of guests. It's now time for a member. State is the member from Sarnia-Lampton. Thank you Speaker. I'm pleased to rise today to congratulate the local business in Sarnia-Lampton on its receipt of a very special award. Sunright markets, which is founded in Sarnia-Lampton some 33 years ago, has grown, become one of the premier grocery destinations in Southwestern Ontario with three locations serving Sarnia and London. On September the 29th of the 53rd annual Canadian Independent Grocery of the Year Awards in Toronto, Sunright, which is owned by Will and Ingrid Wilmanson, was honoured with induction into the Great Canadian Federation of Independent Grocers Hall of Fame. Despite all the success that Sunright has experienced over the last three decades, the business model today remains the same as it was 33 years ago. Employ a great staff and provide customers the choice of the freshest products and finest signature products anywhere. Owner Will Wilmanson can still be found several mornings each week at the Toronto Terminal, Food Terminal in Toronto. Often he is the first to arrive at 3 a.m. It's because of that passion for quality that Will ensures he brings home the very best produce for Sunright customers. As MPP for Sarnia-Lampton would like to congratulate Will, Ingrid and all the staff of Sunright on their induction into the Independent Grocers Hall of Fame. Thank you. Thank you. Further member status, the member from Hamilton East Stony Creek. Speaker, like many communities in Ontario, the City of Hamilton is suffering from a housing crisis. Nearly 6,000 families are waiting for subsidized housing. One in five renter households spend more than half their income on rent and that's a lot of people because almost a third of the households in my riding rent their homes. And now on top of our broader housing crisis we have a pest investigation crisis. There's been a 600 percent increase in reported bedbuck investations since 2006 and there are especially serious problems in the city's social housing complexes. Community groups in East Hamilton have recorded numerous problems with bedbugs, cockroaches, ants and mice. The city's social housing agency is now spending more than eighth of its annual maintenance budget on pest control. The city's Board of Health has responded by creating a one million dollar pilot project to curtail bedbugs across the city. I hope that the provincial government will look favorably on any request the city may make for assistance. Decades of underinvestment in housing has resulted in chronic state of disrepair in the province's public housing stock. The lack of funding for maintenance is a major contributor to public health problems like pest infestation. I hope that the ministers of municipal affairs and housing and health and long-term care recognize the long-term health and fiscal benefits to improving the living conditions in our public housing stock. Thank you. Member Stinges, the member from Bradkins Springdale. Thank you, Mr. Speaker. While many of us consider breakfast to be a normal start to our day, over 1.15 million Canadian students are at risk of starting their day with an empty stomach. That's one in seven children. Newcomer families and Aboriginal children are at even greater risk. Last month, the Breakfast Club of Canada and Minute Maid teamed up to launch the first ever Minute Maid Breakfast Day in Canada. This new national initiative will raise awareness about the importance of a balanced breakfast and how it plays a role in student success. All students should start the day with well nourished and ready to learn. That is why on October 6th, hundreds of Coca-Cola Canada employees launched the Minute Maid Breakfast Day in Canada by volunteering at Breakfast Club of Canada locations in 25 cities across the country. In Ontario, nine schools in seven cities took part in the launch. They included schools in Barrie, Brampton, Hamilton, Kingston, Ottawa, Toronto and Peterborough. Paul Brennan, the general manager of the Coca-Cola facility in my riding of Brampton-Springdale and the five volunteering employees from Coca-Cola helped out an event that took place at Massey Street Public School in Brampton. The Breakfast Club of Canada and Minute Maid, which is wholly owned and subsidiary of Coca-Cola Limited, have been partners for 15 years. Minute Maid donates the juice free of charge, which they serve according to Canada's food guide. Together they support 1,455 Breakfast Club school programs, serving 25 million free breakfasts annually across the country. Breakfast Days in Canada is an opportunity to highlight this important work and inspire others to get involved and make a difference. Thank you, Mr. Speaker. Thank you for the member's statements. The member from Perry Salma's Coca. Thank you, Mr. Speaker. I have a resolution from the village of Berks Falls I wanted to get on the record. Whereas the Ministry of Training, Colleges and Universities, MTCU announced in May that it is ending the Ontario Self-Employment Benefit OSCB program, whereas the OSCB program was designed to provide unemployed individuals who are OR or are recently have been eligible for employment insurance with income and entrepreneurial support while learning to operate a small business. And whereas the program has a 94% completion rate in Muskok over the past five years, and a 76% of businesses launched five years ago remain open, well above industry candidates reported five years survival rate for small businesses. And whereas the 110 new businesses launched by OSCB graduates in Muskoka in the past five years sustain a 142 jobs demonstrating it is successful in giving participants a pathway to self employment and is also an important rural economic development tool. And whereas OSCB programs from rural regions across Ontario have similar reported similar statistics, whereas MTCU's position that entrepreneurial support is available from other service providers overlooks the fact it has not replaced the essential income support component. Now therefore be resolved that the village of Berks Falls recognizes the value of the Ontario Self-Employment Benefit Program to rural communities and requests the Ministry of Training, Colleges and Universities immediately reinstate the program to a formal evaluation which must include input from delivery agencies and participants is completed. And further that the Ministry ensure any changes to the OSCB program retain both the entrepreneurial and income support components that have made it successful. And Mr. Speaker many municipalities in Paris, South Muskoka have passed this resolution. Thank you for their members, Davis members from Bramley, Coromel. Thank you very much Mr. Speaker. Mr. Speaker today the first week of November marks 31 years since the sick genocide occurred across India. This genocide though it occurred halfway across the world many of the survivors fled India and settled in this beautiful country of Canada. In fact many of the survivors settled in my writing and these are their lived experiences. These are their actual stories what they went through and what they survived. The reality is the story and the experiences of these individuals would seem almost unbelievable if it was placed in a Canadian context and I want to paint the picture for you. Imagine elected officials in the outskirts of a capital city hiring goons, arming them with weapons, providing them with kerosene which is very expensive and most common folks wouldn't have access to. Using city buses to transport these individuals to the right residents in areas and neighbourhoods where six reside. And then using elector voter lists to identify which homes are sick residents and which homes are not.  Imagine that an independent government inquiry confirmed that this heinous act could not have happened but for the organization and planning of elected officials. This is not an imagination this is the reality what people suffered and we must always remember this injustice, lest this injustice occur again. Thank you very much. Thank you. Member from Cambridge. Thank you Speaker. I would like to acknowledge the skill dedication and collaboration exhibited by first responders in my community of Cambridge and North Dumfries on Tuesday October the 20th while responding to a fire at a rural residence. A Cambridge fire truck and Waterloo Region police vehicle were involved in a collision that resulted in both leaving the road. While responding to each scene simultaneously stressed available resources two Cambridge fire trucks stayed behind to assist at the scene of the crash while another two arrived to battle the blaze. Despite being in pain themselves from the collision firefighters from the truck involved in the collision worked with their colleagues to remove the police officer from the vehicle. Due to the critical nature of the accident region of Waterloo paramedic service and orange air ambulance were called in to respond to the accident. Police Chief Brian Larkin confirmed that that officer is now recovering at home. At the scene of the fire Cambridge firefighters was successful in limiting fire damage to the residents and later received assistance from Hamilton and North Dumfries fire services which provided more water and staff. Firefighters police and paramedics provide invaluable service to the citizens of Cambridge North Dumfries and indeed all other Ontarians. This incident shows that first responders service is not without risk and sacrifice as they work to ensure the safety of citizens and still come home at the end of the day. We should all be thankful for their work. Thank you. Thank you for the member status the member from Dufferin Callaghan. Thank you speaker I rise today to share a resolution by the township of Malankham that was endorsed by Dufferin County regarding opposing the sale off of Hydra 1. The resolution states whereas the public electricity system in Ontario is a critical asset to the economy and vital to the living standards and well-being of all Ontarians and whereas it is essential that Ontarians maintain public control and public decision-making with respect to electricity and whereas experience in other jurisdictions shows that privatization typically means consumers pay more whereas a privatized Hydra 1 will no longer be subject to scrutiny by the auditor general the ombudsman the financial accountability office or the integrity commissioner and will no longer be required to provide information or services to citizens under the freedom of information and protection of privacy act the public sector salary act or the French language services act and whereas our public electricity system currently generates hundreds of millions of dollars in revenue for the provincial government every year to help pay for public services we all depend on and whereas the sale of shares in Hydra 1 will provide a short-term financial gain for the province in exchange for a much longer-term financial loss whereas the provincial government has no mandate from voters to sell off any part of Hydra 1 therefore be it resolved that the township of Melanchthon call on the provincial government to halt the sale of any part of Hydra 1 and maintain Hydra 1 a public asset for the benefit of all Ontarians strengthen Hydra 1 by investing in the next generation and upgrading our electricity transmission infrastructure. Speaker the financial accountability officer has also noted the sales of that deal. Thank you. Thank you. Members. Members from Ottawa all the way. Thank you Mr. Speaker. Celebrated a milestone from the Francophone community of Ottawa. I had the pleasure of attending Le Bal d'Arribland where more than 400 persons gathered to celebrate the 25th anniversary of La Cité the largest French language college in Ontario. We have to be proud that in 2015 the Collège La Cité has been the home of 30 000 graduates since their opening in 1990. As you are well aware Mr. President French education has always been a priority for our government and that achievement of La Cité makes us very proud. I would like to congratulate the chair of this of this university Madame Lise Bourgeois for her commitment to French education. The night went very well and we celebrate 25 years of Francophone education and we've celebrated all the people who graduated from this Ontario of this university. We have to be very proud that La Cité is a strong open and modern college in our constituency. Thank you. Thank you Mr. Speaker. I appreciate the opportunity to reference climate change. Ontario's action on climate change and how we are working towards a cleaner brighter future is imperative to the future of generations to come. So important is climate change that Pope Francis has written a groundbreaking environmental encyclical Laudato Si and translated that means praise BGU which is a line from the canticle of the sun which is the religious song composed by St. Francis of Assisi which overall proposes a moral framework and a new way of thinking about our relationship with nature. The encyclical is the first encyclical devoted to environmentalism and second it addressed to everyone on earth not just the bishops of the church in third while there are elements of caffeine teaching it is not the focus of the church doctrine the papal encyclical calls on every person on the planet to make safeguarding the environment and battling climate change an urgent priority in the 21st century. I'm proud of our government's leadership our Premier Kathleen Wynn and Minister of the Environment and Climate Change Glen Murray are tackling this climate change head on. A low carbon economy is well underway in Ontario with a coal-free electricity transit electrification electric and hybrid vehicles and emission-free renewable energy and finally Ontario has set a long and short-term goals fighting climate change and has a seeks to be a leader in climate change by building a strong carbon neutral economy communities infrastructure and energy. Thank you. I'm proud that Ontario is working towards that end. Thank you Mr. Speaker. Take all members for their statements.


