 This one's from the University of Alberta. Thank you. I think I'm on. Yes. All right. Well, thank you very much. So, yeah, I figured I would give an introduction to airy structures because I figured most people here probably didn't know too much about the whole story of the post-colonial recursion and airy structures. So as well given introduction, but the way I'll introduce it is fairly new. So airy structures were first proposed by Konsevich and Subelman in 2017. I think as a framework, algebraic framework to make sense of the post-colonial recursion that's been going around for years by Inale and Arante. But I'll introduce airy structures in a slightly different way, which was we wrote in this fairly recent paper with Thomas and our former PhD student, Aniket Joshi. So it will be in the language of the modules and the algebra. So that's that's what I'll do today. So please stop me whenever you have questions as usual. So the starting point from my point of view is Konsevich within, which we all know and love. So this is all about 2D, 2D quantum gravity. So there's like various aspects to Konsevich within. So one way to start is to construct a partition function. But a whole bunch of sums here. I'm just being schematic here because I just want to highlight the properties that you can write down a big generating function here for some numbers, FGNs, which in the theory of 2D quantum gravity becomes they become integrals over the modular space of curves. So the standard intersection number over the modular space of curves. So this is the main object of interest in the study of Konsevich and Witton. And I would call that the geometric side of the story. So it's basically a generating function for some inumerative environment. This gave probably the simplest one intersection numbers over MVN block. And now, of course, the big story of the Witton's conjecture, which was proved by Konsevich is that this partition function is very nice and satisfies some nice intercability property. So in particular, it is a particle et al function of the KDV hierarchy. So the more precise statement would be that if you construct an object U to the second derivative of the log of the partition function, then this is the unique solution to KDV hierarchy with a given initial condition, which is that you set everything to zero, accept the first variable you take just enough. So if you do that, you recover precisely this partition function. Okay, so this is the integrable system, part of the story and this kind of connection here. It's going to be a small arrow is basically Konsevich Witton. So within conjecture proved by Konsevich using magic fun. Yes, great. Now, this is not really what I want to talk about. So but there's a third kind of point of view, I would say, on Konsevich Witton, which is that you can rewrite or you can basically compute that uniquely in terms of differential constructs and the idea is to construct some differential operators. I'm going to call them Li, so I'm not going to write down exactly what these are, but they have very specific formula. So these are differential operators in the variables T here could write a differential operators and they satisfy. So this is for all integer i greater or equal to minus one. And they satisfy the Virazoral algebra. There's no central charge here because I'm only taking this part of the Virazoral algebra. And you can show that's another statement that you can show it's equivalent to this. That's an entrepreneurial statement that it's equivalent to saying that Z is a tau function, is this part of the tau function of the KTV hierarchy. But you can show that and this once you write down explicitly what these differential operators are, that uniquely fixes Z. In other words, it calculates all the intersection numbers via these differential constructs. These are called Virazoral constraints. Usually I'm going to call them differential constraints because we're going to work beyond Virazoral. And one could say that this is another point of view on Witten-Kanzeric conjecture here. OK, so that's kind of the starting point, these three things which are relations between geometry, differential constraints or, you know, differential representations of some algebras and integral systems. Now, one way to think about every structures is as a generalization of this arrow here. But there's many ways you could think of trying to generalize that. So one way is to study things like the Virazoral conjecture for Gromotin environs. So in that in that in that in that trajectory, the idea is that you replace the FGNs by more general and immersive environments like Gromotin environs of some Calabi O3 fold or some manifold in general. And then you can ask that there exists a representation of the Virazoral algebra, which is such that it kills Z. So that's one possible way you can generalize that to the Virazoral conjecture, which has been studied for many years. But there's another way you can try to generalize it, which is to instead of studying with a particular numerative problem, say a particular Gromotin theory and asking whether there are differential constraints, you could say, well, I'm going to try to generalize these constraints such that, so in other words, I want to study more general constraints, but I want to keep the requirement that there always exists a unique solution of that form. I don't know what is going to compute, but I know there's always going to be a unique solution. And that's basically the idea of every structure. So it's a generalization of these constraints, such that they're always a unique solution. Okay, yes. Put respect to the T's, yeah. They're quadratic. So there's basically a linear term and then a quadratic term and nothing. Yeah, I could write them down, but don't have them in my notes. Okay, so that's the general idea. So the goal is to generalize the differential constraints, such that there always exists a unique solution of the form, unique solution Z of the form. Yeah, so what is every about this? So why is it called every structure? So that's a very good question. I