 Rhaen nhw wedi bod gan ydych chi'n adael y ddechrau dechrau? Felly, ddych chi'n rhai'r ddaf, ychydig i'ch yw gyntafol, ddydig i'w ddwych, ddysgwch i'w ddysgu'r ddiwrnod, ac ddysgu i'w ddysgu'r ddysgu, mae'n cael ei wneud o'r sgwm gwylltwch yw rhai. Yna'n rhoi'r gwaith o'r ddweud, byddwch i'r ddweud o'r ddweud o'r ddweud, i'w ddweud o'r ddweud o'r ddweud o'r M. Rwy'n ddim yn dweud, mae'n rhai ddaeth ymddangos o'r hexagum, mae'r ddweud o'r oedd ychydig o'r ddweud o'r ddweud o'r ddweud. Mae'n ddweud o'r latys o'r N. Dyna gwybod i'n ddweud o'r ddweud o'r ddweud o'r ddweud, mae'n ddweud o'r 1 yn ystod o'r 1. Mae'n ddweud o'r ddweud o'r ystod o'r Ystod. Dw i ddy Shiad iddynt ignoredwch i roedden nhw. Rydyn yn rhoi h oes o ddweud o'r min estructur. ein ffactor a Rydyn yn cael eu barberu, si produfestur ar gael gwadau ac mae Ffactor ar ven yn rhyw gilydd er declining o'r ach는데요. olywch chi'n bwysig y selyadau racedo gymfor ar rhedeg 2 o ged daytime. ychydig. felly oeddaeth drws ei hospitals pl Knogwch Ac mae'n meddwl ei gwrs, ond mae weithio gyda'r moddi, ond mae'r gwahanol wedi'u gwahanol, mae'r cymdeithas babigieniaeth, a'n ni dwi wnaethaf i iddi'r pwysig am y bau hyffordd a'r cyfranteun i'r defnyddio ddiad na fydd yn cyfwyr o gynydd y byddai. Yn ymwulech, mae'r ddefnyddio yn cael ei wneud hynny'n cael eu teuluo'r moddi. Yn ymwulech fel amdano i gyd, mae'n gydych chi'n gwybod i ei ddoch. A'u pwysig i gyd yma, I'm going to use a lot of blackboard space for this so I'm going to draw it in it occupies three layers like that one does and on the top it still looks like a hexagon is it now it's a hexagon dilated twice okay then on the middle size where the origin is it just looks like a hexagon and then there's just a single vertex at the bottom so I'm going to try and draw that so I'm first just going to try and draw the top size alright so at the top already another success at the top it looks like this twice dilated hexagon I'll just label some of the vertices so this is minus two minus two one and this one here is two zero one this one here is two two one and well this one here is zero two one and just sitting here in the middle is zero zero one okay now let me just drop down this is where the origin is going to live and then down one more to the point zero zero minus one then just join all these up like say it looks basically the same as that and then just at this middle size there's another hexagon living here where the point I on the so how does this jewel picture of mutation act on it well what I need to do if you remember I need to take the in a normal fan of of my factor so well it's easier to draw the normal fan first which looks like that and then I just need to negate everything so I need to try and draw this in three dimensions so what's happening is there's going to be a linear face through here and this linear face is just the span of my grading vector H and then I'm going to try and draw a fan on this picture so it looks like this I'm going to draw it up here first then just going to draw it down here I'm going to have a go at connecting it up so it looks like okay I don't know if you can see that so I tried to draw that fan in this picture so it's a one of the walls of the chambers is passing through here and coming down and there's another wall the chambers are heading towards you and it's passing through here and then the final wall of the chamber is going back into the black part and it's passing through yn ychwanegu. And then I can see it yesterday, within each of these chambers, the mutation just acts as a linear transformation, I'm just going to write it as a matrix. So here in this back chamber, of it's going to act like is via the matrix 1, 0, 1, 0, 1, 0, 0, 0, 1. Here in this front yn y cwmpdon i hwyl 111ol, nid oherwydd'r cwmpdon, nid oherwydd dda'r ddeithas. Felly,ryddoedd dda'r dda'r dda'r dda'r dda'r dda'r dda'r dda'r dda'r dda'r dda, ond oeddiad sydd wedi'u hwyl ymgylch. Llyfr e polygau oherwydd'r dda'r dda'r dda'r dda'r dda'r dda'r dda'r dda'r dda'r dda. O wybodaeth efallai llawer eu bod hi'n gwybod sut yna'r hyn yn awf. Yma yw'r gём? Gallwch eich bod eto ar y maes yma. Ar y maes yma eich bod eto ar y maes yma. Efallai yn awf dda. Yma eich bod eto ar y maes yma. Trwy hynny'n gwybod ar dyma yma yna'r hynny yng Nghwyrydd. Rwy'n gwybod ar y maes yma. Yma eich bod eto ar y maes yma. Rwy'n gwybod ar y maes yma ar yr oed. Rwy'n gwybod ar yr oed. is having the effect of flattening it out this edge and flattening out this edge and flattening it out it takes it and it just opens it out to make it nice and smooth and so what we get after this cue jewel okay it's just going to be isomorphic to three times three times um no twice the dilation of a cube so and this has volume equal to three factorial times by eight and if I count then so this is equal to minus k x cube and it has so many lattice points what does it have it has cue jewel intersect um so it's got 27 lattice points and this is equal to h naught of minus k x cube okay and this is before maybe you recognise this from you know looking at tarot varieties before but this is the jewel for p1 cross p1 and p1 cross p1 cross p1 so because it works as a piecewise linear transformation it does an immediate corollary and that's basically if we take p to be a phano polytope and if we take a mutation of it the soft minus k numerics are unchanged so we have the degree is unchanged and we have the Hilbert series is unchanged so in other words h naught of successive dilations is unchanged exactly what we'd expect we're expecting the xp and xq are um qg defamation equivalent and so qg defamations uh well i don't know if characterise is too strong word but they preserve the numerics of minus k so this is what we expect all right okay so that's um what's going on in three dimensions in two dimensions as you might imagine the situation is a lot