So last time I stopped with a with the argument for punk a reality We will do this more general anyway soon, but somehow now I want to actually go to the hard left shits here and its applications a little and Then first of all, I will I will cover the classical version Well for Toric varieties at least and then I will go to this Do the left shits version beyond positivity. I will motivate it by questions in PL topology and others But The the meat of the the mini course will be the proof of that Okay, so one three hard left-shits serum One is just the just the kind of the overview without really the meat of the proofs. That's that's one for one So hard left shits and company so Let's do the classical version first so last time we discussed sigma in the case of it being the boundary of Simplicial polytope and Let me fix the dimension. So P is a simple show D minus one polytope. Sorry D polytope so that the spheres of dimension D minus one And then we consider this algebra A of sigma all right we consider Sigma and there were two ways to define it that were Isomorphic, so there was a way of thinking about this as the algebra of cornwise polynomials Model or the idea generated by the global linear functions and then there was a The the this phase ring way of taking a polynomial Just the the free polynomial ring and then taking some quotient and somehow now it's good to remember the definition as A ring of cornwise polynomials because I want to want to consider consider This ring and I consider L in A1 So I think we want one function. So this is just a Function that is cone wise linear all right. So it's linear on every cone of sigma and this should be convex and let me say okay, so the Let me say strictly convex and explain what I mean because the convex geometries in the audience might protest a little Because this is not strictly convex in the convex geometry sense. So let's say I have my I have my sphere sigma right and I see it as a fan and What I want is the function to be well cone wise linear Convex and the domains of linearity should be exactly the cones of the fan. All right, so This I could draw the level set at one and this would be an example of such a function and This All right in red now Would not be an example because it is not strictly convex here All right, so this the domain of linearity is just larger than the domain of linearity there So this this is a fan, okay I think of it right so so I told you think of it for the moment as the functions commerce pyramids on the fan So Now what I draw is the level set at one of the functions Yeah, yeah, I mean I can always assume it because oh well I yeah Yeah, I Take out the global in your functions. I can always assume that it's somehow that is positive everywhere now I take the level set at one Okay Yeah, and so this is the full one this a sigma is the one No, this is a finite dimension this is a quotient. Yeah, but it's secretly depends on some free parameters Yes, yes, yes, but I'm not to mentioning them for now, right? So I'm thinking about it somehow here I have already the simple should be poly top and as I said if I if I have a poly top Then I have already the vertex coordinates which give me the linear system. All right All right, this was I mean here if I think of it as the algebra of commerce polynomials are implicitly already have the linear system All right, I took so here for here think of this here again as P of sigma Model all the ideal of global polynomials Global in your function. Sorry. I mean the global polynomials also, but yeah This ideal generated by the global linear functions So and P of sigma. Yeah, this was the comb wise polynomial functions polynomial functions Okay, this was a coin Okay for for for that for the moment. This is general enough Yes, yes, yeah, remember Yes, yes, it's exactly this condition that each of these cones here is full dimensional Each of these each of these simply sees fill spans a cone. So each K minus one simple expands a k-dimensional cone Yes, but it's a sphere, right, it's gonna call it. Yeah, it's even going strong. That's right So and I have this strictly convex tip convexity, right? This is what I drew here. So this here This is a level set at one The red part is not strictly convex So convex but not strictly convex The white part is Yes, yes, yes Exactly, so these are the ample and models and if it's just convex, it's just Then for all K less or equal to D half We have First the hard left-shed serum. Oh, let me call it the property This is a real space if I have rational coordinates my vertices of rational coordinates, so this is over R Yes, but we don't even need it We will see like a purely a purely convex geometric combinatorial proof of this Okay, we will not go to the algebra at all because I want to get to theorems where we don't even have a variety anymore And we will still prove the left-shed so in the end you should follow I mean for this for for the mini course you should forget that there was a variety because we will not use it at all Okay Is not smooth, yes, yes, it's not smooth, but we will go even further or fuck so it doesn't really matter Yeah, it's rationally If it's rational then it will be toric over fault But if it is somehow if it's just real right now if the polytop is just real coordinates You don't even I mean there are some constructions that kind of mimic This in the real when you have real coordinates, so you can build things that behave like a toy variety, but Then you lose Usually you cannot apply the the classical proofs of left-shed anyway, so we will do it anyway purely combinatorial, okay offer Yeah Yeah kella currents kella currents, yeah This is a moisture zone manifolds these things there are things that exist in this in this context, but it's a Still I mean we will not use it Yes, yes, yes, yes But I mean this leads into the wrong direction. Let me cut it off at this point. So hard left sheds. So All right, so this is the isomorphism between Well, so I mean we already know that these spaces are some of a collector spaces because we have punker a duality, but we want to Realize this isomorphism by a multiplication in the ring so L to the D minus 2k and this is not some office Yes, yes, yes, yes mixed yeah, this is a mixed version of the left-shed theorem It also works. Well with the proofs that we will encounter They immediately imply this mixed version as well. It's immediate. Yeah But let me not state it for now. Okay, so L is this L Okay, so for Maxim this is ample line bubble, right? Right. All right and the theorem does not come alone comes with a Relative that is that we may prove will be proven In company with it. This is the Hodgman relations and for this well, I define the following quadratic form on Degree k and with respect to L and this is just I take ak times ak and I multiply to Um Degree well, I want to I want to have a perfect pairing so I multiply to degree D and what do I do? Well, I take a and B All right, and I Send this to well, okay, so I take the degree of a times B times L to the D minus 2k all right, that's exactly what I want from the left here And so the degree is just a canonical identification of this degree D component here with With the with the reals. We will encounter this later explicitly But for I mean to give it an orientation and you can just say Well, okay, so you just have to say what we would just want to say what is positive and what is negative and Let us come to the convention that somehow the degree of