So, an introduction to cloud native security. It's the maintainer track has the agenda for the presentation. We doing a quick introductions to security and I will be discussing roadmap activities and initiatives and sharing information on how you can get involved. My name is Radna Chital. I'm an executive for cloud security at a financial organization called TIAA. I'm also a technical lead for CNCF security team and research fellow for cloud security Alliance. Thank you. Andres, do you want to introduce yourself? Hi there. This is Andres Vega along with the Radna. I'm also a technical leader for the CNCF special interest group for security. And my day job is as a product line manager for product security at VMware for VMware Tanzu. And we also have Ash Narkar. Hi Andres. My name is Ash Narkar. I'm a software engineer at Styra. I'm one of the maintainers of the open policy agent. And I'm also one of the team leads at SIG security. Radna, so take it. Thank you, Andres. So introduction to CNCF SIG security. What is SIG security? Cloud security is complex. The ecosystem consists of multiple layers of infrastructure, platform components, service mesh, microservices, and so on and so forth. To add to that, there's a constant evolution and ever changing landscape of tools and technologies. This makes for a very complex system to understand and manage for cloud consumers. No complexity is the enemy of security. Threat landscape constantly is evolving in the space of supply chain security, application code vulnerabilities, container breakouts, etc. Not having all the visibility as well as appropriate security tools and integrations into the different layers of cloud native projects. Leaves cloud consumers with a sense that cloud native ecosystem is too complex and is hard to understand. In terms of operating it securely continuously. But that's where SIG security comes into play. SIG security is an open source community that is made up of cloud native security enthusiasts. We want to make it simple to understand how to secure a cloud native ecosystem. And provide best practices, tools, policies, and best practices guidance to the community. You know, cloud native services are deployed in heterogeneous environments and having security controls across the ecosystem requires security policies be deployed in code. So in order to facilitate some of these guidances, as well as help the community. CMCF has major focus areas to support the community. And some of those areas are provide guidance for protection of heterogeneous distributes distributed fast changing systems while providing access to tools and technologies. And provide common understanding as well as common tools for developers to build security into the services as well as meet security and compliance requirements continually. At the same time provide detection tools common audit criterias about the cloud native ecosystem to the community so that they can maintain a sense of security throughout the lifecycle of the services. There are a number of resources that SIG security has published. They provide use cases and personas. At the same time, there are published best practices, patterns, architecture patterns and anti patterns as well. Several other security artifacts like security scenarios, network configurations, service orchestration, application security guidance, etc is also published. Architectures and patterns for providing secure access. In addition to all of this security assessments are specific proposals and projects and several other resources, which can be found on some of the links that are provided in the slide. There's a blog post where there's information about initiatives and any releases or assessments being conducted. We also have a GitHub presence where all the issues are logged and progress is made on those issues by the community and volunteers of SIG security. There's a Slack channel where you can post questions about any specific security issues you may be running in your implementation that anyone in the community has faced could potentially help you with that. We do have regular meetings, weekly meetings in the United States on Wednesdays at 10am GMT-7 and by weekly meetings in the APAC time zone as well, which is on Tuesdays at 1pm GMT plus 11 hours. We potentially will be starting an EU based time zone meeting as well based on the interests of the community and engagement and requirements from the EU community. So please feel free to attend any of these meetings to learn more about SIG security and all the initiatives. I just want to share with you that we just recently got an award for most effective DevSecOps team of 2020 by DevSecCon. And that is Brandon Lum, one of our technical leads who's receiving the award on behalf of the community. So let's talk a little bit about the roadmap and initiatives of SIG security in 2021. Here to our GitHub page where the roadmap is published, I just am showing a snapshot of the same on the screen.  And there are proposals in the pipeline as well. And they're aware of the fact that SIG security published a cloud net as security by paper earlier this year, which was read by the wider community and we'll be talking a little more about that in the following slides. There are security assessment projects in the queue. There is a policy working group where there's much work going on at the same time. There are future projects in the pipeline like serverless security, etc, which will be getting more into in the following slides. So with that, I would like to hand over to Ash. Ash, please take it away to continue the conversation about the SIG security by paper. Thank you. So now we'll talk about the cloud native security by paper. So what is the cloud native security by paper. So it's a SIG security effort to ensure the cloud native community has access to information about building, distributing, deploying and running cloud native capabilities. The papers intended to be an initial starting point for the community in understanding the intricacies of a secure cloud native architecture. It references to other resources that may help in providing in-depth information about the topic discussed in the paper so the reader may choose to research further. The white paper is intended to be a living document created and maintained for the community by SIG security members. So with that being said, we now have an effort to gather feedback around the white paper from readers of the paper and the community at large to gauge the effectiveness of the white paper, how we can improve on the white paper and so on. And to do that, we've started a survey which captures questions like how much of the white paper have you read or what aspect of security would you like the community to focus on. So if you all want to get involved, please go ahead and participate in the survey and also if you want to be part of the team that's leading this effort, check out issue 480 on the SIG security people. And now let's talk about the cloud native security map. So the white paper that we saw in the previous slide, it provides us an introduction of different aspects of cloud native security, but it avoided any specific projects and products that readers could practically use for a specific use case. So for example, you did not see any mention of projects like Kubernetes or Open Policy Agent in the white paper as a means for performing orchestration or cloud policy enforcement and so on. So the motivation behind the cloud native security map is to fill this gap and provide a practical viewpoint of the topics covered in the white paper. As we've seen before, the white paper talks about four phases, you have develop, you have distribute, you have deploy and runtime to create resilient and observable workloads which can be run in your clouds. And so now you have the security map that attempts to provide a mapping of CNCF and open source projects to these four phases in the white paper. The map also provides examples of how these tools can be utilized in a particular category, thereby providing users a one-stop shop to explore available technologies and how they can be leveraged in their frameworks. So if this sounds interesting and you want to be part of the team which is leading this effort, check out the issue 551 in the SIG security people. And then finally, we wanted to provide you a snapshot of the first iteration of the cloud native security map. So on the left pane, you'll see those four phases, the develop, the distribute, the deploy and the runtime. And you see there are subsections for each phase. For example, you have a container application manifest phase or a section which lists all the projects that are part of this section which can perform application manifest security. And then you can also see examples of how you can use those projects to actually inside your architectures to secure your frameworks. So the map is a work in progress and we plan to add more interactive and user friendly features in the map in the future. I hand it over now to Andres. Thanks Ash. That's awesome by the way. Well, security means different things to different people. And you can be secure without knowing you're secure or you might think you're secure without that actually being the case. And saying that it means different things to different people and depending on our background and experience, folks in the group tend to organize based on interest based on areas that they might have the most energy to tackle a problem. Given recent incidents and high profile attacks that have taken in recent time. One of the areas that there's a lot of interest to capture a framework that provides recommended practices and tools is securing supply chains. And the special interest group we can form a working group. As a matter of fact, there is a talk on the main track of the event here at cube con titled protecting ourselves from CNCF gate software supply chain security at CNCF practices and tools where members of the CIG are presenting the work that culminated from folks having gathered for the last three months to talk about what are some of the lessons learned and how to fit recommendations and how to fit different tools from the parts been to fit varying risk profiles of organizations. There's quite a bit of substance in the talk, even though so far the work similar to the cloud native map cloud native security map, it's been somewhat conceptual, we're starting to get to a semblance of architecture and implementation. The work from the group is going to be evolving and taking the content that's been produced to be presented as a white paper that is in the final stages of review. And by the time that we get to the date of the event is we're pre recording the talk, most likely the paper will have been published. But from there on and getting community review, we really want to get to a capturing that framework and producing templates and code samples that you can customize and modify to fit your requirements and adapt to your different processes and tools that you might have in place. So, once again, one of the many projects that are not not touched in the roadmap. This is just to give you an exhibit of things that sick security does. Another great example of the ongoing work that sick security performs is that of reviewing the security properties and the security aspects of the different projects in the ecosystem. The goal behind the security assessments is to provide adopters, as well as newcomers to the technology, a glance at failure