think the reason is, because exactly what you said, so that the prototypical example of an every structure is this, that there are those constraints for intersection numbers. And the way this is proved is by constructing a magic model which is the concept of every major model. So it's an area equation for matrices. So I think that's why it's called the every structure. Okay, so that's the goal. And I'm gonna explain that in three different steps. So the first thing that we wanna do, so we wanna generalize these differential operators. So these differential operators live somewhere in general. So we wanna define the space in which they live. So that's the first step. So this is going to be the completion of the REEs by algebra. So I'll explain what this means. So that's gonna be the first part of my talk which should be fairly quick. I'll just explain what the space is where these operators, these more general operators will live. Okay, the second step of the talk will be to somehow make these constraints a little more, I would say not abstract, maybe a little better way of understanding them. So if you have if the LKs kill Z, then instead of writing it like this, you can basically define a left ideal in your space here generated by the LK. So let's call this that ideal I and then the constraints just become the statement that I times that is equal to zero, right? Because it's a left ideal. So of course everything is at an LK on the right. So it will certainly kill Z. So that's a slightly more invariant way of saying the same thing. And so the second step of what I will do is try to define what this kind of equation means in the more general language of modules or the modules of algebra. So here I want to define, so there's an action here. So I is a left ideal, so it's basically an ideal in the val algebra. So it acts on something, so that's a module. So to define this, I'll have to introduce what we call modules of exponential type. So these would be exactly the type of modules that are generated by such partition functions here. That will be the second part of my talk. And the third part of my talk will be to explain what are the conditions that we need to impose on these ideals here in the read val algebra, exactly such that there exists a unique solution, which in the language of val algebra would be the statement that the quotient of the val algebra by the ideal is canonically isomorphic to a cyclic module generated by this partition function. So that will be the third part and that's basically the definition of area ideals or area structure. So that's the three parts of my talk. Okay, so that's the goal. And at the end, I'll give some examples, probably won't have too much time, but I'll give some, just in words, a few examples of why this is interesting and what kind of an immersive problem do you get out of that, right? Because we're concentrating on this, but the goal is to get something interesting at the end. So get a partition function that is interesting for some reason. Okay, so let me first define what this read val algebra is which is going to be the space where the differential operators live. So now I'm not gonna call them LK anymore, I'll call them HK because they're not the other operators. So just to make it clear that they don't satisfy the variable algebra in jump. Okay, so what is the read val algebra? So let me introduce some notation. So first A is going to be some index set which will index my variable. So that could be a finite index set or it could also be comfortably infinite. Everything works even if it's infinite, so that's good. Now I will introduce the standard val algebra which is basically the algebra of differential operators over the ring of polynomials in my variables X. And also think of it as the free associative algebra of generated by the X and the dels modulo the commutations relations between those. So that's the standard val algebra that you probably know and love. But to make sense of all of that I need to introduce this parameter H bar which is absolutely crucial to the whole scope. So I don't wanna just look at the val algebra I want to make it into an H bar and hence val algebra. And the way to do it is to realize first that this is a filtered algebra. So I need to specify a filtration and ascending filtration on the algebra. So that's basically a sequence of subspaces for the algebra, things like that all the way to the whole thing. So this is an exhausting, exhausted ascending filtration. And there's many filterations you can choose on the val algebra. So the one I'll choose to define every structure is what's called the Bernstein filtration. So what this is is quite simple. So the subspace I is given by, let me write it down but now I explain what this means. Also a sum over indices here. And then you take something like this where this is a polynomial and it should depend on the variable of degree that's equal to half. So what you're doing here is that you're basically kind of assigning a degree one to all the variables and all the derivatives. So your subspace here is going to be all the combinations which have degree less or equal to i which is the index of the subspace, right? So for example, if you're in F2 you would get things like, I don't know, x2, del one, del two, del one, things like that but also things of degree one, like x1 and so on because that's a filtration. So you include the things of lower degree as well in your subject. Okay, so that's the very standard filtration on the val algebra. You can check that it's a filtration. That's a well-known thing to do just like that. I think the group is good. I see a bubble, look, everyone's happy. Good. So if we have a filtered algebra how do we create a graded algebra out of a filter that I think any ID flip classroom? Was that? Introduce a grading, yes. But the val algebra is not a graded algebra. So we have to make it grade. So if you have a filtered algebra there's a very standard way of doing it which is not what we're gonna do. So the standard way is to basically take the quotients, right? You take a