simpler a lot more easy to get a hold of what's going on and prove some stuff i just want to focus on this now so i'll try and be clear about if i say anything that's true in general okay but so take everything that i say is just being true in dimensions so return to the mutation that we have before where we took p2 and we sent it to p114 okay so um let's just recall that p2 um the center is um sent to p114 via um the mutation via a mutation but i need more part space so it's via this mutation so i'm gonna just fix a basis and i'm gonna have h is minus one two and i'm gonna have a is just this line segment from the origin up to um two one the picture in n before but we'll do it again so this is p this is going to be p2 and i start with this final polygon fan is just the spanning fan which is how i know it's p2 and then my h that i've chosen imposes grading on the lattice and we drew them before i'll draw them again this is going to be at height minus one height zero one and height two when we do this mutation when we do this mutation we end up with p11 to p114 which is one two three four five you want to stick the origin here and i can track that and i stick two copies of it at height two it looks like this and may or not believe me but that's going to be a straight line i'm only meant to have this point again you know i draw the fan i want to it looks and the sort of little remark is whereas before all the cones were smith here two of the cones are smith but this cone becomes a singular cone it's a quarter one one singularity so what do we get in the dual picture don't think i'm going to squeeze the dual underneath i'll have to just do it again i'm going to put my p2 here i'll put my p114 here so the dual to that triangle is a triangle you've seen before but in a different context i have less and here's the origin and my h that i've chosen looks like this maybe i should emphasise i'm like are they sure when i write the um diamond brackets i really do mean linear span i don't mean cone okay i think we sorted that out right okay and mutation how does it act on these two chambers well you know i'll just see and write it as a matrix it goes three minus four one minus one on this side and on this side it's just the identity so you know again it's just the identity because this vertex is zero and let me just associate some of these vertices to the edges over there so this edge here i'm going to i'm sorry this vertex here i'm going to label as u1 and through duality it corresponds to this edge here this edge here to this edge here uh this vertex here and so this vertex here correspond to this edge here let me just apply this mutation get my p114 and this is a rather big lattice so i need to draw a seven by seven lattice okay so this is going to be a little break what does it look like in here well i uh you know i suppose pedagogically speaking we should take this and visualize it but forget that i'm just going to apply this linear transformation okay so what are we going to get the origin is here in my picture and i get this see daisy there you go looks like this let me just draw that vector of that um linear space back in so it's passing through here i think yeah again let me just associate these vertices of the dual polytope with my edges um so we've got u1 prime 3 prime there and u2 prime there and then over here what do we have we have e1 prime this must be oops i've joined it on the wrong picture so we're here e1 prime e2 prime and then here is e3 and again notice that just here were um just around here where we had our u1 mutation has flattened it out okay so here mutation has flattened the boundary okay so we saw that in this case it took three edges and it just flattened them out and here it's just taken this corner and it's just flattened it out so now i'll say something that's true in all dimensions it's just a um fact of duality so under duality um let's call it c e which is just one of the cones in my fan that spans facet e okay so over the facet or a facet p of p so it's dual to the matching tangent cone in the dual picture okay so um is dual the tangent cone at the corresponding vertex i could try and give you a definition um i'm not going i think it's just easily illustrated so for example let's just look at p2 again and let me just pick one of the cones i don't really care which one i'm going to choose this cone here so this is my cone c e then in the dual picture well this edge corresponds to this vertex here i hope then the tangent cone is just this cone so it's equal to c e dual and then just translate it and c e dual is just defined to be all those vectors u um that evaluate to a non-negative value against the original cone all right why am i telling you this tell you this because again this is true in all dimensions if the vertex of the dual polytope is strictly inside one of our chambers then is just being acted on by a linear transformation and so the tangent cone is completely unchanged okay and so in other words up to a change of basis the corresponding cone back in the fan is also unchanged by mutation and the only exciting stuff happens on the walls of the chamber which i write that down so f v sorry i'll call them u's u e invert c's is in the interior of a chamber then c e n is unchanged so when i say unchanged there's a change of basis going on but you know the singularity that the cone describes that's not changed coming back to dimension two well in dimension two the only non-trivial fact is you can have uh line segments like this and so we always just have um the dual that is split into two halves and so there's not a lot of scope for excitement basically the excitement happens here and here and everything else just gets shuffled around a bit so in dimension two we need only and the corresponding top and bottom of p because everything else is just getting a little change of