So, we were started discussing about Markov processes in the last class, so Markov process we also called as Markov chains and we are mostly interested in discrete time Markov chains. So, in the last class we defined what we mean by this Markov property and then discussed what we mean by time homogeneous Markov property and then try to connect transition probability matrix with my n-step transition probability matrix. So, in the last class we said that P of n equals to P to the power n. What was P superscript n means? This is the n-step transition probability matrix and what is P here? One-step transition probability matrix or we simply call it as transition probability matrix. And we have also concluded that P of 0 is 1, sorry i, identity matrix. So, in the last class we just shown that P n plus m can be written as P of n plus into this product and at the end we said that we could write it as P of 1 and P of n minus 1 and then we said that so and so P of n we are able to split it in this fashion and we could further split them into this fashion and we said that this can be finally result in P to the power n. Okay, fine. Now, because of this, so I think last time after completing this I also wanted to introduce the notion of finite dimensional distributions which I could not do last time. Let me finish this today before moving to another topic on this Markov trains. We told that any process, a stochastic process to completely characterize this I need to give is finite dimensional distributions. That is I need to for any n, if it is a discrete time process I need to characterize joint distributions or joint probabilities of this format assuming that I am looking into random variables that take, that are discrete. That is let us say, so this is a joint distributions involving n random variables, sorry involving n random variables here and I need to give such joint distribution for all possible collections of my random variable to completely characterize my random process. So here I have taken some particular m random variables at time index n1, n2 and nm. You can choose any time index you want. And this could be again for all time indices. You take any number m and take any time indices m and let us say I have such a distribution how we need to characterize this. I should be able to do then that will give me complete characterization of my random process. And just j1, j2 are the possible realization this random variable can give. So I need to also define this for all possible values my random variables can take. Now let us see what all the things that govern this finite dimensional distribution in the case of a Markov chain. I have this, what I could do is I am going to just introduce this initial state and then keep everything. So if I do this nothing changes, these two quantities are the same like I have just introduced this x0 random variable and just summed it over all possible values it can take. We have said that all possible values are coming from this state space S. Now you can split this probability I call it j1 given x0 equals to i0. And further you can keep on doing this. Here it is xn1. So I just first bought this out and then conditioned up on that then bought x1 out and then in the second time I just conditioned this random variable on that. Now you can keep on doing this. Now let us try to understand this part. So what is this? Now what I am asking? You are in the beginning in the 0th round I am in state i0 and then I am going to jump to state j1 in n1 rounds. This is you are basically jumping in n states from state i0 to j1. What is this probability in our notation p i0 j1? What is the superscript? So it is going to be basically n1 minus 0. Like the number of subs is n1 I will just write it. Now if you look into this, does this probability here depends on x0 equals to i0? No, because if once I conditioned it on something beyond i0 that is xn1 this guy does not depend on this. So I could just write it as when I can do on I can further split it like this. Now what is this quantity here? This quantity I know that it does not depends on x0 equals to i0. So in that case what is this probability then? So is it okay? Is it correct in our notation if I write it as? So this is j1 here right not i1, j1 and this is going from j1 to j2 in how many steps? n2 minus n1 and I can keep on repeating these steps and finally what last step I am going to get is jm minus 1 times jm and what is this guy going to be? Now let us focus on the last term. So whatever this joint distribution of m random variables I am able to now express in terms of this. Now each of the terms here p i0 i1 so superscript n1 this is a n1 step transition probability matrix right and similarly this is n minus 2 n1 length step transition probability matrix. So all of this quantity here they depend on transition probability matrix of different steps okay. But what we have demonstrated earlier is whatever is the step we are looking at all that can be obtained by one single transition probability matrix by appropriately exponenting it right. So now all these quantities a claim is they can be computed from my transition probability matrix p is that correct right. But for the so this quantity how can I write? I can write it as p raised to the power n1 but which term this is a matrix right p raised to the power nm if p is the transition in that which term I will be interested in i0 j1 this is a matrix right I have to tell both row and column indices. So all these things can be obtained from transition probability matrix p. So fine all these things could be obtained from transition probability matrix p but still to complete this finite dimensional distribution I need one more term which is what is the probability that x0 equals to i0 right this is usually called the initial distribution. So x0 right that is your initial step and you want to know what is this probability that my initial step takes state i0 and you want to know it for all possible states. So this is called i0 where your i0 belongs to s is your initial distribution or initial probability distribution. So and often this is denoted as pi. So if you know this initial transition probability matrix and your transition probability matrix p do you think you will be able to characterize all finite dimensional distribution of a Markov chain right. So now this contrast this with iid process then you have iid process to completely define my iid process what I needed to tell you just one distribution right that one distribution that is common across all the random variables. But now here to completely describe a Markov chain I need to tell you two things to completely describe our time homogeneous Markov chain I need to tell you its initial distribution and its transition probability matrix. So as you see like the initial distribution is important because where the Markov chain goes after some round depends on where at which point it started if it starts at different point maybe the future I am going to see may be different. So that is why to completely distribute describe a Markov chain I need to tell you the initial distribution as well as my transition probability matrix. So initial distribution tell you where you are starting and transition probability matrix will describe how you are going to evolve from that point. So Markov chains the way we have scribes it they capture actually lot of things that happen that we face in reality and they give an often nice characterization of those phenomena. For example let us say you want to model some let us say aircraft. So you want to model or you want to analyze where it is going to go and how it is going to go. Suppose let us say its trajectory is mostly defined by what its velocity its acceleration let us say and maybe it also depends on how much fuel it is carrying. So once you give it some initial trajectory and you want to start looking at this position at different different times let us say you are going to look at this position at every second or let us say every minute. Now whatever it reach let us say at certain point of time let us say at every time I am going to measure velocity its acceleration at that time and also the fuel amount of fuel that is carrying that time. If you know that based on that itself you could further analyze further how far it is going to go right. I do not need to know what all the values of the acceleration and the velocity or the fuel it carried in the previous time instances. If let us say if I measured it at let us say t equals to 100 unit then that is enough like all the things that has happened before it is kind of captured there right. The fuel has been consumed till this point but what remains is of importance to me to know the future. So in a sense the current time is kind of already capturing the summary of what is happened in the past. So if you tell me past entire thing I can tell you what is the current thing. But if I tell you what is the current status yes it is a kind of summary of what has happened it may not tell you what all happened how the things happened in the past but it contains the summary of the things that has happened so far that is enough for you to describe future. For example if you just know at time t what is the fuel remaining and its current acceleration and all that is enough for you to see how my aircraft is how far it is going in the future right. So Markov chains are exactly doing this they are just saying that if in your scenario is such that the current state is capturing enough information or kind of summarizing what has happened in the past then you can ignore the past. And your future will be just you can analyze your future based on this current information. Initial X naught right but I am of course it depends Xn1 depends on what happened in the Xn1 minus 1 the previous slot. But here I am just trying to characterize the distribution jointly at different different time instance. But the origin I am not interested here right I am just trying interested in calculating the joint behavior at this point. Indeed what is happening going to happen here we do depend on what is going to from where you are going to start. So that is why this behavior is going to affect what is we are going to see at n1. We are just trying to capture that summary not like what is happening in between. Even when you are going to look at final dimensional distributions right even you need to characterize that aspect also. So for example here n1 could be just 1 I am I am I am considering all possibilities of the indices right it is I am not ignoring that fact. So for example here n2 could be just n1 plus 1 not only one step how it is going to affect all and I am also going to look at jointly all the distributions. What you are asking in this you are asking to find all these joint distributions right you are not asking it just like how it depends on the previous one. What we are trying to describe here is when we have a Markov chain property when you have Markov property all these joint distributions are going to be just captured by my transition probability matrix and my initial distribution. This transition probability matrix is already trying to capture all all you are asking it for from this state what is the probability that I go to next state. But that is not the only thing I will be interested in right may be like to completely understand the process I need to know all these joint process joint distributions. For this we are saying if you want to just what you are asking is okay n1 will only depend on what is happened in the previous step that is fine because I am saying that it do depend on where you are going to start from.  