Questions about that? OK, so let's move on. I gave you a little assignment to complete this exercise. People tried it? Yeah. You tried it? Well, again, excellent. So now I wanted to move on today to discussing the basic formulation level We will study in this course for a while. That's the pattern that defines the non-immediate gauge theory. OK, so let's start. You remember that when we were studying these ambient gauge theory, the U1 gauge theory, Maxwell's theories, the theories that we studied in the last class, we allowed this theory to interact with our scalar field. Such that under gauge transformations, the scalar transforms like prime is equal to e to the power of alpha by 12 times the max, that's going to be lambda. And a mu went into a mu of e to the power of lambda. That's going to be phi as well. e to the power plus del alpha del mu lambda. And then the actual d mu phi once squared plus the kinetic delta of the gauge f squared mu was gauge invariant, where d mu minus i e to the power of lambda plus. Now, you remember also that this is the way of producing an action of a gauge field interacting with other things. It could only work if we had a matter action, which, in the absence of the gauge field, had an invariant under U1, a phase-replacient invariance. Now, given this and given the fact, as you know, that it successfully implemented in nature, or that it is implemented in nature, in the theory of invariant invariance. By the way, phi could have been replaced by a fermionic field, that on the side that we don't have a glass-glass thing would change. If it's at that part of the table. So given that it's implemented in nature, there is a natural urge for a theorist, some of whom I'm speculating for generalizations and structures, to ask the following question. Suppose we had instead a matter action that was invariant under some other symmetry group, some symmetry group that was not U1. Can we play the same game for that kind of matter action? For that kind of matter action. Can we try to devise some sort of interactions with a master's gauge field, and this matter action that knows about the symmetry group of that action? This is a natural question to ask. It may at first seem rather un-motivated. It's a question that historically came almost out of nowhere. When Yang asked this question, for instance, and he gave a talk at the Institute for Advanced Study in a minute, his motivation was that perhaps it was going to describe nucleon, nucleon interactions. And almost had stopped the talk in a little bit, because Pauli kept insisting that it made no sense, because this theory just had a master's gauge field. And the actual motivation to perform this study was not a real phenomenon, because the phenomenology was forced. The real motivation is a motivation that often links to theoretical things. That is, you see a structure, you ask, oh, you guarantee a certain matter action for it. And I ask, is this a natural generalization of the structure? Work out its consequences, and then step after working out its consequences, look back to see if it has good possibly had a reasonable physical application. Physicists function in both ways. One guy can just have a phenomenology, just by this structure. You know, the most beautiful moments of physics happen when these two come together. OK. So anyway, so we're going to try to ask this question. We're going to try to look at some sort of matter action. There is a net end under some symmetric root that's not human. So let's step back for a minute, and think about symmetries. So what is a symmetry? What's a symmetry of an action? This is for some S of phi. The symmetry of a symmetry of the action is some transformational. Phi tilde down is equal to, let's say, F a equals, leaves the form of the action I changed. What does that mean? What it means is that if I compute S of phi tilde down, then I look at this phi tilde as a way of finding it. The resultant is equal to S of phi. E actually can plug in this change of sequence. For most actions and most symmetry transformations, what you get as a function of phi will be entirely different as the function of phi tilde that we've changed up. When you've got some function of fields, you've got some function of fields, and some symmetric transformation rule, that has some transformation rule that has responsibility, we call them a symmetry of the action. We come back to the study of symmetries and the consequences of what are entities. Very soon, that's why they end up in the lecture. But at the moment, what we want to do is to specialize in particular kinds of symmetry. A kind of symmetry that's sometimes called a linearly-realized symmetry. When the fields, say, phi a transform as some matrix. This is a specialization and will not be the most general case, but it is sufficient to motivate the destruction. We're interested in that. Any kind of transformation is a symmetry. But in today's class and for what we're going to do today, we're interested in what are called continuous symmetry rules. So there's symmetry transformation rules that are labeled by a continuous parameter. And I have the property that's some value of the parameter. There's m because I make it. So an example would be the rotation. Rotation rotating by some angle and some axis direction. You've got two continuous parameters where the axis points how much you rotate by. And if you rotate by angle equal to 0, you're doing nothing inside it. But of course, we'll be looking at any chance that I do not space that. But that was just an example to give you the idea of what is continuous if you do this. So suppose I've got some class of matrices, maybe by some number of continuous parameters, which have some values of these parameters. Let's call the value of the parameter 0, dimension. Like I mentioned, this matrix becomes the ideal matrix. And in the neighborhood of those parameters, in the neighborhood of those parameters, let me denote this matrix m by e to the power i with t. In the neighborhood of identity, t is a small matrix, because t is 0 when we have identity. And I want to know what I can say about these matrices t. So now, suppose I have the first thing to say about it is that in the neighborhood of identity, m is equal to identity plus i can speak. So that is an infinitesimal version of the symmetric transformation. The identity just does nothing. And t acts on your field phi infinitesimally. So phi n is equal to phi plus i times phi in the neighborhood of identity. The sequence of sub-transmitters and delta phi if you could write the change in phi in the symmetric transformation is this object. OK. I'm interested in knowing what we can say about these matrices, though, what we can say about the collection of these matrices. OK. So what can we say? Well, one thing we can say is the following. Look, suppose I act with an infinitesimal symmetric transformation. OK. And then I act with another infinitesimal symmetric transformation, the result is. So what is the result? Well, you can use the Baker-Candell House Law for identity difference. So you would be able to write, if you wanted to, as e to phi d1 plus d2, some infinitesimal symmetric transformation times something else. And leading on it, that something else will be up to some number of d1, the combination of d1 and d2. That of d is the fullest set of infinitesimal symmetric transformations. This is an infinitesimal symmetric transformation compounded by something. We know that the product of two little symmetric transformations gives us a symmetric transformation. OK. This by itself is one of these symmetric transformations. So this must be enough. OK. So d1, the combination of d2, must by itself be an infinitesimal symmetric. OK. So we write on the set of all infinitesimal symmetric transformations by dA. This index, A, is a which transformation. We could, in the set of all symmetric transformations, we could write, I times C, A, B, C. OK. Now, I've got an F. Weinberg uses the C for this. Most people use F. Since we've been using multiple textbooks, it's a bit hard to get the information straight. Let me call it F. Is this clear? OK. Excellent. Sir, this would be the point why these two are symmetric transformations. OK.   But the whole thing is just acting by something. We're acting with something else. Look at our definition of the symmetric. It's clear that if you act once and then act again, you get a symmetric. Yeah. Yeah. OK. So I'll take this inverse times this. That's also a symmetric transformation. OK. So this was me. Is this clear? Yeah. Yeah. It is a order t squared because it's the leading order term for this, for this problem. So suppose I take this on this side. Then what you care is order t squared, order t squared plus order t cubed, so on. So it's an exact state.  What I've done is take the right constraint and expand it into power. If you want, you can expand it into power series in t in the coefficient behind this general. This term is the linear term. This is the leading order term at quadratic order. OK. So this is the whole thing. All orders in t must be a symmetric transformation. We're processing that information and order t squared. So I would be writing a commutator as a linear equation. Are we putting in constraint on f or something? No. But then the commutator of a square order product is giving me a linear order. No. You see. What we would have, suppose I put some order, suppose I put some small term type, so on. OK. This would have an epsilon. And this would have an epsilon squared. These are main two series here. They have numbers that are order one. The smallest, we'll go into this epsilon. Right, right, right, right. OK. That's a good question. That's a good question. So we're going to imagine some basis in the space of what's called gender vectors. OK. Instead of all infinitesimal transformations, instead of all infinitesimal transformations, the things that you put in the exponent, that's form of basis. OK. And then to make it near to identity, we'll make the coefficient of those bases like this one. Not make matrices like this one. OK. But the sts and dv's are some fixed bases. They have no spoil. It's not happening. OK, good question. OK. There are many ways of writing the vector character. One way, you could write it as t2, t1 plus t2 times something else. And that something else, it's a leading term, is t1 positive. So three distinct bakel candles are also OK. There's one of which is this one. If you're not comfortable with this, you could just work it out, expand with us, and work it out consistently the same. OK, you'll find that two-second order with some number here, this one. Is this clear? OK. So what we conclude is that the space of infinitesimal selection transformations, these t's, OK, firstly have not changed the form of vector space. OK, because the leading order, t1 plus t2, you know, this has to be symmetry. Leading order and extra order. Leading order is going to be symmetry, t1 plus t2 plus t2 plus t2 plus t2 plus t2 plus t2. So yes. Secondly, we've also concluded that there's a bracket on the space of this, these generators. This bracket sometimes has a liberate. It's a, you know, the form on the space of generators. And it's a metric that maps two elements of these generators to a generator itself. Yes. Is there a significant range of possibilities? How about the up and down? Yeah, I, you know, I would always follow the following in the next convention. I want, at the moment, I, I, I, I make the statement from arbitrary basis in the space of generators.  