J-L Krivine's 'Anneaux préordonnés' (part 2)

0:00 Alex, do you want to see something? and by the way, you can eczematize a real closed field just by saying that it is an ordered field where every bounded set, every defined of a bounded set, that's equivalent. No, it's not surprising. Okay, let me just briefly repeat what we did last time. So we were talking about the paper of Kravine on pre-ordered rings. More precisely, he considers there a commutative ring A with a unit and considers subsets omega, which were called pre-orderings. pre-orderings have properties that they are closed on the addition they are closed on the multiplication they contain the squares and they do not contain minus one okay these are pre-orderings and then he considers maximal pre-orderings over this so you have maximal ones and for the

2:30 such a pre-authoring, first of all, you define the support of the pre-authoring of this intersection of omega, which is negative, which is an ideal in A, so this is an ideal in A, so you can take the quotient, well, rational number of a, ah, yeah, yeah, a minus a is, if the sigma minus sigma is equal to a, then, for example, if the rational result contains, or if 2 is inverted, what, you mean, omega, you mean, minus omega, no, this is all resonative, no, Yeah, I forgot this. Q is contained. Yeah, but you need less. Yeah, one needs less. Yeah, of course. But this isn't very much so, anyhow. Okay, then this is an ideal in India. Could you consider this? In A. In A. yeah right yeah okay now this is a divided by the support of this this was we denoted by a bar it is again a pre-ordered ring but now the pre-ordering has the additional property that its support is just zero and in case you have At a maximal one, this turned out to be a so-called pseudo-field, and he has then used this tool to give some instances of what we now call the abstract Stellensatz, and as an application gives, for instance, the positive Stellensatz and gives the real null Stellensatz. I was talking about this last time. about the envelope and the radical of such a pre-ordering and today and this will be used it's a characterization of envelope and radical will be used in the next talk and in the next talk we concentrate on the on the case where

5:00 this pre-ordering is Archimedean, Archimedean, and this is his main concern, and the other thing was just to give a general approach to this, to the Archimedean ones, and as I will, as I already pointed out, if this is a Banner's algebra here, then every pre-ordering in there is automatically Archimedean so in particular the sum of squares is which is the smallest pre-order okay we should assume already this is a real ring this means minus one is not a sum of squares in the ring and so if you have a real Banner algebra it's very easy to see that some of squares form Archimedean pre-authoring. So this is his main concern. In this next talk, we will concentrate on how he represents a ring with an Archimedean pre-authoring in a certain bound of algebra, namely in the ring of continuous functions on the spectrum, on the maximal spectrum, which he introduced here without topology so far. And he introduces the topology only of Archimedean. And in this case, it will turn out that this AR always is a subfield of the reals. So the evaluation, you could call this evaluation at a certain point, the point is this maximal one, you always get a real number. And then he just takes the topology on the reals and pulls it back. He introduces the topology on the maximal spectrum such such that these maps all become continuous. The maps we are going to introduce then. So let me just recall what this means. This means that for every A in A, there exists an N in N, such that N minus A is an element of omega, or as we have also written, A is less or equal to N. to be more precise with respect to this is defined, with respect to Oneger, of course. And we are now going to present what he did

7:30 about Archimedean pre-orderings and come to this representation theorem. And then I will relate this to later work, in particular to the work of Schmittgen. But let's first do the Archimedean pre-orders. So, with the theorems 7 and 8, we heard the first bang-bang, and we are going to hear soon the second but I will say when we will perhaps we may emphasize that the fact that it's probably the first time that this intersection is of omega and is not assumed to be zero, I believe that it's really for the first time. Stone already only considered this case. So we are on page 340. And from now on, A pi is an Archimedean pre-ordered dream. So, Levin changed the notation, omega is now part. And here on 12, give another characterization of the elements of the envelope and of the radical. We saw last time, Tuberum 9 and 10, where the elements of the radical of the envelope were characterized in the general case, but now we are in the Archimedean case, and there are other characterizations.

