4 this is the neither cosmological constant and it's huge you know the cosmological constant is huge here why it is zero that's a different story okay effective you want to talk about effective you want to talk about i'm talking about in a zero-th order cosmological constant of course there are contributions coming from the vacuum expectation value of the higgs but these are supposed to be much smaller actually yeah okay you know at negative sign yeah that actually but but that comes from the higgs no you mean the d squared the d yeah sigma this is a sigma sigma yeah yeah okay but this takes after sigma would come later because i see sigma lives here now sigma lives here because a priority we don't know okay anyway so the the interesting part actually is this guy so you get actually r over 6 is the Einstein term and then you say trace e now you let me actually give you an idea what trace e is in this case or what e is actually you know e you have to read you have to read by squaring the d by squaring the d and so obviously let me remind you the d you have all this d slash part you have here gamma 5h part gamma 5h dagger you have some distance okay when you look at the d squared yes the off diagonal terms are really going to give you what hh dagger terms which is really nothing but the higgs mass okay so you discover actually essentially and then you trace it and you only get trace h squared so if you really work it out actually in this example what you really get you're really going to get the following actually you get this gives up to a factor so you see this is the guy that would eventually give mass to the neutrinos and this is the higgs mass these are essentially yukawa coupling your square of yukawa couplings square of yukawa coupling so we look at this expression and then we notice the following the curvature term you mean the whole minus sign yeah there's a whole overall minus sign yeah see i write minus okay so if you'd like yeah okay so minus 2 over pi squared if you wanted this f2 lambda squared yes gd4x there's a minus sign here so well the notation that r is negative for euclidean spaces anyway so you see we get the mass term and with the minus sign this is really extremely important if you as we'll see when for the standard model because the standard model is really characterized with a potential v as you see lambda over 4 h of h bar h squared minus m squared o minus mu squared over 2 and then the minus sign that gives the potential the mexican hat it's essential that you get the max minus sign now obviously you are really going to get a minus sign why because if you look at it in the end you are taking trace of e minus tp p has d squared d squared has s squared obviously i'm going to get the first term in the expansion a minus sign the next term will be plus sign it alternates which is extremely nice because of course but you really cannot guarantee that okay in principle before you start we don't know so this actually important points we can point that the curvature which is the einstein term and the mass term re3 is equal footing so h bar h is like curvature of is the curvature of the antenna space and okay you know scale lambda squared what's lambda squared of course actually one would like this lambda squared to rate to be related to eight punch or 16 by g whatever okay it should be related to the so this actually the plank plank squared so from here one would read that if you would like to make sense of the scale lambda that we started with the natural uh interpretation of it that it's almost of course you have numbers you know you have f2 it's a malin transform of the function so you can get factors of 10 or things like that you have the pi squared so it is of the order of the plank mass it's not exactly the plank mass because you have some factors here but at least an order we know that it's of the order of the plank mass the next term day four and according to gilki what's the formula the formula tells me that what is it okay you have uh let me write here okay i'm not going to write the top the surface terms so what reenters is the curvature squared everything squared actually here e squared curvature squared now you know this e squared the omega mu squared they really give me invariance in terms of curvatures square of curvatures what are the squares of curvature of course the r mu nu squared is there but here we're really going to get what we're going to get f mu nu squared which is curvature squared and we're going to get kinetic energy of the x term and the sigma term and things like that so and yeah exactly yeah in mil terms there's nothing but curvature squared terms you know because what's f mu nu you know it's if you take the a squared as you see d plus a squared you know you are you going to get the f the square of the curvature the square of the connection is the curvature and it really it lives here it lives here and one really can compute it some long calculation but in the end is well-defined calculation the important thing there's no ambiguity at any point okay so in this action we're going to get f0 over 2 pi squared y squared douche bunny plus plus conformal factors kinetic terms for the h quartic terms for the potential and similarly everything I write here I have to write also for the sigma field and so on our sigma score anyway see the important thing is that first of all this time there's no scale second we notice that we have a vile squared term now this is really important for you know if you are really doing say the realization of gravity you know this term would make would improve actually the propagator of the graviton it makes a theory of course not unitary but of you know this we don't really worry about because in that case we are cutting you have to cut the expansion and you know if you would like to treat unitary you should not cut the expansion or you have to take the full Fletch theory but at least this conformal tensor or the vile tensor squared is important for the propagator of the graviton and now here we see here actually that we really have a kinetic term for the u1 for the electromagnetism a kinetic term for the sq2 weak and a kinetic term for the color sq3 and exactly and then we look here and then we immediately find a relation that tells me that g1 squared equal g2 squared that's five over three sorry g3 squared g2 squared five three g1 squared which is the famous unification one obtained for su5 or s010 grand unification theory we will have actually we have to talk about this later the next lecture because this is not exactly satisfied it's not exactly satisfied because as we are going to see next time you plot the theorem okay the logarithm of of of the development of the alphas which is g squared over four pi alpha inverse of the alpha inverse then you discover that they don't exactly meet but you know this is a different story you know whether they meet they don't meet they almost meet but in principle you know there's so much in between that we don't really know we don't control and it's premature to say whether this is a bad point or not to get a bad point there you know it's it's almost you know that okay within let me say within 10 percent it works within 10 percent but you have extrapolated 15 or the 16 orders of magnitude actually which is huge okay yeah you know so there is and you know things we have discovered that things have not really been calculated you know