space and quotient by the space just before and that create what's called the associated graded algebra, the filtered algebra. But that's not what we wanna do. What we wanna do is do what's called a re-construction. So what is this? So we introduce a parameter H bar. So the idea is to create a new algebra. So it's not the same algebra now, but it will be graded. And the way you do it is quite simple. You just take the direct sum of all your subspaces. And that becomes a graded algebra by powers of H bar. So it's kind of a very, very simple statement. So in other words, H bar is degree one. Everything else has no degree in this grading. So it's not the same as the Bernstein filtration here. And that becomes graded, which is clear. That's always true for any filtered algebra. If you take the direct sum, the subspaces with some parameter H bar or epsilon, whatever you wanna call it, you always get it graded up. So that's called the re-construction. So we'll call that the re-val algebra. But it's not quite where we want the differential operator to live. We need to do a little more. So we need to complete it. So look at the H-addict completion. But what this means simply is that we're gonna take a direct product instead of a direct sum. So we allow power series in H bar. That's all that we're doing. Okay, but it's not just arbitrary power series in H bar in the sense that for each degree in the power series, coefficient must be in that subspace. So it's going to be a differential operator. So some element of the valid algebra, but in that subspace, so we're like degree less than, oh, that's what we're talking about. Okay, very clear. So that is the re-val algebra. And this is where our operators will live. So now we need to move to step two, which is to now construct modules for this algebra. Right, there's a refined version, yes. So I'm not going to talk about it, but that's certainly a very interesting question. So in terms of every structure, there is no refined version of every structures, but there's been a lot of work on refined versions of the false provocation, which yeah, I won't have time to talk about this, because I'm not going to introduce the false provocation, but yeah, from the point of view of what I'm talking about, that hasn't been studied. So I don't know what that would mean. You'd have to extend, yeah, define the space in a different way to have two parameters here. Yeah, that's actually a very interesting question. Okay, so that was the first step, which I told you would be fairly straightforward, which is the construction of the steps. Now, the second step is to look at modules, and in particular, we'll want to construct what we call modules of exponential type for this algebra. So let's start with the ball algebra, because this is well-known. So let's just start with, there's no H bar, just the standard ball algebra and the variables. So what kind of modules do we know? Well, there's one very simple module, which is the polynomial module. So that's going to be our starting point. So this is basically just a space of polynomial and your variables. And of course, there's an action of derivatives and polynomial, and you get that polynomial. So it's a module for your ball algebra. And it's a very simple module, has all kinds of nice property. It's cyclic, it's generated by one. The unit polynomial, that's obvious. What else at the end of the later? So basically all the differential operators on your ball algebra that kill the generator is basically the left ideal generated by the derivatives. That's also clear, because derivatives are one. So that also means that if you take the quotient of your algebra by this annulator, so by all of this, well, of course, if you mud out everything that has a derivative, you end up with the polynomial module. So that's all very variable. But that's what we want to generalize. So now what we want to do is look at the Re's ball algebra. So we'll have to introduce some H bar in the module as well. And then that will give a natural generalization of that. And then we'll want to generalize it to allow arbitrary partition function, which will be the definition of these modules with natural. Okay, so let's introduce H bar. So if you have a Re's algebra, the natural way of introducing H bar in the module is to construct Re's modules. So in other words, you want to look at filtered modules. You want to look at the filtration on the module that is consistent with the filtration on the algebra. So what this means is that you get an ascending sequence of subspaces for your module, but such that if I take an element in one subspace on my algebra, add on a subspace of the module, I'll get something in the, in I plus J subspace of the module. As a definition of a filtration for the module. Again, there are many choices here we could take, but the natural one is the degree filtration for polynomials. So this is basically saying that the subspace FI of this space of polynomials is set the space of polynomials of degree. That's not equal to five. So it's basically filtered by degree. And it's easy to check that this is consistent with the Bernstein filtration just by action of derivatives which decrees degree of three. Okay, so once you have a filtration on your module, you can define the Re's module. So what you do is the same thing as before, you take now a subspace, some of the direct sum of the subspaces here. And that turns out to be a graded module for the Re's algebra. And you can also define this completion in H bar, which is what we're gonna have to work with, which means you just allow, again, power season H bar, but the coefficients are all phenomics of degree less than half. So it's not just arbitrary stuff. You'll have a very well-defined completion here. You're just looking at power season H bar, not in the bar. Okay, and that's basically a left the H bar hat module. And in fact, it has all the same properties as here. So this is a cyclic left module. It's again, generated by one. This is again, easy to see. The annihilator in the completion here of one is the left ideal