 Okay, cheers. So, before we begin, Lacio and I, we wrote a quick, first draft of an example slash exercise sheet. Okay, so if people want to find it and do it over the weekend, then you're more than welcome. So Google for Lacio's homepage, that's pretty easy to find. And then once you're there, I think you go to the teaching section on his page and it's just the first link on the teaching section. Okay, but it's just the first draft, like I said, and we'll be adding to it and correcting it and making it make sense over time. So today I want to tell you about just one thing, which is mutation, which Lacio alluded to yesterday. So, those of you who've seen it before are going to really hate this action. Okay, so I think there's probably two references that they should point out. So the first one is on the archive. Okay, so it's this paper on the archive. So this is by actor Galkin and myself. And then the second paper is also by Galkin and by Uznitch. And this isn't on the archive. Instead just Google for Galkin-Uznitch and then it's got a preprint number which is IPHU and 0100 Okay, so you can find that. So the material in today's talk is sort of all in those two papers. So I think Lacio had got us to the stage where we were looking at Lloramp polynomials and he sort of passed to the Newton polytope and he said, from this Newton polytope we want to create a target variety. Here we want to take XP, the target variety and this is given by this spanning fan. So in the sort of Toric geometry world this means that P lives in the lattice and so motivated by the fact that we want to turn these Newton polytopes or these polytopes into fans, they have to have some properties for it to make sense. So in particular we need the origin to be in the strict interior of P and we need the vertices of P to be primitive. And if we don't have those two conditions then we can't turn it into a complete fan. So this is why we do that. That's the only reason why we do that. And these are called for obvious reasons final polytopes and true vertex. True vertex. I mean how do you mean by? Yeah, the number of variables can be added. So from our point of view it's easy to make polytopes but it's not so easy to make good around polynomials. And so what we want to do is find a sensible sort of ansatz, a sensible way of you give us a final polytope and we find a way of assigning coefficients to it so that it's a mirror for something. That's basically the moral of what we're trying to do. And so today I'm going to try and move us away from the world of Laurent polynomials and into the world of polytopes. But to begin with I'm just going to recall the mirror for p2 which I've actually did in the very first lecture. So let's take f to be x plus y plus 1 over xy. I should perhaps point out that in moving from a polytope back to a Laurent polynomial there's a whole host of requirements that we're not really sure what they should be so I think they were brought up before like should the coefficients be integers at they are allowed to be negative we don't really know what we're doing just yet but anyway, well we got works which is what I'm going to show you. So we start with this Laurent polynomial this has the period sequence if you remember and I'll actually point you out that you can easily take its Taylor expansion and what you have is the coefficient of the constant term of the constant term of successive powers and in this particular case you'll find that you only get values of k that are divisible by 3 and you have 3n factorial over n factorial to the 3 and here now we have n and t to the 3 n so this is what it looks like in just the first few terms 1 plus 6t cubed plus 90 yeah I think so 90t to the 6 plus dot dot dot we can be pasted in Newton polytope this is just going to be here's my origin this is just going to be this polytope and the associated type variety we just take the fan that passes through the faces of the polytope and so this is our fan and those of you who have seen geometry before will probably recognise this we have xp is just equal to p2 so what we're hoping for is that the torrent variety is constructed this way degenerations as I said of our panel and in this case a torrent p2 is just p2 so the question is are there any ways that we can modify our choice of Laurent polynomial but still get the exact same period sequence but there's one obvious way so I think pretty obviously we can apply a monomial change of variables and this won't change anything because the constant term will be fixed so I'm going to do it concretely in this case I just apply this change of variables then I'm going to get f' which is just 1 plus x over xy plus xy so that's not very interesting and basically we just regard polytopes and we regard Laurent polynomials up to the action of gl dz now I'm going to apply a more interesting transformation to this so I'm going to apply the transformation that's going to leave x unchanged and it's going to multiply y by 1 plus x so doing this I'm going to get g' which in this case is just going to be 1 over xy plus x 1 plus x y squared so this is still Laurent polynomial I mean it obviously didn't need to be Laurent polynomial but I rigged it up so it was still Laurent polynomial and it's harder to work out what the period sequence is for this but if you go away and try it you'll find that you get the same thing so this is also just a mirror for p2 it just gives us exactly the same period sequence and if I draw it's Newton-polyto well let me undo the change of basis first so undoing the change of variables gears via might as well be explicit since I got it written down so x goes to x squared y y goes to 1 over x so what we're going to get is 1 over xy x squared y squared the obvious thing now I guess is to draw it's Newton-polyto and see what sort of torrid varieties come out of this so let's just have a crack at that this is slightly more unwieldy than the