modes and security of a particular project. We produce threat models for the different software. The idea is providing the documentation and the analysis from a security standpoint can really accelerate adoption set cycles when people are evaluating the usage of technology into their production environments. And help them help them be conversant and articulate just, we have a large pool of security experts that are coming together and are analyzing the projects at a very detailed level, taking into account. A lot of different considerations just to make people who come down the line, give them a better understanding of everything pertaining to security. The assessments have been done for a number of projects including harbor key cloak open policy agents spire, I might be missing a few, we have them all available as open source. You can read through the architecture the threat modeling the software development best practices how the projects handle vulnerabilities, etc. For example, but come by this six security repository, get hop.com slash sec dash security. Sorry, get hop.com slash the NCF slash sec dash security, and you'll be able to look at the different assessments that have been performed. One of the streams is given that we've established a bench of security experts really well rounded on on the different areas that that make up cloud security different organizations or the broader CNCF and Linux foundation have come to stick security to tap into that pool of experts. For example, as the recent CKS certification the certified Kubernetes security specialist, Linux foundation learning pulled into six security to get the authors for the curricula, the exam questions, all the content surrounding. It's a great opportunity to contribute your expertise and get, get exposure and visibility and increase the breath of your impact. Really. The last example, I'll be talking about myself is the organization and production of events that we use as a stage and we use as an arena to. To get cloud native security in front of other people and use it as a platform to discuss breakthrough ideas discuss innovation new technologies provided presentations. Parallel to cube con and cloud native con. By the time you're going to be listening to the stock, we will already have concluded cloud native security day and talks will have been made available for posterity, but the program committee from for the event. Came out from six security. So you can get a glance of what type of work and what type of results come out from involvement in the group. And parallel to all the sessions we hosted and they're really good content we got in there. We also collaborated to produce a. Interactive capture the flag activity where different people could come in and get hands on around attack and defense as scenarios. So this is what just one of the many things we do around production of of events. With that said, I will pass it back to or not to exemplify another of the initiatives we do and close us up from there. Thank you, Andrews serverless computing refers to the concept of building as well as running an application that doesn't require server management. That doesn't mean that we don't need to use servers anymore or host and run code. It just means that functionality is offloaded to a provider and the serverless consumer only needs to worry about the application code that they need to run in that environment. There are a number of threats where it comes to serverless deployments today, because of the different components and the integrations that are possible with serverless micro services that are deployed in a cloud native environment. Obviously this is a pay as you go model you have infinite scalability because of the cloud provider capabilities in this space. I know there are a number of serverless platforms out there including open source ones like open VESC AWS has AWS Lambda and Google functions and Azure function platforms are also available in the industry and open source community is also coming up with a few other platforms. They are stateless, easily parallelized, they're highly dynamic, and they can communicate asynchronously with other services as well as components. There's much work that has been done by the serverless security group in CNCF, but the initiative that we are kicking off later this year will be focusing on serverless security the security aspects of deploying serverless functions. What are the threats, how do we mitigate them, what are the best practices at different layers that a serverless consumer has to worry about. At the same time, the persona offer provider and what functionality and what security they need to provide to provide enough visibility to the consumer organizations is also going to be addressed as part of that security best practices. The guidance that CNCF security will be working on later this year the timeframe is sometime in Q3 and Q4 depending on the bandwidth of the participants, but this is going to be an interesting piece of work that we embark on so I encourage you to get involved and participate and get your voice in and also if you are having challenges in the serverless space for securing your platform or services and there'll be a good place to have interaction with other people who are deploying their services as well. So I welcome your participation in this working group and make impact to the industry. Thank you. Awesome. Thanks for that. With that, we just want to wrap up and state hopefully the different work streams and activities that the special interest group for security is driving are meaningful and relevant to you. We want to make this as active, participative and collaborative as we possibly can. We want to encourage your participation. Everyone is welcome.  We want to state that like well the more the merrier and the better rounded and balanced set of skills we can bring in. There's room for it here. There's certainly need. And we would love to have your help. So, great starting point is the GitHub repository for the special interest group, but aside of that, feel free to reach out to any of us directly through GitHub or Slack. And with that said, I hope you keep everything secure. And you enjoy the rest of the event. Ask any parting thoughts. I know it's like Andre said join us. We are open to discussing all things security. We have weekly meetings. So join us for that if you're interested about security and want to learn more. And we are happy to talk to you. Thank you.


To me though, what's most exciting about our universe and also philosophically and emotionally and spiritually most empowering and interesting about it is that through us, our universe has actually become aware of itself. For the first part of the 13.8 billion years since our Big Bang, there was no one home. Our universe is doing all this cool stuff. Potentially no one home. Well, in the earliest moments right after Big Bang, when it was just a plasma, expanding plasma blob much hotter than the core of the sun, I'm pretty sure there was no consciousness in there. But life could be around as early as just a couple billion years ago in other locations. Oh, much earlier than that. Yes. It's quite possible. But very early on, I think it's very likely that it was just very way too inhospitable for any kind of consciousness. And then gradually, it went from being very simple and boring to being more and more complex and first self-aware entities appeared. That's in a sense our universe waking up and becoming aware of itself, right? We know what happened here on this planet. Maybe it's happened elsewhere. Maybe not yet. We can come back to that. Towards complexity. Yes. In any case, I think that's the most important thing that's happened so far because if you look at all these laws of physics, they don't say anything about beauty, meaning or purpose. Beauty, meaning and purpose are all things which can only exist when there is experience because those are in some sense experiences, right? We can only have suffering or joy or pleasure, which are experiences when there is experience, right?  And before it woke up, it was also meaningless in this sense, in this very literal sense. There was no meaning because meaning is something you experience. So this is, I think, the most beautiful thing that's happened so far. This universe gradually got so complex that it woke up. And one of the most striking things I think we've learned through science is that not only has a universe woken up now, but it has the potential to wake up much more than has yet. I was taking a little power nap before this interview and when I first woke up, I just woke up like a little bit and then it became more and more conscious, right? You went through this same procedure this morning. I feel our universe is like that too. We are having very interesting experiences with humans, but it's clear our universe can wake up more. You look around on the telescope and it's clear that the vast majority of space is not as alive as it could be. Some people who listen to this maybe think we live in a, are convinced we live in a Star Trek kind of cosmos where there's all of those cool things going on all over the place and they're just hiding from us. There's not any really scientific evidence for it yet. Maybe there is other life really, really far away. The Zoo hypothesis, there's interesting ones. I think it's much more likely that our universe has much more potential in the future. I think one of the most inspiring things for me personally to think about is what can I do to help our universe wake up more and more? Imagine if it turned out that we discovered that the only life on all of earth was on Rhode Island. There was nothing outside of Rhode Island. No plants, no animals, no nothing. We might even feel then that we have a moral responsibility to help life spread to the rest of the planet so it can wake up more so there can be more joy. More joy, yeah. Interesting. More experiences. More experiences. Of across all different styles of experience. Exactly. Yeah. The opposite thing we could do is. More ideas. Yes. More activity. Yes. Yeah. Exactly.  That would feel like a kind of bad idea. Which is a lot of what you do with existential risk mitigation. But if that's bad then clearly the opposite is a good thing to help life spread. Spread. Yes. And if you look out into the cosmos that's exactly what it is except much more extreme than going from Rhode Island to Earth. Going from Earth to help life spread throughout our Milky Way galaxy and hundreds of billions of other galaxies. It's a good analogy. Would be dramatically more positive I think than just going from Rhode Island to Earth. Yes. And I think most people lose sight of just how much opportunity there is for the future.


It's actually possible to clone your dog, but you shouldn't, and here's why. So you might only be able to think of your perfect pet, but there are other dogs involved in this process. Eggs are harvested from a number of host dogs, and then a cloned embryo is implanted in them. But this process fails 80% of the time, unfortunately leading to many miscarriages and stillbirths. That's why cloning costs so much. It'll run you $50,000 for your exact genetic copy of Fido. But even after forking over that much cash, your cloned pet is more susceptible to health issues. So experts say that 25% of the pet's personality has nothing to do with their genetics. So that means that no amount of money can give you the exact same pet that you want to replicate.


