On a problem of Poincaré: Bounds on degrees of vector fields
CC Attribution - NonCommercial - NoDerivatives 4.0 International:
You are free to use, copy, distribute and transmit the work or content in unchanged form for any legal and non-commercial purpose as long as the work is attributed to the author in the manner specified by the author or licensor.
Polini, Claudia
In 1891, Poincaré asked if it is possible to bound the degree of a projective plane curve that is left invariant by a vector field in terms of the degree of the vector field. In joint work with Chardin, Hassenzadeh, Simis, and Ulrich we address this question. The question can be restated as a problem about the initial degree of the module of derivations of the coordinate ring R of the curve modulo the Euler derivation in terms of invariants of R. We exhibit lower and upper bounds for this initial degree and in several instances we are able to determine the initial degree. Examples will be given to illustrate the situation.
 You know, the organizers, so given the possibility to come once again, it's a fantastic place. And also the conference is going very well, I'm learning so much mathematics again. Hopefully I can climb a mountain, so it will be very, very good. So I already brought up the titles and also the collaborators in this project. They are Marshall Dan, Amida Sundazis, around Simi, Suburna, British. Now, so this goes back to the problem of Pankare in 1891. He asked if we can balance the degrees of plane curves that are left in value by algebraic vector fields of P2. So the goal of the problem, if you want. So first of all, why did he ask that? I mean, the reason is because if you could bound the degree of the curve, then you would have to look for this curve because you actually wanted to find them. You have to look all in a finite measure vector space, so it makes it easier to find such curves. And what is the goal in general? I mean, is to relate numerical invariance of the vector field to numerical invariance of the curve where the vector fields leave the environment. So first of all, the first step I want to do is a translation to commutative algebra because this problem has been studied so far mainly by algebraic geometers and the fresh geometers. So these are some of the previous players that actually didn't pass 20 years quite a bit of progress and they are Campilo, Carneser, Saval, Cruz, Estabes, Galindo, García de la Fuente, Climbán, Luis Neydo, Pereira, Suarez. We started this problem actually in 2001 and that's every project we burn, it takes at least 20 years to publish it. So we had some very good progress in 2001 back in 2001 and then this stuff that we proved was proven by Estabes and Climbán later on. So we had to restart again and try to prove more and hopefully this time we can publish it before somebody else starts. So let me fix the setting. I will have a little simplified setting just because of the talk, so avoid technicality. So K is an algebraic closed field, this unit, and then we assume that the characteristic of K is zero and these absolutely don't need. Of course I've listed the topic, but we assume that C in Pn minus 1 is a curve which is reduced, this we always need, and irreducible, and this I don't need, it's just really doable, it's a good balance. So I have my curve, I'm going to go stick the maker. Now either north with R, do you see if I write down here? Okay, because that's easier given my height. R and S are the coordinates of C and Pn minus 1 respectively. So now what is a vector field? And now I have to start writing here, unless I use this little tool. A vector field X, okay, I think Pn minus 1 of the degree M, we always call it the degree of the vector M, okay, is an homogeneous map that goes from Z to S, okay, and what is Z? Z is the kernel, you know, is defined by the Euler's sequence, is the kernel of the Euler map. So you take the module of differential of S, remember S is a polynomial, so the module of differential is a free module of rank N generated by the partial derivative of Xi. Xi are just, I mean, so the differential, sorry, Xi are just coordinates, okay, and what is the Euler map is the map that sends the Xi to Xi, so this goes to the maximum ideal of S, and this is called the Euler's sequence, because this is the Euler map, and this is just the kernel. Now, another way to think about Z is that if you look at the shiftification of Z, Z tilde is the shift of the module of differential forms of Pn minus 1. Now, so this is degree M, means that this is an homogeneous map of degree M minus 1, the map of degree M minus 1, and any such map is a restriction of a map from the module of differential to S. So such map is just given by a row vector, let's say a 1, a N, so the a i are in S, and they are forms of degree M. Now, first of all, how can you think now this way about a vector field? You know, mind what is a vector field, but this and one and N just define the forms, they define a map from Pn minus 1 to Pn minus 1. So you have a point in Pn minus 1, you look at this image, it's another point, and what's happened? If the two points are distant, you have a line in between, that gives you the direction of the vector fields at the point. If the two points are the same, then you say the field is 0, and if the image is not defined, you know, or exactly where is the 0 degree vector field, that are exactly the similar points of z. Another way you can think, since this is free, you know, another way you can write the vector field, since this is free, you can write it in terms of the dual basis, and this is just the sum of the a i and the partial derivative. Okay, so that's another way you can write the vector field. Okay, now, I talk here about vector fields that leave invariant curve. So what does it mean? I have a curve there. What does it mean that it leaves invariant curve? You just have to write the same sequence now at the level of r, and you have a commutative diagram. So if I speak loud, but I have an ear, it doesn't work because of the altitude. My kids have that problem too. What? My kids have that problem too. So I look at the same sequence, the similar sequence at this level, okay, again you have the map, now this is the image of the outside, okay, and obviously there is this commutative diagram. Now, be very careful because this is our z, usually this map here is not subjective, and what we would do is care is the image, but in my case we assume characteristic zero, that's the reason I assume it, so this is subjective actually. And what is l? You have a way you can think of l. Another way you can think about l is the image of the second map, you know, the theorem complex, okay, so of the second differential in the theorem complex. And that would be very useful. Also, what is the theorem complex? Just the Cousin complex built on this map, you know, on the Royal map. Alright, now, so what does it mean that this leaves the same value? First I want to do a picture, and then I go back to the diagram. So, let me erase this. For the moment I have the Poincare problem, but after I erase it, okay, so C is left invariant by x, okay, or another way that the geometry that's saying is that C is a leaf, okay, I will never use this terminology much. Okay, so what is your image in your head? You have your curve, you want that at every point of the curve, the vector fills the tangent. You have your vector filled, and this should be, obviously the vector fills the find everywhere, but you want that by this tangent to the curve. So that's the first condition. If you take a point at the curve, the vector of the direction is pointing. The other condition you require is also that at almost all point of the curve, the vector filled is non-zero, okay, so the single, it's not single, at almost all points. So there are two different conditions. How do we write them algebraically? The first condition algebraically just says that x here, this was the x if you want, so x was a map here, you see, at the level of s, so you had your C, that goes to s and this is your x, and down here you have your L. Now, you want to induce, x should induce a map mu at this level, okay, so x induces mu, and the second rule is that mu should be non-zero. That just means that the vector fill should be non-zero at almost all points. Alright, obviously this mu will have the degree be homogeneous and you have the degree m minus one, exactly the same degree as x at, okay, necessary. Alright, now we are ready for doing our translation, so to convince the algebra of the problem. So I'm ready for it, so I'll start. I don't know, maybe you are familiar with this, so I try to go slower. So, for the, I look at the, whether we want to, instead to look at the set of curve that are left, at least mu invariance, I look at the set of mu fixed in my C, so what is the set of mu like that, you know? They are just homogeneous, non-zero, okay, are linear maps from L to R, and actually it's easy to see from the diagram, chase the diagram, that they necessarily, you know, can be extended to the module of differential, omega of R, but can be extended. What are maps, homogeneous map from the module of differential to R, are derivations? So what we are talking about are just derivations. So, this is a translation, our object of interest are derivations, and which derivation, but be careful though here, because you want all derivations, but, you know, you want that when you restrict it to L, are not zero, which derivations are zero when are restricted to L? But L is the kernel, if you remember, was the kernel of the Euler sequence, okay? So any derivation that, if you just remember the Euler sequence, just a second here, I wanted, for you, and we have a little bit here. So L is the kernel of the Euler sequence, so if I dualize this, I see immediately that what I have to do modulo, because remember I have to dualize it, I'm interested in the map from omega to R, is exactly the module, M inverse times epsilon. So this is the module of interest, and you're looking at it inside, and to go there, okay? Now, and notice that this is a torsion-free prank, one module. Sorry, epsilon is the Euler derivation, so it's exactly what you correspond to this map, the Euler map, the