 All right, so so now what in this last lecture I want to switch gears and We're going to completely forget about physics and we're going to define some natural geometric objects and I'm only going to be able to sketch it in the in the hour That I have but the point is that that in the appropriate sense the volume of these Geometric objects will end up corresponding to the scattering amplitude But I want to do enough so that you can see how the formula we saw a second ago is the actually the volume of an object that is sort of God-given and Naturally defined even if you'd never heard about it. So let me just give you a sketch of what the idea of the amplitude Hedron is and everything is going to be a Double generalization of the idea of a triangle. So we're really going to start thinking about all the points on the inside of a triangle and There's one generalization of a triangle well to tetrahedra and Simplicies in general so it's triangle to simplex and This has a Generalization into the grass manian. I'll tell you what that is But the generalization of the inside of a triangle into the grass manian is something known as the positive grass mania so that's one set of generalizations and This is something again As I was saying at the end of the last lecture you would have thought I mean the grass manian is just a space of k planes and n dimensions Well, it's projective space is just a set of lines in n dimensions The grass mania is just a set of k planes and n dimensions. Mr. Grossman who was a high school teacher In the 1850s. He couldn't get an academic job So he was a he was a high school math teacher But he invented quite quite strangely from our purposes both the grassman variables that we talked about for supersymmetry As well as the gross money in which is a set of k planes and n dimensions the same. Mr. Grossman did both things But anyway That's just a set of k planes and n dimensions again If you're a physicist you think who gives a crap about any of these things k planes boring, you know What could possibly be interesting about these things? But they're boring, but it sounds like a boring sort of thing that mathematicians do so Surely people thought hundreds of years ago about the analog of a triangle inside the grass manian, but This was not done. It wasn't done until the 2000s is to figure out What is the analog of the inside of the triangle in the aggress money around ten years ago? Sort of amazingly roughly simultaneously with one we are running into these objects from the study of scattering amplitudes It's what our mathematician friends. Well, I mean, I'm just joking. I love mathematicians there It's just a funny. It's just the usual It's the way that we express affection by making fun of each other. So But anyway, so so this this was Not studied till ten years ago. It's already sort of interesting. It took so long to study But and just like triangles are crucial for building up areas, right? That's the basic building block that like builds up areas We found that the pieces of the positive russ manian We're in one-to-one correspondence with the the building blocks out of which you could build up scattering amplitudes, okay? But it was a building blocks and not the whole amplitude itself And so we were struggling for a long time to find what is the object that actually? Builds the amplitudes rather than just gives you the building blocks and well That's the second naturalization we go from a triangle to a polygon the inside of a convex polygon and If we generalize the convex polygon into the grass manian in exactly the same way that we generalize a triangle What we get is the first example of an ample of an amputahedron that we call the tree amputahedron and This is the object whose volume calculates tree amplitudes and There's one more generalization Which is that in this geometric setup you can now ask a natural question of how you can hide particles So you have a you have this big geometry, but you have certain particles You don't want to look at anymore in a quite literal sense. It's a theory of hidden variables, okay? But it's not living in any local space time. It's living in this Funny grass manian space, but in this grass manian you learn how to hide particles And hiding particles corresponds to calculating quantum corrections That takes you to the loop amputahedron and the final up And the final thing to any loop order. Okay, so it's this double generalization and So all I'll have time for in this lecture is To just tell you a little bit about what this generalization is the first thing and a little bit about How we define the tree amputahedron? But I want to tell you enough so that we can see in one concrete example How it is that out of this all of this geometry you get an answer that looks like physics, okay? And where the locality and the uniterity come from out of the positive geometry all right So let's start again. We already talked in the previous At the end of the previous lecture that the points on the inside of a triangle 1 2 3 all the points on the inside All the points y on the inside of this triangle projectively are the form c1 z1 plus c2 z2 plus c3 z3 Okay, so it's c1 c2 c3 now I Can it's this up to any rescaling of c1 c2 and c3 so I'll write that a c1 c2 c3 mod gl1 Okay, but I have to have that all these C's are positive And I'll just say it once and for all here when I say all positive from now on I could mean either all positive or all negative I mean they're all the same sign, okay? That's only the ratios that are gl1 invariant, but just so I don't Say it annoyingly this all the time. I'm just going to from now on say they're all positive Okay, so that's the inside of a triangle c1 c2 c3 That triplet of numbers mod gl1 where they're all positive. What's the inside of a simplex a simplex Is just c1 c2 up the cn mod gl1 with all the cA is positive Okay, tetrahedron and higher simplices but now we want to find the generalization