 So last time we heard, I told you how to construct iteratively left sheds elements using this Kronekalemma. So last time we had this Kronekalemma, which helped us to, essentially what it gave us is that bias pairing property, that the bias pairing property, so the non-degeneracy or the non-degeneracy of the Kronekalemma pairing at ideals implied somehow in dimension one implied the existence of left sheds elements. And then, well okay, so to close the loop we have to discuss how in general this bias pairing property, bias pairing is implied maybe by left sheds and could I mention one again. And this closes the loop and this is what I will, some of the first half will be spent on explaining that. Essentially what we discussed last time was how we get, some of our last time, last lecture, we discussed the bias pairing property for ideals of the form i sigma delta, where delta for sigma is sphere and delta I could I mention one sphere. And I want to essentially now go to general delta in a few small steps and then I want to argue how these things then really fit together, sum it up. And this somehow to give you at least a feeling for how the proof of the left sheds here works using this inductive principle. And then towards the end, so in the second half and tomorrow, I will go over another proof using transcendental theory, which does not really rely on an inductive principle but really just exploits some nice residue formulas. So let's go over this case of these ideals i sigma delta again. So last time we discussed this case i sigma delta, where delta was a could I mention one sphere. Today I want to verify this bias pairing property. And in the case where delta is general, I will argue that it's enough to consider the case where I consider a pair i sigma delta, which is just to remind you, this is the kernel of a map a of sigma to a of e. And here for sigma, and we will look at this in the case where sigma is 2k minus one sphere, over r field k, and e is a could I mention one manifold in sigma. And I will assume that such that e is k minus one acyclic over k, meaning that the homology of e vanishes up to dimension k minus one, and here is homology with k coefficients. And this is the case that I wanted to look at. And let me just remind you what am I trying to do. Well, I'm trying to show that if I look at the Poincare pairing in sigma to k, and I restrict it to this ideal, so this is the perfect pairing, and I want to restrict it to this ideal i sigma delta, then this pairing here should still be non-degenerate. So we want this to be non-degenerate. Well, you only have it in the middle, right? It's k minus one. Yeah. So it's strictly less than k minus one. Wait a second, so let me get to an example again. If k is two, then I'm looking at a three-dimensional sphere. The manifold is of dimension two, and I'm saying, ah, okay, you're right, up to dimension k minus two. Let me say it like that. Yeah, yeah. Yes, thank you, yes. Otherwise, it's just a sphere again. Is this connected or is it k zero? So h zero, it doesn't have to be connected, but I mean the case where we have a one-dimensional sphere is kind of trivial anyway. So the case that we're interested in is really going from k equals two upwards. So three-dimensional spheres are higher. So we don't have to assume connected, but with these assumptions, it will automatically be connected once we hit the interesting cases. But yeah, let's ignore the case of one sphere for now. All right, so we want to show that this is non-degenerate. Okay. And now let me first discuss how the criterion to show that this pairing doesn't degenerate. And once again, what we can look at is the case where, well, we can look at i sigma e, but now let us observe again that, so now it kind of, seems to be a mark becomes a little important. Just a second. This is a k sigma. Okay, so this is, so it's a middle pairing, the middle boundary. Yeah, yeah, so this is to a to k of sigma, which is just k. So e separates sigma into two components, into two components, right? So I have a component m and I have a component m bar. And again, I have that i, that these two components, they span my ideal and they are obviously orthogonal on each other because if I have a monomial supported here and a monomial supported here, then they multiply to zero because, well, they line different components, so they don't form a face with each other, all right? So they're obviously orthogonal on each other. So if I want to prove this, I restrict to proving it for, somehow we can restrict to showing non-degeneracy, non-degeneracy for i sigma m bar, okay? And here's the theorem. So i sigma bar, i sigma m bar satisfies the bias pairing property if and only if. Okay, so now what I do is I look at a k of e. Now, a k of e, this is, okay, so this is just the quotient corresponding to my hyper surface. Now, if you remember in this, when we discussed the partition complex, we remarked that there is a map of the cosmology into a k of e. So specifically we had a map from hk minus 1 of e with k coefficients to the d choose k to this ring, all right? So this came from the discussion of the partition complex in the context of Pankeriduality. And now, well, this here is, okay, this e is a sub-manifold of m, so I can write a map from hk of m, sorry, from hk minus 1 of m with k coefficients to, I can write down a map from hk minus 1 of m to hk minus 1 of e. And again, I carry with me this tensor product that is coming from the castle complex. And now, okay, so now I have a composition of these two maps, and I want this composition here to be an isomorphism. If and only if, let me call this map here star, and I can say star is an isomorphism. All right, the proof for