Euler, through all the map, through all the talk, and this is going to be very important, the Euler derivation. See, you have to modulo it out. So now, go back to Punkarec, that's why I left it here. So what did Punkarec ask? He asked if you can bound the degree of plane curve left invariant by the vector field. And Karneser, and you wanted to bound it using the degree of the vector field, that's why I defined the degree of the vector field. And already the answer is no, due to Karneser, he was one of the players I wrote up there before, and there is an easy example, but I don't know if that's the original example of Karneser, but this is easy. You take the family of vector fields of the green one, given by the derivations. So I write in this way, minus D times Z, the derivative over Z, partial derivative, one minus D, X, the partial derivative with respect to X, okay, for every D in the natural number. And this vector field, each of these vector fields, leaves invariant the curve, okay, CD, because depends on D, of degree D. So the degree becomes as big as you want, you know, while the vector field is degree one, and is the curve just given by one equation, so it's an I prefer surface, defined by, and it's easy to see, you can check it, if you're bored, it takes one second, X, D minus Y, Z, D minus one, just to the derivative, you got that, this is zero, okay. All right, so we cannot expect to find a bound on the curve, on the degree of the curve, just using the degree of the vector fields. We must use other invariants, okay, so other invariants might play a role, and now we look at opposite question, which is a prevalent, okay, and that's why I wanted to look at this old collection, because in Comistival du<|ko|><|translate|> we have a beautiful module, but then it goes all over that. Instead to look at upper bounds on the degree of the curve, I want to look at lower bounds on the degree of the vector fields. So if I want to look at the vector fields as small as possible degree, I just have to look inside this module at the element of small as possible degree. The first non-zero element of small as possible degree. So it's like bounding the initial degree of this vector of this module. This concretely encodes all the information I want. So now I can really restate my goal, and we have a precise module. So now I can really restate my goal, and we have a precise module. So now I can really restate my goal, and we have a precise module. The only reason I present this now, and I start as this guiding thing, is because in this case we can give a complete answer, a clean, complete answer. So the sample here says as always. So remember the assumption, C is reduced to the reduced goal characteristic of K0 and algebraic goals. In addition, in this theorem, I assume that C has a retmetically gorestone. C is in the end, this one is retmetically gorestone. So I is gorestone if you want, I is the term of the thing of the curve, and as at most ordinary nodes as singularity. This is the kind of singularities. This is the kind of assumption that is always present in the work, in the previous work of the people. Then the initial degree of the object of interest. In this case it's actually R, and inverse is R because it has gorestone, okay, is equal on the nodes to the A in variant of R plus 1, if you want to write the Castelmau monoregularity of R minus 1, the curve, remember. In general, what can we do, you know, and actually this will guide the rest of the talk. In general, we can find upper bounds for this, which is not the question of Poincaré, but we can find upper bounds for this module. And for the upper bounds, we have no restriction on singularity at all, but we need arithmetic from a combinatorial. We can find lower bounds, which is the question of Poincaré. We don't need at all any gorestone or comboconinus for the lower bounds, but we need assumption of singularity, not this assumption, but we see which assumption we need in general. We need to control or some things about the singularity will come in the formula, okay, and then really in our work, we can do not just curve, but higher dimensional variety. But for this, I really wanted to focus on Poincaré question. Okay, so I'm going to divide the rest of the talk, and I have one hour because we're at 2.00 an hour. The rest of the talk is two other pieces, and the first piece will be the upper bound and the other piece will be the lower bounds. And I'll try at least to give you an idea of the technique that we use, okay? So in that time, we do prove the sample here. So I completely prove the sample here, but I'll try to do it, give you a normal view of the techniques, which I think is the most interesting thing. I had promised for examples, and we do have a lot, but most probably I'll just quickly tell you then where we are going for the examples. Okay, so let me talk about, you know, I think this was true because one was the translation and three is the first upper bounds. I just talked about one upper bound and three maps, and the three maps are the crucial fit. Okay, so first let me define one object. I try to give very little definition if I can, but I have to give this one. So J is the R idea generated by the Jacobian ideal of, if you want, the expansion of the Jacobian ideal in R, the Jacobian ideal of a complete intersection, a general complete intersection. Let me call it C i of dimension d minus 2 mapping into R. Okay. On to R. On to R. Okay, so what are the three maps? So there exists three more genius maps of the degree zero. The first one, you know, is the one I talked about. I said my L was the image of the second differential in the Delan complex. So it comes from the Delan complex. Okay, so it's just a second differential. That's confusing when you define the L-pile, I thought you defined it before as a curve. It's a curve, but it happens to be. It usually is not equal to the image. But in this case, it's the image because we have characteristics zero. You have to prove it, but it's the image. It requires a proof. Yeah. Yeah. Yeah, it does. I said it usually is not, but in this case, it happens to be. And that's why I wanted to simplify some characteristics zero because we have that in this case. Otherwise, you have to assume something about the characteristic not dividing the degrees of the form. And it all you use the image like we do in our field. So we don't have so much to assume. Okay, so the second map is the map. And that's why I needed J that goes again from this module top differential. Okay, into J, epimorphically, shifted of delta minus two. And I'll tell you what delta is. And what is this map? So this map does not depend on the fact that this is the module differential just depends on the fact that this module is a rank and the definition of J. So let's write here why we have that map. Once I write it will be complete clear. Okay. So remember that the rank of the module differential. Okay, is two and the minimum number of generator of the module differential is N. Okay, and J. Okay, is generated. Maximum minor of the matrix consisting of N minus two. N minus two. Columns of a matrix presenting omega of a presentation matrix of my mother. And what is that the map that goes from the from this guy to this guy, you just take the XI by the exchange and you send it to the minor where you delete that or I and the Roj. Okay, and why the shift, but the form defining the C at the green, let's say delta one, we get many comment that the true to get many quote. And this is the degrees of the forms defining C. And delta is the sun. I'm only minus two. That's why it's a general to start with the largest delta minus one. Okay, so this is exactly what you get when you send that map because the excited to you and you send it to something of the delta. So what is the firm? The firm up there talk about practically because the method goes from the module of top differential forms in this case omega to the canonical canonical module, which is the module of regular differential forms and burn talked about the old time. Exactly. So we know who this guy is and he's called a fundamental class. Now the point is that these two maps are morphism and all these three maps are isomorphic genetically. So what you get is that once you know that torsion you get free beautiful conclusion from these three maps. So if you know that torsion you get first that let me start from L model torsion is isomorphic to J. This is an idea. Okay, and this is a morphic to this guy. The module of top differential forms when I go down torsion. And inside here I find the canonical. Oh, sorry. Sorry. I have this embed the canonical and that if I do I see it. So I go and I see the picture I see that the dual is isomorphic to the inverse the fractional idea to delta is isomorphic to the dual of this guy. And then this is the one I was thinking about the dual of the canonical model sits inside here. And then you get immediately that the module you're interested in. Okay, this module. Okay, it was sitting in here. Okay, so it sits inside this. So it's clear that from here already you can see why this is powerful if you want bounds on the initial degree of this you just have to find bounds on the initial degree of this. I'm not saying that this is easy, but you know, lower bounds, but you can find low. So this is important for all bonds and we use it for all bonds. And then this we use for upper bounds. Okay, and we see actually the upper bounds come immediately very easy now that we have this picture so I can write them directly. So this is the fear. We have more fear than just this one, but this is maybe clean version. And as I said, you have single isotope make up here at all. You assume just that C is automatically the order and really you can do without guards and but you don't get. And I'll talk about this what we write here. And then the initial degree of the object of interest is always less than equal to the A in value of R plus one. Or if you want the regularity minus one. And actually this is completely false. If you take the constant away, you cannot get this and it's easy to show. And you just take an actual perfect idea and you use it easy to show that you cannot find upper bound just