of this notion and You see here We're thinking about them this this as individual numbers mod gl1 But we can equivalently think of it this as a ray, right? It's a ray that passes through an origin a line that passes through an origin So these C's live in pn-1 right in this picture the C's live in pn-1 and they are so there's There are one planes in n dimensions Okay, so what we'd like to do is find some generalization of the notion of the simplex for k planes in n dimensions Okay, so how do I talk about k planes in n dimensions? So the space of k planes in n dimensions is called the Grasmani and GKN and Well, how do I talk about it? I just have to give you k n-dimensional vectors Here's a vector v1 up to a vector vk So these are n-dimensional vectors and they have to give you k of them If I give you k n-dimensional vectors then the span of those vectors is going to give me the k-point Right, so the data is given by giving you a k by n matrix But not quite because if I only care about the plane I should be able to do so this matrix It will be written as c alpha a okay a runs from 1 to n and alpha runs from 1 to k Okay, but c alpha a should be identified with any k by k linear transformation Okay on the rows because all I care about is the plane that's made out of them, right? So it's the space of so so this is the space of k by n matrices the alpha a k by n matrices Mod glk So we see the dimensionality of the space For example dimensionality of g kn is k times n minus the k squared from glk So that's k times n minus k Okay, so that's that's the that's just the the gross money and that's the space of k planes in n dimensions But what we want to do is find the analog of the inside of a simplex in the gross money in so the the analog of simplex is I want to take this k by n matrix and now let me imagine writing this matrix as a bunch of column vectors c1 through cn K by n and I want to somehow say that these are positive There's something positive about them right the positivity is needed to talk about a notion of inside right at the distinguished inside from from outside So what should be positive? Well, I can't say that the entries of this matrix are positive because that's not in very tend to k by k linear transformations Okay, what is invariant under k by k linear transformations are the determinants of this matrix So all I can say is that the determinants the k by k Subdeterminants are the miners. So if I take any k columns ca1 through ca k and I Contract them with an epsilon symbol. This is called a minor. This is the minor a1 a k is a minor of the matrix Then I can say well all the minors are positive So why don't I said all these minors are positive? Well, we just have to be slightly careful because the minor is anti-symmetric, right? The determinants are anti-symmetric So I can't literally say that all of them are Positive because one two and two one can't be both positive So what I have to do to make sense of this is introduce an ordering Okay, I have to order the columns one through n Okay, I order them one through n. That's not something I have to do for the simplex I didn't have to order them, but here I have to order them one through n and say that they're positive so long as a1 They're ordered Okay, so 1 3 5 is positive for example All the ordered ones are positive Okay, no, that's of course if we think back to amplitudes It's already interesting because we have the ordering and the scattering amplitudes and here We see that we need an ordering to make sense of the notion of inside in a simple way in the grass money All right, let's let's let's keep going Now when you have a simplex, it's of course extremely simple To imagine what the boundaries of the simplex look like you say that all these C's are positive And so you just put one of them to zero at a time and you go to the boundaries of the simplex And the boundaries are other simple C's the boundaries of a tetrahedron are triangles the boundary of a triangle is an interval Right and to go to the boundaries these variables that used to be positive you allow some of them to be zero Okay, so they're non negative So we're doing the same thing here. We want the non negative grass money in which I'll call the positive grass money in but But well, you might think well we do the same thing We can send any old minor that we want to zero and so the boundary structure is kind of trivial in the case of the simplex It's trivial. There are other simplices But it turns out for the grass money in the boundary structure is vastly more rich and structure and There's a whole that this is what that sort of ten years of mathematical literature Was about figuring out what the boundaries of the positive grass money and really look like and The reason is that these minors are now non-linear and they satisfy Quadratic relations between each other Okay, so so you can't just set them to zero willy-nilly one at a time instead of saying it abstractly Let me show you what the positive grass money and looks like In a particular case, which will be useful in us for a moment In principle, I should the first simple example is for k equals two But I'm actually going to jump to k equals three because it's even slightly easier to a visualize what's going on So what does I want to show you what does the positive? So there's a positive grass money in we can call g plus kn and I wanted to say what it looks like for k equals three So just for fun imagine that we have this three by n matrix But imagine that the top row of the matrix is positive all the numbers are positive There's some sub region which looks like that and let me rescale all of them to one Okay, if I rescale them to one, I don't change the fact that the minors are are positive, right? So so just just so we can visualize it I'm going to imagine a three by n matrix that looks like one in the top row and then down here There's some little two-dimensional vector x one up to x n. So