generated by derivatives, but now you have to be careful. So derivatives cannot come alone in the Re's by algebra because you have to come in the subspace that must come with at least one power of H bar. So if you look at the left ideal generated by those, this is precisely everything that kills one. So that's the annihilator. So this will play a role in the construction. So I'll call it the canonical left ideal. And just like we had here for the algebra, the quotient by this ideal, you get back to completed polynomial module. Yeah, so that's because of the definition of the Re's by algebra. So you see the Re's by algebra is like this. So a derivative is in F1 or F2, F3, all the high ones, but it's not an F0. So it has to come with at least a power of H bar. It could come with higher powers of H bar, but you only need those to generate the whole annihilator. Okay, so that's the natural construction of the module, but that's a very simple module. I mean, way too simple to get anything interesting. So now we have to construct exponential module or modules of exponential type. So another way of thinking about it, so if you have the module here, that's this module we just constructed, one way to think about it is that the canonical ideal kills one. That's what it says to say that the ideal is one. That's what it says to say that the annihilator of one is this ideal, and it's in fact, it includes everything in the value of what it kills one, but that's what this means. So if you remember what we're trying to do, what we wanna do is generalize this to become a much more interesting ideal than just the ideal that I'd like derivative, which is very simple, and generalize this to become not one, but some interesting partition function. Okay, so that's our goal. So we need to define this kind of module. So that's what we do. The modules of exponential type are going to be just modules, cyclic modules, cyclic left, E, A bar, A, H bar, hat modules generated by an arbitrary partition function which takes the form that we want. So this will be a sum. We'll just make sure I got the, and this is right, yes. So this is basically an exponential of a sum in H bar and these are polynomials of the degree less or equal than K plus two. Now you can, this is just another way of rewriting what we had in the beginning in terms of FGNs. This was just the extended version of the sum. So it's exactly the same as what we had there. Okay, so now this doesn't live, this exponent here doesn't live actually in the sweet module because the degree is higher than the power of H bar, but that's fine because what you do here is you define the module with the action being the standard action of derivatives on the exponential. And when you bring things down, then you'll see that this module is actually has a morphic to this res module times that. So when you bring things down, the things that are brought down are exactly in this response. So it just kind of defines the type of or class of modules that are generated by exponentials. And the one we had here, of course, is the trivial case where all these point are no to zero. So the exponential is just one and you recover it in the original one. Okay, so now we have the space. We have the definition of modules and we can finally ask and answer the question is what kind of left ideals in this completed Ries-Weil algebra are such that if I take the quotient of the algebra by the left ideal, I get a module of exponential for some partition functions and of that. Right, so that's kind of equivalent to or at least it implies that module will kill the partition function. That's by definition because it will be the annihilator of the generator to partition function. And that's really the question we're into. So that's kind of a more demodule formulation of what we started with this condition. And also the generator here will be uniquely fixed if you impose that these polynomials have no constant term. So that's an initial condition. If you impose that, then you can show that it's uniquely fixed as a unique such generator for the module that's connected. So we'll give a unique solution to these constants. Okay, and we will call such modules, such ideals, arey, arey, or arey structures as conceptives call it. Any questions so far? Yeah, so it's exactly the same thing. So it's just a kind of a shorter way of writing it. So what you can do is you can expand these polynomials as some sort of in your variables, something like coefficient like this in your variables, right, with some coefficients. And then you rearrange the coefficients, you just define G and N, so that K is equal to G minus two plus N, and you'll see you get exactly the same expansion. So the FGNs are basically the coefficients of that three polynomial. But now instead of writing the sum in terms of G and N, I'm writing it in terms of the power of H1, if that's really what we'll care about. No, so this was just for intersection numbers here. We're working in more generally. So the FGNs don't have to vanish for the condition. I mean, that was true for the starting point, the conservative within intersection numbers, but here it is. Yeah, I mean, what is important is the degree, though, I mean, the degree is bounded, right, so that imposed some conditions in G and N, but it's not like you can have for fixed power of H1, you can get probably two degrees, but did I do that? I don't know, let's see, K is what? No, it was two G minus two plus N, and yeah, no, it's exactly the same. This is exactly the same as the expansion I wrote before, it was just a matter of defining V and N. I mean, what you do is you define K equals two G minus two plus N, and then you do V and N. Well, this is going to be an example of this. So far, no, I haven't, yeah. Okay, so this is the question, that's the definition of very ideals in some way, which is the answer to this question. So you're trying to characterize the left ideal such that this is true, so that defines the partition function. Okay, but there's a very, very simple way of characterizing these area ideals, which is extremely