E Artin's 'Über die Zerlegung definiter Funktionen in Quadrate' (part 1)

0:00 Before starting this morning, we need to decide what will be the schedule. So the session about the project is cancelled. And the question is, do we want to have a long pause, after lunch, 4 to 3, or 5 to 5, or do we prepare to have a session on TV now, just after lunch, just after lunch at 3, and then to have a longer break at the end of the atom. So, we need to vote. So who is in favor of changing the schedule so that Trivine does instead of the wait sheet? Who wants to chant? Who wants to chant? Who wants to chant? The chant means that we'll have a session from 3 to 6 to 4.30, and then with 3 after. There you go Mr. Turn me back up here that's all I can hear. I asked you what you want downstairs, Sir, Sir, you want me to turn your over. The vote was offending. Changing. That's it. Three at last. Half at last. Two to three at last. Three at last. Four at last. That's what Daniel. We don't move. Four. You don't move. Seventy. No, 17. Five. Okay, so again, for... This is one, this is two. One time. who is the failure of what ? Who is the failure of what ? Who is the failure of what ? Who is the failure of what ?

2:30 Let's toss a coin. On aurait dû commencer quand même. We toss a coin. Which was our new solution from the beginning? Second vote, or to second? Ah oui, l'enfer, on fait déjà une vote comme on complète. Thank you very much. So, today's session's paper is a paper by Real Algebraic Manifolds by Nash. So, in this paper, firstly, he defines this notion of an abstract real algebraic manifold. And then he proves an approximation result. Well, first of all, let's see what is an abstract real algebraic manifold. So it's a real analytic manifold, except that this manifold has a ring of analytic functions on it, and one selects a sub-ring of these real analytic functions, which he calls the ring of algebraic functions. And this subring has to satisfy a set of axioms. So these are axioms that you would think algebraic functions should satisfy.

5:00 So each one of these functions should be real analytic functions defined on the manifold. And if you have a set of functions having cardinality greater than the dimension, then they should satisfy some polynomial relation. And also, there has to be a certain, at least one basic set of functions in R, which allows you to embed, or embed, rather, the manifold in an Euclidean space. And R has to be a maximal set of functions satisfying this condition. So these are the four actions. So a real analytic manifold with a set of analytic functions satisfying these four axioms, this pair would be called an abstract real algebraic manifold. So in fact, every analytic manifold is an abstract algebraic manifold. Because... Excuse me, where did you find the word abstract? Oh, well, not abstract. Okay, sorry. Well, there is actually. Let's see. Yeah, so if you look in page 406, the third paragraph, it says the ring associated with a real algebraic manifold regarded as an abstract algebraic object. So it doesn't call it an abstract manifold, I guess, from the fact that you can define a manifold just by sheath of functions. Okay, so since by axiom number B, a real algebraic manifold, inside the ring there is a basic set which allows you to embed the manifold in an Euclidean space. So such an embedding of the manifold into an Euclidean space by algebraic functions is what he calls a representation of the algebraic manifold.

7:30 And sometimes, by abuse of notation, he will call the image of the representation to be the representation. Okay, so now, as I said, every analytic manifold is a real algebraic manifold, so there is nothing particularly, so the notion has nothing to do with real algebraic varieties. So the connection with real algebraic varieties comes from the fact that if you have this representation so if you embed the function using these real algebraic functions then the image of such a representation would always be a part of a real algebraic variety now what part or what kind of part this is is what leads him to the definition of what you call a sheet so this is in the page 406 the bottom. So a sheet of a real algebraic variety is defined as a certain subset having following properties, namely any two points in this subset should be joined, can be connected by an analytic path lying inside the subset, real analytic path. And this subset has to be a maximal subset of the varieties satisfying this condition. There is a third condition which he says he doesn't know whether it's necessary and I also we also couldn't figure this out that the sheet should also have the property that there should be at least one point on it which is which is a which is a full neighborhood of the variety contained in the sheet. So a sheet, if the sheet is isolated from the rest of the variety, then of course it's a connected component of the variety, but it need not be one. So, what Nash proves in his main theorem, well, firstly, it shows that if you are given a real algebraic manifold, then there exists a representation of it,

