OK, so just as a reminder, we want to prove today, at least give the main steps of the proof with Hohing theorem. We have mu is a simple cycle, minus 1 cycle over k. And then we look at the field extension of k with respect to sufficiently, well, I mean, for us, it's really just the linear system of parameters and L, the left sheds, take the field of rational functions. And then what we want is that B of mu with respect to this theta, which is now, yeah, it's a quotient of a phase ring defined over this k tilde, satisfies our left sheds, the Hohler-Mann relations, and finally, for characteristic equal to 2, total ionisotropy. And to give you an intuition of the proof, let me first argue and give some basic reductions. So basic reductions and lemmas. And the first thing I want to do is, OK, we have hard left sheds. And I already discussed some of the middle hard left sheds. It's the most important one. I want to do this explicitly. I want to show that essentially I can always reduce not only to the middle one, but always reduce to a pairing and to really have a question about the Poincare pairing. We kind of did this already in some cases, but don't want to be explicit in. Just do this. Reduction to the middle. So how is this achieved? So let me just recall, if mu is a d minus 1 cycle, j, an ideal in B of mu and l in B1 of mu, then j satisfies. The Hollermann relations with respect to in B mu with respect to l and in degree in k, if just so that the camera can catch both continue here, if, well, the quadratic form ql, qkl on Bk does not degenerate. OK, so far so good. So now we want to say that really, to prove Hollermann relations in general, and in particular. So g is homogeneous. Gryffindor h, ideally. Yes, so this is a graded ring and this is a graded ideal. So this is a graded ring and this is a. So to reduce this to the middle, I need to introduce a small notion. So if I have mu a cycle, then I can consider the suspension of mu. And this is, if you want, it's just the free join of a zero cycle consisting of a north and south pole with an original cycle. And so it increases the dimension by 1. So the suspension is free join of mu with a zero cycle on vertices n and s. n and s are standard for north pole and south pole for convenience of referring to them. Because if it was 1 and 2, then I would forget which one is which. All right, so if this here was a d minus 1 cycle, then obviously the suspension will be a d cycle. So now let me consider any attenu reduction of this cycle. So consider zeta a linear system of parameters. Yeah? You were hearing the definition of the term of relation? Yeah, it's proper to us or definition. That is, OK, so this is the definition. Or actually, we talked about this again, so it's more recalling it. But I think we talked about it only for spheres. So this is slightly higher generality. Yes, I want to say that if you want to prove this property for ideals, then you can translate this. You can reduce it to a case where k is equal to d half. That's the point, OK? OK, so let me write it explicitly. Want to establish hollaman by only considering the case where d is equal to 2k. OK, so we have a linear system of parameters for the suspension of mu. And let me think for a moment of, I mean, this linear system of parameters, this is kind of a matrix of coordinates. This is a matrix in k, well, the number of vertices, so mu 0 times the call dimension, which is in this case d plus 1. So this is my matrix setter. Now, somewhere in this matrix, there are the coordinates of the vertices n and s. All right? And now, what I do is I take mu 0 plus 2, yeah? So this is the set of mu. You have two more vertices. Yes, but it's just a suspension of the cycle. So it increases the dimension by 1. But I'm going to say the number of vertices.  Thank you, yes, you're right. All right? And now, what I do is I project along a project along n. I project along the coordinate of n. And then what I get is a linear system of parameters setter bar, some more projection along n, and results in a linear system for the suspension. Sorry, for mu itself, for a of mu itself. And this linear system I will call setter bar. I can think of this, actually. I can think of n as actually being a basis vector for a basis vector in the last coordinate direction. And then I also write down, in addition, OK, so in the projection, I have a kernel. I forget one of these elements of tetra. And let me call this, let me call the element that I forget l, OK, the height. OK, so let me write tetra. I write it as tetra bar, together with an additional linear form l that I forgot, OK? So this is just a notation. All right. And then we have the following theorem. And this is the following I equivalent in this situation. And this is first the whole amount relations for the ideal i, or for an ideal j in mu, let me write it with respect to the linear system of tetra bar in degree k, for degree k, for degree k, and with respect to, well, now I have to give you also the thing that plays the role of the left sets element, so with respect to l. And the second part is, OK, I have again whole amount relations for, now what do I have to do? Well, OK, so I take xnj in v of suspension of mu with respect to tetra for degree k plus 1, and with respect to the linear form, with respect to the form xn, all right? So just the variable xn, that's it. J is any ideal. And k, I should say, so k, let me make sure that I don't get, so k is strictly less than d half, all right? J any ideal. All right. Let me just recall the special case of monomial ideals. So if J is equal to i mu gamma, all right, where somehow this here again is, this for me is just the kernel of b of mu 2. OK, so now I have to say a little something. I have to say that, OK, so I take b of gamma, b of gamma in mu, all right, where this object here is, all right? So remember that b of mu was defined as a of the underlying space, of the underlying substantial complex model of the annihilator of the fundamental class. So everything that does not pair to the fundamental class. And similarly, all right, now I have a of gamma, all right? That is a quotient of a of mu. And then I look at the induced quotient in b of mu, and this is b of gamma in mu, in my notation. OK, that's it. All right, gamma is any sub-complex. For gamma, any sub-complex. For gamma, if for gamma any sub-complex, consider the following ideal. All right, then what is the corresponding ideal here? All right, what is the corresponding ideal in the suspension? Well, this is, if you think about it, just then xn is just, OK, so I could take i of the suspension of mu. And, well, here I take the sub-complex that I get now. It's the suspension of gamma union the south pole of the complex with the underlying space of