Having a fireside chat instead of like a regular presentation, but we're recording this so I'll go ahead and like do this like a presentation My name is Eric Chabelle. I work for Red Hat for more than 10 years now Kind of an interesting title I have there's a second person on this Presentation that I put up there. It's a very good friend of mine that I've worked with in the Netherlands. It's where I'm from He spends a lot of time as a solutions architect talking to customers about all kinds of interesting stuff and About two years ago. We came up with this pitfalls kind of theme talking about What customers what partners what people are running into in various aspects of? product development software development Digital transformation all this kind of stuff right a lot of buzzwords involved We did a talk a couple of years ago at Red Hat Summit about three pitfalls of hybrid multicloud And people got so excited about that it was voted like top 10% or something in that conference It was kind of shocking to us because it all started with a phone call between him and me Where he's complaining about how customers keep making the same mistakes all the time You know, I just ran into another one is doing this again And this three pitfalls happened the same way a year ago we started talking about it and We took this Too far away by hand So we did this at Red Hat Summit in Boston this year and what we did if you look really closely It's kind of hard to see but at the top there we put a whiteboard in front of the door So when you came in and we divided we had two arrows We said if you're doing microservices sit on this side and if you're thinking about it Or just interested in it or just here because you don't know what's going on sit over here And this is right at the beginning, but you notice really quickly. It's about 50 50 isn't it? And it was completely full and it was 50 50 and the way they're positioned in the room We got sort of like this thing that one's talking to the other and we did this a little bit as a discussion And we did way more than just three We put up these opinions and and my controversial statements are things that people think Have to do with these microservices and we would let them talk about it And it was really kind of funny. He was at one side of the room and I at the other and you'd see The conversation being taken away from us and they started arguing about stuff back and forth across this So what I did is we took notes during this talk and I picked out three for this presentation and this is the first one How many people in here are developers that everybody looks like it Who here's an architect who here is doing microservices right now for who here has never touched them No idea what I'm talking about Okay How many people here have done services web services rest services whatever services you have a concept of what a service is right? How many people have an actual job And I'm just saying that in a funny way who here's a student So we got one student. Okay. That was that was a little bit of my concern about how many students were going to show up to this because one of the things About working in an organization, you realize that the complexities versus what you program in the university are vastly different. I Remember my very first job they came in here write this web service. That's no problem But writing a web service on your laptop versus writing a web service in an organization is a whole different ball of wax How many people here work in an organization and we use kind of a buzzwordy kind of thing here saying Traditionally siloed organization. So here who here works on applications or Multiple applications in the sense of there's a deployment cycle. It takes two three four six six months to deploy something all the way to production Maintenance maintenance cycles a couple times a year maybe once a year Who here works on the services that are in some kind of service layer?  Hopefully like a g-ray issues or Whatever it is that you're working towards a release cycle Then you make the step towards microservices and for the guys here or the gals here that have done microservices services Does this sound anything like what happened when you guys jumped into microservices and thought wow, this is really great We're gonna make a bunch of microservices and off we go Did your organization change? Anybody have anything that they could share with that So you don't get asleep in this talk today Anybody wants to say anything but What we find and what we see a lot of is people say okay, we're gonna do microservices We're gonna chop this stuff up take a big old monolith. We're gonna break it up into pieces at least some of it and what they're not realizing it and not understanding is that there's a completely different methodology involved with working with this stuff so Instead of being a development team that owns everything Or at least most of it so maybe have a front-end team and maybe have a back-end team or services team But they own all of it now you're splitting it up and you're giving each one of these pieces to individual teams that own that piece So imagine owning that one service, and I don't mean just owning that service like develop it push it into a pipeline, and I'm done You own everything going forward you own the release cycles you own the CVE Security patches you get to decide when where and how you release this and how often So instead of doing your development cycle on your app Pushing this stuff out once every three four six months if you're really good at it. I worked in a bank So it took us Almost eight months just to do one release Now you're talking about all the individual pieces of your application can be released hourly Every half hour if you want to every time you push a fix if you want to Think also about that development team has to own the testing of their individual service They have to own the API front-end of it You have to Document that properly so people understand that your API isn't changing drastically between this version that version of breaking Everybody's use of it Things that people do not think about they're just not ready for that stuff and this makes it super hard So a lot of people are getting into it in the first case start breaking stuff up and then realize wait a minute That's a now start treating each one of these individual services in a team as a business-to-business connection So it's just as if you've now created a little team outside your company And you have to interact with them so does anybody using any kind of piece of software ever from an external party such as Salesforce Some of the Google services, whatever they are they have an API. They have a development team that does stuff completely out of your purview All right, you have to deal with that You don't get a choice. You don't own the API. You don't own the release cycle. You don't own any of that That's something that people are absolutely not prepared for The second one Sounds kind of simple and obvious Maybe you think there's how many people think that just a small service is a microservice you shouldn't The thing behind this is it's got to do with the hints I was saying about having a team and having to treat it as if it's an external party Where this development team owns everything has the interface back and forth like that One of the questions that comes up really quickly after they discover so so imagine the company starts doing this Making this transition bumps into oh we have to have a little bit of an organizational change because you have to give the freedom away to These individual teams to do what they're going to do And play nice and set up the infrastructure to do all that Then you come back around and you start thinking about oh Wait a minute. I have this big monolith. I know performance. I know API calls. I know service calls Are all inside this monolith so I know how this works how this interacts. I know the performance of this I can test this I can put metrics out about this Anybody here attend Christina Lynn's talk yesterday about a reference architecture for microservices She had some really cool drawings up there if you remember now if you look at these two sets of seats Imagine each one of these seats are an individual microservice in your organization And all these are deployed in some data center over here in the world And all these are deployed somewhere else in the world And I'm the application up front here, and I'm using you you and him over there You don't talk directly with each other ever Right. He has an interface. You have an interface and you have an interface. I talked to you I do that through entering this data center at the front end and going all the way to the back to whoever it is and talked I Don't know exactly what the performance in the network is going to be each time I make that call I don't know that If he wants to make a call over and use some service over here He has to come all the way out here all the way in here into that data center, right? So what you're giving up is the freedom of knowing what your Networking calls and speed and the performance is going to be for the flexibility to deploy all these things at any time you want To be able to fix security bugs like that To be able to react to what your customer wants to do I can change my business application up front here to their needs Any time I want because I'm not dependent on all these different services being updated to it's not a monolith anymore So I've taken the monolith broken it up into these pieces and you can now Have a trade-off you have to do deal with how far do I want to go or do I not want to go or can I live with The performance issues that I don't exactly know Anybody go through that when they were doing microservices anybody split up a monolith before Don't promise them anything when you do that So To find that happiness has an awful lot to do with understanding how you're approaching this and what you're trying to achieve and being able to communicate that correctly to your CIOs or to your Lead architects or to whoever is sponsoring this kind of stuff To become that modern cloud native high-performance Quickly agile company being able to react to everything and anything that's happening around you You can't be deploying a monolith Over eight months Trust me. Everybody else is gonna be way faster than you.  That's how that works Any questions so far anything I said sound like something out in left field We just said monoliths are not something you want to be deploying every eight months What if your monoliths is that fast? right So microservices are not always Better than than monoliths What do you think of that? So this is kind of contradicting everything I just said right There's nothing wrong with looking at a monolith and deciding large portions of it are not worth turning into microservices are tearing apart What if I take those three services out of my monolith that I just talked about put them in those data centers and Now I can release my monolith every two months instead of every eight That might be enough business justification to stop right there and don't spend any more money on that project Because I guarantee you no matter what you're doing and how cool you think technology is The business is the one that owns it the business is the one with the money They're gonna look at you and say why are we even touching this shit? I don't want to spend my money on this Great example the bank where I used to work still has in the corner a machine running From when I worked there almost 15 years ago Running the same version of one of our open-source projects and the business unit still says no I don't want to upgrade it's not broken. Don't fix it It's an option So that's what we mean by having like an agile monolith. It's not so much that you're changing everything, but you're making That slight sliding shift over into some microservices. It's a way to start without having huge impact on your organization It's something to think about it So We've looked at the three different pitfalls The organization might not be ready for this stuff Just saying I'm gonna do microservices is not doing microservices and you are gonna run into a wall if you think that's how it works If you try to design your design teams and your development teams the same way as you did for normal monolithic development or Service development up to now. It is a completely different story Your definition of a microservice and understanding what I described with two data centers and how that works and sacrificing The the known factors of how your your your networking is working in your your timing and metrics on your service calls Versus the agility to be able to react to your business and changes Every hour if you so wish without impacting the actual application It's a huge change Then microservices being not always the answer beyond all monoliths There's nothing wrong with identifying that a monolith Can be slightly carved out and left alone and is agile enough and you sped up the development cycle So we're no longer waiting eight months or more. It's now only two So that's three we could stop here, but it wouldn't really be that fun What if we didn't have like a bonus one? So Who here has been attending all these data talks that they've been having here who's a big data fan? Or a data engineer, what are they what are they calling him now data scientist your data scientist? There you go So one of the things that comes up and it didn't make the top three for me, but it's probably a really close for State Applications have state. What about state?  