this is a three by n Matrix and being positive means that one x a one x b One x c that this determinant is positive for a less than b less than c So let's get a feeling for what this means So Imagine the x's are just randomly sprinkled on the plane. These are now just literally two-dimensional vectors Okay, well obviously If I sprinkle them randomly, it's not going to be true that all these determinants will be positive What does it take for all the determinants to be positive? What do you think it takes? What what do you think these points have to look like for the determinants to be positive? They have to be organized into the vertices of a convex polygon x one x two x three and so on up to x n and Why is that because what is this determinant? This determinant is just the signed area of the triangle With vertices a b and c. So this is just telling you that every triangle a b c has got to be oriented the same way and For that to be true. Well, that's it's obviously true that they have to be inside the convex polygon So you see one two three one three four one three five everything is oriented in the in the correct way Okay, but now from this picture we can see why the boundary structure is so much richer Let's say I take so this is a generic point in positive g3 n, but let's say I want to take one of these minors I want to take the minor. Let's say 135 I want to take the minor 135 and put it to zero Okay, now what does that mean for the minor 135 to be zero means that this triangle's area is collapsing to zero Which means just that the points a b and c lie in a straight line Right, so let's say I do it. I take this point x3 and I want to drag it to to live on the line 1 5 Okay, so here I go. I'm going to take x3 and moving and moving oops before I get to hit the line 1 5 It will actually lie on the line 2 4 and Now if I keep going it won't be a convex polygon anymore Okay, so the modular space of points The modular space of convex polygons Is the positive g3 n in this case and It's very rich for this reason because as we start collapsing the sides first you can make consecutive points Lie on straight lines then you can do more interesting things like once Two three and four on a line. I can put four five and six on a line That's one thing I can do or I can take two three and four and I can take three and move it to the point four or move it to the point one and so on okay, so so there's a large set of Moves that I'm allowed to do combinatorially and you have to keep track of The thing that you're allowed to do next depends on where you came from and anyway There's a very very rich structure for what this very simple object of the modular space of convex polygons actually looks like So I'm not going to describe anything about it. It turns out to be All the facets of the positive rest money and are in one-to-one correspondence with certain kinds of permutations So there's ultimately there's a combinatorial object that sits underneath all of this But all I wanted you to see from here. Just is that is where the sort of richness and Complexity in the boundary structure comes from alright So that's the positive rest money a very simple object to define just this k by n matrices with all miners positive And that's the analog of the inside of a triangle in the rest money Okay, but now let's keep going to the second generalization of A triangle to a polygon or the inside of a convex polygon Okay, so So let's say I want to talk about the points on the inside of this polygon Well, all the points on the inside have exactly the same feature. They're all just a positive combination Imagine that there are masses on the outside is the same weighted average center of mass idea All the points are of this form. Okay, the yi are z1 up to zn again This is all projective. So I is running from 1 to 3 here Okay, it's a two-dimensional picture, but it's the vectors are three-dimensional. It's a projective picture and all the C's are all the C's are Are defined up to gl1 and the C's are all positive Okay, but I need to make sure that the Z's are Not random if the Z's are totally random, then I'm not going to get this picture I want to convex polygon not a not a random I want to convex polygon. So that means that I also have to demand we just learned what convexity means I have to demand that the miners made out of the Z's Z a1, Z a2, Z a3 are all positive if they're ordered Already you see that the notion of the inside of a polygon is very slightly Sophisticated right the inside of a polygon is jamming together two different positive structures on the one hand, we say that the external data is positive Okay, and then we take a positive combination of the positive external data Okay So the data is fixed and I scan over all the weight C's Which are positive and that? Carves out a region in the projective space P2 Which is the inside of the a polygon now? I can have a notion of I can triangulate this polygon That's a if we want to sort of cover it. I can triangulate the polygon Now what does a triangulation mean you see? The space of C's is n dimensional or n minus one dimensional when we mod up by the GL one It's much larger than the two-dimensional space of the polygon, right? So it's that's a highly redundant map. It's a highly highly redundant map from the space of C's into the Y space so a triangulation corresponds to finding a two-dimensional cells two-dimensional subregions in positive G1n So you see these guys live in positive G1n The data is in positive G3n, but what I want a triangulation corresponds to For example the inside of this triangle with with corners are 1 3 and 4 corresponds to a C Which is which is some C1 and C2 is 0 and C3 is C3 and C4 is C4 and the rest of them are 0 Okay So if I set everything to 0 but 3 of the C's and they're positive then I cover the inside of this little triangle So a triangulation corresponds to picking certain two-dimensional cells of positive