 Shall I start? Yes, sorry, sorry I was muted. Yes, we are live. Yes, we are live. Okay, so hi everyone and welcome to this colloquium in the Latin American webinars of high physics. So we're super happy to have today Professor Chi-Sheng Feng from the University of Ajif Federal do ABC in San Pablo, Brazil. So Shang, well, this PhD at Stony Brook University, and then he had a couple of contact appointment. He was in Italy in Praskati and then Brazil at the University of Uspi, University of San Pablo, and now his professor since like four years ago or so, as the University of Federal do ABC. Two. We are super happy. Two only. Okay. So, yes, the colloquium will share with will tell us about how to cook the matter antimatter symmetry of the universe. So, please, Shang. So good morning everyone. First of all, I'd like to thank the organizers for inviting me to give the talk during this pandemic. So shall I share my slides now. So can everyone see my slides. Yes, thank you, yeah. Okay, so, so today I will talk about how to cook a matter antimatter asymmetric universe. And so first of all, I would like to give a short introduction about our current understanding of the universe. So current theory is based on the two pillars, content field theory and general relativity. And then, in particular, we have these special model standard models of particle physics. And the general activity we have a specific model, which has the dark energy content, co dark matter. And also we need inflation in the beginning of the universe. So on the standard model side, we have a van the universe is very hot. This is a, we have a gauge theory, which is described by this gauge group, the color, the electric interaction. And as a temperature reduce due to the expansion of the universe, then there's a phase transition. And we have the, the Higgs obtain the vacuum expectation value. And today, we only see the color force. And also the electro magnetism. So basically, the big interaction is broken. So we have a massive weak gauge ball zone. And then in particular, we also observe a neutrino oscillation. The standard model neutrinos are exactly massless. It means we have to introduce a mass for the neutrinos. So at low energy, there are only two possible kind of master we can write down. Once is we use the same degree of freedom as in the standard model. So we write down the so-called myrona mass. So this while a laptop number by two units. And the other possibility we introduce another light degree of freedom. The so-called right-handed neutrino. So this is not charged and the standard model gauge interaction. So this must them conserve laptop numbers. The so-called direct mass. So the main different is basically in the myrona mass scenario. You can have neutrinos double beta decay, which is the beta decay without neutrinos. So you wallet laptop number by two units. So of course, on the cosmological side, there's also many incredible progress, especially about the energy content of the universe. And then this is the menu of the recipe of today. So first, I will try to convince you that our universe is really matter dominated. And then after that I'll present the ingredients. And then I'll spend a significant amount of time to talk about the symmetries of the standard model plus the standard cosmology. And then after that I will introduce leptogenesis, which explain two puzzles, neutrino mass and also matter, antimatter symmetry of the universe. And then finally I will touch upon how to test this kind of models. So this is a brief history of our universe. So we start with, there's a beginning and then you have to go through an exponential expansion, the so-called inflation in a very short period of time. And then after that we have the formation of new clones. At first we have only corks, groans and so on. And then they cool down, they form protons. And then in particular a very important period is when these protons and neutrons, they start to fuse to form heavier elements, but they are still like helium, 3, helium, 4, lithium and so on. So this is an important period. All the light side elements are formed. And then in particular, I just want to highlight the amount of the light abundance depends crucially on the amount of baryons we have, which is the protons and neutrons that we have. And then there's a number of baryons normalized over the entropic density of the universe. So now a day we have measured this precise number is based on the primordial deuterium abundance of the universe. So we get this precise number. And then as the universe continue to cool down. Then at some point, we form a neutral particle. So basically the photons you decoupled from the plasma. So when we have only the neutral protons. So when the universe first becomes transparent during this period. So and then these photons has been traveling since until today we observe them as a cosmic microwave background. From the anisotropy of the temperature of this cosmic microwave. And we can also deduce the matter content of the universe. Basically, we can determine the total matter content and also the in part of the matter content, which proportion are the baryons got baryon as a pressure. So from this measurement, we also get a very precise value determination of the baryonic content of the universe. So one incredible or confirmation of the discount picture is the agreement between these two measurements, which come from a different period and different physics. And then now the question is, are there any evidence of primordial antimatter today. And this is the so far there's no evidence. So on the galactic scale. So we do not see antiproton flux, which is, which is what we observe the antiproton flux is consistent with this kind of secondary production. The PP collision and then we produce antiproton and that antiproton we annihilate with other proton and generate gamma rays. So the amounts that we measure today this is consistent with secondary production. So, and then on the larger scale on the cluster of galaxies. We also do not see these gamma rays, which is produced from the proton antiproton annihilation. So there's no this anomalous signature. So on an even larger scale on the observable universe, there's no distortion on the CNB background due to this additional photons. So we can, at least in our observer universe we know that we only have a matter and there's no antimatter. So this is today. But of course, if you go back to early times, we have a lot of barions and anti-barion. But the important thing is this quantity, the barion minus anti-barion divided by entropy is a conserved quantity. That means this is what we have today. We don't see any significant amount of anti-barion. So this has, if we extrapolate back to early times, they have to be the same number. That means that in the early times, we do have this asymmetry of this order. So the conclusion is we have a, we live in a matter dominated universe. And how easy is it to obtain this asymmetry? In fact, in the early universe, we only need every 10 billion's matter antimatter. We only need one additional matter. So it looks very small, but as we will see later, to explain this access is not so trivial. So now let's look at the, in the standard picture, if we start with the same amount of barion and anti-barion. What happened is we have a so-called annihilation catastrophe. Because of course this annihilation will keep happening. But at some point, this annihilation will be inefficient due to the expansion of the universe. So they cannot, the chance of them meeting each other is very small. And this happened at about, when the temperature of the universe is 20 