You're here. What about the state of my application? How am I managing that? What are we doing with that these distributed data stores? How are you dealing with all that stuff? Most of what we're pushing for when I talk to people about this is there's We use open source technology from Red Hat so that there's quite a few things going on out there in the cloud native world Where they put a layer? Across the stuff that monitors change in your data sources And this is this is the path I would suggest is instead of tearing it all up and trying to store data Locally by each one of these things and completely radically changing an architecture that's already in place you have to realize most organizations have a whole lot of Yeah, technical debt is the word you always hear but it has to do with they've made decisions in the past and you're stuck with So this database is not going anywhere because it comes from Oracle and the CEO made a big commitment to Oracle And that's what's going to be here for the next 10 years. That's just the way it is But to be able to have the state information you want and to monitor that correctly to plug into that correctly There are layers of software that will now monitor those changes and allow you to publish those changes directly to individual microservices and things like that When you look at these slides and I forgot to mention They're already online about 10 minutes before this talk started they published to Chevelle.org my last name is my website That's where you'll find them In the notes of the the talk you'll find links to the very software element you can look at that do that stuff The other option is you know people try to and that's what you mostly see happening It's like a gut reaction from anybody that's done services. It's a tightly coupled the data for each one of these services to the individual Individual services, which means you now have containers out there with stores being coupled tightly to the service in the container Which is not exactly the flexibility you're looking for I got a few references of stuff you can read and look at This stuff is not easy Don't care who you are or what they tell you it's not easy, but Just like anything we do and we're so ready university research it a little bit Look for some of these best practices look for some of these these tips and tricks and stuff And make make decisions that work well for your organization I still I mean one of the most simple things as an architect when you look at an organization or look at stuff That you're trying to do is keep it simple, you know that kiss principle that works all the time that has the longest living Well-functioning execution of a solution is if you keep it simple so if you can take a hint out of this stuff that would probably be the best Anybody have any questions it sound like hocus pocus or can you use some of this down front? You can just talk and I'll repeat it Make some really great points about With the microservice, how do you determine? When to use a microservice and then after that how would you sell it? Okay, so First thing you said it made some good points about microservices. Second thing you said was That's correct me if I say this wrong. How? How do I decide when to make something in microservice and then how do I sell it? Which I don't understand. Oh, well, so let's say you've decided we need to use microservices to solve this problem You Okay, so that he's saying I for some reason in my organization have decided that we're going to do microservices and how do I communicate that to my team? So it has an awful lot to do with the number two pitfall that we talked about You're looking for agility versus It sounds really bad right agility versus stability I Don't mean stability of your application. I mean stability of a known Almost waterfall kind of process of how you're developing stuff from front to back do a release and I go back around I do front to back do another release that stuff takes a long time. It's all tightly coupled We're going for the agility of being able to Expand your organization's Reaction to your domain to your market to your customers a Very good trigger could be that you want to use some service that is already a third party So if you've been interacting with any kind of a sass application or sass service from Sales force is always an easy one to use. Let's say you're doing that or you have some third-party vendor that you're plugging into that process you just replicate for services across your organization and Usually it gets triggered a decision like this is because you're starting to think about doing this modern way of being agile and getting a little bit more quick to respond based on an upgrade or an update to An existing monolithic something where you could carve off a few things and that's what I would always push for and what I always Prefer versus just like a brand new greenfield. Let's do microservices for this project If you do if you see people doing that, that's where all these come from Right because they stumble over all of them You just you're not sure you don't know exactly what you're doing and you it's just logical And you're gonna bump into this stuff Taking things in smaller jumps smaller pieces and smaller bites But you make mistakes without killing yourself or ruining a project or ruining your name or Running budgets up through the roof because you forgot about a bunch of aspects And I spent some time in Holland at one of our The biggest Railway maintenance Organization we have and they're still going through saying this is like a government organization. They're not agile It's not a quick change. Nothing is quick It's filled and staffed by IT people that are really happy with what they're doing for years and years and years and don't want to change Their way of working so you're being brought in to talk about this stuff because a younger somebody is now put in charge to make change and The best way to approach is to ease them into it, right? I Truly believe that Does that answer your question in a long roundabout way? Yeah, there's no single right answer never and never yeah, there's no single right answer absolutely not Like like the third one says there are things that don't need to be a microservice Could because of cost could because of you know the known fact it could because of a customer that doesn't want to pay for it that kind of thing Anybody else Yes, sir So I Okay, I'm gonna try to repeat that for you guys I'm gonna paraphrase a little bit so Say say if I'm wrong How do you ensure that all these individual micro services that are doing data transactions and stuff? Equal the sum of what used to be one great big transaction from your monolithic application close enough Okay, that's that's exactly what I talked about in the bonus one The bonus pitfall There are Various options to ensure that there's a layer between your micro service Architecture and services that are dealing with the state with the database or whatever your storage is whether it could be data lake It could be anything right Data warehousing whatever it happens to be Keeps the consistency and guarantees the the transactions as you're saying and you use that layer to ensure and to Configure for each individual search what they need to know so read write whatever it is whatever data they need to have that's how you how you manage that and That is used also to map the answers back to whatever the application needs to know So it sounds sounds a little bit like you're isolating everything out into individual services and that it's a mess, but It's a bit of magic going on in that That that layer you're putting across it One of the tools is called. I don't even know if I'm saying it right because we say it differently in Holland But debesium if you've heard of that There's links in the in the presentation for that and that specific slide about what these components are take a look at it It's it's about you know putting a layer across to monitor changes and report those back to where you tell it to report it back to So if I'm a microservice that needs to know about an update to a customer ID or something like that or an age because it's a birthday It will push that back you can subscribe to that information. I Didn't say the complexities will go down It's not you know nothing is a magic pill. It's it's it's a trade-off. It's a trade-off between known ways of working versus having much more flexibility and quicker releases I Could even imagine in in in ten years ago pushing a button on my code to commit it to a repository and that triggers all kinds of pipelines That basically can push it all the way to production Without an ops guy touching it Anybody doing that now that's the dream isn't it? So imagine if you have a microservice, it's only about this big right? I mean you didn't change anything you just put a CVE change in it to stop some kind of security thing About I don't know half hour. It's in production I bet most ops guys are scared shitless when they hear stuff like that But if all the automated pipelines are in place of all the testings in place, there's nothing wrong with that. Is there? That's the dream Anything else? Well, thanks Got very crowded