10:00 The envelope, which is the intersection of the maximal preordering containing pi, that That is, in fact, what we call the spectrum of A pi. The spectrum of A pi is, in fact, what you can denote at the maximum spectrum of A pi. that it's a closed subset of the real spectrum. No, no, no, it's not closed. It's not closed yet. No, no, but it's a subset formed by the. It's a subset. It is in the topology we consider now. OK, the topology is not. So, and this envelope is now the set formed by the element a of i, x of a, such that x is greater or equal to minus 1 over n for every positive integer. And obviously, the radical, which is the support of the Engelhoff, is then set for by the element x of A, such that x is between minus 1 over n and 1 over n, for every positive expression. So, we don't go into the details of the computation of this characterization, and I recall theorem

12:30 page 316, it says that if a pi is an Archimedean pre-order ring, then it is a pseudo-field if and only if it isomorphic as an ordered ring to a sub-ring of R containing Q. The if part is obvious, and to prove the only if part, Crimin defines a function, phi from a to r, and for every x in a, phi over x. Can you use the Roche-Ranthaus and r as the R-Ranthaus? Yes, yes. Because I don't see any of this there. I don't see where the notation is. This is not a notation. This is the principle. We are in 64. Tech was not... But it's not so obvious. It's not obvious, but it's obvious in the context. he introduces the rationals in the very second line. So, the value pi of X is the supremum of the rational number such that Q is less or equal to 2x. This result here, to me, is essentially the which says that everything Archimedean can be mapped into the real numbers and the un-bedding theorem was proved in something like 2010 so I'm a little bit puzzled by the fact that he elaborated something. No, there's a subtle point. This is an Archimedean ring.

15:00 A pseudo-ring. It's not a field. No, no, no. The un-bedding field speaks about Archimedean But the definition of an ordered group, of an Archimedean ordered group, is that if you have a positive element, the multiple of this can exceed everything. But Archimedean ring here means the multiple of one can exceed, with the multiple of one you can exceed everything. and if you have a field then the irrationals are dense and then you can prove the embedding theorem this is no longer a field we have only this a pseudo field it is a ring, it's an Archimedean ring in the sense that the natural number goes to the end you may still have infinitely small elements but not in a pseudo field not in an Archimedean pseudo field in a pseudo field you can have Yeah, but this is, okay, if you know this, I have indicated this, then it's just Hahn embedding theorem. I don't know the exact references, but Hahn wrote papers about Archibaldian order groups and fields in around 1910. now I have it here I can let it was one or two papers I think let me see Han 1907 1907 yeah we would need a chemist a großen system I think was it Maybe it wasn't even in an... I'll check it later, yeah. And Hilbert didn't prove in 1899 that every Archimedian skew field came back into a measure. Yeah. But he didn't do it for Bruce. Right, yeah. Or rings. Are the notion of Archimagen field and Archimagen pre-ordering, they are distinct or not? The notion of an Archimagen field is not exactly the same than the notion of an Archimagen

17:30 major pre-ordering. Of course, I mean, this is more general. Yes. But if you, if you specialize the subfields... Yeah, but it requires a proof. Okay, but I just wanted to show the the map from A into R. We consider this, this set of rational numbers and the exists because the ring is, the pre-ordering is Archimedean, so we have a function, and then it's not too difficult to see that it works, and it preserves the order. So, in fact, I wish, what is interesting is probably page 370, where there are no theories. It follows from theorem 13, that if we consider an Archimedean pre-ordered ring, A pi, then for every element omega of the pre-bin spectrum of A pi, Then, we have a map from A onto A over omega. In fact, it's A over the radical, the support of omega. That's proving right A over omega. And then, by theorem 13, we have an inclusion from A over omega into R. And this is an homomorphism of order 3.

20:00 And conversely, if we have an amorphous key of ordered ring from A to R, then the preimage of the positive numbers is the ordering of A, which contains pi, and which is maximal because the image of A by pi is contained in I in the pseudofluene. So, this is the first sentence on page 317, which is in italics. We've been saying that the spectrum of A pi may be notified with the set formed by the homomorphism of ordered rings from A to R. And then, the second sentence in italics is, this set may be embedded, may be endowed but with a compact, with a topology such that it is a compact space. It follows from the fact that And the pre-ordering is Archimedean that for every element x in A, there exists an integer M of x, such that x belongs to the interval minus M of x and plus M of x. So that, for every element of the spectrum, we have killed.

