 Today I want to start with proofs of the hard left-shed theorems Just because we had a little discussion before let me repeat both statements next to each other because it's maybe useful so Hard left-sheds and company And this will be the classical case today So Let me start first. Let me repeat first the the version That is kind of classical or semi classical. This is goes back to McMullen on the paper is on simple polytops Although the language we stated in and we prove it in Follows a paper by Fleming in Karoo from 2009 And this is We have Sigma the boundary of P a deep poly top Simple show Deep poly top And Then maybe I can actually use both in parallel way And L in a one of sigma Convex all right to somebody ever Remember that somehow this this geometry it immediately It immediately implies that we have some linear system of parameters associated to the embedding of P into Rd and Then we have the hard left sheds All right, so we have an isomorphism between degree K and degree D minus K This algebra induced by the multiplication with L to the D minus 2k and We have watch Riemann and Watch Riemann Was that Given the quadratic form q. I think I wrote k first and L then On a k times a k to D To R Given by multiplying these two elements and sending them to the degree a times B times L to the D minus 2k In this book Okay, we can we can we can we can say the what what should be the sign convention it's kind of implicit here, but So in degree zero, all right a small you take a zero What should it be? Well, we want the hodgman relations to be well every element in degree zero Which is a constant functions right is just is primitive clearly because the and so the the the convention is the power of The degree zero function should be Or it should be positive. There's some other sorry the power of a convex function of a strictly convex function should be positive And that is your sign convention Okay, so there's of course this piece is it Although uproarily it is not clear Yeah, okay, really small But is it true that since So of course this is independent of the choice of the strictly convex one because you can connect any two strictly convex one Continuously, but yes, yes, you're jumping ahead. So you're spoilering again Okay, this is something that we will do We will we will continuously move strictly for yes But patience. Okay. Let me just state. Yes. Let me never go to the cinema with you because you will spoil the movies So Okay is definite of sign minus one to the k on P L a k which is the kernel of a k to a d minus k plus one so once further than left sheds map Times L induced by that multiplication with D minus 2k plus one That is The classical version and this is what we will prove or So I will go over the the the main steps and the main ideas of the proof today and let me just compare this Just to repeat to to the to the non-classical version so sigma Okay, let me write on symmetrically 18 and a 21 Is it and I will again restrict to the sphere version. I will not go for now to the cycle version Actually Let me let me say with a serious a version so triangulated D minus one sphere and again Triangulated sphere just means for me homology. So k homology sphere All right, so I'm not imposing any homotopy restrictions here then There exists Theta And there exists L in a1 of Sigma Theta Such that Well, we have our left sheds the statement is symmetric so we're not so This this statement is symmetric And the second part is we have what I call the whole La man Relations which state that q K L Does not degenerate at any monomial ideal at any graph in monomial ideal and Here I should again so I should really say so it's it's critical that We choose theta in a clever way So when we got finally get to to this there and we will see that there are bad choices of theta So you can choose this theta this the system of linear forms in such a way That there will be no left sheds that there can be no left sheds element that can be some of there is no L such that the left No, okay good exact good question. Thank you that's a very astute spoiler, but let me let me go there Let's go to P1 cross P1 Okay P1 P1 What is the simplisher complex or the simplisher fan associated to it? Well, it's just well, it's a fan over the cross polytope, right? It's a normal fan to the unit q unit square. It's a man. It's also the in dimension 2. It's also The cost polytope is also a square, but yes. All right, so this is a fan. All right Let's look at the that let's look at Q1 Q1 L All right Well Q1 L Now this this this really I mean Q1 L is just a pairing of the degree one with itself To degree two, right homogeneous degree two Homogeneous degree one. It's just I mean this is a pairing a1 times a1 to a2 and it just sends a times B to the degree of A times B A B to the degree of a times B So you want to say that this does not degenerate at At square for monomial ideals and let's let's see Well, we take can take the ideal generated by this device out here. So this is V All right Okay What happens so we have I mean so the ideal is just The ideal generic by Kyvee so in order for the parent not to degenerate The only thing the only chance that you have is a for Kyvee to pair with pair with itself Okay, and then of course it is zero. It is zero of course. Okay, so there are several ways to see it Which way did you see? That's correspond to all one one projection of P1. Yes. Yes So the so clearly a pair so this is from an algebraic geometry perspective Here's another just some of the if we want to stain our pictures of comb wise poly of comb wise polynomial functions, right? So Kyvee is a restriction to this upper hemisphere, right? It is non-trivial linear here and zero here, but it also restricts to a global It's also with the restriction of a global linear function All right to the upper hemisphere right because really it is linear at this it is linear at this at this at this ray All right. It's zero here one zero. So it's linear. So it's a linear function All right restricted to the upper hemisphere Yes, that's it. Yeah, so it's so this here Kyvee's but is a square for monomial ideal ideal But Kyvee square Is zero That's it. So often do you see how to repair it? But I could to form this fan, right? I Could I mean to this And in this case at least Kyvee square will not be Zero all right if I would go to the heads above surface here And similarly well, I mean it could so the the point will be okay So now you see that this