America's number one adventurer, K7, form a United States secret agent who operated in 22 countries on land, on sea, and in the air, brings you a story of today. We take you to secret agent K7's headquarters. This is K7 speaking. Send in Agent Z for assignment. At once, K7. Special Agent Z reporting for duty. Agent Z, we have come here to the Far East to locate a spy. He is wanted by a certain nation in connection with a military child of a high-ranking officer, one Major Hurrell. I am convinced the charges against Major Hurrell are false, but only a confession by the spy can save him from facing a firing squad. Out here, he is rumored to be working with a patriot named Wong. I want you to locate him and somehow bring him into the international settlement for questioning. This fool who contains all known facts, keep me informed. The next day, the spy, Dawn Brasai, and Wong talk together in the garden of Wong's home. I have been able to locate six bombers and nine pursuit planes. They will be flown to the city circle on the map tomorrow morning. You are prepared to pay for them, Wong? The money has arrived from my government. We will go to the bank and get it this afternoon. Tell me, at what time will the planes arrive at the field where they are to be delivered? About six in the morning. They are being flown in tonight on the cover of darkness. It is good. If our enemies knew the airplanes were being delivered, they would destroy the field. What time will you be back here with the money? As the sun goes down, you will meet me here. Now I have many matters to attend. The white one will pardon my absence. My servant will let you out. I will be here at six o'clock. The next morning, Agent Z and his assistant, Patricia Norwood, talk together over a late breakfast. Z had just returned from a mysterious errand that had taken him into the native quarter. He told Pat what he'd learned. Pat, I haven't covered a great many things in the last two days. Dawn Brasai has been acting as an agent for Wong. He purchased 15 planes for him this week. Then he's working for Wong's government. Indirectly, yes. And he's also working for Wong's enemies. What do you mean, Z? Yesterday morning, I succeeded in planting a dictaphone in Wong's garden. He follows the usual Chinese custom of transacting much of his business there. I heard Wong arranged to pay Brasai for the airplanes. Late yesterday afternoon, he did pay him. Pat, as soon as Wong had paid him, Brasai went directly to Wong's enemies and told them about the airplanes. How many and what types they were, how and when they would have been flown in last night, and the location of the field they would arrive at this morning. He double-crossed him. Yes, Pat, and that's a dangerous game out here. This morning, enemy bombing planes appeared over the field within ten minutes after the new planes landed. Every one of them was destroyed. Brasai was paid for the information? Yes. And he also collected a big commission from Wong's government for buying the planes. He's making money from both sides, but I have got a hunch he won't be long. Why do you say that, Z? Because Wong also suspects him. When Brasai left Wong's garden last night, he was followed by a man who reported back to Wong. And if we're going to get a confession from him that will clear Major Hurrell and we've got to work fast, I don't think Brasai is long to live. What are your plans? We're going to use direct methods, Pat. I'm having Brasai watched. He's safe with what I'm being. I want you to go to Wong. Tell him everything. Tell him that if he can get Brasai into the international settlement, he'll be arrested immediately. If Wong knows of Brasai's treachery, and I'm sure he does, he'll welcome a method of getting him out of the way. Later that afternoon, Pat faced the patriot, Wong, in his garden. She showed him her credentials and asked his help. Mr. Wong, I've come to you for a favor. If there is reason, it will be freely given. I am here in the east to locate a spy named Don Brasai. In a certain country in Europe, he betrayed a patriot. Unless I locate him and get a confession from him, an innocent man will face the firing squad. His family will be disgraced. Why have you come to me? Because you are the only one who can make Brasai enter the international settlement, where he can be placed under arrest. He betrayed one patriot, Mr. Wong. He may also betray you. You possess much knowledge for one so young. I'm not working alone, Mr. Wong. It is written, there is no traitor like him who plants a poison within the breast that's trusted to his truth. Then you will help me. That must be thought out. Mr. Wong, I'm going to lay every card on the table. We know that Brasai acted for you, that you commissioned him to purchase planes for your government. Too much knowledge is sometimes dangerous. We know that you paid Brasai for those planes. And that he left this garden and went directly to your enemies. This morning, the airplanes you bought for your government were destroyed within a few minutes after they'd been delivered. Brasai betrayed you. He betrayed you just as he betrayed another innocent man. I want a confession from him to clear that other man's honor. Here in the East, we believe in settling our own accounts.  If you do, another innocent man will also die. I must have Brasai's confession. It is true that two should not die for one. You plead the cause of justice. I will help you, but you will have to work with speed. What do you mean? I will send one of my men to the spy Brasai. Tonight, he will cross the bridge into the international settlement at 10 o'clock. That's all we want. Wait. Wong has not finished. He will cross the bridge at 10 o'clock. You will have only one hour to question him. What do you mean, Mr. Wong? I have spoken. 