So, we agreed that today will not be additional lectures, but it will be an exercise session. So, we start by asking you if you want some particular thing explained understood, you know you can ask a question how do you do this and then I may be able to answer I may not be able to answer, but we can try yes oh I thought I will be swamped with all kinds of questions can you explain the very good here is the following in the BTW model or more generally in these kinds of sand pile models you add grains and there are some avalanches and eventually the grains leave the system and you keep on adding and they keep on leaving. So, we ask for what is the probability of there are s toplings some event of size s and what I would like to argue is that the expectation value of number of toplings has to diverge as the size of the system goes to infinity if this is true then it immediately so if then you define this is probability of s for a given size then probability of s l usually is a function like this initially this is log probability divided by log s initially it kind of decays usually it is like some power, but there is a cut off ok, but if you let l go to infinity then the power log goes on for much longer. So, we define probability of s l limit as l goes to infinity equal to probability of s if this limit exists usually it does you know like even if you take the last size limit the probability that you will get no toppling when you add something is finite probability that you will get exactly one toppling is finite. So, this probe s exists even for even when l is very large and it is finite for each finite s then the question is is it a power log. So, the claim is that the first movement of this has to be infinite because what is the proof the proof is that suppose you take very large number suppose you take l is l is a finite number 100 by 100 or some such thing and I keep on adding grains and they keep on leaving ok. So, I add one trillion grains one after another and in each case I measure the amount of toppling done then I can label the grains as eventually each grain will be added at some time and it will leave at some time. So, I ask how many topplings this grain undergo the number of topplings one particular grain undergoes is going to increase with l as at least as l or typically as l squared ok. So, if the number of topplings incurred by each grain is going to diverge then the average number of topplings per grain will also be the same it will also diverge ok. The proof is precisely this number of topplings per grain average number is equal to number of topplings per actually there is a factor here by 4 because each toppling causes in an avalanche when I count one toppling 4 grains count it I got toppled I got toppled I got toppled like that ok. So, if I take the sample of all the trillion grains I added I will get so many topplings were done ok then I say per grain how many topplings were done each toppling in an avalanche will count as each of the grains will say I got toppled I got toppled on that one right. So, the number of topplings per avalanche per grain added average number of topplings found by one grain that found number of topplings under go by a marked grain ok. So, this number is clearly infinite and hence this number is infinite once this number is infinite then the probability of the this probability s the first moment must diverge probe s s summation over s is equal to infinite. So, it is a power level ok clearly I have not used any properties of the particular toppling rule of the sand pipe. So, it is a very robust argument the only point which we have used is that if you add a grain it keeps on moving, but it can only leave at the boundary it cannot disappear in the middle if the grain disappears in the middle then this argument does not work because argument said that it has to move order L in order to disappear right that argument will not work if there is dissipation in the bulk not only dissipation at the boundary ok. In one avalanche? No in the till it leaves from the time it is added till the time it leaves no it is not leaving in one avalanche. So, it is a flip of the you are counting number of topplings, but in two different ways one is per grain and one is per avalanche, but you should get the same answer sorry. Perfect size. Yeah that went as L squared or you know the bound that this number is greater than order L is a much more robust bound it does not even use the fact that the parts of grains are random walks ok. So, you just add it anywhere it has to take order L steps in order to leave ok ok. So, we say so I had this pile and I keep on adding grain grain number one grain number two grain number three all grain grains are marked and all of them leave eventually. So, I add how many grains how many topplings were encountered by grain one grain two grain three. Let us call them E 1 plus E 2 plus E 3 plus E trillion whatever E total number of grains added, but this will be compared with S 1 plus S 2 plus S 3 which is the number of topplings on adding the first grain number of topplings on adding the second grain and so on. But these two numbers have to be equal to a multiplicative constant sorry I beg your pardon I think there is a mistake here it is multiplied by 4 not divided by 4 even E 2 is the number of topplings encountered by grain one till it leaves E 2 is the number of topplings encountered by grain two till it leaves and so on. And this is number of topplings on adding the first grain number of topplings on adding the second grain like that. So, I am just counting all the number of topplings in two different ways I am rearranging the submission and so that is what it is ok it is the proof ok yes ok. So, so there is some I guess for me right now I will write the configuration will be given by the heights know. So, the heights will be 3 2 1 1 2 whatever some heights and the heights are written in the centers of squares ok those are my cells usually we will discuss this in the you know there is a square and the vertices are where the height should be. So, I should write 3 here and 2 here and 1 here and not in the centers. I think this issue is important, but we will stick to this one the I like this easier to draw know because I write 3 in the middle of the square ok. So, then what we have agreed which we will not show now is that there is a burning test which says that if you try to imagine that outside is all burned and then you burn any site whose height is equal to or bigger than the number of unburned neighbors and if everything burns then you are in the recurrent state ok. So, I start by this that I this stuff is all burnt outside or the sink site outside is considered as burnt this is the sink site. Now, what does what can be burnt in this one let us take 3 too many burns, but ok now I burn this this site can be burnt because it has only 2 unburned neighbors and its height is 3. So, I would be typically burning this first and what else can I burn right now nothing in this graph nothing else can be burnt, but now as soon as I come here then this site has height 2, but it has 2 unburned neighbors. So, in the next stage I can burn this once I have burned this then I can burn this and then I can burn this right no no sorry I beg your pardon not this one 2 ok. So, I am stuck. So, this was not a recurrent configuration what is left behind is a forbidden configuration this one because it has 3 unburned neighbors. So, right now I cannot burn it ah ah sorry I beg your pardon you are right of course. So, I should burn this one once I burn this one then this will also burn and now this one below and the one above and then this one as well and then this one as well and then this one as well and then it seems like it yeah this one yeah ok this one this one in the second row in the left one again this one also all right no no very good. So, there depending on the configuration they will burn, but you see there is a sequence in which they burn. So, now I will draw the path the fire took in order to get there yes 0 to 3 ok very good. So, there was there has been sort of flip flop in the notation and so one convention is that the heights go from 1 to 4 and the other convention is heights go from 0 to 3 in this case the toppling rule is that if height is greater than 4 then topple and here the rule is that if height is greater than equal to 4 then topple ok. So, we agreed that these two are actually equal there is no big deal you just have to redefine h prime equal to h minus 1 and then both the rules are actually the same ok, but the burning rule can also be phrased either in terms of h prime or in terms of h ok. And in terms of h prime there is one burning rule in terms of h there is a slightly different burning rule one of them says that burn only if everything if the height is bigger than the number of unburned neighbors the other says if it is bigger than or equal to unburned neighbors ok. So, that is comment it is not just remember that different when suppose somebody defines the model like this then he will use one version if he defines the model like this he will use the other version ok. So, now in our case we found that the fire went like this and then it went here and then it went here and then it went here and yeah then it went here, this fire went from here to here and from outside it went here and it went here and here and then it went here and it went here and so on ok. So, once site is burnt then it can burn and the fire can propagate from there. So, if everything burns then I just draw this path of fire burning and it will burn everything. Let us not write these heights and it will burn everything in some way. So, the burning path defines a spanning tree. Is this, yes and from already burned squares. Yes, but only from one of them. Yeah. Yeah. Yeah. So, the bottom one is a top one. Yeah. It is not possible. Very good. No, no, no. So, the basic philosophy is that the burning path corresponds to a spanning tree. This much is true. Now, is there a one to one correspondence between burning paths and spanning trees? That is the question you have to ask. Each burning path will be a spanning tree because I only burned once. Each site is burned only once. So, it is a tree. All sites are burned. So, it is a spanning tree. Now, we have to ask is there a one to one correspondence between these? That is a little bit trickier and so, what we have to do is to realize that if this site burns, then the height could be 2 or 3 and it will still burn. So, maybe there is no one to one correspondence between the burning paths and trees, but no there is actually because you can burn either from here or from here. So, any site which burns, I can say that the fire came from one of its already burned neighbors, which one? We will have to choose. I have some choice, but the choice is exactly equal to the height allowed at that site. So, I make up some rules saying if the height is maximum, then burn from this side. If the height is less than that, burn from this side and so on. Then there is a unique height association with the burning path. Once I have chosen, decided that if the height is 3, then I will burn from here. If the height is 2, I will burn from here.   We just said burn, but if you want to make a one-to-one correspondence, you add this extra bit and you say burn from this side, burn from this side and then there is a one-to-one correspondence between the burning path and the height configuration. Is that clear? Anything else? Yes?  At this point, one site gets burned, but I will decide whether to say that it burned from top or from bottom, but from not from both. So, then there is no possibility of a loop. Yes, sir? Because given a configuration which I know burns, I just draw the path of the fire. There is a construction. See the point is there are two sets. They are equal in number. You can make any one-to-one correspondence between them in principle. No, actually we first set up the correspondence, then count the number of spanning trees and then say that O, but this must be equal to the other number. But in the end, I am saying the correspondence between the spanning trees and the height configurations.  This is a 2 by 2 graph. There is a spanning tree like this, there is a spanning tree like this, there is a spanning tree like this, there is a spanning tree like this. I cannot make any other trees on a square. Now, the height configurations are whatever they are, 2, 1, 2 and so on. Now, what we have to do? What corresponds to which one? You can make any correspondence. I can say this height configuration corresponds to this one, this height configuration corresponds to this one, this one, whatever. Given two equal sets, you can make any one-to-one correspondence. We are making the one which makes our life sort of easy and makes, you know, because there are big sizes. You make an algorithm to make an assignment. It is a free choice and we chose the burning test to use the assignment. Yes, please. For example, 7 minus 4 minus relations between operators. For example, we have a 1 to the power of 7. We wrote 8 to the power delta i equal to product over j, 8 to the power minus delta ij. So, j is not equal to i. What is the proof of this relation? This is the question, right? No, no, no. I refreshed your question, but you asked first the original question. Delta is some phonemetrics. Yeah, precisely. 