22:30 Not really, if you understand it, if you understand it, if you understand it, if you understand it, okay, okay, the same. And then, we have this. And then, in R, we have this. Because... So, if we consider the set E formed by the product of the interval minus M of X and plus M of X for X in A, Then, an element T, which may be identified with its values at the element X of A, may be considered as an element of the set E. So that the spectrum becomes a subspace, a a subset of the set E, and this set E may end up with a product topology, so that it's a compact space, and to prove that the spectrum with the induced topology is compact, it's enough to check that it is a closed subset, but this is obvious because we just have to right, that this point in this set corresponds to an amorphism of order ring, and they have

25:00 some relation to right, and they are written here, and prevailing. So this prevailing shows that this set is a, this spectrum is a compact set, with this topology. Of course, at this point we may remark that it is not the most general case, because we are in the Archimedean case. In fact, we know now, but only now, that But the real spectrum is a... That's why I'm on. Spectrum. Constellation. Spectrum. You're on the left foot. Spectrum. Spectrum. So this space, even if the pre-ordering is not Archimedean, it's quasi-compact, and then that this is also this is compact it's the same topology yes it's the same We may consider for every x in A, we may consider a map, x hat, on the spectrum, with value in R, defined by duality.

27:30 The value of x up over k is the value of k for x. And this graph is continuous because of the choice of the topology on the spectrum. So, we have this element, X hat, belongs to the ring of continuous function from the spectrum to R. And finally, we have a numomorphism from A to the ring of continuous function. And this map, this is the end of the page, 317, this is an amorphism of rings, and there The kernels are renoted by C. The corner is with the radical because it's obvious an element x at x belongs to the kernel. if and only if k of x is 0 for every k in the spectrum, and the radical is the intersection of the ordering corresponding to the spectrum. And, moreover, the preimage of the non-negative functions is the envelope, and finally, the

30:00 The image of A, like C, is denoted by A hat, set form by the X hat for X in A, is dense the ring of continuous function for the uniform convergence topology. This is just a consequence of the Sturndier's transfer and because this space is compact and the element and the X that separate the points of the spectrum. So, in fact, this is what is now called the representation theorem. And it is worth noting that it is not state as a theorem by Kribben. in the text. And in such, with this, this realm has been proved later, as you know, that You spoke about the real and the proofs, and you spoke also about stone, the previous version of this. The question is if there is an influence from stone on the cribbing. And for me, this is the second bonbonk, from our point of view of real algebraic geometry. And now I come to page 318, with the third and last bang bang, measures on the spectrum

32:30 of an Archimedean real algebra. So I, we can read the Prevence text. We assume here that A is an R algebra with an Archimedean pre-ordering pi, and linear form T on A is said to be positive if T of x is positive for every x in pi. And theorem 14 says that if T is a linear form, which is positive on an Archimedean algebra, then there exists one and only one radon measure, mu, positive on the spectrum, such that the value... It's just the Borel measure which is finite and compact subset. Here the spectrum is compact, There is no problem. Yeah, I've already mentioned it in the Q, finite reality. Or counter reality. Counter reality. Habituellement, you can tell me. So there exists one only one radon measure, mu. That's that T of x is equal to the integral of xs with respect to the measure mu for every x in A. And this measure is positive on the spectrum, that is, P of x is positive as soon as x is positive, that is in phi.

35:00 Conversely, for every positive measure mu on the spectrum, this measure defines a linear form which is positive on A, and we have the corresponding formula. In fact, the second part of the theorem is abuse, what is interesting is the first part. And Krivine showed that the linear form T is in fact zero on the radical. radical. This is not radical. Take an element x in the radical. Then, by definition, we So we saw a characterization of the element of the radical, and it is for every positive integer, here we may use epsilon because we are, I is containing. apply T to this inequality. Now we are in R, so T of x is zero. So that, in fact, we make this linear map, T, induces a linear map, U, on the quotient A over the rest of it. And you know that the quotient, this is the kernel of the map, of the previous map, C. This is isomorphic to a hat. So we may define a linear form U on a hat.