1 24th of the day is all that will be left to Mr. Brasai when he comes to you. Early that evening, one of Wong's men went to the spy Brasai. He told him that Wong had discovered his treachery and that the only place he would be safe was in the international settlement. The warning was well timed. Brasai entered the settlement at 1 minute past 10. A few minutes later, he was brought to Agent Z's rooms. Sit down, Brasai. Who are you? Who have I done brought to you? I think you know. I'm Special Agent Z. I'm working with the intelligence department of the country you left 10 weeks ago. Oh, you followed me here. The man you betrayed, Major Hurrell, faces military court next week. You stole the defense plans from him. He trusted you in his home. Now, I want a written confession that will clear him. I don't know what you're talking about. I stole no plans. The major sold them to me. You're lying, Brasai. Now, there's a pen and paper. Write what I dictate. I wouldn't. You can't make me. It's a lie. I stole nothing. Brasai, I'm going to count to 10. Unless you pick up that pen and write a confession before I finish, I'm going to take you back to the native quarter and turn you over to Wong. You know what that'll mean? Now, one, two. I don't know what you're talking about. Three. Wong is my friend. Four. I've been working with him. I got you. Six. I can't tell you anything about Major Hurrell. Seven. He sold me the plans. Eight. I didn't steal them. Nine. Stop. All right, the confession. Write anything you want. Here's the pen and paper. Write what I dictate. I swear.  From Major Hurrell. Major Hurrell. Major Hurrell. He's innocent. He's innocent. What's the matter, Brasai? My arms. They won't move. My legs. Here, sign that confession.  Water. Poison. Z, he's fallen off the chair. Help him. No one can help him now, Pat. Now you know what Wong meant when he told you that in the east, each man settles his own accounts. Brasai was with us just 1.24 of a day, Pat. What do you mean, Z? Look at the clock. Brasai lived just 60 minutes after he arrived here. Z, repalling K7. A confession from him that completely clears Major Hurrell has since died of poisoning. The case is closed. Thus did an international spy pay for his treachery with his life. It is often the fate of those who work against peace. Listen for my next story. This is K7 speaking.


Airport comes to a standstill as passengers hear unmistakable sound. Airports are often busy as everyone's rushing here and there as they juggle a hundred things while trying to catch their flight. But that wasn't the case in Nashville, where everything came to a standstill when passengers were stopped by an unmistakable sound. Jen Tringale was left blown away by what she saw in Nashville. While traveling through the International Airport, her attention was drawn to a sound that had her quickly pulling out her phone so she could capture the scene unfolding in front of her on video. I witnessed an International Airport come to a complete stop today, she wrote in the caption accompanying the clip as she shared the footage to Facebook. It was a remarkable sight to be sure. As hundreds of people were packed into the airport awaiting their flights, everything came to a stop. There's no doubt that people had places to be and had a limited amount of time to get there, but all of that was put on the back burner for a moment after an airline attendant made an announcement. Then, another unexpected sound soon filled the air. Speaking over the hustle and bustle of the crowd using a loudspeaker, the airline staff revealed that they had some special guests with them that day. The children of fallen service members were boarding their flight as the kids were being flown to Disneyworld on an all-expenses trip, according to Daily Wire. And they wanted to pay them special honor. As the star-spangled banner began to play, everyone in the airport stopped what they were doing to pay their respect. During the performance, soldiers saluted the children as busy travelers stood quietly with their hands on their hearts. For a moment, everything was at a breathtaking standstill. At the Nashville Airport, I walked out into the concourse to this scene, Jen wrote in her post. American Airlines was flying a plane full of children who had lost a parent in combat to Disneyworld on an all-expenses-paid trip and they threw a party for them at the gate, she furthered. But when they announced them over the loudspeaker and they lined up to board the plane, the whole airport literally stopped and sang the national anthem with the military present in salute, Jen continued, describing the moving scene she captured on video. Calling the display, so humbling, Jen Tringale added, Most every person standing around, myself included was bawling at the sight of these kids and spouses who have paid so great a price for our country. She continued, seeing the general public in an airport stand still to honor these kids was simply beautiful. And it seems that social media users agree as the video quickly went viral, receiving the praise it deserves. God bless these families who lost a spouse in the military or these little children who lost a mommy or daddy, one social media user commented on the touching video as they shared the sentiments which were held by many viewers. God bless American Airlines for doing a wonderful thing to send these children to Disneyworld and make them happy. The heartfelt comment continued before concluding, God bless our military for sacrificing for our freedom and God bless America.


Pikeville Medical Center, Kentucky Organ Donor Affiliates and Donate Life Kentucky celebrated the gift of Life Thursday evening with their annual Tree of Life ceremony honoring Oregon donors. We do this every year as a partnership with Between Donate Life Kentucky slash CODA and PMC and the reason that we do this ceremony every year is to truly honor and you know provide remembering and the opportunity to just come together and celebrate the legacy that Oregon donors provide to those who are recipients in years past and to just you know provide support to those in the community who are walking that transplant journey. Tech the halls with Appalachian Wireless offering you up to eight hundred fifty dollars off sunlight devices extended through Christmas Eve with no activation fee up to eight hundred fifty dollars off the latest and greatest devices we are you we are Appalachian Wireless. Currently over 100,000 people are on the national transplant waiting list with another person being added every 10 minutes. We really just want everyone who has been touched by donation and who has been organ donors in the past year or just in years past to understand that you know you have a community rallying around you supporting you through this walk and you know if you're on the transplant journey know that people are you know still in the giving spirit they want to save others they want to provide that gift of life. There's so many people here in our eastern Kentucky community who are in great need of life-saving transplants and this is just a way to celebrate those who are providing hope and life to continue on to those people who need it. Reporting for Mountain Top News I'm Breonna Robinson.