last one 1, 2, 3, 4, 5 so what do we have we have 1 over x squared so I'll stick the origin here and then if I go up to y I need a line that goes across twice so I have I'm going to get this and this okay so it's still it's implicit but just notice in here this is no longer a smooth this actually has a singularity of quarter one one and so in this case x of what's called this Q this is just weighted projected space P114 so this is an example of a mutation for us mutations we regard two things that are the same to change of basis monomial change of basis have been the same anyway so for us mutations of these birational transformations that pretty much look like what I've just written that present the period okay so slightly more generally sorry if you feel that you've already digested exactly what a mutation is but they're so important to the whole of the rest of everything that we do I think it's worth just spending a time on this so slightly more generally so that is what I sell some Laurent polynomial in you know an arbitrary number of variables but I'm just going to make the last variable distinguished just for this case and I'm going to work myself a second Laurent polynomial but this one is going to be in just the first b-1 variables and I'm going to define basically exactly the same map so nothing is going to happen for the first d-1 variables and then y is going to get sent to a let's suppose that when I apply that I do end up with a Laurent polynomial because in general of course it's just rational now there's some details I'm not going to go into but basically then applying the change of variables formula to the period integral you'll see that essentially nothing changes and we see that pi of f is just equal to i g I mean it pretty much is just change of variables the problem is like the domain of integration but that's not very bad but of course in general this isn't Laurent so what do we need to do to rig it up so that it is Laurent okay, what do we need to assume in order for g to be Laurent rational is it you know it simply would be well again there's literally no mystery about this whatsoever I basically need to be able to divide my factors the make up g by the say okay so let's just write f is just equal to some sum basically graded by the final variable so here we have the pi's of Laurent polynomials in d-1 variables and only finitely many of the pi's and on zero and let's just suppose that there exists some i in d-1 variables okay such that, now what is this it should be that pi is equal to r i a to the minus i for all h i less than zero so in that case I can just write down what g is so then it's just given by well the negative bits so i for minus infinity to minus 1 and this a is going to go and we're going to give ourselves an r i and then the positive bits just multiplying there so that's just all that it is that's just what we need to do so let's define what I mean by mutation so definition so let's just award ourselves a lattice of rank d let's swap it to z to the d and let's award ourselves a grading on this lattice so we're going to let h be an element in the dual lattice and it needs to be primitive and so h induces ok now we're just going to award ourselves an a so let's a and we want a to be in the zero piece of this grading so I'm just going to be in h perp intersect so this per h and a is the data that defines the mutation so it defines a map the rational functions to the rational functions and what do we have and we say oh and you tell it the map is sorry just give them by is going to go to x to the a a to the h and here we have a is n so we say that some Laurent polynomial f is mutable with respect to this data if the image under this map is also a Laurent polynomial so if g which is just going to be this is Laurent and we just call g a mutation and because it's just kind of nice to give it a name we call a a factor right so just a few little observations so the first one a yes is if we just set a to be a monomial actually this isn't very interesting all we're doing is applying a monomial change of basis to f so we kind of don't want to worry about cases like that so we'll think of that as just a sort of trivial mutation so just extending that what I can do is I can just take my factor a and I can just multiply it by a monomial and that's just going to give me the same mutation too so h a and h a times by x b give isomorphic mutations so we try to only regard a factor a as being defined up to this translation which part from here oh h a and h a times x to the power of b just s give isomorphic mutations yeah it's excellent introduced underlights so these just give isomorphic mutations so in other words that's probably too long so we think of a as being defined only up to these translations and I said we don't really worry about a monomial change of basis we just regard those is being the same so if you give me two polynomials that are connected via sequence of mutations then the period sequences agree so ng a sequence of mutations plus change of basis they have the same period and the question to which we don't know the answer is the converse truth so is this an effinol it is any yeah absolutely no no no we've done exactly that yeah we've done exactly that and they've always been connected like it's very common so think about it this way there's roughly speaking 4,000 reflexive free topes and you can put a lot of polynomial on them and these will be roughly speaking mirrors to the 105 final manifolds so there's like a 40 to 1 duplicity and you'll find that actually whenever you find two things that have the same period sequence made this way then they are connected via mutation never found a counter example and have not always just done it in the naive so after after going on about how we just regard two polynomials that are isomorphic under a change of basis has been the same and I want to say that actually it's sometimes quite