basis so here's the top and here's the bottom of my polytope and there's just a bunch of stuff happening between the top and the bottom sizes corresponding to the two points in this are you know i'm just going to put there and this is a picture in n i should say oh here's the origin this is in n and i'm going to use my h to be grading in that direction i've rigged it up so that meditation says i need to remove a factor from here and add a factor in down here and everything else just moves around the corner let me just draw my cone that i want to think about so if this is my edge e i want to look at this cone and this is my cone and well i should have said i want this to be a funnel polytope so i'll just say p funnel and with respect to our grading this is at a negative height let me label these row one row two so this is at a negative height so i can just write it as minus r and then this is at a positive height so i'm just going to call this s and i've called this little edge e and just going to call this edge here an f no i'm drawing an edge down here it might just be a single vertex it's not going to make any difference to what i'm doing but i could only sketch what i'm doing i can't really give it you as a brief it's not worth the time and i'm just going to look at the length the lattice length of this edge okay so i'm going to call this l and it's just going to be the number of lattice points along this edge subtract one this value r here is just the gorenstein index of the singularity corresponding to z if you know what i'm talking about so r is the gorenstein index the singularity the apologies in advance i'm just going to confuse the distinction between a cone and the corresponding singularity that's my setup and my aim is to understand what happens when i mutate something out of this and put it down to that i'm going to assume that my factor which is contained in h perp so my a is contained in here um it's just a line segment of length one lattice length one and it turns out that whether or not a mutation exists given data h and a is purely a local property of this cone and really the whole of the rest of the polytope is irrelevant now i have to emphasize that is absolutely not true in high dimensions but two dimensions is nice and it's true so whether um gives a mutation of p or not um it turns out local property of c e or of e there's no distinction and so just a warning not true i just want to sketch for you what's going on let's let me convince you in words remember this has been sliced up into a bunch of um slices strips and we're allowed to mutate if each of these strips i can remove an appropriate multiple of a where that multiple is just the height of that strip okay so here i need to be able to remove our copies of the line segment and then at the next height down any to be able to remove our minus one copies our minus two copies et cetera et cetera okay and it also if you remember in the definition i said it just didn't matter how we chose our remainder terms and so it turns out that a lot of the time we're basically free to choose a bunch of empty slices the only problem is if there's some vertices i'm not going to go into any of those details i'll just point you out a paper where you can read the details if you want okay but the upshot of this is basically we can do the mutation so long as this length l is bigger than r okay just what you expect so i ff and only f is less than or equal to and for obvious reasons i'm going to call this the length and i'm going to call this well the height or the garden-style index anchor and what happens when i need that well the mutation well it shortens the length by r okay and then down at the bottom i increase the length of f by s moved it from upstairs and i've shoved it in downstairs and everything else in between just gets shuffled around a bit but basically doesn't change it's um what we saw the the rest of them just lie strictly inside chambers of the dual picture and so this means that if i basically figure out how long each of my edges are and work out how many times the height goes into the edge and do that for every edge all the way around i've got a mutation invariant because when one of them shrinks by its height one downstairs grows by its height so for each edge e let me just write no this is a little bit up and down to it i'm sorry but i'm just going to write the length of e as um so many multiples of the height of e plus some remainder and so what do we have here we have c e k e they're in the non-negative integers and i want zero to be less than or equal to c e to be less than all right okay then i'm just going to write oh it's a shame let me call it n n equal to the sum of all these k e's is invariant and so is these collection of the little remainders that's left over let me try here and sort of give a toric view of what i'm saying here is my cone c here is two rays and it's at height and i'm saying that i can subdivide it into k cones that are all of the same length and height so i can just go chomp chomp and there'll be maybe a little bit left over at the end but what i've done here is i have k cones equals height r and then here we have zero or one depending on whether c is zero or not remainder cone so this has length c e but height c and height r and so i'm going to call these i'm going to call these p cones probably a bit low for some people i'll just state the definition over here we call c a t cone the length is equal to the height the Garnstein index we call the r cone is less than the height why have i bother doing this what the point is that mutation is just shuffling around these t cones and leaving these r cones untouched so if i take the set of all the r cones i end up with after doing this subdivision so let's let b be the set of the r cones of p and this is after doing that subdivision then oh and transpose