mu. And so what you see is, OK, so if I start and want to prove the hard left sheds here, all right, then I would, then gamma would just be empty, right? I want to prove the whole Laman relations for the empty complex. That is just, all right, so then I just have the hard left sheds left. And so I can just, every time, so let's say I want to prove the left sheds from degree k to degree d minus k, then in the next step, applying this lemma want, I get, I prove the left sheds, I have to prove the whole Laman relations with respect to a monomial ideal from degree k plus 1 to, well, OK, so I have d plus, OK, I have d plus 1, I increase the so called degree by 1. But then I also minus k minus 1 because, right, minus k plus 1. And so it's really just, so the top one is the same, it's just d minus k. And so you see that, iteratively, I get closer to the middle until I'm saying something just about the Poincare pairing restricted to the ideal. All right, so nothing fancy. So you can see what's after the suspension of gamma union, is it s join u? Yes, s join s. It's just basically the southern hemisphere, all right. Yeah, that's it. Excuse me, what was the d and n cycle in your definition? I think you mentioned the homology, the well, simplificial manifold, it's on previous talk, but maybe you... Well, this is more general, right? So OK, OK, so a simplificial cycle for me is just, OK, so let me just repeat it, OK. Also, you can just come down, I assume that you're in your office, no? So a simplificial cycle. Mu of dimension d minus 1 is a pair of, well, I take a simplificial complex, mu bar d minus 1 dimensional simplificial complex, all right. And mu an element, well, OK, it doesn't matter whether I take the simplificial cycle now or an element of homology, because, all right, this is the top dimension by definition. So a mu in, I could say, let me just, a cycle in this underlying space with respect to k coefficient, all right? Well, it's a cycle. It's a z, all right, OK, maybe, OK, so, yeah, yeah, so. I mean it's the same as a homology, yeah, so. OK, and the proof, proof is rather simple, it's really just a diagram chasing already, so proof. OK, so let's consider the left sheds isomorphism in B mu zeta bar, right, l to the d minus 2k, d minus k of mu zeta bar, OK, and then I will write down some maps, so let me just write down, OK, so, so let me note that xn plus xs plus l is in theta, all right, is an element in theta, OK, so let me write it in the following way. So what do I mean by that? l restricted to the vertices of mu, all right, because I took l as exactly the element that I lost in the projection, right, it's a kernel of this projection map. So this was an element in theta, all right, so this is in theta. And now what I, OK, so what do I do? So now I take this vertical map here, so just a multiplication with xn, and so where do I land? Well, I land in the ideal of xn, and this I can describe rather simply by taking the suspension of mu and then taking this relative to the south pole, all right, I see that I wasn't very wise with the, OK, so electronically I could move this now, I will not be able to here, so I will write down, OK, let me write this down in a code, and then in another row explain it, and then I will take this vmu s join mu bar, and I will explain what this means in a second. This map here below is a multiplication by x to the n, well, where x to the n is supposed to act as a left shift element, so this here is in degree d minus k, so what do I do? So what I take x to the n to the d minus 2k minus 1, and this is a commutative diagram, it's all good and nice, let me just really explain what this space here is, and this is really just the dual under the Poincare pairing to this here in B of the suspension of mu, and I will describe it directly for instance by saying, so this is some B of mu s join mu bar is B of mu modulo the annihilator of, so this is isomorphic to xn B of mu theta bar, and I was also stupid, I have to write the suspension of mu, the suspension of mu modulo the annihilator of xn B of mu theta bar, which I can also write as well, I can basically write it as the image under the partitioning map or B mu modulo the kernel of the following map to the direct sum of vertices in mu, and then I take B of the star of the vertex intersected the star of the north pole in mu as a sub-complex in the cycle, and this is a commutative diagram, the top and the bottom are, well the top one is just the left sheds isomorphism that you want, the bottom horizontal map if you think about it, this is just the Hollermann relations for this ideal, so I want to recall that somehow, if I want to prove that a pairing doesn't degenerate, really what I only have to do is, I have to prove an injection or in this case an isomorphism because they are really Poincare dual from the ideal to the ring modulo the annihilator, and then that's just equivalent, so this is just the Hollermann relations, so I have this nice commutative diagram, and now really somehow restricting to J gives a desire, so restricting to J gives a desire, okay? And that's it, so there's nothing fancy to this lemma, but now we can reduce to the middle, and we are happy to reduce, we can kind of ignore actually left sheds elements, some of our left sheds elements don't actually enter our calculation anymore, we really just look at an even dimensional cycle of dimension 2k-1 and we look at the middle pairing, we want to say something about that, so it's one less variable that we have to care about, one less issue. No, no, this is just general for any tether that you want. And so the next, the second lemma that I need is that something that I hinted at earlier, and Ofa actually asked so, this is also for him, so at some point I claimed that all these objects are generated by square-fee monomials, in particular when I deduced, for instance, at the beginning the Greenbaum conjecture form from the left sheds property, back then it was kind of a side remark, because it plays an important role here in the proof, let me just do it. So now we need the second ingredient, and this is square-fee monomial generation, and this is, okay, so I give it a simple complex, so delta is a simple complex, the claim is that A of delta, with respect to any linear system of parameters tether is generated by square-fee monomials, and since I, okay, so this holds for any linear system of parameters, okay, holds for any, holds for any linear system of parameters, but in this generality I will not do it, I will leave it as an exercise to you, it's not hard, basically I will, since I work with generic linear system of parameters anyway, I will just give you the proof of generic linear system of parameters, okay. Here. So then it's going to generate the