10:00 which is an isolated sheet of a variety. so what I mean is that the image of there exists an algebraic embedding says that the image of the embedding is an isolated sheet of the variety and it calls such a representation a proper representation so that's in fact not stated as a theorem but it is part of the proof of the main theorem the main theorem theorem, which is theorem 1 on page 409, says that if you have analytic embedding or a differentiable embedding of a manifold in an Euclidean space, but for the purpose of this paper, whenever I talk about manifold, it's assumed to be a closed manifold, so a compact manifold without boundary. So if you have, and real analytic. So if you have a differentiable embedding of such a manifold in the Euclidean space, then this embedding, so, I should write something. So, you have your abstract manifold N, and there's an embedding into the Euclidean space, which is differentiable, then of course the image of the embedding need not be an algebraic variety, real algebraic variety there, or has anything to do with real algebraic variety, but there exists and there's a real algebraic embedding of M into the Euclidean space C, and you can choose C to approximate phi as closely as you wish. So this means that not only does C approximate phi as a map, it also, the derivatives of phi and C are as close as you want. So this is the notion of every order of every order yes the the approximation is uniform uniform yes so but this the image of this algebraic embedding K is only guaranteed to be a

12:30 sheet and it's not necessary I mean it's you cannot prove that the this sheet is an isolated sheet. But in course of proving this, it does prove that there exists another embedding of N, let's say, C1 into some bigger dimensional space, and in fact this R here is just the dimension of the manifold. And this, so this is no longer any approximation because it goes into a different space but the image of this embedding is an isolated sheet of a certain real algebraic variety. So this the second statement is something that is proved inside the proof of theorem 1. Okay so are there any questions of what the so this is an approximation result that if you have any embedding there's always an algebraic approximation to it and but the image of this algebraic embedding is just a sheet of variety and doesn't claim that it is isolated so it might there might be other components of the irreducible variety containing that sheet that meets it so there's no guarantee that's isolated but of course approving this it does prove that you can algebraically embedded into some other Euclidean space of bigger dimension but the embedding, the image of the embedding is isolated sheets, so in particular, it is a connected component of a realized pipe set. Okay, so, in course of proving theorem one, he needs some, need some auxiliary facts from theory of manifolds. So the first fact that it needs is that if

15:00 you have analytic manifolds which is embedded in some Euclidean space, then you can, there There always exists a small enough, so again, manifold is always a compact manifold without a boundary. So there always exists a small enough tubular neighborhood of it in the Euclidean space such that inside this tubular neighborhood, the nearest point map is well defined and is in fact an analytic map. So there already exists a small enough tubular neighborhood such that inside this neighborhood the map sending a point x to the nearest point y of x of this manifold is well defined and is an analytic map. So essentially the idea would be that you want to approximate this analytic math by polynomials. and the so that's a well-known fact which improves with lemma 1 and lemma 2 is is the fact that you can approximate analytic functions inside some compact the domain by polynomials and approximate in the sense that you approximate the derivatives uniformly. So, these are lemmas 1 and 2, which I think are quite classical results which one studies in differential geometry, for instance. And before we talk about the proof of the main theorem, there is certain other auxiliary facts that he needs, which he has put in a section in the end,

17:30 calling some preliminaries. It's kind of strange to have the section on preliminaries to be the last section of the paper. these results in this section has nothing much to do with approximation so theorem 5 for instance says that so if you have an algebraic representation of a manifold. Oh, by the way, so probably I should verify a few things. So, first of all, if you have an algebraic manifold, namely an abstract real analytic manifold and with this ring, so this pair. Now, this ring is certainly not unique. So, of course, if you have a ring of algebraic functions, then if you take any dipheomorphism of the manifold to itself and then compose the functions R with these dipomorphisms and you will obtain another ring. So this ring is not any, there is no uniqueness except that if you have two such rings then there always exists the dipheomorphism from M to itself, which induces an isomorphism between the two rings in the other direction. So these chosen rings are unique up to this sort of dipheomorphisms. And in fact, it shows that if you have two manifolds which are, so two real algebraic