So, okay, so, so last, so if, so RMK as you remember is a Noetherian local ring with residue field K. If this is a complete domain, then there is a constant EHK of RI. So remember I is always a ideal of finite co-length such that length of R mod IQ and here, okay, let me I'll just say it clearly. So this I'm just recalling the last time's theorem. So D is a dimension of the ring. So we are still quite have not proved the existence of the limit for arbitrary module. So this is what I'm bringing. So here it is. This we have proved as a complete domain. So remember, I'm sorry, I think you forgot to share the screen or do you plan to share the screen? We don't see a screen. Oh, okay. All right, perfect. Oh, yeah. So last time we were on the way to prove the existence of Hilbert-Kunj-Madeep city. So we had proved the result when R is a complete domain. And so now this, in fact, this is a, we can prove it for arbitrary module. So, so this theorem, this is terrible. So what is happening? So let M be a finitely generated module, R module. So here the D was a dimension of the ring. So again, there is a constant depending on M and I. Now I'm saying it is positive. And here, in fact, it was greater than one such that length of the length function M upon IQ M is equal to EHK of MI, Q to the power D. Here D is again the dimension of the ring actually. So, so we have all ingredients to prove the result now. So what we'll do again, you can assume R is complete local ring and K is a residue field, which is algebraically closed. Then consider the prime filtration. So let 0 equal to M0 containing M1 be a prime filtration of M. So I'll quickly recall what is the, how do you, so that means the definition is that means MI plus one of MI is isomorphic to R mod PI for some prime ideal. So how do you get it? It's very easy. Just you are saying that, let me quickly say M has an associate prime. So there is a ideal P1 container. This is, sorry, this is ring R inside M. So now you can take this to be your M1, this one. Now go modulo M upon M1, repeat the same process. So there will be R mod P2 sitting inside this. Now pull back P2 all the way to M. And this is the way you get the primary prime filtration. Okay. Now we have already proved the additivity of the length function. So that implies that your length of M upon IQ M should be equal to summation length of MI plus one upon MI into IQ MI plus one upon MI plus some element of order D minus one. Okay. But MI plus one mod MI is a, is a, is just a complete local domain. So applying the previous result to what you get is it is equal to E summation I running EHK of MI plus one upon MI at IQ to the power D plus order Q to the power D minus one. Here if you have MI plus this R mod PI is of dimension less than D, then this will get absorbed here. So I can, without losing anything, I can just take the elements where MI plus one upon MI is equal to D. If they're less than D, they'll get absorbed here. So it's okay. So such for MI plus one upon MI for such thing, which is isomorph to RPI, it means that PI belongs to lambda. So if I look at R local as a PI, then it's a field. So the prime filtration gives me just M zero PI and PI. So many of them will actually become zero. So it means the number of times the models which are isomorphic to this one will occur in this filtration is exactly equal to the length of the model MPI. So I just look at this again. And it tells me this is summation of EHK R mod PI length of MP where P is in lambda into Q to the power D plus order Q to the power D minus one. So you have this length formula. Now in this formula, if you go modulo Q to the power D, so it just tells you EHK of M with respect to I is same as summation EHK R mod P into length of MP. So this is the final thing you have proved the length formula as well as or the second assertion. I didn't write it anyway. So this associativity formula, this is called so associativity formula. Okay. So this associativity formula has a nice corollary. So what it says is a following. Suppose you are is a domain. So you're again, the condition is RMK where R is a domain, then your multiplicity of M for I is nothing but multiplicity of the ring R at I into the length of or rank of this thing. So this is rank of M as R. So that means basically studying the multiplicity for module boys don't study the multiplicity for the ring. So you can always assume your ring to be a module. You can just study for rings and another thing if I take any other ring, which is a finite module over R, then I can recover the multiplicity back and forth. So let me say it. So R to S finite map of rings where M is local, then EHK of S of extended ideal that is the problem is same as EHK of the ring R at I into rank of R at S. So if I have this ring, which I understand well, or this ring, which I understand well in the sense, I know what is the multiple HK multiplicity of this or this, then I can recover the another one merely by knowing the rank. So I have given you exercise. So this one example was a Verones subring inside a polynomial ring. So remember polynomial ring or power series ring, we understand very well, because the length of the each length function is actually equal to Q to the power D. So is equal to one the quotient. And then then the ring of invariance if I take so G is a fine, suppose G is a group of order finite order acting on a polynomial ring on a polynomial ring ring R, then RG to R is a finite extension. So I give you another exercise, but that is before proving the existence of HK multiplicity. So it says, suppose you are in the situation RMK, where characteristic is positive, then multiplicity of the ring R with respect to I, mod D factorial, D is the dimension of the ring, is less than or equal to EHK of I, is less than or equal to E of I. So is everything is okay so far? So so immediately tells you something, it says if dimension of your ring R is one, then your EHK of I is same as E of I. So it's an integer always, it's