useful not to do that so I'm going to start with again the mirror of p2 so x plus y plus 1 over xy and I'm going to mutate this in three completely different ways ok so the first thing is I'm going to notice that I can write this as x of 1 plus y over x plus 1 over xy so this means that I can pick some mutation data as follows I can pick my h to be minus 1 minus 1 and my factor a to just be 1 plus y over x and if I apply that mutation I'm just going to get x plus x plus y squared over I can also write this as x times by 1 plus 1 over x squared y I could award myself the mutation data minus 1 2 for my h and 1 plus 1 over x squared y for my a and this is going to give me the Laurent polynomial x plus y times by 1 plus 1 over x squared y ok and then the third way is I can write this as 1 over xy times by 1 plus xy squared plus and here I can just give myself my h to be minus to be 2 minus 1 and my factor to be 1 plus xy squared and this is going to give me 1 over xy plus x y squared so here I've got three fundamentally different mutations coming out of the mirror for p2 but in fact the three Laurent polynomials they're the same under a change of basis so the point of this is sometimes you have to remember that you achieve the same polynomial in different ways so all three mutations isomorphic three different mutations and here I guess I should sort of just stick in a little reference to paper by Paul and Paul hacking in Prokhorov so on the archive it's 08 08 5 5 0 so this is a beautiful example it turns out you can keep mutating this beautiful trivalent graph of mutations and it has connections with the Markov equation and all sorts of things and so it's really described in some detail there all right so this is all at the level of Laurent polynomials and I kind of said that we wanted to move away from Laurent polynomials because although it's easy to write phanol polygons or phanol polytopes it's not easy to know how to pick the coefficients of them to turn them into Laurent polynomials let me basically just reinterpret all of this in terms of combinatorics of polytopes so definition so we just begin by fixing our lattice and we're going to pick a primitive vector in the dual lattice and we're going to award ourselves this time convex lattice polytope A inside I'm just going to say inside H perp just to be clear about what I'm talking about there I'm taking that to be all those points A in N tensored with R such that H of A is 0 so pretty obviously A is of code I mentioned at least one and I'm not insisting that it's a phanol or anything like that and now I'm going to award myself what's basically my Newton polytop so let's let P in N tensored B A convex lattice polytope actually it doesn't even need to be a polytope we're going to work through this definition and you'll see that this definition would apply even if it was a non-compact object so this can be a polyhedra rather than a polytope and I'm not even insisting that it's a phanol polytope although typically this will be a phanol polytope and then to just apply those conditions that we saw before for other amp polynomial that allowed it to be meetable and need to apply those conditions to P in order to do that I'm just going to have to take the sizes given by the grading so let's let P I to be given by the convex hole of those points lattice points in P that are at height you know only finitely many of these are non-empty and then I need that little condition that I had for the Laurent polynomials that just allowed me to do the division and stail around so phrased into combinatorics it reads like this and suppose that there exists these convex lattice polytopes contained in H perp such that before it was multiplication and of course here now it's going to be Minkowski sum such that this Minkowski sum is contained in P I and on the other side we also need it to contain the vertices of P at this height here I is negative so once I've got that then I can define a mutation so the mutation here it requires knowledge of H and A and it's just going to be given by you know this looks a bit ugly but if you just think about what we did with the Laurent polynomials it's completely natural so I'm just going to remove these A's so I just get the remainder and then a positive height add on copies of my A's so it's going to be P let's call this a J PGI plus so this is exactly analogous to that formula we had before I guess the only thing that's slightly different is yep this is my next remark but before I say that you know the only thing that's sort of obviously different is the fact that here we have to include the vertices but that kind of makes sense because we think of this as coming from a Newton polytope and the Newton polytope has those vertices precisely because there's a non-zero coefficient there and so it's important that we preserve the vertices when we do this okay so that's why the vertices appear okay so a little remark so although these RIs these remainders have to exist for the mutation to exist in fact the particular choice of them it turns out doesn't change what the mutation is okay so I'll say that so the mutation does not depend on the choice okay I'm not going to prove that at all but it's essentially due to the convexity of this thing because we take in a convex body different choices of the Rs they just get subsumed into a convex hole so it really doesn't make any difference they are identical yep you're completely right okay yeah I mean so you're right of course that these little RIs they just correspond to the different remainders that you could have got with all the different yeah absolutely yeah yeah so um out of time today but I'll show you the dual picture and in the dual picture basically the Rs don't even feature and it's completely canonical basically just marry what we said before if R is just a point oops not R I'm calling it A if A is just a point then the mutation of P is just