20:00 manifolds. So, in fact, this is the notion of equivalence. So, we'll call two real algebraic if there exists some dipomorphism from M1 to M2, which induces an isomorphism from R2 to R1. So this is a natural notion of equivalence. And it also shows that if you add two such real algebraic manifolds, which are diffeomorphic to each other in the ordinary sense, then they are also equivalent as real algebraic manifolds. So, the real algebraic structure does not produce anything strange, so if two manifolds are diffeomorphic, then they are also equivalent as real algebraic manifolds. Okay, so let me mention two of the auxiliary theorems that are used in the proof of the main theorem. So there's, I think, this is called the theorem 5, which is that if you have representation of a real algebraic manifold, so you have an embedding of M into some EN using the algebraic functions of this ring, then the image of the embedding must be a sheet of an irreducible variety having the same dimension as E. So if this is a representation, then C is a

22:30 of dimension, equal to the dimension of the manifold. So probably I should sketch the proof of this fact. So, okay, so the fact that this is a representation, so first of all, let's say that the dimension of M is R, and the fact that this is a representation of the real algebraic manifold means that if you look at the coordinate functions here, x1 to xn, then every R plus one of these functions must satisfy some polynomial relation, non-trivial polynomial relation. Every such r plus 1 tuple of these coordinates choose one of these relations, so some polynomial, and consider the variety, the real variety defined by all these polynomials. So for every r plus 1 tuple, you choose a polynomial relation that must be satisfied on P and you let V0 be the variety that is defined by all these relations so clearly V0 contains the image of P . And now so either V0 is irreducible or If you can decompose it into two, you can express it as a union of two varieties, and see if since phi of m is closed, it's not too difficult to see that if v0 is equal to v1 union v2, then phi of m must be contained in v1 or v2.

25:00 And so you can go down, and since you cannot have an infinite chain of... How do you see that? You say it's not too difficult to see. Let's see. I can't see it immediately. Well, last night I convinced myself it's not too difficult to see. Analytical continuation. Yeah, I guess so. So suppose, so suppose P1 is a, so let's look at the difference, points of P of M outside like V1. If this is empty, then it's contained in V1, or, let's see, if this is the point of V1, how to... If there is a... But if it cuts the two, it is... No, no, I see. I probably should look at the points of... so this is some open so if this is empty then we are done so if this is not empty then it is some open subset of V1, it is some open subset of V2, and so then any polynomials that vanish on this open subset of V2 must also vanish there, right, by just analytic, some kind a penalty continuation so this means that so what does that mean that means that every polynomial that vanishes

27:30 and that also and that was the continuation and so then it any polynomial that let's write it down so if P is 0 on V2 then Yes, whenever we are talking about manifolds, I've only been connected manifolds. Well, so P is 0 on P of M. Okay, so this will give you some variety, irreducible variety which contains P of M. and now we have to show that the dimension of this variety is the dimension of the manifold well if the dimension of this variety is bigger than the dimension of M then this means that there must exist some so let's say the dimension of M is R. So if the dimension of the variety is bigger than the dimension of M, then there would have to exist R plus 1 independent algebraic functions on this variety. But this is not possible because we chose V0 to be the variety defined by every, by the definition of V0, every R plus 1 on

30:00 dependent so you cannot have more than R of this so then the dimension is of the variety of which P of M is a part is the dimension of R and to see that in fact P of M is a sheet you need to look point of P of n. So a general point is just a point which has local dimension the same as dimension of n. So if you take a general point on P of n then there exists R plus, well, so you can find R algebraically independent functions which serve as coordinates of some neighborhood of x in phi of n. Now, these functions should also satisfy, we should also form a parameterization of a possibly smaller neighborhood of the variety. And so, first of all, if you have an analytic path, that connects X to some point on the variety, then this analytic path has to stay inside this smaller neighborhood of P of M, which is also a neighborhood of B, and so since the path is analytic, it cannot leave this neighborhood. Sorry, since its content is never, since the part is analytic, it cannot leave B, since C of M is an analytic manifold. So that means that if any point in V that you can connect by an analytic path starting

32:30 an X can, in fact, be connected by an analytic path that stays inside the field. The X is, you can choose X to be a general point, so X is, oh sorry, but you have to proof that X you choose X relation but in order to prove that it is a sheet you need to to prove that you can connect every point of... It's connected, you have to prove that it's maximum, and that's because the back can be hard to prove or not for it. So this I think is here in five. I think that in the definition of shift points A and C are in the arrival of each other, So let's give an analytic two points to the pi of M.