37:30 by view of x hat is t of x. Then this linear map view from a hat to a is continuous. For uniform norm on a hat. And we just have to use the fact that the norm of an element y hat is, by definition, greater than y hat, quite by definition. So, using this and some, we obviously obtain very easily the fact that u of y hat is less than the uniform norm of y hat times k of 1. And by symmetry, we have this. That is, in fact, the fact that we have a continuity. So, since we have a continuous map, it can be extended to the... you know that this i-hat is contained in the ring of continuous function on the spectrum. So you may be extended by continuity to this ring of continuous function. This is, in fact, that with a measure on the spectrum. And then we have to check that this measure is positive, and to check that this measure is unique then there is theorem 15 I will not speak about it because it seems to be a little technical a technical theorem but theorem 14

40:00 and 15 are used in the last part to be the important part for Cribbin. This is the application to analysis, functional analysis, to real banal algebras, and as Alex said, in this case, all pre-ordering are Archimedean. So that for Cribben, what was interesting was Archimedean pre-ordering in order to use the results to functional analysis. And then this theorem I used on examples, but I, yes, I forget to speak about the extension of the representation theorem in the other paper. probably you will speak about because I think for for us it's most more interesting to to go back to theorem 14 and to read it probably in with an historical point of view but an historical point with a story which go So after, it was after Privy, this story after Privy. And so I think that Alex will speak. Yeah, I will continue this now a little bit. I will say a little bit about the representation theory, what came later. But let me just mention that the existence of a measure here, this is just Reese representation theorem. So this is a well-known theorem. So once you can extend the linear map to this ring of continuous function, you know that there is a measure. This is Ries' representation theorem. He doesn't mention this. For him, it's so clear that he doesn't even write it. This is the main point here. And this representation theorem and this application of the representation theorem is quite important because much later, this is, in a sense, as we know now,

42:30 this is the solution of the moment problem for semi-algebraic, bounded semi-algebraic sets. I'm going to explain this. He had it already there, but he didn't know that bounded semi-algebraic sets give an Archimedean pre-ordering. So let me explain this. Actually, he has one example in a second paper, which we do not discuss, but he has a little extension, just one page in Crass, in just one year later, and there he extends a little bit his observation about Archimedean pre-orderings, I will mention it also, And there he gives an example of a bounded semi-algebraic set where the pre-ordering is really Archimedean. So he has one concrete example also. Okay, now, as I mentioned twice already, over Banner algebra, every pre-ordering is Archimedean. This he proves in this paper later, but the proof is very easy because every element x is bounded in the sense of the reordering by its norm, and its norm is a real number, and so any real number is bounded by a natural number. so that's it so in particular sums of squares uh this was his main concern but and then last i remember last time when i talked here uh there was the question uh are these all the uh this the only interesting uh archimedean pre-orderings and as we know now there are many other interesting and they come from basic semi-algebraic sets so let me talk about the ring I'm going to talk about now is the ring of polynomials or you could just take an affine ring but I stick to the polynomials and we consider certain polynomials h1 up to hs and we define the set the basic semi-algebraic set Let me just write H, which is the A's from Rn, such that H1 from A is greater or equal to 0, and so on, and Hs from A is greater or equal to 0.

45:00 Now, this is a closed, basic semi-algebraic set. It looks like this. Yeah, this for instance. And sometimes it may also be, the polynomials may also be linear. It looks like this. and I have already drawn it in a way so that this is not only that this is found a bounded so it's a compact set and actually we are assuming that the now that this is compact now let me okay let me first say for such a semi-anthropic set what what is the pre-ordering we are considering the pre-ordering here are source functions of the set of polynomials which are which so we could consider we are interested to know which polynomials are greater or equal to zero on this set. And there is a canonical set. It's of course the h's and the sums of squares. So we consider the pre-ordering t of h, which is then just defined to be, let me write it this, the sums h1 to the nu1, h s to the nu s where nu i is just zero or one and some sums of squares these are sums of squares now this is sums of squares so sigma nu is a sum of squares in the ring of polynomials here and if it's I am assuming this set is non-empty and if it's non-empty then this is a pre-ordering the minus one cannot be in there so this is a pre-ordering this is a pre-ordering as we have used the notation just now, on the ring of polynomials A. And there is, now we can formulate the following theorem, and it has a long history, theorem, this preordering, T of H, is Archimedean.