G1n Right any two-dimensional cell positive G1n has an image with a triangle and a triangulation is choosing a particularly nice combination of these cells that covers the space so a triangulation for example Is the sum of 1i i plus 1 these triangles like 1 2 3 1 3 4 1 4 5 And so these are a bunch of triangles that cover the inside of the space Okay Now in this case this is all It's all very visual and we can see it we can see it all in action just by drawing pictures In a moment we're going to have to learn how to do this without being able to draw pictures because we're going to go outside Things that we can visualize in two or three dimensions So let's just talk about how we could discover some of these things sort of purely algebraically Okay, how could we discover purely algebraically? For example an important feature of the polygon is that the boundaries of the polygon The boundaries of the polygon Correspond so why is some point inside the polygon the boundary corresponds to when y lies on one of these lines Right, so I want to know what corresponds to the boundaries Well, it's when y lies on a line like one two or two three in general It's when y z i z i plus one goes to zero these are these are the boundaries Okay, now again, it's obvious from the picture of the polygon that it's one two two three three four and so on But how could we discover that if we were blind? We couldn't draw a picture and see it Well, let's say we want to investigate whether a given z a z b is z a z b is Z a z b a boundary of the space The a z b being a boundary means that y z a z b Would have to be always have one sign Let's say always be positive or greater than or equal to a zero somewhere, okay? And so the boundary would correspond to it hitting zero So just by contrast if I look at this example of a square one two three four You see that why one two is always positive, but why one three? See why is in here that why one three is positive? But if y is here when y one three is negative and that's because as I move in the interior of the space in the middle somewhere Why one three goes to zero, but it's in the middle of the space. It's not on the boundary. So it can have either sign okay, so What it means for it to be a boundary is that you find some some line segment where y z a z b is always positive Again here, it's all trivial because we see the answer, but let's see how we discover algebraically that this is where the boundary is Um Well first why is there any hope that there are boundaries at all? Well, let's let's let's expand it So we said that y looks like c 1 z 1 plus c n z n So that y z a z b would look like c 1 1 a b plus c 2 2 a b plus dot dot plus c n and a b and Now you see there is a hope that this might be positive because the C's are all positive and Also, these brackets are all positive if the Zeds are ordered if the Z's are ordered right if they're correctly ordered So there is a hope that sometimes they might be positive But now let's see when they are positive and let me draw a picture this circle is now just a circle of indices one two Three four somewhere here. There's a somewhere here. There's b. Okay, and then we're back to n Okay, so the C's are all positive. That's good. What about these? Brackets well, so one a b is positive. Okay, that's fine if a is less than b. That's fine Well, maybe two a b is positive But you see eventually someone in here is going to be negative There are some indices that might be there between a and b If there's any space between a and b there's always some bracket in there That's going to be negative because it'll be out of order right if there's some k here There'll be something that looks like k a b which is negative a k b and so this is negative and If it's true that there's any one guy in there that's negative We're screwed because there's always the C that just makes that one more important than the rest of them And we can make y z a z b arbitrarily negative So we can see that the only way we can have a situation where this always has the same sign is if there's no space at all between a and b Okay, and that's how we discover that the boundaries are Of the form I plus one right which of course again we see from the picture So we discover that why z i z i plus one is positive and now now let's say we want to triangulate the space Well, what does the triangulation mean? It means we want to cover it with a bunch of triangles such that the boundaries of the triangle is just the union of all these i i plus ones Well, so then what are you going to do? You know that you have to have these i i plus ones So the simplest thing to do is to have some triangle that has i i plus one in it and Just has any other point just put any other point that you want in there and Algebraically, how do we see this is the boundary? Well, I just think the boundary of this region and This is just in the usual sense of boundary. This is the sum over i star i minus star i plus one plus i i plus one That's the usual anti-symmetry of the boundary operator and this sum cancels telescopically over i Okay So that's how I see that the boundary is indeed the sum of i i plus one So that's why that's such a simple canonical triangulation and the example where I put it at one Is just you know one where where the triangulation point I chose to be One of the points of the polygon to be as efficient as possible. Okay All right now, let's generalize Away from a polygon the first thing I'm going to do is I'm going to stay in projective space even I'm not going to go to a Higher space and I'll stay well actually let's let's jump first straight to the full definition Okay, so now I want to generalize this notion of the inside of a polygon to the grass manion So what I'm going to do is the following so instead of having So what we had here polygon Was something that lived? it was a one plane in One plus two dimensions This is a space of y Okay, why I which is just this is