But I that is not going to completely describe my process right I may be interested even to know what was before that that is what the meaning of FDD is all possible description I need to give about the process. We are just trying to model this okay somebody has given you and then I want to analyze the system. How to obtain this transition probability matrix is a different question for that one has to I mean really do big big experimental setup do real measurements. So in this course like right now we are putting all these things under carpet okay how we are going to compute this initial probabilities how we are going to obtain this transition probability matrix this is all there is a separate thing how we are going to find this like you have to do simulation and then try to estimate this parameter. So these are all parameters for us P is a transition probability means it is a parameter it is a matrix with so many parameters I do not know those parameters what I am just saying that see if you pass on these parameters to me I am going to analyze and tell this is how your system is going to behave okay and so for example that is how right like if you want to evaluate 2, 3 aircrafts each aircraft will commit its own parameters then as a system analyst you will see okay take all these inputs and then you try to analyze the system and who is going to give the parameter for this system select the vendor will give the system he would have said okay these are all my system so it is fuel efficiency is this it can sustain this much of turbulence and it can go at this speed and this it can take this much of acceleration and all using those things you will come up with this set of parameters and then use it to understand your future okay let us understand another example all of us know queues right like we wait lot of time in the queues so let us say we have a queue and there is a server let us take a simple idealized case where time is loaded like it every time a guy may come and join the queue so in every slot there is with some probability a new guy will come and join the queue it may happen that in that particular slot nobody will join but with some positive probability he may also join now and also meanwhile there is a server here which keeps serving you let us say you are just in a going to watch a movie and you are in the line to get the ticket so somebody is preparing a ticket for you and as you get the ticket you will move out and lines may either grow longer or grow shorter depending on people are joining or people are exiting the system now let us say at some point let us say time xn so n is going to tell you how many people are there in the queue at time slot n so how you are going to define time slot is up to you you may want to count the number of people in the system at let us say every second or maybe at every minute it depends on the I mean it depends on the scenario like if people are coming very very frequently you may want to take up smaller time scale if your people are coming slowly you may want to take a larger time scale so time being let us say this is just n equals to second n a second one second like that every second I will join the system with some probability and you want to see how many people are there in the system okay now so time being assume that you know this service rate the way this guy is preparing the tickets then by that you can kind of understand at what rate people are leaving the system getting out of the system okay so the guy who get the ticket like he will just rush inside the theater and it may go down but that it could be that that process that the rated or the rate at which this guy serves the people it could itself be something stochastic for example some guys will be very annoying may ask the receptionist the ticket guy a lot of questions and keep him engaged so that guy will take more time to get served or some guy is just want to grab the ticket and go so that guy may take less time so there itself some stochasticity can happen and also we are saying that when the people are joining there itself is a stochasticity every time guy is going to know if you know all suppose if you know the description of this stochasticity at any time if you know how many people are there in the system can you at least give probabilistically how many what will be the next state of the queue here for that time being let us say people are entering with probability q in every slot and every slot a guy gets out and leaves the system with probability q okay so if your system currently at n there are n guys waiting in the queue in slot n can you say how many people what are the possible values for Xn plus 1 so what are the possibilities the next guy so the new value of this number of people waiting in the queue will increase by 1 if a new guy joins nobody leaves it is going to decrease by 1 if nobody joins but one guy leaves it is going to remain same if either nobody joins or nobody leaves or one guy joins and one guy leaves right so to describe this new number of people that are waiting in the queue does it matter like starting from the beginning how many people are there till around n or only the previous one previous lot is enough there right like if I know what is my status currently I can describe what is going to happen in the future in a way the number of weight people waiting in the queue is already kind of summary of the system right what has happened so far it is capturing the summary of the system and that summary is itself enough for me to define the future I do not really need to know what has all happened it till X1 X2 all the way up to Xn minus 1 all that is enough is Xn so you see that Markov change is exactly capturing such kind of system where my current is already capturing or kind of summarizing all my past that is enough to describe my future and in many cases this will arise for example this queue model and the aircraft example I gave like just that is a toy example but many realistic examples can be thought of that follow such kind of scenario yeah yeah in that case if I am going to think of this Xn as a process Xn process then this is going to follow so right now rather than understand rather than seeing whether it satisfies the Markov property what I want you to understand here is the motivation in this example is it enough for me to know the current state of the system to know the future or is it like I need to know entire thing that has happened so far to know the future that is enough right in a sense the current state is kind of capturing the summary that is enough to describe my future so that is where we want to start using this Markov change it can change but say this is what like I am not saying you without knowing P and Q you give me P and Q and then I am going to as an analyst I am not so there are parameters so there are two things there are parameters given this parameters as an analyst sitting and analyzing and telling what happens who is going to give this parameters this is a different question so for example if say if you are a planner you want to plan something somebody will say I have 10 manpower to do this somebody says I have this much budget to do this and somebody will say I have only this much time to do this so this has been given to you and now you will be coming up with a planner analysis exactly like that so here this description is given to you and now you are analyzing what is that if somebody is going to say okay that time I have is less and somebody say okay money I have is less and somebody say I have less manpower then you are going to reanalyze with this new parameter and say okay this is the performance right now I am just seeing like a big boss okay you tell me what is all happening what all the things you have and then I am going to plan and or analyze what is going to happen okay so fine now the way we have defined Markov property it says that I tell you okay this at this index I know the status of the system then from that point onwards I will to explain my system I do not need to know anything about my past but most of the time it so happens that this time index that we are going to deal with may not be deterministic itself could be a random thing okay for example I want to see how much my aircraft goes further when it is fuel tanks become a half okay so now the point where the fuel tanks become half itself could be a random thing right because the fuel consumption that is going to happen it depends on so much of environment that you are going to face that you may not have control that itself is a stochastic thing so your fuel tank may get half in just 2 hours or 1 hour 50 minutes or maybe 1 hour 55 minutes whatever it could be a random time and from that point onwards you want to understand you want to understand how the things behave in the future so there this n here could itself be a random quantity right so for example you want to understand so let us say you just randomly dropped into some cinema hall and at random you dropped in it is not necessary that when the counter opened you may have after that you may enter the queue after 10 minutes or 1 hour or whatever so the time when you are joining the queue itself could be a random time now you want to see okay once I join this how the things I am what are the things I am going to see like what is going to happen for example in this case let us say you may want to analyze if I join the system at some point how much time I need to wait before I get my ticket okay so to analyze how much time I need to get a ticket you want to understand okay what if if I my entry time is not exactly deterministic but that itself is a random right so all of us like go and join queue at some in rush and at some point we join and from that point we want to analyze yeah this is an arbitrary time but this is a deterministic this is all index right this n I said this is some n and now I could make it random by saying that okay this is x t t is a random variable now t is going to decide which index I will be looking at okay so repeat just this example you can imagine you are joining this queue and the amount of time that you need to wait in this queue before you get a ticket that is a random thing right because it depends on how many people were before you and also depends on the people before you how fast they are getting served how much time they are getting served now if you know that you have exactly entered at a particular time you know this but you may not know you may be entering the system at random point from that random point now you want to analyze you want to understand how much time I may need to get served okay so but so I am saying apriori if you know exactly at what time you entered it is fine but if you do not know this time of joining that itself could be a random how that that is going to affect your future so now the question is the Markov property we have discussed so far does it carry over to the case when I have such a random times so I want to basically we want to basically analyze suppose let us say t is a random number t is a random variable which is integer valued now I want to know so now I want to know instead now see like I have now put this I want to ask you if I tell you the this the observation of the states of the system till this random time t now the next state it is going to take in the the state j1 it is going to stake in the next round can I write it as simply this is one step right p i j1 so if this t is some fixed n this is what our description right p i j so this guy is independent of everything else except this xn equals to i and that is the definition of p i j1 but now I want to ask the question what happens if this time is not deterministic is random so notice that when I make it random this t could be taking any possible values that this random variable can take okay so now let us make this notation a little bit more clear what I mean by xt how to interpret this term xt