mega electron volt. So if we do this calculation, we end up with 10 to the minus 19 of barion and anti-barion. And this number of course, first of all, we don't observe the same amount. Second of all, this is really nine of the magnitude too small from the observation. So this could not explain what we observe today. And then of course the other idea is we can impose an initial condition. We have an initial asymmetry. Now of course, due to inflation, whatever input, they will be diluted by exponentially diluted. So this idea also, besides not elegant, it also doesn't work. So we conclude that in these windows after inflation and before the beginning of Big Bang nuclear synthesis, we need to have a mechanism to generate these matter, anti-matter asymmetry. So what are the ingredients we need? So the ingredients is proposed by Shakharov in 1967. The first one is kind of straightforward. We need to violate barion number. So after inflation, if we start from zero, we need to violate barion number such that we can generate a non-zero barion number. The second condition, we need such that the interactions distinguish between particle and anti-particle. In particular, we need both charge violation and charge parity violation. So why? So this is one very simple way to see. So you can have x which decay in a barion number violating way to left-handed barions and right-handed barion. So if charge is conserved, so we take the charge conjugate, we get this b left-handed bar. And right, we take a charge conjugate, we get the right-handed bar. So the rate will be the same. That means we cannot generate an excess of particle over anti-particle. If Cp is conserved, so this will be transformed to be anti-b but right-handed. And this will be transformed to anti-b but left-handed. Again, the total rate is the same for the decay to particle and anti-particle. So that means we need to violate both. And then the last condition is we need to deviate from thermal equilibrium. So an easy way to see is in thermal equilibrium, we know all the particles will have this distribution. For fermions will be plus sign, for bosons will be negative sign. So this is either fermi-derac or boson sign distribution. So since b cannot be conserved, it means that the chemical potential will be zero because this is not a conserved quantity. So now from CPT, we know that baryon and anti-baryon have to have the same mass. So then you can integrate this phase-space distribution over three momentum. So since mass is the same, this is the same distribution. So we end up with the same amount of baryon and same amount of anti-baryon. So we need to deviate from this condition. So do we have what we need in the standard model? So baryon number violation is there because baryon number as you see in more detail later, it's not a symmetry of the standard model. And especially in the early universe, this kind of violation is in particularly very fast. And then c and cp violation. So basically standard model violates parity because only the left-handed particles feels the weak force or equivalently the right-handed anti-particle feels the weak force. So that means the parity is violated but charge is also violated because if you transform left-handed particle to left-handed anti-particle, this doesn't participate in the weak force. So that means c is violated maximally. And then cp violation is also violated. This was due to the unremovable cp phase in the Yukawa interactions. So this was detected later. And in particular, it's due to this unremovable phase. It's due to we have a three generation of quarks. But unfortunately, if we include this phase, so this phase enter in this so-called yawscode invariant. So though this phase is order of one, but the problem is we have effectively a tight hold of gym mechanism. So it's suppression due to the quark masses. So this comes into the calculation of generating variant asymmetry. So due to this kind of suppression, in the end, we have a number which is 10 of the magnitude, which is small to a number that we would like to explain. And the last condition is our thermal equilibrium. So in principle, we can have this kind of condition during the phase transition because we know that electrolyte has to be broken. So there's a phase transition between the electrolyte symmetric and electrolyte broken phase. And then this calculation from this work. So it turns out that this phase transition is smooth because the Higgs mass now we know is right here, is 125 GV. And in order to have the first order phase transition, the Higgs has to be lighter than 70 GV, actually 70 something. So beyond that, the transition is second order is smooth. That means we won't have an out of equilibrium condition. And let's look at, I also take this chance to discuss about the minimal supersymmetric extension of the standard model because we know that the Higgs potential will be different due to the existence of second Higgs doublet. But even in this case, the only survivor scenario is when we have a very light top scenario, like right-handed stop. So right-handed stop is basically a scalar partner of the top. So this you come into the Higgs potential, this stop. So basically we need the stop to be below 115 GV to have a strong first order electrolyte phase transition. But right now, essentially this idea is ruled out by the large-hardened collider data. So at least in this minimal set up, even in the supersymmetric extension to the standard model, we cannot get the first order electrolyte phase transition. But of course, everything is not so set because we know that the universe is expanding. That means that if it's possible that we have certain interactions which are slower than the Hubble expansion rate and which means that these interactions or the particular which interact through this interaction will not be intermal equilibrium. So it's possible that we have, if we introduce certain new physics, we can make use of the cosmic expansion to have this kind of our equilibrium condition. And let's discuss a bit about the symmetries of the standard model and cosmology. So the standard model is dictated by this gauge symmetry. And then we know there's a phase transition. And in particular, if we focus on the Yukawa interactions of the quarks and the laptops, we see that there's additional symmetry. You can rotate all the quarks field. There's the left-handed quarks and the right-handed up and down quarks by a common phase. And these terms will conserve this rotation. Of course, the kinetic term will also obviously conserve this rotation. That's why you only focus on the Yukawa terms. Similarly, for the laptop's field, if you rotate the left-handed laptops and right-handed laptops by the same phase, this will be a symmetry of the standard model. So of course, this is without neutrino mass. Because without neutrino mass, no, sorry, this is... So of course, now we are talking about without neutrino mass. Once we introduce... And sorry, what I want to say is without neutrino mass, actually we have more symmetry because this is a neutrino mass. Without neutrino mass, we can actually rotate left-handed and right-handed laptops' fields, such that this becomes diagonal. So that means that we can actually rotate each family of the laptop's fields. So we actually have more symmetry. We have three modes for each family, electron, neon, and tau. And then in the standard model, so we know that we have electric charge conservation because electromagnetism is not broken today. And in particular, this kind of interaction, which will let the charge shall not be observed. And there's a bow on the lifetime, which is much longer than