isomorphic to P so we regard that as trivial and we also have that if I translate my factor so by any element then again it's just isomorphic factors these these A's are really only defined up to translation and that's kind of useful because it's often handy just to insist that the origin is one of the um the vertices of your A and then obviously now we have a slight notational nightmare um suppose we have that f and g of the ramp polynomials with um g equal to the mutation with respect to only if um only if A is off code dimension one but there's two edges but I mean yeah that's true except you know here I'm taking the slices through P as determined by the gradient page so I mean it definitely appears in every single one of these choices also you know just flipping the sign of H I mean that's not something you could do arbitrarily it really matters which sign of H is so even if it's equal to code dimension one you still have a side but you know the choice of grading really does um matter in these slices that you're taking okay anyway the point of this is you know if I just do a mutation of a ramp polynomials and I mutate a new polytops I get a mutation of um polytops um what do I need to say I need to say that the Newton of g is equal to the mutation with respect to H and the Newton of A but the converse isn't true so what I mean by that is if you've already fixed your f and u g and you find the mutation of the Newton polytops it doesn't mean that you can mutate your f and u g okay the particular choices of coefficients really do come into play into all of this but um you can always pick some coefficients on your polytops so that it lifts to a mutation of the ramp polynomials that's not a problem and you know that's gonna be important in the rest of the stuff and then one final remark um again I well maybe I can sort of sketch a proof to you um so let's let q be a mutation of p then we have q is phano if and only if p is phano let me just sketch in words why that's likely to be true yeah okay so maybe I should point out that yeah you can always invert mutation by flipping the sign of H okay yeah so why is it that q is phano if p is phano well I mean the first thing is that p has the origin strictly in its interior at size zero so you see from this definition that for sure q is also gonna have the origin strictly in its interior that's alright the real question is why are the vertices of q primitive okay so the vertices of p are primitive why are the vertices of q primitive and the answer is maybe I regret having said this okay if we look at our q and I just look at a vertex of q so here's some vertex of q okay where does it come from well it's come from the Minkowski sum of I mean this is a sketch it's come from basically the Minkowski sum of a vertex of v and a vertex of a so this vertex let's call it a no let's not call it a let's call it v so we have v is equal to a plus u for some a vertex of a and b a vertex of p now if v whoops you a vertex of p now if you wasn't primitive then the b of course some other point was the origin inside the polytope and that point I could be written as I suppose it's going to be okay it's going to be more precise it's going to be h u plus a if I look at this point here this point is given by its height that's what I wanted to say if he's small I'll just write that one bit bigger point is that if it was divisible then I could just divide this and I'd get a point that was inside my interior so if v was divisible have I'd have u minus then I'd have u as divisible which is another out this is not meant to be a sketch so this is kind of important for us in the sense that once we're in the category of phano poly on polytopes we never leave the category of phano polytopes when we do mutation so I guess what Alex was saying is if we have our two phanos that are rising under mutation that's kind of tautological I guess the point is that we want mutation to give us a defamation from this phano a qg defamation and so if it was possible for mutation to a phano polytope we'd obviously have problems so the fact that this is true is sort of a good hint that these two sort of varieties defamation so the last point is that mutation so if we have q is equal to the mutation okay then we believe believe, prove, Nathan proves it yeah, hear that we have I guess by paper Nathan Elton I guess it's in 2012 we have that the two tarot varieties xp and xq defamation equivalent you know if this wasn't true that would be a ludicrous thing to believe alright I just want to wrap up with one final example I'm just going to do p 2 again I'm going to do p2 at the level of the polytopes I'm going to start with p q-gornstein q-gornstein q-gornstein yeah q was in the rational what is it? yeah I don't understand that either alright so I'm going to pick this mutation data cos this imposes a grading so what the heck if we draw it so this is the grading imposed by h so stuff down here is at height-1 this is at height-0 this is at height-1 and this is at height-2 so basically say that I have to subtract off appropriate copies minkowsky subtract whatever that means appropriate copies of a at negative heights and then at positive heights I just need to add on minkowsky copies of a so if I do that it's that big example we've already seen 1, 2, 3, 4, 5 1, 2, 3, 4, 5 alright let me just draw the grading back on this just cos it helps so the origin is here so down here I subtract off 1 copy of this a so I end up with just the point and then as I work my way up well up at the very top I need to add on 2 copies of the a cos it's at height-2 so I end up all the way out over here and then the whole thing it's sufficient for me to just take the shortcut so as we move up through the grading we're removing copies and then we're adding on copies and you get this so this just gives me a map from b2 to b11 so you know I'll stop there and do some higher dimensional examples next time