Okay? Suppose I enter a similarity transformation and lower index would transform, let's say like this, the upper index would transform the case of this. Always, I'm keeping consistent track of the transformation properties and the change of the basis of the definition. Is this clear? Okay, excellent. Now, as we've explained the symmetry transformation that is now done next is matrices acting on freelance. Okay, so these these are some matrices. These are finite-dimensional matrices. Now, it's an obvious property of finite-dimensional matrices that, okay, cyclically summed over x, y and z is zero. Why is this true? Let's check it. Okay? This is just true for matrix multiplication. Let's check it. What is this quantity? This is x into yz minus zy minus zy times x. We write this as xyz minus yz plus xyz but I'm writing this as yzx. I'm hoping to be allowed to do everything for me. Okay? Plus zyx minus xz. One of these one. The cyclically summation this guy is equal to this guy. Because I take the x round here. This is x, y, z. Okay? So these two cancel out the cyclically sum. Okay? Similarly, this is equal to this other cyclically sum. Okay? So I take the x then, I take the x then. Just as a trivial identity for matrices, it's true for any operator product that is associated. In particular, it's true for pangolin. Here I identify that it has nothing really very much to it as a fancy name. It's called the Jacobian. Okay? It's a good name.  It's true. This is what I did. Okay? So let's do something deeper. This is obvious. Anyone can do this. However, for this symmetry, that's very important. Let's try to work out the consequences. It seems to be true for any three matrices. It's true for any matrices that can be multivariate. It's true more generally for any all I've used here is associated with the product. It's true more generally for any associated product. This is a very, very elementary statement about product. The reason it's not so familiar to us is that when we grow up in school, we learn about products that are commutative. So this is of course true. Well, it's stupidly true because every term is zero. Why do we need it? It's not so familiar because we didn't grow up as children or anything. But in a happy little time. I mean, it's about as simple as a b is equal to b a for numbers. It's that kind of idea. But you see, even very simple facts become sometimes a very deep concept. Okay? That's one of those. Let's try to understand. Actually, before we do that, I'm going to introduce a notion of representation. First, we've got this abstract idea. T-A commutative, T-B, T-B is whatever. Any group of matrices in any Hilbert's face or any group of operators in any Hilbert's face that obey this identity are called a representation. A representation of the central. We will usually be interested in finitely I mean to representation. Finitely I mean to representations. Now, usually. At least for what I'm talking about now. Okay? So finite I mean to representation of this algebra. Just some finite dimension matrices. Very good. We know most of this. It's not your best you do. It's your mind. Okay? Just very definition. The very, you know, notions that go in making up the free algebra already supply you with one representation of the algebra. Okay? So this representation that I want to that I want it to link. See, suppose we've got an infinitesimally symmetric object here. Act on it with another infinitesimal. Okay? Discuss. Give me another infinitesimally. At least. But is it a representation of the algebra in the sense that we have talked about? No. Because the kind of representations that we're interested in are representations in which identity and attribute. Okay? So, of course if we look at delta t let me know. Let me just say this. Suppose I just suppose I look at the following at the following. You could try and make what I said correct by looking at what you get but as that minus something else. Try to make that correct. But let me forget. Sorry, let me erase what I was trying to say. Suppose I look at this this object here. So I've got an infinitesimal group and I act on it from the left by one there. From the right by its inverse. Okay? What does this generate? You know, what do I get from this? Okay? Now an infinitesimal order. So, an infinitesimal order what we get from this we can use beta, capital R. Maybe once we should just let me see what I get at order if I call this A and this B. So suppose I put some epsilon 1 here epsilon 2 here. Let me take this and expand it to you know what is epsilon. I'm just actually keeping the term for the epsilon 1. Here is the leading term in epsilon 1. The leading term in epsilon 1 and then also expand that to leading order in epsilon 2. So what do I get? So from here I'll get 1 plus epsilon 1 G 2 by i plus i epsilon 1 by 2. Okay? From here I'll get 1 plus i at 2 and back at epsilon 4. Okay, good. Let me work at the moment you see, suppose things would complex the epsilon's and also the things. Then I'd break that up into two different selection generators. The real thing actually. So for me these epsilon's are always going to be real. So the most general generator times A5, expand this out first. Let me keep terms that are linear order in epsilon 1. The terms that are linear order in epsilon 1 are what? The terms that are linear order in epsilon 1 are this guy and this guy, but they just can't solve it. There's simply a flex of fact that I didn't have this. This is a symmetric generator times its inverse, so that just gives us k. Okay? The term that's linear order in epsilon 2 is simply this object itself. So that's what we started with. I'm interested in the change. So the leading order in the change is proportional to one factor of this and one factor of this. If we did that one factor of this there would be no change and if we didn't have any factor of this these two things can't solve. So the leading order in that is what? It's T2 times T1. So I have to take all these products. So one term is T2, T1, I did it. I did it. The other term is T2, T1, I did it. And now in terms of identity T1, T2 is that it is proportional up to 7. Now first, to T1 combinatory action of the group on itself infinitesimally generates infinitesimally generates a linear action on the space of generators. You see, we started with some generator here. Again we acted with it acted on it with a symmetric transformation and we got something proportional to what we started with. Now it's this something proportional to what we started with. That wouldn't have happened if we just looked at this simple action. Because the leading term would have been T1 plus T2. So there would have been part of delta T which is just T2 without a factor of T1. This is not a linear action of the group. It would not have come from an infinitesimal version of the matrix delta. You know, phi is equal to matrix times phi.  So the simple action of the group from the left or from the right at the infinitesimal level does not generate a representation of the group in the sense that we define it. But this combined action of the group from the left or from the right defines at the infinitesimal level a representation of the group. And the representation is such that T1 acting on T2 is given by the top detail. So let's look at this representation in a little more detail. Suppose I had suppose I had right, so let me let me write this T2 but we did a T1 switch the sign. And suppose I want to know how T2 Tb the logical role of these two objects in the expression I am writing is different. This is the basis vector in the vector space that hosts the representation. This is the selection generator. Is this clear? You know what we get? We get it. By our definition I, F, A, T, C times Tc. This is the fact that we've got a linear action because the action of the generator on a basis vector gives us a linear combination of basis vectors. So then we've got some basis vector say Phi V Phi M. And we have the expression Delta Phi as a column. Delta Phi N is equal to I times Ta M Nm Phi M. We note this as the representation matrices Ta in the Rth representation. In a particular R, there are some labels on this. This is the matrix. And we will read out this matrix as a matrix with the lower index as the first index and the upper index as the second. So that's why it says N. Representation. Every basis element of the algebra has an associative matrix. By how it acts on the basis vectors the name of the basis vectors of the vector system. Another question I'm going to ask is what is the representation matrix for the symmetry algebra? Acting on the basis vector of generators. How does T act, T B act on the basis vectors T in? We've got the answer yet. We've got the answer yet. You see? So this particular representation has a name. It's called the adjoining representation. The adjoining representation. The adjoining representation. A. The T adjoined B. Thank you. Thank you. Thank you. Yes. Now T adjoined B. Now I need to make sure I'm using the invention that the reference uses. By the way, the reference of R. I never used the reference in today's lecture. Chapter 16 in Weinberg's point of view of theory. We have come to the chapter when he used in years when the it's the convention of most natural matrices in the moment. So I can write this as minus F A B. By the way, these F's are obviously anti-symmetric in the two lower indices. Because the commutator is anti-symmetric in the two lower indices. So I can flip the positions of these two lower indices just by changing the scale. So this is equal to minus F A B. That's F B C. And this is T B. Acting on T A as an A C. And this is given by minus I F A B C T C. The only thing difficult about this whole A is to keep track of these things. Yes. Previous or like very old that's a little better. What? A small parameter. No. Okay, I'll compare. It's true to first order entity. That's it. Okay, good, good, good, good. Is this clear? Okay, look. We have this into the bi-tube and some element of it we put the similarity transform separated another element. It's the change in the infinitesimally and we're working infinitesimally. So we want to know what is the change infinitesimally? What is the change infinitesimally in this group element? We found that it was proportional to T B combinator T A. Because of that we have a particular representation of the group where the group elements T B on the representation vector T A. You see, what's confusing about this is that both the role of the symmetry group and the role of the representation vector is being played by the same object or equivalent objects. But in our mind we have to distinguish. So this T B is the generator. This T A is the basis. Is this clear? I've decided. Now in general in general the action of T A R Let me just look at all this the action of just T A. It's given by this form. It's from both sides. So the in general the action of T A R is given by this. And so we identify and so we identify now there's an issue of eyes. Already I I I I I I That's right. You're right. So the action of T itself one way of thinking is that I just look at the action of T itself that would not be an idea. The symmetry change is e to the power density but the matrix thing is just okay then I'm consistent with my notation then I work. This times this necessarily has a line. Why does this have a line? Sir, that means that it also has a line. Because you have been symmetric. Symmetry transformation is equal to right. Well, let's meet it. You're right that the symmetry transformation is equal to the point. This symmetry transformation generates the change i times t on 5. Okay? But now when I define the action of t, there's room for this. Okay? Sir, there's some matrix t here. Okay? Whose action on t a? Well, you say it does have the action t. Maybe it has. Let me do this. Let me do this really carefully. So first let's actually find the change in t. So suppose I add here tb. I add here t a and tb. Okay sir. Wherever I see 2 it's a b. Okay. So now let me see what the change in this object is. So what I get here is i t a. And I'm sending these epsilons. Let's add the epsilons to 1. So you get i t a times i squared into tb. This whole thing we want to write as i times t something. So that tells us that... So let's write this as i times t plus i. Good. So this tells us that tb commuted at t a and better be a linear combination of t's with an additional i. So better be a tb