is somehow it has some to do with this with with the linearity here We will later see That more precisely This this this pairing property is true if The linear system teta acts as left sheds elements on certain sub varieties here, okay, but this is something We will do not not today But next next time we meet All right So do you see that even somehow it's kind of transversal even the smell very nice and smooth varieties Don't satisfy this principle It's a somewhat strange principle So I don't know whether I mean I haven't seen it an algebraic geometry Considered yet because maybe it's also not really I mean it's not about something about a fixed algebraic variety Doesn't really make sense All right, but today I want to prove The classical case So Which is for me, I mean The semi-classic okay, so it's not the classical proof of the heart of sheds here, but I want to do this classical combinatoria All right Mmm, and for this we will need A double induction so we will have two sections That Well You can imagine going from going to the mountains for so section one will be The climb and then Section two will be well if you don't climb you have some horizontal stretches. So this will be The hike Okay, so Let me do the climb first so What do we need well? We want to have a lemma that in some way implies Well in in induction and dimension in some conditions on on on polytops of dimension D minus one From some sort of condition there. We want to conclude The left at least some some something about left sheds or Ho Chi-man in in dimension D and This is the lemma. It's it's a it's it's Surprisingly surprising in its ingenuity. It's an amazing lemma So we consider sigma The boundary of P The depolytope Let me not say every time that it's a pleasure because it was space Simplisher is such a long water Then we have L in a one of sigma strictly convex And now we make an assumption Assume that the Ho Chi-man relations hold in co-dimension one holds in co-dimension one Yes, of course, I mean I mean Ho Chi-man that what does I mean Ho Chi-man says that this form does not degenerate Which is stronger than saying that I mean it says some it says in particular that this form It does not degenerate which in particular says That small doesn't degenerate on the primitive subspace. Yes But then you have it for every primitive where you have it for all the K All right, and then you have the left sheds decomposition implicitly built in Yeah, there is also the left sheds decomposition statement which which follows from our left sheds, but it can be stated there separately But if you just know Yeah, it depends what the formal if you just know Ho Chi-man in terms of just you define the primitive by kernel of something But of course just we need we need to know Okay, so let's just say I mean that's okay, so let's let's not We have hard left sheds and Ho Chi-man fine. Okay. Ho Chi-man plus out left sheds I Think yeah, you're right. You can state it in some way that it's independent. Let me let's just be on the safe side It could I mention one? All right, so for D minus one polytopes Then hard left sheds is true is true for Sigma, okay proof Well, let's assume the contrary assume there exists an alpha not equal to zero in a K of Sigma such that L to the D minus 2 K times alpha is Equal to zero And now what we do is Is What we consider is The restriction to the star of array so we have a star of a vertex which corresponds to the star of array in the fan picture consider a Restricted to the star of array corresponding to the vertex D All right, so this is a restriction to a star All right, and this naturally so let me make a picture All right, so this year would be the Ray and this would be the star now I can naturally project this down All right and get a fan of code I mentioned one This is again project if it's again has Again has the restriction of the strict the restriction of L To this fan is again strictly convex So this is naturally if you think about it, this is naturally a structure for the link All right, this is again the this is naturally the link if you think of the link here in this complex All right the link the link was all those faces that do not Form a face together that the form to get faced together with my with my ray But do not intersect it except trivially All right, and now I mean so this was supposed to be blue and this is a projection All right, so I will I will denote this because it's supposed to be a co-dimension one I will denote this by a The link of the vertex It's it now We need two facts One Alpha Okay, so we had last time we had when we discussed punker a duality. We noted that Being a punker a lot of loyalty what means just that it I pulled back to some star the vertex non trivially Yes If you want to Exactly, there is no choice of tether here, that's the point Otherwise, yes, that is right, but later we will discuss this In more generality and then we will explicitly see what is done to the tether, but it's here. It is so alpha right, I can map alpha into the direct sum over The stars or the links of the vertices it doesn't matter link of the vertices and All right What do I know? This is it has what do I know? Well, I know that this has non-zero image all right, so This here is something not equal to zero because remember this map from the UK of sigma This was an injection. All right. This was punker a duality All right, this this map here. This is an injection That's the first fact Which I need and The second fact is just that well so alpha If I restrict it to the link All right, so let me say alpha link V Sigma Well, what does it do? Well, I can also multiply it with a pullback of the left-hand class Let me just write restricted to V then I don't have to write link V every time this will be zero Corollary consequence or whatever is that alpha V is in P L V AK of link V C. Oh, yes It's primitive. All right Yeah, yeah, okay, so let me delete the classical they're not the non-classical version for now Just to finish my P L P and degree No, P the index To the two indices in north of which index as well P. Ah, this is just a restriction to V. So There's a pullback to the link of V Do you mean this the subscript here? Ah, yeah This is L restricted to V which means restricted to the link of V. Yeah. Yeah, I wanted this I did want to write link V every time because anyway, but to make you happy And okay, that's not a judge. Yeah, you're right, sorry Okay, and now let's compute