boring, but we're gonna start with that. So if you want to kill this partition function, well, there's very simple way of doing it, so you define operators that look like this. Make sure I get the power, right? Again, this, right? So all I'm doing here is if you take the derivative and exponential, you just bring that, bring down the argument and you think it's derivative. So of course, this will kill that, but you can do that for all variables. So for all indices in your index set, and you get a collection of differential operators that kill that partition function, right? That defined basically in terms of the data of the partition function. Okay, and then you can define the left ideal generated by those, we called it I bar, and it's clear that this kills Z. And now you can ask, well, is it's everything in the value of what it kills Z? So if you fix Z, it's a partition function. Is that everything? Is that the annihilator of that partition function? Or is there anything else? Turns out there isn't. This is actually the annihilator. So one way to prove it is quite simple is to show the following. So these operator are obtained from the canonical operator which were the derivatives by an atomopism, which we call a transaction. So you can define a very simple atomorphism that takes the generators of your completed degrees of algebra and map them to see, don't change it bar, you don't change the variables, but you map the derivative to these. So you basically kind of deform the derivative by this infinite sum of polynomials. You can show that this isn't an atomorphism. So you have to show that the commutation relations are preserved, but the H bar commute with each other because of the way these polynomials are. They're derivatives of polynomials. So it's just commutation of partial derivatives. So it isn't an atomorphism. So because this isn't an atomorphism, then you know that indeed everything works out, right? Because you can define whenever you have an atomorphism, you can define a twisted module, which is basically, I'm gonna write it as this. So what this is, is the same space. So it's still polynomials. So it's cyclic generated by one. It's still the same space of polynomials, but you twist the action. So what you do is you define the action here, of your val algebra, you define a new action on this space, on this twisted module, whereas the action is actually on the polynomial, the space of polynomial completed with these algebra polynomials. And the way you define the action, you take a differential operator, you act on a polynomial, and the way you do it is by twisting, right? So that's the standard construction of twisted module. And because of that, then you see that indeed, this is the left ideal here is the annihilator because the left ideal is the image of this kind of nickel ideal, which was the annihilator of one. So everything's still true for the twisted module, so you get that indeed. The twisted module here, sorry, I missed one step. So what you get here is that indeed, I mean, this is true because you get a transaction, but the one thing that I missed here is that you can also realize that this one's that transaction is just conjugation by the partition function, right? Given any differential operator, if you only shift the derivatives by polynomials, what you're doing is just conjugating by the exponential, right? That's a standard calculation. So in other words, what you get here is that you can also think of this action here, instead of thinking of the action as being twisted, you can think of the action as being the standard action here, but where you do not act on polynomials, but you act on polynomials times the partition function. And that was the definition of these exponential, these modules of exponential. So anyway, this might have been confusing and is not actually important for the rest. So he can ignore that the end upshot of this here is just that the module of exponential time generated by Z is a cyclic module is generated by Z. And indeed it is the quotient of the read variable by this area ideal is isomorphic to this module, which is what we're doing. There's no another, this is just a fancy way of saying that there's no other differential operators in the read variable, but are not in the left ideal that I tried to write this, that killed the partition functions, that includes everything. That is also unique if you impose that the polynomials here don't have constant. But this is, as I said, this is, I mean, what could say is the answer does characterize these left ideals, but it's very, very boring because you're characterizing the left ideals by giving a generating set, which contains exactly all the information of the partition function. So it's basically you, right? And you're just specifying, well, it's not useless, but you're just specifying the ideal in terms of the partition function of vice versa. So it doesn't give you any interesting information about the partition function itself. And so what we want to do, and that's the idea of every ideal. So I guess I have to have what? Five, what's that? 20, no, it's the best. I don't know. You're not the chair. Oh, that's perfect, yeah. 10 more, I don't know. 10, okay. That's also perfect. Okay. Whatever you say is going to be perfect. So yeah, so, okay. So this is a characterization, but that's a very boring characterization. Now, the interesting thing is the following result, which is due to conceivage sublament. And in this language, we kind of reproofed in our paper from last year. The interesting result is that you can characterize these ideal in a different way, which is much more kind of more invariant. And the key idea here is that if you have a left ideal in an algebra, then of course you can find many generating set. So one generating set is those, and that's a boring one because it includes all the information of the partition function. But if you can find another generating set, which is more interesting, that might actually give you some information. In fact, it might allow you to completely solve the partition function