okay so I know that what is xn of omega my the value taken by my n xn random variable at sample point omega now what is x of t means so this t here here and we are going to denote it as x of t of omega so t itself is a random variable so on that sample point just first tell me what is the index and from that index what is the value taken by this xt of omega on that sample omega yeah they will be same so when I write x of t the way you have to read it is so this is a random variable right xt so this is again so t is from omega to 0 1 2 all the way and I know that x of n is also from omega to r this is for all n all my xn's are defined on the same sample space right now when I say x of t it is just that I am not specifying you which index n I am just specifying you a random index so when I write so this is basically xt of omega when I write this means first look at the index at the sample point omega and look at its value at this omega okay so okay another example okay so just let us see if this fits well here so let us say that I am interested in 100 stocks okay or whatever number you take in the stock market and what I am interested in is how the values of the stocks change on every day okay so let us call this stocks 1 2 up to 100 I can think of this as in my sample space omega is this stocks now on each day I can focus on this stock and then see how its value is going to change if you are going to fix a stock let us say that corresponds to omega then you know the value it is going to take on different different days that is one sample path you have now what I want is I would say that okay the value of my particular stock when its value becomes less than certain number on the day when its value goes below certain number let us say so now what is t is going to be t is going to describe you that day how you have said that particular stock when its value become smaller than some number right so that its value may become smaller than some number in some day you do not know that could be a random quantity right you go to that day and then look at the value taken by that stock on that day t of omega so the value that stock has so in this case this t of omega basically gave me the day on which the value of that stock omega went below certain number and then this is going to tell exactly what is the value of that stock so you realize that why this random I may be interested in knowing my value process value at a random time because you may want to put some conditions like that for example you may want to in the in the in the stock case you may say that okay what is the value of my stock when it is exceeded 500 rupees okay so you will look for that time when it exceeded 500 rupees 500 what I am just saying is it is exceeded 500 but its value itself need not be 500 right it may be 600 but what I was interested in when it exceeded 500 it may have exceeded 500 rupees on the 100th day but when it exceeded what is the price that will be given by this and when it exceeded 500 that day is given by this t of omega so and here omega corresponds to that particular stock that I will be interested in x will go from no x it is not like x exactly so x has actually basically I can think of x as index by and with index and sample so this same thing we have written as x and omega so you understand the meaning of this random variables here for every n I am defining x n taking value on omega I mean assigning values to omega right in on real line so instead of fixing n I can just make it a joint one so I can think it as so okay so I think yeah yeah n belongs to capital N n is the set of natural numbers okay so either so you first so to define a random variable the process what I asked give me an index and then give me a sample then I will tell you what is the corresponding value so here I am instead of separately defining for each index then I can just say okay give me the sample point and give me the index I will tell you what is the value that it takes so in this case this x is a process we have discussed this right when we are talking about a stochastic process stochastic process can be thought of as a collection of indices sorry collection of random variables where every index corresponds to a random variable or alternatively we can also think it as random process x as a map which gives value r for a pair omega and n so same thing in that case we can think this as if this n so here this can be thought of as for so if x of omega in this case can be thought of as x of t omega and omega so sorry yeah x of t as this we will go into tell you what is the index we will be looking at and what is the value so for example when I say stoch right so the t is defining so on a particular stop omega let us say you have defined when it exceeds value 500 so t of omega will tell you that time when it exceeds 500 that that index so that is fine I mean that that ambiguity is there for time being for our example let us assume or maybe we can refine it as the first time when it takes value 500 we can do all such refinements then it is going to be unique in that case and then we can define like this


In 2015 an Australian mother gave birth to triplets 12 weeks early. This couple couldn't wait to have one more baby, which they hoped would be a girl, to complete their family of five. But when mom's belly wouldn't stop growing, their lives changed completely. This young family of five were living the dream. Yes, residing in Perth, Australia, parents Chloe and Rohan Dunstan, and their three energetic boys, Evan, Otto and Felix, were happy with their large family. But they believed that a little girl would be the perfect addition to the clan. So where did it all begin? Well, Chloe first became a mother at the age of 19 when she gave birth to Evan, who is now five. Then 11 months later, a 20-year-old Chloe delivered Evan a brother in the shape of Otto, who's also currently five. And shortly afterwards, the now three-year-old Felix joined the party, but the Dunstons weren't finished there. Indeed, the couple tried for a fourth baby, and Chloe soon fell pregnant. However, this pregnancy was completely different from her previous ones. You see, her first three pregnancies had all been fairly easy, with Chloe having only experienced a little tiredness from time to time. But even from the start of this pregnancy, everything appeared to have intensified when compared to her previous experiences. She complained of feeling nauseous for one thing, and she was also constantly exhausted and invariably hungry. But perhaps oddest of all was that Chloe's belly just wouldn't stop growing. To begin with, then, Chloe was convinced that she was going to have a daughter. She even wrote on Facebook, they say you get sicker when you're carrying a girl. However, the size of her tummy led her to wonder whether she might be expecting twins. Whatever it was, though, her bump just kept on expanding. So, to put their minds at rest, Chloe and her husband decided to have an early scan. At this stage, Rohan was certain that they were only having one baby, but the couple made contingency plans just in case. They even joked that if it turned out they were expecting twins, they would head straight to the shops to buy some matching outfits. However, neither of them could have anticipated what would happen next. In a Facebook post about the experience, Chloe recalls being off with the fairies as the ultrasound technician started the scan. After a short while though, the technician asked the young mom to go and empty her bladder, so they could carry out an internal scan. Still, the young couple couldn't have prepared for what the technician was about to show them. First of all, the technician pointed to the couple's baby on the screen. Then, before they had a chance to digest what they were seeing, the technician moved the scanner. Suddenly, a second shape appeared before their eyes. And although Chloe had thought that there was a chance she could be carrying twins, it was still a shock to them both. However, the technician wasn't finished there. After waiting for the pair to calm down, the technician dealt Chloe and Rohan another surprise. Yes, in between the two babies was a third. Stunned, Chloe began to cry freely. She later claimed on Facebook that she didn't stop shaking for the rest of the day. Unfortunately though, the technician had some bad news for the Dunstins. The technician explained that she was concerned about baby number three. After all, it was considerably smaller than the other two and might therefore face a battle to survive. Regardless, she told the young family to return for another scan in a few weeks. Hopefully, by then, all three babies would be healthy and growing at the same rate. Chloe later recalled that she initially thought it might be best if the third baby didn't survive. But now she has said she hates herself for even considering such a thought. However, at the time, the idea of doubling the number of children they already had when she was just 22 years old petrified her. Especially as this pregnancy was so high risk. Following the ultrasound, then Chloe and Rohan headed straight to the shops as planned and bought three matching outfits. However, neither of them said anything to one another. After receiving such unexpected news, they were completely overwhelmed at the prospect of bringing up six kids. Eventually though, the couple came round to the idea of having three more children and the thought of building their family became a blessing. But they were about to be dealt a devastating blow. Unfortunately, it became apparent that the smallest baby was being starved of oxygen and nutrients. It wasn't growing as it should, there was a real chance that it wouldn't survive. Ultimately, doctors informed Chloe that she had two options. She could either terminate the weaker baby, thereby getting the other two a stronger chance of survival. Or she could take the risk and continue with the pregnancy to offer all three a shot at life. Choose the latter, however, and she would have to deliver the babies prematurely at 28 weeks. The couple were faced with a difficult decision. And although the pair seriously considered letting the weaker baby go, they ultimately decided against it. However, Chloe felt guilty for putting the other two babies at risk, especially when they were growing so healthily in the womb. In spite of this, the decision was made. And even though Chloe's pregnancy was not an easy one, she determined to enjoy it. Then on July 3, 2015, Chloe delivered the triplets 12 weeks early. Furthermore, with two boys and one girl being born, Chloe finally got her wish for a fellow female to join the family. However, the newborns were not out of the woods yet. All three triplets were born dangerously under weight. Henry was the heaviest at a little over 3 pounds, Rufus was next at 2 pounds 10 ounces, and Pearl tipped the scales at just 1 pound 8 ounces. Out of the three, the little girl was in the worst shape. So as a result, all three babies had to undergo emergency blood transfusions and were put on ventilators. Poor Pearl was born very weak though. She had a heart defect and problems with her liver, and the little might also suffered a pulmonary hemorrhage and a minor bleed on the brain. Regardless, she survived, and so did her brothers. Indeed, after nine weeks in hospital, the two little boys, Henry and Rufus, were strong and healthy enough to go home. And two weeks later, their little sister was able to join them. Now the triplets had a new adventure to tackle, three older brothers. Today, the two-year-old triplets are thriving, and the family has set about sharing their story with the world. To this end, the Dunstins have launched multiple social media accounts, including a YouTube channel, to document their day-to-day existence.