Let's compute. Well, not quite all dream about in your form But let me compute minus one to the K times the hurt your own by linear form of Alpha Alpha, all right, which is just minus one to the K of L to the D minus 2k To the D minus 2k times alpha squared Well, I see it is Yeah, degree in Sigma and now minus one to the K Yeah, okay In the in the in the small in the link No, no, no, this is still the global one. This is still the global one All right, so I'm still computing it globally. Ah, this is K. Okay, so maybe I should write okay So this is the degree in Sigma and this is in Sigma. All right. This is still now I'm still computing in Sigma Okay In the full Sigma. Yes. All right, and let me just come to a convention. So let me write Let me write L As a sum of li X i where the li positive. All right, it's convex Okay, so this is the sum Okay, minus minus one to the K times the sum of and now I basically just okay So now I I I take one of the one of the one of the L's right and split up. So I take degree in Sigma still of Li X i and then I compute Well, I compute alpha squared times L to the D minus 2k minus one All right, I can put this out minus one to the K again And then I okay, so now I Okay, so the ally is a scalar doesn't matter X i is a pullback to the vertex so this is degree now in the link of the vertex of Sigma of Okay, so what I have left what I have just in them. I have Alpha squared L to the R again. I yeah, so now I have to restrict to V L V D minus 2k minus one I Know sorry, sorry now I pulled back. Sorry. That's right. I pulled back. Yeah No, but you need some Information you said that you had to choose an orientation on the top Which is you said before that you do it using the top power Zero Okay, but I mean if I if you think about it if I if I look at it in degree zero I really only induct on degree zero all the time So it's not so there's no cyclic logic here. All right, you are looking at The degree zero function Yeah, you are looking at the degree zero function The unique element in degree zero and you can pull it to pull the whole arguments who in degree zero or you can define the degree directly Or you couldn't actually you can actually verify directly that there exists a that there exists a strictly convex function Whose power is non-negative So Equalities of degree and not up to some scarlet which is also surprising I mean because the degree was just a choice of the Of the top dimension. I mean it's You didn't normalize the degree enough to argue the quality Okay. Yeah, I'm I'm I'm I'm cheating a little here, but I'm just giving you the idea of discuss a cup roof. Okay That's right, I mean there is a little more to discuss and there is also a little more to discuss with respect to this projectionism of this But let me just give you the idea Because what happens now right So this is of sine minus one to the K if Well, what's happened well under which condition if alpha restricted to the link of V is not zero Because it's primitive All right Because it was primitive here, but then this some okay, so we know that one of these alpha V Must be non-trivial because of this injection here All right In particular This whole thing All right. Okay, so now minus one to the K minus one to the K so this must be strictly larger than zero All right, and that's it Yes, yes, yes, yeah, okay fine. I wanted to make it. Yeah, you're right, but All right, so that is the Vertical induction Yeah, we don't have HR you don't have the Hodgman relations that way So now we prove the Hodgman relations. All right Mmm. All right. Let me find something to So now we go horizontally we have the hard left sheds for deep polytops hard left sheds for deep polytops want Hodgman not human resources All right So, how do we do this? Okay, so all right, so we have sigma and This is the boundary of some polytop P We have no clue how to get the Hodgman there So what do you do first if you don't have no clue you do an example So here's sigma Let me Did you see that I do a good notation? Let me let me do a sigma tilde Which is the boundary of Of the de-simplex yeah, it doesn't matter What is the algebra a Sigma tilde Well, maybe actually I want to let me call it sigma zero Anticipating it a little Well, I mean the variety is just all right, it's just projective space So the ring here is just I Take the polynomial ring in one variable So this is really just one variable this time And then I truncate it Indegree D plus one, okay, that's it So the only I mean the only non trivial primitive Classes live in degree zero All right, and okay, so now Okay, so there's only really only I mean there's any only one there's only one One L in a one All right, it's automatically convex if we I mean if we do choose a sign right and then Well, we just multiply it To the d power Compute the sign or make the sign convention in this case and then we're done. That's good So this case is fine. So but this is okay. This is sigma zero. We want to go to sigma one right How do we get there? It's not the smartest use of space today Well, we want to stay within the realm of convex polytops we want to go from the polytops that we understand to a different polytop if we may not understand And what we do now is we deform This polytop into this polytop in a continuous fashion By doing the following I introduce new vertices All right, so I have some I have some mystery locations That I want my new vertices to be in so draw them. Let me draw them here All right, so these are the vertices that I want to have in the end These are the vertices I have at the start. So I introduce a new vertex in the interior All right, and then I continuously move it in general position to the vertex corner that I want Okay And at each stage I take the convex hull this gives me a family of polytops going from P zero to P one okay, so family the one parameter family PT Connect family of polytops of polytops connecting P zero And P one right I take the convex all at its at each stage now If I do this in general position at almost every point in time, right? Except that the finite number of points in time. This will be a co-simplisher convex polytop right At almost every point in time almost at every point in time This is simply show all right now Notice that When I move From one simplisher location in this family to another simplisher location All right, so let's say in the time interval from from T to from from from TA to TB All right in this interval P