commuted at t a is something of this form with an i. Okay correct. And now how? And now what? Because I'm writing. You see. I'm sorry sir. Just a minute. So the last one is delta t a is equal to f every c. Absolutely. Yeah absolutely. Thank you Magesha. Yeah. Magesha is a beautiful thing. So this delta t a is f a b c. A b c c. Okay. And let me compare with this formula. Sorry. Both of you right. Delta phi is equal to i t r a. And then we get the additional minus. Okay. Thank you. Sorry about the confusion. The confusion. Okay. So this is the formula for the i t r a and then we get the additional minus. Okay. Thank you. Sorry about the confusion. This was. A joint action of b is defined by this matrix t b a c is equal to this. A is to understand what is to try to understand what implication you see delta t. Delta t. Okay. See we've got the space of t. One particular generator of it. Okay. So suppose I have you know I've got a vector of possible generators t 1, t 2, t 3. Okay. So t a is a basis vector the space of generator. Now how do we write these the action of a symmetry? We write that acting on a particular basis vector we get m n phi n. What we want to do is to identify this matrix. This matrix you know in the matrix we don't write the fine set. Okay. So what we got here was that delta t was equal to minus i f a b c t c. So this t c here is playing the role of this phi. Okay. So the matrix here just has this. This is here. So we've got we've got this this object here. And we have this abstract way of thinking that that seems to guarantee that seems to guarantee that this action give us a representation. Of course it's not satisfying until we check the substance. You know check it by hand. Now this action actually give us a representation of the root. What we have to check in order to check that we have a representation group. Now we have to check is that t b a joint or t b c indeed satisfied the group a joint. Okay. So is indeed f b c a t a joint. Now this has no way. This was just a definition of structure. Okay. So with this definition that identity should be true. But now we can just plug in what this is. Okay. We can just plug in what this is here. So some bilinear identity at some identity of bilinear quantities at the f's. Now you can ask where from can I get some identity of bilinear quantities from the f's.  Okay. So of course I get t a c t from here that we see. Okay. I'll just cyclically sum this.  And then the second comment that I give is f d. I get the expressions in the f's cyclically sum. So the three of them. If you plug this into here. Every term is a bilinear expression in the f's. Because each t is an f. So this is t times t minus t times t plus f times t. Every term is a bilinear expression in the f's. Okay. And I won't try to do it on the board because it will take me 15 minutes and I'll hear the sides wrong. But I'll request you to check that this expression is precisely the Jacobian identity. You get it from the same point. The fact that on general grounds the group itself generates a representation of its own algebra is born out of algebraic grounds. Okay. By the Jacobian identity. And the action of the group on itself in the so-called adjoint representation. The action of the group on itself in so-called adjoint representation is generated by the adjoint matrices objects here. This is how the group acts on its own general practice. Also very abstract. Let's look at an upload. Let's look at one familiar example. Familiar example is that of SL3. Okay. SL3 is generated by angular momenta. You know momenta are by themselves vectors. So the action of angular momenta on the generators is the adjoint action of SL3. And so in this case the adjoint action. Okay. The adjoint action is the action. The adjoint action is simply the spin one vector representation of SL3. Okay. This is very abstract in general for any group. But for a simple familiar group like SL3. In some language you know this thing. Okay. Excellent. So this is all the abstract. All the abstract lead group theory I wanted to do. Oh, there's one. There's one more thing. Let me get one. Okay. Now there's one other construct that will be very important for us in producing this object. This construct is that of an invariant tensor. So we will sometimes be interested in taking two generators and making a group invariant. Okay. So we will sometimes be interested in taking two generators and making a group invariant of them. That is multiplying by some object. Yeah. Such that the transformation of this whole object under an infinitesimal generator is zero. Let's try to see if we can find such a group invariant for any representation of the object. Let's study this object. Let's study trace of T8. Let's study what happens when we look at trace of T8 in the representant. So now I take this object and I act on it with the action. Okay. I act on it with the action of an infinitesimal generator. I know how generators act on piece. Okay. So what do I get here? So delta of this is under let's say Tb. Under the action Tb. So this is T8, T8, everything is summed. So here I had plus here I had T8 and then the same thing as this. As we would probably expect. So what do I want? I want... No. Actually, this is actually what I wanted. Sorry. Okay. So this object is this. Now let's look at the trace of this object. Sorry. Let's look at the trace of this object here. This object here has vanishing trace if the following is true. If... Oh. And where am I? Did you see? Sorry. Just one minute. Yeah. Let's see. Wherever we... I'll go to F, A is downstairs. It's good I just compute this. Have I got my message right? Yeah. Yeah. Right. Right. This is not just possible. Yes. There are no two places to place it. I'll tell you what. Let me do the following. Let me do it. G A. Let me put an arbitrary metric here. Okay. That's actually... Suppose I've got G A here. Okay. Now what do I get? I'm going to say this. What? C. Ah, you don't want to be... Yes. C. Sorry about that. How does this look? Sorry. Sorry. Sir, just to introduce you to the GACs, if you are into this 19... What now? We don't yet know what up and down is. Okay. Let me just put an arbitrary GAC here. I'm sorry. I shouldn't... Let me come in. Let me just work out the strategy. Okay. So what do I get? So what do I get here? Well, here I get TA, commutator, TB, TC, GAC. Okay. Plus TB, commutator, TA, TC, GAC. Okay, good. This object here, up to some over-on lines and so on, is TA, F, B, C, C. D, T, D, GAC. And this object was plus F, B, A, B. Because the following was true. Suppose the following was true, that there existed a metric, such that if I use that metric to raise all the decisions or to lower all the decisions, then F was not anti-symmetric, just in the two-bottom indices. But in all three indices. Suppose that was true. Okay. Then this object, then the trace of this object will be zero. Because under trace now, so now we raise this object here. Okay. We raise this CA with an A here. And we've got TA, TB. And here we've got, we raise this object as well. So we've got, we've got this object here. Yeah. Let's say that all of these are mixed. Okay. Then we've got TT minus TT. Because of the anti-symmetric, because we assume the anti-symmetric of the Fs. So this would be the trace of commutator. And therefore we've got that. Is this here? When raised by G. Or it's equal to the F with lower by G. Okay. Suppose I exist in a G. Okay. That such that F when raised to the lower by G. Okay. Would make this F completely anti-symmetric. Okay. Then the trace of this object would be invariant under group I. Probably to try to prove here, but I'm just going to cite about the theory of groups. Now, the following is group. You know, all groups in the world can be divided into compact and non-compact groups. Groups are groups such that the group manifold itself is compact object. It doesn't go off to infinity. So an example of this is SU2. It's a space of all rotations, which as you know is a three-spill, or double-double. I mean, any other case, SU2 is a three-spill. On the other hand, there are non-compact groups. Where the group manifold itself is non-compact. Example of this is the low-edge group. But the set of boosts, you know, increases without a part. Next group is like SO2, 1. Sorry, SO3, 1. And if you draw it to some sort of group manifold, um... I think it's precise. But anyway, yeah. If you draw it to some sort of group manifold, um... This group manifold will just... If we draw it to some sort of, you know, some sort of space, this would be more like hyperboloid than a sphere. The result I'm going to code for you is the object. Therefore, every compact group is always possible. Okay. It is always possible to find the basis. Um... It's always possible to find the magic. It's always possible to find the metric. Such that, by raising with this metric, F becomes completely anti-scientific. And when you do this, when you do this, okay? Uh... For such groups, it's also always possible to find a basis for the space of the algebra. Such that the metric is just... The GAB is proportional to that. No, I'm not telling you why they should be. I just quoted the result. This is a result from mathematics. Uh... There is a nice, simple proof of this result in the appendix to, uh... a Weinberg book. Okay. Uh... I'll ask you to read it. We won't go through the proof here. Okay. So, don't worry about trying to understand why this is true. First, make sure you understand what is true. The class of symmetry generators is called compact groups, lead groups for compact groups. Okay. For which this statement is always true. Okay. That it's always possible to find, uh... to find a metric. And moreover, it's possible by a choice of basis to put this metric into the form delta A. In fact, the more general state that's always possible to find in such a metric is more generally true. Okay. So it's true, not just the compact groups. The thing that is special about the compact groups is that it's always possible to convert the form delta A. For the non-compact groups, it's always possible to put it in some similar form, but where some of the identity of this delta are minus. So it's always possible to make it, you know, 1, 1, 1, minus 1. But for a compact group, it's always possible to find... find such a... you know, find a metric and find a basis in which this becomes identity. This is the claim. It's a claim that I'm not going to try to establish in this class. Okay. This will be very important for us when we start building our theories. Okay. That's it for my incompetent foray into the theory of the algebra. Let's move on to physics. Any questions or comments on this? So now that we've tried to understand these symmetric groups in some way, now that we've tried to understand these symmetric groups in some way, when we're going back to the problem that we started up, maybe we've got this phi and we want to understand how to build a theory of phi that has some... that has invariance under the symmetry algebra does not just be 1, but we'll do some master's teachings. Okay. So now let's imagine that... let's imagine that the kinetic term of our theory took this form. Before we got the teachings. GAB phi B del mu phi where G is the same metric that we discussed before. It has to be the same metric that we discussed before in order for this thing to happen invariance. Oh. This would be the same... No, sorry. G would be the same metric we discussed before. We've got these things, but in general, these are any kind. This is just some GMN such that this whole object was invariant under the symmetry transformation of interest. For instance, the symmetry transformation is SO3. Okay. And these phi's were in the vector representation of SO3. Then this GMN would just be exact. There is some action that is invariant under the symmetry under the symmetry algebra, the global symmetry algebra that we were interested in. And now we want to try to promote this action to an action that is invariant under the gates of the symmetry algebra. So, now, what we want to do is to look at the first, the transformation to a for the global group as we discussed into the bar I that's called this lambda times phi. The lambda or let's in general say some u. So the finite one. U times phi. U times phi. Where u is some symmetry generator which does not depend on the points in space. This is supposed to that was the situation. Now, we want to know, suppose I take u and make it u of x. Actually, that was the case in the theory of the electromagnet. Okay. Suppose I take this u and make it u of x. How do I preserve it? So let's see how the important question is what happens to d mu u of x. Okay. So this object here is equal to u of x plus the second term plus d mu u of x. Object here was invariable to u being independent of d mu phi goes to u of x times d mu phi. This would preserve in that because you pull the u of x outside the denominator. It doesn't matter that it's function of x. This term however is dangerous and in general will break the invariance of the action under the symmetry group. However, what we're going to try to do is to imitate what we did with the electromagnet. We're going to try to generalize this derivative to covariate d mu minus i a mu. So imagine that we take the action to involve not little d mu but capital d mu acting on phi. Obviously, a field should transform in order that it cancels this transformation problem, this funny transformation term in the transformation of the derivative. It's sort of clear that suppose a goes to universe phi del mu is del mu u of x c, yes, u of x times u of x and i of x. So I'm going to cancel this and minus what do you think is this to be the transformation problem here?  