nothing makes, there's no changes. So now we are looking at the some properties of the HK multiplicity. So what, so you have this number. So what does it mean? I mean, what kind of properties of the ring it reflects? So the first thing first, so if you know regular multiplicity, it is one, if and only if your ring is regular, of course, provide the ring is unmixed. So here also the same thing, it's a formally unmixed, suppose this is an important assumption, formally unmixed means if I take a completion, all the all the prime, all the irreducible components are of same dimension. So EHK of R is one, if and only if your R is regular. So we are not going to prove anything. So of course, if R is regular, then EHK of it's, you can assume it's a power series ring for completion and it's easy to check that this is one for the converse that takes bit of a, that's a lot of work and it's given by Watanabe and Yoshida, I think. So remember that if you recall the first result we stated of Kunz was saying that length of length of R more, this is with respect to maximal ideal, so I should say maximal ideal. And Q is equal to QD, which is that means I'm saying that length of QD is one, if and only if R is regular. So this, so this, this, this assertion is certainly stronger. So this is saying I don't have to look at one length if I take a symptotic length, which is happens to be one, then it implies that R is regular. So here I would like to mention that this, in fact, this length, this limit function of this length, where Q tends to infinity was actually thought by Kunz and he somehow he thought this limit doesn't exist, but he did think that this limiting function will be something very important. So many years later in another contest in the context of number theory, Monsky considered this limit and he proved it existed when he came to know that Kunz also has thought about it. So in his honor, instead of putting say Hilbert Samuel multiplicity, he called it Hilbert Kunz multiplicity. That's one thing. Another thing here, I would, you may have wondered why we have a formally unmixed condition. It's important because lower dimensional component don't contribute anything to the multiplicity. So let me give an example. Suppose if I take a power series ring in three variables and I look at x, y comma x, z, this ring, suppose this is R, then EHK of R, or maximal ideal I mean, is same as EHK of the ring K. That's what in associativity formula has proved is x, which of course we know one, but certainly your ring R is not regular. So are there any questions so far? Okay, so now perhaps this following lemma might be one of the reasons it really caught on with community algebraic community as the following. If you remember, if I have two ideals I and J, so this is let me recall here now. He's containing J, I mean their M primary ideal. Again, I'm in a local setup such that multiplicity of I, so this is your hypothesis says that E of I is same as E of J. This implies that J is contained in the I closure. So here perhaps you had to assume your ring is again formally unmixed just to avoid the unnecessarily lower dimensional component. And in fact, it's a converse is obvious if so. So this is not a thing. So the integral closure versus usual Hilbert-Samuel multiplicity, here the tight closure versus HK multiplicity run exactly parallel. So it says if so consider I containing J and are formally unmixed, then J is contained in the tight closure of I star if and only if E of HK of I is same as EHK of J. So most rest of my talks perhaps I won't give much of a proof. So let us see the proof of at least easier side. So suppose your J is containing I star. So I want to prove that in that case, HK multiplicity is same as of I same as HK multiplicity of J. So what you do for this, the very J is contained I star and so for any element of J I would find X belong to J. For example, there exists some C X as that C X MC, I mean, this is not a multiplication, just a subscript X to the power Q belongs to I Q for all Q. Since J is finitely generated for each X we have. So we just collect the generator C X for each generator and pick the product. So which of course will belong outside the minimal prime. So belong to R is zero. I forgot up or down for such that C into C into J of I Q is contained in I Q. So this implies that I look at this module J Q mod I Q. It is a R mod I Q colon C module as a finitely generated module. So in the sense generator of J Q will suffice. But remember if J is generated by M elements, J frobenius Q will also be generated by M elements. So in fact, there exist M copies you can have or such that direct sum of R R I Q colon C subjects to J Q upon I Q. So this implies length of this one is less than or equal to length of this. But what is the length of this?  This this one this length which is equal to M into which is less than equal to M into length of R of I Q C. So this this module remember this is over R mod C where C doesn't belong to any minimal prime. So dimension of R mod C is less than dimension of is strictly less than D. So this is a more of order Q to the party. So here in fact I can write it. So here I don't have to write it. So I go more low Q to the party and take the limit. So which implies your EHK of I minus EHK of J is zero. Okay, so this is one thing what how about the converse converse? So suppose I'm not going to prove but I would like to mention something. Suppose your EHK of I colon one more element say XR is same as EHK of I. So you would like to say that X belongs to the tight closure of I. So this is what you want that will prove the converse. But what happens here? In fact, the result, this is a theorem of a Hawks-Renek. It says if suppose X doesn't belong to the tight closure of I star. You remember if the X belongs to tight closure, there should exist C. So in particular C should exist some Mk minus Mk plus one independent of Q says that C into XQ, I'm sorry here I'm writing it belongs to IQ, etc. That