does not PS stays simplisher All right, then the combinatorics of the fan does not change hard left sheds does not change hard left sheds is true All right by induction All right so so P in PT Simplicial in interval The geometry changes all right the geometry of the whole thing changes the set out changes with the polytop geometrically You have a convex all of some number of vertices and the number of other things you see yeah, the the combinatorics does not change. Yes When the combinatorics changes, so like when this point goes out of the Is it in the boundary and the limiting point is it the case that You don't have a simplisher Yeah, in general, I don't want I mean okay dimension to you only see some pleasure But then I'm a higher dimensions. We will see it. Yes There will be change in the combinatorics, but first of first I want to say what happens if there's no change in the Then then you can go from one to another by easy Yes, yes because now we know that the hard left sheds is true, right? Therefore the hodgepun by linear form does not degenerate. Therefore the signature stays the same in this continuous motion. Yeah That's just it. So so just let me just say again because not everyone might have already Figured out as fast as offer. So in this interval hard left sheds remains true hard left sheds is true Hence her dream and is preserved Okay, so we have to discuss what happens in the situations, right? when hot way when When there's a transition in the combinatorics, right? This is what we have to discuss Okay Hmm now what happens in these cases where? The combinatorics changes Yes, yes, yes, that's what is happening now So the points are still moved in general position. So when When there's a transition in combinatorics, then there are d plus one points instead of the normal d points d Being the simplex inside the boundary hyperplane, right? So Think about it like this. So if I have if I have the these two triangles I'm looking at the three-dimensional polytope right to make it more interesting. All right at some point. They become flat And then they go into the other direction. So what I'm encountering is d plus one points in a boundary hyperplane Supporting hyperplane hyperplane at Some Yeah, at some point at some point in time t zero at some point in time T zero and let me just make an example of what this looks like. So in a three-dimensional polytope So D equals three. I would have four points in one hyperplane a two-dimensional subspace all right and Before the transition it might look like this All right, so it's bent in this way all right some of this here in this the hyperplane spanned by this That's higher than this one and then after the transition it looks like like this All right the outer points The outer points they are in change, but some of the the transition lies inside Inside this quadrangle inside this d plus one poly this this d polytope on d plus one vertices Okay, this sorry the d minus one polytope on d plus one vertices All right, so this is a transition. This is called the partner move right partner move No, but I want I'm a little bit Concerned because you start with a point in the interior Yeah, okay, okay, there's another partner move. There's another one that can happen in dimension three Yeah, this is one of them the other one in dimension three is Right the vertex moves out through the This is another one Okay, and these are the two the two moves In dimension one in dimension three in dimension two, there's just this one move and In general you have roughly d half All right So we have to understand what this does There are several ways to do it. I will go over over one I mean one way would be to see to notice that These are some of that there is a common refinement here, right? That is coming from a blow-up, right? I could Blow up this point There's a common refinement here All right, and then I can construct a map between this and this I will now follow I will basically follow the paper by by Fleming and Carol Actually the reference I deleted to construct a map between One part and the other right before the flip and after the flip and for this let me come to the following convention right so in total the d plus one vertices here And let me call the one so let's say this this this part here has m vertices and and This part here has d plus one minus m vertices All right, they add up and I will make the assumption that m is less or equal to Well, I want to to have exactly Don't die I Want to have Want to be below the threshold so d plus m should be less or equal to d plus one half And my notation will be That this here will be sigma plus and this here will be sigma minus All right, so the flip that introduces The lower dimensional phase right the lower cardinality phase this will be but just by convention sigma plus The other one will be sigma minus just for notation. So in this case here again, this here will be sigma plus And this here would be sigma minus Because I want to construct a map from sigma minus to sigma plus All right Yeah, that's it. That's that's what it was really happening Alright and now we want to construct a map and actually Um Yeah, it will not be It will not be a map of algebras So it will not unless unless you're in this case where it's a refinement It will not be a map of algebras It will be a map of graded vector spaces that behaves nicely with respect to the left shat element and told you about in your form But otherwise All right, so I think that's this a good time This is a good time to take maybe 10 minutes whatever what do you want? It's again Let me actually make the non-symmetric situation even though it's a little boring So M the cardinality of the minimal simplex and do for us minimal simplex introduced simplex Introduced used Sigma M less or equal to d half Sigma plus After The flip Delta plus will be the flip locus so Just this after the flip. All right, so just the Where the the combinator exchange is the d plus one vertices is involved And the simplex on them Sigma minus analogously before the flip Delta minus Before the flip and then Gamma Will be Sigma plus Without the flip locus All right, so the The complex on the remaining facets And this is of course the same as sigma minus without the corresponding flip locus here Yes, yes, yes Variations And let's understand some apps So let me introduce for a second and we will also need it later Yes, it's sub complex It's exactly the star of the minimum face