and this before is just equal to the number of t cones actually this pair n and b is a mutation invariant and we call this the singularity content right okay yeah i i mean what this cone is as a singularity okay or what this cone is but don't care about how i've put it into the basis okay but it's slightly more information than just the c e okay cool so geometrically what's going on so um these t cones uh basically the same thing it's the slight white line going on but basically these t cones but the same thing is t singularities which are all of the form one over r squared one um d minus one where gcd i d is equal to one and these things were studied by collar shepherd barron back in 88 and the point of the matter they're precisely the smearable two-dimensional torrent singularities so these correspond to the smearable so she was a quotient so from our point of view this begins to sound kind of believable and the qg deformation what we're probably doing is we're smearling away all of the t cones and we just left with these this collection b of um r cones of singularities and indeed that's the case p with singularity content equals np um so so that's that then we have xp qg um the forms to a del petso or before del petso i guess island number n qg rigid singularities b that's what's going on and i should probably give you some references before we just finish off the references all on the archive i'm going to point you at three papers so one four zero one five four five eight and then um okay i said but unfortunately we're not writing down the rest i'm going to point you at one paper okay and the rest you can google for all right okay so we got this beautiful um invariant and so it makes sense to say let me try and classify all polygons that have given singularity content so we can try and classify them all and we're going to have to do this up to mutation and this is doable okay this is really something that you can do and it turns out that there's only finitely many and there's only finitely many mutation equivalence classes i mean i don't know if you realize quite how cool this is so one can say go away and find me a bunch of toric models for del petso that have bloody bloody bloody singularities and you can do it okay there's an algorithm that just lets you go away and produce them all and there's just going to be finitely many answers and the proof is constructive and so it makes sense to um just start working your way through the simplest possible rigid singularities arcones so the simplest or smallest arcone is a third one one let me draw a picture the cone that goes like supporting hyperplane is just passing this is what a third one one singularity looks like and so you can go away you can find all the final polygons up to mutation that have third one one singularities and everything else just the t singularity so um i had a reference for this t but i don't apparently so we can classify all final polygons up to mutation with singularity content equal to we don't care how many t cones and then some number he's doing a lot of work um some number m third one one singularities and he turns out that there's only finitely many mutation equivalence classes there's exactly 26 of them you can see those 26 mutation equivalence classes on the problem sheet and so this is really great let me draw a few of them because i want to carry on studying them tomorrow i'm going to draw three of them chosen pretty much at random so these are just three representatives in three different mutation equivalence classes that is the origin remember these are pictures in n so it makes sense for me to just draw the spanning fan just to remind you so here is one of them okay now as a total variety this has got a singularity here that's not third one one the point is that i can smear all of that away in this way to give me something that's just a third one one okay um let me draw another one so this one like this it's pretty similar but it said now i've moved the origin to here so the fan this case looks like this then one more this one's a bit of a beast so what does it look like okay so again it's seven across and actually i'm going to need the diagonal i'm going to need the upper triangular path and it looks like this the origin now is sitting here so the fan looks like that is apparently on that side of that array and this is apparently on this side of the array and i guess it's kind of an exercise for you to go away and calculate the singularity content in three cases but i'm just going to write it up for each one so this one has got five t cones and then it's got a single third one one singularity and in fact i can see that third one one singularity it's this cone here this one has got six t cones and they're also just has a single third one one singularity again i see that third one singularity it's just this cone here and then finally this one well this one has also got just six t cones and um you might sort of at first just think it's only got a single third one one singularity there but in fact there's some more hidden in there if you go through the um decomposition of the cones it ends up having three a third one one singularities so just to finish off this is really good we've got 26 um mutation equivalent polygons which are the only ones in the whole universe that have these third one one singularities but how can we understand them as del pencils right how can we sort of smooth them and the obvious answer is we want to be able to make taric complete intersection models from them so just to sort of finish the question now is how can we make um taric complete intersection models um or x one of these p are a closely related question if you give me one of these p how can i put all the romp polynomial on it in a sensible way so that i recover the period sequence for x how can i do all that all right so stop there