the lifetime of the universe. But this is okay because this is protected by the U1 electromagnetic gauge symmetry. And then we also haven't observed this kind of interactions. Bion number conservation, a relation interaction. So basically, we can have a neutron, which, so we have, basically what we observe is only bion number conserved interactions. In principle, we can have proton decay. So this and laptop number one, this final state have laptop number zero, so this wallet laptop number. So this is not observed. So because this symmetry is not protected by gauge symmetry, in principle, we can violate. So we also have a bow. So this is a lot longer than the age of the universe. So then we also observe that in the standard model, a laptop number is conserved. So this is a laptop number conserved interactions. But due to tiny neutrino mass, we have observed laptop flavor violation because neutrino can oscillate to different flavor. But all these are not protected by gauge symmetry. So what we have observed is a violation of total laptop number. For example, here, this is neutrinos double beta decay. So this is a neutron in a nucleus. So they convert to proton in a nucleus. So basically the nucleus change. Then emit to electron without any neutrino. So here you have a zero laptop number. In the end, you have a laptop number of two units. So this wallet total laptop number. And this cannot be observed. And the lifetime is also weaker than this. But yeah, I did not code the number. But then this process which wallet laptop flavor, but not the laptop number. So this is in principle there. Because if you calculate with this tiny neutrino mass, but this is very suppressed. Which you cannot see in the experiment. But what do you see if the energy that we consider is much higher than neutrino mass? So basically we can consider this laptop flavor as a good symmetry. So in fact, a barion and laptop numbers, they are not the symmetry of the standard model. This is first pointed out by two. Because if you calculate the divergence of the current, the barionic and the aponic current. So they have this so-called chiral anomaly. So basically they come from a quantum correction. So basically we see that at the Lagrangian level, this classical level, this current are conserved. But when you include quantum corrections, they are no longer conserved. So if you look at the barionic charge or leptonic charge. So you look at the variation in time. So you see that they are not zero. So basically which means that this barion and lepton charge, they can change with time. And how do they change? So they are changed. They are related to this. They depend on the SU2 gauge field configuration. And it turns out that there is such a configuration where in the vacuum, we have many equivalent SU2 vacuum. But this vacuum, so this you can show that is related to topological charge, the change. So basically this vacuum, what they differ is barion and lepton number. So the difference is by three units. If you go from this vacuum to this other, the two differ by barion and lepton number by three units. But the thing is today this vacuum, to go from one vacuum to the other, you have to go through this barrier. And then this barrier is, if you estimate, it's about 10 TV. So this tunneling process is very, very suppressed. If you calculate the tunneling probability. So that means today we will never observe this kind of process, barion and lepton number relating. But then it was realized back then, if the temperature of the universe is very high. So basically this kind of barrier will no longer exist. So this kind of process, barion and lepton number relating process, will be very fast and it's proportional to temperature. If this big coupling constant, or big fine structure constant to the five phase power. So that means we see that the symmetry is actually temperature dependent concept. Once we put the standard model into an expanding universe, because the temperature of the universe is changing. So sometimes, for example, barion and lepton number at low energy is a good symmetry. But a high energy is not a good symmetry, because it's violated in a fast way. Even if the standard model Lagrangian is not changed. So notice that this barion minus lepton charge is exactly zero. The time variation. So if you look here, if you take the different because this is the same quantity. So this is exactly zero. That means that in the standard model, this barion minus leptonic charge is exactly conserved. It's not broken even by the quantum effects. It's exactly conserved. And then, but then at a temperature we are usually interested in, which is in the early universe. So when this kind of lepton flavor violating process is between a mass is very suppressed. So essentially the lepton flavor is conserved. So all we can say that this we have three B minus L charge, because in each direction of the leptonic flavor. So this is the average of barionic charge minus lepton flavor. This is effectively we have three charge, which is conserved in the early universe. We can represent this kind of B plus L violating process, but B minus B over three minus L conserving process, which this operator. So you can imagine if as we go from one vacuum to the other. So for the vacuum, you generate three quarks and one leptons. You can imagine you have this kind of process. Keep happening in the early universe, but this operator conserved this charge. They only violates barion and lepton a number, but not the B minus L. And this actually is a perfect source of barion number violation. In the early universe, we can make use of this barion number violation to generate a non-zero barion number. And now I will spend one brief slide to talk about the assistant of this symmetry and the possibility of generating an asymmetry. So if a symmetry is an exact, so basically we cannot generate all the time. It's an exact all the time. That means it's not broken. We cannot generate even a tiny amount of asymmetry in the corresponding charge. So that means, but after it's generated this assistant asymmetry, they can actually protect this asymmetry in the charge from being erased. So they have two rows. So basically we need a slightly broken symmetry such that we can generate this asymmetry in the corresponding charge. And then after that, this symmetry better to be effective again, such that this charge will remain until today. So the early universe is expanding soup of many particles, even in the standard model. In particular, these particles are interacting with each other. So how do we organize this kind of system? It's expanding and with a lot of particles. So we use the so-called early universe effective theory. That means we just study at each temperature that we are interested in. For example, we have a model of biogenesis, which occur at certain interval of temperatures. Then we study this regime and see what are the effective symmetries that we have. And basically how do we organize one key parameter? One key parameter is the harbor rate. So to know if the symmetry is there or not, so some interactions, so whether they occur very fast compared to the harbor rate. So if the first start, if this interaction is very fast, that means we know that all the particles which participate in these fast interactions, they will be in kinetic and chemical equilibrium. Kinetic equilibrium meaning they have this phase distribution, the equilibrium phase distribution. And chemical equilibrium meaning that for whatever particle that they scatter initial to final state. So the sum of the chemical potential has to be equal to zero. This chemical equilibrium condition. So