volume, and each degree is generated by square-fee monomials. Yeah, exactly, each degree is, okay, so okay, in each degree, in each grade component, thank you, that's right, I mean otherwise it would be trivial, that's right. And here I will really just do the generic case, so here, tether-generic. So I assume any kind of condition, like if you take tether 0, are you empty? Well, if tether is 0 then it's not a linear system, what I mean? So you have to have somewhere like that. I mean A for me is always called dimension 0. Yeah, in fact it's called dimension 0. Yeah, so it, yeah, so I will just do it for the generic case. Okay, so what's the trick? Well, let me construct a larger complex where it is true. So consider k, a simplisher complex, where it is true, all right, so k, a simplisher complex, and it's larger than, it's larger than my complex delta. I always have problems with somehow which directions in this, in subsets I always have problems with which direction is larger because I mean it's the inequality, but then it kind of looks as if this eats that, so whatever. So this is why you see me hesitating when I write subset signs on some occasions, and I might do them wrong at some occasions. So tether-generic, okay, so such that claim is true, and in particular it's obviously true for, then it's a particular, obviously true for delta as well because well, right, if a larger complex, a larger ring A is generated by square for monomials, then the smaller ring will be, that's fine. So what complex k, then it's true for delta, and so what larger complex do I take? Well, I take, okay, so if, let's say delta is, delta is of complete dimension d-1 on n vertices, then what I take for k is just the d-1, so k for me is just the d-1 skeleton of the simplex on n vertices, right? And now because I took tether-generic, it will be, tether will be a linear system of parameters for k as well, delta is linear system for k, okay, very close to be politically inquired. All right, tether is a linear system. I think one can do easily is a general case like this, you want to make the support of a monomial bigger, so if you have a given monomial, so the value is corresponding to some simplex, then by the condition of linear system of parameters there's a linear combination of the system of parameters that it is, let us say, a given variable plus some combinational variable is not in this simplex, and then if something occurs with an exponent bigger than one, then you can use this equation to move, to get something where you lie outside the gate so you have a bigger support and then you increase the support in this way until you get, you must get a square. Yes, yes, yes, yes, that is, okay, that is, yeah, it's correct, that's also another way to do it, yes. This is the very lazy way to do it without any work because now the point is that k is shellable, so this complete simplex complex is shellable, and for k shellable complexes we already saw how to give a basis in terms of, in terms of faces when we attach new simplices along the shelling, so k is shellable, right, and then remember that somehow we described exactly what happens in the shelling step, so if I have a shelling step of a simplex complex, let me call it s, and then I have s without the facet that I remove in the shelling, then I could precisely describe the kernel here as basically k times x, x minimal or something like that, so let me call it something like that, I think I call it sigma f where sigma is the minimal face of f not in s without f, I think I wrote s without f, like this, and that's it, all right, and then you see it also immediately, so what alpha set is also correct, it's just the amount, okay, so you have to think for a few seconds about the linear algebra involved, but that's also correct, all right, and these are the two, these are the two ingredients, that's the two kind of basic ingredients that I need, and now I need to basically do a little analysis of the degree function, so let me go there and let me start with it a little before the break, so recall that, I should have deleted somewhere else, but recall that if f is a facet of the underlying space, so underlying complex of nu, then, well, I mean, the degree function is only defined up to a multiple, but what I can do is I can just, choose one that is somewhat canonical, and here's my choice, the degree of xf, right, of this monomial of the maximum dimension is equal to, I take mu, so this is a simplification cycle, I take the oriented coefficient with respect to f, I somehow consider my vertices ordered in some arbitrary way, and I divide this by the determinant of this matrix theta restricted to f, okay, that's an identity I need, and you can equally also write down from this, once you have this, you can write down formulas for the degrees of square full monomials, so let me in particular write down one formula, so the following, I need the following formula in particular, and this is I want the formula of the degree of x tau squared, and now, okay, so remember, I really only care about the middle degree, so this, so mu for me is mu, a cycle of dimension, 2k minus one, equal to d minus one, and really I'm interested in the middle pairing, so tau, the cardinality of tau should be equal to k, all right, so now this works out, and now I have to sum over f, the facets of mu containing, containing a given, my given elements containing tau, and then I have the degree of xf, and then I have to multiply the following, so products, products, and then I will, so the product is over all those f without i, let me call this, theta of f without i, okay, so not theta of f without i, what do I mean by that, right, it's no longer a full-dimensional, this is no longer a matrix of rank d at a minor, so what I mean by this bracket is I take the volume element corresponding to it, and I do this for all i contained in tau, and then I divide this similarly by the volume element for theta of f without i, where I now go over all the i's, or maybe I should do j, or you go over all the j's in f, but not in tau, okay, and this is just a nice formula, you see that somehow, it's nice and well defined, right, I could, I mean if you don't want to compute the volume elements of a lower rank matrix, so just add another generic vector, and you see by the scaling it just doesn't, it doesn't depend on the other generic vector that you use to compute the volume here. Anyway, so that's the formula that I need, and now I want to define some differential operators, I want to define some, I want to in the end study, so what you're doing is this is the volume form, right? Okay, so think of it just, I take another matrix, so I take another vector, yeah, you can think of it as a contraction, so think of it as