Now this d mu, as we've done before, we break it in two terms. One that goes through on the d mu, on the phi and the one that acts hits the u. So we get a total of three terms.  So what do we get? One term is when this d mu goes over to the phi. That's just u times d mu. One term is when we get this object acting on u phi.  Okay? And then there is the last two terms. The enormity is done from here and the term with the d mu hits the u. And you can see that they are against each other. They are designed against each other. So this norm of d mu is done. It's the u, the u inverse is killed. We've got minus i times minus i so minus sign times d mu of x. So that's minus d mu of x which is precisely the plus d mu of x when the denominator hits the u of x. Okay? So under this transformation rule, under this transformation rule cover the rule that d mu phi of x goes to u of x then d mu of x. We had gauge field along with if we had gauge field along with the we are not gonna do that. So if we had a gauge field along with our actual depuis and the gauge field had this transformation problem. Then if we had gauge field along with the curriculum the gauge field has this transformation problem.  Provided that the original action was gauge invariant and the global transfer. Okay, that's nice, but it's not sufficient to construct, to construct the first of a full interacting field theory based on the symmetry group. Because the full interacting field theory needs a dynamic term, not just a matter of fields, but also and most importantly for the gauge fields. So what we need is some sort of dynamic term for the gauge fields that's also invariant under our transformation. Since you will say it in general, well, yeah, what is it? Well, some of these phi m should be some of the phi n stuff. Now, all I've said is to say that this object here, now I'm saying that we will find the whole thing that will be used in general complex. Well, used in general complex, yes. No, no, no, no, it's true, it's true, it's true. But you see, okay, this is a matter of notation. But you see, suppose you take phi and phi star as the two sets of different variables, then the generators act on phi star with a minus sign compared to the generators acting on phi. Okay, so that will come. Well, look at, let me look at specific theories between them. So this is like, this is an implicit variable. This is implicit in the statement, that this whole action was invariant under global transformations. Okay, so I'll start by writing it. That's it. Not that, that is good. Okay, let's move on. So, now what we want to see, I don't know how to go forward with what an electromagnetic measure was, if me and you are doing something like that. Okay, and what do I originally think of trying to do? It is just defining an f mu, which is d mu a mu minus, by the way, notice that this, notice that this a here, under global symmetry transformation, transforms in the adjoint representation of the h group. And the u's that are not functions of x, transforms in the adjoint representation. This, if you make a statement, would be transformation under the normative. So a is, live in the Lie algebra of the h group, and under global transformations, transform in the adjoint representation of the h group. Is this clear? So, in particular, this transformation rule implies, that if we specialize it to, to, to u of x's that are not functions of x's, in particular it implies that a is transformed in the adjoint representation of the, of the h group.  So each x should be replaced. At least the term B-mutor half-concepts are joined in representation of the value of the representation. Exactly. And that's what we've been doing. Exactly. So, now, Mangesh's question is, how does D-mutor solve it? Now, exactly as Mangesh said, you see, the transformation that we, the transformation law here, was that D-mutor transforms as a U-demu U-inverse. This is acting on U-tensile. It would transform like U-tensile. So, this object here has been devised so that this covariate derivative transforms in the actual representation. So, the field strength would play the role of the F-smallness that we add into the covariate. Yes, yes, yes, yes, yes, and all that. So, but suppose we try to develop it. Suppose we try now to generate, to define a field strength that was just D-mu-a-mu minus D-mu-a-mu. And check if it had some nice transformation property. The answer will be that it does not. But we would like some homogenous transformation property somewhat like this. But this does not have a nice transformation property. Let's check it. So, under the gauge transformation, this goes to D-mu times U-a-mu-inverse minus i-D-mu-mu-a-mu-inverse minus D-mu- It happens in this expression. See one good thing that happens is a derivative of this symmetry. But there is the other actions. There is the action of D-mu on this map. There is the action of D-mu on this U and this U-inverse. And of course, D-mu and A-mu is what we would want. is what we want. So that's like the homogeneity stuff. But there are three other objects.  And you can try your best to fool around with it, but you see nothing particularly nice. This is the end of trying to construct a field strength just by the simple generalization of the URI construction. It doesn't work. However, we don't need to try to guess to generalize it to something that works. Because we've implicitly already found the answer. We've implicitly already found the answer from Mangesh's observation that operate a DMIU acting on any object already transforms into ice. Now, since DMIU transforms into ice, the correct way to generalize the U1 thinking goes as follows. See, suppose I had DMIU minus IAB. Suppose I, let's go back to the U1 case. Suppose I DMIU minus IAB acting on a field, think 5. Think of it first by DMIU and then think. Suppose I act with DMIU minus IAB times DMIU minus IAB minus the other way around, minus DMIU minus IAB. Think of what I get when I do this. Just check the answer. So, there are two terms that obviously give us DMIU. There's DMIU DMIU acting on 5, that's here because it's electric. So, this gives this. And there's AMIU in U acting on 5. That's also 0 because in the U1 case, there's two numbers, it doesn't matter what we get. So, of the four terms here and four terms here, two of them cancel. Two of these cancel and two of these. What remains is the constants. We don't have all derivatives acting on 5 and you don't have no derivatives. What remains is the term in which, what remains is the term in which, okay. And then you can also look at terms when you have two derivatives acting on 5. Those also give you 0. Can you see that? You can have AMIU tell you and it appears in the same sign but minus because of here with AMIU. Let's go back. Okay. Suppose I have, I look at all the terms with one derivative. We get here minus I AMIU tell you and here minus, let's say minus I plus AMIU tell me. Okay. And then clearly the same thing from this side. This is a method. So this cancels that. So what are the terms that cancel? Any term which has both derivatives acting on 5 cancels. Okay. Any term that has these two A's with no derivatives taken anywhere that cancels. So what's left? What's left are the terms with one derivative acting on, with one derivative and one A. The derivative of acting on A, those are the only terms that are left. Okay. Yeah. I'm dealing with the U1 case. Okay. So those are the only terms that are left. You see two derivatives acting on 5 which act cancels. One derivative acting on 5 which act cancels. So the only term that's left is 5 itself with no derivative acting. Okay. If the derivatives don't act on 5, then what could happen? Okay. It must be that this derivative acts on A and this derivative acts on A. That's the only remaining term that's left. And you can see that that is proportional to, that is proportional to Fb. So another way of thinking of the field strength in the U1 theory is as the commutator of two covariant derivatives acting on a field. It's a wonderful fact that the commutator of two covariant derivatives acting on a field is simply proportional to the field strength acting on that field. Okay. So since we have a nice covariant derivative in the non-immediate theory, since we have a nice covariant derivative in the non-immediate theory, that is the way we're going to generalize our field strength by generalizing this acting. What we're going to do is to say in the following. You know what I'm saying? Given this covariant derivative which was d mu minus i d mu. Okay. We're going to define our field strength by the following identity. d mu minus i d mu times the commutator of d mu minus i d mu. Then we're going to form some field phi. It's going to by definition be F mu mu acting on phi. But this definition makes sense. It must first be that all the terms with both derivatives acting on phi, or terms with one derivative acting on phi on the left-hand side of that object, vanish. But the check of that is exactly the same as what we did. The algebraic check that we did for this term vanishing in the Rewind case also works in this case. Okay. So this definition is sense of the derivative. So firstly it's sense of the derivative. Secondly, it's guaranteed to give us an object that transforms in the correct way. Because we know that each of these d's transform under wave transformations like u, d, u inverse.  