35:00 There is an analytic pass, but because you just take an analytic pass on M and push it by V. The property holds on M. Any two points on M can be joined by an analytic pass. And then you can just push this pass by P, so that's OK for A. That's okay for A, but one needs to show that it's maximal, so... The problem is it's maximality? No, if you have a part of a path inside B, then it can never leave B because the path is analytic and B is an analytic in B. So if a part is inside B, then the whole path should be in B. That proves maximality. in the paper, I should have used this. There's a point, yes. Nice job. Yeah, and so since, so what is confusing me though, oh and so and such a analytic part should always pass to a general point it's because your points are Okay, so, so that's the argument that the one paragraph on top of the page, which page? Yeah, yeah, that's the...

37:30 Yeah, it's used that... It's actually a compact analytic manifold. Well, so B is a compact analytic manifold because it's an algebraic embedding of something that's already an analytic manifold. So that if there is some interval of the part that's lying inside B, then the whole part. and so this so now we know that if you have a representation which is not so P might be other parts which are singular so all we are saying is that the M inside V is a sheet, but we are not, and we are saying that the dimension of V is at most R, but V might have parts which are singular, because we do not know about anything else about V. The sheet is, of course, a compact manifold. The sheet that we are still bending is a compact analytic manifold. Yeah, and that one is . That one is . So that's the problem with, well, not a problem, but one should be careful about the proof that the approximation theorem are in the same dimension. and that's one, says that the image, so if you have an analytic embedding of M into an Euclidean space, then you can choose arbitrary, close to it, an algebraic embedding, but the image of the embedding is only guaranteed to be a sheet. So it's part of a real algebraic variety, which can be singular. Okay, the other part that we need that so this says that if you have an algebraic representation then what you get is a sheet and the other part is that so this is theorem 6 which says

40:00 So, if you have an analytic embedding, Embedding. Such that T of M is a sheet So, if you have an analytic embedding, then you know that the image is a sheet of some real variety, then in fact, this is an algebraic representation. Thank you. So I guess, so the fact that you know that phi is, the image is a sheet, tells you that that if you take the coordinate functions on here, then, then every, and suppose that then every r plus one of these functions are dependent on the on P of m and you

42:30 So I guess you can use the pullback of these functions on M via phi to define the embedding, which will give you the algebraic representation. So, in fact, if you take P of M and the algebraic functions on P of M, which are restrictions of algebraic functions here, then this gives you the real algebraic structure on it. So if you have an embedding such that you know that the image is a sheet, then you already have the algebraic representation. So for whom we should discuss now the theorem and see where these two are applied. So I'll show the proof of theorem 1, which is theorem 1 is on page 409, where it says that any differential embedding d of a closed differential manifold may be approximated by an algebraic representation of the manifold.

45:00 So far, the ring R, which makes the manifold a real algebraic manifold, is not chosen. But as he said, in the last part of the paper where the preliminaries are, it says that any ring I can put as assigned to the manifold makes it the same real algebraic manifold. So the ring is not important. The theorem is done for a convenient ring. And that will be OK. So the situation is, let me do some pictures. Suppose, for instance, this is the manifold D, the dimension of D to this number, it's called R. And it's invariant in this number is n. I have my neighborhood, a dead neighborhood, the tubular neighborhood n. And for each point x in the neighborhood, there is a point y of x in the manifold such that this is the closest point in the manifold to x this point should live should lie in a in a line this line is called K. This line is normal to the manifold. This is the situation in this case. Let's do a picture. If the dimension is 1, I have the tubular neighborhood. For each point here, x, there is a closest point, and this point, well, it's true again that it lies in a line perpendicular to the manifold passing through x, but we will work this time with a plane