we can also have interactions which occur very slow. So this can be due to, that's the exact symmetry in the model. For example, high per charge. There's no interaction which wallet had per charge, so that means zero. Or we can have some interaction which is suppressed at certain temperature compared to the harbor rate. So we can get an effective symmetry. So if we have these exact or effective symmetries, we get the conserved chargers, the conserved nodal chargers. The final one is so-called relevant. So when this interaction is, it can be faster and then after that it will be about the same as the expansion rate of the universe and then become slower. So when we have this kind of transition in between, so we will have a so-called approximate symmetries. So right now the nodal char is no longer conserved, but they will evolve in time. In particular, we have to describe the evolution of these chargers. Let's say it's starting from zero, then in the end what they will evolve as a universe expanding until we get some certain final value. So to describe this kind of non-equilibrium dynamics, we have to use the so-called Bosman equations. So this is the relevant. So in the end, what we need to do is to identify all the effective symmetries. And then in particular, then in the end what we need is just to describe the evolution of these approximate nodal chargers. So this is the way to think of how do we organize this kind of biogenesis in the early universe. So now I would like to spend some time to talk about the relation between the U1 symmetries and the chargers. So given an interval of temperature, so we identify all the U1 symmetries. So for each complex particles, which means that it can distinguish between particle and antiparticle, for example, complex scalar. And so in all the fermions, the wild fermions, if they are in kinetic equilibrium, we know that they have this kind of face-to-face distribution. So in particular, one important key point is, although this assumption is not really needed, but this is whole truth for all the standard models of particles. So for particles and antiparticles, because they can annihilate to be gauge boson. So basically, you know that because gauge boson, the number is not conserved, so they have zero chemical potential. That means the chemical potential of particles plus antiparticles will be zero. That means the antiparticle will have the negative of the chemical potential of the antiparticle. And for any reaction of type 1, which is very fast, as we discussed earlier, we have this condition of chemical equilibrium. And then if some U1 is an effective symmetry of our system, for example, B minus L, that means that by definition, for any interaction of the type 1, if we sum up the charge of each particle, they have to be sum to zero because this interaction conserves the charge by definition because if this is the symmetry of the system, that means we can actually solve the chemical potential in terms of charges because you can see that if we sum over all the particles which interact in these reactions, since the charge is conserved, so this sum OI, this will be zero. So that means the sum of mu, all the chemical potential will be zero. So that means it's possible instead of using chemical potential, we can describe in terms of charges. So these are some constants which are not determined, but we can actually solve it in terms of charges itself. So how do we do that? So basically, first of all, we rewrite in terms of number density asymmetry because these are things that we are interested in, the asymmetry in a particular species of particle. So in particular, if you calculate this value, we can expand in terms of chemical potential over temperature. And this is much smaller than one, and this is true because we know that the number density of the particle is much higher than number density asymmetry that we observed today. So it's 10 to the minus 10, this asymmetry that we would need. So this expansion is always good. So we just keep the linear order. So that means we can, now we rewrite the chemical potential in terms of the charges that we have earlier. This is a generic solution. So that means we represent now the number density asymmetry of the particle in terms of only the charges of the system. But we still have this constant. But what the trick we would like to do is, now we would like to solve this constant in terms of the charge density of our system. So we can construct the charge density, so u1x. So the charge density will be some of the all the particles which carry this charge. And then of course this charge in 9-0 only when this particle has asymmetry. So if we sum up, then we put this relation. We have this combination which only depends on the charges of the particles in your system. So this we can represent it as a matrix where the index represents the u1 charges of the system. So if we have two u1 charges, we have a 2x2 matrix and so on. So now we can solve this constant in terms of the nodal charge density and this j. So this j is completely determined by the charges of the system. And then we can put this back in. So this constant is solved. So finally you see that we have successfully written the number density asymmetry of each particle in terms of the nodal charge of the system. That means we can have as many species of particles in our thermobuf as possible. But all of them can be represented as a linear combination of the nodal charge you have. So if you have three conserved charges, that means this is what all you need to study. How these three conserved charges evolve if they do or if they are conserved, they are 0, they set to 0. So the whole thing is simplified. And so now this is the quantity we are interested in, the bionic charges. So we do the same thing. We calculate the total bionic charges. So for all the particles which carry bion number, the quarks, they will contribute. So if they are corresponding bion number. So now we can rewrite this as this form because we have solved this in turn of only the nodal charges of the system. So that means the bion asymmetry can also represent by the effective charges of the system. So in this way, all the fast-reaction, actually they are resummed. In some sense, they are taken into account in terms of these matrix, which depends only on the charges of the particle. And then the type 2 and type 3 reactions are also captured by these charges. So for type 2, which means that these charges will not evolve in time if they are so slow compared to hardware rate. So that means they are constant. So if these charges, this symmetry are never broken. So they are 0, for example. And then what we need to study the evolution with Bosman equation is only of the type 3, where they will evolve in time or temperature. And so biogenesis could work. So we can see, we can generate non-zero bion number only if there are some approximates u1 symmetry, such that we can generate a non-zero charges. So now I will go to the example of the standard model. I think I'm a bit behind in time, but I'll try to speed up. So for example, we consider this high temperature. We have four charges, the hyper charge and the three b minus l charges that we have talked about earlier. So now we also have to consider the Euclid interactions, because the interaction mediated by the Euclid can be fast or slow depending on the temperature. So this is due to Euclid and you insert the gauge boson. So you can estimate this scattering rate, which depends on gauge coupling and Euclid coupling. And they are proportional to temperature. So then you get, if you want this rate to be in thermal equilibrium, so