These two cancel, both the transforms like a u, d, u inverse. Okay? So this definition is guaranteed to give us a definition such that under gauge transformations, F mu mu goes to u, M mu mu mu mu mu, is the right way to generalize the notion of the field strength to the non-qualities. Now, all that remains is to work on algebraically what we get, okay? At first you might think we're going to run the contradiction because all we're getting is what we got in the U1 case, but that's not true. It is true we get the terms we got in the U1 case, okay? It is true that we get the terms that we got in the U1 case. I think I will... Yeah, please. I mean, I need a minus i. That's the commutator. Now, let's look at the logic again. What I wanted to do is to find a notion of the field span in the Abelian theory. A definition of the field span that would nicely generalize to the null. Okay? One definition was d mu a into minus b, d mu a. That was a nice notion but it didn't nicely generalize. So we looked for a slightly better notion. So we found the most sophisticated notion of the field strength in the Abelian theory. We found that in the Abelian theory it was true that d mu commutator of d mu was the Abelian theory. And then we said, well, this is a notion that will nicely generalize because the covariant derivative in the null Abelian theory has by construction been something that transforms nicely under each other. So if we generalize this definition, then that's guaranteed to give us field strength that will transform homogeneously. In fact, in the actual representation of the gate rule. That's the definition of the field strength. There's the object. Okay. And now to complete our construction, now to complete our construction of a real theory based on this non-Abelian gate rule, what we did is very simple. We simply look at f mu mu, let's say, e f mu mu b g where g was this invariant metric that we invented on the space generators. And add that with the usual minus whatever. Okay. Now we put a minus here. Here we had a minus also. Yes. Oh, please, please. I'm sorry, I'm sorry. I don't know how to read this. Okay. This now gives us an action. That's a problem problem. It's invariant of the gate transformation. It has a kinetic term for all things. Okay. So after this, Okay. This is strictly the adjoint. This is strictly here. Yeah. Yeah. These are adjoint indices because we know that the gauge fields in a particular field strength transform in the actual representation. Okay. So any field strength f can be written as f mu mu a times t a. Okay. And we write in the adjoint representation. Okay.  These are just generators of heat. Yeah. So f mu a times t a. Actually in India, it doesn't matter. Okay. Yeah. This is not specific to the entire principle. What? This is not specific. This is not. But if this object here f mu mu can be written as f mu mu a times t. It's really because both a and computer a can be written as f mu mu. Okay. And now the action will be generated by commutation in the gauge field. The action of transformations will be generated by commu... You want to know how does the infinitesimal generate a t b action, so to speak? It acts on f and t b commutator with f. I don't need to say what representation this t a is written this way. I have not chosen, you know, the coefficients of elements in the actual representation. So let me say that any object that is in the actual representation is by definition alpha a times t a. That statement puts it in the actual representation. Puts the space of alphas in the actual representation. I don't need to say anymore about what representation those t a's happen today. Is this correct? Yeah. Written this way, there's no t a. These a's are just indices that tell you about the transformation properties under the statement. Is this correct? You mean t a squared with the derivatives of what a's happen? What? What a's. These could be generators. These are generators of this group, but it could be in any representation for the states. It could be in the fundamental representation. It could be in the actual representation. It could be in any representation. You see, these f's, f a's are defined to be the coefficients of t a in the expansion f is equal to f a times t a. I don't need to say what representation. Because the action of f is communicated by t b. In whatever representation, write f a. To say it another way, this is now an abstract group theory. Is equal to f a t a because the generator is in the group. And the action of the group on this is in the actual representation by combination.  I don't need to say anymore. This could be just an abstract. Is this correct? But since f mu over here has been built up of the derivatives of the A mu and A mu use themselves which are the transformational representation. No, wait, wait. These A mu's here. Acting on any particular matter field. Had this expression t A mu A t a. But t a is for the representations for that matter field. But you see, we could have a theory with three different matter fields. Transforming three different representations. So if we were constrained to build a kinetic term for the gauge field in which the representation in which the matter appears, it would be in trouble. We would not be able to write such a term. But we love those terms. This is a very important point. The gauge field itself transforms in the adjoint representation. Independent of what the matter is doing, the gauge field can be written as f a t. It is then a certain transformation. Transformation by commentator with t b. And this transformation property is independent of what representation the t a is going to be. So the gauge fields transform in the adjoint representation. Therefore, gauge fields can be written as f a times t a where that t a isn't absolutely any representation. Is this clear? I will very quickly clean up about the work. So where are we? So now we have got some sort of non-imperial gauge theory that we can that we can that we can use as a basis for some mathematical decision. So what is our action? What is our mathematical? As is the aim. And by the way, now that we've got this theory we can look at it even in the absence of matter fields. Just like because the actual aim is in the absence of matter fields. We use matter fields as a motivation to build up the statement.  We usually use another dimension. That's 2 by 4. And just let me look at it again. Let's start with our path integral and once again try for a tablespace interpretation of this part of the general of j times y squared. So we're going to do the same things as we're going to do basically I mean exactly the same things as well. But there are two interesting twists to what happens on the inner of the case. The first twist goes as follows. There's this metric GAB. Remember the idea that for compact gauge groups it was possible all the ways to make GAB problems. Or it was possible to make GAB a diagonal matrix with 1's or minus 1's. But for compact gauge groups it was possible to make it with just 1's. Now what happens when you make it with just 1's? What it means is that all kinetic terms have the correct size. All terms have a i dot a dot in a square with a positive plus. If this GAB has some negative signs then for those bees we will get the wrong sign ahead of it. Now this is exactly the kind of thing that we discussed in the last class. It's not good. It leads to unbounded energies from leads to problems with unit activity. Again this whole trick of generalizing to the study of non-annual gauge theories only works. We're able to find a metric of of of ABs that is positive difference. And as we discussed before that only works for compact groups. Though for formal purposes it's sometimes useful to study gauge theories based on compact groups. At first approximation, at first unless you do something specific you don't think, if you don't think too much certainly for applications. We should restrict our attention to gauge theories based on compact groups. That's the first thing. The second thing is more simple which should also be okay so let me now imagine that I'm working with a compact group and that I'm working on a basis such that GAB is proportional to LTIP okay and F squared means F mu nu A F mu nu A everything contracted to it. Just to make like an approximation. What about the next? The next thing is the Hilbert-based interpretation of this model. So we want to imitate our discussion of the of the of the millions. In the imitation of our discussion of the millions theory what do we find? We find that the Hilbert-based interpretation is given by the following hybrid object. The Hilbert-based interpretation is given by a path integral of A0 this is A0A times e to the power minus I I'm HD What is H? H is having to interact with a Hilbert space that is made out of functionals square integral functionals A, I for all things nice but what is H for particularly? H for particularly as before so more particularly what is what is this object? Well first I got the Lagrangian the Lagrangian what is F0I F0I sum over implicit the whole thing squared by 2 minus FIJ FIJ whole thing squared and whole thing by 4 no change here from the Lagrangian we'll just write it in the series okay and then what is the canonical momentum with respect to A0 IA the canonical momentum is simply F0I as before with more momentum so the Hamiltonian as before is equal to AI dot F0I minus the Lagrangian before what we're going to do is to write this in terms of the canonical momentum the canonical momentum was F0I F0I was del A0AI minus del IA0 minus IA0 this is what this was our definition of the page now 300 in terms of majority as it starts to everything has a right right thank you so this was now the only question is in plus or minus ah yes so if we written it in terms of matrices it would be this I just have to pull out the coefficient of I mean, it's a difference it was del mu A mu minus del mu A mu minus I A mu then we substitute in this A mu is equal to ah A mu A T A and then we use the commutation relations so this guy these are all probability A this guy now is here so this would be A mu C and then this is F V C A A with an I plus plus plus F V C A times A0P is to write this object as this plus this minus it so once we do that this F0I F0I will combine with the F0I F0I and the Hamiltonian so our put Hamiltonian here will be F0I ah squared by 2 F ij by 4 ok and then we will have we will add it and then subtracted this object so plus ah del I A0A minus F V C A A0B A is A I C ah times F0I and let's raise the commutation what do we get is it now the integral of the integral over A0A other than action wave functions ok is the generator of shifts in A I so all of this does is shift A I with this quantity and integrate over that what this does is takes psi of A I A and then has that quantity plus del I A0A minus FV A A0 ah B A I C so this will just be like a shifting of like when we do a Gaussian integral put another like we do just do a shift like change of variables to ah this is our change of value it's like the projection for we have to identify what this projection is. Now this you can check is simply A I A times plus Di of A0B with the covalent derivative acting in reaction representation I see reaction of an infinitesimal gauge transformation on the A I fields how do you know that when we know how A I transform under finite gauge transformations we know how A I transforms under finite gauge transformations all you have to do is to take that finite action and specialize it to the infinitesimal case and you pass once again the inverse base interpretation of this problem is very simple once again the inverse base interpretation of the problem is very simple one it's projecting the wave function onto the gauge invariant sector it's projecting the wave function onto the gauge invariant sector my gauge invariant is now defined by gauge invariants the correct gauge invariant for the non-variant any questions or moments of this I'll let you go 3-4 minutes any questions or moments of this ok now in the last 2-3 minutes 3-4 minutes of today's class we discussed let me say it anyway we discussed an issue concerning these pathet techniques these pathet techniques involving non-variant gauge theories an issue concerning them that came up last class I think if you were uncomfortable about the gauge invariant sorry do you have something going on here so in the last bit of today's class I want to discuss the pathet-proport procedure let me do the problem with this with this so we got this pathet technique actually ok so where are we so what we've done is very briefly survey the 3 basic kinds of pathet techniques we'll encounter in this class pathet techniques of scalar fields pathet techniques of gauge fields a billion is where there's not a billion and pathet techniques are over for me ok problem set for you by next Friday give you some basic manipulations with non-variant gauge theories to try to overcome the gain also with some programming techniques ok so we've spelled out the basic grammar of what's going to happen in the class we need to do a little more formal analysis of the gauge theories I want to discuss this pathet-proport method I want to put a little bit of calculation as possible using pathet tables in gauge theories and also the BRST point quantization the BRST Hamiltonian gauge theory after we we go through that we start actually we start using this physics we start doing more physics oh yes, this Friday is in the magazine can't we be arrested for doing that what is your schedule if we need beta classes when can you we may have a afternoon what about Tuesday afternoon Tuesday afternoon what about Wednesday afternoon Wednesday at 2 o'clock I'll have to check availability of this this class would be really useful in this class could somebody take charge of making a beauty list what if you are in the student room who's not in the theory student room who here does not see okay it's a one person somebody who sits in the theory student room can you create a beauty list on the edge of it okay, so just so that we have to change something it's easier to send out so this class no, I'm this class we couldn't try this class, what is the case this class I don't want to say it I'll have to confirm availability Wednesday afternoon Wednesday afternoon I just have to check various things that is the same we'll provide it we'll try to