47:30 k which passes through x and y of x, and it's normal to the manifold. Okay, so the dimension of this plane is always I minus R. Okay? I should make a remark that he also makes this remark that the proof is much simpler if you assume that the normal bundle of this given manifold, suppose you have given an embedded, already embedded manifold, such that the normal bundle is trivial. So you can choose a global system of frames which gives you coordinates of the normal bundle. In this case, you'll have, say, a global set of functions, y1 to yn minus r, which gives you the coordinates of the points of a particular neighborhood. So this gives you a frame of the normal bundle, which you can choose globally. And then of course, you can approximate these. So the manifold itself is then defined where Y1, Yn minus 1, these functions are zero. So in this case, the manifold is the intersection of these many hypersurfaces. and so the natural thing would be and these functions are globally defined in the tubular neighborhood so if you can approximate these functions by polynomials then you would probably get and you look at this set then you would expect that this will define something that's close to the manifold and so this is I think the proof of this restricted class of manifolds by Seifert that he refers to this he says on page 407 paragraph 3 or 4 a paper 2 by Seifert that this condition

50:00 Conwell is trivial. I think it's a strong topological condition. It would imply some banishing of some characteristic Euler class or something, so it's a much restricted situation than what he deals with in this space. But because of the fact that we are not assuming this, we have to deal with matrices and so the The proof becomes a little more complicated, but the idea mainly is this, that you can approximate these functions by polynomials. But when you say that the line between X and Y is a normal choice. So X and YX belongs to the same normal plane. So the nearest point map has the property that if Y of X is the image of X, then X belongs to the normal plane at Y of X so what here there are some objects we have the manifold so there's the embedding D of the manifold point X point y of x which is the nearest point we have this vector v equal y of x minus x and we have the plane k and we have let's define the the linear function k which is the project the orthogonal projection on the plane k okay so these are like objects that appear in the drawing but now we are gonna we're going to approximate each of these objects by an algebraic function so k can be thought of as a matrix in the global coordinates of here yes it's an n by n matrix which is a projection matrix the matrix of the projection so first we have this vector

52:30 And we approximate it by a vector u of x. These are the approximations, which will all be algebraic functions. Polynomians. So this is a polynomial. Then there are going to be some algebraic functions. so here this vector of each coordinate of this vector approximates each coordinate of these vectors and it approximates it approximates as much as you want and also the derivatives of U approximates the derivatives of B So, first I have to say that we can suppose that D is analytically embedded, because otherwise there is a previous theorem, due to Whitney, which says that you can suppose so, and if This analytical embedding approximates the differentiable embedding, and now we approximate the analytical embedding by an algebraic embedding, then our algebraic embedding will approximate the differentiable embedding. So, did you ever look at the output? No. No, I couldn't get the output. Well, I didn't, but it's, well, he says that it's the same idea of approximating a differentiable function by an analytic function. So now we know that there is a partial result like that, and now there is also a general result that is made for the going from the . Is it possible to identify the main varieties in Ashford? Well, as opposed to Seifert, he is that he deals with the case when the normal bundle is not necessarily orientable, but compared to Whitney, I don't know. That's so interesting.

55:00 I can't say it's unlikely though that deals with in this generality. Well, I shouldn't say that. But I think it's unlikely that witness paper deals with in this generality, that all manifolds can be approximated. so this vector approximates this vector and now we want to approximate this operator this matrix so this is a symmetric matrix because it's the matrix of a orthogonal projection so if I approximate each of its coordinates I'm going to get matrix L which is again symmetric but as this is an approximation then this matrix maybe not have may not have the same rank as matrix k so we have to do something with this matrix to get a matrix with the same rank rank why do I want a matrix with the same rank because the image of this matrix is k and what I really want to approximate is the plane k so I want to get to get an approximation of this plane as the image of some matrix approximating this matrix K? Is it clear?

57:30 The point is that K is a projection matrix which projects into a plane of dimension N minus R, I guess. So that means it has R eigenvalues which are 0 and let's say N minus R eigenvalues is 1. But if you approximate K by some polynomial, which is close to K, you are not guaranteed. So it will have R very small eigenvalues and N minus R eigenvalues close to 1. But these small eigenvalues are not guaranteed to be 0. So L, even if it approximates K, need not be a projection. So then he has a trick that makes, takes L and makes it into another matrix that is also very close to L, but now it has, it will have been R eigenvalues to be zero. Well, the trick is the following one. We take A of lambda as a characteristic polynomial of monetary escape. OK, so this polynomial can be factorized as V of lambda equals lambda to the r times, sorry, C of lambda equals lambda minus 1 to the r, OK? So n minus 1. Ah, n minus 1. Yes, yes. Thank you. So, the characteristic polynomial of matrix L, we'll call alpha of L, no, sorry, alpha, while this here, this is a polynomial on lambda, again, it can be factorized in two parts, this part, and the other part approximating this other part. So this is V of lambda times gamma of lambda. So here is, this is the polynomial. I mean, the coordinates of this vector are polynomials. The coordinates of these vectors are polynomials. Of these matrix are polynomials. The coefficients of alpha are polynomials. the coefficients of beta and gamma are not polynomials anymore,