a monomial here Right to face a monomial of a facet is positive, and that's it and Now what we want to do is Well, we want to make a non-trivial statement about this form, right? So the hard left side just says this will be perfect But we want to say a little more and this is if we look at the primitive forms on the L AK Exactly Where was I I want yeah, I wanted to say what this is I want to say what this is yes So I take PL a K and this is the kernel of the map from a K to a D minus K plus one of sigma induced by taking this element L, but multiplying one to far And so L to the D minus 2k Plus one and then how are they connected in the Hodgman relations? Well, we want this to be side then Ql qkl is definite of sine minus 1 to the K on The subspace pkpl a K in a K of sigma All right, and that's it That's all human relations Oh too far Okay Minus one to the power K on the subspace that I defined here. This is a kernel All right, and now let's talk about some Applications, so remember so for corn Macaulay complexes For Delta corn Macaulay The The H vector which was No, no for code for Delta corn Macaulay what does mean that Corn Macaulay's property of the ring. Yeah, well, it's no no no no, but it's a common code But we had horse's theorem, right? We had also a theorem But also theorem it's not it's you can it's a property of the ring But which is also a property you can also formulate it as property, right? So over Yeah, yeah, well, I mean yeah not not not necessarily it's a Homology-wise, it's a wedge of spheres right with respect to the k homology the H vector was a right which was the dimensions of the graded components right of the of this ring this was an m vector all right, so we had that That this was characterized as The the Hebert functions of polynomial rings right so this was just what we did last time and now with this With this theorem what we can say is That the following vector so the g vector the gi Vector which is the successive differences of the h i hi minus hi minus one Ah Yes, yes, we don't have conquerority, but now we have more right we have the symmetry The m vector this was the hebert's These were the coefficients of the hebert's series of a polynomial ring right a graded commutative algebra generated in degree one All right, it's clear that now we have this inclusion here from here to here because this is already a graded But this is already a graded point. This is already a polynomial ring Yeah, yeah, yeah in McCauley's theorem just said that you could go in the other direction then but now we can say more All right, so now we have this so we have punk a red royalty So we have the symmetry, but we have even more we have this left-shed property And what what it says is that if I look at this vector gi right consisting of the successive differences then well, I can look at The g vector all right, I can look at the g vector It's it's no longer well it is up to the middle right G vector which is the dimension well, so okay, so this is the dimension for of of a i Modular the left-shed element a i So we let's just let's for simplicity for simplicity Let's just define this for i less than less or equal to d half. All right, and then it is positive This is an m vector All right, this this is the implication of the hard left-sheds here What we will not do In this next job, but what is also true is that there is also a reverse construction So for every m vector that you write down like that you can find a polytope Thank you. Yes. Yes So this this inclusion here. This is done by bill iran Lee But this is a combinatorial construction that we will not spend too much time But if you want I can yeah in the coffee break or afterwards. I can explain it all right So that's one application, right? Um Yes for every m vector there is a there is a simplisher polytope whose g vector, right? So successive differences of these of the h vector Realize it This No, no, no, no, it will only be right tomorrow. We are talking about polytops, right? We're only talking about polytops Well What do yeah, so what do you mean the more general objective that I don't consider Ah, yeah, okay, so here so now we are here in this situation we should say for for Sigma the boundary of a polytope. All right, so this is a titiation The common code is too weak, right? It doesn't have conquered it. Yeah. Yes. Yes. Yes Exactly, we will go we will see that this works essentially whenever this whenever this phase 3 is going down So can you detect the dimension of the polytope so you are this h vector So the dimension of the polytope can be seen from yes Yes, it's just the dimension of the fundamental degree of the fundamental class Okay, okay Yes, but polytops it's always in the case of polytops we are always have a sphere all right No, no, no any m vector, but we are somehow by passing to the g vector. I lose a symmetry, right? All right, I'm only looking at the g vector up to the No, no, okay, so if I just give you the m vector you want to you have to give me in addition the dimension of the polytop You do want that's right You have to in addition give me the dimension of the polytop or you have to say, okay The m vector goes up to some entry and then the polytop will be of double the dimension right this here only the g vector here It only goes to half the dimension You cannot say that because coning preserves it in Macaulay's theorem you can always take a cone Right, so if you give an h vector you could always take a and you could find a superficial complex at rear Sorry, you give an m vector find a superficial complex a common Macaulay's superficial complex that realizes that m vector as its h vector Then you could take a cone over that and it would still be the same h vector So then you cannot recover just in the goldenstein case if you have the h vector All right, if you know that it is okay, even then you can say okay Even then you can call but if you say it is a sphere then it must be then it's the dimension of this first victim Yeah, yeah the coning that's We will see later that it space a space a special role so coin Macaulay will it will remain in the background a little Okay, so the from the g vector you if you give me the g vector you have the dimension because it's just double. Yeah, yeah All right, all right, and let me just briefly mention Okay, so this was an application of the hot left sheds. Let me briefly mention an observation I think it goes back to Timorin, but maybe it's also Earlier Application of the Hogema relations and this is kind of important if you if you if you want to go back to the Locom cavity That I mentioned at the start for for characteristic polynomials. So we take alpha and beta to convex elements in a1 of a1 of sigma all right and Yeah, let me let me stay let me say that No, it doesn't have to it doesn't have to be for whatever I okay, so for the moment, let's say strictly convex, okay? But we will see in a second that we can delete it So then you can write down well, okay, so let's let's write down the Hogema bilinear form in degree one But let's not write it down completely But let's just restrict to the subspace by off spanned by alpha and beta. It's a too much. It's a two-dimensional subspace Now you just write that down All right, so then what do I have? Well, I could So if if I think of so let me think of beta as my beta as my L and alpha is just another form And then I could write down degree of alpha times beta and then I have to Multiply with beta to the D minus second power or some other element. Ah, let me let me just say okay So L is let me just let me just L is any other element. All right to the D minus two And Actually, I want in this entry.  