you can see that the temperature has to be below certain threshold. So for example, if biogenesis occur in this window, that means all this Euclid scattering mediated by D, U and E are not in thermal equilibrium because the temperature is higher. And this, which means that if we rotate the right-handed downcoat, upcoat or electron, so that means because this Euclid essentially you can ignore them in the Lagrangian. That means we have this symmetry, rotation of the right-handed, the field. So potentially we can have this 3U1 corresponding with this rotation. But actually up and down, they are not symmetry. This is broken again by quantum effects. So this is, they have a mixed SU3 anomaly. So we are end up with, in this regime, we only have the rotation of the right-handed quarks. So this has been considered, if we consider this range, that means we have to include U1 in the right-handed leptons. So it is solved for the number of baryons. We have an appearance of, it's a hypercharged density, B minus L, plus the charge density in the right-handed electron. So in principle, if we introduce a model which violates this slowly, we can generate a non-zero discharges and they will be, or we can assume if there's a model which generate pre-assisting a symmetry in the right-handed electron. So hypercharged is not broken. That means we can reset this to zero. It's never broken until an electric scale. But then at this, below this temperature, this electron you cover, get into equilibrium. That means this is no longer a symmetry of the system. That means we can only represent into the four charges that we talked about. Since this is not yet broken, it's zero. And if our theory never breaks B minus L, that means this has to be zero. So that means you cannot actually store the asymmetry in the right-handed electron. If your B minus L is never broken. So then you have a standard model, plus plus, and in some extension. So in this consideration, in this paper, they consider B minus L is never conserved charged because it's broken, for example, by this kind of operator. This you see later is called Weinberg. Then they assume there's a pre-assisting asymmetry in the right-handed electron. That means you can represent a baryon electronic charge in terms of a two-symmetry of the system. Again, this you set to zero because hypercharge is never broken. But then they assume that after that, this becomes slow. B minus L becomes conserved again. So below this temperature, B minus L is conserved. So you can represent bionic charge in B minus L, and okay, this is zero. So you notice that B minus L is no longer zero but proportional to this asymmetry in the right-handed electron. So you can have this possibility. But the key is to introduce this kind of interaction. And also, after that, this should become conserved again. That's the key point. So how many, what are the time we have? So this is another point. I cannot see the chat, but... So this is a standard model as well called plus, plus, plus. So again, we extend this to study with additional E1 charges. But maybe I will skip this slide because I still have a more important point to cover. So these are the summary that we have, that we mentioned earlier. I will also skip. So now I would like to discuss killing two puzzle with this so-called leptogenesis scenario. So this is a brief summary. We know that standard model neutrinos are massive. At least two of them are massive because we know the... for neutron oscillation, we know the mass differences, although the absolute scale is not known. And also there's a phase which violates Cp. If it's zero, if it's not equal to zero or pi, that means Cp is violated. But of course, the measurement cannot pin down this thing, whether it's violated or not. And then type of neutrino mass. So first of all, if standard model neutrinos are only like degree of freedom, as we have seen earlier, we can only write down the marijuana mass B minus L, is violator in this case. And so if... since this doesn't conserve standard model gauge invariant, so we can construct the operator-wish-conserve standard model gauge invariant. This is a so-called Weiber operator. So they are non-renormalizable. And in particular, this require a new physics which comes in at the so-called seesaw scale. And if there are more than one... there are more light degree of freedom which are stirred out under... they do not interact under standard model gauge interactions. So we can actually construct the so-called Dirac mass. This is a new light degree of freedom. In this case, B minus L can be conserved. And of course, for standard model gauge invariant, we can have this kind of operator-wish-conserve, which is the Yukawa term. This is renormalizable. Or we can also have the seesaw mechanism at play. So by introducing the new singlets under standard model, or some new doublets, but under some new SU2. This is not the standard model SU2. So this is nice because we can still have the seesaw mechanism to explain the light neutrinomas. But for this kind of mechanism, we need a very, very small 10 to the minus 12 of Yukawa coupling to explain the observed neutrino mass scale. But additional symmetry is required. For example, if you want to have this mechanism, we have to forbid this term. So normally, people introduce new symmetry. For example, this right-handed new carrier Z2 and 5-carrier Z2. So you cannot have this operator. You can have this operator. And this is naturally. If you have a new SU2 doublet, this is naturally forbid by this new gauge symmetry. Of course, we can also have, in this scenario, so-called quasi-derug. That means B minus L is still violated, although by only a bit. That means this derug master is split into two Majorana states. But these two states, their mass is, the splitting is proportional to B minus L relation. And so the seesaw mechanism, if you look at the seesaw operator, it's nice because it solves the first problem. The light neutrino is very light. If this seesaw scale is very heavy, because it's very much proportional to the wear for Higgs square over this big scale. And we can also have this operator, also B minus L by two units. And we can have a new Cp faces in this coupling. And in principle, in this lambda, is hidden some new heavy degree of freedom. So in particular, this new heavy degree of freedom can be responsible for generating a cosmic bio asymmetry through the so-called leptogenesis. So of course, the bonus, if you consider this operator, we can have neutrinos double beta decay. So this is just to mention this mechanism is proposed even before the discovery of neutrinomas. And what they depends on the B violation, where does it come from? It comes from the B plus L violation in the standard model that we have seen earlier. This is the so-called electrolytic spiron interactions sometimes. And this is the windows of genesis we have. You have to be after inflation. And before BBN, these are biogenesis windows. And for leptogenesis, since it's realized on this B plus L violation interactions, so you have to happens above 130 GB. So the windows is slightly smaller. So the dish we would like to cook is, how are we supposed to generate this bio number in this regime before the BBN? Okay, so for the Myrana seesaw model, so at a three level realization, as we go above in temperature, above this scale, we can probe into the content of this operator. What is hidden in this lambda? We can have three possibility of fermions singlets or fermions triplets or scalar triplets. So these two have SU2 gauge interactions, but this will be completely singlet under the standard model. So we have all the ingredients