Hey, everyone. Welcome to theCUBE's presentation of the AWS Startup Showcase, cybersecurity. This is season three, episode three of the ongoing series that features exciting startups from the AWS ecosystem. I'm your host, Lisa Martin. Pleased to be joined by Chris Lan, the Chief Product Officer at Halcyon. Chris is here to talk about defeating ransomware in the real world. Chris, welcome to theCUBE. Great to have you on this very important topic discussion today. Thanks for having me. I really appreciate it. Of course. Before we dig into the current state of the cybersecurity landscape, tell the audience a bit about Halcyon. This was formed only two years ago, back in 2021. What do you guys do? What were some of the specific gaps in the market where cybersecurity is concerned that your co-founders saw we could fill this gap? Absolutely. So it started out really around the time of the Colonial Pipeline incident, which a lot of folks, whether they're in the industry or just follow current events are obviously very aware of. And that was kind of a wake-up call for the founding team at Halcyon around, just the landscape was changing dramatically with geopolitical events, with economic, macroeconomic events, as well as we saw a very acute problem that was starting to rear its head. And we also noticed that that viewing kind of the ransomware risk through kind of the traditional lens and tools that existed out there in the market was insufficient in that. And what I mean by that is not that there aren't great security products, technologies and services out there that solve a lot of cybersecurity issues, but in the particular context of ransomware that because of these geopolitical events, because of these macroeconomic events that aren't transient in nature that are kind of the new normal as the globe realigns, as the globe kind of digests everything that the world's gone through the last three years, first with the pandemic and then with kind of coming out of the pandemic that conditions had changed in the risk landscape, motivations and the business model underlying ransomware was a big kind of missing piece that people weren't thinking and talking about and that this was really kind of how we talk about it at Halcyon and it's like non-governmental advanced persistent threats, if you will, where the outcome and the desire wasn't to achieve undetected access to a business and organization to steal intellectual property over years and or decades. It was, hey, how do I achieve the maximum amount of leverage as quickly as possible so I can drive real demonstrable economic returns as a threat actor. And that was very different. Yeah, you talk about the colonial pipeline it sounds like being really a catalyst for Halcyon forming back in 2021. And I think that's the timeframe. We've been talking about ransomware for a very long time in the cybersecurity landscape and how much is changing but I feel around the colonial pipeline time that's when ransomware became a household word and I started hearing about it in non-technology conversations where people were starting to become aware of it and aware that the bad actors out there had access to emerging technologies and that it was so lucrative for the bad actors. Talk a little bit about the strategy and the vision of Halcyon given the fact that there's so much, you talked about the geopolitical climate, the macroeconomic climate and the fact that there is so much return that these guys are getting. What's the vision and strategy of Halcyon to help organizations defend against that persistent threat? Yeah, I think you're asking a real fundamental question to our story and just to what we observed in the market and that is because of the nature of what is driving this threat category or this scenario of cyber risk that a couple of things really crystallized and that is for us, there is no perfect security and I know that I have immense amount of empathy for customers and partners. A lot of my being in the cybersecurity industry for 30 years personally, I am very good personal friends with a lot of chief information security officers Fortune 500s and I have a lot of empathy for the job that they're responsible for and the job that them and their teams do but there's not an easy but or there wasn't an easy but and there wasn't an approach to this particular crystallizing form of cyber risk that was really taking a resiliency first mindset a failure is inevitable mindset and working backwards from that worst case scenario if you have a ransomware event or you are targeted that may be one of the worst days if you're a CISO or you're on a security team that you're gonna find coming into work and having that working backwards from that inevitable failure, that assumption position that no matter how sophisticated your security program is no matter how many frameworks you followed or regulations you're adhering to that you can't put enough gates in the system and that products that existed out there that were designed to try to help mitigate some of what was unique and different about ransomware whether it's the business model or whether it's the technology involved wasn't taking that approach and so when we created the solution in the company we really started from that failures inevitable mindset let's work backwards with the goal being that we're gonna detect and protect as much as we can and surgically focused on ransomware but we're also gonna go beyond that and have our technology and our solution really assume that recovery mindset so that if and when you have that ransomware event that impacts your organization that recovery is that first order goal and doing it safely, doing it efficiently and doing it quickly because in most of these scenarios like with customers and prospects we've helped is that downtime isn't oh our employees can't be productive that downtime is I'm losing $2 million a day in revenue because I can't quote sales deals because my ERP system is encrypted and not accessible. Those are the things.  Not an are we gonna get hit? It's really no organization globally is safe from it. Share a little bit of the stats of ransomware and specifically ransomware as a service with the audience so they really understand the gravity of the situation and then we'll kind of dig into why there's some vendor confusion out there. Yeah so you know I will say as somebody that you know has been a part of solving really hard cybersecurity issues for the majority of my career is first and foremost, you know the cybersecurity industry can be a victim of its own habits to some extent. And I would say the first thing that you hit on that is vendor confusion. Cyber security has become a multi-billion dollar industry. There's products and services for everything for specific cybersecurity risks. And so it creates a lot of just you know if I'm a practitioner, I'm a chief information security officer it creates a lot of noise. You know just being able to sift through and understand what is actually something that'll help me mitigate this risk. What is a control that I can put in place is actually effective because we as an industry you know want to elevate cybersecurity to more of a business conversation which is you know in a risk conversation at the same time, not all risks are created equal not all risks exist the same when you're theoretically talking about business risk and then you transition to what happens when that risk happens in the real world. There's a disconnect there. Additionally you know in our quest to try to make cybersecurity more accessible whether that's to you know bring more people into the field and make it a more accessible career choice to demystify a lot of the technology because it's a very technical field. You know and to meet, you know we've pushed out and the motivation was great. Things like you know NIST standards and MITRE you know frameworks for understanding the problem space and we get so fixated on are we following the playbook? Are we following the standards that somebody has laid before me? We're not questioning and we're not taking a step back from those and de-confusing the problem that you have you know that these organizations face and how is the most systematic way of going about preventing and responding to these types of problems like ransomware. And so it's created the kind of this perfect environment of high noise, low signal when it comes to understanding the risk but more importantly what you can do about the risk. And so that you know that is probably one of the you know one of the unique aspects of where we are today. And so you have you know thousands of cybersecurity solution providers out there saying this is how my phishing solution stops ransomware. This is how my XYZ solution or service stops ransomware and it's just the reality is it's not as you know at least with broad based security controls that are there just to you know to try to give you as much coverage as possible. You know they don't really take into consideration the unique attributes of ransomware of the business model of and you talked about it you know the difference between the ransomware operators the threat actors that are engaging in the actual you know ransomware event things like initial access brokers it's just a very complex topic. And when you're bombarded as a practitioner and as a CISO with all of this you know you want to make it easy you want to try to kill as many birds with a stone so to speak when you're deploying product or you're engaging vendors and it's unfortunately it's a dynamic you know threat it's a dynamic risk it's a very complex kind of you know attack or supply chain and as a result of that it's created an environment that's made it really hard for you know for organizations to get their arm around it effectively so that when they do and if they do actually have their event they know how to respond they know what the playbook needs to be and we can actively you know actively start to see progress in shrinking this average time to response of being 24 days now to being four days. You know not being you know in that you know hypothetical example where you're you may be you know not able to recognize or to book you know over a million dollars revenue a day you do the math on that's 24 million dollars if the average recovery time is 24 days so you know that's got to be shrunk and in order to shrink that that was really kind of the call to action that we saw when colonial first happened. Yeah that the response time the recovery time is absolutely critical because what you talked about that even the reduction from 24 days to response time to four is huge because ransomware is I read a stat I think it was from Cyber Ventures in the last year or two where once every 11 seconds a ransomware attack happens. So the threat is there but it's an existential threat to organizations globally ransomware as a service is very lucrative. How should organizations be thinking about ransomware with the failure is inevitable mindset? It sounds like to me it's a mindset shift and a tool shift to help start dialing down the vendor confusion but how should organizations really be approaching ransomware from a response and a recovery perspective and how does Halcyon help them achieve that? Yeah so I mentioned this in my answer a few minutes ago first is viewing it more through the lens of ransomware is viewing ransomware is more of the next evolution of advanced persistent threat. So APTs which is the acronym for advanced persistent threat I don't wanna get too jargony given that we're really good in the cybersecurity industry at clinging onto our jargon but was really back in 2008-ish when Google Aurora published their research on basically foreign governments getting persistent access to networks whether it be defense industrial based suppliers or whatnot to steal intellectual property.  