7 equals 8 to the power of 4. I think you see in the columns and write the algorithm. No, I did not write any columns. I did not write any algorithm. So, let us go through this proof. It is important to be a sure of the argument. So, this relation is true for all configurations with positive heights. So, suppose there is some configuration or the other. I do not know what it is. So, this is called c and I would like to know what is a i to the power delta i acting on c. c is any configuration. The statement is that of course, you can add one and then topple and then add another and then topple and do all kinds of stuff. But the result will be the same as first I add all of them together and then topple. In that one, actually c can have some sides which are unstable also. It does not matter. Now, I will say that oh, but I am allowed to topple any order I like. So, first I will topple at i. So, the statement is only requires that at side i, now that I have added delta ii, it must topple. Once it topples, this is the same as acting minus delta ij particles at j acting on c, product over j. End of proof. No, no, no. It should be clear that the argument is only one line argument and it impose it uses the abelian property. It uses the property that once I add delta ii particles at the side, it must topple. That is all it uses nothing else. How do you what? So, forbidden configurations, but which? Yeah. Yeah. Ok, very good. So, the statement, no, no, it is very, it is all right. Suppose I am in the configuration, suppose my c belongs to R, suppose assume, suppose there is a stronger relation. The stronger relation is that product. So, let us write it like this 2, 1, 1, 1, 2, 1, 1, 1, 0 acting on c is equal to c. It says that take whatever configuration you have, that is the recurrent configuration add to it this stuff 2 here, 1 here, 1 here, and topple and see what you get. Then you must get back the recurrent configuration. If I do not get back the recurrent configuration, then what I do is that suppose I get this acting on c is equal to c prime which is not recurrent. Then in c prime they must be, it is then I look at c and I try to burn it and you will stop, the fire will not reach everywhere. Then whatever remains is a forbidden configuration. The forbidden configurations were sort of found first and the burning test was invented later based on generalizing the forbidden configuration, but now you can use the burning test to define whatever the fire cannot burn is a forbidden configuration. Yes, please. Is that clear? Yes, please. There were four spanning trees that is equal to the number of recurrent configurations. Ah, ok, very good. So, I think one should distinguish between these two things. There is a spanning tree like this and there is a sink site which is a, let us draw the sink site like one site, but this site is connected to this one with two bonds and connected to this one with two bonds and connected to this one with two bonds and connected to this one with two bonds.  That number is 192. You have to draw the edges in the suitable way and yeah, everything can be modified. What is the, there is some Latin word for this which I have forgotten, but anyway, ok. So, the proof to, so if there is a forbidden configuration, so you imagine that there is fire outside everywhere and you try to burn it and you would not burn it. You would not be able to reduce any further. So, then it is a forbidden configuration. Yes, please. Yes, yes, yes. You mean you add some grains. Yeah, yeah, yeah. Is there a configuration? Yeah, see eventually, yeah, yeah. It is a recurrent configuration. Recurrent means comes back again and again. Yeah, yes, yes. Ok. I have explained it three times now. Let me see. Can we discuss it later? I am saying the proof of the forbidden configurations was that if you don't start with two 0's like this, you will never get there. You cannot create this configuration by toppling process. That was the argument. Either it is there in the initial configuration or it won't be there because if you try to add something, you will never get 0 0, ok. So, once you are in the recurrent, in the initial configuration, you may be starting with everything 0. That is allowed. But as you keep on adding grains, the number of adjacent 0's keeps on decreasing and this number can never increase. And so eventually, they get lost and they are never found again. So, in the recurrent states, they don't occur. If you start from recurrent state, then this is never, it's not there and it cannot be generated, ok. Yes, please. So, we wrote E alpha is equal to product over j A j to the power B alpha j, right. This is what you want to understand, ok. So, firstly, this is an operator. This much is clear. Now, I want to show that E alpha raised to the power D alpha equal to what? Show is identity operator on the set of recurrent configurations, ok. So, what is E alpha to the power D alpha? E alpha is equal to product over j A j to the power D alpha E alpha j, right. This can be written as equal to product over j alpha j A j. D alpha is a big D. We wrote delta is equal to A D B. So, D was a diagonal matrix. So, D B is a diagonal matrix and D B is the same as D alpha B alpha beta, ok. So, now, but D B is equal to A inverse delta from the previous equation. So, this is equal to product over j A j A inverse delta alpha j, ok. But this can be written as product over j A j to the power delta alpha j A in A j to the power delta beta j A inverse alpha beta, ok. I wrote on the right, but it is the same because I have written indices. So, now, I do not have to worry about matrix, but this stuff is identity and so, this becomes the whole thing becomes identity, ok, end of proof. Yeah, precisely. See, if I have a group element, what power do I raise it to to get identity? If you raise it to enough power sometime or the other you will get identity, but these have been constructed. So, that this one raised to the power D alpha will give me identity, but I have different choices of D alpha and I get different generators and then this generator of the different group, ok. Yes, please. Yeah, of course, the set of recurrent configurations where the group works, if they you know we are determining the structure of the recurrent group. So, it is only in the recurrent group, no problem, no other questions, very good. So, I made up some questions because I thought maybe you will like to have a sample of the kinds of questions you might be asked or you can test your understanding on what you have already understood so far by seeing if you can answer these questions. So, for each question I will ask you, do you think you understand how to answer this? Means, maybe you cannot figure out the answer in two, three seconds, but you know how to go around it, given ten minutes you will be able to work it out or not. If you can answer it then you say yes, otherwise you say no, then I will sort of have a feeling as to how much of my effort has been successful in communicating some ideas, ok. So, here is a problem, the problem will be written down here, ok. So, it says define, some of the stuff will be written on said only in words. So, I define a sand pile with the following rules, on toppling Z j goes to Z j plus Z j 1 plus Z j 2 plus Z j 3 plus Z j 4 by 5. I did not write it very well, let me say it like this. There are these, there is a site if it says if Z is bigger than 3, then you just take the average of all these five numbers and give it to each one, each one gets the average of the five heights. If the height is to be then you just spread it out equally amongst all the neighbors, the heights are real numbers not integers. Is the definition of the model clear? Yeah, yeah, yeah. In the toppling process the value of the neighbors is also modified, ok. Sorry, it is precisely the same, all of them get the same value.  Is the model clear? It is one of like one of those sand pile models. I did not say it, but yeah, you work with positive numbers. Sorry, average and? Yeah, yeah, yeah. So, I should say Z j, I go to this, all the five neighbors go to that value. Now, yes sir. No, they are not integers, they are real numbers I said, ok. So, it is clear, the definition of the model is clear. They are real numbers, they are real numbers. It is not, yes sir. The neighbors are zeros? Yeah, it does not matter. You just average and give everybody the average. Yeah, yeah, yeah. So, it says on the borders this guy will get some number and it will leave. Initially the height was zero then it gets something, but it is reset to zero. Heights outside are always reset to zero. The definition of the model is clear, ok. The question is only this. Is this model a billion or not? They are not getting one. They were, neighbors are not getting one, but they might get something, you know. It is, ok. No, no, no. So, let us not jump to conclusions. How many people think they answer, they know the answer. Just raise your hand. Yeah, the question is, is this model as defined a billion or not? Do you know the answer? I do not know whether it is yes or no, but do you, can you figure out the answer or you have no clue about what is the answer? That is the question. If you think you know the answer, just raise your hand. A little bit higher so that I can see. I see a substantial class of, substantial set of people have not raised their hand, ok. So, then there is a problem. See, I define the model clearly. Then I am asking, is this model a billion or not, ok. So, we have to check, you know, if the, if you add at i, add at j, this is the same result as adding at j and at, at i, you know, that was the test or that was the characteristic. So, can I just check it, you know, can I add this, start with some configuration, add at i, add at j, see what happens. No, so the answer is, they don't come here. No, no, so you are able to see or you are able to check that this model is not a billion, that is the answer. We keep on adding, no, you know, in that model, in the sand pile, there is no depth of house and grains. You keep on adding one after another and you reach the steady state, you keep on adding and so on and so forth. Yes, but the node is unstable if I add four grains. No, no, so in the beginning, there are some configurations where you add nothing happens.  But the a billion property requires, therefore, all possible configurations, the same thing happens. If I have, for example, this, four, zero, what happens? So, this one will topple now, by, oh, so we said, yes, so we will imagine that there is a zero here and zero here and zero here and then divide this four equally and then erase all the boundary sites. So, I said that it's a real number, we said it tries, you are not paying attention. Okay, no, it's okay, the heights are real numbers. So, it's four by five, the height is, later it becomes four by five. This is four by five, this is also four. Yeah, I asked if it was a fraction one. Yeah, yeah. So, all I have to do is I have to find one configuration where you first add here and then add here is different than when you add here and then add there. That's all. Okay, so I will take a graph like this, put this, we said if z is bigger than three, so this is 2.5 and this is two. And let us say all the neighbors are one-one. Please, can you, please, let us take this configuration. I'm just made it up right now. Consider a big lattice and I construct this configuration where these heights are 2.5 and 2 and the neighbors are some 1.1, whatever you like. Now, first I add here an average over everything and then I add here an average over everything and then I do it in the reverse order. You will not get the same value. Okay, and so then the model is not a million, end of proof. So, we say, yes, we said 1.1, 1.2, 2.1, 1. So, we decide to add one right here. Yes. So, we go there and we take the sum over m divided by 5, each one of these distributions. We get this is stable. So, we want to prove that it's not a billion, we then add. Yeah, you added i and g. Yes. Construct a c, i, j such that a, i, a, j, c is not equal to a, j, a, i, c. Once you have constructed one such counter example, then the result is not a billion, end of proof. No, because we, in the proof of a billion, no, it is because you can start with this configuration and a, i, a, j, c is not equal to a, j, a, i, c. This we can construct. And then, but for a billion property, it has to be true for all configurations. Okay. So, it's not true. Okay. So, you can check. Again, you can construct this. Yeah. Of course, it is different. The point was only to see that given a model, you can check does it have a billion property or doesn't have a billion property, full stuff. It's more or less the definition of what is a billion property. But we are applying it to a context which was not recently discussed in class. This will work. Yeah, yeah. Sorry, can you repeat? Hmm. No, no, no. I thought it is true for all configurations. That's what we proved. Okay. Yes, sir. We can find the inverse. No, no, no, no. It's too far. Inverse, we don't, the a billion property does not necessarily require an inverse to exist. Because sometimes we work in this extended space of configurations where heights can be negative and all kinds of stuff. Let us not go into the group. Of course, if all operators are commuting, then they will form a commutative group. If they don't commute, maybe they form a non-commutative group. What do I know? Yeah. Yeah. Yeah. Yeah. Yeah. All I can do for sure, given a configuration C, which is by definition set of z i's. I know what is a i c is equal to z i prime. This I can construct. So the operators are well constructed. You can check if they are commuting or not. So it's not clear that there is such a relation. It's not clear that there is such a relation. But this happens because we have a real number of brains that we can have on each side, right? Because if I am in chairs, I can always write a relation. No. Because the a billion property may not still hold. It may happen that, see we said that a i to the power 4 is equal to a i 1, a i 2. That property also need not hold in general, okay? I think these are all special cases. Okay. Is this end of this discussion for this question? Yes, please. Yes. Necessary condition is just a i a j equal to a j a i. That is necessary and sufficient. No, no, no. So I understand a i a j equal to 0. Now you can deduce this from some other condition. Yeah. If that other condition uses the toppling rules in some way, then okay, I mean you have shown that that particular toppling rule is commutative. Do you want to prove that? Sorry, just let me answer this one.  