Apparently, before we start, I wanted to make an announcement. Apparently, there was a mishap with the grading on L2. I'm sure that everybody's aware of it by now. I will figuring out what to do. I'll make sure everybody gets the grades that they're supposed to get. Nothing's going to happen with whatever the grade said. If you did it all and the grade said you got a 40 or something like that, it's no big deal. We'll make sure that it's all taken care of, I promise you. I know you guys got a lot to worry about without having to worry about all that. It was my fault, apparently, and I apologize for that. Just to let everybody know, there is a mishap with the grading on L2. I'll figure it out and we'll get it all straightened up. Other than that, quiz two was passed back today. If you didn't get it, like I said, you could come up and get it after class. There's the key up here for quiz two if you were wondering why you got anything wrong. First thing I would do is add up the points that I took off and if it's different than what the grade says, make sure you come and bring it up to me because I have obviously been known to make errors and grade it. That being said, I guess I wanted to finish up chapter three. I know we kind of sped through at the end talking about polyatomic ions. Of course, this wasn't on the test because I thought we went through it rather fast. Let's just go back to this, 3.6-compounds introduction to Bonnie. Monoatomic ions, hopefully by now you're fairly familiar with, these are ions that consist of one atom, like H plus, Li plus, Ma plus, K plus, Mg2 plus, S2 minus. These are all the common main group ions and the transition metal ions that you are familiar with. We contrast these to polyatomic ions and these are ions that are composed of two or more atoms together with an overall positive or negative charge. I'd just like to compare monoatomic ion with polyatomic ion first. A common monoatomic ion that you may be familiar with is Br minus, of course. A polyatomic ion would be something like this. I'm just going to draw the structure of it just to emphasize, even though we don't know structure yet. Just to emphasize that you can see it's more than one atom. It's one, two, three, four, five atoms. All five of those atoms, we say, have a single positive charge. In fact, the way I've drawn it, I'm saying that the positive charge is actually residing most on that nitrogen. In fact, it is. The way you'll first learn how to do this is to take this and put brackets around it and say the whole thing has one positive charge. This is a conglomerate of atoms, if you will. That has a positive charge amongst them. If we look here, it's only a single atom with a charge on it. That's the difference between a polyatomic, poly meaning many. Polyatomic meaning many atoms. It's a polyatomic ion and a monatomic. Some polyatomic ions I'd like you to be familiar with are these pink here. Ammonium, nitrate, sulfate. This table isn't found in your book, unfortunately, but it is in the notes. Go to the notes, find this table 3.3, and look for the ones in pink. Those are the ones I want you to be familiar with. Ammonium, nitrate, sulfate, hydroxide, cyanide, phosphate, carbonate, bicarbonate, and acetate. I don't think any of the other ones. If we use any of the other ones, I'll specifically tell you what their names are. I definitely want you to know if I were to give you one of the other ones, you could identify it as a polyatomic ion. If I gave you MnO4- you could say that's a polyatomic ion instead of a monatomic ion. I wouldn't expect you to remember that it's the permanganade ion. It's got a negative one charge. Here's just another list of polyatomic ions. It, again, takes some things about polyatomic ions. You can think of polyatomic ions as charged molecules. A molecule consists of more than one atom. In this case it would be like a charged compound. A compound consists of more than one type of atom. Polyatomic ions are covalently bonded to one another. What we're going to learn today is that these lines represent a chemical bond between these two atoms. We kind of started talking about this in lab last week about bonds, breaking bonds, making bonds, and that type of stuff. Those lines in between those atoms represent bonds. If we believe that then we can say these polyatomic ion atoms are covalently bonded to one another, but the overall particle has a positive charge. It's an ion. Even though you've got covalent bonds in it, it's an ion. The whole thing will form ionic bonds. If I got these two guys together, what we would actually form is NH4, because it's got a positive one charge and it's got a negative one charge, so those will cancel out. We've got NH4Br. This would be the formula unit of these two combines. This name of this compound is ammonium bromide, and we'll get to that in a second. If you look, this compound here, or this ionic substance, this salt, has both covalent bonds within it and the cation. It has covalent bonds. Between the cation and anion has ionic bonds. It's got both types of bonds in it. Here's some common, we can get to it, common metal polyatomic ion compounds. Calcium hydroxide, OH- is hydroxide. How do we know it's OH-? Because we know calcium is 2+, and if we've got a formula like that, 2+, then since we have two of those OH things, then they both must be minuses to cancel it out. This is how you actually will predict if you don't know the charges inherently on these polyatomic ions, strongly urge you to memorize. But there might be on a test something that you forget, or something like that. You can go back to what you know about the main group elements and infer, if you will, what the charge on the polyatomic ion is. Some other ones, strontium nitrate, sodium bicarbonate, sodium hypochloride. These are all common compounds that you find. Here's another picture of some ionic compounds. Most of these have polyatomic ions associated with them. So cobalt nitrate, cobalt 2 nitrate is this red solution. Potassium dichromate, of course, is the orange solution. Cobalt 2 sulfate is this blue solution over here, and potassium permanganate. Okay, so to name these compounds, I know that you guys haven't memorized the polyatomic ions yet, but to name these, you will have to know all of those polyatomic ions. So this one, of course, would be ammonium chloride, barium sulfate, iron 3 nitrate, copper 1 bicarbonate, and calcium hydroxide. So notice I was putting the, for the iron and the copper, I was putting the Roman numeral in the name. You have to do that with transition metals. Okay, let's look at ionic bonding, strength. What you find is there's another periodic trend here. Kind of, you've got large ions that have 1 plus and a 1 minus charge, right? These are not going to be attracted to each other as if you have smaller ions that have a full positive 1 positive or negative 1 charge. And the reason is because if you think about these two, we can say this is a mon... and this is another monatomic ion. Okay, this would be like comparing, I don't know, rubidium and lithium, we'll say. Okay, what you find is that lithium is a very much smaller ion, and rubidium, due to its placement on the periodic table, but they both have the same amount of charge, 1 plus, okay? So rubidium can actually spread that charge out, and what we'll say is delta plus, okay? Delta, so that means partial positive, okay? So this thing can spread its charge out all the way across its surface, yeah? And so can this one. It can do it too. But of course, it's got a smaller surface to do that. So if you were to think about it in a different way, how many little delta positives could you write on the surface of each one of these things? Okay, of course you could write more on this one than you could on this one due to the fact that this is larger, okay? Does everybody get that? Okay, so what happens is or what's the case is that atoms don't like to be charged in general, okay? They do it to get more stable, but even though they're getting more stable, they're still not like completely stable because they've got this positive or negative charge on them, and they don't like it, okay? Which is the reason why they form ionic compounds, okay? So rubidium or lithium, they don't like to have this positive charge, so if they see some chlorine over here, they'll stick to it, so it'll cancel out that positive and negative charge, okay? So what you find is that when you've got two real small ions, of course with the same charge, and compared to two real big ions with the same charge, what'll happen is that these two... So this is negative, negative, negative. These guys will be very much attracted to each other. These guys will be less so because they don't mind having that charge distributed about their surface area as much as these guys do. They still don't like it, so they'll still be attracted to each other. They just won't have as much strength or force of attraction that these guys do. Does that make sense? Because these ones are small, okay? And then, of course, when you increase the charge amounts, so if you don't increase the size of it and then increase the charge, then it makes it even crazier. It's like, okay, so instead of lithium, we'll do beryllium. So beryllium 2 plus really can't stand it because it's got two pluses on it, right? And it's got to distribute amongst the same amount of surface area. So if you find oxygen 2 minus put it in the presence of beryllium 2 plus, they'll really stick together because they're both so small, okay? And then, of course, when you get bigger 2 plus, of course, the attraction isn't as much. And you can see that by looking at the melting points of these different compounds. So lithium fluoride or lithium oxide, the melting point of lithium oxide is much bigger or higher than the melting point of lithium fluoride. Why is that? Because oxide has the 2 minus charge and fluoride only has the 1 minus charge. And then if you look at calcium sulfide compared to potassium sulfide, again, it's the same sort of instance where the... So they're about the same size, calcium and potassium, right? But look at the difference in melting point. It's fairly dramatic. It's due to the fact that calcium has a 2 plus and sulfur has a 2 plus, whereas potassium only has a 1 plus and sulfur has a 2 minus. Okay, so covalent bonding. We talked about covalent bonding. But again, we went through it kind of fast, so I didn't do it on the test. Covalent compounds form from sharing electrons between one another, so they don't transfer their electrons like we talked about with ionic compounds. So they typically form when two or more non-metals react. We'll show more about this next statement in the next slide. But the bond results from the electrostatic attraction between the nucleus of one atom and the electrons of the other atom. Covalent compounds exist as discrete molecules that behave as individual particles, as opposed to the kind of extended networks that ionic compounds result in in compounds here. This ionic compound is actually a polyatomic, monoatomic ionic compound. So this would be like calcium nitrate or something like that. So you can see the polyatoms here with the mono atoms. Just a S2 monoatomic, and hopefully see the difference between those. And then hopefully also see the difference between these and this, right? This is just one little particle. Same thing, same thing, same thing, same thing, same thing. Let's show another covalent compound. What's the difference, right? Very small, discrete individual particles relative to whole conglomerate of things. Very small, discrete particles. Okay? So when we name covalent compounds, usually this only applies, I know you're going to start wanting to name very, very long covalent compounds, but this usually only applies to when you're talking about covalent compounds consisting of two different types of atoms, okay? So you can start naming covalent compounds, but don't try to name them if they've got more than two types of atoms because they have a different name. They won't be named like this. But if they do have two, only two different types of atoms in them, what you can find is that you can name them. So the names of the elements are written in the order which they appear in the formula. Usually they're written with the atomic number of the first atom being less than the atomic number of the second atom, but that's not always the case as you can see. It's not always the case. They're all written very strangely, but if you see the formula written like this, you're going to want to name it in the same way that the formula is written. So there's not really a rhyme or reason the way that they are written. You can kind of say this may be the case that the least, but we haven't learned really about electronegativity yet, but you could say the least of electronegativity is named first, but again it's not always the case. But what I would always do is if I get something like this, SO2, what you're going to want to name it is the first atom first and then the second atom second, no matter what. So this compound, SO2, is called sulfur dioxide because it's only got one sulfur atom and normally you want to put a prefix in front of everything except if the first atom only has one unit to it. So in the case of sulfur dioxide or carbon monoxide, since the first atom only has one, you don't put the mono. So the prefix indicates the number of each type of atom. If only one atom of a particular element is present, the prefix mono is usually made from the first element and the stem of the last element is used, is changed to IDE. So you can see nitrogen triiodide, not mono-nitrogen triiodide. Notice this one has four phosphorus in the first atom that's listed, so we call it tetraphosphorus and five oxygens pentawson. These names, these prefixes, you'll have to remember because they'll be used very commonly. So most of them are pretty straightforward. If there's any you have any trouble with remembering, maybe I can help you thinking of something that reminds me. So these are some very common covalent compounds.  