this way, so you take another vector, a generic vector, and once you remove i, you add this other generic vector in its place, okay, this has to be the same for all those, all of those that you take here, and then you compute the volume of this new, now you have again a D times D matrix, computer determinant, that's it. All right, all right, and now what I want to, let me use this one, it's the most comfortable, so now what I want to consider is the degree function of the element u squared, right, where u is in vk of mu, and I want to understand, I want to evaluate this in a nice way, and the trick will be to consider, to define a differential operator here, and I will define this first, I will first consider this differential operator depending on tau, and in particular I will define it, so given tau of cardinality, of cardinality k, and sigma in mu another phase, u in of cardinality k as well, and let me assume that such that tau intersection sigma is empty, so they don't intersect, and now what I will do is the following, for rho an element in, okay so what do I do now, I take the rational functions, I'm working in the field of rational functions k tilde, which is the rational functions where the variables correspond to the entries of the matrix theta, and now I want to consider them as rational functions, and want to compute, yeah for now I change my perspective, so now these are rational functions, and I want to take derivatives with respect to the variables here, okay so this is somehow my short end of studying this function in a nice way, so what do I do, well I want to say that, okay so I have my matrix, alright theta, so let me write it outside because I want to do some things inside, and then I have inside I have, I have sigma which is sigma 1 dot dot dot sigma k, and then I have tau which is tau 1 dot dot dot tau k, and now I have a rational function in these variables, and I want to consider what happens if I take my rho sigma 1, and add to it a perturbation in direction of tau 1, alright so I want to take this differential, so rho differentiated in, by taking the rho sigma 1 and perturbing it into the direction of tau 1, and dt, and then the next thing I want to do is, okay so this is my first differential that I take, and then I want to perturb d sigma 1 into its own direction, so d plus t sigma 1, dt and so on and so forth, and what I end up with is a differential operator that I will call, that I will denote by this, so partial sigma tau tilde, and then I will just normalize this, and say that partial sigma tau, this is just for me the whole thing normalized by the degree with x sigma x tau, okay alright, and that's it. Now it sounds a little weird, but it turns out that this is, that this gives a nice formula, let me go here, because for instance I have the following lemma, if I apply this to this formula here, which by the way I should say, this is to l, then what I get is that d sigma tau of the degree, alright so now I think of the degree as a function in these indeterminates of x tau squared, this is just the degree of x tau, x sigma, you will see later, so it sounds like for the moment that this normalization somehow is a little bit stupid, but you will see later why this makes it a very nice formula, because after the break you will see that this implies that the degree of u squared, so at least okay, so let's say in characteristic two, this will imply that if I take this differential and it does not matter which phase in the star of tau I take, and apply it to the degree of u squared, then this will be just the degree of x sigma times u squared, which is a very nice formula, and it's a little more complicated in positive, not in positive characteristics, in all characteristics that are not equal to two, but it's still a nice formula, so there's one correction term that we will need, and then this we will discuss. Also I will discuss the derivation of this formula after the break, but now let's take a ten minute break to ten minutes. So let me just clarify something, so the degree of x tau squared, this depended, maybe I shouldn't have deleted the formula, but this depended, I see this as a rational function, this depends on theta, this is a rational function in theta, and this is what I mean when I write, I take the differential of this function, because it's really a function that depends on theta, and then again let me write down, this was the degree of x sigma, x tau squared. Let me notice also that if v is any vertex not in the star of tau in mu, then degree x tau squared is independent of the rho of theta corresponding to v. I was very stupid and deleted the formula, but you remember that it's summed over the degree function, it was a degree function of the facets, which depended on the vertices of the facets, and then I took the sum over all these facets containing tau, but if v is not in the star, it does not form a face, then it's not in the facets, in particular it will be independent, blah, blah, blah, blah. All right, so this is the formula, now it gets actually very simple, so let me indicate the characteristic 2k's first, and then I will go to the more complicated formula. So what do I want to prove? Well, in the characteristic 2k's remember, I wanted to prove that actually something stronger, I wanted to prove that u squared is not equal to 0 for all u in vk mu. So in other words I'm saying that degree u squared is not equal to 0, that's another way of saying it of course, same thing. Potato, potato, yes, tomato, tomato. All right, so how do I do this? So by Poincare duality, and because v mu is generated by square frets, square free monomials, there exists a sigma of cardinality k, such that sigma x sigma times mu is not equal to 0. Let's just Poincare duality, nothing fancy. Now let me go one further, so let me look at the ideal of x sigma in b mu in degree d. I mean this is just isomorphic to k, so this is one dimensional vector space. This is isomorphic if you think about it, this is just the star of sigma in mu, or the link of, if you want, it's the link of sigma mu, so if you think about it, if you have a cycle and you have a vertex or a face, then the link of it will again naturally be a cycle. So this is again a cycle, again a cycle. And this in degree d is actually isomorphic to the degree k component, because I took the link, and this here, in particular this here is isomorphic to k, so it's a one dimensional space. In particular it is generated by a single square free monomial, the degree k component of this is generated by a single element x tau, generated, and in fact it doesn't matter which one you take, but this doesn't matter, generated by x tau, by tau of cardinality k, of cardinality k in u, well actually in the link of sigma in