1:00:00 because they are like the factors of these polynomials, or some factorization of these polynomials. But then it can be proved that, in fact, the coordinates of beta and also gamma depend algebraically on the coefficients of beta and gamma depends algebraically on the coefficients of alpha. This is because as alpha equals beta times gamma, then each coefficient of alpha is a product of coefficients of theta, a sum and product of coefficients of theta and gamma. Okay? So, there is a polynomial relation between the coefficients here and the coefficients here. I mean, this coefficient depends polynomially on this coefficient. What? Polynomial? This depends polynomially on this. Because the living term is the product of the living terms. The second term, this is the first question. So alpha is the polynomial? Alpha is the characteristic polynomial of matrix K. Of L. Of L, sorry. In fact, all these polynomials are monic, so here I should put something there. alpha 1 is 1, alpha n minus 1 equals b n minus y times 1 plus 1 plus alpha n. So these coefficients depend polynomially on these coefficients.

1:02:30 So the idea is to look at the inverse function to look at the coefficients of beta and gamma as a function of alpha. This can be done if the Jacobian matrix of this mapping is non-singular. But it turns out that the Jacobian matrix of this mapping is what is usually called the Sylvester matrix of these two polynomials, and its determinant is the resultant of these two polynomials. But these polynomials do not have a common root, because beta is approximating B, and all roots of B are close to zero, and gamma is approximating C, and all roots of C are okay so beta and alpha have no common rule so its resultant is different from zero so this means that the jacobian of this mapping is not singular so you can get the betas and the coefficients of beta and of the coefficients of gamma alpha, and as the function is algebraic, its inverse is algebraic 2. Well, I mean, so in modern language, you should probably say semi-algebraic, even though this does not... It's a Nash function. Yeah, so it's... So he says it's analytic, and so it's... That's exactly where this is where the Nash function is. Sorry. It is a Nash function. Yes, but he hasn't called it. This is one of my functions. Well, so, analytic and similar.

1:05:00 But everything is happening in a tubular neighborhood of the manifold, so there is nothing global here, so it's always inside. Because, but I don't understand that, then maybe some, the roots which were zero, will turn... They are supposed... Are they going to stay real, right? Sorry? Are they going to stay real, all these roots? I don't understand. Are they going to stay real numbers? Yeah, yeah, yeah. So when you choose the matrix L, which is the approximate in K, you might as well choose it to be symmetric. Yes, as K is symmetric, and you approximate each coordinate, L is the same. So, this polynomial approximates these polynomials. And, oh, I erased the other one. Sorry, I'll do the small list again. Does it say that it's the, what do you think? Yeah, the Jacobian turns out to be just the resultant. But he doesn't mention, he doesn't call it Sylvester. So, so far this is the situation, these are the objects, and these are the approximations. So matrix L approximates matrix K, but also matrix beta of L approximates matrix K, this This is because beta of L is approximating B of K, because beta approximates B and L approximates K. This is K to the R, and since K is the matrix of a projection, this is, again, K.

1:07:30 But now, the Asian values of this matrix are the polynomial beta on the Asian values of L. So, but here, sum of this alpha is, this beta is a factor of the characteristic polygamium here are some of the relation values when we do our roots of B, of beta, OK? So this matrix has R and zero relation values and N minus R and zero relation values, real relation values. Is it OK? So the matrix need not be... It's not a... It's not a... What? It's a projection. It's not an octagonal. It's a projection. It's not octagonal. Okay? But the distortion is very small. I mean, the other... The other eigenvalues need not be 1's. Are close to one, close to one, not one. So it's a projection, but not an orthogonal projection. So it's image. Projector. Projector, it's there only one. Oh, okay, well, okay, sorry. It's close to a projector. Close to a projector. So the plane, its image is a plane, it's called P, the plane K let me see ah, so this matrix is called P sorry Wait, C2 lambda is lambda 1 swap to R. No, no, sorry, sorry. I'm going to start. Sorry.