So notice this one only consists of three atoms, this one consists of five atoms, this one four, this one three, instead of a million. So if you want to try to name these compounds, maybe we can by now. What do you think about this one? What is this one called here? Silicone? Dioxide, right? Si is silicon, for those of you who didn't know. What about this one? Dioxygen? Penta-oxide.  Okay? What about this one? Carbon tetrachloride, carbon tetrachloride. I have to fluoride, yeah. What about this one? This one? I don't know what we'll do. Okay, so I guess you guys kind of got it. Try some more. Try to write the names of the formulas of these things. What would be the formula of this thing? N-O. N-O, yeah. What about this one? N2-O4. N2-O4, N2-O4. And then this one? P2-O5. P2-O5. And then you could probably do that last one. Okay. We're going to go over this at the end of the next chapter, so I'm not going to go over it right now. On over covalent nionic compounds, we're actually going to talk about the natures of the particular bonds that come together to form these different peptides of compounds, covalent nionic. And in fact, all the other types of forces there are between particles. The main ones we're going to focus on are covalent bonds and nionic bonds. Okay, so how are atoms held together? So remember I showed you that picture earlier. Here's another demonstration of atoms being held together. So if we think of this as being a hydrogen atom and this being a carbon atom and this being some other type of atom, I don't know. What we can do is stick this carbon atom to this hydrogen atom by putting this little gray thing in between them. So I guess I... So this carbon atom and this hydrogen atom, right? We can stick this little gray thing here in between them. We call this thing a bond, okay? And that kind of sticks these two atoms together. Like that, okay? So you can kind of think of bonds like glue for atoms, okay? So they'll stick the two atoms together. So the more official definition is the force of attraction between any two atoms in a compound. So this attractive force overcomes the repulsion of the positively charged nuclei. So of course both of these nuclei here are positively charged, right? And when they come to really close together, positive and positive, they repel each other, right? But the bond can overcome that repulsion. Yeah, and interaction... And these bonds are just interactions of valence electrons between the two different two elements that are composed in the bond, okay? So we need to remember that valence electrons describe the chemical reactivity of any atom, okay? So it's only the valence electrons. That's why we had all that work on learning of valence electrons and noble gas configurations and all of that business. So it's going to come into play. Okay, so remember the noble gas configuration. If we look at an element, it's conventional or expanded electron. Configuration is for lithium, for example, it would be 1s2 to s1. And then the abbreviated form would be just showing the noble gas plus the valence electrons. Okay, so when you look at this, this really does emphasize how many valence electrons lithium has. But this, again, is kind of a cumbersome way to think about it when you're trying to stick atoms together, okay? So, but we do want to emphasize that it's the noble gas configuration and the valence electrons that really pushes all of this reactivity upon these different elements, okay? So instead of representing it like electron configurations like we've been known to do in the past, for the past two weeks we've been doing, let's go backwards and just look, since we know valence electrons now, we can just look at the periodic table and instead of writing all that electron configuration business, we'll just write dots that represent the valence electrons. And I kind of alluded to this last Wednesday. But what you'll find is that you can represent these atoms plus their valence electrons just through the atomic symbol and dots. So remember, we said Li, we could write as 1s2 to s1, or we could write it as helium to s1 or we could represent it much easier, more concisely, as lithium plus 1 valence electron, okay? So that dot represents its 1 valence electron. And what you'll see is if you look at this kind of makeshift periodic table here, we can see as you go up in group, right, you get one more valence electron, one more dot added to you until you get to the neon or argon configuration, that's the noble gas configuration, they're stable because they have 8 electrons around them. So this representation of these dots and the atomic symbol is known as the Lewis structure named after some guy named Lewis, okay, long since dead guy. As with only valence, since only valence electrons participate in bonding, it makes much easier to work with the octet rule, of course, especially if you're looking at, you know, I don't know, selenium or something like that, you don't want to have all these numbers and letters written out, you can just put the atom, put some dots around it and say you're finished, okay? So remember the number of dots directly corresponds to the number of valence gels that each atom has. So how do we represent these? You put the atomic symbol, you place one dot around the symbol until there are four dots, so this is the way a lot of people like to do it. So for lithium, so let's do boron, for example, we would start here, put one, two, like that, so we're going around the atom, carbon, one, two, then nitrogen would be one, two, three, four, and then pop, like that, okay? So it would have what we call a pair of electrons and three unpaired electrons, okay? So this is the way you want to start drawing your Lewis structures, just one dot, one dot, one dot, one dot, until you get to four, then you put up the next dot with the first one, okay? Remember, two electrons, it's okay to have two electrons, okay? Yeah, and each unpaired dot, so like in nitrogen here, it's got one, two, three unpaired dots, each unpaired dot is available to form a chemical bond, okay? So if we look here, lithium's only got one unpaired dot, and in fact, lithium only participates in ionic bonds, and this is really going to apply mostly to covalent bonds, so you're going to want to really concentrate on the non-metal up here, okay? You can still represent what's going on with ionic bonds, and we're going to a lot, but Lewis structures really help you with covalent a lot more. Again, notice the Lewis dot structure of all of the group one elements here are all the same. It would be the same for all the group two elements, same for all the group three elements, so on and so forth. And here's a good representation of that. So notice transition metals. Again, they're very strange, so we only really want to concentrate on the main group stuff. And again, all of the metals below the staircase and in the group one, group two, those will all be ionic bonders, and everything else up here will be covalent bonds. So remember an ionic bond is a transfer of one or more electrons from one atom to another, okay? So let's form an ionic bond. So one thing, let's draw, so we've drawn the Lewis structure of lithium here. Let's draw the Lewis structure of chlorine. So it would be Cl plus how many dots? One, two, three, four, five, six, seven dots, right? So how do we draw that? We go one, two, three, four, and then now we've got one on each side, so we're going to jump back to there. Five, six, seven. Notice we've got three paired electrons. These paired electrons, I'm going to start calling them lone pair electrons, okay? Because that's what I always call them. They're called lone pair electrons. So anything that's paired up like this. So lone pair electrons are not used in bonding. We've got one unpaired here and one unpaired here, right? What's going to happen? Those two are going to be used in bonding. So what actually happens with a ionic compound which lithium and chlorine will make is that the metal will lose its electron, right? So this one's one electron. This doesn't want to have it. So what will happen? It will transfer its electron there. So notice my arrow only has, and this is different than what they do in the book, okay? But my arrow only has one side to it, so it doesn't look like that, okay? A two-sided arrow, this really means to transfer two electrons, okay? A one-sided arrow means to transfer one electron, okay? So I want you guys to get familiar with this. Even though in the book it really states that two-sided arrows are okay to use for two electrons, but it will become really confusing if you guys try to do any more chemistry than this class, okay? So I want you to use one-sided arrows to show the transfer of one electron. Two-sided arrows will be the transfer of two electrons. In fact, in this class, you'll mostly be using the one-sided arrows, okay? So don't get fixated on them too much. So notice what happens here. We transfer, so we got electron transfer here, okay? From one atom to another. So when we do that, so we're kind of forming a rudimentary, if you will, chemical equation. If you remember, chemical equations from what we talked about last, well, Monday, if you were in my lab, but in any of the labs you probably talked about chemical equations last week. This is a rudimentary one, and again, we're not really anti-chemical equations yet, but what's going to happen here is when that lithium transfers its electrons there, it's going to become positively charged, of course, and now we're going to have a chlorine with a full octet. And since it's gained that electron, it becomes negatively charged. So now we have two ions. Then what happens? Well, these two ions, now they're one of them likes to have a charge, remember we talked about, so they want to stick together. That's called an ionic bond. So does that make sense? So there's kind of these three steps to it. The transfer of electrons, forming of the ions, and then sticking together of the two oppositely charged ions. That's how you make an ionic bond. If you go back to the practice test that I gave you, number five on the practice test is essentially doing this thing with, I don't remember what it was, oxygen and iodine I think, but actually in that case we're talking about covalent bonds. So this is different, so I do have a couple more minutes there. This is different than forming a covalent bond, so notice this is the transfer of electrons. A covalent bond is due to the sharing of electrons. So let's draw, hopefully we get it. Okay, so let's just go over again what I talked about in ionic bonds, and once we finish the ionic bonds then you guys can go. So ion formation takes place by electron transfer. You can see the transfer of the electrons so the sodium plus the electron, the electron then goes to chlorine, forming the full octet here, so both of them gain the noble gas configuration and then they stick together. Again here's another representation of it, but notice you don't have just two atoms sticking together. You've got this whole conglomerate here. Okay, you see that? So ionic bonds form between simple ions when representative elements lose valence electrons, the electrons are gained by other representative non-MLF atoms. Both atoms are changed into ions with noble gas configurations and the resulting ions are attracted to each other sticking together to form this conglomerate. And then you can see the arrangement and the crystal if you want. And then you can see, remember, ionic charges, please. So we'll stop there. We'll stop there for right now. So remember all of that stuff. It's kind of the review from last Wednesday and next time we'll start with the electron negativity and do some in-bonding. That'll probably take up most of it. Do not slime the sheets. If you need to slime the sheets, they're up here. If you didn't get your quiz to, come up and get it. There's a problem with quiz two.