this fact, it turns out to be just a little diagram chasing in the end. So what I do is, okay, so I write down a short exact sequence like this. So let me write down, okay, let me start with 0 up here and then try to make the arrows short enough so that I don't waste too much space. So I have ikme, which is defined for me as the kernel of a of ak of m to ak of e. All right, and this is a sub-complex, this is a surjection, that's all good and nice. And then into ak of m, well, again, I have my partition complex. So I have the map from hk minus 1 of m with k coefficients. Let me just not write down the k coefficients to the d choose k. All right, and what is the core kernel? Well, I mean, let me not describe it for now for the moment. Let me just call it b of m, bk of m for the moment. And now notice that our mystery map, right, this mystery map also appears again here. All right, so it remains to understand this map. So let's understand this a little. So this here I can think of as mapping, so this is included into i sigma m bar, which is isomorphic to a sigma m bar, which we discussed last time. And then, OK, so what is b of m, bk of m, let us discuss this. So b of m, this is a of m modulo. Well, now I look at the image of the direct sum over vertices in m, a star of the vertex in m. All right, so this was our partitioning map. In particular, this is just the kernel under the partitioning map. All right, so sorry, this is just the image under the partitioning map. In particular, this is the same as looking at a of sigma and modding out the annihilator of i sigma m bar. But as we discussed last time, having an injection from here to here, all right, this injection, all right, so now this is an injection from an ideal to the punctuality algebra to the string modulo, the annihilator of the ideal. So this is, the injection here is equivalent to the bias pairing property, bias pairing property. And in fact, it's an injection if and only if it is an isomorphism because, well, these spaces are Poincare duals. So this is an isomorphism if only if it's an injection. So now let's look at this. So this map here is an isomorphism. Okay, so if and only, okay, so this map here is an injection if and only if this map here is. But then these two spaces are of the same dimension. So if this map is also a surjection, this here will be an isomorphism. So these two spaces will be the same. In particular, I get this here is an isomorphism if and only if this here is an isomorphism if and only if this is an isomorphism. Which is exactly what we wanted, right? So isomorphism here is exactly the isomorphism here, which is exactly the isomorphism here. It's really just diagram chasing. So diagram chasing gives us a result, gives the isomorphism. All right, and that's it. So it's really just a diagram chasing. All right, okay, so now, okay, so now I want to, okay, so now I went to, I discussed hypersurfaces. And now I want to discuss general complexes, i delta, i. So how do we prove how to prove that i sigma delta satisfies the bias pairing property in general? So how do I do that? So I'm again in this case, okay, so I'm looking at a simplicial complex, right? A simplicial sphere and a subcomplex in it. So we are considering, so sigma is of dimension 2k minus 1 and delta subcomplex. So really, I can assume that delta is of dimension k minus 1. Why? Because I'm caring about this ideal, which is, all right, I'm caring about this ideal in degree k. So I'm really only caring about, I'm really only caring about the monomials up to degree k, I'm only caring about the faces of the simplicial complex up to cardinality k, which means that I'm only caring about the simplicial complex up to dimension k minus 1. So the idea now is construct hypersurface e containing delta such that a of sigma, sorry, a of e is isomorphic to a of delta and in particular that i sigma e is isomorphic to i sigma delta. And we will cheat a little, we will not achieve this in sigma itself, but we will achieve this in a subdivision of sigma, but we will see that this doesn't matter, that we do it in a subdivision. Okay, so let me explain that and I will mostly restrict to the case where sigma is a PL sphere just because I want to get to the transcendentality argument. So I will sketch the argument and this construction and why this is enough. And then I will kind of, this will be the, then I will summarize the, what we did and summarize the proof and then I will go to the transcendental proof. And so what, how do you do that? Ah, okay, so. Are you going to construct e and such that in the cycle? Yes, yes, I will, okay, so I will construct e such that it is a cyclic. So this here will be k minus 2 a cyclic, thank you. I will, yes, thank you. So, okay, so let me go over this construction. First of all, let me, let me note that the decomposition theorem implies something nice about some invariance property of this, of this, of the bias pairing property. Of the invariance property of the bias pairing property, there's too many properties. So an invariance of the bias pairing property under subdivisions. And for simplicity we will mostly assume, assume that sigma is actually a PL sphere. Because, well the construction is a little simpler in that case. Okay, so first observation, subdivisions of sigma outside that do not affect that do not affect delta preserve the bias pairing property. This, so subdivisions here, for us we will take stellar subdivisions. But it's essentially every, what you can take is essentially every simplicity map that preserves the fundamental class. I will give an example in a second. So for us, for us here stellar subdivisions, for us stellar subdivisions survives because I restrict it to