means a letter of IQ colon XQ is intersects this number. And here I'm just going a bit fast. So this one says it's even better if X doesn't belong to I star. So the colon shrinks like anything. So that's the main point here. So it says the set of C says that C XQ belongs to IQ. These kind of elements in fact belong to MQ over KL for some constant L. So that means the letter of this IQ colon XQ, this thing shrinks like anything. It means it is proportional to the power of Q. So that will contribute the length function to be some epsilon to the Q to the power D something like that. So this is the hardest part to prove actually. And there they use like an element of a small order and the valuation. So they have kind of quote a weaker criteria for tight closure, but they say these are equivalent. So this is a tricky argument, but it's very intricate. So you can look up. So this is very important point that your annulator is actually keeps shrinking rapidly at the power of Q. So that is the property of HK multiplicity. Now I would I am actually not covering all or I cannot cover all the important results, but you would like to know what else kind of what in the okay tight closure is one thing, but there are other characteristic P singularities of the ring, which EHK tells. So let me quote a few of them. So one is, I think EHK of M first of all. So the philosophies as follows is smaller the multiplicity, EHK multiplicity, better the singularity will be. So already you have seen one sample. If smallest is one here for M. So if it is one, the ring is regular. Nevertheless, suppose your ring is not quite one, but this is still smaller than one plus one upon D factorial. D is the dimension of the ring again. Then R is when we call a F rational. So I think this result was proved in several steps. First some bound involving characteristic P was the Blickeler Inescu and then this one, this one I am not sure maybe Inescu can tell me. So, so the here the point is not here, but generally, unlike usual multiplicity, EHK multiplicity is not a discrete set. I mean, so it can take many arbitrary value. So one question is suppose your ring is not regular, then how big the EHK multiplicity can be. So the answer is this is greater than equal to one plus one upon D factorial into D to the power D. This is just these are fun facts. So, sorry, D to the power D. That's what you said? Yeah. Okay. So I have not shown you any non-trivial examples of EHK multiplicity. So that so far cases I have listed like where on is ring or the ring of invariance, etc. Since they come from the polynomial ring, which is doesn't involve any characteristic turn on in the EHK multiplicity, you won't see it there. So, so let me the famous example of Hansen. So, when Monski had this introduced later on, he computed the multiplicity for the next non-trivial class of ring, which is say plane curve modulo, you can go a cubic equation. Quadric is nothing not so spatial here right now. The cubic equation he did compute for, I don't know which one, one set of people have proved for characteristic two and other people for not characteristic not equal to book wise and Chen. But their EHK multiplicity is not interesting for us because it doesn't reflect any characteristic. And that I will perhaps tell you later the reason for that. So, let R be this x4 plus in the node sided x4 comma y4 that's that's nonsense. So, this is x. So, you are k x y z modulo, this equation x4 plus y4 plus z4. Okay. And characteristic of k is p. So, what they have shown is EHK of R is equal to three plus one upon p square. If p is congruent to plus or minus three mod eight and equal to three if p is congruent to plus or minus one mod eight. So, this example is actually interesting for several reasons. So, one reason is it tells you something. If I look at EHK multiplicity, the deformation things are not going to work for the computation of EHK multiplicity because suppose there is exist x in R. Remember, the usual multiplicity we always come one way of computing Hilbert-Samuel multiplicities, you go modulo general hyper plane section or general element. In fact, that's the one definition of your usual multiplicity. You keep on going D number of elements. The ring becomes quotient ring becomes zero dimensional. So, just count the length of the ring. That's one thing. So, or what we call reduction ideal. But here, this kind of induction is not possible. Because suppose there is some element x in R or maximal ideal such that EHK of R is same as EH, I don't know why this is doing this of R mod XR. Then this now your ring is of, ring is two dimensional ring, cool dimension two. And this is cool dimension one. And we have just now seen that for one dimensional ring, this is same as R mod XR. Now, for Hilbert-Samuel multiplicity, there is no characteristic involved, right? So, this is a characteristic free thing. So, obviously, this cannot be equal to this. So, this is not possible. Now, I would also now, I'll try to give some, basically, many I'll talk here, is the following. What happens? You look at this, this, this is called in geometry, we'll call this as a plane curve, because if I look at proge of R, it will become a one dimensional projective variety, and which is sitting inside the plane, which is P2. So, for these kind of curves, we have a notion of a locally free shape called vector bundles. So, I'll, I'll very briefly touch on that. So, so slightly generality, not the example of Monsky. So, you take k x y z modulo h, h is a homogenous irreversible primate polynomial. Suppose you have this, then the theory is the following. Suppose and degree of h is d. So, sorry, this is h, d we have been using for the dimension of the ring. So, here I'm using for the degree of the polynomial. So, the general theory tells me the following. So, this, this is my S. So, call it SS. This is 3d mod 4 plus L square. I'll just explain all these terms 4d P2S, where P is the characteristic of the ring. So, P