So first and foremost viewing it as kind of that next evolution where it may not be governments that are directly engaging in the activity but there is relationships there disparate plausibly deniable relationships there. And so they're very advanced they're very focused on the outcome and the big differences whereas maybe with APTs of old it was governments or directly affiliated government groups trying to get persistence trying to steal intellectual property the goal with ransomware is pure dollars and cents and it is with a secondary being maximum leverage to be achieved so that they can reach their goals. The second piece beyond kind of just that threat actor is understanding that the business model has a big big difference here. It's not just trying to disrupt operational continuity to target you it's not just about trying to get access and stay beneath the radar and not be too disruptive and hope you don't get found. It is much more let me get in, let me get out let me get the keys to the kingdom from a data perspective or lock up your operation so that leverage is achieved. And then finally as I mentioned earlier is you got to really kind of security organizations and security leaders got to really change their mindset and it's really hard especially when you're problem solving but work backwards from what type of outcome if you're gonna approach this as failures inevitable and that you don't have 100% of control if you're gonna have a really bad or really one of your worst security response days because of ransomware but you can control, hey, when and if that day happens here's what I have in place proactively that gives me mitigating controls or visibility and operational the tooling so that I can move from detection and protection immediately into response and recovery seamlessly and that you have not only the tools there but you have the partner there that, you know there are little things that we look over when we were talking about cybersecurity like, oh, there's gonna be other stakeholders involved I have a ransomware event the first call I'm making is to my cyber insurer once the cyber insurers involved you don't get to call the shots anymore cyber insurer calls the shots I don't care whether you're CISO or Coca-Cola or another enterprise not to, you know that's just a random example but it doesn't matter how big the organization is you, you know, now you layer on top of that the new SEC reporting requirements that were just rolled out recently in Black Hat and it's a very dynamic environment with a lot of stakeholders where a lot of the control and a lot of, hey, this is the most important goal everyone else we're working backwards like can get really lost in the shuffle because you have other players involved that are worried about other things not that they're not in valid things but they're not the same things as the organization that's dealing with a ransomware event they're thinking how do I safely get back up and operational as soon as possible and that is not necessarily how the cyber you know, the number one priority for a cyber insurer or the number one priority for a retained general purpose incident response partner that you've pulled in. Wow, Chris, this has been a fascinating conversation thank you for first of all telling us about Halcyon the catalyst to launch the company being Colonial Pipeline makes perfect sense how you're really helping organizations to defeat ransomware in the real world which is as we talked about an existential threat it's a mind set shift, a tool set shift we appreciate your insights and your time Chris we're gonna have to have you back on the program because I feel like we're disrupting the surface here Yeah, it's an interesting topic and I really appreciate the time sorry we couldn't go a little bit deeper but happy to be back anytime. Next time, all right Chris thanks again we wanna thank you for watching the AWS startup showcase Cyber Security we appreciate your time and your insights for Chris Lam I'm Lisa Martin, we'll see you next time.


In response to the charts that describing the benefits of statin drugs only in terms of relative risk reduction, as a statistical deception created to give the appearance that statins are more effective than they really are, it was pointed out that the problem with describing things in terms of absolute risk reduction, or number needed to treat, is that they can depend on the duration of the study. For example, say there's some disease that has a 2% chance of killing you every year, but there's some drug that cuts your risk 50%. That sounds amazing, but then you realize that at the end of the year, your risk will only have fallen from 2% down to 1%. So the absolute reduction of your risk is only 1%. If 100 people were treated with the drug instead of two people dying, one person would die, so you'd have to treat 100 people to save one life. So there's like a 99% chance that taking the drug all year will have no effect on you, either way. So to say the drug cuts your risk of dying by 50% seems like you're overstating things, but think about it. The benefits accrue over time. If you have a 2% chance of dying every year, year after year, after a few decades, the majority of those who refuse the drug would be dead, whereas the majority who took the drug would be alive. So yeah, maybe in the first year you took it, there was only like a 1% chance it would save your life, but eventually you could end up with a decent chance the drug would save your life after all. This is the very reason why using relative risk makes sense. Absolute risk changes depending on what time frame you're talking about, but with relative risk you know that whatever risk you have, you can cut that risk in half taking the drug. Now statins only cut your risk 25%, but since cardiovascular disease is the number one killer of men and women, if you're unwilling to change your diet, that's a powerful argument in favor of these kinds of drugs. You can see the same kind of dependency on trial duration looking at the postponement of death by statin use. How much longer do you get to live if you take statins? The average postponement of death has some advantages over other statistics because it may offer a better intuitive understanding among the general public, whereas a stat like number needed to treat has more of a win-or-lose lottery-like quality. So when a statin drug prevents like one heart attack out of a hundred people treated over five years, it's not like the other 99 completely lost out. Their cholesterol dropped too and their heart disease progression also presumably slowed down, just not enough to catch a heart attack within that narrow time frame. So what's the effect of statins on average survival? According to an early estimate, if you put all the randomized trials together, the average postponement of death was calculated that it may be three or four days. What? Who would take a drug every day for years just to live a few days longer? OK, well first let's try to put that in context. Three or four days is actually comparable to the gains in life expectancy from other medical interventions. For example, that's nearly identical to what you'd get from highly effective childhood vaccines, because vaccines have been so effective in wiping out infectious diseases. These days they only add an average of about three extra days to a child's life.  That's why we vaccinate. It just seems like such a small average benefit because it gets distributed over the many millions of kids who get the vaccine. Is that the same with statins? An updated estimate was published in 2019, which explained that the prior estimate of three or four days was plagued by important weaknesses in the actual average postponement of death was actually ten whole days. Headline writers went giddy from these data, but they didn't understand as that this was only for the duration of the trial. So if your life expectancy is only five years, then yes, statins may only increase your lifespan by 10 days. But statins are meant to be taken a lot longer than that. What you want to know is how much longer you get to live if you stick with the drugs your whole life, and in that case, it's not living 10 extra days, but living up to 10 extra years. Taking statins can enable you to live years longer. What's the cause? For every millimole per liter you drop your bad LDL cholesterol, you may live three years longer, maybe even six years longer, depending on which study you're reading. A millimole in U.S. units is 39 points. Drop your LDL cholesterol about 39 points and you could live years longer. Exercise your whole life and you may only increase your lifespan by six months. Stopping smoking may net you nine months, and drop your LDL cholesterol about 39 points and you could live years longer. You can accomplish that by taking drugs or achieve that within just two weeks of eating a diet packed with fruits, vegetables, and nuts. Want to know what's better than drugs? Something important and fundamental has been lost in the controversy about the broad expansion of statin therapy. It is imperative that patients be informed by their physicians, and what about drug labels, too, that not only their cholesterol levels, but also their cardiovascular risk can be reduced substantially by adoption of a plant-based dietary pattern and without drugs. In fact, dietary modifications for cardiovascular risk reduction, including plant-based diets, have been shown to improve cholesterol status, as well as obesity, hypertension, stomach inflammation, and on and on. The importance of this plant-based approach is magnified when one considers that in contrast to statins, the side effects—weight loss, more energy, and improved quality of life—are beneficial.


If it wasn't written down, I simply would not believe it. These stories we're gonna talk about today are, in my personal opinion, the height of hypocrisy. And the first round to go over is, unfortunately, an ETF approval. I say unfortunately because this is a leveraged futures Bitcoin ETF. And I know people are excited about the potential for a spot Bitcoin ETF, which will come through hopefully with BlackRock or Fidelity or Bakery. The thing that shocks me and the whole thing about hypocrisy here is the reason that Gary Ginza of the SEC says that he is not going to approve the spot ETF so far is because he's afraid of manipulation and he's trying to protect the consumer. And when I take a look at this, I just think to myself, is this really protecting the consumer by doing a leveraged futures ETF? Well, whatever it is, it's gonna happen.