for the leptogenesis, biogenesis through leptogenesis. We can have this new particle, they can decay. So this, they can decay in a B minus L violating way to lepton or Higgs, or this can decay into two leptons, for example, or we can have a mixture of the cases. And in particular, all these decay can be CP violating. That means the decay to particle, to leptons and anti-leptons is different, the rate occur at different due to the complex space that we have here. But in particular, we need at least two generation of these heavy guy, or we can have only N and Phi because we need a relative phase to have a CP violation. And here, I say has to be smaller than zero. In fact, we need to generate more anti-leptons than leptons to get a correct binary number violation. And then we can have a late decay. Late decay meaning that this particle, you stay long enough. So when their lifetime is equal to the hardware rate at this temperature, so this temperature is much below their mass. So that means they decay very late. So that means they are out of equilibrium because you have some overabundance compared to their supposed to be equilibrium density. And then finally, what we have is the binary number will be related to some coefficient about one third to the B minus L charges of the, this is the conserved charges. But since it's broken slightly, so we can generate non-zero charges here. And it's equal to the negative of this, because notice that B minus L, because they have the opposite side is the lepton number. That's why they actually need more anti-leptons than leptons to obtain a correct sign for baryon asymmetry. So then this ATARC capture these dynamics. How late is the decay, for example? So it can be one is the most efficient and zero is a very inefficient. So one, that means that you can have a super late decay. That means that you don't have the wash out. So everything, all the X decay region, you'll be responsible for this baryon asymmetry. Okay, so the question's next is what is the lower bound on this scale? We have upper bound from the neutrino mass. Neutron mass have to be smaller, and this C has to be perturbative. But from leptogenesis, is there any lower bound? So I'm way behind time, but I will speed out a bit. But here, I think this is important point I want to stress. There's actually a restriction come from neutrino mass and the required CP violation. So we can imagine CP violation come from the loop collection and the tree level. So this is a phase. This guy have to go on shell in honor to generate the so-called phase for the CP phase. Besides the coupling, and also we need to have this so-called canopy phase. So if we look at the CP violation for leptogenesis, we notice that there's a neutrino mass diagram hidden under the CP violation. That means the CP parameter has to be proportional to neutrino mass. So if we require to generate sufficient biome asymmetry, that means we consider the most efficient case and this is what we need for the CP violation. This implies a bound on the mass scale we need. This is so-called Davidson-Imbraa bound. So this mass scale of this guy, which decay cannot be very, very light. Of course, there's a caveat. So actually there's another contribution to this CP violation. This doesn't violate the total B minus L, but only violate the leptom flavor or B minus L flavor. So if you represent this by the effective operator, this is a dimension 5 operator for neutrino mass. So this dimension 6 is not constrained by neutrino mass. So they come in from here. So what I changed on the diagram is the arrow. So this is B minus L violating. This is B minus conserving. And this operator contribute here. So it's not constrained by neutrino mass. Although this contribution, if you sum over all the flavor, it goes to 0. But this flavor is important because tau, you cover interaction, get into equilibrium at this temperature. So that means we cannot sum over flavor because the thermal buff distinguishes the flavor. So we can contribute to the CP violation. And it has been shown by this study that we can actually reduce the scale to 10 to the 6 in the Type 1 C-Storm model, including this contribution. So, okay, another caveat is to say that if this state are not very heavy, in principle, we cannot, we have, they will play a role. In particular, even if they are decay, we play a role. The decay of the Ni, which is lighter, Nj. So we have this additional contribution. In particular, if they are much different, is of the order of their decay, then we can get a super big enhancement in this CP violation. This is from the perturbative bound. We have to be smaller than one half. But in general, they can be a lot, they can be much bigger, but still smaller than one half. And in principle, in this scenario, there's no lower bound. It's only lower bounded by, before the B plus L interaction frees out at this temperature. So this slide, I will skip. And there are other ways for leptogenesis. For example, if you suppose that in the windmill operator, instead of the standard model Higgs, what is the second Higgs doublet, which will obtain a small wave, small wave compared to the standard model Higgs wave. You see that this CP parameter is enhanced by this small wave. So you can actually reduce the scale. So you can see the scale is actually reduced. Yeah, basically it can reduce to up to even 10 to the 4. 10 to the 4. And then of course, you can also think that neutrino mass, instead of coming from 3, it comes from the loop. That means translate to the CP parameter. That means the CP parameter is enhanced by this loop factor. So at one loop, for example, one of the models, so called scotogenic, where neutrino mass comes from one loop. So the same thing, you can enhance the CP parameter. You can reduce the scale for leptogenesis. And then there's also a very big industry where they are studying light end scenarios. Sorry. So in this scenario, the physical mechanism is completely different. For example, in this scenario, you are having an end oscillation, this so-called the right-handed neutrino oscillation. So in this case, the end is actually at GVV scale, but they start to oscillate much higher than the 130 GV. So that means they can store asymmetry. This oscillation conserves a total B minus L, but they violate individual B minus L in the flavor end. That means the other sum is 0, but the individual, this doesn't have to be 0. That means the B will be non-zero due to the B plus L violating interactions. And there's also an idea called the Higgs decay. So in principle, Higgs can decay to this GV scale, right-handed neutrino and leptons. So this decay also had to happen above the 30 GV because to generate leptons asymmetry, which can be converted to binary asymmetry. But this, you need a thermal effect to have CP violation. But this, of course, this effect is there. So this mechanism is also possible. And so please interrupt me if you think I should stop, because I think I am... You can start wrapping up. Sorry? Should I continue? To start wrapping up. Okay. I will try to mention give me 10 minutes for everything. So for another idea, this I will just mention the basic idea. Because we have looked at the Majorana seesaw. What about Dirac seesaw? In this model, it's also very more exciting because in principle you can... B minus L is conserved. So the compensating B minus L is actually stored in the dark sector. So it's possible that you can generate together the asymmetry in our sector and also in the dark sector. So you can naturally explain the similarity between barrier and dark matter energy density. So how to test this model? So in the high scale, we