With toppling rule. With toppling rule you can just check. Oh, I don't know. Yeah, no, I understand. No, no, no. These questions, they can be set in some more general setting and you can think about them. They are supposed to a check for comprehension and allow you to extrapolate further. We which we are not doing. Yes, please. There was another question here. Yes. Yes. Okay. No, they all become equal. That was the rule. All the heights become 11 by 5. So people didn't even understand. See, I asked three times, have you understood the definition of the model and people didn't understand. Okay. You serve. Is this argument clear to you so far? Okay. Okay. Yes, sir. Depends upon your neighbor. Yeah. No, it's not a proof. Depends on how it depends on the neighbor, no? Maybe, maybe not. I don't know. I'm saying that without giving a toppling rule you cannot give a proof that everything is a billion. There is a counter example.  Other set of toppling rules, they are not a billion. Yes, sir. Sorry? When a system is finite, and it seems to write the relation between operations like toppling in one side or in the other. Yeah. We can say that once I found the identity operation of the current state, then the system is a billion for sure, right? Because I am moving in a torus in one dimension or in another. Yeah. I don't know if there is some identity operation which is a non-trivial identity operation. You know, if you don't touch anything, then of course it remains unchanged. We are doing something, we say add all this stuff and then you relax again and you will still get back the same stuff. I don't know for an arbitrary toppling rule this result is true. Okay? Yeah. Yes, that is what I want to do, but I am sort of not being allowed. Yeah. Anything else? Okay. So, can we go on to the next question? Yes. This is not a proof. This is a reference to authority. No, no, no. I am asking for a proof, not saying in some textbook or the other, this is written like that. Can you give the proof yourself? Yeah. Yeah, yeah, yeah. The statement there is correct. Looking for a counter example, I agree. Ah, so, okay. Everybody satisfied and happy? We can go to the next question. Okay. Okay. So, this one is still referring to my first lectures. So, I want to have a profile H of X. You know, this was this profile, like this mountain profile. And I want to construct H of X, H of X plus delta, average value. Actually, I will write like this. H of X plus delta minus H of X squared average goes like delta to the power y for all y, for all delta. Okay. You cannot see. Okay. Construct a profile H of X such that H of X, H of X plus R, let's write minus squared goes like R to the power y. So, this was, you know, in the beginning, we said that there is this mountain profile. And if you calculate height difference at two sides, it goes like power y. And can you construct? Can you write a Monte Carlo program which will generate these height profiles? Given y. I give you y, you generate a program and give me how to give the output. It will give random height profile, but it will have this property that on the average H of X minus H of X plus R goes squared, goes like R to the power y. Average over, it's an ensemble average, but if you feature long configuration, you can average over X. Yes, please. It can be, hurry. Yeah, yeah, yeah. Yeah, yeah, yeah. But it should give you a random answer each time and the ensemble should have this property. Yes, that is the question. Has everybody understood the question? No, no, no. So, I am giving, put of the program is a random profile which has this ensemble property. Can you generate a random profile which has this, from this ensemble which has this average property? Yeah, H of, I only give you y, for all R it should work. Yeah, yeah, yeah. Trial and error. I think that will be very tiring. You write a computer program, let the computer do the trial and error. No, because I want 1 million outputs and you are going to do trial and error, you will keep me waiting for very long and that is, I cannot afford that much time. It's a question clear. Okay, okay, whatever, next. Anybody else has a better idea? Yes, ma'am. I would like to say the origin and then discretize the space for the smallest value of delta through the new height according to that distribution. So, taking the square root is distributed as H of X plus R is equal to H of X plus R for y equal to 0.7, can you work, will your program work for y equal to 0.7? If it only adds incrementally, then some kind of central limit theorem usually works and it doesn't generate long range correlations of the type I am looking for. Yes, there was some other question. Only one point. This is varied for all R. Okay, yes, please. Yes. Yeah, precisely. That is the answer. Okay, so the answer is, this is all questions depend on some pre-knowledge. So, I assume that you know this or you know this or you don't know this and I also know that all of you actually don't end up knowing that pre-requisite, but this is a test of that pre-requisite. So, it is well known that if you want to generate something like this, a random profile, the way to do it is like this. Suppose there is an H of X, then you can generate H of X, H of X plus R average. This is called C of R, defined. We are giving you a particular form of C of R, but you can choose other forms if you like. Okay, but then how do you generate something with C of R? So, it says that given C of R, define tilde of K is equal to C of R Fourier transform. C tilde of K is the power spectrum of the thing in the wave number K. So, you take the signal, break it into Fourier mode and the power in the Kth mode is called C tilde of K and if the power in the Kth mode. So, then if I want to generate H tilde of K, is a random variable with variance tilde of K. So, you pick H tilde of K as a random variable, but its square average is C tilde of K. Then you get all the set of H tilde of K, then take Fourier transform, you get H of X, because these numbers are random. You will get random answers each time, different answers. This F H now will be Gaussian random variables. Okay, so I didn't insist, but you know I just said make a construction which will give me this and this one works. Okay, so that is an example of such functions and by you know because this C tilde of K can be chosen to be whatever I like. I can in this case it is chosen to be 1 upon K to the power some number which depends on y and then when you Fourier transform this will give you some function and now you adjust the beta so that the variance comes out right. That step is not being done here, but the point about the question is only this that the answer is that you construct a function with a given Fourier components and then add up the random components and then you get the full spectrum and the power of the Kth mode is pre-specified. Yes of course, yes. So beta is also known. I didn't work out the relation between beta and gamma that is left as an exercise. Okay, yes please. This depends this is a translational invariant system and it has a stationary Gaussian process. It's working. Yes, yes. So there is a well known theorem which is called the Wiener-Kinchin theorem. We say if you have a stationary process then the Fourier transform the correlation function is the power spectrum. So if I give you the correlation function I am giving you the power spectrum.  No, this is the expectation is in the so this Wiener-Kinchin theorem is not known to everybody in the Wiener-Kinchin theorem. It says for a stationary process h of x Fourier transform of correlation function is equal to power spectrum. Okay. This was assumed to be known. Okay, so you know the power spectrum of a process then you construct a random signal with that power means quiet is fixed but the values need not be fixed then take Fourier transform and you get the random signal. Sorry, yes average. Okay. Can I go to the next question? Okay, so very good. So h x minus h of x plus r full squared average is equal to h squared average plus h of x plus r squared average minus h h but this is equal to this and it is independent of x because it is a translational invariant process and so it is just a number. So, h h correlation function is equal to constant minus h minus h squared average. Okay, so it is up to an additive constant these two are the same. Okay, next question I will erase this part it says show that directed asm expectation value of s I will you know is equal to constant time say it is not goes as it is equal to let us define the model. So, the model was defined like this there was some square lattice with periodic boundary conditions this length is l, this length is m and when some particle leaves from here it throws two particles down. Okay. And you add only at top and leave at in the particles leave at the bottom and then the question is that what can you show that the number of top links per particle added will be constant time set. So, if you remember the proof it said that on the average you know every particle will travel length l but for a directed pile every particle will actually travel exactly length l in order to leave. So, the argument works without any change is this clear yeah, yeah of course limited value of s by definition is in the steady state and there is a conservation of particles and the particles are added at top and they are removed at the bottom see nothing this proof does not require anything beyond what you have already learned but it might just require restructuring your thought a little bit to get the answer exactly right. Okay. Okay. So, this one find the steady state probability different configurations compartment plus engine train. So, to recall the train model was defined like this there is a train and there is some spring in between and there is a compartment one and there is a spring and there is a compartment two and this engine moves at rate one every time step but if this spring becomes of length more than two it becomes three then it resets to either length one or one two and it pulls everything behind so it just moves this compartment but then this spring will become longer and then maybe it will also reset and that compartment will jerk forward and so then it stops at some stage and then the engine moves again and so on and so forth and the compartment behind keep on moving in jerky motion. So, then there is a steady state now the question is can you if I take the steady state and I look just before the engine has moved further are the after the configuration has relaxed before the engine has moved one more step then what are the allowed configurations of the train there are different configurations and what are the probabilities of different configurations in the steady state how many of you can do this problem one half hand ok ok no no no so firstly once everything has relaxed what are the lengths of springs they can be one or two that is what we said nothing else so a configuration is specified by l 1 equal to 1 l 2 equal to 1 configuration is specified by l 1 l 2 which is the length of spring 1 length of spring 2 and the allowed configurations are 1 1 1 2 2 1 2 2 is this point clear now in the steady state these configurations will occur with and we want to know what are the probabilities of these different configurations ok so how do I do this well I have to find what what happens if the configuration now is 1 1 and the engine moves then what will happen to it 1 2 actually maybe I should write it like l 2 l 1 l 1 is to the right and so it will become 1 2 l 1 will increase and the other configuration will remain there right this is all that will happen but suppose the configuration is 1 2 then what happens so it will become 1 3 but 1 3 has to reset so this will reset to 1 or 2 but depending on that value this next value will also change so 1 3 once the engine moves further will become very good it goes to 2 1 or 1 2 with what probabilities we said with equal this was our rule with equal probability this probability 1 by 2 this is also probability 1 by 2 what happens if the initial configuration is 2 1 and the engine moves 2 2 what happens if the initial configuration is 2 2 and the engine moves so very good so the point is the engine will