OK, so let's first summarize what we did yesterday, also to somehow re-fix notation. We had this motivic ring, which will be our coefficient ring for all the computations we will perform today. And this I wrote as R mod, which is the Grottenich ring of complex algebraic varieties. So free-building group in all varieties, up to isomorphism, modulo, cut and paste relation, and then localized at motive of the affine line, which we want to have invertible, and we want to have a square root of it. And we want to have naturally denominators, 1 minus L to the i inverse, for a greater equal 1. And this is a ring where we have virtual motives of varieties. That's where the virtual motives live. OK, that was the one thing I explained yesterday. And the second thing was, if q is a quiver, we can attach to it a category of finite dimensional representations, which has very nice homological properties. One homological property was even something like a sadduality, which I only briefly mentioned. But the most important thing was that we have an explicitly computable homological Euler form, a category of representations. It is hereditary, so of global dimension 1. And the homological Euler form is something we know explicitly. Maybe I will redefine it today. And then finally, we consider the stacks of isomorphism classes. So the modularized stack of isomorphism classes in this thing is nothing else than, well, we have seen a certain affine space with an action of a nice algebraic group. And we can take from the quotient stack and take the disjoint union over all dimension vectors. So this is the quotient stack. This was an affine space, and this was a product of general linear groups. And using this, I then defined this motific generating function, which I will define in a minute. But that's the summary of what we did yesterday. And this is always an irritating moment for me in a lecture series that I summarized what I did in the previous talk, and it's just like five minutes. And then I always think, OK, I could have done this in five minutes, but it took me an hour. So what was going wrong? I, OK, it was the explanations missing and trying to give you a feeling for what all this is. OK, so I hope this whole hour yesterday wasn't useless for you. OK, so anyway, now we will come up with a formal definition of the motific generating series in the motific quantum space. So it's a generating series. Some of it feels like a zeta function, but very important is also the coefficient, the ring in which this generating series actually lives. And OK, so I will define the following. Let me, so this will be the motific quantum space. And this is the following. Well, this is a formal, it's like a formal power series ring, yeah, with as many variables as you have worthy C's in the quiver. And the coefficient ring is our motific ring R mod. And everything is slightly twisted by the quiver structure. And this we'll see now is the complete R mod algebra, complete because we are looking at formal power series, where the base is t to the d, where d is a dimension vector. So this is this one discrete invariant of a quiver representation, the dimension vector, encoding the dimensions of the vector space is involved. And multiplication, and the multiplication is defined as follows. So this is something like a quantum version of a formal power series ring. We take minus l to the 1 half, anti-symmetrized Euler form, t to the d plus e. That's our base ring for all the computations which we'll follow. So it's formal power series in as many variables as we have worthy C's in the quivers. And instead of writing down ti to the di, a product over all ti to the di, we formally work with such monomials t to the d, which is just short. It's a minus. Yes, it's a minus. It's the anti-symmetrization of the Euler form. It's the anti-symmetrization, yes. So you can see this in today. Exactly. Exactly. It's the anti-symmetrized Euler form. And this we take as multiplication. So it's like. Excuse me? And it's associative. It's associative, unital algebra. The unit is just t to the 0. And it's complete with respect to the, well, the augmentation ideal is the ideal everything generated by t to the d for d non-zero. And everything is complete with respect to this ideal. We have formal power series, which we'll use. There's one question. Yes, please. The introduction of this. Can I vote you? Well, I think yes, yes. Well, the most tricky point is somehow why do you anti-symmetrize the Euler form? And well, then this anti-symmetrized Euler form is like the honest Euler form of some three-calabi-yau situation. So you're somehow imitating, kind of imitating a three-calabi-yau situation with a quiver, which is hereditary of global dimension one. That's somehow philosophically the explanation. But the practical feature is that just the formulas are as smooth as possible with this twist. Yeah? It's a logic of quantum torus. This is, yeah, I call it the motivic quantum space because for quantum torus, I would like to have inverses of the t to the i maybe. So it should be over some Laurent polynomial ring. But motivic quantum space. So and inside this, so very pragmatically, the motivation is the formulas are very smooth when you work in this ring. For me personally, for many years, I worked with a different ring. I just worked with a twist just by the Euler form, not by the anti-symmetrized Euler form. And I always had problems with the formulas, which were not really nice. But this is the way the formulas are easier to remember, although you always have this ugly little minus square root of the left sheds motif inside. This you cannot get rid of. OK, so and now let me finally introduce this motivic generating function. And then explain why it somehow knows the quiver. So aq is then defined as sum over all dimension vectors. And then we take the virtual motif of this space of representations divided by the virtual motif of the base change group and form a variable t to the d. So this is the motivic quantum space. And the series is the motivic generating series. Instead of an a, you could also write a z. So that really feels like, well, whatever you prefer, a zeta function or a partition function. It's really somehow of this kind, as we will now see. Because that's the series we will try to factor. It's actually being smarted by some follow-up. Excuse me. The virtual motif, yes. OK, the virtual motif was you to normalize the motif by an appropriate half power of the left shift's motif for x irreducible. So that, for example, Ponka's duality manifests itself as invariance under exchanging L and L inverse. This was the example we have seen for projective space. That was the twist which is necessary. We will actually use this twist now and make this motivic generating series more explicit. Yes, let's do this. That's maybe the perfect point to formulate a little lemma. And you know the proof of this lemma when you have done this exercise which I gave you yesterday. The exercise was to compute the motive of GLN, of the group GLN, yeah? And so let's compute this thing here. Let's first. So this active ring does not depend on the quiver but the function does, right? No. OK, so we will see after this calculation that the function depends on the symmetrized Euler form and the ring depends on the anti-symmetrized Euler form. So altogether, the datum of this series in this ring recovers the quiver. We will see this in a second. Now, the multiplication in the ring, the multiplication in the ring, it really depends on the anti-symmetrized Euler form. And this function only depends on the symmetrized Euler form which we will now see if I formulate this lemma. OK, so the lemma just is an explicit calculation of that. So this Rd of q, if you recall the definition from yesterday, maybe I should recall it, why not? This Rd of q is just an affine space, yeah? Rd of q is just direct sum over all the arrows in the quiver of the space of linear maps from a di-dimensional to a dj-dimensional space. So this is just affine space of some dimension and its motive is just the power of the left-shed motive. The group Gd, which acts on this, is a product over general linear groups for the vertices. So its motive is a product of the motives of general linear groups, which we have calculated yesterday in this exercise. OK, and if you just plug in the result from yesterday's exercise into this, then you see that we can reformulate this Aq just as sum over all d into the q0. And then the numerator is minus l1 half to the power minus the Euler form applied to dd, d to the d. And then the denominator is just a product of Pochamer symbols, where you plug in the left-shed motive or the inverse of the left-shed motive.  So this is something like a l-hyper geometric series, like in q-hyper geometric series, except that our quantum parameter is now always the left-shed motive. So this really feels like a l-hyper geometric series. OK, and so this lemma is just really direct calculation with the motives of this affine space and this group. And from this we can now draw a remark that the datum of this series and the ring knows the quiver. So this motivic ring, so this motivic quantum space, knows the anti-symmetrized Euler form, because it's really in the definition of the multiplication. OK, the motivic generating series only involves the quadratic form defined by the Euler form. It's only about the quadratic form. But from the quadratic form you can recover the symmetrized bilinear form by polarization, the usual polarization trick. So Aq knows about the quadratic forms, and then after polarizing it knows about the symmetrization. So both together know about the whole Euler form. So both together know about the Euler form of the quiver, because, well, we can standardly decompose into a symmetric and anti-symmetric part. So we know the Euler form, and from the Euler form we can directly recover the quiver, because the Euler form just encodes the vertices and the arrows. OK, so this answer to this question from yesterday of how much of the quiver is seen by this. So Aq itself only sees the symmetrization, but the important thing is to have it in this ring and then we really recover the whole Euler form. OK, so time for first examples, namely for examples where you really don't see all this quiver machinery. Yes, Fabian? When you say that it knows about the quadratic forms, I mean, when you take the coefficient of this power series, then I mean, you want the exponent of the L part, right? But the coefficient ring is probably not like a unique factorization domain or something. So how can you get this out of this? Well, take the coefficient of t to the d and develop it into a formal power series in L inverse. Then the lowest degree coefficient is this one. And from the exponent you can read off the Euler form. Then you can read off the quadratic form, for example. I mean, knows really means, well, you can just read it off from the definition. So how much of the quiver is contained in the definition of the quantum space and the motivic generating series? OK, now we have to do two classical examples. There's a question? Yes, please. Is there some difference equation satisfied by Aq? Yes, yes, there is. I once used it in another version. So where I have different twists. But yes, yes, you can characterize it by some nice Q difference equation. And we will see other Q analysis business popping up in a minute. For example, taking the logarithmic Q derivative is something which plays a role