Great pleasure to introduce Kirsten Wickelgren. Let me just say very few words about her. She obtained her PhD in the year 2009 from Stanford University on the supervision of Gunnar Karlsson. And right after that, she obtained a prestigious American Institute of Mathematics Fellowship, which she spent at Harvard University for five years. She went to the Georgia Institute of Technology afterwards. And since 2019, she's been professor at Duke University. She's been working on the intersection or union of arithmetic and homotopy theory and has a long list of impressive papers and collaborations, and especially on what she's going to talk about today, which is an introduction to A1 homotopy theory using enumerative examples. Please go ahead. Thank you very much. And also, there are now notes for these talks. And I do realize they were posted a little bit late since tardiness is an almost universal human trait. You all probably know what I mean. It does mean that I was up pretty late last night. So if my energy starts to flag, it will be very much appreciated to have feedback and help and corrections. It's really a pleasure to get to be here and so the more I get to interact, the better. And thank you very much, Oliver. So we're going to use A1 homotopy theory to do some enumerative geometry and enumerative geometry. Let's say that what it does is it counts algebraic objects over C. And an example to say what algebraic geometric objects means, we can ask how many lines meet four lines in P3. So a nice picture of this is that inside P3 of C, there's R3 where we're all in our own separate rooms. And inside R3, we could take four lines in our rooms that don't intersect and actually just ask any friend, even who doesn't do math professionally, how many lines intersect all. So the answer, if you allow the lines to be complex, is two. And in your own little room, it might be two and it might be zero. The goal here is to introduce A1 homotopy theory and have fun. And but we could also make a concrete thread to guide us is that let's make a goal to record arithmetic information like whether those lines are over C or over R about geometric, those geometric objects that we're counting over some field K. And the number of these objects, these lines, is going to be fixed over the algebraic closure. But it may vary over the field K. And we'll use as a tool the A1 homotopy theory that we were discussing in Danny Krashen's taught lectures and Frederick Glees' lectures due to Morel and Belvatsky, as we've seen, but that's the homotopy. Let's write it down to homotopy. And let's start with some classical homotopy theory. We have a sphere in dimension M. So we could take real numbers whose, the sum of whose squares is 1. And another useful way to write this is it's equivalent to PnR over Pn minus 1, the R points of Pn over the R points of Pn minus 1. And there is a degree from the homotopy classes of maps from SN to SN to Z that roughly speaking counts the number of free images. And we were trying to count. So given F SN to SN P and SN, let's suppose that we don't crush the whole sphere or one dimensional part to this P.  And the degree of F, we'd like it to be something like N capital N. It is. It's the weighted sum of a local degree of F at these inverse image points. And the local degree can be defined in terms of that global degree again. It is the local degree. So we'll let the be a small ball.   And it doesn't have any of the other QI. It's just QI. So then we can map U to V. And we really want to map the boundary of these small balls. This looks like the over the boundary of V or SN. And this looks like U over the boundary of U, which is also SN. And we have an induced map. And the local degree is the degree of this induced map. And we have a formula from differential topology telling us that instead of sort of in some sense recursively defining the degree of F in terms of another degree, we could make the local degree with calculus or compute the local degree with calculus. So we have a formula from differential topology. So we can pick local coordinates near our points and express our function then as a tuple of polynomials. Or a tuple of smooth map if it's smooth. So let x1 to xn be oriented coordinates near QI. Let y1, yn be oriented coordinates near p. So then our function becomes a tuple of functions in the nxi of 1, m, rn to rn. And we can take this Jacobian determinant, we'll denote it by j, to be the determinant of the partial of fi with respect to xi. And then our local degree depends on this determinant. So the local degree at QI of f is plus 1 if the Jacobian is positive and minus 1 if the Jacobian is negative. And a great special case is if f is actually a complex analytic. So we've forgotten some extra complex structure. Then f is going to preserve orientation, putting us always in this case. So let's say that f is a polynomial map over c. And well, suppose it doesn't have a multiple root. So this Jacobian, it's not 0. And then because there's this multiplication by i that tells us which directions we're going, this has to be 1. And we do get that the degree is counting the points in the preimage. And this might be the number of solutions to something we were interested in. So it's the number of solutions to f1, f2, fn equals 0. And a great example is the fundamental theorem of algebra. So if we have a polynomial with complex coefficients of degree n in the usual sense, then its degree is also n in the topological sense. And we have n solutions if they're all simple. It's the number of solutions to our polynomial. And this is fantastic. The f1 were these folks here. So if f is encoding lots of equations at once. Now those f, they might have sort of a twisted target. So we can also count if the f actually corresponds to a section of a vector bundle, then we're not taking maps to some a1 or to some line. But that line is twisting over the space of our possible solutions. So remark, we can also count solutions to f equals 0. Or f is a section of a vector bundle, vector bundle. So here's our vector bundle v with its canonical map to its space x. And here's a section f for function. And then the sum of local degrees of this f is another invariant from homotopy theory. It's the Euler class. So the Euler class of v can be computed with any section. And it's a sum over the zeros of our section of a local degree of f. And in order to define this, we need to view f as a function. We can take a local trivialization of our vector bundle and some local coordinates. And we really must have an orientation so that we don't wind up changing the value of our Jacobian as we change our choices. So rank n, and we'll talk more about oriented vector bundle. So we can also count the solutions to such equations. Here's an example of doing that. So let's let our x be the gross monion. The indexing convention going on here is that it's parameterizing the p1s and p3, like in our initial question. So that's also dimension two subspaces, w and c4. Or equivalently, the p1s and p3s, like the lines, the projectivization of w. This is where we take out the origin and mod out by scalar multiplication. And w and lambda and c star. And this sits inside the projectivization of all of c4, which is p3, or the c points of p3. Over this gross monion, we have a tautological bundle. So here's a rank two vector bundle of the tautological bundle. And it's defined so that over the projectivization of w, over this element of the base, we put w. So the vector space over this line here is the rank two vector space, w. So let's consider that question we started with. Let L1, L2, L3, L4 be four lines in p3. How can you tell if another line L meets, say, L1? It's that in the intersection of the corresponding w's is not empty. So if you take the wedge product of the forms defining L1, it's 0 on L. And there's nice exercises that Sabrina Pauley has that's made that say a little more on this, but that means that the way we can characterize the lines meeting our four lines is as the zeros of a section of a vector bundle. So the lines meeting L1, L2, L3, L4, they're the zeros of f for f, a section depending on the LI of the following vector bundle. It's four copies, one for each line of wedge two of a stool over the gross money. So our section above a w, say, the first copy of this, it puts the restriction of the two forms defining L1 to w wedge together. And that means by the discussion that we were just invoking having that the number of lines is this order of their class or their number is E of the same bundle. So if you believe that the many points of view on the order of class from algebraic topology includes one where we can compute it with any section. And in particular, this means that this number is independent of our choice of lines. Of the choice of lines, as long as the associated section really did have isolated zeros, which happens when they're generally chosen or when they don't pairwise intersect. And for that, we needed that this was always one so that we were working over C. Then we can compute at least over C. But the answer is two, using homology of gross monions, the splitting principle for computing Euler classes from ways we'll get to later and tools of that sort. So we've been using a degree to count algebra geometric objects. And we'd like to do this over fields that are not R or C. So Lan and Morrell came up with a notion of degree for an arbitrary field k that reported more than what was happening over the algebraic closure. So an evocative place to start is for a rational map from P1 to P1. And above, we had this formula here. And we remembered the sign of the Jacobian at each point. And Lan and Morrell are going to allow us to remember more than just the sign of the Jacobian. And they'll do it by landing the degree in this group. So the degree of F will be valued in the growth index bit group of k, which we've seen from Danny Cashion's talks or the bit group, which recovers the growth index bit group. And this is defined to allow for formal differences of symmetric non-degenerate bilinear forms. So we'll group complete, meaning we're allowed to subtract even though we only started with an addition. And the addition that we're going to start with is perpendicular direct sum of bilinear forms. And we can also take the tensor product of bilinear forms of non-degenerate symmetric bilinear forms. And we've seen the quotient of this on Monday and Tuesday. So the bit group of k with the fundamental ideal and the Milner conjecture with the relationship of those quotients, say, tall cohomology and Milner k theory is also in the growth index bit group and also applies to give us tools to understand elements of the growth index bit group. And the bit group is what happens when we quotient the growth index bit group by this hyperbolic form 1 plus minus 1. So a useful invariant of bilinear form is the dimension of the vector space that's on. So this