u. All right, fine. So far so good. In particular I can write u, u is an element in b mu, but I can write it in various ways. I can write it in particular in the following form. I can take the component that pulls back to the link, and this is lambda tau, it's some coefficient times x tau, plus some lambda omega x omega, where this omega are not in the star of sigma mu. All right, so a, a, a. All right, so I am generated by a single monomial in that star, so all the other components come from outside of the star. All right, so I should say I mean, it's a class of that element, right, it's a class of this element, but that doesn't matter. I just wrote a specific form. And now let me compute, actually compute, compute now the differential of sigma tau of degree u squared. Well, this is, okay, so now, all right, so I won't, so it's a sum, it's a degree of this boy here squared. All right, we squared this little boy. What happens here? Well, it's, I mean, we are in characteristic too, so all the mixed terms vanish. Let me not write it down, but somehow all the mixed terms vanish. What is left is just the sum of the squares of the individuals, so it's just degree of lambda tau squared x tau squared plus the sum over this omega, I will not write down, all right, so they are not in the star of degree, degree lambda omega squared x omega squared. All right, and the differential, of course, differential sigma tau, sorry, differential, all right. Okay, so now, now let's look at Laptitz's formula for a second because it's just so easy and nice. So if I have, well, okay, so I can pull out this, I mean, it's linear, right, the degree is linear, so I can pull out the lambda t. So now I have, so I have something of the form differential of something f squared times g, all right. I'm in characteristic too though, so remember, okay, so, okay, just Laptitz's formula, this is two times f times, oh, sorry, it doesn't matter, actually, what we have here, right, this is zero, and so this is just really just what is left is just, sorry, two times f times f prime. So what do I have left is really just f squared times the differential of the g. So what I have left here is really just, okay, so what do I have? So this is just lambda tau squared times the degree, okay, so now I take the degree, okay, now I take my lambda, so I take the degree of x sigma x tau squared. What happens to the other ones? Ah, okay, so again differential, ah, let me write it here, sigma tau. Well, each of these are outside of the star of tau, each of them, in particular, if I differentiate after varying along the vertices of sigma, then, ah, sorry, these are not in the star of sigma, so each of these omega, if I look at the star of phase omega, they will miss one of the vertices of sigma, otherwise they would be in the joint star. In particular, this here will just be zero, all right, because if I take any, so if I take star tau, if star omega in mu, right, contain all the vertices of sigma, right, right, contain all the vertices of sigma, all of sigma, sigma, then it's not hard to see that then, well, then I would be, then this here already would be zero and I'm done. So, in the notation t double sigma, you add t times sigma to, you add t times to the column. Yes, yes, I add to the column of sigma, right, I have the column of sigma and I add t times tau, right, that's it. But here, because of that. So here, right, I differentiate them, it's the columns of sigma I modify, all right, yeah. All right, so this is just zero. Okay, so now I can pull this back in, this is just degree x sigma x tau times lambda tau and the whole thing to squared. And if you think about it, now then you can compute that this is actually degree of x sigma times u, all right, because the only, okay, so x sigma times u, the only times it was non-trivial is if sigma, well, the only time that this, that x sigma times this here is non-trivial is if sigma and the face of the coefficient are in the common simplex, these here are not, so only this term remains, so that's what I get. But this here, by assumption, is not equal to zero, all right, because that's the way that I started. So now, and now I'm done, all right, now that's it. So this finishes characteristic too. All right, okay. So now I will do the case of general characteristic. Actually, I should have kept, yeah, okay, you remember. Okay, general characteristic, five, four, okay. So let's consider an element u in, okay, so let, I want to say, okay, so I want to consider elements in u, in bk of mu with the special representation. So what I will look at, so at the start is u in k tilde of the underlying space, and sigma, I will take a facet. So, sorry, sigma, a cardinality, so this is in degree k, this is cardinality k face. And I will call u compatible definition, definition u is compatible with sigma if, and now I look at the support of u, okay, so the monomials, okay, so I can assume that u is some other, all right, so this is the sum of square free monomials with some coefficients, that's how x times, maybe I should, let me use lambda A xA because I want to use tau for something else in my notation. If first, the intersection of u with the star of sigma in mu, of u with the star of sigma in mu, if this here is of cardinality one, so is a unique face tau, and I will, yeah, is a unique face tau, and the coefficient, so u at tau, well, I will just normalize it as one. Second property that I want is, if I have u tau on, so then I want, the second property is that if I have a face, if tau prime is any other face in the support of u, then the star of tau prime in mu intersected sigma is a simplex, all right, is a face of sigma, is a simplex sigma prime, and it's a face of sigma, possibly empty, and now what do I want? Well, I want to say, oh, ah, is a face of sigma prime or empty, and the coefficient of u in tau prime is independent of one of the variables or actually for me it's enough, it's independent of all the variables of sigma without sigma prime. Okay, so this is my definition of compatible, and now I will consider a slightly larger field extension, and what I will do is I will take my field extension in k tilde, all right, k tilde this was k, and I took the field of rational functions with theta in it, sigma with respect to theta, and then I add another, these were just the entries of this matrix, I add another copy, theta prime, okay, and now what I will do, I will associate to u an element u prime in k of theta prime obtained by replacing an entry theta ij with the corresponding entry theta ij prime, and now what I will do is, okay, so if, so theorem or lemma, lemma for u compatible, I will now write down, I take the differential of sigma in tau, where tau is a unique face in the support of u that is also