the PL sphere case and I will just indicate what other marvelous things you can do if you want more general. And let me make a picture to illustrate this. So let's say we have our sum potential complex delta. All right, so this is delta and it sits inside our sphere sigma. This is delta and it sits inside sigma. And then, now perhaps for some, for some gods forsaken reason, we don't like sigma so much. For some reason it looks ugly somewhere and we want to refine it. So let's say here there's a triangle somewhere and we want to really look at the blow up. And so we want to take stellar subdivision here. So we have this and we want to replace this triangle by its stellar subdivision. All right? In any case, whenever you have a subdivision of simplicity complexes, you can induce a map. So from the complex before the subdivision to the sphere after the subdivision. And this here will be an injection. It depends on which point to take, which point of view you take. But if you think about it on the level of converse polynomial functions, it is obvious, right? Because a converse polynomial function before the sub-division will be converse polynomial after. Right? Somehow you just have more, right? You just have more space, but somehow you just don't have break points at the point of the subdivision. That's fine. Okay, and then if you think about it, what you can show is that, well, that the subdivision here is really the image under this map. It includes the converse polynomials before and after. So this is just the pullback map plus a component that really comes from the subdivision. So this is really the image of the geese in. So let me just call it G, not G, G. And this here is orthogonal under, this decomposition is orthogonal under the Poincare repairing. All right? And now let's think about it. Let's look at i sigma delta and its subdivision. So i sigma prime delta and i sigma delta. So i sigma delta really consists of all the monomials that, somehow this here, all monomials not in delta. What does this mean? Well, these are all the monomials outside of delta. And now if I, okay, so this year all the monomials in sigma prime outside of delta. So really, I have the same decomposition here. So why does this mean, okay, so again, so I have this decomposition, this is orthogonal. Why does this mean that the bias pairing property, so the non-degeneracy of the pairing is true before, if and only if it is true after? Well, it's kind of clear, right? Somehow this, the splitting here is orthogonal. So I really only have to, so here I have it, okay, so if I have it before, then I have it here. This bearing, this here is the image of the geese and on which I have Poincare reduality anyway, right? Because I have it, right? Otherwise, I would already fail here. So I have Poincare reduality here. In particular, I have it here. The other direction, right? If it fails here, it must fail on one of these components. And that's it. There's nothing fancy about it. So this subdivision really preserves the pairing. Now let me note that really there was my, so now I did stellar subdivisions. I could have been more fancy. I could have said this triangle is really subdivided into something more fancy like attaching a torus. So this looks like a little minion now to this, but I mean, so really I could have been more fancy and more general, but let's sum up. So for the purposes of PL spheres, I really don't have to go into any fancy kind of subdivisions. But this is really, this is a more general principle. Whenever you can define this map on the convex polynomials, right? Whenever you have a nice substantial map, then you can define the map on the convex polynomials. Then you have to make sure that this map preserves the fundamental class, so the Poincare pairing is preserved. And then you get this decomposition. All good and nice. So now the next step is something that is kind of very nice and old in the anthropology is this folklore fact that if I have delta, a k minus one dimensional complex, and delta lives inside sigma, a 2k minus one sphere, then delta embeds into the boundary of its regular neighborhood and regular neighborhood of delta. All right? So example, if I have a graph in the three-sphere and then I take the neighborhood of the graph, then I can just by just by basic general position, I can move this graph into the boundary of the regular neighborhood. In particular, there exists a refinement, a refinement sigma tilde of sigma such that, well, okay, so I should say refinement sigma tilde of sigma not affecting, not affecting delta such that delta lies in partial of the boundary, somehow delta is a sub-complex of the boundary of the regular neighborhood, which is again a sub-complex of sigma tilde. All right? Okay, so how does this help me? Maybe I should move this up because that's better. So now, okay, so now I have delta sub-complex of some closed hypersurface, right? Delta boundary n. So this is what we have arrived at now. Now we are, I mean, careful. Now we are in sigma tilde. But as we observed there, the Poincare, but we can look at i sigma tilde instead of i sigma tilde delta instead of i sigma delta. So we are fine. Okay, and how do you move delta inside the boundary of the regular neighborhood? The thing is that you got the delta on the boundary. Yes. That is because, okay, so what you can do is take the regular neighborhood, all right? Let me try to, I mean, I cannot draw an example. I'm a very two-dimensional drawer, so I cannot really draw it well in dimension three, but here's the gist of it, right? So here's your regular neighborhood, all right? And now push your delta into general position, all right? Push it into general position. So if you