is the rank homomorphism. So the rank of a bilinear form, the vector space v over k cross v to k is the dimension of v over k. And we can recover the growth index bit group from the bit group. We have a quotient to the bit group. And we also have a rank. And given a rank and its quotient to the bit group, we get the element of gwk that we might want to pin down. So this is a pullback. We saw some useful generators for the bit group, which I'll put the notation for to get us all on the same page. So stably, meaning after you group complete any bilinear form symmetric non-degenerate can be diagonalized. In fact, in characteristic 2, you can diagonalize even without group completing. So that means our rank 1 forms are going to be generators. So here, this is the element of the growth index bit group associated to the bilinear form on k that sends xy to axy. And over a field, we have this nice presentation. So we take the tensor product of the two rank 1 forms corresponding to a and b. And we get the rank 1 form corresponding to ab. And an elementary diagonalization shows that this diagonal bilinear form is the same under a basis change as this one whenever u plus b is not 0. And these two together imply that for any non-zero u, u plus minus u is 1 plus minus 1, which is this hyperbolic element h. And as an exercise, we can show that 2 implies 3. And this form here is also the one whose brand matrix looks like this, which Danny Kruschen wrote. And you watch in characteristic 2. But again, the group completion is OK. Let's get some examples. So if this group is trivial, then all our generators are the same. And the rank gives you an isomorphism. For r, we have 1s and negative 1s for this group. And the difference between the number of 1s and the number of negative 1s is called the signature. We have a map, rank, comma, signature. And they have to have the same parity that this induces an isomorphism to z cross z. This is we diagonalize the form 1, 1, 1, minus 1, minus 1. And signature equals the number of 1s minus number of 1s. We saw in the Milner conjectures that the atolicoamology with Zeeman-Tucco coefficients gives us some useful invariance. And one of these is the discriminant. And so for fields where we can compute atolicoamology, like fq, we get the discriminant. And these are enough to determine the entire bilinear form. So this is an isomorphism to z cross fq star over fq star square. And we have another interesting term here in addition to our z. And we have another interesting z in addition to the z we started with. We can keep going and get more interesting extra terms. So if we had qp, we get another one. And more generally, for a complete, discreetly valued field, we can compute the bit group as two copies of that of the residue fields. We've got k complete, discreetly valued field, k residue field. So for example, we could take the p-addicts or the Laurent series in a formal variable. And we have that the residue field is this finite field. And we'll assume that the residue characteristic is not 2. And then we have that the bit group is two copies of that of the residue field. We're going to have transfers in a way lining up with Frederick DeGlese's talk. But I want to introduce notation for them and a formula now to be able to talk about them. So let's give ourselves a separable field extension. Let's finite extension of fields. We have a trace map or a transfer from the bilinear forms over z. And we need to get down to a bilinear form over k. And it will compose with a trace from Galois theory. So the sum over all the embeddings into an algebraic closure. So if we have a bilinear form over e, we can view the now as a vector space over k and apply the sum of the Galois conjugates to get down to k. And this is still non-degenerate by Galois theory and defines the wanted map. OK, so we have the growth index bit group, the bit group. And let's get back to Lan and Morrell's formula for the degree of a map from p1 to p1. So back to formula. So we'll have f from p1k to p1k. And we'll pick a rational point at the base. And we'll assume, again, that it's non-constant. So we just have finitely many pre-images q1 through qn. And for simplicity, we're going to suppose that the Jacobian, which is just the derivative, is not 0. Then the degree of f says Lan and Morrell is don't just take the sign of the Jacobian.  And we'll take this transfer. So this f prime is sitting in the residue field of qi. And we take the sum. So we're also counting the pre-images. And they have a weight that's picking up information about their field of definition and the function's behavior there. And then we have the very lovely fact that this does not depend on our choice of p. And there's some exercises in the notes to play around with this. Now, more generally than just a map from p1, Morrell is going to give us a degree map from the sphere, which, like in the beginning, we'll view as pn over pn minus 1, so the top cell of pn. And its value will also be in the growth index group of k. So this is going to mean here this will be unstable a1 homotopy classes of maps, which we didn't define, but it's just let's say the analog of homotopy classes of maps. Like we started with our degree function from the functions from sn to sn, but they were homotopy classes of pointed maps to z. This degree has the pleasing property that it interacts well both by taking real and complex points. So if we took the degree, let's say k over k is r, so we have pn over r, let's just say k is r here. And we have this degree, the gw of r. An element of gw of r was determined by its rank and its signature. And this gives the degrees of the maps on the real points and the complex points. So we could take c points and have pnc over pn minus 1c is s2n to s2n and the degree in algebraic topology. And we could take real points and get homotopy classes of maps from sn to sn and take the degree from algebraic topology. And the a1 degree simultaneously captures both. There are formulas to compute this that doesn't involve summing up a lot of terms. So Christof Kazemov and Thomas Brazilton, Stephen McKean, and Sabrina Pauley give formulas for the degree of lists of polynomials in terms of positions. OK, so we're going to say more about where this degree map comes from. And for that, we're going to also say more about pn over pn minus 1. This quotient doesn't exist in scheme, so it's already an achievement of a1 homotopy theory that we are allowed to do this and get something like a sphere. Now, let's talk about that. So the smash product is the quotient of the product by the, I should have said these are pointed spaces. So we're going to have a favorite point in y, and that's what that one is supposed to mean. And the favorite point in x, that's what that one is supposed to mean. And we have a smash product like so. And an example is that from classical typology that if you smash an n-sphere and an m-sphere, you get an n plus m-sphere. And in particular, if you smash s1n times, you get sm. And the suspension, which it turns out that there's a lot of phenomenon that sort of behaves similarly after sufficiently many suspensions, and this is the smash with s1. So p1 can be formed by gluing together two copies of a1. So p1 is a standard a1 and then a1 containing infinity. And together, their overlap is this gm, so it's a push-out. And the a1-homotopy theory comes from forcing a1 to be contractable. a1 is like a line. So as part of our introduction to a1-homotopy theory, we could make a lot out of the analogy between a1 and a point. That might have been a good thing to do, but let's do it now. So we're going to declare that a1 is homotopy equivalent to a point. c goes to 1 over z, c goes to z. And then it follows from the formalism of homotopy push-outs that when you take the homotopy push-out of a diagram to points like this, you get the suspension. So the homotopy push-out of this diagram is the s1 suspension of gm. And so we get that that's another way of writing p1. So we'll have several spheres. We're going to have s1. And half gm is spec a1 over z, or a1 minus the origin. And we'll then be able to talk about sp plus q alpha, or s1 to the p, gm to the q. And some other notation for this is sp plus q q. So another example that will be handy to have is that an minus the origin is s1 to the n minus 1 smash with gm to the n. So to see this, we can use induction. We're going to cover, say, a2 minus the origin by its ax, by the complement of its axes. So and in general, we can take a1 minus the origin, cross with an minus 1 minus the origin. This produces a cover, an minus the origin. And this is also a push-out. So we're gluing a cover together. And this an is contractable, and this a1 is. So we have the opportunity to replace this with, well, this one, say, with an minus 1 minus the origin. And then we're looking at a push-out along the projections. And it's a good exercise with push-outs to identify this as a suspension of x smash y. And in particular, we get that an minus the origin is the suspension of these two pieces here. So in induction, we know what these folks are, which shows the plan that we started with up here. So we also have the pn over pn minus 1. We could start moving this pn minus 1 in from infinity until it's the pn over pn minus the origin. And then we could get rid of the pn minus 1 at the end at infinity and get that quotient. And then this is contractable here. And it follows that this is the suspension. So this is s1 to the an or gm to the an p1 smash with itself and times. So that was our sort of summary of ways to think about some spheres. It turns out to be very useful to allow an algebraic topology desispending. So the stable homotopy theory in algebraic to classical algebraic topology permits us to desispend our spaces. And this gives us duality statements and allows us to represent co homology. And there are a lot of phenomena, which sort of repeat themselves in high enough dimensions or with enough suspensions that are then computed very effectively. And in a stable homotopy category. And that is also a useful thing to do in a one homotopy theory. We, we can desispend, we can unsmash with p1, which means we can also unsmash with with s1 and GM. And we get a stable homotopy theory that will, will, will denote it as shk, which is standard. Um, so morels degree homomorphism comes from a theorem of morale and morale hop and Hopkins morale. And it says that in the stable homotopy category, the maps from s not to s not. And remember we had some other ways to talk about these which which looked. A lot like spheres. This is this growth in the group of K. Moreover, we can in fact have a beautiful computation of the stable homotopy classes of maps to from s not to all of the GM smash ends. So whenever they're the same number of copies. Where did our spheres go. Whenever the same number of copies of s one. And we want to compute maps from a sphere