1:10:00 Thank you. But now, So, for example, when you say that it's a resultant, maybe I know, I know, but maybe it's not the case for a resultant. Okay. So the definition of resultant? We can take that to be the definition of the result. Well, I don't know. When I read this, I asked myself something like that. It talks of it with certain naturalities of how. It says, it turns out to be the resultant, since the The result, banish is only when they have a common root as if it was something known. Fermat is the same. Fermat is the same. For the resultant? Yes, of course, because it is... It was known for the 19th century. No, the resultant is known for the 19th century, but the fact that it appears as this, It's determined exactly that when you look at the mapping, which maps too many polynomials on their products, and then the result of the Jacobian matrix. And so, of course, it's a natural that vanishes when we are in Citroën. But it is sometimes defined this way in modern books though, like in Wonsarganten's books.

1:12:30 Okay, so, but, well, so what happened with this vector, P of U, P, like, u it's approximating k times B okay but k was the projection on the on this normal plane so and b lies in the plane k because it's this is x this is y of x this is the the closest point it being it should be in the normal in the normal plane okay so this is me so So this vector, V of U, is also approximating the vector B. But what happens if x actually is in the manifold? If x actually is in the manifold, then the closest point, y of x, equals x. So B, which is the difference, should be 0. so in fact B equals 0 is the condition defining the embedding the image of the embedding D of the first embedding so the condition defining the approximation to to D will be exactly this, P of U equals P. P of U equals U. So the set satisfying this of part, I mean part of the set of points. So satisfying this equality will be the approximation to D.

1:15:00 This vector is called P. P equals P of P. So, let me see. But this is... Well, this is an equality. That means... No, this is not exactly this way because, I mean, part of this, of the points that is this equation is going to approximate the image of the Well, I think this condition should characterize the whole approximation. Of course, if you write it down in terms of some defining polynomials, you can get extra. Yes, yes, yes. So, I mean, this so far is not a variety because phi is not a polynomial. Phi has to do with phi, phi has to do with beta, and the coefficients of beta were not polynomials. They were just algebraic functions, OK? So the functions here, these are as n minus r equations because phi is a vector in e to the I minus 1, I minus 1. So the functions here are algebraic. So each of the coordinates of vector phi, it's of the sort of, I mean, there is a polynomial of the coordinates, and 3, which is identically 0 everywhere, because this function is algebraic. So the set of points where this vanishes, this defines the approximation, is the set well, this polynomial vanishes ok, well, split out so, this set is contained in the set in this variety

1:17:30 ok, so the variety is not I mean, this is not a variety, but we have to look, in fact, to where this polynomial vanishes. So I think this is the reason why the approximation, you cannot say that it is a connected component of a real, so phi equals zero does define the approximation, but all you can, he can prove is that what it defines is a sheet of a real algebraic variety. Because if you look at this kind of smallest algebraic set which contains this, then it might have other parts. It's okay. Okay. because this trick of replacing L by beta of L you can also because if you make other what you plane p of the same i mean this plane p of the same dimension of k approximating this plane okay okay so this is the variety suppose this is x this is y this is the the normal plane okay where we have a plane we want a plane p approximating this plane okay So, the trick to get that is you have the projection matrix to K, to claim K, you approximate, this is not a projection matrix, you do this trick, what you get is not a projection matrix because it's not orthogonal, but as it has R0 education values, its image is a plane of the same dimension of K.

1:20:00 Because after all, the plane is changing analytically, right? So you cannot choose your matrix to be an orthogonal projection and also expect it to be a polynomial. If it's OK, well, so we have that part of the variety defined by these polynomials satisfies that 3 equals 0, and it is an approximation to d. So this is the set B. This is in page 413. In the middle it says phi equals P of U. The idea is that P is P of U, which is approximating K of B, which is B, and this is the set he calls B, the script B. No, B. That's in the middle of page 414. 414, the third paragraph, or the definition, I think, is in the second paragraph in page 414. Well, in fact, we'll see that B is defined somehow else, and it coincides with this. Well, so we work like this. V approximates V, and V equals 0 and V. So, in our approximation, we want that t equals 0.