This is degree of alpha times beta times L to the D minus two and Then I have here the degree of beta times alpha times L to the D minus two and Then I have the degree of beta squared Times L to the D minus two Right and I have this this is the matrix that I get for the Hogema bilinear form restricted to this to buy This two-dimensional subspace And what do I get? Well, okay, so now let's look at The signature of this matrix so that I have one positive eigenvalue Coming from degree zero right. This is minus one to the zero All right, it's this one positive eigenvalue and all other eigenvalues on a one, right? So now then we consider consider here q L In degree one in which order that I write it q 1 L, right? This is the Hogema bilinear form on degree one Right all the other all the other eigenvalues are negative So what happens to this matrix? What is this? What is What is it the definite this is signature of this matrix? Well, it's an easy argument for matrices that The the signature somehow the the signature can we be neither smart can it be neither definite positive definite Nor can it be negatively definite. So there's a positive eigenvalue. It's a negative eigenvalue So meaning that in particular if I compute the determinant of this Right, it will definitely not be definite. So the determinant will be negative So what do I get? Well, I get that this times this so Degree alpha squared and then L to the da da da L to the D minus 2 times degree beta squared times L to the D minus 2 and Then what else do I have what I have this times this right? So I know just compute the determinant All right. So I have minus degree alpha beta L to the D minus 2 and This I have squared and this here is less or equal to zero All right So what does this mean? well If I if I pull this to the other side of the equation Then this suspiciously looks like the Alexander-Fenton inequality and it is This is Alexander-Fenton So you look at the mixed volumes of Convex bodies, so you take the Minkowski sum of convex bodies So a convex body B convex body and then you take some other convex bodies and compute the volume As as a function of the dilation. So let's say you have You have the function It's compute the volume of T little a a Plus t little b B plus Some other convex bodies, okay? And then you measure you want to look at the coefficients here of this X-volume and you want to look at the mixed coefficients, right? Ta times to be which is exactly this coefficient And this is larger equal to the product of the two adjacent. So ta squared But times to be squared. That's a point. That's Alexander-Fenton Sorry, may I have a question? Yes Can you please explain why why this matrix couldn't be negative definite? What if alpha and beta are both primitive that is negative definite, isn't it true? So the Ho Chi Riemann Ho Chi Riemann form is negative definite on the first primitive source Yeah, this is why I assumed that they are both convex. I see, okay. Yeah Yeah, and of course some of the point is now now that I know this inequality I can remove strictly All right, so now because a strict any any form is a any any convex form is an approximation of strictly convex forms And then I don't do no longer need convex. I don't know longer the strictly convex on the fan. Okay. Yeah Yes So Alexander-Fenton, I think you only obtain if you use the mixed version of the lab Yes, I mean I cheated here. I took the same I took the same L. You're right, but I won't go there Yes. Yes, I cheated but don't tell anyone. Okay. It's a Yeah Yeah, you're right So this is not quite the most general version Yeah So And that's all good and I so we have we have many applications of these theorems But There are some questions. Well, it's about this this version of hard left shit and odd Riemann is just not enough So let me let me Just give two of them that are interesting The first is kind of immediate. All right. I mean So we characterize the g vectors for the boundaries of substantial polytopes, but We know by by also theorem that The Panker a reality extends more generally right the Panker a reality We mentioned last time it works for general general triangulations of spheres even homology spheres All right. Also, we worked now over the reals, but I could go to any other characteristic. All right, so Well, that's that's that's a theorem extent that's its characterization By the way, I should say that this is due to Stanley this observation This is that's this observation of Stanley extent. All right, so question one Does send the extent does the g theorem extend Well to okay, so let me spoiler the the most general version we could do at least for Homology spheres would be yeah, so K. So K homology spheres for K homology spheres We will even go more general but somehow for the more general I have to explain a little for to to to Triangle a yeah triangulated to yeah, so to simplisional. Yes. Thank you. All right. I mean And this is really just one direction the somehow if we prove the hot left-shad theorem for The a-rings of simplisional homology spheres then in particular we get this automatically because the billiard billiard already small gives us the other direction Okay, that's that's one theorem or one one question for now it will be a theorem and Well, I don't I don't know whether it will be small and whether we get done this week, but Probably not but by the end of the Hadam lectures we will be done And the my second favorite question in this context Let me start a new blackboard because it's it's it's my favorite question. I need small my favorite result Because it's just so it's a beautiful and this was the green bomb conjecture right some of this green bomb problem of We look at a simplisional complex Embedded into our 2d And we assume that for us the embedding is PL and then we want it alright, so we wanted the question and As I will explain the theorem It that's the number of I mean The number of k phases of Delta is that knows k plus 2 Times the number of k minus one-dimensional phases of Delta That's the theorem that we will prove How does this okay, so I we already explained how this g theorem follows from the left sheds property How does this follow? well, so to start with Let us Instead of just embedding it into R2k we could also embed it into S2k obviously right we could just compactify What we also could do instead of like just embedding it into s2k we could just think of Delta as a sub complex of Sigma a triangulation A triangulated S2k Alright, and this is the only case where we would only point where we use a PL nurse to extend Delta to a triangulation of the sphere and Then what we can do is?  