do this in dimension three and not two, what you will have will not intersect the original delta. Push it into general position. Let me be like this, all right? So the issue is here now, of course, yes, yes, you do just do a radial projection to the boundary. All right, all right, all right. Where was I? Ah, yes. Okay, so we have delta in the boundary of n, okay? So we have, let me draw delta and hyper-surface. So we have delta and then it lives inside this hyper-surface here, all right? The issue is, of course, so what do I want? So I wanted that A of E, so the quotient of the phasoring, the quotient of the phasoring of sigma corresponding to E was actually isomorphic to A of delta. At least I need this in degree k, all right? And again I should write only in white. But in general, at this stage I only have a surjection, all right? I have a larger complex. So I can only say that I have a surjection from this larger manifold to delta. So what do I do? Well, if this is larger, then there is some monomial outside of, then there is some other, then there is an element of, then there is an element of A boundary n in degree k that lies in the kernel of this restriction map. In particular, there is a monomial, right? Some of this is generated by some monomials. And there is a monomial in this, there's a monomial supported outside of delta that generates this element in the kernel, all right? So there is a monomial somewhere, perhaps here, that generates an element in the kernel. So what do I do? Well, what I can do is I can just remove this phase and its neighborhood, all right? I remove it and I leave it a little hole here, okay? And I can do this and repeat this several times until I have a surface with holes such that this is an isomorphism. So kill elements in the kernel, in kernel, by introducing holes, okay? Now, so far so good. Now the issue is, okay, so it's not so hard to see that we can make this surface sufficiently connected, right? If there were two different components, all right, so if boundary n had two different components, then what, just because maybe our graph had two different components, right? If delta had two different components, then what we can do is we can just attach a handle inside Sigma and make them connected. So the connectivity is not an issue. And us drilling holes does not affect the connectivity, but of course now we have a surface with boundary, all right? So e is now not closed. Problem, e is not closed. And so now what we have, so we have an e that is sufficiently connected that in degree k precisely encodes delta, but it's not closed. But now it's really simple. There's a simple trick that we can use to actually produce a closed hypersurface that does a trick for us that encodes a bias pairing property nicely. So let me use this backboard to finish at least that part. So we have e k minus 1 acyclic, but it has bound, it's now ak of e is isomorphic to ak of delta, but e has boundary. And now basically what we are doing is just we double e. So we look at e solution, so solution, well take e, double it, compactify and consider the resulting ideal. So let me explain what I mean. Okay, I will explain what I mean by compactify. So let's look at e on its own, all right? So e is my surface with boundary, all right? And then, okay, so as Pierre already says, okay, I can just double and obtain, all right? I take e, another copy and identify them at the boundary, the canonical way, all right? I have another sheet on the bottom, like this. So there is no compactification here, you're all right, but e lives somewhere, all right? So I should have said this construction that we did is obviously orientable the way that we did it. So orientable. So now, okay, but so I have, how do I deal with sigma outside of, right? So this is here outside, I have sigma tilde, all right? So really there are some phases here intersecting, so phases of sigma outside of e. And now I have an orientation and I have an upper part and a lower part. I have simplices intersecting my e from the top and simplices intersecting my e from the bottom, all right? And when I, okay, so now I want to see this in a new, somehow I want to look at a compactification of sigma tilde without my surface e as amount, right? So I want to compactify this such that the boundary compactify, such that the boundary of the compactification is exactly the double of e, all right? And really what I do is I really just encode what phases do I intersect from the top, what from the bottom, and that's it, right? I get a new superficial complex. So I get, all right, so this is my sigma, sigma hat. And now I'm done. Now I have a closed hyper surface in a new sphere, so sigma in a sphere sigma hat. And what remains, and this is somehow, okay, this is a diagram chasing, I will skip, is to show that i sigma tilde e satisfies the bias pairing property if and only if i k sigma hat double of e does, all right? And now I can go back to this theorem that I did for closed hyper surfaces and apply it here, and I'm done. All right, okay, so let me summarize. There is one more caveat that I have to go over, and then after the summary. If I identify this two copies, I get back my homology, yeah. I mean, so to fill it in here, fill the resulting antibody in, this will be our homology sphere again, yeah. All right, so for the summary, so what do we have? We have the Konek Alema, and what did the Konek Alema give us specifically, right? It told us we can construct left shits elements, provided we can prove the non-degeneracy of pairing, not, let me say tomorrow, maybe in sigma, we can show the non-degeneracy of pairing. What pairing did we look at? Well, we looked at, I mean, not the pairing, what spaces, what