in the star of sigma, so this is just before, so degree of u times u prime, and this evaluates to the degree of the partial differential sigma tau of u times u prime, and now plus the degree of x, okay, so x sigma times u, that's it, okay, so it's a slightly more complicated formula, now substitute, right, so notice that the differential, the differential here does not affect u prime, it's a differential after the variables of theta, not after the variables of theta prime, but after the differentiation I can substitute u equal to u prime, and I see that, so I have sigma tau degree of u times u prime, all right, I compute the differential and then I substitute u equal to u prime, this is equal to degree of the differential tau u times u plus, and the last term is unchanged, sigma u, I didn't need that. So why is this good? Well, okay, so now recall again by Ponca-Riduality, we may assume that x sigma times u is not equal to zero, in particular this here is not zero, in particular one of these two terms here is not equal to zero, but if u is in a monomial ideal, right, if u is in a monomial ideal k, so before the acting in reduction, in k of the underlying space, then okay, so u prime is just obtained by substituting the coefficients, so this will still be in the monomial ideal, and this here, okay, so here I have the coefficients in some derivation, right, I will take the coefficients and I apply some operator to them, but it will also still be in the same monomial ideal, so it will still be somehow, then so on, then so u prime and the partial differential of u that we defined. So why is this good for us? Well, remember the Hollermann relations, right, some of the hard left sheds in the Hollermann relations, they state that somehow I want to have the non-degeneracy at a certain square-fee monomial ideal, now I reduce this iteratively to saying something about the square-fee monomial ideal and proving just the non-degeneracy in the middle where there's no left sheds, right, so now okay, so I want to prove this for things in a monomial ideal represented by some monomials and I cannot quite say that u pairs with itself not trivially, but at least I have two other candidates that pair and one of them, right, if this is non-zero, one of them has to be non-zero, right, cannot be, I mean, you cannot have a sum of three terms only equal to zero, only one of which is not equal to zero, yeah, that's just the basic it, okay, so now, okay, so now let me, let me, let me finish by giving you the indication of why we can reach a compatible element and then we're done, all right, so why we can assume that if I have u represented in some way, then I can make it compatible and that's it, all right, I think this is characteristic two, no? Yes, this is characteristic two. So in our case, for us, what we are interested in is proving, right, so now we have an ideal of the form i mu delta, where delta is some sub-complex, delta in mu sub-complex, and again, let me just write down the parameters again, mu of dimension to k minus one, delta, yeah, delta is just some sub-complex, doesn't matter, and we're looking at considering, and we consider the pairing, right, k times ik to our ground field, well, now this is a ground field theta, and now consider, okay, so now consider u in k, k tilde of mu, well, what do I want really? I want k tilde of mu relative to the complex delta, all right, represented, okay, so I can already start with saying that it's represented by square free monomials, and now I want to say that in B mu I can find another representative that is compatible, and how do I do that? Well, first of all, notice that if u times x sigma, okay, so I know that there exists x sigma such that u times x sigma is not equal to zero. Well, no, exists x sigma such that u times x sigma is not equal to zero. If sigma is in the support of mu but not in delta, then we're already fine, all right, then we're already in the ideal. Hence, we consider only the case where sigma is a phase of delta, okay. Can you say again what you said you wanted to find a representative in B mu that's compatible? What does it mean to be compatible? It's essentially, it's a normalization on the phases that, okay, so compatible means that there is a unique phase that, the support of mu, of your element u intersected with the star of sigma is essentially just one phase. Anyone's sigma to be in delta or away from delta? So if sigma is away from delta, then we're already done. We want to prove the bias pairing property. We want to prove that there exists, so we have u in our ideal I mu delta, right. We want to prove that there is another element in this ideal that we pair with. That's a non-degeneracy of the Poincare pairing at this ideal, right. That's the case that we reduced to. They do think that are outside of delta. Ideas spanned by one of those cases that are outside of delta. The ideal is phases that are outside of delta. Yes, yes, exactly, yes. But it might be that somehow, it pairs only with a phase in delta, right. We know that it pairs with some, with some monomial of degree k, but this monomial might be corresponding to a phase in k, otherwise the whole thing, the whole exercise would be trivial. And indeed, if we have already, if sigma is already in the complement of delta, then we are done. So the only case that we have to consider is that sigma is a phase of delta.  So let me draw delta. All right, so this is my delta. And let me add sigma. This relies somewhere in delta. And now, okay, so it, but somehow the complex, the support of mu, this extends beyond, this extends beyond delta, so it goes on in some way, all right. Okay, so now I basically, I normalize iteratively. So first of all, I consider B of the link of sigma in mu. And I realize this again in degree k is just a one-dimensional vector space. In particular, I can assume the first, I can assume the first condition of compatibility. We can assume that the class of mu in i delta, i mu delta, is represented by an element. Well, okay, so what do I mean? So it's by an element. So if the dimension of the link is one, then we can assume it's generated by a single phase tau. In particular, we can assume that by an element, let's say, okay, so let me say, the first iteration is having the element u prime, or maybe I should use u prime. Let me say u0 such that the support of u0 intersected the star of sigma in mu is a unique phase, is equal to tau. And this is the first condition of compatibility, all right. Well, almost. I mean, I want the coefficient to be one.  