recursion. And that's exactly what happens for conceivage as well. So that's the idea. And that's the term. So the term is the following. So let I be a left ideal in this completed resalgebra. If it satisfies two condition, so if I is generated by a collection of differential operator, let me call them H, not H bar, such that they all start with the derivative plus higher order terms. But this is much more general than the H bar. So remember here, all the higher order terms were just for anomalies, especially for time. Now you can get the derivative. You can get anything you want of higher order. It doesn't matter at all. You can get anything you want. You just need to make sure it starts with the linearity. So in some way, you can think of it as some sort of deformation of the canonical one, the derivatives. But that's the first condition. And the second condition is that if you commute ideal with itself, you get something that lies in H bar square by the left ideal. So if these two conditions are satisfied, then I is an area ideal. So in other words, then all the properties are true. So that means that the quotient of the algebra by I will be isomorphic to a modular exponential type. So it will be completely, it will fix a unique partition function as being killed by the element of the left ideal. So that's a much more general statement because, well, you can get, I mean, this generating set is much more general than the one I had before. So that's kind of interesting. Now I'm not gonna talk about the proof here because I don't have enough time, but maybe I'll just mention the two main steps of the proof. The proof is actually quite interesting. But the two main steps to prove this, oh, before I talk about the proof, I just wanna highlight a few things. Yeah, first I wanna highlight that this condition is non-trivial because there's a lot of confusion sometimes. So one thing is that, if you commute two elements of a left ideal, then you always stay within the left ideal. That's obvious because just right down the two elements of your commutator, there's always something on the right which is generating the ideal. So you always stay within the ideal. That's always true. And it's always true as well that if you commute two things in the left ideal in the ball algebra, you always get something that you can always bring down an h bar square that just comes from commutation of derivatives and variables, which degrees degree by two. But it's non-trivial to say that this thing you'll get here is in the left side. That's a very non-trivial statement. And most often it's not for general left ideal that won't be true. So that condition is very non-trivial. Okay, so, but that was just a remark already. Okay, and basically just the proof here just to show you how it goes in two quick steps. The first step is to, what you want to do is show that this is an area ideal. So what you want to do is show that you can find another generating set here which consists of operators like this. And that will tell you that it's an area ideal. And one way to do it is to construct these h bar first explicitly. So what you want to do and that's where the first condition comes in. You wanna use this fact so that if you extend this h a for every time you can transform the right most derivative by an h a again up to higher order terms. So in the end, what this means is you can construct some, you can write h a as something like that plus an infinite sum and h bar whose coefficients are polynomials plus something which is within the ideal. And then you define that as being the h bar and that within the ideal. Now the only reason you can do that is because you're working in the completion. You really have to have infinite power series and h bar otherwise there's no way you would be able to do it. So it's very important that you're working in completion. But you can do that and you define the h bar as being these objects and then you see that they are in the ideal. Now you have to check that they actually generate the whole ideal but you can do that. And then the second step is you want to show that they commute. So you construct those as being the h bar and you have to show that they commute because if they commute, that means these polynomials you can provide them as derivatives of single polynomials that's Poincare-Islamic. So you have to show that they commute to do that and to do this, this is where you have this condition. You have to use this condition. This condition in fact imposes that there is no non-trivial Poinomial within the ideal. That's the upshot to that condition. And that's the key condition here to show that this is equal to zero. Okay, anyway, this is the main step. If you look at conservative troublemen this is not at all how to do it. So it's completely different from what they're doing but it's completely equivalent as well. So that's just a different formulation of what they're doing. Okay, so I guess I have to stop soon, do I? So what I'll just say here is a few things. So I want to say why this is interesting. So the main reason why this is interesting is if you can find h a that's our h bar Poinomics. Okay, so what's the point here? So remember what we're doing here is we're constructing an ideal such that there's a unique solution to the condition. So in other words, we're constructing an ideal generated by these h a so that they uniquely kill the partition function. Now, if they are h bar Poinomials and all of the bounded degree then that gives you a recursion and recursion for the coefficient of the exponential. Because this is the coefficient of the exponential is an infinite power series in h bar and then you would have only a finite number of conditions. I mean, everything is infinite if your number of variables is infinite but still less number of conditions in h bar. So that gives you a recursion for the coefficient. So a recursion for these Poinomials Q or a recursion for the FDNs if you would spend them in terms of coefficient. So that gives you a recursion in powers of h bar. So that's what you get and that's