ideals did we look at? Well, we looked at the kernels of these generic linear combinations of X v, right? But this is not tomorrow, it's not so important for the moment. What currents we looked at? We looked at the kernel of the already constructed map, right? Our candidate for the left shits map that came close in A of link of the vertex w in sigma. All right, remember somehow there was this, okay, so we wanted to apply this basic representation theory of the Konek Alema. All right, so we noticed that we have to show that the kernel of the old map mapped under this model, the perturbation, the map that we want to add, the divisor that we want to add. And we want to say that this does not intersect the image, all right? And then we observed that this is equivalent to just saying, okay, the pairing on either one of them does not degenerate, and then we were done. But so this we have to show, all right? So this here, all right? So this is, notice that this is in co-dimension one. So we really gained something here in terms of induction. And now then the second ingredient was bias pairing property, right? So it's about non-degeneracy of the pairing at ideals at, okay, so at i sigma delta, well, is implied by left sheds properties at hypersurfaces for hypersurfaces, all right? So this was exactly, we identified, we looked at A of E, all right? So why was this a left sheds property? So what we looked at here, all right, was we looked at A of E. And this was, okay, so this was a co-dimension one manifold inside sigma. So sigma was of dimension 2k minus one. This E is of dimension 2k minus two. So the co-dimension of sigma of K of sigma is one higher than the co-dimension of K of E. So we took out this additional element and we wanted the kernel under this element to be prescribed. All right? So now we wanted to have the kernel and the co-kernel under this element, under this multiplication to be prescribed, which is a left sheds property. All right? So this here, somehow, this was K of E, right? So we took out some of our linear system of parameters, theta for coming from A, coming from sigma. But this was really, somehow, it was one too long. So we think of this as really of K of E modding out linear system of parameters that was one shorter. This here is the attenuary reduction and then we said something about how this last element, L, it's nice that last and left sheds have the same letter, that this last element, and then we said something about how would this last element act? All right? And this last element acting, this is really something saying about middle isomorphism, which is saying something about the left sheds. All right? So this is a left sheds property. All right? Okay. So the non-degeneracy of the pairing at ideals of this is applied by the left sheds properties for hyper surfaces. And so we see that we actually, we gain in dimension in both steps, all right? Seemingly, somehow, we are in a very nice position. The only issue here, this is why somehow I said there's a caveat and we have to want to explain a little how to get around that caveat, is that these ideals here and these ideals, okay, so let me just say it like this. So these ideals here are not in general of this form, all right? So these are not in general monomial ideals. So let me explain briefly how to get around that issue. And then after the break, then we will do a break and then we will go to the transcendental theory proof. So issue, these ideals here are not monomial in general, not monomial in general, or orthogonal complement to monomial. All right? So the issue is not, so if I just looked at this here, this would be the orthogonal complement to a monomial ideal, that would be fine, but I'm pulling it back to the link of a vertex, which I'm intersecting then to ideals, and that is not so easy to say that this is a game monomial. So I have to say something. So intersection of monomial ideals in general is not monomial. So how do I deal with this? So let me make some space somewhere. Let me make some space here, all right? So maybe it's not so clear that this is not in general monomial ideal. That's the camera. It's in general just the orthogonal complement of the monomial ideal, but I could just as well look at the orthogonal complement, which would be the image of this generically in your combination, and this would be clearly a monomial ideal, but the image, if I already satisfy the transfer of the prime property, is just the span of the images of the individual maps. But this is just generated by the monomials of the individual v and the index at w. Okay? So how do I remedy this? Well, somehow, let me say that this here is the orthogonal complement. So let me repeat. It is orthogonal complement to a monomial ideal if the union of the vertices star v and sigma, where I go over v in my index at w, and I take the union of the vertices again, star v sigma, with the star of the new vertex star w and sigma. If these two objects here are sub-manifolds, sub-manifolds, okay? So if this is the case, then these are nice ideals that I can deal with in a nice algebraic way. So how do I then... Well, let me give you a situation where this doesn't happen, and then let me describe how I get around it. Yeah, it's really just the neighborhood. I take the union of the stars, right? So I take all the faces that intersect my vertex set, and I draw it in a larger closure, right? The issue is that tomorrow... I could have some picture that looks like this, and then maybe another vertex here, and this could have some nasty singularities here. So this here could be... Let me draw it nicer. So this here could be another one, another vertex in my index area, it's somewhat degenerate, it's