1:22:30 So, now, what he does, he considers this function, zeta. This function is defined in the tubular neighborhood of v. Ah, OK, I have to define something else. He defines c equals k of phi. This is not an object, and it's not an algebraic approximation, an object, because it involves, I mean, real k is not, I mean, k is the plane, and its coordinate may not be, might not be polynomials, nor algebraic functions. So this, it's in the middle here, it's approximating b, but it's not an object, not an algebraic approximation. Let's put it here, somehow, somewhere in the middle. But the idea is that the condition phi equals 0 is the same as the condition c equals 0. In one direction, it's obvious. If phi equals 0, then c equals c. This is obvious. and the other way is if c equals 0 then by definition k of p which is k of p of u equals 0 this says that p of u lies in the image of Paulina Well, P of U obviously lies in the image of P, but also lies in the orthogonal plane to k but as here it's k and p was a plane approximating this is plane k and this is plane p

1:25:00 plane p is the image of p and this plane was almost the same in some sense one was approximating the other one so in the there is nothing except the zero vector in the intersection of plane p and the orthogonal two k okay so if phi equals zero then p of u which is phi c equals 0. Is it OK? So c equals 0 is, again, part of the set of points that is fine. This is an approximation to d, which in fact is the same equation. But put in that way, it involves not only algebraic functions. So now, let's take the function theta equal x plus b minus c. As c is approximating b, this part is almost 0. okay and again as approximating was good in the sense that also the derivatives were approximating and so so this map is approximate zeta is almost approximating it is approximating identity map and also the three of the derivatives of the identity map in particular it's jacobian would be non-vanishing, OK? So now the approximation B is defined as the points zeta minus 1 of B. I mean, are the points when I do this approximation, when I apply this function lies in the original embedding of the manifold. I want to say a little bit more, I guess,

1:27:30 because by choosing a slightly smaller tubular neighbor, the inverse would also be defined over the... So, why is it so that B equals the set of points satisfying phi equals zero, or, which is the same, satisfying C equals zero? Well, in fact, if c equals 0, if we are in a point of this set, then zeta equals x plus b, which is y of x, and y of x always lies in d. so it is true that for an x where C vanishes when I do theta of x I like this lies in D okay in the other direction if I have a point that when I apply theta all this expression lies in D Well, we know that this part, which is Y of X, always lie in D. So if this lies in D and I mean, OK, all this line D because of the definition of B. And wait, let me see. So which direction are you trying to prove? that if x lies in b so if x lies in b ah, okay, if this lies in b so this means that zeta is a point in d and it has to be close to x because this part is almost zero, okay? because this is approximating this so this is a point in d but the closest point to x in D was by definition this part, x plus b, y of x, okay?

1:30:00 This means that this part is vanishing. So if a point satisfies that it's in B, then this vanishes, so it is c equals 0, okay? So this is the same. So what do we, what have we got so far? So far, we have the variety defined, but not these functions equals zero, but the polynomials equals zero, has a part which is called B, where between D and B, there is this function Zeta, which is a diffeomorphism between D and B. In fact, analytic. Sorry? In fact, analytic. Analytic, yes. And the distance between, for each point, from here to here, is this, which is almost, I mean it's as small as you want as much as you did your approximation so if we can now prove that B the sheet then this is an algebraic embedding by theorem 5 because we have now an analytic embedding of the manifold into something which is clearly algebraic because defined by F equals 0 and we need to to show that it's a sheet. So now we are in the middle of the page 414. We still must show that it's like a formal definition of a And I guess that it's time now. So maybe I can complete the proof next time. So, in fact, the way he shows this is an algebraic representation is we will construct another V star, which is in the next page, and V star by some trick, so that there will be a natural correspondence between V and V star. B star is obtained by adding some coordinates to B and this is an isolated sheet which you'll be able to prove.

1:32:30 So in fact you'll get embedding which would be just, so in fact B star is a graph of some function, another function of B, so you will get an embedding from D into B star and this is an isolated sheet so we know that this is an algebraic representation from which we'll get that this is also an algebraic representation At this point, what we have constructed is what we call now the Nash-Mathipal to the same Euclidean space, approximating the...