that was introduced Minus Okay, this is okay, so again These are all the this is a simplex complex on all the facets that are not part of the flip locus All right, I take the down closure of all the facets So, okay, you take all those guys which are not in the But you since you need to take the boundaries So you have to add the stuff that is yeah, I take the closure. I take the simplest of those are yes Okay, the closure Geometrically the closure of the difference which is triangulated. Yeah And it is a simple complex on the set of vertices that could be smaller This is could be small. Yes, right. I mean there's one less vertex here in this case Right this vertex does not appear outside, but in most cases, it's the same I mean in most cases, I mean the you introduce a minimal phase of cardinality m All right, and the minimal phase that is where the combinatorial difference starts and now also the edge bright difference, of course right Let me just briefly introduce the relative objects, okay, so if I have two simple show fans one contain in the other All right so a So be sub fan of a of a Then the space of this this here will denote the space of converse polynomials on a that vanish on B Similarly, I could do the phase string picture and say if I have two simple show complexes containing each other All right Then all right, so remember that the phase string of a simple show complex. This was defined as a polynomial ring Model of the non-phase ideal of a and now what I do is I take The non-phase ideal of B of the smaller thing. All right. I'm vanishing there model of the non-phase ideal of a All right and then Well, I mean this this this I mean you can think of it as a module over huh, this is a non-phase ideal of B Divided by the non-phase ideal of a Remember as well the non-phase ideal of a simple show complex I delta All right, this was The ideal generated by those x to the alpha So that the support of alpha Was not part of the simple show complex Right and you can think of this as as a k a module So inside which ring Inside okay, so we fix the number of vertices of the bigger complex. All right. That's our polynomial ring right In some polynomial ring. Yeah, so we take the number of vertices of the vertices of the bigger complex And for the smaller complex you take the Monomials which are non-pacers but including those which contain value. Yeah, exactly. These are somehow these are not faces exactly. Yes Yes, thanks. Yes So these are small these are not vertices of B Therefore, I take them in the non-phase ideal. Yes That's a natural way to do it. Yes. Thanks and here somehow here. It's more natural, right? It's just You have a big fan. We have a space of common polynomials. Just declare those you look at those that vanish on On on on the sub complex right on the sub fan sub fan So a B is a sub fan of an over fan a yeah, you have the common polynomials on a And you look at those that vanish on B. All right I'm just giving you once again. I gave you two ways of thinking of this ring Now I give you two the two ways of thinking about the relative object. Okay, okay? And now I want to write down Some Short exact sequences Oops, so what I could write down is for instance, I can look at the shortage that sequence a Delta plus minus Boundary Delta That's minus You just plus minus just means it works for the pluses and for the minuses, right? So a sigma Plus minus to a gamma To zero Why is this a short exact sequence well one way to do it would be Well, we could we could look we could we could use the fact that we are shellable And then you can prove exactness along a shelling or you can use the fact that gamma is Again or homology disk, right prove this before the I think in reduction were just trivial and then you prove this Then you then you take out the linear system, which are complex thing in built inside when you verify because you do it Before quotient thing and then you apply the causal complex. Yes Information that there is only a zero Yeah, so the number of elements in the cousin, which is the same for all three. Yes. Yes, okay Right, so all they are all of the same dimension. That's right and Then I also have the Reverse Exact sequence so now I take gamma as a relative object plus minus To a delta plus all right All right, let's look at The top map let me I think blue is not so Visible let me go Let's look at this map here All right, my goal is Let me write it on the other side of the blackboard my goal is to construct a map from a sigma minus to to a sigma plus And in fact, I want to construct an embedding Again, that's great at vector spaces first of all All right, so This is exact And now the question is Where does this where where does where does this here become non-trivial? Well, it becomes non-trivial In the case of sigma plus Right, what is the minimal generator here? Well, it's a minimal phase right? It's a minimal interior phase so the generator Generator for a Delta plus on redelta plus is in degree is in homogeneous degree M In the case of Delta minus This is in degree in homogeneous degree D plus one minus M By my assumption that M was less or equal to d half either way I have an isomorphism All right, I have an isomorphism from From a a sigma minus to a gamma And from a sigma plus to a gamma for Homogeneous so for 4k in for degree k for 4k Less than M. Okay, that's an isomorphism What happens for the other degrees? Well Let me use Green So for degree at least M. I Want to look at This map and what do I get? Well, so now Okay, so I always have an injection All right, so I have a Gamma Only of gamma It's always an injection All right, so let me let me rotate it or write it on both sides So I have my map to a Sigma plus and I have my map to a sigma minus. Was there some sound? Okay Okay, so for a sigma minus Okay, so a sigma minus I Have to look at the closed star of Delta Minus at the closed so small the delta minus I have to look at so now I have to look at What are what what are the non trivial and entries in delta minus? Well, it turns out. Okay, so I Clearly have a non-trivial component in degree zero and a degree and in degree Okay, so degree zero comes from the constant then I have Okay, so if I have if M is strictly larger than one then I have another entry But if I think about it really this is this is really just projective space again All right, so and it goes up to well it goes up to the cardinality of the simplex So what I have is that this here