47:30 if and only if this set W of H is bounded. So you have a lot of Archimedean pre-orderings. If this is bounded, then it is Archimedean. An interesting consequence, but actually this consequence was why people like Becker and me became interested in this, was is the following when you have a fuck we from this representation okay now we could apply the representation theorem from Korean at this point and then we know if we have a function which is say strictly pop I have okay what the what let me just mention if we have this situation then this is essentially the Now, you can identify the maximum spectrum with this set, with the induced topology from Rn. And then, if we have a function then, which is strictly positive on this set, then we know first of all we know that it is because it's compact we know that also f plus 1 over n is suitable n is still is greater equal to 0 yeah at least I mean minus yeah thank you I know this is even greater equal to 0 for suitable n and now we use the fact that this means this function is in the envelope but yeah greater equal to zero on this set which is a spectrum means it's in the envelope but then from the representation theorem which was just written here and there you know then this function plus if you add any one

50:00 over m it's in the pre-ordering yeah so from this we get f minus 1 over n plus some 1 over m and then here I just take the same which is f then belongs to the pre-ordering which I have now denoted like this so what do we get from this we get, if we have a function which is strictly positive on this set, which is strictly positive on this set, then it belongs to the pre-order, and this means it has a representation like this, which in a sense generalizes Hilbert's 17th problem, but without denominators. You You do not have denominators, here you have sums of squares of polynomials, yeah. So this is a very strong version, and this corollary, which I have just indicated here, this is in the paper of Schultgen in 1991, and the way he proves it is that he solves first the so-called moment problem, the W moment problem, and from this solution, he gets as a corollary this fact that if a function is strictly positive, then it has such a representation. Okay. Okay, and once you know this, so he proves, whenever, Schmidchen proves, whenever this set is bounded, the moment problem is solvable, I'm willing to explain it in a moment, then it's solvable, and if it's solvable, you get as a corollary this fact here, yeah, and And in particular, you can deduce from this that this is Archimedean. So he didn't use the expression Archimedean. He started with this assumption and used functional analysis in order to prove that the moment problem is solvable and then got this.

52:30 But then, this is clear, because if you have given, what do you have to show for Archimedean? You have to show whenever you are given a function, n, you have to find a natural number such that this belongs to the preordering. Okay, whenever you are given a function, this is a function, that's a polynomial, on this set, which is compact. so it takes its maximum so we take a natural number which is beyond this and then you have that n minus f is strictly positive if you choose the natural number not too small and it's strictly positive and then by this observation it belongs to see to this period so using using functional analysis he proved in the sense that from boundedness this is Archimedean but and later this was a starting point when when when Becker asked the student of him whose name is Werman to try to find an algebraic proof for this thing and he succeeded to do this so there's nowadays a quite easy elementary algebraic proof from here to there that if it's bounded the other way is is almost trivial so this is now a theorem and if Craven would have known this that a bounded set it's not medium pre-ordering you could he would have had many more applications of his theory but he of course also this is quite elementary it needs some dirty tricks and it's it's not so easy and it's also maybe unexpected so nevertheless he would have had them more examples he had one example in this his example is when is the unit cube. To be more precise, okay, let me erase this now.

55:00 So, he considers the unit cube. So in R2, we take just, this is one, just, it's not really the unit cube, but it has length 2 of this cube. To be more precise, he considered the basic semi-order z, defined by the polynomials 1 minus xi, 1 minus xi, and 1 plus xi. yeah so the Omega we are considering is not very sloppy this yeah though the XI is the absolute value of the XI is less or equal to one because the remember that w was defined that this is greater equal to zero now this is a compact set And in this case, he really, as he says, the pre-ordering generated from this, so this would be the T generated from 1 minus 6i and 1 plus 6i, is easily seen to be Archimedean. It is actually really easily seen to be Archimedean, because, see, this contains, of course, those elements, because it's generated from this. This means you can exceed every generator, plus or minus one, you can exceed by one. but if you can do it for and you and you can of course exceed every real number by a natural number so you can exceed all the generators of the ring are x1 up to xn by natural numbers now up along the the construction of a polynomial you do it by induction that if you have two polynomials bound uh so uh exceeded by natural number the the sum and the product also do so but this also an easy induction you see that every element every polynomial can be exceeded

57:30 by natural number so this is really and this is it's here it is easily seen that this is uh an archimedean pre-ordering and this he uses in this paper which i mentioned in this one page paper which perhaps one should include to the other one there he gives this as an example that and then he states that every strictly positive polynomial on this set on this set here has a representation in in this pre-ordering exactly what we have in the general case that this he does of this paper is a little bit more. In this paper, again, he's more interested in the Banach algebra. And the main point of this paper is to show that when you are dealing with Archimedean pre-orderings, you can throw away the squares. So this means he defines, this is what later Harrison has called an infinite pre-prime so you actually need only omega plus omega contained in omega and omega times omega contained in omega and you do not need the squares to be there sufficient to have the 1 is in there and the minus 1 is not in there now this is less than a pre-ordering it And then he proves in this paper, if this is maximal, then A square belongs to this. So the maximal ones, the maximal such objects are in reality are maximal pre-orderings and everything works as well. and because of this stronger and then he applies it to to a banner algebra again but also in this case you can apply get a little bit more because you get whenever you have a function f which is strictly positive on this set here it is in the pre not it not it is in the pre-crime it is in here generated from not involved squares so it is a poll then you can say it is a polynomial in