This is p2 which is the same as g1 3 and And and and we defined a region inside g1 3 which was the inside of the polygon and How do we define this region? It was that why I was ca Z a I summed over a right so I runs from 1 to 3 a runs from 1 to n so a Little bit more loosely. We are just saying that y is equal to c dot z Where c is positive and z is positive. Okay, so that's a little mnemonic for what what what we did External data was positive and the C's are positive So now we're going to generalize this we're going to have y's that live as k planes and K plus m dimensions m is just going to be any integer 1 2 higher Here was the case m equals 2 and k equals 1 And so we're going to have external data The external data is going to be just n points d a I in K plus m dimensions Okay, so I is going to run from 1 to k plus m And now I have to tell you how to carve out a region in the space of k planes in k plus m dimensions So so the y's there's going to be k K plus m dimensional vectors so alpha is going to run from 1 to k. I want this to be a linear combination of the positive external data once again that runs from 1 to k plus m So this is fixed this data is fixed But it has to be positive positive in the sense that z a 1 these minors z a k plus m are positive This is fixed data External data and the C's have got to be positive in the sense of being in the positive grass manian gk I so that So that all these ordered minors of the C's c a 1 through c a k are Are positive? Okay, so that's it this this this defines the analog of the inside of the poly of a polygon in the grass manian and That's the amputated. That's the tree amputated. So this is the tree amputated for Tree amputated and that's labeled by n K and this an m and it lives in the space of y with the z's as external data so that's what the object is just a Geometrically now for physics it turns out that actually all the m's are important But but but most obviously to begin with for physics. We have m equals 4 so I'm telling you that the external data is K plus 4 dimensional K plus 4 dimensional vectors now the 4 you saw a second ago that that was these bosonic Momentum twisters we're talking about are just that four dimensional Four dimensional vectors for each external particle There are these extra k directions here though, right? What are these extra k directions? We didn't have that before that's the part that I don't have time to explain But in this story, there's no supersymmetry obviously. There's no adas. There's no super variables Everything is bosonic the geometry is completely a bosonic All of the super stuff is encoded in those extra components of the Z's okay bosonically encoded in the extra component of the Z's and actually geometrically what's going on is you have this k plane So if you imagine you have this this is a schematic picture. You have a k plus four dimensional space Here's why there's a there's why is a k plane in this k plus four dimensional space and now the Z's are points, but Some pieces of these vectors are in the direction of the y and some are not and the ones that are not in the direction of the Y these are the momentum twisters That we talked about before and the parts that are in the direction of the Y's is where all the super stuff goes here But is encoded in a completely bosonic way So in the end it's really true that if you hand to me the kinematical external data I associate it with it canonically This configuration of points in the in k plus four dimensions Okay Now something else that I haven't told you so this is just a space We're how am I getting functions out of this scattering up to their functions. They're not the they're not just just spaces but Once you have this space if I go back to the polygon If I go back to the polygon What I have is what I have is here's why and here's this region. There's this natural region associated with it Now it turns out this is a non-trivial fact, but it turns out that for these kinds of spaces Once you specify a region there is a certain So here here I have I have a polygon There's a certain differential form in in Y space There's a certain differential form on Y that depends on p and this differential form has the property with the property that it has it's fixed by the geometry by the property of having poles or More generally logarithmic singularities when you start taking higher residues other than just the code of mention one singularities logarithmic singularities on all the boundaries p of the shape What this means is that in the neighborhood of any place where the form develops singularities There is a choice of coordinates where in that neighborhood in the neighborhood of any Locusts of singularities the form looks like dx1 over x1 up to dxd over xd if it's a d-form Okay, that's what it means of having logarithmic singularities What's important about this kind of singularity is that if you take residues if you just take one complex variable dx over x means that you surround it with a circle and when you take a residue you get one And this means that this generalizes so the singularities as a complex object This thing has singularities where you take tori and surround all the possible all the possible Lower dimensional boundaries of the object that you're talking about and all the residues are one It's a simplest possible singularities you can have But that's how associated with the geometry you associate given a geometry There's a certain differential form with logarithmic singularities on the boundary of that geometry and That's how given the geometry we get a function and that function is the scattering amplitude Why do we know it's a scattering amplitude because The properties of the geometry force this form to have exactly the singularities of the scattering amplitude has to have Which involves putting particles on shell and all the rest of those things that we give all these physics Interpretation to in terms of