here and leads to Hilbert schemes. Yes, we'll see these in a few minutes. Yeah, yeah, so I mean, this is some of the right keyword. So this brings you into Q or L analysis. And you can think about all possible tools from Q or L analysis, which you might want to apply to the series to study it. OK, so example. But I think I need a larger blackboard for this. So in this example, I will do the case of the quiver, which is just a vertex. And a vertex with a single loop. Because these correspond to classical things. Namely, let us recall the so-called Q binomial theorem. And it's a theorem which exists in two versions, luckily for us. One version is, let me first write down the classical version. So you take the following hypergeometric series, sum over all d, t to the d divided by the Pochamer symbol 1 minus Q times 1 minus Q to the d. And this summation factors into a nice product, product i from 0 to infinity, 1 over 1 minus Q to the i t. This is usually proved using just partition combinatorics. So on the left-hand side, you see a generating function for partitions by weight. And this is one way you can prove this. And another nice identity, which also goes under the name of Q binomial theorem, is if you have some quadratic form in the numerator, which is Q to the d, d minus 1 half t to the d. And then again the Pochamer. And this is product over all i from 0 to infinity, 1 plus Q to the i t. And now if you think, well, OK, this might generalize. Maybe I can put some other Q to the d squared here in the numerator or not, then number theory people will tell you no, that's a completely different story. If you don't put Q to the d squared half here in the numerator, but Q to the d squared, then you are in the realm of Roger's Roman Nujem identities and all sorts of really difficult number theory stuff. So there are just precisely these two values for the numerators where you have such a nice factorization. OK, so from this it follows quite easily that we can factor the generating series for the quiver, which is just a single vertex, into, I have to read this off because I have to be extremely careful about signs in this whole theory. Product over all i from 0 to infinity, 1 over 1 minus l to the i plus 1 half times t. This corresponds to the first identity and corresponding to the second identity, which is the quiver with just one loop, you have a Q equals product over i from 1 to infinity, 1 minus l to the i t. OK, so yeah, a good advice to get into this motivic invariant business is to do the calculation. I mean, everything's here. So for example, OK, let's look at the second case here. For this one loop quiver, the Euler form is actually identically 0 because you are taking a contribution d squared for the vertex minus d squared for the loop. So the Euler form is identically 0. So this term vanishes, and then you have just this quotient. And then you rearrange, bringing, so pulling enough powers of l out, and then you'll see you arrive here. And it's really recommended to do the calculation. It's completely elementary, but you will see that you have to be extremely careful about all the signs. And this is a typical phenomenon for this theory of motivic invariance, that all the signs and these ugly little half powers really are really important, so you should do this calculation. I mean, it's completely elementary, but we should do this. We will need this later. There's one more case of a connected quiver for which you can write down such a product factorization of the motivic generating function in elementary fashion. Namely, the third case is this quiver. And if you know about quiver representations, these three, the trivial quiver, the one loop, and this one, these are the only three symmetric connected quivers which are not wild. Because this is an a0 tilde, this is an a1 tilde, and this is an a0. And all other symmetric connected quivers are wild, and it's really a tame wild thing that you can write down the factorization. So there exists an explicit factorization for this, which I don't want to recall now. But I will recall the categorification of this maybe in the last talk. And OK, so these are the only elementary examples of these motivic generating functions. What does wild mean? What does wild mean? Oh, what does wild mean? OK, good point. This is minus. Surprisingly, it's minus. Although here in this identity, you have this distinction between the minus and the plus, but here it is minus both times due to this funny little twist by minus square root of left-shut. Wildness. OK, let's just check for the time. OK, a category C, a billion finite length, very nice category, like modules over some algebra, is called wild. Well, to be precise, strongly wild, but let me not explain this distinction, if you can embed the category of representations of a free algebra in it. If there exists an embedding of the category of representations of a free algebra in two variables faithfully into C. So it means the classification of isomorphism classes in the category is as least as difficult as classifying representations of the free algebra. And it is known, for example, that if A is any finite dimensional algebra, you can embed its representation category inside the representation of a free algebra. So having solved the classification problem for the free algebra, you have solved the classification problem for arbitrary algebras. So this is some kind of recursive thing, and that's what it's supposed to wild. Geometrically, you could say objects appear in arbitrarily high dimensional moduli. So equivalently, moduli are of arbitrarily high dimension, not moduli of all representations. So for example, if you look at smooth projective curves and the category of coherent chiefs, then you have two cases which are not wild, namely P1 and elliptic curves. Well, because for P1, the classification of coherent chiefs is discrete. Everything is direct sum of torsion chiefs and ONs. And for elliptic curves, you take the ATIA classification, and the moduli are harmless. The moduli of bundles are only the elliptic curve itself, basically. But starting in genus 2, you have moduli spaces of coherent chiefs of arbitrary high dimension. That's called wild. And for example, such an explicit embedding of representations of the free algebra into coherent chiefs on a smooth projective curve, that was really used by Sechardry in studying the biorational geometry of the moduli spaces. So this implicitly really appeared in vector bundle theory. Markus, we have a question in the chat. For a thin category, is the dimension of representations always bounded by 1? The dimension of, well, in the composable or stable representations, the moduli. So yeah, OK, well, you have to be precise about what moduli you consider. Usually, you consider stable objects. And if you're strict to stables, then for a tame algebra, you only get one dimensional moduli. Yes, indeed. That is the famous, I mean, infinite dimensional algebra theory. There's the famous tame wild dichotomy. Either your algebra is wild, moduli of arbitrary high dimension, and you can embed the free algebra. Or you are tame, which means moduli spaces are only one dimensional. Maybe with many, many different irreducible components, but only one dimensional. Where can we find this equivalence? Sorry? Where can we find this equivalence? Ha, ha, OK. I will try to find a good reference. Yeah, I mean, there are many technicalities around this. There is really a distinction of being wild and being strongly wild. And what I call wild here is really strongly wild. But OK, I'll check for some nice references. And I also maybe slightly really question. So I look at these two functions, and I think both on partition function and fairly on partition function. Why should I think about the point of being both on and the loop between them? OK, well, it's an accident, actually. There's nothing behind this, I think. So, namely, what you can do is you can exchange L and L inverse in the generating function. And this changes the Euler form to what? To the diagonal form delta minus the Euler form. So usually, if you take the Euler form of a quiver and instead compute the diagonal one minus the Euler form, you no longer have the Euler form of a quiver. Because negative numbers of arrows don't exist. The only exception is that you interchange the trivial quiver and the one loop quiver. So it's an accident. So I was always hoping for some boson fermion principle. Well, we will see this boson fermion thing also when we categorify all this in the homological whole algebras, because then we will get free bosons, free fermions as the homological whole algebras. But it's just a coincidence in this very special case. No more questions. OK, let me check the time. Yes, OK, wonderful. So let's start with actually working with this generating function, a of q. And yes, it's always very interesting to see, for me, that usually you, meaning the audience of some lecture series, is more interested in the qualitative aspects, like this wildness thing, whereas I'm really into this quantitative aspect. I really want to do some hardcore calculations with this series, what I really like, what the audience usually doesn't like so much. So that's always a conflict, and we have to navigate through this. But I'm always happy if you ask lots of questions and I can explain different things. So that's actually fine. And if I then have to skip some of the dirty calculations, then that's fine. OK, let me give you something which is some more halfway between qualitative and quantitative. Let me give you, OK, now this is all about factorization, part one. Factorization just means I want to do some multiplicative things with a q. So for example, I want to factor it into a product. This we have seen in these two examples that we can factor this into an infinite product. We will later see how to do this in general, but I first want to motivate. We can also do something else. This is a form of power series whose constant term is 1. The constant term is just for t to the 0. And for t to the 0, you just have a constant term 1. Any form of power series with constant term 1 is invertible. So aq is not only an element in this motivic quantum space. It's even an invertible element. So we can look at its inverse. And so here's a theorem. And later, when we have seen how to define dt in variance, then this theorem is exactly on a very formal level the dt-pt correspondence. Although I can't call it dt-pt at the moment. So it just says, OK. So let's take this motivic, which we're around. Let's take this motivic series, not in the variable l, but let's twist the variable l slightly a little bit. I will explain this notation in a second. Let me twist the variable t, the formal variable t, by minus l 1 half to the n, where n is a vector. And I have to explain what this notation is. And let me multiply it by its inverse with a slightly different twist, minus l 1 half minus n t inverse. I don't want to write this down as a fraction because I'm still working in a non-commutative ring. This motivic ring is still non-commutative because we are working with this anti-symmetrized euler form there. So I shouldn't write it as a fraction. If you calculate this fraction, then you get a motivic-generating series for Hilbert schemes, the virtual motive of Hilbert scheme of the quiver t to the d. So that's the formula. And now I have to explain all the terms. And actually, I will explain the idea of the proof, which is essentially really simple. And the rest is just dirty calculations in this motivic ring. OK, if you know Tom Bridgeland's papers about motivic-haul algebras, this is what he calls the quotient identity in the motivic-haul algebra. So let me first explain the terms. What does it mean to plug in not t, but minus l 1 half to the nt into this? Well, try to guess the definition. And your first guess is the correct one. f of minus l 1 half to the n times t is defined as, OK. Let's assume f of t is a series with coefficients ad, t to the d, just some formal power series. And then plugging in this, this we define as sum over d.  