This is really No, no, no, but here in this thinking of this as a triangulation All right. This is a simpler circle. There's a simple shell complex Right that contains Delta as a sub complex You have to yeah, there's a there was an argument that you can make for PL embeddings That is for instance, I mean, it's it's written down in Binks Yeah, it's it's not it's not obvious, but it's also not too spot. It's yeah It's written down in Binks note on on on three manifold I think it's called topology of three manifolds and Pancari conjecture something like that maybe Johanna, can you look up whatever I cite in the I will I will I will leave it for the moment No, no, no, it's not an odd thing It's an old thing. Yes It's not but it's not an odd dimensional thing. It's not something about the three here. See three sphere. Yeah Exactly Yes, yes, yes, yes, that's the point. It's somehow As long as you can extend it trying to a triangulation Of the sphere this bound applies But it's not obvious that you can always do that Magic Thanks Alright And so How do I how do I go? From here to here? Well, here's the observation. So let's look at we have a sigma All right, and we have it's quotient A delta, right? That's just the restriction map to Delta to the faces of Delta So we have the Subjection to a Delta Actually, let me anticipating a little Let me write it like this. All right What was Delta any simplification complex that embeds into the 2k sphere All right, and I extended it to a triangulation It's actually doesn't have to be tomorrow. It can be actually I don't actually care whether this is PL equivalent It can be a home. It can be just a just a triangulation a non-pl triangulation.  So this the theorem here applies whenever Sigma where they never Delta can be realized as a sub complex over homology sphere of a k-homology sphere Yeah, yeah, yeah, but I mean the the the the original question here it starts out by a PL embedding, right and to R2k So I have this ejection. That's good. And now I can I can look at The dimensions of the graded components and try to I can try to bound them a little So for instance, I could look at The degree k component of Delta and Well, let me try to give a Let me try to give a generating system right that always embed this always estimates things from above So I I try to give a generating system. So well, what better than the cardinality k faces So this is at most the dark cardin the number of cardinality k or dimension k minus one faces of Delta All right And right tomorrow if I if I estimate something from above using the number of generators I can do a similar thing For an estimate from below, right? I can I can just take the generators and then estimate the number of relations So let me do this in a specific dimension So the dimension of a k plus one of Delta All right, so I write down the number of generators So the cardinality k faces of Delta and then I have to work a little and think a little about the number of relations that come in, right? So if I if I was talking about torque varieties now, I would look at rational equivalence equal to zero, right? and Okay, so this takes a little thought maybe we'll Somehow we will later see a model of this ring where it is kind of obvious but for now, let me just say that this is K plus one times the number of faces of Delta so for every for every For every cardinality k face, right? It's the ring condition, right? It comes from one degree below I get k plus one No, but you divide by when you divide you cause that a of sigma yeah, you divide the ring of sigma which is in dimension 2k by You have to divide by linear form by but the number of them is a dimension Is that is something like to kill? Yes? Yes, but I mean So there are some there are some elements that just take out the the the square full terms, right? Some other the monomial somehow I restrict as well I will take out some elements, right? So some more monomials that are that good that contain squares right that are not square free So the first if you think about it if I am looking now at the cardinality the cardinality k plus one faces So the first k plus one linear forms They will just kill off. They will sorry the first K linear forms.  What is happening? All right, it's a free polynomial ring. So it has a lot of square full terms All right initially Square full mean meaning here Not square free. I don't know whether this is a standard terminology But and only after then I will take out the square free I will we will go later to a model where it is obvious, okay Yes, yes, but the the trick is here we will later see a model where it is obvious Okay, we will go to the Ishida complex and then it will be obvious Doing it in this model is kind of tedious Yeah, oh meet Yes, yes, yes, yes, I'm still working with a right tomorrow. Yeah, that's right. It doesn't have to be generic Yes, yes, yes, it's actually I cheated a little right I didn't say that is generated by square free but tomorrow tomorrow because A is generated right As a k vector space by the square free elements, but we didn't actually do that yet. You're right No, no, no as long as a screen as long as a linear forms Reduce some of the dimension to zero reduce the cold dimension the cold dimension to zero. This is true The the ring will be generated by square free elements as as as a k vector space Yes, but again instead of Right, this was there some of the overview the introduction section So instead of explaining this now we will later see a model where it is obvious Okay, so we will introduce another model for this ring and that will be obvious for now Take it as a mystery But later we will see it in detail. Okay, okay Yeah, that's right, that's right So right some of the number of Thetas, so the length of the system of Thetas is the cool dimension of K sigma which is in general All right, this will be larger than the cool dimension of Of K Delta All right It's any time you have any Delta and you take out the linear system of parameters So you take it out enough linear forms to make this to make the cool dimension zero. It will be generated by square free Okay, we will see it. Okay. We will We're getting distracted from this from the occasion because now you see now the inequality becomes even nicer Right, this equality was already already very nice, but what I now have to show is just that the dimension of the k graded component of Delta is Larger equal to the dimension of the k plus one graded component of Delta All right, and now why do I do this? Well, this is the reason that I wrote things like this so I have Well, I have an isomorphism between degree k and degree k plus one of sigma They are Poincare dual all right the spheres of dimension 2k So the fundamental class lives in degree 2k plus one But if I have the left sheds property, so if left sheds is true If left sheds property price is true Then what I can do well, I can look at the following quotients. So a k Delta a k plus one of Delta All right, and then I have above here I have the on top here of the left sheds element and then of course I have the induced map here But this will be an isomorphism if I have the left sheds property. These are subjections by construction So this year will be a subjection. Okay, so this year. So this year will be subjective implies this subjective Which implies this inequality? Which then implies my desired inequality, you know, that's it. All right And now let me Let me go to the theorem and let me use a big block for that There's a similar argument for for for Degrees less than strictly less than D over 2 right so now we are at the middle But some of the middle is the most interesting inequality in the end at least for me because the Intersections if they exist right if Delta so dense we expect some of transversal intersections Oh Yeah, but the maximal objects are really tricky to understand in higher dimensions, I mean you can you can always more you have you have edges that you can So here's a here's the issue that you can that you can that you'll run into in higher dimensions So if you start with a graph in our for all right, then you could ask well How many how many how