is isomorphism for K larger equal to M In the other case, I only have an injection which together Right now I have covered all degrees Gives me my desired injection Because Delta minus vanishes above degrees above these degrees Okay, Delta minus here. Okay, so this year is a All right, I'm looking now at this trot exact sequence to contact this map Okay, so I'm saying that Delta minus All right this year vanishes from degree M on Yes, it's a flip locus before the flip right I always someone by convention my flip introduces a lower dimensional The the lower dimensional phase Okay, so let us say in this example. Yes Yes, so what is what is Delta minus what is a of Delta minus In this object here. Well, this is just a constant. This is just the reals. All right. It's just the simplex Yeah, it's not a projective variety right This is a cone right. This is a cone in a simple shell fan. All right. It's a simple. Yeah, it's one simple shell cone It's just a fine part. Yeah Yes, I don't know Yes, yes, yes, it is but and I think there are some people calling one direction flop and one direction flip That's I that part I don't remember that might be some people call one direction flop I Call both directions flips Alright, I actually wanted to go a little further today so now If you think about it a little more Then we can actually describe the co-kernel here rather explicitly So if you think about it the co-kernel comes exactly from here All right, the co-kernel of this inclusion comes exactly from here, but it is truncated in degree n minus m so what I have is That Finally, I have this injection and finer I have that a sigma plus I can write as a sigma minus plus Now I take a delta or Delta plus boundary Delta plus But I truncated in degree d minus m so truncated After let me be specific. I truncate everything after degree deeper d minus m after D minus m so I All right, so this is if you think about it. It's just a polynomial ring in one variable Up to degree d. I truncated it already in degree d minus m all right And this is nice a composition I Claim that this decomposition is orthogonal is orthogonal under the punk array pairing under the punk array pairing and For this let's Let me call this here All right, so let me call this here. Okay so let's let's consider beta in Beta in beta and K Of degree lowercase K larger equal to M All right, and then let's look at alpha in Well a sigma minus of degree Well, what is it? What is the degree? It's D minus K Right and since K is at least M right, so this here is most D minus M And Now what I do is okay, so in this degree. I could look at well, I can look at once again at This Part of the exact sequence this part vanishes in the sigma minus Therefore this part is an isomorphism right, so this is something that we already had discussed before so therefore alpha is in the image of the inclusion of gamma boundary gamma D minus K to a sigma minus D minus K because in these degrees I'm an isomorphism All right right, so this is isomorphism in D degree D minus K Therefore, I'm taking alpha Which vanishes right now. This is an element alpha In sigma minus that vanish it that is not trivial just in gamma right which is the component Right somehow gamma was the component outside of the pick locus here. It is not trivial. It vanishes here It is and I multiply it with something in beta Which is not trivial in the flip locus and vanishes on the boundary therefore the product is zero Product is zero. All right That is it so this is punk repairing Now if I'm orthogonal punk repairing and will also be automatically Orthogonal under the hodgman pairing so I really only have to say something about the signature on Well on the generator of this co-kernel right the the primitive parts All right the primitive part will become purely from here There will be one generator for the primitive part so beta Generator Let me call it just beta again or maybe beta bar generator for a delta plus Boundary delta plus All right, and I want I just have to compute the signature of Of the hodgman bilinear form on beta On beta bar this will be in degree m right degree m and Specifically this is a characteristic function of The minimal phase that I introduced right so in this case it would be the characteristic function of this ray So beta The Yes, okay, so now I'm looking at this at time at this at this transition time right When I have both fans together, all right, and then I have this the linear function now will be The the the function L will be linear on these d plus 1 vertices and now if you think about it This implies the compatibility, okay Yes, exactly the fundamental classes coincided the person that's the point. Yes, I wanted to yeah Let me okay, so let me try to give the rest of the intuition so beta I Should know I shouldn't rush then you get it gets chaotic, but we try to give them give the rest of the Intuition So now I have to say why does a right? Why do I have the right sign? All right, so beta bar Is the characteristic function of the minimal phase introduced introduced, okay Which is the characteristic function again is the product of the minimal of the characteristic function of the vertices of The minimal phase introduced, okay So now okay, so now we look at the characteristic function of this vertex here All right in this case and we notice that on this flip locus. It's a concave function Right on this call it is zero outside. It is one here. It's exactly concave right if I have another flip locus like this All right each of these vertices in the characteristic function of this vertex will be concave on this region and Then I multiply with this characteristic function, right? KW which is all again concave so every time I multiply with the characteristic function here of a vertex That will become the multiply with a concave function Okay, so in the end I have a concave function to the M and Then times a left sheds element So the sign that I get is exactly minus one To the M for the month for the product of the concave functions and Then I get my left sheds element therefore the sign is minus one to the M Yeah Okay Somehow that's the concave function concave Linear To the M times the left sheds element times the L to the Well times the left sheds element to the D minus 2 and D minus 2M that's it And this gives me the sign this gives me the sign minus Yes, yes, that's another way yes, all right, and this is a proof all right so Basically, I Flip I control what