what's called the Polish goal recursion. I mean, it's not usually formulated like that and that's equivalent to the Polish goal recursion. So what this means is you get if you write down these differential operators very explicitly then you can write a very explicit recursion for these FDNs and that's what is being known as Polish goal recursion. But the key here is just that if you can find a generating set for your ideal that is not in the completed algebra but just in the just h bar Poinomials that's where it's interesting. Otherwise, these are fairly boring structures. And an example of that is of course the one we started with to conceive it with. So in this case, we define these h a to be just the visual operator. Oh, I mean, we have to shift here if you, this doesn't matter. So these were something like that plus terms which were actually quadratic. I didn't write them down explicitly but if you look at the definition of these conceivage within operators, the visual operators there's a linear term and then there's a term here which involves second order derivatives and which is also that three two Poinomials in the X. So that's an example of differential operators the h bar Poinomials. They do satisfy this condition by construction here. And it's easy to show that, well, I mean, that's just the property of the visual algebra. So because this is true, that implies condition two. Which is not too hard to show. In fact, condition two is completely equivalent to the statement that if you take the generators, the generating set and commute them then you should get a combination of generators but the coefficients in general don't have to be constant. They have to be in the algebra. So they're differential operators. In the case of the visual algebra the coefficients are constant. So it's a simpler case. In general, this is completely equivalent to condition two. That's another way of casualizing the idea. But these visual operators are an example the left ideal generated by those is an example of an area. And there's a unique solution, unique partition function associated to those and that's just the concept with the generating function for intersection numbers. Yeah, this is just saying that these coefficients here because we're just conditions about the left ideal. So what you wanna make sure is that everything stays within the left ideal but these all have to be constant. You just have to be within the algebra. Okay, so that's the main constructions. I'm just gonna end by just telling you a few examples that if all we were doing was reproducing consider it within that be very boring. But it turns out that there's a lot of, lots of problems that can be formulated in this language. Yes. Yeah, so that's a very good question. So can you, could we kind of, I mean, in some way that's a definition of these ideals but can we kind of understand better if you fix the degree? Suppose you fix the power of age. Well, can you understand better? I think, yeah, that's a good question. So I don't have an answer, I haven't been but there's a natural classification problem if all like these kind of demodules or ideals in the completed reeve algebra. The fact is that this completed reeve algebra hasn't been studied that much, I think. But for us, I mean, the val algebra is extremely study but with this reconstruction but for us it's absolutely key. So if you don't introduce H bar and complete in H bar none of this, oh. So this is a very, very important. Okay, but what people have done though is more like try to construct examples of various structures or area ideals and then show that the Z, the partition function has an inertive mean. And there's a lot of such example. The first one, as I said, is conceived within that was the starting point. But the second one, which comes very naturally would be what's called the bears and grots with an BGW example. So what happens here? So you take your operators to be, again, the reservoir operators, but they look slightly different. They still are just a linear term plus a quadratic term but it's a different representation of the reservoir algebra, which we could write down. And then what you can show is that it is an area ideal and Z is a generating function for intersection numbers but not the standard one. So now you have insertion of a class here which is what has been known as the nobri for Kyoto class. So an example of Kyoto classes are much more general in this. That's an example of it. And this has now been proved as a theorem. So you get this and also in this case, it's proved that the Z is also a tile function for KDV but with different initial condition is the PGW tile function for the KDV hierarchy. But that's another example, but then you can do much more. So what's key here is that this, because of this, this doesn't have to be a Lie algebra, right? Because these don't have to be constant. And in fact, a lot of examples are not Lie algebras and W algebras turn out to be a very fruitful sources of area ideals. So what this means is you pick AW algebra, for example, W of GLR and you can construct, you look at the generator of the W algebra, you can construct a differential representation for these generators and you can construct many of those. And you can show that they satisfy these. So we have a paper with Thomas and there's been pretty papers on this where we kind of classify those. And then the inumerative meaning in general is still unknown, but in some cases, we get some inumerative meaning. So we get, for example, the R spin intersection numbers, the section numbers over the modularized space are current with R spin structure. We also get, so there's a recent paper by Thomas Newton and his collaborators where they show that you can also get Kioto classes or some Kioto classes for some R spin intersection numbers. And but most of the cases are still up in the air with the inumerative information. So that remains to be studied. And there's lots of other example, we can construct area structure for all kinds of W algebras. So SO2N, E6, SB2N, which we did in this paper with Aniket