many two-dimensional faces could I add or how do I embed it optimally and in general? This does not really exist What is I mean there is I mean as I said you can if there is an isomorphism here Then you send to a triangulation in a nice way in a certain sense in a But otherwise That is not true right all right So here's the theorem that we Prove and There's Two versions that I will so there the proof of the hard left sheds The the hard left sheds theorem has two proofs One from to the 18 and then we have recently another one Together with papadakis and Petro to And this is 21 That gives a slightly different flavor of of It gives a slightly different frame of argument that nevertheless relies on the same intuition and the same on the same Basic idea that I will explain so we have we start with sigma K homology sphere and this here for me is really just okay. So once again, so this is a k-homology manifold All right, such that it globally also has a homology of a sphere So it's really just if you want to think about it Gornstein Gornstein I think it's Gornstein star so the dimension the dimension of the complex is the same as Other the the the cardinality of the maximal simplex is the same as the the degree of the fundamental class so Cardinality Let's see say dimension of the of the complex is the same as degree of the fundamental class class Minus one all right and Then the statement is as follows so we have a Sigma right a sigma Specifically it came it came with a with a linear system tether right so this will here was a Linear system of parameters And what I can now do I can look at the modular space of these ace of these a signal right I can I can vary the tether All right, there's there are many choices for the tether. And so what I take is a generic tether For generic tether Okay, and then once I took generic tether. I took it like a generic L in a one generic a one of signal generic And then we have the following Theorem well the first thing is the hard left set serum Yeah, there's an open dense set of linear forms tether all right that I can take that support yes And the generic tether and L means a generic in the set of pearls tether L generic or generic Yeah, yeah generic in the pearls. Yeah. Yeah generic also in the past. Yes Oh hard left shits. All right. So now I have a k ah Then for all then for k less or equal to d half Ah, I should have said of dimension of dimension d minus one I Have the isomorphism between between degree k and degree d minus k Induced by Induced by the multiplication with L L to the d minus 2k and this is an isomorphism and That is somehow the that's the analog of the hot left shits and again somehow the classical of left shits It came with this cousin of our dream on relations, and it also also this version does come with the cousin That Equally concerns is by linear form q, but in a different way and there are two versions Um, so Here is the 18 version What it says is that I look at q Qkl right this was from a k times a k to ad All right as before that's the same by linear form and now what I do is I restrict to certain subspaces especially I Basically restrict to the tolls equivalent Invariant subspaces if I want to think about this neotoric geometry or if I if I want to think about this Just in the terms of the ring then I'm saying that this per this formula is perfect when Restricted to any Um square free monomial ideal, so I can no longer say anything about the signature right how dream I said something about the signature on Certain subspaces I had no longer have any any Handle about the the signature in any way or form here But I say okay, so this form at least doesn't degenerate at many subspaces and subspace is specifically motivated by the combinatorics Okay Yes, Maxim Ideals So these are these are just the ideals I Take a I take a sub complex delta of sigma and I take the kernel of the restriction of these are the ideals I'm looking at yeah, of course you have to I mean to for the statement to be non-trivial. They have to witness something They will have to witness something from the low degrees. That's right From the care from degree K. Let's say quote well, yeah, and that's that's the statement about one Huh, ah, yes, thank you. Yeah, yeah, yeah, okay Oh fair Okay, there is some some complex Delta. Okay, no for take any subcomplex. Yeah, okay, so for any ideal of this form, okay Sorry my ask a question and regarding regarding the signature of despair in do you have some counter examples or you just don't know how it behaves In general, there's okay, so you can give examples where the signature where you just have no chance of getting the right signature Notice that this does not even depend on L. All right. So now if I have a Sphere of odd dimension. All right, so let's say 2k minus one Then the fundamental class is in degree K. So it's just the punk array pairing on degree K And already the punk array pairing in general for for for for general linear system of parameters just has a wrong signature I mean you can even I mean even for a one-dimensional sphere. You can show that there are linear system of parameters Where you don't have a signature of what positive eigenvalue and all the remaining ones negative So signature of plus minus minus minus, right? Yeah Of course Yeah There's a geometry in this motorized space where I control it. I mean I Don't think I don't think it's so easy to understand the geometry of this motorized space So in particular there's more the left sheds locals if you want right the space of theta And L where you get left sheds. We just know I mean the proofs just give it for generic We will see why but I mean I don't have any control of it No, I don't think we have a I don't think Yeah, I don't know whether this is feasible All right, let me state the Let me state that there's about this this new version and let me in the interest of saving a little time. Let me actually Just State the characteristic to version and this came comes from so this is a version of 18 So now we go to the paper by Papadokos and Petrotou in 20 and then it's more general version P 21 and I will just state the characteristic to version So if the characteristic of K is 2 Then There exists a field extension K tilde of K such that Q Okay, so now I take the small and then I take theta theta in K over K over over this field extension and I take L also in this field extension All right, so I really just take a nice transcendental field extension in the end such that Q Q Qkl Right, so I there I said it doesn't degenerate a monomial ideals and now I'm saying it never degenerates Even if you just take any some of any linear form Sorry, any any any degree K element multiplied with itself in this bar in this form Q The result will be not zero. Okay never degenerates Never degenerates And this is yeah, so Q K L alpha Alpha will not be zero for all Alpha defined over the also the larger field. Okay So right, so this is now I should maybe say so now we are looking at a tilde to make clear that we are In the larger field. Okay, so a tilde just says okay, so we have a sigma and theta But all everything is now over the larger field. Okay, and this will never be zero for all alpha in a tilde Not zero. Yeah. Ah, yes. Yes. So for all for in a signal that are not zero. That's right All right and These are the two versions. I think Now is a good time for a short break All right, because now we will go some others. That's the end of the statements of the theorems If we want