happens with the signature at the flip and every other time my signature is preserved Therefore I end up with a desired hodgkin relations by a continuous deformation Yeah Yes, so Yes, there's also there's exactly so there's this there's a right down of Proving some of the classical algebraic geometry language The hard left sheds on hodgkin on this way. This is the paper by the Cattagio migrini that I mentioned that does this way But now I followed basically just the convex geometry version all right Now I wanted to say a little about They they do essentially something very similar to this I don't quite remember the generality, let's look later. We know it also leads away from what I want to do actually so Let's let's discuss later Thanks. All right, so I wanted to at least Give the intuition for one case there's Where we go now beyond classical variety is really somehow this year's I mean okay, it's close enough right it's it's okay It's a it's over the reels, but still It's a complete fan, so you can kind of imagine that it is true So I wanted to at least do one actually I wanted to do two cases But let me know to restrict to one and this is the case of May choice If you want I will talk about The other case over coffee so the alias Williams work all right, okay So 3.3 beyond beyond polytops, but still convex so this is We will not even have so we will not have we will discuss fans now that are not complete so That is that is point. I mean this you're right tomorrow. There is another way of talking about not simply show polytops But then we have to go and introduce some combinatorial version of intersection co homology. I Won't I won't go into it again. This is something we can discuss more in private so a may try it for me is a combination of a ground set e finite and Let me just come to the convention that it's just numbers from zero to some number in For now and L is a subset of the power set Such that Well, I want the empty set And the ground set to be an L I Want in addition that if I have two elements in my lattice Then the intersection should be and I want so-called covering action so if I have an S in L and I look at the set CS of Sets of T without S where T is an L and T covers S in L That this partitions the complement of S and E Want it to be Bigger than S strictly bigger and there should be no two no album in between them in the order, okay? So these are ordered by inclusion, okay L order by inclusion I mean there are several different versions of Define Yes, yes, yes, this is not independent set. This is by by the flats, okay I will give you the intuition in a second what's your fight once over is happy Yeah, so this is okay normally people talk about independent sets All right, but this is okay, so and then find an abstraction for them right exchange action Now I want didn't want to because I wanted to to to to go to an object in algebraic geometry object more immediately And here what I did is I basically look at the flats So basically you look at the subs the subsets of a vector configuration induced by subspaces Okay The covering covering That's I mean Depends on what tastes of not river you have but that would be the non trivial one Okay, but if the intersection is that if they are not equal then the intersection is strictly smaller than one Because they cannot contain and then it must be L Are you then it must be a you took a okay, so that's clear. Yeah, so the covering for Yeah, if you want it, that's it won't give you one. Yeah, and And from this you get also the notion of the dimension of Yes, yeah, yeah, okay, okay You're you're already stepping a little further. That's again Okay, this is the intuition for for for subspaces and now I want to define a fan associated to this I want to define a fan B of M and I should try this is a matron M Associated to be associated to M to M And this will be a fan in our n minus one or Right, so it's yeah, it's yeah Sorry, actually I'm not with my notation. It's just all right. Sorry. I have n plus one elements So I have The numbers are in right so I have the numbers one to All right Okay So I want to define this fan and so the way that I do it is I take E I I in the ground sets a generating system for My matron for for For RN such that The sum All right, so these are these are n plus one vectors right because I started with the zero vector and I I'm generating so there's one linear relation and I'm just demanding that this linear relation is that the sum is equal to zero Okay All right Then to every subset F For for F a subset of the ground set. I define EF To be Well, I define you have to be the sum of the EI over our INF And Then finally what I define is for Flag of subsets, so subsets all up by inclusion fi in Flag in The ground set let me just let me just exclude from this the empty set and the whole ground set I take I Define E of fi to be the cone Over EF one so the cone over the elements EF two and so on Okay, so this is a scope and Now I Can finally associate My fan associated give my I can I can finally associate the fan to the matriarch. So B of M this is The union is the cones Cone let me just I introduced the notion so EFI Where Fi is a flag in L Okay, and that's my object Um, let me just give some examples just two of them. So for instance If my matroid is Well the numbers from So to to And then I have to give you a subset of the of the power sets. Let me give you some Okay, so I don't have to give you the empty set and the total set. So let me just give you one and Just something boring zero one and two then B of M Well, this is just a one-dimensional fan in our two Okay That's it In Equivalently I could Yes, yes, and once again topical variety. Yes, that's that's topical currents Um If you don't know tropical Again, somehow this is a tropical hyperplane, but it doesn't matter for us now. Okay tropical hyperplanes Tropical. Yeah, it doesn't matter for us now. What you were talking is a fan in the usual sense. Yes Yes, it's a fan in the usual sense. It's just a non-complete one Right, that's a fan here So the fun is the I I just the one those ones and not the not the space in which we So you've got Zero one to the maximal kind of So here's I declared the elements in L, but I left out the empty set and the total set Okay, so two you have two To what is to one? Yes, this is E1. This is a tool and this is zero Okay, maybe maybe let me do another example. Maybe it's What is the matter? That's a matriot on the