1:00:00 those generators yeah it's you take sums and products of such things and just put positive coefficients there so it's not a polynomial it's a positive polynomial a polynomial is positive coefficient so you get a little more And so this is already in Green's paper, this is also from 64, also from 64, this paper. And I should think, and yeah, as we know, we have overlooked this, but also other people have overlooked this. people who have been interested in such objects here. For instance, much later, Handelman gave a proof not just for this, but for polyhedron, polyhedron which have an internal point. He proves that every polynomial which is strictly positive on this type. This was done much later. Let me see when this was from Handelman. This was in 88. So, 1988, there's a paper of Handelman where he treats the polyhedron case and proves this by completely different methods, which are more in the spirit of the people working in the area Nils was mentioning. Do you know him? No. Then it must be still another group. And he was very proud about this. He was very proud about this and he was selling it as great news, but of course it also follows from Krivine. But he also overlooked that somebody else proved it before him, still in another group, and this was somebody from France, it was Cassier, also somebody from functional analysis, proved this for polyhedron in 1984.

1:02:30 1984, but a long time after Krivine. So this is just a little bit about the history. Maybe I should just mention what is the, now at the end, what is the moment problem. Now let me return to what I have used before, W I used before. So, we have this W of H1 up to Hs, and now we assume that this is compact, and we have the pre-ordering, now I'm back to the pre-ordering in this situation, and then Schmittgen has proved by means of functional analysis that whenever you have a linear form, as Grevin is considering in theorem 14 yeah but also it's t okay so we have a linear form oh no the piece already used so let it be l linear form from the polynomial ring which is now our ring a from there to the reals and and then The question is, when does the moment problem is? The moment problem is when does L come from a measure, come from a Borel measure? mu, and by this I mean it is that L from X is just the integral over W, over W from,

1:05:00 for F, L from F, F, D mu. This, of course, is a linear form, and the question is when, which linear forms come from this and it is obvious that and this should be a positive measure so it is obvious when this function here the necessary condition is when this function is lies in the pre-ordering then this it takes on that it takes not negative value on omega. And then, of course, this integral has to be non-negative. So a necessary condition is that the L is greater or equal on the preorder, on T of H. This is a necessary condition. And the question was, for the people working in this area, here, is this also sufficient? And Schmitt can prove the convents. He proved when if L, so, from the L is an integral, yeah, that you write it this, this of course implies that it is non-negative and Schmitt can prove the convents is also true. So it comes from a measure if and only if it's non-negative But now look at the paper of Graveen. The condition in CRM14 is exactly this, that the T is non-negative on the pre-ordering. And when you have that it is non-negative on the pre-ordering, you have it comes from a measure. So this is the solution of the moment problem much, much earlier. The only difference is that he didn't know, Graveen didn't know that this is Archimedean. This we now know. So also the solution of the moment problem, in a sense, is over there. OK, I think this relates Green's work to later work.

1:07:30 And now it's your turn to ask questions. Yes. Looking at the end of Criving's paper, there's page 322. There's actually an instant moment problem in one variable, in the classical moment problem, given the characterisation when a sequence which is from the boundary, the sequence of the moment of minus 1, 1 of a positive measure. So the actual moment, the integral of the power of x, the integral of a minus 1 1 of x n would be mu x. So the condition for a sequence of numbers to be the moment sequence of a minus 1 1 is that you have some, well, actually a collection of infinite collection of form to be positive semi-definite. There is a condition which is written with the sum of t of i plus g, i, alpha j is a positive form. So that is a quadratic form. For every end, this quadratic form is positive semi-definite. Okay? So this is, so my question is, I've seen such conditions in papers about moment problems by people from functional analysis, so what is the position of this result with respect to what was known at the moment concerning this moment problem? Because this is, I've seen this condition in more recent papers, So, to see what is the position of this result. Krivine mentioned no, nothing in the literature about this normal problem that's not a curse. And also, this is, of course, related to the counting real roots. because here you can see the two of n, you can see it as Newton's thumbs.