and so on. We can show that we get representations of those that are area structures, but we don't know what the inumerative interpretation is. And finally, there's many more. One of the recent paper we wrote is we can use that to construct with the curve vectors for W algebras, which are related to gauge theories and symmetric gauge theories and Kioto vectors in all this construction. So that's another thing we can construct those as basically being partition functions. So generators of these modules are a financial type. So there's lots of, actually, there's many more than that. There's also all semi-simple co-homological field theories fit within this frame for it as well. So whenever you have a semi-simple co-homological field theory, you can write the FGNs as correlators and you get an area structure. Nothing variants also appear in this. I mean, there's lots to take on. So it seems to be a fairly general structure that is worth studying. So yeah, I guess I'm gonna end here if I should go on the end right now. Let's thank the speaker. Thank you very much. Are there any questions, comments for the speaker? Right, so that's a very good question. So if you have an area ideal, you know that there's a unique partition function that is killed by the ideal. That's the definition of an area ideal. So the point here is that if you're, the case where it's interesting is if you can find a generating set for your area ideal, which are free nominal, because then you basically just do standard differential equation. Bring nominal operator acting on a series so you can get a recursion for the coefficients and you can solve recursively. That's exactly, I mean, consensus within is the example to keep in mind. So that's exactly how you construct all intersection numbers from the ideal constraint. You just have quadratic differential operator acting on Z and you can solve recursively and that gives you the standard recursion. Yeah, so it depends what you're interested in, but yes, too. But in fact, in practice, how we construct area ideal is by constructing two general and set equation. So like all these examples, you start with some algebra. I mean, at least as the point of view that has been taken so far, you start with some algebra, you construct the representation, some of the differential operators should have satisfied that and then you have four three and that would be an exclusion. You can calculate it, decrease it. Right, so FTRWU, you mean, should be in the list. It hasn't been, I mean, it certainly is. It just hasn't been. So I mean, yeah, the answer is, as soon as someone does it, it will be in the list. No, I mean, my expectation is yes. I mean, they should be typically in this framework, but yeah, exactly, that's what I'm saying. So locally, that's why I'm, I mean, so that, I mean, yeah, one case I didn't write down here is also all Gommelton theory of Tori Calabio threefold. So all of those are also calculated by this and also hermit's numbers. So all kinds of flavors of hermit's numbers, like, all the false 10 hermit's numbers, all of those also fit with that. So this construction here, if you look at local behavior that look like R10, that should become FTRWU, but yeah, so for compact manifolds, I thought that doesn't really seem to fit in there so well. Yeah, so there are a case, I mean, semi-simple always work. There are cases where it's not semi-simple that works, like the R10 case, but why? No, it's not that it will not work, it's just that. As I said, there are cases where it's not semi-simple and it does work, like the R10 and FTRWU. I mean, originally there was a- So there are counting tables in the room that wouldn't fit in there? No, I mean, I don't know. No, but I mean, yeah, I don't know. T-center sections first. Yeah, so you see something really interesting on that note is that, so there was a paper of a Shadwin and collaborator like a few years ago where they studied it's called Atlantis Hervis Numbers, which is kind of a different flavor of Hervis Numbers that look like the Orspan one. And they were really excited that they found a spectra curve and they realized it was the same spectra curve with Orspan Hervis Numbers. So they could basically, they argued that this was a case that did not fit in this framework, but we just wrote a paper as long as it shows that it does fit, but you have to extend the framework a little bit. So that's it. So it seems like it, whatever you deal with intersection numbers of sunsets and MDN, it seems like it, but I mean, I don't. I mean, the VRSR conjecture was proved only in a few cases if you consider the most stable maps to something not only curves. Do you have any other example beyond those that were known? No, but I have to say VRSR conjecture is quite different from that. So in the sense that if you take the VRSR constraints for general boom with inferior, it doesn't fix partition functions like already for boom with inferior C1, but you have VRSR constraints, but it doesn't take it, it only takes the descendant. There are these operators which annihilate all the generating series. I was wondering because you said I mean. I don't, I don't, I don't know about phone. Okay. So do you have any examples of modular of stable maps to something which apart from Tori Koloviartri, which you know, your every structures annihilate? Yeah. So boom with inferior C1 is another example. And it's an interesting here. I mean, that's the interesting thing. So there is an area structure that boom with inferior C1 and it completely fixes the whole going to the stationary set. So, and the operators are very different from the Okunkov and the VRSR operators, which are, if I understand well, they don't fix the station into the only fixed descendant, I forget. But, but there is an area structure here in a completely fixed theory that was done by Novorin and proved by Shadrin and collaborators. So of course, Tori Koloviartri, what else? Well, I mean, there's all kinds of all the phones of those as well, all the phones, all those that support. For golden theory. Okay, now I can ask you to assist.