not a manifold, so I'm not in a good business, okay? So that's a good position. It's not a manifold. So how do I get around that? So let me sketch off how is this circumvented? So, yeah, sketch of proof. So, conventing the caveat. So here's the trick. So let's say we want to prove... we want left sheds for, let's say, for explicitness, let me say, for sigma 2k-sphere, 2k-dimensional sphere. And let's say it doesn't have an order on the vertices that satisfies this property. So you can show that most spheres that you will be looking at in daily life, they will not have an exhaustion by vertex of the vertices in some order such that at every intermediate step you have a manifold. So let's say this is one of these nasty spheres where you don't have an order that is nice. So let's say we just fix an arbitrary order. So arbitrary order on vertices. And now the trick is the following. So we look at sigma, and this is a sub-manifold. It's a co-dimension once here. In this here. Yes, yes, yes, exactly. It's a weaker property than shellability here, but still you can find many examples where you cannot get it. That's right. So this is a weak form of shellability. Weaker than shellability, but still much too strong. But what we do now, so sigma we can think of as a co-dimension one sphere in a bigger sphere 2k, in a bigger sphere 2 sigma bar. So this is then a 2k plus 1 sphere. And then we saw that observe left sheds for sigma. This is what we discussed on Monday. This is equivalent to saying that the ideal i sigma bar sigma satisfies the bias peri property. And now this is equivalent. Well, now I can again reduce to the k-skeleton. So this is in degree k plus 1. So this is equivalent to saying that i sigma bar, and I take the k-skeleton of sigma satisfies the BPP, the bias peri property. All right. So now let me call this an empty composition because everything intermediate is a manifold. Maybe m is not so nice, but an empty composition. So there's an order on the vertices such that every intermediate step you are a manifold. So now what I do is I find a refinement sigma hat of sigma bar such that there exists a hyper surface with boundary such that... Well, what do I want? Such that a k plus 1 of e is isomorphic to a k plus 1 of this k-skeleton of sigma. In particular, the hidden motive here is that the ideals are the same. Sigma hat e isomorphic to i sigma hat e, sigma. All right, that's the motivation, the hidden one in degree k plus 1. So I find this and such that e has an empty composition. So it turns out that in this code I mentioned, you can actually construct e in such a way. You can construct many hyper surfaces with the first property. And it turns out that basically by doing some surgery and twisting this hyper surface a little, you can ensure that this decomposition is actually a nice empty composition that you get. So now... So now what do we have? Well, then the bias pairing property. Then the bias pairing property for i sigma hat e in degree k plus 1 is equivalent to a left shift property for e. All right, this is what we observed. But now this e is nicely decomposable and we can actually imply the induction. So now the e is nicely decomposable. But to construct the left shift elements here, well, we apply the conical lemma, but now the kernels here, they are nice. In e, they are nice. They are nice and orthogonal complements to monomials. So let me... I have this somewhat bad habit of if there is not enough space and I will just squeeze it, so let me not do that. But in e, the kernel of this direct sum of an initial segment v initial in the order under the stars of vertices in sigma pulled back to a link of w in e is monomial and so we can complete the induction now because now we have a monomial ideal and we have gained the dimension. So e is of the same dimension at the starting sigma. So the dimension of e is the same as the dimension of the sigma that we started with when we wanted to prove the left shift. But now we actually gain a dimension because we are looking at the link here. So now we have the dimension. This is the dimension of the link of the vertex in e plus one. But now we gain a dimension and therefore we can complete the induction. And that is the overview of this argument. All right. And this is somehow where I want to end with this argument. Then after a 10 minute break, I will go to the transcendental theory argument. All right. So for the final section, so what we will do is we will actually, we will talk about slightly more general objects and we will give a new proof based on some very nice residue formulas. So beyond positivity again, the cycles and transcendentality. Not in kind of a new age way, but in the sense of transcendental extensions. Okay. We have what is our object? We consider mu as a simplicial cycle, k-cycle, which for me is a pair of a simplicial complex, mu. So this is a simplicial complex, mu in these brackets, a simplicial complex, which I will call the underlying complex or support or underlying complex. Maybe let me specify the dimension here for simplicity of dimension d-1 of dimension d-1. And I have mu of the underlying complex and this is also of dimension d-1. And then mu is an element in the homology dimension d-1 with k coefficients of this underlying set, of this complex. And this is what I call for me a simplicial cycle. All right. Now let me consider a of the underlying complex. This is always in a hidden way. There's also always a linear system of parameters here. And this is just the phase ring again. So this is really just, again, k of the polynomial ring modulo i of. All right. And now what I will think of, well, is, well, I already, when we discussed this last time, I explained