to go to all right, so now we will go a little to some homological tools some very basic homological tools and This will be called section on the partition complex started feeling kind of naked without a mask but Okay partition complex and Conquer a duality or if you prefer gonstein as a bunch of property go duality Okay And this will have two sections So we will do to one The works and then we will do to one to to the cheats so So once we will go All I mean we will explain why why certain why certain algebra are Pancredority algebras in this context in particular Phosphorus in general will explain it or two just because it will we will encounter a tool that is just useful in all kinds of Context also later and then we will see that Well, if you're just interested in Pancredority, there's a cheap way to always get it for me for any simplificial cycle So let's let's start with the works Work art and then party art So how do I prove Pancredority? All right smart In a polynomial ring right, I mean how would I prove that? How would how would you prove that? a sigma right so a polynomial ring is a Pancredority algebra Pancredority algebra. Well, I mean Here's the thing it is equivalent so Pancredi in this in this ring generated in degree one is equivalent to saying that the so called is of dimension so called So the so called it is Is all those elements that I Nailated by every element in the in the of of of of of of of positive degree So ever and ever you multiply with everything anything of of of degree At least one you get sent to zero so the so called of the algebra should be of dimension One all right, so there's just one element that gets killed with under any multiplication that is not to go just a constant Right, so it's just not not just a company Not just a multiplication with a say with a scalar All right, so and this is kind of obvious if you think about it, right? What does it say right now? I mean in this direction is obvious because All right, what does Pancredority algebra states? Well, I mean it says that every elements Of degree k that is not the maximal degree. I can multiply with some non two with some elements of Of another a higher degree such that I End up in degree D and I am non trivial All right Right because this power is perfect. That's what it I mean and the other direction is well Okay, so if I if the circle is of dimension one then if I have some element x in degree k Then at least I can multiply with some element y To get x times y in degree right so why because I'm generating degree one I can say that it's okay, so it's of degree one so it says some element y times x in degree k plus one and then I can Y prime times y times x in degree k plus two Until I am in the until I am in degree D. And this gives me the perfectness of the pairing. That's it Yeah, yeah, that's just what I'm saying. Yes. Yeah, it's I mean it's true I mean yes, it's true here, but sometimes people smile I sometimes people didn't really don't realize this because I think if they're thinking about Pancredority and for manifolds and Then it's yeah, you're right. It's obvious, but you have to say the obvious thing And the greatest go well. I mean, okay Okay So what do we need for for the ingredients well to to prove Pancredority Well, okay, so now to prove Pancredority for sigma Triangulated Sphere Sphere of dimension D minus one dimension D minus one and Yeah Yeah, it should be dimension D minus one The day is a special yet mother the dimension as a simplisher complex is always one lower because the the the degree of the fundamental Yeah, yeah, yeah, we are not talking about torque varieties and so So what do I need well, I need two ingredients so a What I need is that for all k less than D if I look at the degree k component of My ring well, then I want to be able to pull back to One of the prime divisors. So what I need is I want to look at this map Restricting to the stars of the vertices that we explain. I don't think I introduced what the stars yet And I want to say that this map here Over all the vertices in Sigma That this map is an injection is Injected all right. What you said last time is the shelling stuff This was what it only a special case of this Yeah, yeah, this is only a special case now we are this is a triangulated sphere And for me from now on triangulated sphere will always k homology. This will always be a k homology sphere Yeah, so this is this is much more general All right, so triangulated sphere for me will now be k homology sphere always. I don't have to say it again Yeah, the key the links are all again k homology. I've said again the k homology of a man or of a sphere Okay, yeah, I what I didn't say what the star is so the star of a face in a simple to complex is Defined as those faces inside my complex With a property that tau union Sigma is also inside the complex No, no, no that would be the link Yes, thank you. Thank you. Yes, you're right Yes, so for every individual right if I ignored This are some it would be just a surjective district restriction map But now I take it all together and then I want this to be injected Is Sigma V is V is a vertex V goes over the vertices so the zero-dimensional simplicity of Sigma Right I want this map to be injective We will see yeah, okay, it follows from punctuality, but we want to prove this to prove punctuality. Yeah Okay, and B B. Oh, well, okay, so this is the pullback and then what we want to say is that I Have the star of a vertex in Sigma and I'm sitting in the UK and then I multiply With the corresponding variable and this will create an ideal Inside actually the ring Sigma And I want this map here to be injected as well. So these are the two properties Okay, so for now I just said what we want to prove and this will be true. This is true This is true this is true For all Linear systems of parameters that are well that I mean they have to be linear system parameters So they have to reduce the code dimension to zero. Yeah. Yeah, so it has to be a linear system of parameters For all that is so for all that is for all that is So Yes Again, yeah, we won't go we're into our job into the algebraic geometry So, let me let me give you let me prove B for you first because it's because it's much easier And the trick is okay I Mean the trick is always With these kind of things Look at it before you do the attenu reduction look at it before you take our tether Okay, so then I have Star of the vertex in Sigma Going to K of Sigma and okay, so here's the multiplication by Xe And what is the co-kernel here? Well, this is the restriction map to K of Sigma without B. All right Okay Goes to zero. I mean, okay, so this year before I do this attenu reduction It's it's it's more or less clear that this is a short Exact sequence All right, and now what do I do? Well, okay, so Sigma for me was a sphere right Sigma without V is there for our homology disc. It's still called Macaulay All right, and now I mod out by tether, but because it's called Macaulay All right, the the linear system will be a regular system for for For for for Sigma without V, but then the sequence says exact. All right Okay, so Tether is regular Well, it's not only regular here, but it's also regular here because it's a homology disc and Then this is then I can mod out and I get the same system. I get the same exact sequence for the ace All right for a of this, but then I in particular have this injection. That's it Yeah, yeah, this is just yeah just exceeded you know