OK, let's see if this makes sense. So let's assume it's l to the n over d. Our usual variable minus l to the 1 half. That's always the standard variable here. So assume f of t is a formal power series, sum over all d, a coefficient ad, t to the d. And this d is always a vector of length q0. And now I want to define what it means to have this twist in the variable t.  And this is just the new scalar product, sum of the product of the entries. Yeah? OK, so that's a way to twist a formal power series in many variables. OK, and that's what we do twice in defining the left-hand side. And we use the fact that aq is actually invertible in this ring.  And I have to explain what Hilbert schemes are. Now, again, think about coherent sheaves on a curve. And think about how you define the moduli spaces of vector bundles on curves. This you usually do by GIT. So first, you find a large enough Hilbert scheme which parameterizes all your semi-stable bundles over fixed rank and degree. And then you're mod out by some further structure group, sl, and do the geometric invariant theory. And this kind of Hilbert scheme is precisely what we need here. So to define the Hilbert scheme, if we have such a dimension vector n, we have an associated projective representation of our quiver q. More precisely, this quiver q has one projective in a composable representation for each vertex, which is usually called pi. And then you just take the direct sum over all pi to the ni. And that's what you call pn. Well, we'll not see these projectives. And we don't really need to see them so, but there is a projective representation. This will somehow play the role of O of 1 in the vector bundle case. And now in the vector bundle case, to realize all semi-stable representations of fixed rank and slope, you present them all as a quotient of some large O n or large power of some O 1 or O n or whatever. And so this is what we do here. I define Hilb dn of q as a set as follows. It is the set of all presentations of a representation of dimension vector v as a quotient of a fixed projective representation up to a natural notion of equivalence, which you can already guess. Namely, two representations are, two such presentations are equivalent if you have a diagram intertwining the two presentations. That's what you would call a Hilbert scheme also in the coherent sheaf setup. And that's what you call a Hilbert scheme here in the quiver setup. And this exists as an algebraic variety using GIT. So you can realize this as a set of stable points in some large vector space. And then you're mod out by the structure group. And well, now we have such a nice identity. This is the full Quartz scheme. Is this projected? Yes, yes, the full Quartz scheme. Yes, yes, yes. Yeah, we're not taking care of any more numerical invariance like Hilbert's, yeah, yeah, OK. Well, OK, to call it the Quartz scheme would somehow be a better analogy. Yes, that's right. Yes, OK. So we're performing a very simple algebraic operation here. And in terms of q analysis, you would call this logarithmic l derivative. That would be the slogan. Well, in q analysis, the logarithmic derivative would be f of q times t divided by f of t. You take some twisted version of your former series and divide it by that. That's some more analog of that. OK, OK, so that's our first identity using this ring. And actually, it tells us that what is important about this is that the left-hand side is something very formal. Aq is defined in terms of quotient stacks. So they don't have a direct geometric meaning. But the right-hand side is something totally explicit. It's really about motives of actual varieties. In particular, I noticed this, just to finish the sentence. On the left-hand side, you're not a priori allowed to specialize l to 1. We have talked about these motivic measures yesterday. One motivic measure was the virtual Hodge polynomial. This is always OK. But to take Euler characteristic requires specializing l to 1, because the Euler characteristic of the affine line is 1. And a priori are not allowed to do this if you localize, like we do in our motivic ring, for localize by terms like 1 minus l to the i. So we are usually not allowed to take Euler characteristic. And this you can really see from this explicit form of the motivic general rating series. Better not specialize l to 1 here, because of these denominators. But taking this logarithmic l derivative, somehow mysteriously cancels out all the denominators. And what you're left with is these Hilbert schemes. And that's the honest motive of a variety of which you can take the Euler characteristic. And these are very interesting numbers. So the left-hand side is completely formal. And the right-hand side is something concrete. And that's the surprising thing, that you're doing very formal things in this motivic ring, but you get out some actual geometry. And that we will now see in factorization part two in another case. But first, Fabian's question. What do you mean we take the left-hand's motive to the exponent of the managing vector? Not relation. Yeah, yeah, yeah, OK, this notation which I defined here. So this is really the formal definition. So this power only makes sense in this formal substitution of the variable t, which is also nonsense. There is no single variable t. There are variables ti for the vertices of the quiver. But this is defined as twisting the coefficients of the formal series by minus square root of left-hand's to n times d, where this is the naive scalar product of two-dimensional vectors. So this is really just a formal definition to arrive at smooth formulas and to make the analogy to q analysis a little bit clearer. OK, so that was this first instance of this factorization thing. And let's do a second one. So this black one will reappear in a second. So usually when you want to define some modular spaces, you have to choose a notion of stability. In this factorization part one, we were quite lucky that we didn't have to choose stability. It's somehow implicit in defining the service scheme. But so there's no stability involved here. And now in factorization part two, we will involve a stability. Factorization part two. And so the usual modern formulation of stability would be with a central charge function. I will do the old-fashioned slope stability because this may be closer to the coherent sheave intuition, which some of you might have in mind. So let me just assume we are given a slope function mu from non-zero-dimension vectors to the reals. And this should be axiomatically. It should be something which looks like the slope function for vector bundles. So it's of the form degree divided by rank. So mu of d should be theta of d divided by kappa of d. And this should somehow play the role of degree and rank for vector bundles. So it should be linear functions on the Grottenig group. It should be linear on dimension vectors. So theta and kappa should be real-valued linear functionals on the quiver. And of course, you need some more condition. Namely, this denominator should never be 0 for a non-zero-dimension vector. So we require that kappa of nq0 except 0 is contained in the positive reals. OK. So whatever is formally necessary to define something which behaves like a slope function. So that's our abstraction of slope function. And if you just take theta and kappa as the real and imaginary part of a complex-valued function, then you have a central charge. So if you take theta plus i times kappa, which is a complex-valued function, then this is what is usually called central charge function. And this is some of the modern formulation of stability, which usually works with central charge. But I will keep this old-fashioned slope function. OK. Just as an aside. OK. Now we are given a slope function. We have a notion of semi-stability of quiver representations. As usual, a representation v of q is called stable or semi-stable if the slope decreases or really decreases on sub-representations. If the slope of any sub-representation is less than or equal the slope of the representation for all non-zero proper sub-representations. This is exactly the notion of semi-stability, which you have seen from in vector bundles or coherent sheaves in whatever setup. Just take the abstraction of not choosing the slope function a priori. And of course, well, if you make the wrong choice for theta and kappa, then this notion might be trivial. It might happen that any representation is semi-stable or just no representation at all is semi-stable. So all the subtlety is in choosing these linear functions for a given quiver. That's a completely different story. And we'll not touch upon this and just say we want to prove a result which does not use any special feature of this slope function. It should be just true for whatever slope function you have, even if it produces a trivial notion of semi-stability. All right? Oh, OK. So finally, then, inside Rd, inside the space of all representations, you have a semi-stable locus. This is the set of all representations which are semi-stable. And because semi-stability is a generosity condition, this is a Zariski open subset. You're just trying to avoid certain bad dimension vectors of sub-representations. You're trying to avoid dimension vectors of sub-representations which do not fulfill this property. And avoiding certain sub-representations is a generosity condition. So it's open. OK? Final ingredient is that, so now that we have this relative version of this representation space, we can define a relative version of the motivic generating function. So if x is any real slope, we define the x-semi-stable motivic generating function. No, this deserves a larger space. Sorry. So if x is a fixed slope, we can now define a relative or semi-stable generating function, stable motivic generating function, which is aq, but now x semi-stable of t. And this is, well, you can already guess what it is. Namely, you take the virtual motive, not of this whole representation space, but just of its semi-stable part. And you divide by the same structure group. OK? But now we have to bring the concrete slope into the business. So we are doing the summation only over those dimension vectors where the slope is x. So where d is in mu inverse x. So where the slope of the dimension vector d is x or 0. All right? And then the result is what is now called a wall crossing formula. And this is somehow the, it's the prototype of all wall crossing formulas you can encounter in this theory. And I will tell you what it formally amounts to, namely almost nothing. The coefficient d in mu. Yes, OK. So OK, so I should write this in a different way, equivalent but different, because this was not nicely written. So I would like to take the sum only over those dimension vectors whose slope is x, because I want to do everything local in the slope. But then I have a problem, namely the slope is only defined for dimension vectors which are non-zero. And I definitely want to have my motivic generating series having constant term 1, because I want to do multiplicative things. But OK, so let me just formally write 1 plus. 1 plus all the dimension vectors which are non-zero, but have this fixed slope. So that's maybe a more honest way of writing this down. OK, and then the result is the wall crossing formula. And this is a big advantage of this, of this motivic ring that this has an extremely smooth formulation. Aq is the product over all reals of these local functions. OK. And? Ordered product. Ordered. It's an ordered product, and it's ordered by a decreasing slope. And while you have to think a few minutes about well-defining ordered products over the reals, obviously, but the important thing is that all these are power series on the right-hand side. All are power series starting with 1, and everything is graded by the dimension vector. And then if you think about it for a minute, you can easily see that such an ordered product over the reals in descending order is actually well-defined. Yes? The slope only takes various interaction among both