evolution Unitary evolution in space-time is interpreted here instead as a consequence of what the of what the boundaries of this positive geometry looks like Okay, yes That that's that no no it's a well a particle going on shell in this picture corresponds to moving y to the boundaries Okay, so so so exploring the singularities of the function means you see This this I don't have time to explain in more detail But in the usual picture you would take the data and you move the data until you encounter a singularity That's dualized here with the data staying fixed and you move y around until you hit the Until you hit the singularities and so the singularities of the amplitude corresponds to just hitting boundaries of the amplitude But the fact that the fact that the amplitude factorizes when you sit on a pole is the is reflecting the Geometric fact that if you take this c dot z space and all the magic is what those determinants those determinants being positive It has the feature that when you go to a boundary you discover the the Amphitheidron breaks up and on the boundary What the geometry looks like is the two lower Amphitheidra? glued together with the addition of one additional Piece of apparent external data corresponding to the internal line, okay But that's that's a feature of the that's a feature of the a geometry. Yeah Yeah That's right that the in this language the actual residue would would correspond as usual and we have an amplitude the residue would Residue of the amplitude would correspond to the amplitude that you made it multiplied by the amplitude that it decayed on the other side Yeah, okay, so So just in the so anyway, that's so that's just the definition of the tree Amphitheidron I don't have time to define the idea of hiding particles in the loop Amphitheidron All I want to do in the in the last few minutes is go back to the simplest case of k equals one m equals four Where it's we're not even seeing the fancy gross money and structure yet here This is still just going to be living in a projective space. So it's just just a polytope But where we can Where we can see how we make contact with the physics that I told you about so So we're doing k equals one m equals four so the y's are Just the sum of or just ca zai But these z's now are five dimensional and they satisfy z a one through z a five are positive when the a's are ordered and The C's are positive That's the space that we're talking about so now this is living in G one five or p four so it's a four-dimensional shape not a two-dimensional shape Okay So it's harder to visualize But let's try to figure out Where are the boundaries for the boundaries? So let's just mimic the exercise we did for the polygon We want to ask so the boundary is now going to involve four z's rather than two z's But whereas can I have z a z b z c z d can I have why is the ab this all be positive? Is that possible? Well, it's exactly what we did before it will be c one one a b c d plus dot dot plus c n and a b c d should be positive and Let's again draw the circle of indices one two now. I have an a a b a C and a d Then down to n okay, and Just like for the case of the polygons We find that it could be has a chance to all be positive because these minors Ordered or positive and the C's are positive But if there's any space between a and b if there's any room there that'll get something negative If there's any room in any of these places that'll get something negative There is only one way that it's always positive and that's when it looks like I I plus one j j plus one Only when it's the boundaries are when y z i z i plus one z j z j plus one goes to zero And then you can say that all everything is positive because there's no room Between you only ever have to jump over an even number of guys in order to reorder it And so it's always positive So that's that's quite striking because it's remember we're asking on general grounds We have a function of a bunch of z i's why would it care about i i plus one and j j plus one? That's that's locality right that's that the poles should know about when x i minus x j squared goes to zero Right why even though it's a function of the z's does it care about the lines i i plus one j j plus one Well, that's a consequence of just the of the of the positive geometry in this case so we discover that the boundaries so we discover that that the boundaries are Just the sum over over all i j of this i i plus one j j plus ones And so how do we triangulate the space? It's exactly the same So the space is triangulated By the sum over i and j of some star i i plus one j j Plus one and by exactly the same argument as for the polygon you can discover that the that the boundaries of this Everything that doesn't involve i i plus one cancels Telescopically and you're just left with with the boundaries that that we want so so So So that's that's the geometry in this case It's triangulated by exactly the objects that we're talking about for the amplitude and the final part Is if you just work out what is the form of logarithmic singularities? Which just turns out to essentially be the area of these the volume of these? Simplices it's the area of the triangles in the previous example That that you precisely get to the to the formula for the scattering amplitude so So that just that's a little taste of what it looks like In the more general cases in the more general cases, it's much more intricate because the geometry is not linear So the boundaries are not defined by linear equations. There's a much richer There's a much richer structure involved But here at least you you see the sort of first taste of how it is that that the positivity is enforcing these physical properties in this case Most clearly locality, but when you go to the more the richer examples You see both the locality and the uniterity at tree and loop level coming out of this coming out of this picture All right, I'll leave it at that. Thanks