that there is a, that there is a canonical isomorphism between hd-1, this underlying complex, and ad of the simplicial complex. So this is a canonical isomorphism here. And what can I say then? Well, now what I can consider is a dual to some mu b. This is a dual to mu in the homology. All right. And what I can in particular do is I can think of mu b as a quotient, mu dual as a quotient of ad. Sorry. I don't get what you're doing. You do the slope on creditability because it seems it's good. It has come over. You don't have any class in commons. This is just a top degree. All right. It's just, I'm paring on the final. Okay. So it makes no sense. We have kind of like fundamental chain. Yes. But you don't have canonical commons to class. Maybe you can say that you have canonical chain. Well, I still have a paring, right? I can still just... Oh, maybe you're right. Because you have not only commons to class but canonical chain, which interprets as a quotient. Yes. Exactly. All right. And then what I can do is I can consider b of mu. And this will be the smallest quotient of a mu, such that b of mu in degree d is exactly this class in degree d. All right. Kill everything, yeah? Maybe get it lost. Yeah. You said your order doesn't... It's ad of mu. Yeah. Maps to... It's co-homologer. It can pair with this homologer class. It can make to your field. Yes. But not to m check. Yeah. Because... So I want to... I would just consider this as a one-dimensional... All right. No, it's nonsense. What is mu check? The mu is... You arrange your problem in... OK, OK. So... So mu check is... OK, so now I have a pairing of degree... of degree d to my field, right? To k, to my ground field. Yeah. You get a function of all your things too. Yeah. So I... OK, so now this pairing will give me a one-dimensional quotient of this top degree. All right? Of ad. Yeah, or you get one-dimensional quotient of co-homologer, yeah? Yeah. It's not a class in co-homologer. Yes, I get a one-dimensional quotient. Yes. It's not a class. You say that mu... Maybe you want to write with mu check. It's a dual to mu in h-dimensional. It's a quotient. Yes, yes, yes. That's what I mean. So it's a quotient of hd minus one. Therefore, I think of it as a quotient of ad. No, because literal and corrective. Like, you get elements here and... Ah, OK, so yeah, OK. OK, so quotient, OK. So it's quotient, yeah. OK, thank you. Of hd minus one. Yeah. All right? And what I get here is essentially the... The quotient of A of bracket mu with the fundamental class, right? With the prescribed fundamental class induced by mu. All right? So, and again, this still depends, right? So B of mu, still in a hidden way, depends on theta, right? So this is still theta inside. So I should write, right? There's still theta encoded inside. All right? And this is a Poincare duality algebra. It is a Poincare duality algebra. Duality algebra. The fundamental class in degree d. OK, so now I can ask again... OK, so I can ask again, does B mu have the left sheds property? And again, the theorem, maybe I should not start on the bottom of this to write the theorem. Yes, sir. Where do you stop? You have something, you have a manifold. You leave these inside the manifold. It doesn't have to be in the manifold. It's just a simplification complex. Yes. And then you have a homology class in the simplification complex. There's no manifold here anymore. OK? And so the theorem is as follows. This is going with Thomas Papadakis and Valisirica Petrotou. And for this, I will not just say a generic element. So I will be a little more specific. So mu d minus 1 cycle over K arbitrary. This is an arbitrary field. And now I take a field extension and what is the field extension I take? So while I have the matrix theta and its entries, and then I have the element L, the left sheds element L and its entries. And I see them all as some algebraically independent numbers. So each of the entries is a variable and I think of my new field, I give a field extension where each of the coordinates is an algebraically independent variable. OK? So I have all my entries here. And then basically I take the field extension and I take the field of rational numbers with respect to all those variables and this here will be K tilde. OK? So K tilde, rational field of rational functions of rational functions over K and I extend by the individual variables theta and L. And now I mean immediately a generic element. Then B of mu satisfies the left sheds property. So IE, so let me B mu K to B mu to D minus K L to the D minus 2 K is an isomorphism. It satisfies the whole amount of relations that is, right? So the hot dream on bilinear form QKL does not degenerate at monomial ideals. And then let me give one additional property that is very nice, know it in characteristic 2. So if the characteristic of the ground field is 2 then we have something even nicer and even more beautiful. This is that Q, the hot dream on bilinear form never degenerates. So QKL of U, U is not equal to 0 for all U in B mu K. And here I should say K should be less or equal to D half. So that's kind of the strongest form of not degenerating anywhere. This is somehow no ideal. I degenerate at no ideal. Surprising. But that's what happens here. All right. Okay. So now we are all over this field extension. Maybe I should say, right? This is now over this field extension K tilde. Okay. So why is this surprising? Go to the classical hot dream on relations. All right. You have the signature plus one and minus one. So definitely there will be one point where it will just be zero, the pairing with itself, just by intermediate value theorem. And you can show that tomorrow, I mean, if I pass to the, I mean, if I pass to