sides? No, because I allow theta and kappa to be real valued. A priori. Yeah? You gain anything by doing that? Yes, because later on I would like to make a generosity assumption. And sometimes to make things as generic as possible, I want this to be real valued. So that, for example, the fiber for fixed slope is just one ray. And if I just make these reals which are independent over the rationals, then that's definitely fine. So I gain a little bit. OK. Time is almost over, but let me just show you the proof of this. That's not my terminology. Well, OK, so it's OK. Yeah, before explaining the proof, let me show you what is the wall crossing aspect. The wall crossing aspect is not this identity, but on the fixed category of representations of the quiver, you can look at different notions of stability. So you could have two different slope functions, two different slope functions, giving you really, really different stables or semi-stables. But this formula tells you that this ordered product is always the same. So the ordered product over this series with respect to mu is the same as this ordered product with respect to mu prime. So instability space, in the space of all possible stabilities, you have many walls defining the transitions where changing the stability function really changes the set of stable or semi-stable objects. And it means that crossing a wall in stability space might drastically change the class of semi-stable representations, but there are certain things which are unchanged, namely these ordered products, because they just evaluate to the same motivic generating function. So that's where the name wall crossing comes from. And well, formally, two minutes, the proof is formally equivalent to the existence of the Harden-Arrow-Siemann filtration. So every object v admits a unique so-called Harden-Arrow-Siemann filtration. And this is incredibly strong. I mean, if you look at things like the Jordan-Hölder filtration, they are not unique. The sub-factors of the Jordan-Hölder filtration, they are unique up to isomorphism and permutation. But here it is the filtration itself which is unique. And if you define it correctly, if you index the terms in the following filtration by the reals in a tricky way, then it is even functorial. You get a functorial filtration. There exists a unique HN filtration. v equals 0 equals v0 in v1 in vs equals v. It's defined by two properties, namely first property. All sub-factors are semi-stable for your fixed slope function for some slope. And second property, the slopes are decreasing. So this already appears in Work of Harden-Arrow-Siemann on calculating the number of points of modular spaces of vector models on curves. And it's really an incredibly rigid structure on the category because it's a filtration which you can functorially attach to representation. OK. And this basically proves this motivic identity. And namely, these decreasing slopes you see here in this decreasing order of the product. And everything you have to do is you have to stratify your representation space. So your representation space, Rd, is a union of strata of a fixed Harden-Arrow-Siemann type, where Harden-Arrow-Siemann type just means you record the dimension vectors of the sub-crochets. This is a finite filtration, a finite stratification by locally closed. So you get a corresponding identity in the Gotten-Dickringer varieties. The motive of this is just the sum of the motives of these. And these are, well, they're isomorphic to base change from the group Gd to a certain parabolic group. And then you have a vector bundle over a product of semi-stable loci for smaller dimension vectors. Vector bundle over a product of semi-stable loci for certain dimension vectors, smaller dimension vectors. I don't want to be too technical about the notation here, but that's the principle. You stratify this affine space into locally closed things. Each of these locally closed things is a vector bundle over the product of semi-stable loci up to some change of group. You induce from a parabolic group to our base change group. And this is a very simple geometric fact, which is the consequence of the existence of the Harden-Arrow-Siemann filtration. And writing down what this means in the Gotten-Dickringer variety immediately gives you this product identity. So that's basically the whole proof. And while the most important feature is buried in here, the fact that this is a vector bundle, that's equivalent to the category which you're considering to be hereditary. So this is more or less equivalent to the global dimension of your category of representations to be one. So that's the limitation of this. The global dimension of this category of representations is one, global dimension of rep Cq equals one. So hereditary means global dimension. Yes, right, global dimension at most one, exactly. OK, so that's basically enough for today. And next time we will start with this identity and explore these local contributions, these functions, and relate them to modular spaces of semi-stable representations. And then finally define the DT invariance. OK, thank you very much. Are there any questions? Yes. Of course, it is a standard key, the scheme of the plane. Of course, it is related to the simulation for an experience. So it is also given to them. Yeah, yeah. So that's not related to a quiver, but to a quiver with relations, you can consider this Hilbert scheme, which I introduced for the quiver with one vertex and two loops. And the dimension is D, and this extra datum M is just one. So this parameterizes left ideals of finite co-dimension in the free algebra in two variables of co-dimension of, exactly, thank you, of co-dimension D in the free algebra. And now inside this, you can consider those points. I mean, formally this is represented as equivalence classes of two linear operators, A and B, and a cyclic vector, V. V is cyclic. And then you can consider the closed subset, which is defined by the equation AB equals BA, commuting linear operators. And that would be the, and that shows that inside here is a closed subset. You find the usual Hilbert scheme of points in the plane. This has actually much simpler structure than this one. Related to same with the term of definition, where you think that this is the same. Ah, OK. Ah, yeah, OK. The reason why this is the same as my definition is, well, the free algebra is one of these algebra which has the property that any projective representation is already free. So as projective representation, we just take the free algebra itself. And then let's have a look at what is the surjection from the free algebra to V. Well, let's take the image of one. And the image of one is then a cyclic vector. And X and Y are mapped to two linear operators, A and B. Such that V is a cyclic vector for this. And that's the connection to this other way of writing the Hilbert scheme. There are some questions online. Yeah. Can you please get a bit more detail on what the Haddon-Arceeman is defined to be, Haddon-Arceeman type? Type, yes. OK, yeah, OK. For time reasons, I tried to avoid this. Thanks for the question. And now I can give you the definition. OK, so I'll just elaborate on this thing here and do the precise notation. So the union you take is over all decompositions of your dimension vector into non-zero dimension vectors for which the slopes are strictly decreasing. Because that's the one condition which appears here.  D dot is this decomposition. And hn d dot, that's the set of all representations V such that the dimension vector of the i-th sub-crochant in the Haddon-Arceeman filtration is precisely di for i from 1 to s. And this is the i-th Haddon-Arceeman sub-factor, sub-crochant. OK, so from the Haddon-Arceeman filtration, I just record the dimension vectors of the sub-crochants. They have necessarily have to fulfill this condition. And this is, in fact, a locally closed subset. So existence of such an hn filtration, that's, in fact, a locally closed condition. That's not difficult to see. And then what you do is you have a, well, OK. And then you identify this with an associated fiber bundle. Namely, inside the group Gd, you find a parabolic subgroup corresponding to this d dot with respect to this decomposition. That's like the standard parabolic you find in a GL for any decomposition of your dimension. And then you have a certain vector bundle of known rank. So the rank is easily computable. But if I write it down, I make a sign mistake. Of known rank over the product i from 1 to s of the semi-stable locus in the representation variety of y. And the basic idea is that if you have something in here, then to the representation v, you attach the tuple of the hardener simmer and sub-crochants. That's the basic idea. Or that's what you want to do. But you can't do this literally on this level. You can do this literally on the stack level, but only literally, but only tautologically. So you cannot attach to the representation sub-crochants because the sub-crochants do not have a preferred base, linear basis. And that's why you have to perform some base change here. But that's essentially what's going on. And from this, you just compute the motive. Motive of rd is the sum over all these. Motive of gd divided by a motive of the parabolic times the motive of this product times a power of the left sheds motive for the vector bundle. And if you just write this down and simplify everything, then you get to this product formula. Another question. Is there a connection between the variation of stability x and the monotomy of aqx? The monotomy of aqx sounds great. I have no idea what this could mean, but it sounds great. It would be wonderful if it's true. So if the one who asked this question could give me some details of what this monotomy could be. It's a anonymous participant. OK, sorry. So if this anonymous participant can hear me now, if you could just send me an email, anonymous if you want, of what this monotomy could be, I would be happy to think about it because it sounds like a great suggestion. Yes? There is one more. Could you explain how to obtain the answer for aq? For the one point quiver without an arrow directly from the definition of rdq, if it isn't rdq trivial. Yes, it is. Yes, it is. OK, yeah. Let's do this calculation. Yeah. I was hoping for something like this for the question and answer session this afternoon. But if we have time, I can do this calculation now. Then they'd be also good idea to do this. Let's do one or two of these tricky calculations using all these signs and half powers of l in the Q&A session this afternoon. I ask myself one question. Are you going to prove the first structuralization? If you are interested, I can try to squeeze it in. Yeah? Fine. OK, can do it. I have a philosophical question. So can you do this for the stack of player chiefs on a MOOC project card instead of the quiver? Ah. Hm. How many defines a genetic series? Hm. No, there are tons of convergence issues. I mean, just in defining the modular space, you have all these problems of boundedness. For the modular stack, it's not finite type. Exactly, because it's not finite type. All these boundedness problems which you have will appear here. And things which you write down naively are usually not convergent because things are not a finite type. So it has to be very careful and just first approximated by restricting to bounded values of the slopes and then study the convergence for a slope going to plus minus infinity. But it has to be very careful. Any other questions? Yeah, just like the first formula. So the second formula geometrically means that you have to start on a semistrification. And does the first one have some geometric meaning? Yeah. Yeah, yeah. OK. That's also more or less your question. I don't know if I should do it now or? Do you think you can squeeze this in the question? I will do it in the Q&A session. Yeah? I'll give the proof of this and the calculation for the trivial quiver and maybe some other calculation. We'll do this. Yes. Perfect. Well, let's stop this again. Thank you very much.