is characteristic of S. So, what are these L? So, L is an integer. It's important. It's bounded by d minus d3. That's important. And S is also an integer. I'll explain the significance of this one. So, here do stop me and ask the thing. So, what happens? You project if I S. So, it becomes take approach of S. So, now X is a curve here. So, X is a curve. So, you have a locally free sheaves on that. So, one is a structure sheave. And then line bundle non-trivial one, OX1. So, what do you do? OX1. So, this is OX. So, this is locally the ring sheave on X. So, there are three things. This is just given by the elements X, Y and Z. This three copies of OX. One goes to X. One goes to Y. One goes to Z. So, degree 0 goes to degree 1. That's why there is a shift here. So, this is a locally free sheave on X. This is a locally free sheave on X. So, the kernel has to be locally free sheave on X. So, this is called vector bundle here. So, coming back to the... So, wait a minute. So, here, if you look at the people, when the people who work on vector bundle theory, they often like to work on something called semi-stable vector bundle. So, that means that by definition, if a W is a sub-sheaf of V, here it is a curve. So, everything is a locally free. Otherwise, you have to stick to our torsion free. So, mu of W, which is degree of W mod rank of W should be less than or equal to, I'm sorry, mu of V. So, this is some definition. So, this definition tells you that lot of homology of V will disappear if V were semi-stable. So, that's the one reason people in vector bundle, we would like to work on semi-stable because it also forms a bounded family if you restrict some parameters of the vector bundle. So, and on the other hand, any vector bundle have a filtration by say, say something like a V0 and continue V, Vn equal to V says that VI plus 1 upon VI is semi-stable. So, you can break it into this kind of factors. Otherwise, so these are called hardener cement filtration. So, that's one thing. So, that's useful for the formulation of this one. But one other thing is a problem. If characteristic of k were 0, then if I take pi to be any finite map, y to x, then pullback of, so pullback here will be a tensoring with a ring. So, pi upper star is semi-stable. So, V semi-stable implies pi upper star V semi-stable. However, that's not the case when you come to the characteristic P and the major obstruction is the Frobenius map, which is the finite map. And you start with the semi-stable bundle, pi upper star V may not be semi-stable. So, you may start, okay, I'll start with the hardener cement filtration of V where every subcocean is semi-stable. But when I pull up that hardener cement, still I don't know it remains hardener cement for the pull up. So, these problems are there. So, this is the theory of semi-stable vector bundle. So, here what is happening? My rank is of V is 2. Then, perhaps I'm not saying it right. So, if Fs upper star V is not semi-stable for some S, then by hardener cement theory, there is a line bundle sitting inside Fs upper star V. And this is the hardener cement filtration. Okay. So, okay, having talked about all these things. So, what is the coming back to the HK multiplicity? So, this one is telling me if I have this L and S. So, you forget about this generality. So, and consider the Suzuki bundle V for that. So, V F upper star V, S minus 1 iterated of V, they're all semi-stable and Fs upper star V is not semi-stable. So, that's N, the hardener cement filtration which is here, Fs upper star V. So, that means mu of L should be greater than mu of F, this one. So, this is equal to L mod 2. So, I come back to looking at this number. So, this is telling me at which stage your bundle will destabilize. Till then, it will remain semi-stable. If S is this one is 0, that means your bundle is going to remain semi-stable for all S, that means. So, this S is telling me in the place where it's a provenance iteration where the bundle is not going to remain semi-stable and this number L is telling me how much it has destabilized. So, these are so HK. So, if you have these two information, you know HK multiplicity and compute it right away. So, generally this is that's not easy. So, on the other hand, suppose if I know the HK multiplicity of SM by chance, this is there some number then and P is greater than D into D minus 3 then recover L and S. So, for P bigger than this number HK multiplicity will give me the information about the hardener cement filtration of your vector bundle. So, this is the crucial point here. So, coming back to our example of Monsky. So, D is actually 4. So, it is 3 plus this thing. So, this is telling you something. So, HM example. So, it is telling you if P is congruent to plus or minus 1 mod 8 then FS upper star V is semi-stable for all S for all SM and if P is congruent to plus or 1 minus 3 mod 8 then F upper star V is not semi-stable. So, this example is a bit of a surprise for vector bundle people who work in vector bundle because the thing is generally if suppose V is semi-stable in characteristic zero then V reduction mod P is semi-stable for all P positive. So, the semi-stability condition is an open condition that means but this one is saying however if I look at the Frobenius pullback condition this is not an open condition. I may start in characteristic zero the bundle which is semi-stable. So, of course it is going to be semi-stable for all finite pullbacks but this is not an open condition. So, just a computation of HK multiplicity gave you this example. So, so, are there any questions now? Okay. So, talking of coming back to this example of Monsky. So, it tells you that for infinitely many P things can it doesn't achieve minimal and for infinitely infinitely many P it can differ. This is no open criteria for this HK multiplicity but there is something nice. I said that this L is bounded. So, as P tends to infinity this number is going to be zero. So, this leads to the open question. So, you have