nothing. Yeah, they're just a relative. Yeah relative homology. Yeah, that's it All right, it's just I mean you take a triangulated sphere remove a vertex what you have left as a disc No, no, no because I multiply with xv Yeah, I Save myself the relative stuff Okay, so that's that's that's parts the simple parts so now for Part a we will need what I call the partition complex So once again The trick is so I will do this actually in the generality of Let's say Delta is a corner Macaulay complex. Okay Delta is corner Macaulay corner Macaulay of Dimension D minus one and Yeah Yeah Yeah, this is just the restriction to a restrict. Yeah, I quote a lot by the back speed. Yeah, that's right so and the trick is once again, I look at the unreduced version K Delta and then I map to the direct sum over the vertices and Delta zero and And Take K of this the phasing of the star of the vertex in Delta and then next step I go well, what would be more natural to then to go to the edges next right the edges in Delta So these are the one-dimensional faces and then I take K of the star over the edge in Delta and Then I go on Yes, yes, yes, you have a sign if you think about it. Well, how do you how do you choose a sign without actually? Well with being a little lazy look at this in degree zero All right, look at K. Look at these rings here, but only restrict to degree zero Then what I want is to be in degree zero is naturally the Czech complex Given by covering sigma with the open stars of vertices right to the interiors of the stars And now I take just the Czech complex in degree zero. I have the natural choice of signs Yeah, well, I mean, okay, so I can order my vertices and then give the sign. That's that's right and It turns out if I give the sign this way then naturally this will be exact in positive degree so Degree zero component is Czech complex Wait again, so this guy I think Luska will frown. I think it is this this version or is it I don't know I don't remember this axon. I think Czech complex and in degree Larger than zero. This is just exact and let's let me call this Partition complex of Delta All right, there's no no theta yet If you I don't know whether you want to call it augmented check in the green degree you mean Degree I mean the degree of polynomials here, right? This is our graded rings. Okay So the conflict is concentrated in the time of the group the complex that you have to do the convention like you start with Zero minus Yeah, yeah, yeah, yeah, yeah, yeah, there's yeah, yeah, there are too many degrees. That's right So there there is some oh, huh? Yeah, a polynomial degree or polynomial degree. You're right. Thank you And the polynomial Yeah, yeah, yeah, yeah, yes So the Czech complex calculate the reduced homology of Delta. Yes Yes, yes, so it has something in dimension D virus one Yes, yeah, and only in this Okay Okay, that's that's that's that's My my my first complex and then okay, so now what I do is I take the course through complex coming from theta, right? I take my theta Let me see where I've space Right, and this is some other kind of probably one of the first things that you Ever do in this now when you do homological algebra with respect to sequences. All right, I take take care I don't know cause two complex Huh? From Czech language to Hungarian language Okay, yeah, that's okay. This one. I didn't know it. Okay Yes And now now we marry the two so so we take P tensor delta tensor console complex Now we have this double complex. We take the total complex of it Consider the total complex and now what happens right somehow now, I mean, so I have this this double complex, right? Which somehow again, this is just the direct sum of The ith components here with The jth components Here Where I plus j equal to a given constant with a given to a given k And now well, okay, so what do I have right somehow in the direction of this complex P What do I have well in positive degree homogeneity or polynomial degree? All right, this is just exact All right, so I want to compute right tomorrow I will bore many people probably but that's why if this is a direction of P and This is a direction of K So in positive degree This is this is exact, right? So positive Yes, it positive homogeneity degree Yes, it positive. Yes. Yes. Yes. This is why I don't will not write this down. There's too many degrees So there's a homogeneity degree in positive homogeneity degree P is exact So we can push down until we end up in degree homogeneity degrees zero Okay Similarly in in in the causal direction. We are exact Because we are called Macaulay right all the stars Delta is called Macaulay all the stars of vertices are called Macaulay the stars of edges are called Macaulay The ends where exactly the causal direction and I can push it I can push it into to the boundary here right the standard the standard homological algebra what we what I get is then exactly that Oh and again, there's too many exacts I Get that Oh The desired property that a delta to the direct sum over the vertices in Delta of a star of the vertex Right because the homology vanishes This will be exactly the cost of all the costulomology This will be just the homology coming from the Czech complex and this will vanish in degree K less and D This is an injection and now let me state the more general version so if Delta is books bomb and this this is just saying that For all vertices for all vertices in Delta the star of the vertex Is is is called Macaulay Maybe do any purity Let me Maybe if it's disconnected is okay, so So then Delta pure pure and For all vertices in Delta the the star of the vertex is called Macaulay Then I can say something finer then What I get here is that Well, I can look at the kernel of This map the kernel of a K of Delta To the direct sum over the vertices, let me just write it again a star of the vertex in Delta And this is isomorphic to well the Comology and I should in degree K The Comology in degree K minus one With with K coefficients obviously of Delta and then Because the Kossu complex right I mean the Kossu complex is really implicitly Right if I once I go to the modernity degree zero I will have some powers of it coming from the Kossu complex and this will be exactly The D choose case power of this vector space Yes, yes, so this is our refinement. Yeah Well, but yeah, the top is not trivial, but I'm restricting to K less and D Yeah, yeah, yeah, because again empty sets. Yes, I know the empty set is kind of your thing, but yes All right, okay, so let me delete that one and then okay. Well, okay, so What I can draw from that is I can do the analogous version of manifolds All right, so Delta So or let's say if M is a closed orientable manifold I'm going to say okay triangulated manifold official Off dimension D minus one off dimension D minus one Well, okay, so you will see that I will not be a punker a duality algebra this a of a of M will not be punker a duality algebra But what I can do is I can look at So a of M is not PDA in general but What I can do is I can take a of I can take B of M is not Thank you. Yes, I said it but I did write it but B of M Which is the quotient of a of M by this kernel here right by HK minus one of M To the D choose K For all K. That's why I do this all K less than for all K less than Less than D This is a punker a duality algebra is and now really is a punker a duality algebra. All right. That's it The idea because of you know, I mean that I mean it's it's an ideal in a trivial way because These