can... That means that experiment cannot reach the energy of this operator to see what is inside. So we can at best have this so-called circumstantial or supporting evidence. Laptop number violation or CP violation. So I just want to say, usually in the news, okay, they are measuring this HIN at 9-0. Of course, as you see later, this is not directly connected to the CP phase for leptogenesis, but still it's very important HIN. That CP is related in the leptom sector. And then out of the green condition, basically impose the upper bound on the treatment mass. This is consistent with cosmology. So basically in the dark one seesaw, if we look at the composite cover in the... This is a leptonic mixing, which we can measure. And the neutrino mass, which we can measure. We still have this high scale parameter. This is the right-handed neutrino mass. And this is the so-called complex rotation matrix. They have three moduli and three phases. So all these six plus three parameter in principle cannot be probed by experiments because they are at a very high scale. So that means that in principle, even if we see some CP phase in this mixing matrix, so all we don't see CP phase, it's possible leptogenesis can occur because there are phases in this R matrix. So there's not actually a direct connection. But if we consider leptogenesis below this temperature, we can see that we cannot sum over flavor. That means a U matrix actually will appear here. So that means the phases that we measure can actually contribute to the leptogenesis. But still to decouple the phases here and here is not trivial because we have no idea about the phases here. So can we probe all these... In principle, at low energy, we can probe all of them. So in principle, these two operators will be generated in the type 1C. So once they are modeled, these two are connected, are generated from the same model. So we can measure this. This will be in the neutrino mass. And this you come into the so-called deviation from a lepton unitarity because we have these new states. So the leptonic mixing matrix won't be exactly unitary. But this violation of unitary in the type 1C depends on the scale of the C-solve. It will be very suppressed. And the current bound that we can bound is very big. That means in order to probe all these... Yeah, it's probably not so soon to probe this operator. Or probably very difficult. So we can also go on a more fundamental model where we have more correlation between the parameters because we have less parameters. Then leptogenesis can be an output or constraint for these kind of models. Skip here. For low energy scale, I just want to stress that which means that experimentally we can reach this C-sauce scale. That means we can study this lepton number violation or C-pullation in collider. And finally, I will talk about there's also an idea that we can see the gravitation away from this B-minus L violation. If they are violated spontaneously at high scale, we can generate cosmic string. And this cosmic string will lose energy through emitting gravitation away. So depending on the breaking scale, we might be able to see it. And this is an idea if this B-minus L breaking is still at high scale. But this scalar response here for breaking is a very long list because they decay only to this light. For this, they consider light random neutrino. So we can have some modification to the signal, to this signal. But still it's possible to observe. And this is another idea that we have recently about quasi dirac C-sauce model. So I want to say that in this model, it's also possible to test this in neutrino oscillation because this is a parameter space which is consistent with biogenesis. And they can be probed because this is a mass splitting. But each standard model neutrino is no longer dirac, but they will split into quasi dirac state. And this splitting is very small, but they are in the range which is possible to see in solar oscillating experiment or atmospheric. So this is my summary slide. So I will leave it at that. I will be happy to answer questions. Great. Thank you very much, Shang for this super complete. So are there questions from the audience? I can see one from our YouTube channel. Someone is asking whether it's possible to distinguish the three different C-sauce models in the near future experiments. The three C-sauce model, let me go back to the slide and measure the differences. So in principle, if this happened at high scale, I would say that it's very difficult to distinguish because the only difference in this three type of model is they have a gauge interactions. So that means the parameter space will be different because they are very close to equilibrium. So you need to have the parameter space will be different. You need to have when the decay has to be faster than this gauge interaction. So basically the parameter space will be shifted, but this happened at very high scale. So in principle, it's very difficult because the shift is in the high scale parameter space. So it's not easy to say it's not possible unless you have a realization of these three, this model at low scale, for example, using resonant enhancement, then it's possible if you can produce these states, they are light enough, then of course you can produce them, then it's a different story if they are light enough. And another thing is of course, at the lighter scale, since this guy carry SU2 gauge interaction, so you can produce them through gauge interaction as opposed to this single fermion. It's even harder to produce, even though they are at light scale because it only depends on the Yukawa coupling. There's a second question by Marcio. He's wondering about the connection between dark matter and the matter-to-matter asymmetry. So his question is how much dark matter models can affect these matter-to-matter asymmetries and that they are important for the models or we don't even need them at all? I think this is really a very model dependent question because at face value, these two scenarios are not related. But of course I think one of the ideas is usually because since we observe the dark matter density very similar to the baryon density, baryon energy density. So that's why usually we try to have a model which to link these two scenarios. So one idea that I show on this slide is basically to link that such that when through the N decay, you generate asymmetry in our sector and the equal amount in the dark sector. So in the mirror sector, that means you can also have dark baryons and so on. Then in mirror sector, we expect these dark baryons to have a mass very similar to our baryons which means that today, if the asymmetry, they all come from the same amount, it's a number asymmetry. So which means that naturally, the energy density will be very similar. So this is one idea to explain this. This is a puzzle. But in general, they are not related. So yeah. Okay, thanks. That's another question now from Carlos Alvarado. He's asking, so if you can go back to slide 43, oh, you're already 43. So yeah, he's asking, why do particles go on shell when thermalizing? Why do particles go on shell when thermalizing? I guess the question is at thermal masses. Yes, I'm not sure if I understand the question. Maybe Carlos, if you can develop a bit more. He mentioned slide 42, not 43. Slide 42, okay. Ah, yes, 42. Okay. So about this effect, I guess. Ah, so basically, if you do the kinematics, because this, I mean, basically, these two particles cannot go on shell if you don't include the thermal effects. You cannot have, because if you look at the kinematics, because this end is very light, but if you include thermal