move then the first spring will relax and the second spring will also relax and then you have different possible outcomes so with some probability this will happen with some probability this will happen there are transition probabilities of different configurations so you can write in the end that p 1 1 p 1 2 vector at time t plus 1 is equal to w times p 1 1 p 1 2 p 2 at time t then I know I know the matrix w I just worked it out then I find the largest eigenvector of this w and that is the steady state yes please with equal probability so the relaxation occurs from the engine side the first spring just changes and it pulls the compartment behind once it pulls the compartment behind then the next chain will become long then it will pull the compartment behind and so on there is the dynamics as defined okay so the matrix w is a 4 by 4 matrix so it can be worked out without too much hassle that is how you determine the steady state of this system yes yes of course once you define the model the matrix w is well defined okay yes yes I guess so but so in the definition of the model we said that the first spring relaxes the next one relaxes yeah okay that you know you can ask what is the definition of the model which order do springs relax and so on and so the answer is that it may happen that you know this spring relaxes but then this one relaxes this relaxes again so we said no no no start from the right end and then only propagate towards the left yes please sorry yes I hope so no no no I am asking is this fact well known to everybody that the steady state will be the largest eigenvector of a Markov matrix yeah at least some people know it know it is clearly not known to everybody but I am asking that I we actually use this property sometime you know we said that there is a evolution operator w which we wrote down w is this matrix it is the probability evolution matrix yes the largest eigenvalue is one in this problem because the steady state the vector does not change all other vectors have eigenvalue less than one that is a general property of Markov processes yes of course this w is also w gives you one step evolution in time and then if you want 10 step evolution you write w to the power 10 value will be the raise to power 10 but one raise to the power 10 is still one all other numbers are smaller and they decrease so the proof is that if you write a Markov process in which there is the matrix w then it has one eigenvalue at least which is one and all other eigenvalues have to have modulus less than or equal to one they cannot be bigger than one because if it is bigger than one you raise it to power 15 probabilities this will become divergent which is not allowed so the Markov matrix cannot have eigenvalues with modulus bigger than one that depends on the details of the Markov matrix so then it is a degenerate Markov matrix in which nothing evolves it remains the step put then the steady state is not unique if the eigenvalue of the Markov matrix one is not unique if the eigenvector corresponding to eigenvalue one is not unique then the steady state is not unique okay yeah it is not okay can I ask a question about the previous exercise do we have to find C or do we have to prove it is you have to determine the eigenvector that is what it says determine the probabilities of different configurations exercise the previous seven do we have to find C in a question about yeah yeah yeah you have to find it it is an arm what is it I would say one I would say half actually we gave the same please remember we gave the argument each particle of course takes L steps but when I ask how many toplings occur in each topling two particles move so there is a half if you don't put down a half you will not get full credit for some such question clearly okay yes I would only look for if you have got the general idea I will not look for a very detailed and careful proof so long as I see that you have got the idea I will give you credit okay this is a low test low order test in the sense I am not really distinguishing between a person who gets the Nobel prize and the person who doesn't get the Nobel prize we are looking for people who pass the course and don't pass the course so the threshold for checking is also lower I will not ask very subtle questions I will not ask please excuse me when I talk I prefer that other people don't talk okay so I am saying that in the exam we are not really it is not a very high grade exam so we are only checking for basic understanding like if I ask you this problem if you somehow I am able to convince myself that you understand what is the W matrix and you have tried to write it you have tried to write the eigenvector you made some mistake in the algebra determining it you will get partial credit clearly you will not get full credit but a person who doesn't even recognize that this is a Markov process and there is a transition matrix then of course that doesn't get any credit whatsoever okay you are saying something namely you will not bad one I see okay I hope not my attempt has been to make clear okay okay so here is a suppose I write the same pile on a 4 by a lattice okay so the sides are labeled I guess this one is called well we will count x this way so this is called 1, 1 this is called 1, 2 this is called 1, 3 4 this one is called 2, 1, 2, 2 2, 3, 2, 4 is this point clear the coordinates are defined x this way and y this way this is consistent with the quantum mechanics is done and operators are written and so on and so forth or Urdu or Arabic is written right so we count from right to left and so there is an operator called Axy or there is a set of operators called Axy and then there is this operator algebra we already wrote down I do not know a to the power 4 equal to a with some indices below okay so the question is use the operator algebra express Axy in terms of a x equal to 1 y okay can one do this Axy in terms of put x equal to 1 and y can take value 1 to 4 so can you write a to 1 in terms of a 1 1, a 1 2, a 1 3, a 1 4 you think that is nice but can you do it no this is an interesting question that I cannot okay I think I am going for okay so here is my problem these operators can be expressed in terms of a 1 1, a 1 2, a 1 3, a 1 4 they are them what about this one a 2 1 well I have an equation which says a 1 1 to the power 4 is equal to a 2 1 a 1 2 right so why just invert it I write a 2 1 is equal to a 1 1 to the power 4 a 1 2 to the power minus 1 they are the addition operators no they are these matrices of size 4 to the power n by 4 to the power n actually working on the recurrent set now a operators are the operators which were defined in the first lecture and they are the addition operators they act on the space of configuration give you a new configuration but think of them as matrices and then this equation holds a 2 1 is equal to a 1 1 to the power 4 a 1 2 to the power minus 1 what is a 2 2 one of you is brave enough to give the final answer yes yeah you speak out then I will write a 2 2 equal to a a 2 2 to the power 4 no I want to write a 2 2 in terms of other things a 1 1 inverse that's it change the order but this commutes that sounds good to me is this clear so I can write also a 2 3 and a 2 4 there will also be some powers of a 1 1 a 1 2 a 1 3 a 1 4 once I have done this then I can also write a 3 1 1 a 3 1 y in terms of these powers ok you always get integer powers of the basic operators positive or negative I can do at least one more problem ok then yeah so we said express x y in terms of x equal to 1 y so it's clear they are just integer powers of it so all these operators they are looking like many but they are only expressed in terms of these 4 basic operators then you can ask further there are these 4 operators the generators of the full group or not so that question is a little bit you know difficult to answer not immediately clear not difficult to answer not immediately clear so that's not part of my question the question stops here if you recognize the basic operator satisfied by the algebra basic equation satisfied by the operators write the rearrange them in some way you get the result it's not a big deal the only thing which is required is a little bit of ability to think about the problem directly ok question number 10 is like this that I have a sand pile like this which is L by L but L is big then I add at some site at the boundary you cannot see anything particle so I will make the boundary like this this is my pile and this is empty so it is half space x bigger than equal to 0 ok now I add a particle at the origin here a is 0 ok and then you calculate how many toplings occur at distance r I think we showed earlier that g of r which is the number of toplings at r went like 1 upon r to the power d minus 2 ok this was shown and for site adding site at the boundary boundary g goes like 1 upon r to the power d minus 1 so I am just saying that I add a particle here how many toplings will occur how does that vary with r it varies with the different power then if this was in the bulk if it was in the middle not away from the boundary so this power is changed from d minus 2 to d minus 1 then can we show that it goes as r to the power d minus 1 ok now the proof will be suppose you have a point charge and I calculate the potential due to a point charge potential due to a point charge in 2 d goes like log r but suppose the potential is near a conducting wall I am sorry suppose the charge is near a conducting wall then the corresponding potential behaves differently what is it one word one word ok does everybody agree so I think the only thing to realize is that we were doing the electrostatic problem all alone nothing very profound the propagator g is just the electrostatic potential we did the register problem where we got some log r we solve the Laplace's equation to get the potential the Laplace's equation is the solution of an electrostatic problem this is the discrete analog of the electrostatic problem but there is there still dipoles and the potential at large r will go like 1 by r in 2 d 1 by r squared in 3 d multiplied by cos theta there is a angle dependence also to the potential right that is all there is sorry sorry please one at a time basic operators are the operators a then what of the group corresponding to the strip problem not for the universe ok yes yeah ok can you repeat kind of a general argument I didn't get it no that is why I am asking you to repeat ok the power r 1 over r to the d minus 2 but we are adding particles to the whole boundary and say like it's an easy plane right so we can say just we can do like integration on the boundary yeah I cannot do that integral in my head no no no you can do the integral but is it clear what is the answer of the integration that is proportional to r to the 1 over r to the d to d minus 2 times r but it is times 1 by r it's not times r it is times 1 by r the answer the first problem was log r the new answer is 1 by r if the potential dies faster not slower 1 r to the power d minus in 3 d the ordinary potential goes like 1 by r the potential near the conducting wall goes like 1 by r squared it goes faster goes down faster your integration would suggest the opposite ok no no but the argument is actually good what you have to say is that if I want to solve the problem with phi equal to 0 on the some line then there is an image method of images you take a positive and you take the image charge which is negative and add them up and that gives you the dipole potential and dipole potential then as a 1 by r extra factor added this argument is continuing to work for the discrete Laplace's problem can you give me the the argument is that both of them satisfy the Laplace's equation with the point source no that is the end of the argument it's a correct what an rigorous argument see if there are two different problems one is an electrostatics problem one is a sand pile problem what is the relation between these two problems the connection is that they satisfy similar equations so I already showed that the g function which I define for my sand pile problem satisfies the Laplace's equation Poisson equation and so there is problem is like this electrostatic problem of some sort suppose you are doing a real experiment with sand then what do you expect you know all the electrostatics you have read all of Jackson then what it depends on how much you are able to connect different problems ok so if I am an experimentalist is not an excuse if you should be an experimentalist with some imagination and some head then you are able to make the connection putting the machinery very good when I add a particle here things leave and then some toplings occur and things leave from here so fewer toplings occur because particles leave ok is that clear I think we should stop because I am getting tired ok


So, this idea that you can look at the company in this simple piece-part way and run each piece-part well, and it'll add up to something good, is just not showing itself to actually work. And I think it's becoming less effective as corporations get bigger and more complicated. Corporations now are of a size unimaginable even 50 years ago. I think that if we switched from the notion that all jobs are flat to a whole set of jobs are actually project-oriented and organized around projects, I think we'd have higher productivity, higher satisfaction, and greater adaptability in the corporate world.