using the regularity of the invariant. You need to involve other invariants. For example, we have one using the type we have and the invariants we have one using the please T and the violence, but other things have to come to play a role. Okay, so we see where God is that is used here and what you would need to do in general. So the proof becomes actually quite easy. As I said, you always start from the universe sequence. Let's take our starting point. And if you do allies that. Okay, we already saw that our module. Okay, sits inside. And this is ours. I just got our senses, come on. I call it sits inside a star and what the cook on a citizen side, but the cook on a citizen side, the X one of this and are against. But this is the only as a more effective. If I go to X two of K and stuff. Why do I want to do this because this is a sock. And this is the soccer of our moments one X two shift. And this X one X two are generally. So what's the fun that for you just use coming calling us. Okay, nothing more by general this module that we care about is concentrated. And this is the real important thing. In the grease less than equal with a variant of up. But if it's course then that's not really important. But because this is just needs coming closer. So I don't use course. From yes, this is concentrated by the degree if I pass that degree is true or the same. So the initial degree. Of the module of interest. Is less than one. The maximum of initial degree of L star and a of R plus one. And now you also know. I didn't want to erase that sequence. So maybe I erase and do something. I try to do something. That's okay. Now if you look at that sequence here, I have this omega is inside. I'll stop. So omega look at the dual of mega sits inside. And stuff. So if I want to. Bound the initial degree of L star something inside it. I better just bound what I am inside it. Now if I go to then this is our shift. So this is in the degree R and it starts in the area. So this guy, this is. The initial degree. Of L star is less than. Okay. That's it. Now if you didn't have course then. You see you had to bound that initial degree and that's where. Types coming in and stuff. Okay. Let's do the last part of the talk. And hopefully I don't know so much over time. And I want to talk about the interest of one career and where the history actually the work of that are people lays on which is lower bounds. And here we have to somehow use some idea from geometry and we'll use general protection. Okay. Okay. Okay. Okay. So you have your beautiful curve and P-1 and what you want to do you want to project it down. To project it to the plane. You just have an item surface and usually the problem is not easy to compute the you know the not not so much easy. To compute your module in terms of the boundary of the initial degree in terms of inviolence when you're playing curve. The problem is and that's the problem with geometry. How do you translate your bound that you're typing the plane to the bound from your curve as 6 and P-1. And there are lots of difficult to do that and they do as you come across in a very complicated way. We don't get some such easy translation and we are able to do that and we don't need to call it. And it becomes very beautiful to go up and forth. That's why you can get up and down the lower bound this way. Okay. So let me write our statement and then the problem is just to compute the module of interest for a play for a night surface. So this is the last chapter of the talk and this actually we have to thank over both of because some of these three came where the last three weeks. So let's say lower bounds and general projections. So here this is about the general projection. You have your ring R. Remember your ring R is the coordinate ring of C which is a curve in P-1. And what do you do with this general projection? You just consider three forms, general linear forms in R. Or mine. It's not mine. General linear forms in R and we can see that the algebra generated by this, this six and side R, obviously, we call A. This map is a finite, birational, okay, homogeneous extension. And then what you have is that A is the coordinate ring of a curve in P-2. So A is got a stem, the nitrous surface, okay, of the same degree as your curve C, okay, the finite value of the same degree as the curve we started with. So what is the connection between the two module of interest, the scenario of two module of interest? The derivation of A, module of R, excellent, A, and the module of derivation of R, module of M, M plus, fancy. There is a spalling, let's try to do this, when the initial degree of our module of interest, let me put an iron, the above all the iron is needed, okay, is bigger than equal than the initial degree of the module of derivation of A. Now this is the order, oh, no, sorry, A, sorry, and I need A not a inverse because it is the same, and then I have minus the A in variant of A plus the A in variant of R. And it is also less than equals to the same degree of the same module of the A in epsilon and then plus A of A minus A of R, okay, so you get this beautiful thing, and then you can move the stuff up, stuff it up, you can get up and lower, they are really interesting now, lower, you can do both. Correct, so how the proof, the sketch of the proof is, but you consider the conductor of A of R is A, this is the column, yes, and this is nothing else than the column, and this is nothing else than the column, 9 to 8, but remember that A is cost, so this is the canonical module of R, shifted by the invariant of A, and now, so what is the initial degree of this conductor, is that this is degree of R, so this A of A minus A of R, and this is epsilon degree, okay, so this is this quantity that you see here and here. Now why there is that quantity, because you can embed, you have to pass the third module in the middle, so you can take the module of derivation of R into K, and you have a restriction map here, you can just restrict all this derivation to A, so you get the module derivation of A to R, as you restrict them to A, okay, so this is the restriction map. And then you can also embed, obviously, the derivation of A, these are just particular derivation. Now the point is that remember we have to be quotient, our object of interest is quotient, but the Euler derivation goes to the, this is Psi, let's call it, the Euler derivation goes to the Euler derivation, and what's more important is that once I mold out this guy, this is torsion free, so what you get is still an inclusion, and this is epsilon R, and this is still the same and inverse epsilon A now, as it goes to the derivation of A, and this is R, A, sorry, epsilon. Now, so what does, furthermore, remember these are all torsion free now, they're all modules, furthermore you can easily show that with my conductor kills both co-currents, that's not difficult to see, and this implies immediately that if I take the conductor and multiply it by an element, here it goes to finish here. So this tells me that if I take the initial degree of this module, plus the initial degree of the conductor, because I take the element of the conductor and multiply it by an element here, I finish here. It ruins your element because these are all torsion free, so the initial degree of the conductor, this is, I get an element there, so this is bigger equal than the initial degree of this guy, because I get an element there, and then you can play the same role with the other one, and then you get the balance you want. Now, we are left now, okay, to look at the, and I still want to do that, to look at the case as I said, where it took an hour, so I can do it, I knew I would get to solve the case of plate curves, but it's very interesting, so I would like to describe that also because I can show you how the results of previous people have been generalized, and that's really what we want to get to. So, this is really the plane curve case, so here, plane curves, okay, so just that, now we're currently at the two of the degree D. Okay, and the initial degree of our module, we're interested in this case, it's odd because it's an hyper surface, as I call it, is bigger than equal than the minimum of D minus 2 and 2 D minus 2 minus A of r mod j, the A variant of r mod j, j is the Jacobian ideal of r, the fungial Jacobian ideal of this case, and this is bigger than equal, if you estimate this A variant of D minus 2, this notice that is what we had before, is the regularity minus 1 of the A variant plus 1, plus, and here it comes, the singular locus, the singular locus of C, this is the cardinality, so the singular points, the number of singular points, minus the levy multiplicity of this r mod of the Jacobian. Now I'm going to define what the levy multiplicity is, and if you have any module M, you can consider, I mean, graded or whatever, you can consider the associativity formula for multiplicity, and you know that the multiplicity is the sum of the multiplicity of r mod p times the length of M localized at p for every p in the support of M of maximum dimension. Now this one is bigger than equal, that is, in place of the length, you put what, you know, she talked about, the length, so you still have this form, I say, please, but instead of length, you put the length of this card, okay, so at this point, and remember this is just the smallest power of the maximum ideal, this is a module of finite length, so it's the smallest power of the maximum ideal, it kills the module, and this is what's called, this quantity is what we call, I think, I think Vasconcelo called it like this, the L multiplicity or the levy multiplicity of the module, in our case the module is a module, okay, so let me just really give you a sketch of the group that comes, as the color comes, immediately, you know, what you want it, but generalize also, put the previous work in the literature, okay, and then you use this general projection to put it inside, okay, the corvette, okay, so the proof, and this actually, the first type of the proof, if you want, was really inspired by work of Eisenberg-Urch, you know, you have a, F is the form that defining C, remember, C is an Eiffel surface, okay, and you can see that now this J is just the ideal