the two loops in the model that we have because this field actually turned out to play an extremely very important role the sigma which a priori we ignored at one time we even make a prediction that the Higgs mass is given by a certain number and this turned out not to be correct because we have ignored actually this this this field however if you take this feed into account then you know you're completely in agreement with experiment and in addition actually the romanization group equations taking the sigma into effect has also only been carried to one loop order and nobody has carried it to the border so we really don't know so maybe it would change little bit actually you know my guess is that it will change but I don't know whether it will change in the right amount in the right direction because this is up in the air so so let me summarize the game so we started from a very general idea the general idea that space time can be approximated by a product of a continuous time discrete manifold the real answer we don't know you know it may be just completely discrete but to a very good a very good start to say it's because we know that generativity works very well so we know the continuous part up to the Planck scale works very well so we can up to that energy level or a little bit below we know that the uh you know Raymanian manifold part works so to assume that we have an invisible extension which makes it almost commutative but in reality non-commutative that we can extend it by a finite space we entertain this idea and we discovered that to a almost uniquely we can predict the structure of the model we know the Fermini representations we know actually why there are 16 fermions we know what are the representations of these fermions we know the gauge group yeah this actually yeah is this i'm saying that what we can that things what we cannot do is much more of course because what we cannot do for example predicts say the electron mass and you know all the masses because in principle you know it's the if there is more restrictions on on the one should be able to for this remember the finite space contains all the Yukawa coupling so it contains actually information about so if you know that for example that k is like 10 to the minus 2 or something then you know that the ratio of the electron mass of that top quark mass is like 10 to the 4 or something so the minus 4 so in that case like this information we are putting in by hand we are not you know we are not predicting but if one really is completely successful you would everything comes out but you know this is really far-fetched for time being because you need some extra structure which we really cannot put our hands on not not yet you know so there is i think there is another aspect the criteria and the properties that we verified do not really still tell us actually why the why there are three generations this we don't know this must come out somewhere action you know so this aspect we have no idea one doesn't need explanation for that we need an explanation for this there is not even a guess whether they should not be a false generation yeah this section almost rolled out now actually no but in fact there are two things to say yeah there is one or these two Japanese who got the Nobel Prize and the reason why we needed more than two yeah and that's the category for Kobayashi Masukawa because it involves a complex number only if you have more than two sticking so I mean three is the first number of generations for which you get a complex number so for which you get violation of charge conjugation but then okay I think now experimentally it's known that there is no faster yeah yeah so why three no actually it's not that we don't know nobody knows ourselves it's up in the air why the masses of the firms are the way they are you know the standard model after all has what 17 or 18 parameters sure but there you can say one thing which is it's exactly you know like if somebody would come and would tell you why is the metric of space times the way it is okay you could say well uh can play out some guests but you know you could say well it's like this it's because of circumstances so the issue is what is the geometry of the finite space yes is it due to circumstances or is it does it have a deeper meaning right and this we don't know one would hope actually see that look if maybe well it 20 years ago you know okay you know before say the advent of locomotive geometry if somebody asks you why the standard model you say you could have answered the same but now actually I think we are very close to knowing that it is it must be like as you think across as you too it must be like 16 so we must have a hex field so this actually have been answered satisfactorily but we are far from you know the full answer and this is this is not here actually here okay if later if you know we take off this condition and we really want to push everything to grand unification then of course you have many more hexes actually not only one we are going to get but that actually will be relevant at higher energies and we will not be able to see what really goes on at lower energy except for the six doublet you know it's not clear but that's that's another story you know anyway I would stop here I think because next time I will I will continue by analyzing that quantum aspects of the model and maybe you know theorizing a little bit about so what you have so far is the first characterization of course yes just to answer what Thibaut has last um right but what is quite a one should stress one thing you know which is that there are seven or eight sides which have to be correct and they are all correct and and moreover you know it encapsulates this enormous complexity of the Lagrangian into something which is I think you know this thing was done before you know this would have been like and then people come and verify it it would have been like in a revolution but the thing actually lost the luster because people didn't think that okay you know the answer you are fitting it and this is the you know biggest thing that one has to fight in the misconception that there are billions of possibilities and simply were given an interpretation of these of the one possibility which is correct and many other possibilities which is not really correct because you know according to this specification that I went through and that there are not really many possibilities you know it's almost that comes uniquely you know but then one would have to explain why if you want the correction by the finance base improves the ordinary four-dimension and continuum and what it improves it changes the cure dimension yeah changes the cure dimension from four and and that's a big improvement because then you can write this very well it's the most economical to be finite I mean it could be that the splitting is actually more complex but I mean the simplest way to see it is that you have a foreign national space and then you make it two-dimensionality dimension without changing the metric dimension which is remaining but if you have a you know if you start finite then you'll have to fight all this can lose a client mode says I would say you know we have many extra degrees of freedom which we don't see so so new dimension by something final exactly by something final which doesn't have this infinite number of modes like it goes I mean that's a very very basic