to a sphere that have the same number of copies of s one. We lined up, we wind up in this, in this group. It's a group because stably, this will get a suspension in it, and you can add maps from from the suspension of a space, you can smash maps together. So this is actually a ring. And that's it. It is the combination of the bit ring and the Miller K theory ring we were looking at on Tuesday. So they combine to this ring here, which is Milner bit K theory so came Milner, which was the Milner K theory. And w of K, combine to this, but K theory. And let's define it. So it's a graded associated algebra, it's generators. So let's look at them as as maps from some spheres to itself. In a minute, but first let's get an abstract, just algebraic definition. So generators. We have you came over bit one K for you and K star and Ada corresponding to the hop from that. Subject to the relations. Yes. Okay, as a field, an arbitrary field. We have the Steinberg relation says you times one minus you as trivial. We have the relation that bracket AB is bracket a plus bracket B plus Ada bracket a bracket B. And Ada commutes with the hop map commutes with our generators, and the hop maps times an element that will correspond to our hyperbolic element from before is zero. And the hyperbolic element in terms of these, these generators is written Ada minus one plus two. So let's relate this to the growth indeed fit group. So remark, the growth indeed fit group of K is isomorphic to K zero Milner bit, and the bracket a element corresponds to the hop map from topology is like an S three to S two. So it's it's going down in the number of like in actually we're gonna see it in a second the number of GMs. We'll get to that. Because it's like an a two minus minus zero to you. And so the angle brackets a corresponds to one plus Ada a. And so our H. This was one plus minus one. This corresponds to one, then the one here is actually the additive identity so it goes away, and then plus another one, minus one. So, let's let's talk about how, how these correspond to maps between spheres. Here's some things about morale and morale Hopkins is proof. The map bracket a corresponds to the map from S zero so S zero is two points. And we're working over K so we, this is spec K, just went union spec K. And the bracket a corresponds to the map from S not to GM has to take the base point to the base point but the non base point it will take to the point a base point to a. And now, Ada is the map from a two minus the origin to P one that takes X comma why to the point in P one determined by by X and why. And on the C points, take C points, we get C two minus the origin which looks like a three sphere. And that's mapping to CP one, or the two sphere, which gives a half map, but on our points. It's a map from the circle, the circle, and it's multiplication on by two, in a particular. Over R, or if you're inside our data is not no point. And compared to classical algebraic apology. And the sheet is no potent. This is, this is different. This is a new thing. All right, let's introduce a little bit about the relations and and and and see a little bit more about about this, this degree. So, to do that. Let's talk about multiplication on GM, and we'll need a little bit of homotopy theory. The smash product was this quotient of the product by X times a point union point times why, and we could glue these points together, and the wedge is where you glued them together, blue base points and and the stability, the product is the sum of these folks. So if we suspend once meaning smashing with S one. There we go. Then, we can, we can make a map backwards like this, how do we do that, we have a map from X cross Y to X, which gives us maps like this. And the suspension of the projection of X. And since this is a suspension we can now add them. So what we'll do is we'll add this suspension of the projection onto X, and why, and we'll give us an expression for the product to be as some and morale shows morale and Hopkins show that the product on on GM is related to the hop map. So in our stable homotopy category. The multiplication map on GM that takes a rational point a and a rational point be to the rational point a be so gets called. It's all so motivation. And it can be expressed in these terms so stably we have product looks like that that some. And a map from a wedge product is just just maps from each of the wedge factors. And this is the identity on GM, the identity on GM and the hop map. So, this lemma allows us to justify the way we were looking at angle brackets a. So, we've got a map from P one to P one, sending Z to to AZ. If you remember the way we defined the degree was we took the inverse images of a point and looked at the derivative. So the degree of this map here is we can take the inverse image of zero, that is zero. Look the derivative that is a this has degree a by our by Landon Morales intuition that we stick the Jacobian inside brackets. And, but this is also, and we wanted to see that this corresponded to one plus a times a, so that we would have that GW sitting inside came on their bit of zero. So now, we can look at this map. This is the suspension of in GM, sending Z to AZ. And we can express Z to AZ is the product of a with Z. So, we take GM cross that K. And we get this map by the identity cross a GM, cross GM, and then multiply. Then we get to break this map up into the identity, and a constant map which is trivial, and Ada, and we get, we get, we get that. Similarly, this lemma. Here. This lemma also gives the relation that a B. So what would you get if we, where are we, if we, if we took the non base point to a times B over there is what we would get going this way. That's B. Ada lemma can be used to show the second relation. And there's a lot of great things to say about the relations. We'll, we'll end with that. And this will define a map from our generators and relations description of came on or that K theory of K to our ring of these homotopy groups of spheres, and then morel and Hopkins show that this is an isomorphism. It's a beautiful computation, or they, or they do. And in particular, we get the degree map that we've set about to count with particular. We can go from unstable homotopy classes to stable and then get a degree, even though in fact we are our isomorphism was really sort of defined the other way. So, I, this was very successful. So today about other homotopy groups of spheres they encode a lot of geometric information. It's not going to make it into this talk. But while while we're here, let's talk about some some big problems. So, let's do do the notation. So, this is the zero line of the stable homotopy groups of spheres. Another notation is these sheets of homotopy groups and we take global sections to get to get these these groups. And this is all the zero line. And then we have all the other lines. So the R line. So this is the sphere spectrum, just sort of all the suspensions that he wants suspensions of s not. And we know a lot of things about homotopy groups of spheres that imply very interesting geometric consequences classically. And we are beginning to know a lot of very interesting things in a one homotopy theory. But maybe fewer than you might expect. So what we've got, we've got an annals of math paper by Oliver Rondig's Marcus Spitzveck, and Paul Arnaud Auspair. I'm computing the one line in terms of her mission k theory and number k groups. And they know some things about the two line. So, what about over 2019 characteristic k, not equal to two. And we're, we're left with some, some questions. So what about over more general rings as you remarked this this was over a field. So, over more general rings. And then the analog of the growth indeed fit group over Z. So over more general rings, and then this is the group that will be using most just in these talks. What about that one. There's a result from this year by Tom Bachman and Paul Arnaud Auspair. So, once we localize. It's the growth indeed fit ring of the adjoin one half, which is the group completion of the symmetric non degenerate bilinear forms over the adjoin one half. And watch out over over non fields your presentation is not the way to define. And what about over more general data can domains and the and what do we know about that we can immediately feed that into problems that then then use this to get to get a new murder results.  What about the rest, maybe with a particular eye to the successful consequences and algebraic topology of knowing things about these groups there they're powerful. Now, once we have the stable groups, what about the unstable ones so it is not known. What kind of stable groups are equal to what kind of unstable groups so what is there a growing ball suspension theorem, which stable by star star correspond unstable groups. We're going to use such computations to count things, but there are other uses. I'm tempted to say something about vector bundles but I don't think I don't think I want to try to add live. Let's keep with the goal that we arbitrarily chose for ourselves and get back to counting things. And for this. Let's talk more about oiler classes. Barge Morel defined one right around 1999 and chow bit that Jean Faisal has a lot to say about, and Morel writes about one as a principal and obstruction in his book. And Ashok and Faisal compare them. And Mark Levine, as well as Jesse Cass, and myself use them for enumerative purposes. And dig lesion and con gave a beautiful functorial approach to please. And Tom Jushan in a deal con. Tom Bachman and I did a bunch of work with a growth indeed Sarah, which was also in work of connections with coherent duality in work of Levine racks that but actually my cockings was. I was calling one time when I gave a talk but they do a lot of other things like the much of the gasp and a theorem and they're there. They're many interesting points of view on on this class and start with one of them. So, let's give X smooth case scheme of dimension D. Let's take a rank, our vector bundle, and we need an orientation so that well from a certain point of view so that when we choose local coordinates we can't accidentally swap an X one to a minus X one and change the sign of our determinant. V is going to be oriented by the following data, a line bundle L and an isomorphism phi, where is a line bundle, the determinant of the is identified by our particular isomorphism phi with the square. And then we can talk about X being oriented when it's tangent bundle is and what we'll need to take an Euler number. Relative orientation so V to X is relatively oriented when the bundle where the Jacobian lives, the Jacobian of the zeros. It's a determinant of a bunch of derivatives that's a map from the determinant of the tangent bundle to the determinant of the where the determinant is the highest wedge power. So, for example, if you have X is PN, or the gross monion. No, Kirsten you didn't finish