Here are five totally made-up crises Republicans have invented to distract from the real crises facing Americans today, the growing concentration of wealth, the worsening climate crisis, and the undermining of our democracy. Fake crisis number one, anything they claim is woke. And we reject woke ideology. Don't ever apologize to the woke mob.  Although Republicans struggle to define what woke even means. Woke, woke, woke, you know, it's like just a term that you use half that people can't even define it. They don't know what it is. They're constantly using it as a weapon to combat anything that seeks to foster tolerance and acceptance. Pride flags, woke, Rosa Parks. Green M&Ms, the wokeest. Fortunately, most Americans think being informed and aware of social injustice, which is what being woke really means, is a good thing. Fake crisis number two, the panic over trans people. Transgenderism is in fact satanic. Trans people just want the right to exist safely as their true selves, like everyone else.  But they become easy scapegoats for the GOP who vilify them and threaten to criminalize their very existence. Fake crisis number three, critical race theory. Critical race theory in education.  Now in reality, critical race theory is mostly taught in universities, like quantum physics or philosophy. It's really not taught in K through 12, nor is it dangerous. It's merely a framework to understand the role that race and racism have played in shaping America's laws and institutions. But Republicans have deliberately turned this obscure academic phrase into a weapon to silence any discussion of race they don't like. Unfortunately this includes teaching many basic historical facts. Fake crisis number four, couch potatoes. Republicans are whipping up anger over welfare recipients supposedly abusing the system. Well, I don't think hardworking Americans should be paying for all of the social services for people who could make a broader contribution and instead are couch potatoes. The reality is most people who collect benefits already hold jobs and work exceedingly hard. Like Ronald Reagan's claim about so-called welfare queens, the couch potato myth is a cruel racial dog whistle. In fact, the vast majority of Americans who receive government benefits are white. We should be asking why so many jobs pay such low wages that workers need government help to get by. Fake crisis number five, out of control government spending. Another lie, apart from mandatory spending like social security, Medicare and Medicaid, government spending has actually fallen more than 30% in the past 50 years as a percentage of our total economy. Yes, the national debt is a problem, but in recent years, among its biggest drivers have been the Bush and Trump tax cuts, which have added nearly $10 trillion to the debt since their enactment. All five of these so-called crises have been manufactured by the GOP. They're entirely made up why? To deflect attention from the near record share of the nation's income and wealth now going to the richest Americans. As the wealthy pour money into politics, largely the GOP, they don't want the rest of America to notice they're rigging the economy for their own benefits, that their greed is worsening the climate crisis and they're undermining our democracy. So the game of the Republican Party and their major donors is to deflect attention to use fake crises to disguise what's really going on. Let them get away with it.


Okay, well hi everybody. Welcome to our all-day movie workshop. I'm Peter and we've got a great movie coming up today. So I'm going to be the host and I'll just give you a run-through of what's happening today. So we're going to start with the movie with David and then afterwards we'll have a 10-minute break. Then we're going to have our breakout rooms where you can share any emotions or thoughts that came up during the movie. And then we'll have a 45-minute break and then I'll be holding a closing session. So I'll pass it over to David now. Hi everyone, welcome. We're in for a great adventure today. It's going to be a deep one, a very deep adventure.  This is a classic awakening movie and I know a lot of you have not seen it so you'll be thrilled at seeing it for the first time. It's going to take quite a bit of commentary and there was a movie some weeks ago, maybe even months ago that I did called Grace and Grit, where I had to pretty much do a big setup and then kind of chop the movie and to segments to actually show the movie as a teaching device because this is the kind of movie if you just have never seen it and you just play it straight through without a lot of commentary then your mind can start to question what is happening. Kind of like maybe the first time you watched Mr. Nobody and you watched the movie and you were five or ten minutes in just going shaking your head like I don't get it. So with a lot of quantum movies they go so far beyond the linear time frame that the ego offers us to look through at the world and they're really about going way beyond what seems to be the accepted meanings and interpretations of the world. What do I mean by that? Well if you talk to most people on Planet Earth if you said yesterday, today, tomorrow people could relate to that in whatever language or culture you were sharing but there's something much, much deeper that is available to us if we want to really get the point. What is


In this presentation, we will apply direct labor to our jobs. Let's get into it with Intuit's QuickBooks Online. Here we are in our job costing dashboard to get a concept of our objective. We're going to go to our Excel worksheet. Within the Excel worksheet, we're going to be now on the EX1.12 tab. If you want to record the journal entry, you can be in the blue item. I'm going to be here in the green one for the example of it. We have the next item, which is going to be typically increasing the jobs that cost a good sold account in our practice problem. That's going to be the direct labor. We have the direct labor, which is going to be totaling up to 218,000. That's going to be applied out to Job 14, 15, and 16. When you think about the labor component, whenever we think of jobs, note that it's going to consist of basically materials, labor, and overhead. When you think about the labor component, then you could think about, well, what's the format of our labor? Is our format in the form of contract labor, in which case we just pay someone in 1099 at the end of the year that we can apply that information to the job? Or are they employees, which is going to be a bit more complicated because then we have to process the payroll and think about how we're going to allocate the payroll to basically the jobs. Now, if you want to run the payroll within QuickBooks and use payroll in order to allocate, then you can use the time sheets in order to do that. We may get into that at a later point in time. We're not going to get into that at this point. We do have a bit of a... We're going to talk at this point about if it was a contractor that you're having, that you're going to contract out so we could see the transaction. And we'll discuss a bit of a workaround if you want to enter the payroll, but don't want to go through the kind of tedium of entering the time sheets as the method that you're going to be using to allocate the time to the jobs through the actual payroll processing. So let's think about that now. We're going to go in same kind of journal entry, first of all. If you didn't have any kind of payroll issues and if it was just basically same process here, we would be paying the employees, meaning cash would go down or if it's a contractor, we would be paying the contractors. Then we would be debiting the cost of good sold account for the job. So they'd be going into the cost of good sold account. Then we got to think about the cost of good sold account and break that out to the actual jobs to allocate them to the jobs. We have to track in some way which jobs these would be allocated to. So for example, we said that we're going to allocate 30,000 to 14, 15 and 15 is going to be 68,000 and then 16 is going to be the 120,000. Those three then, of course, adding up to the total of the total of... So I got distracted. I had an irresistible scratch or itch in my ear. But in any case, that's going to be the 218,000. So those are the 218,000. So we're going to do a similar process when we go over to our QuickBooks file. So let's go into QuickBooks, how we're going to enter this into the QuickBooks. We're going to use a similar process. We're going to go to the plus button up top. I'm going to enter an expense. Now remember, we're thinking about it as a contractor, but you can also think, well, what if we processed payroll and what if I just wanted to process payroll and then use some other method like a billable type of method to allocate the time that is spent for employees to the jobs. Well, here's how we can think about that as well. We're going to use the expenses in a similar type of way. If we're actually paying the contractors at this point in time, then of course the cash would be going down. If we're not paying the contractors, we could use a similar method. In other words, if we processed payroll and then want to do an allocation, we could use a similar method that we did with the material. So let's first, I'm going to close the categories or item here and we're going to go down to the items. Now we could do this like one item at a time. If we so chose or we can put them all into one item and also note that if you had different kind of labor, different kind of contractors doing different types of things, then you might set up different items for those different contractors as well. And then you could set those items up to go to different kind of costs to good sold accounts in a similar fashion as we had done with the materials accounts. But we're just going to put everything into direct labor here. So it's a broad category for direct labor. So I'm going to say direct labor and we're going to be picking that up. And the first job was for 30,000, I believe. So that's the full amount that's going to be going to job number 14. I'm going to make it billable. We're going to say that's the full amount for job number 14. We could do this with three separate expense accounts like we did with the materials, but we can also use the one expense account and just simply apply them to two different jobs as we will do. So I'm going to say direct labor again, but this time applying it and this one's going to be for 68,000. That's different. That's different as well. And