in S, you know, given by the partial derivative, you have three partial derivatives, and what you can prove, this is an ice ideal, okay, set the coordinate of S, the polynomial, and the point is that, and that's what comes, inspired by work in a certain sense, is that the CZG, module of CZG, so this is CZG conference, of the crucial part of this ideal J, okay, maps on the module of interest, that's really the point that happened because of the hyper surface, okay, so if I want to bound the initial degree of the module of interest, then I just have to bound the CZG, obviously we know very well, even though this is a CZG conference, the bounded CZG is not easy at all, however, in this case, this is an hyper surface, so you can bound it with link catch, because you can write each, you know, row of the matrix just as a link, so this is the initial degree of delta F over delta X1, delta F over delta X2, column J, and now because this is a link, shift it to one, because if it's a link, you can relate it to canonical module, and that's how the A-variant comes in, but now remember we want to bound the A-variant to get to the second formula, and again this is inspired by work of Eisenberg, who reached out to me about the A-variant, and that's how the information of the singularity comes in, so let me name at least here because I don't want to erase the thing there, so we want to bound this A-variant, and we use actually work also, Cavillier-Schindler, to do that, so we have to bound A-variant to bar 1J, we can take them mixed, and now if we want to bound the A-variant, we can just take something that is containing J-variant, and for this, we can look at the intersection of all, this PI, this prime, corresponds to singular points of the curve, so the PI corresponds to the singular points of C, and SI are exactly the levy lengths of our mod PI, our mod J, the Jacobian, general localizer PI, this is the Jacobian ideal now, of the J we were finding, this is the Jacobian ideal of our curve, but now you can consider to bound this the radical of the Jacobian, the intersection of the PI, when you are subtracting one from each one of them, and so, but now the Jacobian gives you the localizer where the thing is not smooth, so what is not normal, so you can add here the conductor, that's what I want, I want the conductor, and the advantage now is that you go and product of ideals, and not just product of ideals, these ideals are ideals of dimension one, you can list it to the regular rate, and then there is this beautiful formula due to Schindler or due to Cavillia, sorry, a Syedman, a Syedman, where the regularity is just, you know, the regularity of the product can be estimated using the sum of the regularity, okay, so, in practically, I don't do all the steps here, but you have to pass a by the regularity which you get, and this is the regularity of R over the sky, this F, and if you do the old path that you had to get with R over the conductor plus, but what do you get? The sum, actually the minus, okay, the sum of the SI minus one, and this is obvious, you get T, that's the cardinalism of the singular locus, minus the L, of R over J, is the sum of the SI, so this one is that, you can use, you know, now the integral closure, and again, this was, that was the aspiration from the work of Isaac Louridge to estimate this, and you can see that this is actually, this term is actually equal to minus the initial degree, first of all, you pass from R to the conductor to the dual of R mod of the conductor, the omega dual, so the appropriate X in omega, and then to approximate this, you go to the integral closure plus D minus three, and the point is that, and here, we need the reducible, otherwise we have to do something else, because you have a field of characteristic, it is algebraic closed, and R is the domain, you know, R bar is only the positive, and R is the negative, and the negative degrees, and degrees zero is equal to R, so this thing is bigger than the problem one, so this is less than equal to minus one, so this is less than equal to minus one, and now you get your formula, so I'm really, I have three minutes according to my clock, since I will be 60 minutes established by girl, and I can finally state the main theorem, that I want to state, so hopefully you can follow this, I may go too fast, but, so you just put together the general projection, that formula, so what do we assume in this case, so the main theorem, I call it the main theorem because this is really in the spirit of Poincare, and so you assume, you see, we don't have any, you call it a goal, you got an assumption, nothing, only reducible for the curve, but we assume that C, it leaves in PN minus one, but has only plain singularity, because you want to do general projection, and you know, and keep track of the singularity, so it's only plain singularity, so this is really what we should now, in the future, get read of, okay, and J now, I call it JR, he distinguished it from before, is the full Jacobian idea, no projection that we had in the first part of the talk, okay, and then the initial degree of the module of interest is bigger than equal, but aim variant plus one, but I promise, plus the cardinality of the singular locus minus the l-mult on R, J, and notice that this part is equal to zero, if and only if, are as C as only, no, ordinary nodes, as singularity, okay, because each one of these counts one, for each singularity, they'll count this one, okay, this generalize the work in the literature, because they did it only for ordinary nodes, not for plain nodes, but they assume first, I think the first thing, first result is really due, I think, is a survival of Lin's nato, I put the name up in the board before, this is the 90s, and Van Buren was the free guy, the free sea, plain curves, Van Buren was the complete intersection, and that was done by Carnichere, Campilus, and D'Afuente, around 2000, and Ben Esteves did it for Comacoli with the general projection, but atmetically Comacoli, and all of this, I don't know the singularity, and we get rid of all this, and we have plain singularity, the point is that, you see, when you do the general projection, because you assume all the plain singularity, you can keep the singularity, you may get more, but the singularity you get are all the ordinary nodes, so the difference here doesn't change, and all, that's a problem, so you have control over that, and then you get the formula for here, thank you. I think it's okay, that case, because that's a 9th of surface, but I, ha! We did check that, but we should, because we can, that's an easy case, and we can go, and the interesting, I think next one, what should one should do is really look at least for plain curves, start with the general zero, because in that case, we have a very good idea of the singularity, just looking at the hyper-verge of the parameterization, you know, of the curve, and that's the work we did with Andy and Cox, that just an hyper-verge called general information, then we can relate it to general cognitive, and we have done many examples, it looks like in that case, you have class because of the number, you have another number, so this number comes down to quality, and exactly in terms of the singularity you get, so that's the first thing we should really pursue, to get the exact formula for the singularity, and then try to be, you know, higher genius, and it's really about, now, finding what the quality working with the plain curve is first, and then we do the general direction. Obviously, if we want to get rid of planes, you know, of this assumption, we need to think about something as a general professional. I think we've already stuff on the very vector fields as being real, that in the complex, did you ask him about the complex number? I think so, yes, I think so, he asked it over the complex number, he's original question, the takers over the complex number, when he talks about these fluids and the things about it, as the curve keeping the actually fluid inside, that's how he talks about it, you know, that once is a violent, disdainful thing inside, so, yeah, I think that's, yeah, because that's what you want to keep until the day of the, yeah. So a lot of this ties into hyperplanaration, so in case with your plain curve, it's actually a lot of mention, or if you were working with hyper surfaces that are hyperplanaration, I'm curious about whether this says anything new in that specific context, it would be interesting, because a lot of these things are very, very well studied in that community. So that would be, and it would be interesting, especially if there's lots of conjectures, so the, you know, the module derivations of hyper surface, the whole industry, the folks who work on that, and Sim has certainly worked in that for a while. The other question I wanted to ask is, do we have that case where you had an aromatic limestone curve, so suppose that curve is smooth and canonical, so when you're looking at that, this must be some sort of classical result, so what are the vectors that make you have a canonical curve, can you say what the vector fields are, could be that curve invariant, and we think that maybe your results say something there, and it should relate to something. In fact, the first thing we did in generalization 2001 is look at the smooth state, but for higher dimensional biotech, because for curves it's all known, and we looked at higher, and we could assume that it was smooth, and some, we could have intersection as smooth, and so far as then, with some assumptions, but they could occur for the smooth state as smooth, and you get this kind of balance, thermo-guide, and then the other thing, also just not for hyper plane dimensions, just for hyper surfaces you have this sort of, longer than the vector fields, so you think of the hyper surfaces that advise around trajectory space, and then like Robert Deets's Algebraic Gram Theorem relates the topology of the complement to the Schoen-Polbann, or the corresponding longer than the tangent bundle, so there's I think a lot of classical algebraic channels that we live in, and we also nicely did the smooth, so, you know, coming from in a gallery. Alright, so many questions, thank you.