1:10:00 So the quadratic form which appears is an ankle matrix, is an ankle form constructed with a Newton's thumb. And the fact that it is a positive measure means that you have actually, if you think of the sum of Tirak's majority points, it's actually the fact that you have real roots in there, and the fact that all roots are real corresponds to the fact that the form is the signature of the form, So that's a correspondence between also this result of the moments and the fact that roots are real. First of all, I must say that I did not look at this in detail, but uh so i cannot really answer the question but i can say something about this first of all people working with moments uh moment problems don't state the question with linear forms they usually state it with sequence with moment sequences which is which is just instead of taking a polynomial here you take a monomial sequence of monomials yeah you take the values And then, for instance, the square of a function, something becomes a little bit difficult. So the condition that a square gets no negative value, then I think this is just expressed in this double sum here. I would have to check it. I cannot say. It is expressed as a quadratic form. Yeah, and here he is dealing only, he's working in the Banach Algebra. And now recall, in the Banach Algebra, we know that every pre-ordering is Archimedean. So in particular, it's a real Banach Algebra. In particular, it's a sum of squares. It's the smallest pre-ordering. So, looking at theorem 14 tells you that every linear form which is on the pre-ordering, non-negative, that's the condition, it should be non-negative on the pre-ordering, and then it comes from a measure.

1:12:30 So being, in this case, for linear form, being non-negative on the pre-ordering means being non-negative on the square. Because, and being non-negative on the square, I think it's just this double sum condition. For a banal algebra, this is the only condition you require. Whenever you have a linear form where squares are mapped to something non-negative, it comes from a measure. That's the only thing I can say now. I cannot say anything about it. It's an epimorphism, probably, but it's maybe not necessary. We just have to look at the kernel. Yeah, forgetting about the kernel now. I mean, in fact, without the kernel here, too, it's so theoretical. But module of the radical... Yeah, then it's an isomorphism, yeah, I think so. But then... But it's a real Barnach algebra, yeah. Yeah, of course. I think it's clear that everything that's non-negative is just a square. Right, yeah, yeah. So if you work modulus is the kernel, then everything non-negative is a square, of course, yeah. So you really don't have a choice of what is pre-organization? Yeah, modulus is the kernel. Sorry, when does the green differ start to appear in your bibliography? Pardon? When these differ start to appear in the bibliography of...

1:15:00 The earliest place, I think, is in, Max has once written in some model theory, Springer lecture notes. There I saw it, yeah. I knew it before because it happened in century before. But it's also in your book. When was your book? You have it in the first edition. I think so, yeah. He told me about that in a seminar in Palestine, what Ligua was and Savard was. It was a lot of junk, not in Savard, but all this has been done by bringing back to the world. Because I gave the authority of this to Ligua, etc. That's what I knew at the time. And Savard is junk for self. No, it belongs to the unit, so I went and looked at it. About these more problems. Interactive measures, only when it's . Yes, so it means that this, because it says that there is an infinite number of quadratic forms, when you are dealing with the roots of the polynomial of infinite degrees, there are some recurrence relations. I feel, Robert, I'm getting interested here because of the infinite, but of course it's only a finite one, and it's only a finite one, and that's also what people discuss in the global problem, when is it a finite one, when is it discrete, when you can really concentrate on the finite one.

1:17:30 So the way the people don't want to check information, they want to check on the subject of the population, and then they restrict a given sign of the information, and they will have to check on the information. The notion of Archimedean is not the first order notion. Yes, but it does, actually it's not very far, maybe, but it does significantly difference because, for instance, in these, like Schwedgen-positive-scherensatz, as we saw, you cannot have uniform bounds in the representation, although for the positive-scherensatz which doesn't use Archimedeanity, so that's well. I will explain in the text by Alex, the same first order feature means that you have uniform bounds for this representation. So that's a huge difference that you have, maybe not so important, but actually that's a huge difference. But you can add a constant to the language and an actual machine saying that this constant is bigger than any integer, so it's no large imagine it's possible. Yes, but still you are still there. You cannot negate an actual machine of action.

1:20:00 But you are still there that you have one case you have uniform bound, then the other case you cannot take uniform bound. Okay, if we don't have any questions, then we are through.