the sentence when home is what is oriented. Sorry. Thank you. So, the gross monion. It's tangent bundle is home from the tautological to the quotient. And if you remember our conventions, this is the gross monion parameterizing because the parody is what matters. PMS in PN so let's be particular about which which parody we're having this one kind of agrees with that. Then the determinant of the tangent bundles dual. This is the cotangent bundle is O of minus minus and minus one. So X is orientable. If and only if and as I and on P one is relatively orientable if and only if and is even. So things are positive over so there's a question in the chat about a square root of the determinant. If we take our intuition from from the reels. We, we want to know when the determinant of our coordinates doesn't doesn't change sign. So if we if we took a new basis, we want the determinant of the basis change matrix to still be positive if we're going to get the same sign of the of the Jacobian. So, the being positive is the same thing as being a square P one is relatively orientable. And is even. All right, so here comes, here comes our, our number in GW okay, and this this number is the analog of viewing a homology class in top dimension as something in Z, except now Z is replaced by by GW okay. So, let's suppose, again, X over K is a smooth, proper scheme of dimension D be to X with F is a rank D vector bundle is relatively oriented by this line bundle on this isomorphism. And let's give ourselves a section that has isolated zeros. In fact, so such that let's have the isolated zeros actually have multiplicity once we're imagining our Jacobian is not zero. So such that way to say that is at people zero consists of multiplicity one zeros. And equivalently, we have that this this map from the tangent space to V. So let's say for all x that is a zero, when when you have a zero of a section you get an induced map from the tangent space of x to the tangent space of f of exit V but since this is x zero, this using the zero section, this is just the tangent space of acts at the point, plus the fiber of the vector space. And so we get, we get a map from the tangent space to the, the fiber of the vector bundle. So this is what we have when we have a zero. And this map has non vanishing determinants of the map on tangent spaces as non vanishing determine. So then we're going to make our order number be the sum of all those Jacobian determinants over the zeros. So, the number of our vector bundle and let's remember its orientation, or forget it if it's clear. And with respect to a section, let's call it any of those sort of E as well. So let's pass this, this order number, the sum over the zeros, the degree of f used as a function. And so your two points of you want it so the degree X F can be computed choose local coordinates, and that's really any tall map from an open neighborhood to choose local coordinates. And we want it to be an isomorphism on residue fields at the point we're interested in, let's call this this navy coordinates on X, and local trivialization of the, which are compatible with the orientation to submit the definition of that with, there's the relative orientation, then you can express it as a function this. Now, X doesn't have to locally be some sort of a fine space, but nonetheless, you can jiggle it a little bit, and it won't be any longer. So then you can write F as a function from the dimension from a to the dimension of X so X is looking in these misnavish coordinates, it's like an open set of that to a to the D, and we can then make our determinant J F is the determinant of fi over X J. We had this assumption that our zeros were multiplicity one. So we get that this is not zero. This is in the residue field of X. Thanks, everybody. In the residue field of acts, and then the degree X F is the transfer of the Jacobian acts. And since it's, it's, it raised suspicions to say this although it's actually okay. There's here's an equivalent point of view. So or equivalently, we have this, this map, the induced map on tandem spaces from our section F that we were mentioning above that starts looking at like a map from the tandem space of X to the fiber of V. And then we can define J J of F of X to be the determinant of this map, which is then a map from the determinant of this vector space to the determinant of this vector space. We have all the same dimension now so the determinant that the dimension of X is the rank of V now so the determinant is the art or the deep right. The deep wedge power and our orientation precisely gave us information about this. So the orientation gives an isomorphism between this home space containing the Jacobian and the square of a line bundle. So if we pick any basis for this, our Jacobian just becomes a number. But if we switch basis, the corresponding number changes by a square, since this is the square here. So, this actually gives so here's J of F of X is sitting in here, but this is induces a well defined element of K of X over K of X is unit square. induces a well defined element that hey will also call J F of X of K of X is like trivialization of L of X over K of X is units squared by choosing a trivialization of L of X and this agrees with the above. Okay, so we have, we have four minutes. I had planned to talk about what happens if the zeros are not simple. But that's, that's a longer discussion and it's going to fit here. So, question happens. Well, we'll set it will last the question this time I'll answer it next time. If the zeros. Multiplicity one, we were missing this in differential topology as well. So CF, we can said that in differential topology, the degree was one, if the Jacobian was positive and minus one, the Jacobian was negative. So then, what happens when the Jacobian is zero. But, you know, something happens. So, we can use this to answer a further question that that sort of came up at the beginning of the talk. So answer will be next time. And the fact that it's possible to answer, and there's a good way to interpolate between the quadratic form that's giving the Jacobian giving the local degree. As, as you as your zeros come together, and gives us some leverage on this question. So why is the Euler number, the F independent of the section. F. In our four lines example, that's analogous to saying that the answer to works for any of those four lines that you chose. So the answer one coming from this is that sections with isolated zeros can often be connected by families of such sections of sections whose zeros are isolated but they might be higher multiplicity. So, our sections will be parameterized by an a one, and it'll lead to an element of the growth in the big group of the functions on a one or K of X, but this by a one invariance of the growth in the big group, or by harder serum will give us a well defined Euler number. Or as a pain in the neck, and I wouldn't know. So, alternatively, we can introduce the Euler class, inspired by DeGlesion cons, and and Mark Levine's and John fizzles perspective. So, another way to get rid of this dependence on the section is to have an Euler class and a co homology theory number is push forward of an Euler class. There are values and interesting formology theories. Like one Frederick de Gles ended with in this talk. And then we'll have for free and much more useful techniques for computation to control this expression. We will set that up. And get this control. Those those those wonderful powerful tools that I'm over. So, so let's end here. Thank you very much. Yeah, thanks a lot, Kirsten for a wonderful lecture. Let's thank her. Either by clapping real or an icon. Is there any questions. I had a question. I might have just over read into like an offended remark you made but when you introduce our notion of transfer here you said it was related to the stuff that DeGles talked about and this is all very new stuff to me so can you say more about how those two things are related to what happened in the last lecture. Yes. So, and the stable homotovic groups of spheres have to have a whole bunch of transfers, and which transfers you have is an interesting question. We saw some transfers that you that you have to get. The transfers here it's here they're coming from the looking at this, the sphere is going to map to Hermitian K theory and Hermitian K theory has as it's as it's K zero of these filenier forms on on vector bundles and has a nice point of view. So I'm going to give you to Mara, Yackerson, and Mark Boy, why an Elton, Elmanto and says, no, no, and Khan that that gives you a way to look at Hermitian K theory is this kind of uniform, you sort of universal thing with transfers when you have coherent duality sort of push forwards. What is done by this trace formula is sort of looked we looked at that vector space over E, and then we get the vector space just over K, it's kind of like pushing forward. Just just looking at it like that and composing with this trace sort of coherent duality thing. And there's a very lovely characterization of a permission K theory as this thing that has transfers when you work on recognition principles for for for cosmology theories and some. Well, that's one answer it's probably a little convoluted in this. He could, he could say a lot about about that answer maybe there's. I'll stick to that. Okay, thank you. Sorry, sorry, I want to advertise one. So that's for these infinite p one loop spaces and connected to that big open problem of we don't have a friend fall suspension theorem is, do you want to characterize for finite p one loop space is another very exciting sort of possibility in general field. Any further questions related to this question of all suspension, do we even know if, you know, smashing something a sufficiently large number of times ever stabilizes. So, is there a finite number but that finite number might grow arbitrarily with the connectivity, such that you stabilize all over. Can you can you take that one. So the problem is that we don't know a finite generation of, of the groups involved there. But for otherwise you, you, you would get, you would get stabilization of these coordinates. There are concrete computations done also by, by Kirsten and co authors, which indicate that stabilization occurs and behaves rather similar to, to what happens in topology if you combine it. What happens at sort of complex and, and real points, but yeah, it's, it's, it's a very interesting open, open problem. Any other further questions. Yes. Do you manage to recover these very classical results like 27 lines on a smooth hypersurface v3. I'm so glad you asked that question. We will Friday. Absolutely. Okay, so I think this is a good teaser for Kirsten's next lecture on Friday morning afternoon evening night wherever you are. Let's thank her again but don't forget there's a problem and discussion session with experts in the field, starting in about eight minutes. Thanks again.