 You know, the organizers, so given the possibility to come once again, it's a fantastic place.  And also the conference is going very well, I'm learning so much mathematics again.  Hopefully I can climb a mountain, so it will be very, very good.  So I already brought up the titles and also the collaborators in this project.  They are Marshall Dan, Amida Sundazis, around Simi, Suburna, British.  Now, so this goes back to the problem of Pankare in 1891.  He asked if we can balance the degrees of plane curves that are left in value by algebraic vector fields of P2.  So the goal of the problem, if you want.  So first of all, why did he ask that? I mean, the reason is because if you could bound the degree of the curve,  then you would have to look for this curve because you actually wanted to find them.  You have to look all in a finite measure vector space, so it makes it easier to find such curves.  And what is the goal in general? I mean, is to relate numerical invariance  of the vector field to numerical invariance of the curve where the vector fields leave the environment.  So first of all, the first step I want to do is a translation to commutative algebra  because this problem has been studied so far mainly by algebraic geometers and the fresh geometers.  So these are some of the previous players that actually didn't pass 20 years quite a bit of progress  and they are Campilo, Carneser, Saval, Cruz, Estabes, Galindo, García de la Fuente,  Climbán, Luis Neydo, Pereira, Suarez.  We started this problem actually in 2001 and that's every project we burn, it takes at least 20 years to publish it.  So we had some very good progress in 2001 back in 2001 and then this stuff that we proved was proven by Estabes and Climbán later on.  So we had to restart again and try to prove more and hopefully this time we can publish it before somebody else starts.  So let me fix the setting.  I will have a little simplified setting just because of the talk, so avoid technicality.  So K is an algebraic closed field, this unit, and then we assume that the characteristic of K is zero and these absolutely don't need.  Of course I've listed the topic, but we assume that C in Pn minus 1 is a curve which is reduced, this we always need, and irreducible,  and this I don't need, it's just really doable, it's a good balance.  So I have my curve, I'm going to go stick the maker.  Now either north with R, do you see if I write down here?  Okay, because that's easier given my height.  R and S are the coordinates of C and Pn minus 1 respectively.  So now what is a vector field?  And now I have to start writing here, unless I use this little tool. A vector field X, okay, I think Pn minus 1 of the degree M,  we always call it the degree of the vector M, okay, is an homogeneous map that goes from Z to S, okay, and what is Z?  Z is the kernel, you know, is defined by the Euler's sequence, is the kernel of the Euler map.  So you take the module of differential of S, remember S is a polynomial, so the module of differential is a free module of rank N generated by the partial derivative of Xi.  Xi are just, I mean, so the differential, sorry, Xi are just coordinates, okay,  and what is the Euler map is the map that sends the Xi to Xi, so this goes to the maximum ideal of S,  and this is called the Euler's sequence, because this is the Euler map, and this is just the kernel.  Now, another way to think about Z is that if you look at the shiftification of Z, Z tilde is the shift of the module of differential forms of Pn minus 1.  Now, so this is degree M, means that this is an homogeneous map of degree M minus 1, the map of degree M minus 1, and any such map is a restriction of a map from the module of differential to S.  So such map is just given by a row vector, let's say a 1, a N, so the a i are in S, and they are forms of degree M.  Now, first of all, how can you think now this way about a vector field?  You know, mind what is a vector field, but this and one and N just define the forms, they define a map from Pn minus 1 to Pn minus 1.  So you have a point in Pn minus 1, you look at this image, it's another point, and what's happened?  If the two points are distant, you have a line in between, that gives you the direction of the vector fields at the point.  If the two points are the same, then you say the field is 0, and if the image is not defined, you know, or exactly where is the 0 degree vector field, that are exactly the similar points of z.  Another way you can think, since this is free, you know, another way you can write the vector field, since this is free, you can write it in terms of the dual basis, and this is just the sum of the a i and the partial derivative.  Okay, so that's another way you can write the vector field.  Okay, now, I talk here about vector fields that leave invariant curve.  So what does it mean? I have a curve there. What does it mean that it leaves invariant curve?  You just have to write the same sequence now at the level of r, and you have a commutative diagram.  So if I speak loud, but I have an ear, it doesn't work because of the altitude.  My kids have that problem too.  What? My kids have that problem too.  So I look at the same sequence, the similar sequence at this level, okay, again you have the map, now this is the image of the outside, okay, and obviously there is this commutative diagram.  Now, be very careful because this is our z, usually this map here is not subjective, and what we would do is care is the image, but in my case we assume characteristic zero, that's the reason I assume it, so this is subjective actually.  And what is l? You have a way you can think of l. Another way you can think about l is the image of the second map, you know, the theorem complex, okay, so of the second differential in the theorem complex.  And that would be very useful. Also, what is the theorem complex? Just the Cousin complex built on this map, you know, on the Royal map.  Alright, now, so what does it mean that this leaves the same value? First I want to do a picture, and then I go back to the diagram.  So, let me erase this. For the moment I have the Poincare problem, but after I erase it, okay, so C is left invariant by x, okay,  or another way that the geometry that's saying is that C is a leaf, okay, I will never use this terminology much.  Okay, so what is your image in your head? You have your curve, you want that at every point of the curve, the vector fills the tangent. You have your vector filled, and this should be, obviously the vector fills the find everywhere, but you want that by this tangent to the curve.  So that's the first condition. If you take a point at the curve, the vector of the direction is pointing. The other condition you require is also that at almost all point of the curve, the vector filled is non-zero, okay, so the single, it's not single, at almost all points.  So there are two different conditions. How do we write them algebraically? The first condition algebraically just says that x here, this was the x if you want, so x was a map here, you see, at the level of s, so you had your C, that goes to s and this is your x, and down here you have your L.  Now, you want to induce, x should induce a map mu at this level, okay, so x induces mu, and the second rule is that mu should be non-zero. That just means that the vector fill should be non-zero at almost all points.  Alright, obviously this mu will have the degree be homogeneous and you have the degree m minus one, exactly the same degree as x at, okay, necessary.  Alright, now we are ready for doing our translation, so to convince the algebra of the problem. So I'm ready for it, so I'll start. I don't know, maybe you are familiar with this, so I try to go slower.  So, for the, I look at the, whether we want to, instead to look at the set of curve that are left, at least mu invariance, I look at the set of mu fixed in my C, so what is the set of mu like that, you know?  They are just homogeneous, non-zero, okay, are linear maps from L to R, and actually it's easy to see from the diagram, chase the diagram, that they necessarily, you know, can be extended to the module of differential, omega of R, but can be extended.  What are maps, homogeneous map from the module of differential to R, are derivations? So what we are talking about are just derivations.  So, this is a translation, our object of interest are derivations, and which derivation, but be careful though here, because you want all derivations, but, you know, you want that when you restrict it to L, are not zero, which derivations are zero when are restricted to L?  But L is the kernel, if you remember, was the kernel of the Euler sequence, okay? So any derivation that, if you just remember the Euler sequence, just a second here, I wanted, for you, and we have a little bit here.  So L is the kernel of the Euler sequence, so if I dualize this, I see immediately that what I have to do modulo, because remember I have to dualize it, I'm interested in the map from omega to R, is exactly the module, M inverse times epsilon.  So this is the module of interest, and you're looking at it inside, and to go there, okay? Now, and notice that this is a torsion-free prank, one module.  Sorry, epsilon is the Euler derivation, so it's exactly what you correspond to this map, the Euler map, the Euler, through all the map, through all the talk, and this is going to be very important, the Euler derivation.  See, you have to modulo it out. So now, go back to Punkarec, that's why I left it here. So what did Punkarec ask? He asked if you can bound the degree of plane curve left invariant by the vector field.  And Karneser, and you wanted to bound it using the degree of the vector field, that's why I defined the degree of the vector field.  And already the answer is no, due to Karneser, he was one of the players I wrote up there before, and there is an easy example, but I don't know if that's the original example of Karneser, but this is easy.  You take the family of vector fields of the green one, given by the derivations.  So I write in this way, minus D times Z, the derivative over Z, partial derivative, one minus D, X, the partial derivative with respect to X, okay, for every D in the natural number.  And this vector field, each of these vector fields, leaves invariant the curve, okay, CD, because depends on D, of degree D.  