the algebraic closure here, right, if I take K tilde but take the algebraic closure, then definitely there will be points where this is zero. So this is really something that only works in this. Okay. Yes. Yeah. I mean, you can also achieve this. I mean, okay. So let me open question. What happens in other characteristics, right? Other characteristics. But the point is this will never be true if you pass to the algebraic closure, right? So you take the extension of the complex numbers by these transcendental variables, then it might be true. But if you take the algebraic closure of K tilde, it will not be true. That's the point. Okay. Okay. So let me give like the first few indications of what we will need. I mean, we will end in like 10 minutes. So I think I will have to repeat anyway. So maybe let me give you some corollaries and then give you an idea of what we will do. All right. So corollary, well, what are some of the things that we need to do? We need to do something. We need to do something. We need to do something. So corollary, well, what are nice cycles, right? So if sigma is a sphere, right, then you take just the fundamental class as a cycle. Then B of, or A of sigma is just B of the fundamental class. Similarly, if M is a manifold, an orientable closed manifold, then B of the fundamental class is really just, well, it is A of M modulo, well, the kernel of the partitioning map, right? The kernel of fM to the direct sum of the vertices in M, A star of the vertex in M. All right. This is just the kernel of the partitioning map. I mean, but somehow, what else is encoded by these? Well, for instance, pseudo-manifolds. If you have a pseudo-manifold, if P is a pseudo-manifold and it's orientable, then B of mu satisfies left sheds. So B of fundamental class satisfies left sheds. So these are nice examples, but these are, one has to be a little bit careful. So this B of mu here, you could look at its Betty numbers, right? You could look at the dimension of the K graded component and try to extract some combinatorial meaning. So for instance, for manifold sigma, or for manifolds of spheres, and you can also show that for pseudo-manifolds, the dimensions of the graded components are, in these cases, the dimensions of the graded components here of the Bi is independent of the linear system of parameters. So they have a combinatorial meaning, even though in this case, we don't have a closed formula, no closed formula. So in the case of cycles in general, one has to be careful in the sense that there is no combinatorial meaning immediately. So it's really somehow a rather interesting property to have left sheds for cycles. And perhaps if I'm done with this sketch early tomorrow, then I will give some applications of this. But really there seems to be no algebraic geometry analog. So this is kind of left sheds, but really there seems to be no good algebraic geometry interpretation of this left sheds. Because it should depend on the coefficients of the cycle. Yeah, yeah, it depends on the coefficients. It's kind of, yeah. Constructible function. Yes, yes. Yeah, but it's a kind of, it's a, it's small. It's, yeah, it's a left shed theorem that is very far from any kind of geometric interpretation. All right, so what will be, what will be the idea, what will be, what will we work with? Well, we will basically look at residues of the degree function and work with them. And so the main trick is, well, first let us, let's define the degree function, right? So the degree function is just the map, right? It's just a way of identifying B of mu with K, right? B mu on degree D with K. So how do I define this degree function? Well, here's a way to write it down explicitly for facets. So define, define this, or kind of, if you want, just normalize this by defining it on a facet. Define this by, okay, so I take a facet for F facet of the underlying complex. So this is again a D minus one-dimensional face, D minus one-dimensional, simplex. And then we can evaluate the degree of corresponding monomial XF in the following way. I consider the vertices ordered, and then I look at my cycle, right? And I look at the ordered, the oriented coefficient of my typical cycle. So remember, mu, this was an element in HD minus one of the underlying complex, right? And then I divide this by the determinant of theta restricted to the minor corresponding to F, okay? This is the determinant of this. Remember, theta was this matrix, right? I think of theta as a matrix with some entries and then face F cuts out a minor, and I compute this determinant. And I can prove actually that there is, well, actually, yeah, let's not do this today, right? So we will tomorrow prove that somehow we will see that this is consistent. Maybe I will give you just intuition why this is consistent. And then we will look at this function, and in particular, not for function for faces F, but we will look at faces. We will look at, we will try to evaluate this at monomials of the form X tau squared, tau is a face of cardinality, cardinality d half, or at most d half inside, yeah, of cardinality d half inside my complex. And then we will see that this actually, this function really has very nice poles, and then we will compute some, yeah, we will do some basic analysis with this, prove some nice identity, and then this anisotropy, this total anisotropy, the anisotropy everywhere will just fall off, and then we have to modify a little to do this in general characteristic, and we'll prove just the Hollermann relations there. All right, and I think somehow it doesn't make sense to start with the proof now, so let's finish here, and then, because I would have to anyway, repeat a lot tomorrow. All right, thank you.