ground side zero one two and the elements in L are just Kids and non-trivial elements in L are just zero one and two just the one element subsets Okay Yeah, empty as I didn't total set, but they're always And similarly I could just take okay another major that always works right I could take L to be the the the entire power set All right, I can I can take L to be the entire power set so every subset is a flat Yes, so think of this somehow this here corresponds to just vectors in our two general position vectors in our two Three general position vectors. All right. That's what this corresponds to All right, so now the subsets induced by subspaces are this this this and then the entire thing And Now I could I could also take three vectors in our three in general position and then what I get is essentially Well, I get the same ground set All right, and then I get well the entire power set All right, and then the Fan that I would associate to this Would be so I have I Have my vectors is a row e1 e2 I okay, I did it the other way round there to one and then I have well corresponding to the two element subsets I have e1 2 This inclusion is 0 to and the inclusions and I have e 0 1 inclusions Right, so now this is a complete fan again. This by the way some Either you want to call it the free Metroid Or the lattice is just a boolean lattice Right, that's it Well, ah, you mean you want to think of it as a tropical limit of Yes, yes, I mean that now we can think of this. I mean these are these are hyper planes right now. This is a hyper plane All right, actually somehow okay, so now I wanted to let me say a little bit and tomorrow then we can talk after okay So Let me say the main theorem here is the theorem myself Jimmy Ha very cats and Let me write it in a very brief way so I can look at BM All right, and I can look at a of BM Right again, I have a geometry given like this is a space of converse polynomials on this fan All right, so this is again the space of converse polynomials model of the ideal Generated by the global linear functions All right And this satisfies satisfies Well one punk a radio alati with the so called degree equal to the longest chain longest chain In L and now let me take the proper part. So which is L without The empty set and the ground set and it satisfies our left sheds And it satisfies our dreamer Yes Yes, we can construct a non-compact or a variety over over this over this fan All right, that's a here's a here's an interesting all right somehow Here's an interesting fact and maybe okay, so maybe let me first say what L convex means here in this way in this content What what convex means in this context? Okay But for this year we need a strictly convex home But you notice that every L here every matroid in embeds into the into the Boolean lattice Which is a complete fan where we have a notion of convexity All right, just restrict those Another way of saying it is that the coefficients of your L All right from a from a strictly sub modular function. That's another way of saying it okay The coefficients of this L right write them down as a linear combination over the variables they form a strictly sub modular function or A sub of an element in the interior of the sub modular comb the sub modular comb sub modular Well, okay, so modular function is okay So notice that somehow here that the variables occur the index by subsets All right, and now I can compute VA union B Somehow I can I can compute a function on the union of of two sets I can compute a function on the intersection of two sets and I can compare it to the function of on on the individual set the function is sub modular if what you have what you have on union plus Intersection is less or equal to what you have the sum of the individual parts And that's a problem. Yeah, and now you have a cone of such functions taking an interior point now in geometry, so there is this condition for unfairness in terms of line but in terms of the Corresponding combinatorial But this is sometimes stated for projectivity with complete funds and then if it's not a complete fund Well, this is this is what I said first Think of think of your metroid as a right So think of your fan as a sub fan of the of this complete fan here, right? Every every elder is a subset of the power set take the complete fan This gives you a notion convexity this gives you a notion of ampulness Restricted to the sub fan coming from your given metroid That's another way of doing it yes Yeah, that's right. That's right. For instance, you can if you want to look at the ample cone of M zero n bar Then you can ask well, okay You can ask how it how it how it would look on there is a metroid corresponding it This is just a graphic metroid of the complete graph and then describing this ample cone is actually an interior problem It might be larger. You're right It might be larger than just a restriction Right, it might be different things might be I mean there might be more There might be more ample functions than just those coming from the restriction here Depending on how you define ample. Okay, maybe we should I mean, okay So maybe we should finish for today because I already I mean some of the remote people might have other might have other appointments so Okay, so okay, so it was towards the end was a little rushed. I'm sorry I will try to repeat a little in the beginning of next time Before I go to the kind of the the really interesting stuff, which is the left sheds beyond positivity Here we still have a notion of convexity, but somehow I will explain a little more about the context of this about the ideas behind this theorem Next time but maybe we should go to questions and now Have a discussion Yeah, and do you have a question or Yeah, I wanted to ask you so the drink you did you define for for this matter Is it a show ring or is it at something else? Yeah, it's sorry. Yeah Okay, and regarding the proof of this theorem you mentioned that I mean at least for the how much Riemann Is it also built on this on this inductive framework? Yes. Yes. That's exactly the same. Yeah Okay, interesting. Thank you What is the last question whether how the proof works, but now I didn't get to it Yes, yes, if you think about it, this is a this is a normal fan of other promoter you don't yes Well, I mean that's okay. There are several things, but maybe we can let's talk about them. Yeah, are there more remote questions? Okay, so then let's unplug and let's have coffee