then I'm going to make it billable. That's the same. But this one's going to be going to job number 15, job number 15. And we'll pick that up. And then we're going to do this again one more time. As you could probably guess for the direct labor, it's going to be going for 120,000. And that is going to be billable. And this is going to be job number 16, job number 16. So we've got the 30,000, the 68,000, the 120,000, those three tying out to the 30,000, the 68,000, the 120,000 in our job sheet. If we were to add them all up, they add up to that 218,000, the 218,000 being the bottom line number on our expense item here. Now note, if we're paying, if we're actually writing a check to a contractor, someone outside, then we would simply see the checking account go down by that 218,000. That's what we will do here. What if, however, what if you had processed this through the payroll account and now you just want to allocate this information to the job? Well, these items would be going to wherever you want to go, the direct labor. And then you can do the same kind of method. You could say, well, wherever you allocated the payroll before, you could basically make a negative amount here. So if you put it into wages expense, you could make a negative wages expense. You could pick the payroll account, the payroll expense here, salaries and wages, if they went into salaries and wages, I can then make a negative amount to the negative 218,000. And what that'll do is it'll basically, these items will take this amount out and record it as an expense, right? It'll record it as an expense under the cost of good sold for basically labor. And then this item will reduce the expense that was recorded in wages when you process the payroll. So the net expense account will not be changed from when you process the payroll, you're just reallocating some of the expense. So that's what you might want. That's one kind of workaround you could think about with the payroll.  So in any case, I'm not going to do that now. I'm going to say as if we were paying a contractor that we're allocating out. So I'm going to remove that. And so there we have this. So there's the 218,000. I didn't put in the payee and the date and whatnot. This is going to be on the 12th, it looks like. So let's make this the 12th. So I'm going to go plus, plus, plus, plus, and that brings it up to the 12th and then I could have put in contractor one. That's going to be the contractor. That's their name. It's kind of obviously somewhat of a generic name. And it's going to be coming out of the clearing account. No, it's not coming out of the clearing account this time. This one's actually a check that will be coming out of the checking account. So this will be coming out of the checking account this time. And the method then is going to be, let's just say cash. Probably it would be a check. Actually, should probably have a check and then we would need. All right. So in any case, there we have it. If we had a check, we'd probably have a reference check number. So, okay, that's it. That's all I'm going to do. So what's going to happen? Then the labor down here is going to be applying it out to the expense account. And then the other side is going to be coming out of the actual checking account as we would expect. And then it'll be applying to the job or project 14, 15 and 16 and these amounts respectively. Let's save and close and then take a look at the reports. So save and close reports on the left hand side. We'll open those reports up. We're going into our standards. That being the balance sheet first, let's open up the balance sheet. We're going to be in 2020. So as long as you're up in 2020, the dates should be okay. You can see the checking account went down this time. Was that, well, now it's at 82,000. I'm going to close the hamburger. And there's our transaction. There's the 218,000. It's a expense type of transaction. If we were to go into that transaction by zooming on the transaction, there it is. There's the activity. Going to close this back out. And do you want to leave without saying? I didn't change anything, I hope. So I'm going to leave without saving it. I'm going to go back to the balance sheet. It got me worried. I'm like a little worried now that I messed something up, but I don't think I did. So then if I go back down, it looks good. Now let's look at the other side. So I'm going to right click on the tab. Let's go to the tab so we can duplicate it. Right click on the tab up top and then duplicate it. So now we've got like the balance sheet on the right and we could do stuff on the left. So we're going to go back on the left, open up the hamburger. We're going to go on down to the reports on the left side, open up the P&L, the profit and loss, otherwise known as the income statement. Let's open that up and then close the old hamburger again.  I'll make this one a little bit larger. I like to see it at that 125, the 125 zoom. And this is going to be, we're in 2020, so that should be good. So now we can see all this in just cost of goods sold, now adding up to that 568,000. And you can see this is the same thing we had on our Excel sheet. Now the cost of goods sold, adding up to that 568. So there's the 568. If I minimize that, that looks normal. And then if I was to break that cost of goods sold out, we could then see it by our major categories, that being the labor and the materials at this point. But that's not all the labor and materials because this is only for the current time period 2020. Labor we only put into one account. Notice we could have done the same thing as we did with the materials. We could have broken out the labor into various components of the labor so that we can have different cost of goods sold expense accounts breaking out the different components of labor if we so choose. So that is an option for you. If we then go down, we're at the 568,000. So there's the 568,000. If we go to the balance sheet and we check out the old balance sheet over here and scroll down, let's close the hamburger here too. There's no reason for that burger to be open. Then we see the 568,000 in the net income here as well. There's the 568,000. Now if we go back to the income statement, we also note that that's just for the current period. But these jobs crossed over the current period. So we could then see, let's take a look at the old dropdown here instead of totals. I want to see this by customer. Let's check it out by customer and then run the report again. And then we got 14, 15 and 16. And these are going to be our open jobs now. And we can see they're at 130,000, the 238,000 and the 200,000. So if I go back on over here, note that would just be the current activity. So the 130,000 for job number 14 would be these items, the 100 and the 30. So that's going to give us job number 14 for the 130,000. The 238,000 would just be the current items that happened in the current period, which is going to be the 170 and the 68,000 adding up to that 238,000. So that adds up over here for the 238,000 job. 16 only happened in 2000 in the year 2000 and after and therefore it's all this, it'll be there. It's all there. Now, if we wanted to see the full job, then I could bring this back to 2019. So let's make it 10119. Then we can take a look at our open jobs and see the lifespan of the job. And then if we scroll on down, we could say, okay, here's job number 14. If I bring it on down, we're at the 171,000. So if I go back on over here, there's the 171,000 and there's the 280,000. There's the 280,000. And here is the final item being, it won't let me go back the 200 and then the 200,000. 200,000 down here for the total of the 651. So the 651, if I go back on over here, the 651 is going to be here. So that's going to give us the full job. Remember that reconciliation of being able to take a look at the full open jobs and then breaking out what's in the beginning balance. It's going to be a tool that can help us if we need to do some kind of conversion at the end of the year to a completed contract method or a percentage of completion method if we needed to record the work in process account or something of that nature. So that could help us out with that process as we go. And we can also then track by the project. So let's right click on the tab up top again, right click on the tab up top, duplicate that tab, go back to the tab to the left. Then I'm going to open up the hamburger. I'm going to hold control and scroll back down to get us back down to like 100%. So QuickBooks doesn't do anything like weird when I go into the reporting. Then I'm going to go into the projects and then we can close up the hamburger once we're in the projects to give us a little bit more room. And then down here, I'm going to hold down control and go even a little lower than the 100% down here so I could see my projects a little bit better just the way my screen manipulates. Then I can go into any one of the projects. Let's look at 16 this time. I don't think we've done that yet. So we look at number 16 and there we have our materials and we've got now our labor item, which is our new item. And of course, this will give us another kind of breakdown of our items. Now also note that if you were to be doing this information with the timesheet information through payroll and using timesheets, then you can get added information of time tracking type of information within there. You might use T sheets as well within that process. So we're not going to get into that at this point in time, but just be aware of that. So then you've got your reports, of course, on the right hand side for that tab. So we're going to go back to the balance sheet. We will be printing out the trial balance whenever we remember at the end of these so that you can then use them and go back to them as a reference. That's going to be it for now though. Let's get out of here.


And our crews with the news giveaway is back again this summer as well. Every weeknight between now and the middle of August we're announcing five potential qualifiers for our crews giveaway. So if you see your name tonight, give us a call at the number on the screen before 5 p.m. Friday and you'll be entered in the grand draw. Tonight's potential qualifiers. Helen Ann Zuztec of Baxter, Fred Underhill of Pierce, Barbara Maxwell of Bemidji, Karen Rongstead of Bemidji, and Daniel Brown of Bagley. Call us at<PHONE_NUMBER> before 5 p.m. Friday to qualify and have a shout at winning that grand prize. If you've enjoyed this segment of Lakeland News, please consider making a tax deductible contribution to Lakeland Public Television.