All right, so now we can just assume the coefficient is one, is one without loss of generality. All right. And now I have to look at another phase, any other phase, sigma prime, okay, I have to look at phases tau prime in the support, all right, I have to look at phases tau prime in the support of u, so let me write it like this. Here's another phase tau prime in the support of u0, my already modified element. And I know that the intersection of star of tau prime in u0 with sigma is a phase sigma prime. So this might be, in this case, a phase instance like this one here. So this is sigma prime, all right. And now what I can do is I can look at sigma prime and extend it to the interior here by multiplying with a monomial corresponding to an interior vertex. So a vertex of mu not in delta. And I can assume that u0 times x sigma prime union, this vertex in the interior, is equal to 0 because otherwise I would be done because then I would already pair with something in the ideal, i mu delta. So I know that this pair is 0 and then you can just write down, okay, so furthermore, furthermore, we know that in, you can show that x, that sigma prime union v, again, if you look at the link of this, v of link of this, is again generated by a single element, is generated by a single element in the top degree. And then combining these two facts, we see that the coefficient of u0 tau prime differentiated along sigma must vanish, in particular, it will be independent of one of the variables of sigma, not in sigma prime, and hence we are compatible at this coefficient, and this we can just repeat. Okay, so this is for all the phases, tau prime, whose star intersects sigma, and that's it. So this combined gives compatibility, and this, then we are done, that's it. So now we have compatibility, we're done. I mean, there was one place here where I cheated a little, and I said, because I said that u was just in k theta, just right, I said that u was, I started out in k theta, right, but strictly speaking, I have to look at u defined over k theta theta prime, right, but since theta prime is a transcendental extension over k theta, right, k u k theta k theta prime is a transcendental condition, is a transcendental extension over k theta, if I have the property over all k theta, over all elements in k theta, I have it over all elements in the transcendental extension, and I'm done. Okay, that's just a transcendentality argument. That's it. And now this is, that's it. So this is the end of the argument, or at least the sketch of the argument. Let me sketch one thing that was gained out of this exercise of why would we want to extend this to cycles, and this is that we can treat complexes, that we can treat, call it complexes, that are not only spheres, but wedges of spheres essentially, so things that are called level. Okay, so let me give an application, let me think what kind of, what I can delete. So let me delete this part here. How this transcendental, can it get by specialization? Sorry? Can it get by specialization when you use theta? It's the same kind of... You're asking whether I can get anisotropy? Yes, or even a whole lemon, but for not transcendental extensions. I mean, once you have whole lemon, in some transcendental, in some field extension, you have it for any infinite fields over... If k is infinite to begin with, then you have it automatically over k, because it's an algebraic geometry, it's an algebraic condition, and once you know that there is a point where it is non-zero, then neutral normalization tells you that over k it is. The only issue is that if k is finite, then you don't have neutral normalization, and that's it. But otherwise, yes. Otherwise you get whole amount immediately. The only thing that doesn't go through is the total anisotropy that doesn't go through. Because, right? So the scrappy monomial ideals, they are defined over the smaller, they're already defined over k. But an element... To say that an element is square free, you're really small, or that an element squares to not something non-zero, you're really telling something about an element defined over a larger field, and then this is different. You just add elements that are not defined in the original k. That's the issue. So application. So that's the G theorem for two Korn-McCauley complexes. So what is a two Korn-McCauley complex? Well, that's a Korn-McCauley complex such that if you remove any vertex, it's still Korn-McCauley. So delta two Korn-McCauley, that's the definition. If it's Korn-McCauley, and for all vertices in delta, if it's Korn-McCauley, and let me write it in the next row, and for all vertices in delta, taking the complex delta without this vertex, I remove the vertex from circulation and take all the other faces that don't contain this vertex, is also Korn-McCauley as well. Right? So for instance, triangulations of spheres are Korn-McCauley, but some of our triangulations of disks aren't. Can easily show, right? Because if you have a disk, ah, maybe I, perhaps I have to assume that they are of the same dimension, of the same dimension. Let me assume it for safety. All right, so if I have a disk, then I could just remove some interior disk that intersects the boundary, but then here I will not be pure. I will be of a lower dimension. That's the issue. Okay. A Jabraic characterization, Jabraically, to Korn-McCauley, complexes are level, are level. That is, okay, so, I no longer have Poincare duality, but I have the following statement. So, um, for all K, so, for, for, for, that is, for alpha in a K of delta, there exists beta, ah, too common of dimension, let me say dimension d minus one, exists beta in d minus K of delta, such that, alpha times beta is not equal to zero. That's, that's the property of being level. Notice that I did not, it will require, that ad is isomorphic to the ground field. All right, so it's not one-dimensional. Okay. Corollary of Leschitz for cycles of Leschitz, huh? Ah, okay, so let me say, this is the theorem of Stanley, I think, or Hofstra. It's at least contained in Stanley's book. So I want Leschitz for cycles. I want to state the corollary of Leschitz for cycles. And for this, I introduce, so remember, remember that B of B, for this one you need also to impose some, I mean, you take a general, you take a general, it's just a linear system of parameters. It doesn't have to be generic. So for now, there's no generic here. All right? So I introduce B of mu, this was a quotient of, all right, this was a quotient of A of the underlying complex of mu. And in particular, somehow it doesn't really matter what complex I take. So I take, I could have taken any simpler complex, but then what I took is basically the smallest quotient of this that has mu at its fundamental class. All right? And now what I define is B of M. So M is just some quotient of the top degree of my, of M is just some quotient of the top degree of my, of my itinerary reduction. So let me define B of M. All right? So what