So the degree becomes as big as you want, you know, while the vector field is degree one, and is the curve just given by one equation, so it's an I prefer surface, defined by, and it's easy to see, you can check it, if you're bored, it takes one second, X, D minus Y, Z, D minus one, just to the derivative, you got that, this is zero, okay.  All right, so we cannot expect to find a bound on the curve, on the degree of the curve, just using the degree of the vector fields.  We must use other invariants, okay, so other invariants might play a role, and now we look at opposite question, which is a prevalent, okay, and that's why I wanted to look at this old collection, because in Comistival du  we have a beautiful module, but then it goes all over that. Instead to look at upper bounds on the degree of the curve, I want to look at lower bounds on the degree of the vector fields.  So if I want to look at the vector fields as small as possible degree, I just have to look inside this module at the element of small as possible degree.  The first non-zero element of small as possible degree. So it's like bounding the initial degree of this vector of this module. This concretely encodes all the information I want.  So now I can really restate my goal, and we have a precise module.  So now I can really restate my goal, and we have a precise module.  So now I can really restate my goal, and we have a precise module.  The only reason I present this now, and I start as this guiding thing, is because in this case we can give a complete answer, a clean, complete answer.  So the sample here says as always.  So remember the assumption, C is reduced to the reduced goal characteristic of K0 and algebraic goals.  In addition, in this theorem, I assume that C has a retmetically gorestone.  C is in the end, this one is retmetically gorestone. So I is gorestone if you want, I is the term of the thing of the curve, and as at most ordinary nodes as singularity.  This is the kind of singularities. This is the kind of assumption that is always present in the work, in the previous work of the people.  Then the initial degree of the object of interest.  In this case it's actually R, and inverse is R because it has gorestone, okay, is equal on the nodes to the A in variant of R plus 1, if you want to write the Castelmau monoregularity of R minus 1, the curve, remember.  In general, what can we do, you know, and actually this will guide the rest of the talk. In general, we can find upper bounds for this, which is not the question of Poincaré, but we can find upper bounds for this module.  And for the upper bounds, we have no restriction on singularity at all, but we need arithmetic from a combinatorial.  We can find lower bounds, which is the question of Poincaré. We don't need at all any gorestone or comboconinus for the lower bounds, but we need assumption of singularity, not this assumption, but we see which assumption we need in general.  We need to control or some things about the singularity will come in the formula, okay, and then really in our work, we can do not just curve, but higher dimensional variety.  But for this, I really wanted to focus on Poincaré question. Okay, so I'm going to divide the rest of the talk, and I have one hour because we're at 2.00 an hour.  The rest of the talk is two other pieces, and the first piece will be the upper bound and the other piece will be the lower bounds.  And I'll try at least to give you an idea of the technique that we use, okay?  So in that time, we do prove the sample here. So I completely prove the sample here, but I'll try to do it, give you a normal view of the techniques, which I think is the most interesting thing.  I had promised for examples, and we do have a lot, but most probably I'll just quickly tell you then where we are going for the examples.  Okay, so let me talk about, you know, I think this was true because one was the translation and three is the first upper bounds.  I just talked about one upper bound and three maps, and the three maps are the crucial fit.  Okay, so first let me define one object. I try to give very little definition if I can, but I have to give this one.  So J is the R idea generated by the Jacobian ideal of, if you want, the expansion of the Jacobian ideal in R,  the Jacobian ideal of a complete intersection, a general complete intersection.  Let me call it C i of dimension d minus 2 mapping into R.  Okay.  On to R.  On to R.  Okay, so what are the three maps?  So there exists three more genius maps of the degree zero.  The first one, you know, is the one I talked about.  I said my L was the image of the second differential in the Delan complex. So it comes from the Delan complex.  Okay, so it's just a second differential.  That's confusing when you define the L-pile, I thought you defined it before as a curve.  It's a curve, but it happens to be. It usually is not equal to the image.  But in this case, it's the image because we have characteristics zero. You have to prove it, but it's the image.  It requires a proof. Yeah. Yeah. Yeah, it does.  I said it usually is not, but in this case, it happens to be.  And that's why I wanted to simplify some characteristics zero because we have that in this case.  Otherwise, you have to assume something about the characteristic not dividing the degrees of the form.  And it all you use the image like we do in our field. So we don't have so much to assume.  Okay, so the second map is the map.  And that's why I needed J that goes again from this module top differential.  Okay, into J, epimorphically, shifted of delta minus two.  And I'll tell you what delta is.  And what is this map? So this map does not depend on the fact that this is the module differential just depends on the fact that this module is a rank and the definition of J.  So let's write here why we have that map.  Once I write it will be complete clear. Okay.  So remember that the rank of the module differential.  Okay, is two and the minimum number of generator of the module differential is N.  Okay, and J.  Okay, is generated.  Maximum minor of the matrix consisting of N minus two.  N minus two.  Columns of a matrix presenting omega of a presentation matrix of my mother.  And what is that the map that goes from the from this guy to this guy, you just take the XI by the exchange and you send it to the minor where you delete that or I and the Roj.  Okay, and why the shift, but the form defining the C at the green, let's say delta one, we get many comment that the true to get many quote.  And this is the degrees of the forms defining C.  And delta is the sun.  I'm only minus two. That's why it's a general to start with the largest delta minus one.  Okay, so this is exactly what you get when you send that map because the excited to you and you send it to something of the delta.  So what is the firm?  The firm up there talk about practically because the method goes from the module of top differential forms in this case omega to the canonical canonical module, which is the module of regular differential forms and burn talked about the old time.  Exactly.  So we know who this guy is and he's called a fundamental class.  Now the point is that these two maps are morphism and all these three maps are isomorphic genetically.  So what you get is that once you know that torsion you get free beautiful conclusion from these three maps.  So if you know that torsion you get first that let me start from L model torsion is isomorphic to J.  This is an idea. Okay, and this is a morphic to this guy.  The module of top differential forms when I go down torsion.  And inside here I find the canonical.  Oh, sorry. Sorry.  I have this embed the canonical and that if I do I see it. So I go and I see the picture I see that the dual is isomorphic to the inverse the fractional idea to delta is isomorphic to the dual of this guy.  And then this is the one I was thinking about the dual of the canonical model sits inside here.  And then you get immediately that the module you're interested in.  Okay, this module. Okay, it was sitting in here.  Okay, so it sits inside this.  So it's clear that from here already you can see why this is powerful if you want bounds on the initial degree of this you just have to find bounds on the initial degree of this.  I'm not saying that this is easy, but you know, lower bounds, but you can find low.  So this is important for all bonds and we use it for all bonds.  And then this we use for upper bounds.  Okay, and we see actually the upper bounds come immediately very easy now that we have this picture so I can write them directly.  So this is the fear.  We have more fear than just this one, but this is maybe clean version.  And as I said, you have single isotope make up here at all.  You assume just that C is automatically the order and really you can do without guards and but you don't get.  And I'll talk about this what we write here.  And then the initial degree of the object of interest is always less than equal to the A in value of R plus one.  Or if you want the regularity minus one.  And actually this is completely false.  If you take the constant away, you cannot get this and it's easy to show.  And you just take an actual perfect idea and you use it easy to show that you cannot find upper bound just using the regularity of the invariant.  You need to involve other invariants.  For example, we have one using the type we have and the invariants we have one using the please T and the violence, but other things have to come to play a role.  Okay, so we see where God is that is used here and what you would need to do in general.  So the proof becomes actually quite easy.  As I said, you always start from the universe sequence.  Let's take our starting point.  And if you do allies that.  Okay, we already saw that our module.  Okay, sits inside.  And this is ours.  I just got our senses, come on.  I call it sits inside a star and what the cook on a citizen side, but the cook on a citizen side, the X one of this and are against.  But this is the only as a more effective.  If I go to X two of K and stuff.  Why do I want to do this because this is a sock.  And this is the soccer of our moments one X two shift.  And this X one X two are generally.  So what's the fun that for you just use coming calling us.  Okay, nothing more by general this module that we care about is concentrated.  And this is the real important thing.  In the grease less than equal with a variant of up.  