effects, basically, these Higgs and these Laptons, basically, you can imagine they can, they can get energy from them because they interact, basically interact with the thermal bath. So they can be, you can have on shell state even in this so-called loop diagram. You can imagine this is, you have the emitting of the Higgs absorbed by the thermal bath and then thermal bath absorbed by Higgs, which gets absorbed, same for the Laptons. So basically, to have these two particles to go on shell, such that we have the CP phase, the kinetics CP phase, we need to consider the thermal effects. Okay. There's another question now from Devasis Bora. So he's saying, for quasi-direct neutrinos, the Laptons number violation will be very small. Do you necessarily need resonance enhancement there, or there is some other way as well to get the correct Laptons symmetry? Okay. So I guess it's, yeah, I'll use the slide that I have in the end. Yeah. So in this case, I guess he's asking about this case. It's true that this is very small. We introduced a small b-minus L violation. So that's why, so basically, even if we are at resonance, so this is basically maximal, meaning a CP violation, we already consider at the maximum, the resonant. And as you can see, it's still suppressed by this small violation, because in the action of b-minus L violation, there's no splitting. This will be exactly zero. There's no mass splitting. So you don't have CP violation. So it's true that it's suppressed. But the important point is, we know what is the CP violation that we require. And in terms of this, you give a constraint on the splitting that we require. So that's how we get this relation between the splitting and successful, the parameter space where you can have successful leptogenesis. Okay. Now, Juba is asking if you could explain a bit more, how do you get the Quasi-Dirac mass range allowed for leptogenesis? Quasi-Dirac mass range allowed for... Ah, this range. I see. If you mention this range. So basically, this range is just from... Okay, there are probably... The first question about this range is obtained basically from the... Basically, the mass splitting can be anywhere. But then this is the... We superimpose the mass splitting on this parameter space. On the parameter space where we can see that the leptogenesis is successful. And it's from here we can see that these are the mass range we require for successful leptogenesis. And then now, the other question is whether this mass splitting can be observed by neutrino oscillation. And this will come from the... Yeah, I did not have the reference here. But this will come from the... Basically, if the mass splitting is too small. Smaller than 10 to the minus 18 EV. That means this mass splitting will be responsible. The oscillation length will be very, very long. So this kind of... We can only probe in the cosmology scale of the oscillation. But in this mass splitting scale, the oscillation scale will be... It will be the one which we can probe in the solar experiment or atmospheric experiment within this scale. But in any case, for successful leptogenesis for the scale smaller than 10 to the minus 18 EV square, it's not rare ever anymore that we cannot get successful scenario. But the probe will be cosmological probe. But it happens that for leptogenesis, the scale, the mass splitting scale, which we can probe in the neutrino oscillation experiment in the solar and the atmospheric. Okay, thanks. Are there any further questions? Joel, Roberto... Yeah, I have a... First of all, very nice. The colloquium was super interesting. I have a comment. I have a question. I have a doubt about when you mentioned the scotogenic models. In general, the question is, can you produce the same effects, for instance, with models for neutrino oscillation with higher loops? Because if I remember, you can generate neutrino masses up to four loops. Let's say there are three models. Are those models possible to have asymmetry? It's possible. Sorry. So, yeah, it's possible because the thing is, at three levels, we don't have this loop factor is directly connected to neutrino mass. But if you go to higher loop, that means... Because what comes in here is this diagram. Which is no longer neutrino mass because we have to include the loop factor for neutrino mass. So that means, as a result, the CP parameter is enhanced with respect to the neutrino mass, by this loop factor. If you consider a higher loop, like one loop, two loops, if you go to higher loop, that means the skill of M becomes too low. At this point, maybe it becomes too low that we cannot have successive leptogenesis because the required skill of M has to be so low, then that means... Because we know that the decay has to happen before this B plus L interaction goes off equilibrium. Probably not so favorable even if you consider too high loop. Maybe two loops is okay. Four loops is probably not so good. And also the mechanism will be more complicated because they are not... Yeah, they will decay in principle. Two depends on your particle in the loop. So this is just estimation of what is the CP parameter you can get. But in principle, it depends on more couplings. Okay, thanks. Just a very short question because also with the time. When you mentioned, there were one question about dark matter and generating dark matter also with the same mechanism. Is it possible... Because you mentioned here in all the talk that mostly you need... you want symmetries in order to produce the... or global symmetries but continuous symmetries. Is it possible to generate the same kind of effect with discrete symmetries? For instance, in the typical scotogenic, you have a set two inside the loop. Can you try to generate kind of similar effect with discrete depth of number or something like that? Yeah, I think this usually... Yeah, yeah, Z2 because usually you want to quantify asymmetry of a particle. So this is a U1 charger. So Z2, I don't think it's related. It can relate it to asymmetry of the particle. So you can have Z2 in the model. For example, these two particles can carry Z2, no? So they can be absolutely stable. So... But then the thing is when you're talking about the asymmetry carried by this particle, you still think about U1 global charges. So for example, in this case, you can imagine when you are generating B-L here and some B-L compensating in the other sector. But since these two sectors, they don't equilibrate. So that means you can think of this... If they equilibrate, you get zero again. So that means you can think of this quasi-conserved charge as a U1 store here. So as long as they equilibrate, if they don't equilibrate, basically you have this quasi-conserved charge we can evolve from zero to non-zero and then you remain there. And if this remain there and then the standard model charges compensating B-L, we also remain there. So... But Z2 is... can be part of the model, but they are not directly related to the number density asymmetry or the charge asymmetry of the model. Okay, thanks. Any further questions, guys? So if not, I think it's a good time to stop. Thank you very much, Sheng. We're super happy with your colloquium. So before closing, I just want to make an announcement. Yes. Is that we'll have a super special webinar on September 30. So it will be our seminar number 100. So that's why we have a very special invite. It will be Ray Weiss. So if you remember, he's Nobel Prize laureate. So we expect you guys to attend. So we expect you guys to attend for this super webinar. So it's September 30. So basically a month from now. Okay, so thank you very much. Sheng again. And see you all next time. Thanks. Ciao.