a RMK. So, R is the noether and local ring and residue field here and characteristic of K is zero and now RP MP and suppose your ideal is I this is let the question is does EHK RP IP limit P times to infinity exist. So, see the answer to this question will give you a notion of Hilbert-Drucking-Multipstick characteristic zero or well-defined notion and also for at least for M primary ideal suppose we know the equivalence of tight closure of an ideal in terms of the HK multiplicity perhaps it will give you a notion of tight closure in characteristic zero but of course for the ideal which are M primary. So, I really don't know the development there. So, talking of these notions there is notion of epsilon multiplicity. I think people can look up. So, this is Katkowski and all and then there is a generalized I think that's by Neil Epstein or something. I don't know HK multiplicity. So, these are another different notion of this one minute. I'm just I don't know what happened to my screen. Can you see the screen? We cannot see the screen but it's 950. So, thanks. So, should I stop? Yeah, I mean we can have a 10 minute break and come back. Thanks. Yeah, sure. And then hopefully we can resolve the screen issue. Yeah, thanks. Thanks a lot. Let's see if are there any questions for Vijaya for this lecture. I don't see any questions. So, listening to the speaker. Thank you. And we'll take a


Sonic Superstars doesn't feel right. And I'm sorry, I'm really sorry. I don't like to make and play any videos, I don't like to grumble, but I had the chance to play the Sonic Superstars demo this week, and I know that if I hadn't played it and if I were on the fence about it, I'd want someone to tell me their honest opinion on the game, especially given how on the fence I've been about Superstars since it was first revealed. So, a quick bit of context. This weekend was EGX, a gaming expo in the UK that I attended along with a few friends. There were a smattering of indie titles and a few more well-known publishers, but the two big booths were Sega and Nintendo. When I asked if I could record gameplay footage, the staff at the Nintendo booth were tremendously helpful, even going so far as to get me on the booth with the best filming angle and to adjust the camera for me. When I asked if I could film gameplay at the Sega booth, I was told to film around the booth generally, but that I could not film any actual gameplay, which is why this footage is a bit odd, sorry. Sonic Superstars was naturally the first game that I queued up for, because I am a Sonic fanboy at heart, and this was the game that I cared most about experiencing myself. If you watched my reaction video when the game was first revealed, first off, thank you for coming back for some reason, but secondly, you'll remember that I was worried that Superstars would end up being the new Sonic the Hedgehog 4, an attempt at a retro revival with more modern graphics that ultimately fails to nail the old-school feel of a Sonic game. A lot of people misunderstood that video and assumed that calling Sonic Superstars the new Sonic 4 was a compliment. Sorry, I should have made things clearer. My concern with Sonic Superstars upon seeing the first trailer mostly came down to the presentation. Takashi Izuca has spoken about how the developers chose the game's art style because they worried that pixel art will data game make it old-fashioned, an opinion that I flatly refute. The look of this game made me think mobile game, and I've not been swayed from that opinion by seeing more footage of the game that's purportedly leaked in the past week, although I would also believe that the supposedly leaked footage I saw was made by a fan as a joke, which, well, says a lot about what I think of Sonic Superstars art. But, as much as I love a really good-looking game, such as to single out my favorite retro revival, Wonder Boy the Dragon's Trap, I'd be very happy to get on board with Sonic Superstars if the game felt like a classic Mega Drive era Sonic. I've said that from the start. If Sonic's momentum is right, all is forgiven. I just want to curl into a ball, roll down slopes, whiz through loops, and have a good time. And, well, it's clear that that's where they were going for with Superstars. The vague mechanics I want to hear, certainly more so than Sonic 4 or Sonic Generations, it's close. Close enough that more casual fans may be perfectly satisfied with it. But it didn't do it for me. I'm sorry. Really, I am. I didn't want to say this. As someone who plays Mega Drive Sonic games as a kind of meditative experience at this point, as someone who's won several Sonic speedrunning competitions, although I have been beaten repeatedly by Webster, check his channel out, Sonic Superstars just did not feel right to me. I've got several theories for why that is. Having only played the demo at a convention, I can't speak for the whole game, but I can try to explain why it didn't click for me personally.  It won't surprise regular viewers to know that I chose Amy. I've been campaigning for her inclusion in this kind of game since before Mania launched.  