But if it's course then that's not really important.  But because this is just needs coming closer.  So I don't use course.  From yes, this is concentrated by the degree if I pass that degree is true or the same.  So the initial degree.  Of the module of interest.  Is less than one.  The maximum of initial degree of L star and a of R plus one.  And now you also know.  I didn't want to erase that sequence.  So maybe I erase and do something.  I try to do something.  That's okay.  Now if you look at that sequence here, I have this omega is inside.  I'll stop.  So omega look at the dual of mega sits inside.  And stuff.  So if I want to.  Bound the initial degree of L star something inside it.  I better just bound what I am inside it.  Now if I go to then this is our shift.  So this is in the degree R and it starts in the area.  So this guy, this is.  The initial degree.  Of L star is less than.  Okay.  That's it.  Now if you didn't have course then.  You see you had to bound that initial degree and that's where.  Types coming in and stuff.  Okay.  Let's do the last part of the talk.  And hopefully I don't know so much over time.  And I want to talk about the interest of one career and where the history actually the work of that are people lays on which is lower bounds.  And here we have to somehow use some idea from geometry and we'll use general protection.  Okay.  Okay.  Okay.  Okay.  So you have your beautiful curve and P-1 and what you want to do you want to project it down.  To project it to the plane.  You just have an item surface and usually the problem is not easy to compute the you know the not not so much easy.  To compute your module in terms of the boundary of the initial degree in terms of inviolence when you're playing curve.  The problem is and that's the problem with geometry.  How do you translate your bound that you're typing the plane to the bound from your curve as 6 and P-1.  And there are lots of difficult to do that and they do as you come across in a very complicated way.  We don't get some such easy translation and we are able to do that and we don't need to call it.  And it becomes very beautiful to go up and forth.  That's why you can get up and down the lower bound this way.  Okay.  So let me write our statement and then the problem is just to compute the module of interest for a play for a night surface.  So this is the last chapter of the talk and this actually we have to thank over both of because some of these three came where the last three weeks.  So let's say lower bounds and general projections.  So here this is about the general projection.  You have your ring R.  Remember your ring R is the coordinate ring of C which is a curve in P-1.  And what do you do with this general projection?  You just consider three forms, general linear forms in R.  Or mine.  It's not mine.  General linear forms in R and we can see that the algebra generated by this, this six and side R, obviously, we call A.  This map is a finite, birational, okay, homogeneous extension.  And then what you have is that A is the coordinate ring of a curve in P-2.  So A is got a stem, the nitrous surface, okay, of the same degree as your curve C, okay, the finite value of the same degree as the curve we started with.  So what is the connection between the two module of interest, the scenario of two module of interest?  The derivation of A, module of R, excellent, A, and the module of derivation of R, module of M, M plus, fancy.  There is a spalling, let's try to do this, when the initial degree of our module of interest, let me put an iron, the above all the iron is needed, okay,  is bigger than equal than the initial degree of the module of derivation of A.  Now this is the order, oh, no, sorry, A, sorry, and I need A not a inverse because it is the same, and then I have minus the A in variant of A plus the A in variant of R.  And it is also less than equals to the same degree of the same module of the A in epsilon and then plus A of A minus A of R, okay, so you get this beautiful thing,  and then you can move the stuff up, stuff it up, you can get up and lower, they are really interesting now, lower, you can do both.  Correct, so how the proof, the sketch of the proof is, but you consider the conductor of A of R is A, this is the column, yes, and this is nothing else than the column, and this is nothing else than the column, 9 to 8,  but remember that A is cost, so this is the canonical module of R, shifted by the invariant of A, and now, so what is the initial degree of this conductor,  is that this is degree of R, so this A of A minus A of R, and this is epsilon degree, okay, so this is this quantity that you see here and here.  Now why there is that quantity, because you can embed, you have to pass the third module in the middle, so you can take the module of derivation of R into K, and you have a restriction map here,  you can just restrict all this derivation to A, so you get the module derivation of A to R, as you restrict them to A, okay, so this is the restriction map.  And then you can also embed, obviously, the derivation of A, these are just particular derivation.  Now the point is that remember we have to be quotient, our object of interest is quotient, but the Euler derivation goes to the, this is Psi, let's call it, the Euler derivation goes to the Euler derivation,  and what's more important is that once I mold out this guy, this is torsion free, so what you get is still an inclusion, and this is epsilon R,  and this is still the same and inverse epsilon A now, as it goes to the derivation of A, and this is R, A, sorry, epsilon.  Now, so what does, furthermore, remember these are all torsion free now, they're all modules, furthermore you can easily show that with my conductor kills both co-currents, that's not difficult to see,  and this implies immediately that if I take the conductor and multiply it by an element, here it goes to finish here.  So this tells me that if I take the initial degree of this module, plus the initial degree of the conductor, because I take the element of the conductor and multiply it by an element here, I finish here.  It ruins your element because these are all torsion free, so the initial degree of the conductor, this is, I get an element there, so this is bigger equal than the initial degree of this guy,  because I get an element there, and then you can play the same role with the other one, and then you get the balance you want.  Now, we are left now, okay, to look at the, and I still want to do that, to look at the case as I said, where it took an hour, so I can do it, I knew I would get to solve the case of plate curves,  but it's very interesting, so I would like to describe that also because I can show you how the results of previous people have been generalized, and that's really what we want to get to.  So, this is really the plane curve case, so here, plane curves, okay, so just that, now we're currently at the two of the degree D.  Okay, and the initial degree of our module, we're interested in this case, it's odd because it's an hyper surface, as I call it,  is bigger than equal than the minimum of D minus 2 and 2 D minus 2 minus A of r mod j, the A variant of r mod j, j is the Jacobian ideal of r, the fungial Jacobian ideal of this case,  and this is bigger than equal, if you estimate this A variant of D minus 2, this notice that is what we had before, is the regularity minus 1 of the A variant plus 1, plus, and here it comes, the singular locus,  the singular locus of C, this is the cardinality, so the singular points, the number of singular points, minus the levy multiplicity of this r mod of the Jacobian.  Now I'm going to define what the levy multiplicity is, and if you have any module M, you can consider, I mean, graded or whatever,  you can consider the associativity formula for multiplicity, and you know that the multiplicity is the sum of the multiplicity of r mod p times the length of M localized at p for every p in the support of M of maximum dimension.  Now this one is bigger than equal, that is, in place of the length, you put what, you know, she talked about, the length, so you still have this form, I say, please, but instead of length, you put the length of this card, okay,  so at this point, and remember this is just the smallest power of the maximum ideal, this is a module of finite length, so it's the smallest power of the maximum ideal, it kills the module, and this is what's called, this quantity is what we call, I think,  I think Vasconcelo called it like this, the L multiplicity or the levy multiplicity of the module, in our case the module is a module, okay, so let me just really give you a sketch of the group that comes, as the color comes, immediately,  you know, what you want it, but generalize also, put the previous work in the literature, okay, and then you use this general projection to put it inside, okay, the corvette, okay, so the proof, and this actually, the first type of the proof, if you want,  was really inspired by work of Eisenberg-Urch, you know, you have a, F is the form that defining C, remember, C is an Eiffel surface, okay, and you can see that now this J is just the ideal in S, you know, given by the partial derivative,  you have three partial derivatives, and what you can prove, this is an ice ideal, okay, set the coordinate of S, the polynomial, and the point is that, and that's what comes, inspired by work in a certain sense,  is that the CZG, module of CZG, so this is CZG conference, of the crucial part of this ideal J, okay, maps on the module of interest, that's really the point that happened because of the hyper surface, okay,  so if I want to bound the initial degree of the module of interest, then I just have to bound the CZG, obviously we know very well, even though this is a CZG conference, the bounded CZG is not easy at all, however, in this case, this is an hyper surface,  so you can bound it with link catch, because you can write each, you know, row of the matrix just as a link, so this is the initial degree of delta F over delta X1, delta F over delta X2, column J,  and now because this is a link, shift it to one, because if it's a link, you can relate it to canonical module, and that's how the A-variant comes in, but now remember we want to bound the A-variant to get to the second formula,  and again this is inspired by work of Eisenberg, who reached