elements here they die under any multiplication of positive degree All right, but this is this here The homo the homology they they they they are exactly the so called elements All right, you see that some of these actually this year applies in every degree But I don't want to kill the top so called element But all the other ones I kill right sent to the graveyard and in fact what we prove what I what I prove is That the the left sheds theorem holds for this more generally punker a duality charge this more general punker a duality algebra All right, but it we can do it even more general and this is now. So this is the The the End of the section that the works and now we'd go to the cheats, which is the final part for today So let me see this one I can read it so now to the cheat Let's say I take any Any simplisher complex delta Simplisher complex of Dimension D minus one And then I take mu In the degree D component of delta I take mu in the degree D component any I I want to actually take this as a quotient somehow To mu any one-dimensional quotient dimensional quotient So it's a one-dimensional quotient and in fact there's something nice here, and we will see this later Is that this here really this is always isomorphic to the D minus first coromology of delta. So regardless of manifold property, we will see this with k coefficients And then what I can do is I can always generate a prankariduality algebra From a from a of delta which has this as a fundamental class Okay, so then B of mu Is the maximum quotient? Sorry, it's a minimal quotient minimal quotient of a of Delta that Well that contains That makes maps would not really to such that such that B of Mu in degree D is Equal to mu isomorphic to mu So okay, so I just basically force a prankariduality algebra by kicking out anything that does not pair to mu Yeah, yeah exactly exactly Yeah, think of it like this I take out I look at the pairing right and whenever an element Goes to zero the cost goes to it goes to zero under this couple right I pair any element in degree K to degree D and if it goes to zero under this quotient I kick it out That's my ideal All right, that's clearly an ideal Okay Describe it like and So this is actually a nice object I mean it only depends on on you right so you can add faces of Delta to Delta that are not a mu the algebra will be the same Here's a nice way of thinking about the degree map in this context and this will be the end for today Let me see so here's a check complex Let me let me finish here All right Hmm All right, I can think of I mean I can look at the dual right new V Dual to mu in in the homology of My complex All right, so this is just a simple cycle right really this homology cycle And then what I can do okay, so What does this give me the simple show homology cycle right? I mean okay, so implicitly I have the pairing but I mean Well, the nice thing here is I can write down the degree map until now this really again the pairing degree explicitly in terms of this mu so then the degree Then the degree of a monomial x tau is Where tau is of cardinality of cardinality? Yeah, so it's a simplex of Cardinality of cardinality of cardinality D and the degree of this monomial is then just mu tau Divided by the determinant of The matrix I look at the matrix. I look at the main minor Corresponding to tau and here I'm explicitly I'm orienting so I have an order on the vertices I mean taking the order oriented coefficient and I compute the determinant of this matrix Yeah, it's a monomial so x tau This is the product of so x tau This is equal to the product of the x v where v is in town Yes That's the oriented coefficient of Of of mu in this in this of of of the face tau in this in this in this Homology cycle and I divide it by the order by by the determinant of The minor in theta corresponding to tau again oriented. All right So they are implicitly have an order on the vertices to make this work. Yeah You know it's not zero Well, it might be zero right if I if there is a face of Delta that is not supported Right, it's not supported in the in this official cycle there will be zero But then the degree will be zero on that face Delta as a simplex complex mu is a homology mu is a homology cycle mu v is a Dual homology cycle Yeah, yeah, but I mean I could add faces that don't lie in the cycle and they just don't appear in the ring that's right So this is the most general version and then of course I can okay, so now you can start with a homology cycle right and you can start with a homology cycle that any cycle that you want and you can look at The algebras be mu Center All right with it Ad ad is always isomorphic to the core homology a upper D is always isomorphic to the core homology In degree D in dimension D minus one of the complex. All right and There is also a dual model that we will encounter soon For for the homology that works out. That's the same. This is exactly the issue. They're complex And That is yeah, okay now we can look at right I can you can fix a simple homology cycle mu and you can look at all these algebras be mu theta right in particular you couldn't look at generic one and a generic Generic element of this. All right, so I take fix fix mu fix mu and take a generic in terms of in terms of The possible theta set you take All right And this will satisfy the hard left-shed serum will satisfy satisfy such as That is the most general version of the of the left-shed syrup that we have in this context hard left-shed Yes, yes the generic small and a generic l in a in B1 Well theta is just I mean so you have you have choices for theta right you have As many choices I mean you do You fix some yes Fix a fix mu fix a special homology cycle mu and now you have an algebra and now you have many many algebras Coming from these from from from modding our different systems theta. Okay, then you take that a generic enough Yes, and this will satisfy the hard left-shed property with respect to a generic element l in a l in B1 so the Theta a generic yes, and this is a generic in the You are more of a wish field now over any field or well, I mean any infinite field any infinite field Okay, so generic in the sense of this is a risk to the power of yeah Then the generic apple is a risky of the generic point even just Yes, that's a theorem yes, I won't write it down again now because we are already over time but We will repeat it next time it's time I don't know what August told you but okay, we can try I She did a complex. Yeah, I mean there is a way to think about yeah, we will we will go over this next Delta is a too complex in S4 does the PM bearing have to be locally I'm not it to extend to a triangulation. No, if so No, it is my if you look at the reference of Bing Then any any any it doesn't even matter. There's not even the dimension condition that is relevant. So If you have any P L embedding of a simplicial complex into the sphere, then you can extend it to a triangulation up your triangulation even so it doesn't mean it doesn't Depend on unnoted unnoted this but you can write me an email and I will if you don't remember the reference I will Send it to you again Yes, yeah, that's exactly the idea you shield it off. Yes. Yes, that's right You basically look at the links of vertices and say okay by induction I can I can I can do it in this in this vertex that then locally you you build like a neighborhood around it And then outside of this they bought you with whatever refinement. That's it. That's the idea