do I do? Well, okay. So what I can do is a following. I can take this as A of B of M, A of delta. And I take A of delta modulo, the kernel of A of delta to B of mu direct sum of B of mu. And where mu goes over where mu is is basically a generating system for generators for M. All right? So what do I mean by by a generating system, right? M, I mean these mu are quotients of, so, but so these mu V what do I mean by a generating system? I mean that if I take the direct sum over these and I take the direct the induced restriction map, so direct sum over the restriction map that this is injective. So mu V such that this is injective. And then what does level mean? Okay, level means that A of delta is the same as B of the top degree A D of delta. And now what do I get? Well, I get the following variant of the left shed theorem, right? So I have, okay so let's look at B of let's look at B of M in general. All right? In degree K and in degree D minus K. Now what can I do? Well well I could restrict to any single one of these B of mu. That would be a suggestion. But I can also just look at the direct sum over all these mu V in B of mu direct sum mu V in mu V as above. Okay, now what do I have? Well if I have the left sheds here, so these are the vertical maps of injections if I have the left sheds here, so L to the D minus 2K is an isomorphism which is exactly the left sheds here for cycles then in particular the map up here is an injection. Left sheds here implies injection here. So implies injection here which means that L to the D minus 2K is an injection here which means that for instance in particular that so for instance the dimensions of the graded components B of M degree I is less or equal to the dimension of the graded components in degree I plus 1 for all I less or equal to D half limited. Alright, so you have a monotonicity. So what does this mean for the level complex? So in the level case in the level complex for level complexes or for 2K Macaulay complexes for the level complexes the H numbers of delta are monotone increasing up to I up to D half right up to the half point that is one of the corollaries and that somehow I think where I should finish because I'm already overtime. That's why cycles are important. You can now go a little beyond you're just way more flexible than than just spheres and manifolds. That's nice. Alright, thank you.


Okay, this is James P. Madonna in Lodi, New Jersey, and my buddies came to say hello to me. Wild Maladux.  Okay, Wild Maladux is the male. That's the female. They all live here. It's like a bird sanctuary for Wild Maladux. And they came to greet me. They kind of made their home in this creek, this babbling brook. I really don't know what the difference is between a brook and a creek, but hello. And that's that. See them all in the creek? Wild Maladux. You can see the jade green head on the male as opposed to the drab females, right? The ducks that are in luck, actually. These ducks are in luck because they get fed. They get fed and people don't hurt them, which is nice. I would say it's the Autobahn Society should make this a bird sanctuary for Wild Maladux here in northeast New Jersey. And I guess the cold weather don't bother them, being that ducks have a thick layer of fat on them. And there they go back in the icy cold water. All right, that's enough. That's enough of the Maladux. Let me get indoors visiting my friend, the Reverend Dr. William J. Eisen.


Welcome folks, we have the down industrial trading down 137 Nasdaqs, up 37 S&P's down 4. Gold, gold contract down $5.20, trading at $19.67 an ounce. We have a light-sweet crude up about $76.83, $39 a barrel, notes and bonds. A 10-year note, up $6.6, trading down $11.01, a 30-year up 14 at $1.21, $22, and Kingdoll. Now check this out, Kingdoll's up $454 ticks. And we'll go through this when we're on the show, but the bottom-on you broke the upturn of the conviction. You're going back up to try to test the under of that channel line. And what's so intriguing here is that with the dollars, the up to $454, the S&P's are still only down $2. So that's telling me that you're going to get a failure on the dollar once again, in particular because you still have notes and bonds at a high or price-low yield. Euros at $108, he ends at $145, British Pounds at $126 to $1 U.S. Dollar. Well, you know, let me take a look at the spy. What you have, this spy still wants the highs. That's the bottom line. You know, we'll see how this shakes out, but the bottom line is that you're right into it right now. You're at $451, you know, $451 is the bottom of the low of the high of the same day. I suspect, you know, coming into tomorrow, you know, we have jobs numbers tomorrow. Bottom line is that you had the inflation numbers coming today. They basically came in, you know, in line. And when I say in line, you know, if you take the .02, well, you're dealing with bottom line, like about a 2.4% inflation. So 2.4 is pretty amazing actually when you think about it. So that's where that seems to be going. We take a look at the NDX100, the 3Qs, same type of setup in the 3Qs. The bottom line with the 3Qs right now, you know, the Qs are up about $70. You have a contraction of volume happening, no doubt about that. But now you're back inside that higher range. Once you get inside the .370 full, well, guess what? The top is game, which is the .387. And more than likely, it's going to be on that jobs number tomorrow. Gold, you take a look at the gold contract. You get a sideways move on gold out here today.  You only get $120,000 contracts traded. You know, gold's moved higher with a lot more, there we go, with a lot more contracts traded, just to give you an idea. Yeah, it's coming into $175,000. Stay right there, folks. Come right back.


Today we are doing live fire revolutions for our certifications as Airport Rescue Firefighters. We are first response to any off-station aircraft emergency where we support everything that happens on the airfield. This is extremely important class for our guys to understand the importance of firefighting and saving government aircrafts. Of course, most importantly, saving the Marines that are inside the aircraft. Our firefighters have been extremely excited to be out here with this training. Anytime they can come out here and do live fire burns, they're always extremely happy behind it. We would like to continue to build upon this training, doing bigger, more complicated exercises and responding to drills onto the airfield. I feel that our firefighters here are extremely motivated. They're really happy to be here. They are enjoying the fact that they're coming out here and doing these live fires. It's hot out here and they're still going into these hot temperatures and just enjoying lives.