out to me about the A-variant, and that's how the information of the singularity comes in, so let me name at least here because I don't want to erase the thing there, so we want to bound this A-variant,  and we use actually work also, Cavillier-Schindler, to do that, so we have to bound A-variant to bar 1J, we can take them mixed, and now if we want to bound the A-variant, we can just take something that is containing J-variant,  and for this, we can look at the intersection of all, this PI, this prime, corresponds to singular points of the curve, so the PI corresponds to the singular points of C,  and SI are exactly the levy lengths of our mod PI, our mod J, the Jacobian, general localizer PI, this is the Jacobian ideal now, of the J we were finding, this is the Jacobian ideal of our curve,  but now you can consider to bound this the radical of the Jacobian, the intersection of the PI, when you are subtracting one from each one of them,  and so, but now the Jacobian gives you the localizer where the thing is not smooth, so what is not normal, so you can add here the conductor, that's what I want, I want the conductor,  and the advantage now is that you go and product of ideals, and not just product of ideals, these ideals are ideals of dimension one, you can list it to the regular rate,  and then there is this beautiful formula due to Schindler or due to Cavillia, sorry, a Syedman, a Syedman, where the regularity is just, you know, the regularity of the product can be estimated using the sum of the regularity,  okay, so, in practically, I don't do all the steps here, but you have to pass a by the regularity which you get, and this is the regularity of R over the sky, this F,  and if you do the old path that you had to get with R over the conductor plus, but what do you get? The sum, actually the minus, okay, the sum of the SI minus one,  and this is obvious, you get T, that's the cardinalism of the singular locus, minus the L, of R over J, is the sum of the SI, so this one is that, you can use, you know, now the integral closure,  and again, this was, that was the aspiration from the work of Isaac Louridge to estimate this, and you can see that this is actually, this term is actually equal to minus the initial degree,  first of all, you pass from R to the conductor to the dual of R mod of the conductor, the omega dual, so the appropriate X in omega, and then to approximate this, you go to the integral closure plus D minus three,  and the point is that, and here, we need the reducible, otherwise we have to do something else, because you have a field of characteristic, it is algebraic closed, and R is the domain, you know, R bar is only the positive,  and R is the negative, and the negative degrees, and degrees zero is equal to R, so this thing is bigger than the problem one, so this is less than equal to minus one, so this is less than equal to minus one, and now you get your formula,  so I'm really, I have three minutes according to my clock, since I will be 60 minutes established by girl, and I can finally state the main theorem, that I want to state, so hopefully you can follow this, I may go too fast,  but, so you just put together the general projection, that formula, so what do we assume in this case, so the main theorem, I call it the main theorem because this is really in the spirit of Poincare,  and so you assume, you see, we don't have any, you call it a goal, you got an assumption, nothing, only reducible for the curve, but we assume that C, it leaves in PN minus one, but has only plain singularity,  because you want to do general projection, and you know, and keep track of the singularity, so it's only plain singularity, so this is really what we should now, in the future, get read of, okay, and J now, I call it JR, he distinguished it from before, is the full Jacobian idea,  no projection that we had in the first part of the talk, okay, and then the initial degree of the module of interest is bigger than equal, but aim variant plus one, but I promise,  plus the cardinality of the singular locus minus the l-mult on R, J, and notice that this part is equal to zero, if and only if, are as C as only, no, ordinary nodes, as singularity, okay,  because each one of these counts one, for each singularity, they'll count this one, okay, this generalize the work in the literature, because they did it only for ordinary nodes, not for plain nodes, but they assume first, I think the first thing, first result is really due, I think,  is a survival of Lin's nato, I put the name up in the board before, this is the 90s, and Van Buren was the free guy, the free sea, plain curves, Van Buren was the complete intersection, and that was done by Carnichere, Campilus, and D'Afuente, around 2000,  and Ben Esteves did it for Comacoli with the general projection, but atmetically Comacoli, and all of this, I don't know the singularity, and we get rid of all this, and we have plain singularity, the point is that, you see, when you do the general projection,  because you assume all the plain singularity, you can keep the singularity, you may get more, but the singularity you get are all the ordinary nodes, so the difference here doesn't change, and all, that's a problem, so you have control over that, and then you get the formula for here, thank you.  I think it's okay, that case, because that's a 9th of surface, but I, ha!  We did check that, but we should, because we can, that's an easy case, and we can go, and the interesting, I think next one, what should one should do is really look at least for plain curves, start with the general zero, because in that case, we have a very good idea of the singularity,  just looking at the hyper-verge of the parameterization, you know, of the curve, and that's the work we did with Andy and Cox, that just an hyper-verge called general information, then we can relate it to general cognitive, and we have done many examples, it looks like in that case, you have class because of the number, you have another number, so this number comes down to quality,  and exactly in terms of the singularity you get, so that's the first thing we should really pursue, to get the exact formula for the singularity, and then try to be, you know, higher genius, and it's really about, now, finding what the quality working with the plain curve is first, and then we do the general direction.  Obviously, if we want to get rid of planes, you know, of this assumption, we need to think about something as a general professional.  I think we've already stuff on the very vector fields as being real, that in the complex, did you ask him about the complex number? I think so, yes, I think so, he asked it over the complex number, he's original question, the takers over the complex number, when he talks about these fluids and the things about it, as the curve keeping the actually fluid inside, that's how he talks about it, you know, that once is a violent, disdainful thing inside, so,  yeah, I think that's, yeah, because that's what you want to keep until the day of the, yeah.  So a lot of this ties into hyperplanaration, so in case with your plain curve, it's actually a lot of mention, or if you were working with hyper surfaces that are hyperplanaration, I'm curious about whether this says anything new in that specific context, it would be interesting, because a lot of these things are very, very well studied in that community.  So that would be, and it would be interesting, especially if there's lots of conjectures, so the, you know, the module derivations of hyper surface, the whole industry, the folks who work on that, and Sim has certainly worked in that for a while.  The other question I wanted to ask is, do we have that case where you had an aromatic limestone curve, so suppose that curve is smooth and canonical, so when you're looking at that, this must be some sort of classical result, so what are the vectors that make you have a canonical curve,  can you say what the vector fields are, could be that curve invariant, and we think that maybe your results say something there, and it should relate to something.  In fact, the first thing we did in generalization 2001 is look at the smooth state, but for higher dimensional biotech, because for curves it's all known, and we looked at higher, and we could assume that it was smooth, and some, we could have intersection as smooth,  and so far as then, with some assumptions, but they could occur for the smooth state as smooth, and you get this kind of balance, thermo-guide, and then the other thing, also just not for hyper plane dimensions, just for hyper surfaces you have this sort of,  longer than the vector fields, so you think of the hyper surfaces that advise around trajectory space, and then like Robert Deets's Algebraic Gram Theorem relates the topology of the complement to the Schoen-Polbann, or the corresponding longer than the tangent bundle, so there's I think a lot of classical algebraic channels that we live in,  and we also nicely did the smooth, so, you know, coming from in a gallery.  Alright, so many questions, thank you.
201806261031-Polini_lrv_0_95.png 201806261031-Polini_lrv_96_167.png 201806261031-Polini_lrv_168_935.png 201806261031-Polini_lrv_936_9215.png 201806261031-Polini_lrv_9216_9323.png 201806261031-Polini_lrv_9324_15479.png 201806261031-Polini_lrv_15480_17003.png 201806261031-Polini_lrv_17004_17339.png 201806261031-Polini_lrv_17340_19343.png 201806261031-Polini_lrv_19344_21731.png 201806261031-Polini_lrv_21732_22535.png 201806261031-Polini_lrv_22536_28643.png 201806261031-Polini_lrv_28644_31679.png 201806261031-Polini_lrv_31680_33791.png 201806261031-Polini_lrv_33792_35399.png 201806261031-Polini_lrv_35400_38411.png 201806261031-Polini_lrv_38412_38951.png 201806261031-Polini_lrv_38952_43259.png 201806261031-Polini_lrv_43260_43595.png 201806261031-Polini_lrv_43596_56963.png 201806261031-Polini_lrv_56964_57011.png 201806261031-Polini_lrv_57012_57167.png 201806261031-Polini_lrv_57168_57971.png 201806261031-Polini_lrv_57972_64751.png 201806261031-Polini_lrv_64752_66347.png 201806261031-Polini_lrv_66348_75863.png 201806261031-Polini_lrv_75864_82379.png 201806261031-Polini_lrv_82380_82439.png 201806261031-Polini_lrv_82440_82799.png 201806261031-Polini_lrv_82800_90107.png 201806261031-Polini_lrv_90108_91091.png 201806261031-Polini_lrv_91092_93947.png 201806261031-Polini_lrv_93948_95111.png 201806261031-Polini_lrv_95112_98417.png