Her acceleration is a bit laggy. Immediately, I felt that this was a game that needed the drop dash, a move that Amy lacks. Faced with the prospect of playing the whole demo like this, I couldn't do it. I restarted, picked Sonic, picked the beach stage, presumably the first stage in the game, and hoped for a better experience. It was better, definitely. But it still felt like the muscle memory within me was fighting this thing. It was kind of like, okay, as a European, I first played Sonic in PAL format, which is slower than NTSC. Then when later ports came out, I experienced Sonic at the NTSC speed he was supposed to be played at, the speed that he's played at in all of the versions of the game released in the United States and in Japan. Now, going back to my original PAL Mega Drive in the cartridges feels a bit sluggish. That's how superstars felt to me, like going back to the PAL version of a game that should be played at NTSC speed. Smarter, more knowledgeable Sonic fans than me will go through this game with a fine-toothed comb once it releases and provide details on Sonic's momentum versus the classic games, and if they come back and say that my complaint is all in my head, I won't actually be surprised. Because in part I feel that the problem is with the level design and the game presentation. The camera zooms in and out as the level progresses, and at times it feels too cropped in, too claustrophobic. It makes it hard to see what's coming, which increases the chances of hitting a road bump or a set of spikes or a badnik. What's more, the level design is dense and cluttered. I don't know if it's just because I know that Naoto Oshima worked on this game, but it put to mind Sonic CD, and my endless desperate attempt in that game to build up speed, as is required for the central gameplay gimmick, only to hit an obstacle and have to start all over again. It's a lot like that, except when the game says, okay now go fast, at which point you're shot through a series of predetermined ramps that you have little to no control over. It's like the game doesn't have space to breathe, there's too much going on, and the result is I didn't feel the satisfying dopamine hit that comes from getting up to a solid speed and cruising through levels on mental autopilot. The special stages and minigames don't help. By this point I'm the kind of player who doesn't usually jump into a star ring above a checkpoint unless I need to, because it slows down the progress of the game. Naturally, though, for this demo I wanted to see everything there was to see. One minute Sonic's doing his best Spider-Man impression, which is basically just a series of homing attacks with delay, the next he's bouncing through special stages reminiscent of Sonic 1. If you like these diversions, there is a lot to enjoy here, and when you get an Emerald you get a new power, which impacts the regular gameplay in... marginal ways. The problem with all of these additional game modes is there's a noticeable loading lag when transitioning between them. It may only take a second or two, but it adds up when you're coming in and out of these special stages regularly. It just feels like another impediment to what I want to do in a Sonic game, which is build up speed and zoom around fast. It's particularly worrying that I was playing this on what I assume was a PS5, given that there have been early rumors that the Nintendo Switch version of the game isn't exactly in great shape. If even the top-of-the-line version of Sonic Superstars has this admittedly extremely minor loading issue, I worry for this game on less powerful hardware. Do you know what usually loads a bit faster than 3D character models? Pixel art, but maybe that's me just being bitter. But okay, hang on, time for a tangent. Here's a quote from Takashi Izuka about why they chose this particular art style for Sonic Superstars. Quote, After we had made Sonic Mania, we wanted to take the side-scrolling platform action gameplay in a new direction. In the early days, we did have an idea of maybe having hand-drawn animations as a concept, but when we thought about it, the 3D models we could really get to mimic the pixel animation and the pixel art. So we set out with the 3D models to really animate the game and have the game look and feel and animate just like the classic, the original Sonic the Hedgehog game. End quote. Which...ugh. We were so close to getting a hand-drawn Sonic game. Something like the fan game Sonic 2 HD or Sonic Freedom, which just look gorgeous, just brimming with character and vibrancy and life. But what Izuka is not saying here is that throwing in a 3D character model from various potentially murky sources is an awful lot cheaper than hand-drawn animation or pixel art, which both take more time and care and attention. That's why Sonic Superstars has all those extra skins, all the weird LEGO ones or Amy in a slightly different dress. That, I suspect, is the real reason for this particular art style. Budget. Oh, and those new powers granted by the Chaos Emeralds? They're tremendously clunky, infrequently used and just a bit of a distraction. I got the Emerald that lets you jump up waterfalls, forgot about it entirely until the game prompted me, and then found myself clicking through a radial power wheel with the triggers before I could use it. Was there really not a better control scheme for that? But I digress. My point in all this rambling is that I needed to know for certain before buying Sonic Superstars whether it would feel the way I wanted it to, and while I've only played a limited demo, it's enough for me to feel like this really isn't the Sonic game for me, and that's a genuine shame because I wanted it to be. I wanted to love this, I wanted to get as much out of it if not more than I did from Sonic Mania. There is hope, I suppose. I'm very interested to see what modders do with this thing, as is often the case with Sonic games. I wonder if a few tweaks to the physics will make this feel a little more like I wanted to, or if it can be used as a launching off-point for fan games that are somewhat more to my personal liking. My experience in playing the demo then confirms the thing I've long suspected. Sonic Team and Sega may be doing their best, but they're not the experts on this particular subject matter. The people who have the best understanding of how to make Sonic games are Sonic fans. That's the moral of the story, I'm sorry I don't come with better news. But, little teaser for you here, I did also play Super Mario Wonder, and it doesn't feel fair to give Sonic a kicking when I was also left feeling like I didn't need Mario Wonder in my life. So, next time, hopefully coming in the next few days, the second part of this video. My problem with Mario Wonder, and why I won't be buying that game at launch. It's only fair, right?