Hello and welcome back to The Craft Box, my name's Brinton. Today, we're getting Scarecrow. Level 48 spell, let's go. Greetings, young wizard. You're puttering around adventuring, but it's time to learn a new and powerful spell. I speak of the Scarecrow, the original from which all less a copy sprung. He's the Lord of Night, the King of Dread, and a Herald of Death. I can't just teach you this spell, however. You need to demonstrate your commitment to your school and to the forces of death magic. First, you need to venture to Dragon Spire, to the Necropolis, a place of utter darkness and despair, and recover from there a dram of the quintessence of fear. Liquid fear itself, from the coldest streams of nightmare. Speak to Dimitri Ironchains in Dragon Spire. He'll explain everything. Okay. Let us go and head over to... the Labyrinth. I'll be back when I am in. The Labyrinth. Okay, I am back. I can't believe what I'm hearing. Did that mad old hunchback really send you here for a dram of the quintessence of fear? He's... well, he's mad, I say. Simply mad. Something must be done. But it's not my place to say. As for the dram, I'm afraid you're out of luck. There's a small fountain connected to a well, deep within the Necropolis. In that fountain flows the quintessence of fear, but it's well guarded. The quintessence of fear is potent stuff. Even a drop of it splashed on your skin is enough to cause terrible nightmares for a year. Do not, for any reason at all, drink it.  If you choose to brave that dark and terrible place, I wish you all the luck in the spiral. You'll certainly need it. You've come here when first, you'll need to get across the bridge. After the raid, the key to the bridge was left somewhere in one of those houses over there. Thought I'd just get that, actually no, I didn't even get the XP from that, which is a shame, but oh well. Oh well, that's fine. That's fine, yeah. Where are we actually going? A little of fear down the bottom here. Oh god, this is... yep, it's contagious. Nope, can't go this way. The rhyming is contagious, there's too much rhyming in this. And it's contagious, man. There it is, hello. Hello. Um, I assume we can't let other people in here. Yep, I mean it's fine. I'll be back once I've defeated these guys. Okay, just finished off that. Let us do this. You carefully fill a small vial with a quintessence of fear, making sure not to get it on your skin. Instead of splashing, the water whispers terrible secrets. The vial is dark and icy cold in your hand. You tuck it away in a safe pocket. Okay, let us head back to D'Worgan and see what he wants us to do next. This is a very long side quest. Not side quest, spell quest. Nope, let's go back again. I think it's four quests long. It's not actually too long, but it's four, like, three step things, I believe. So, yeah. Um, yeah, but let's talk back to D'Worgan. Talk back to, go back to D'Worgan. D'Worgan, hello. The next necessary component is a seed of death. What's it for, you ask? Patience, young wizard. All will be revealed at the proper time to tell you now would be premature. To gain a seed of death, you've got to go and fight the powerful demon creature known as Kajamusha in Mushu. Kajamusha carries a seed that is the source of his strength, a seed from which can only sprout death, not life. Journey now to Mushu and fight Kajamusha and return it once with the seed of death. Okay, I'll be back when I'm at Kajamusha, who is, I believe, at the end of the dungeon, so that's fantastic. Okay, I am back. We are finally at Kage Musha, who is at the end of the dungeon. Second last boss, second boss, so the middle boss. Yeah, so let's do this. Probably going to hit next round, possibly. Yeah, double blade, faint, I should kill. What, how much health does he have? Not much, hey. Yeah, 1,900, not much, cool. Um, actually don't have another faint, so another thing, so that's awkward, we're just going to hurt, you know, that's fine. If we don't kill, we can always wait a bit for things and stuff to happen, and yeah, pretty much that. Oh, okay, all we fizzle. It's obviously a sign. Um, yeah, let's try this again. Try again. What's this, ice shield, peasant? What's this, ghoul? Nope, not even a hit, that is awkward. Nobody's hit us, I don't think, which is kind of disgraceful. Boop, okay, and this should kill. Critical, very much kill. 4468, nice. How much is this going to do, like, 800? 700? 7000, sorry, not. Kajimusha falls back into feet. From his body emerges a tree. Too late, the tree of life is already doomed. You cannot stop the tide of darkness, surrender, and know the cold hand of despair. Don't actually need to defeat you, sorry. Let's go back to Dwarven. Dwarven. Hello again. You seem pleased with yourself.  Seed and water, just one more step and that spell is yours. Now, with the seed of death and the quintessence of fear, you can make your scarecrow. What? You thought I'd just give one to you? Oh, no, that wouldn't be appropriate. Each must be crafted by its user. There are a few items you'll need to make your scarecrow with. Wood from the death tree in Wizard City for the framework and some sackcloth from Gloria Crendl. Go speak to the death tree and see if you can get some twigs for the skeleton. Then Gloria Crendl for the sackcloth. Bring them back here and we'll see what we can do. Mm-hmm. Let us do this. Mortar cello. What are you looking at my twigs for, little death wizard? Just their twigs? That shouldn't be a problem. Here you go. A few twigs fall from the death tree's limbs onto the ground. You gather them up. What things do we have to train? We've got all of those clothes. 10 training points. That's a few. And let's go talk to Gloria Crendl, who is an old town. Haven't seen her in a while. Don't need that. That's for sure. Do-do-do-do, do-do-do-do-do-do-do. Do-do-do-do-do. Hopefully I'm not on woo. Don't wanna be banned. My own woo. Nope, not on woo. Not getting banned today, yes. Yes. Mmm-kay. Let's do this. Gloria Crendl, where are you? Are you? You are here. Hello. Hello. Sad cloth? That won't be comfortable at all. Why would you want some of that? You need it for a scarecrow? Well, that makes sense. I have some here. Hope you can put it to good use. Gloria gives you a length of rough cloth. Greetings, young wizard. Just, oh, begin by bringing me some shrouds from haunted minions on Triton Abbey. Okay, bye. I'll get to work once I have those. Okay, let's go back to Dwargan. Dwargan. Dwargan, hello. Seed, twigs, sackcloth, and water. All of the pieces are in hand. Let's make that scarecrow. Let me show you how to make a miniature scarecrow to use as an offering. Dwargan helps you craft a tiny scarecrow using sackcloth and twigs. You place the seed of death in its chest where a heart goes. Excellent work, young wizard. You could have quite the career in dollmaking if you choose to give up all this magic business. Now you must go to the yard outside Lord Nightshade's Tower in the haunted cave off Triton Avenue. There you must plant the scarecrow in the ground and water it with the quintessence of fear. Oh, yes, it'll be quite the fright, so brace yourself. Okay, let's go and do that. I'll be back when I am at Nightshade's Tower. Mainly because it's at the end of Triton Avenue and haunted cave. Okay, I am back. You followed Dwargan's instructions precisely. Setting the scarecrow on the ground and watering it with the quintessence of fear. Can we just take a minute to, like, respect how cool that looks? It's so funny. Don't feel scared at all. Young wizard, you have summoned me from the marsh where I stood watch for so many years. I will serve you and drive our enemies from this land. I shall show them fear in a handful of dust! Yay. New spell. New spell, baby. Let's go. Let's go, baby. Let's go. Dwargan. Da-da-da-da-da. Da-da-da-da-da-da. Hello, give us a spell. The scarecrow appeared to you, did he? Very well then. He found you worthy. Here's a spell to summon him with. Yay. Master of Death. Yass. Favorite badge, best badge, amazing badge. Um... Death skull spells, there we go. Look at that. Yass. Basically, I was checking Mortis before just to check that I had all my spells because I knew I hadn't got that one yet so that's the best badge to get. Um... Basically, every time you... Well, basically once you get all of the spells up until level 48 it seems. So all of those secret trainer spells as well. Um... I just finished my report on the famous grab of Sunken City. Can you please take it to Professor Drake for me?  You get the Master of School spell. Unleash your myth and then you get Master of Monsters which is pretty cool and I think life is Master of Nature as well. So, yeah. Um, but otherwise it's Master of School. Which is really cool. Um, but yeah, we have Scarecrow. Yass, Scarecrow. So, 400 to all enemies. Swap half to health. Which is actually pretty good. We have this, so that means 575. Um... Plus blades means it's pretty good. So yeah, thank you for watching. If you liked it, subscribe and if you haven't subscribed, press the little bell to know exactly when I post another video. And remember to drop that inside the box. That was a very, very quick outro. Sorry.


All my accomplishments, everything I've done in my life, nobody does great things by themselves. Every talent needs great people to be successful. And that's one of the things that I am aware of and I know that I am very fortunate that somebody gave to Purdue so that I could have a scholarship that changed my life. That changed my life. What a lot of people don't know is I grew up in a single parent household. I am the oldest of four. I'm dyslexic. I have ADD. I've never met my father. We grew up in a tough neighborhood and that somebody took a chance on this kid who had all these red flags and said, you're okay to come here.