Combinatorics and Geometry of Fundamental Physics and Cosmology

So, uh I think we can get started. So, welcome back everyone. Um we're now going to have uh Nema's final lecture of the school. Actually, the the final lecture of the school. >> Oh, I'm very honored. >> So, uh yeah, please go ahead, Nemo. >> All right. Well, I hope you guys have uh have survived and your brain isn't totally fried. Um uh but hopefully it is in a good way. So, all right. Um so, today I want to give you uh uh uh uh some idea uh I want to do two things. Um uh ultimately I want to give you some uh uh some idea. I'm not going to be able to do it all in detail, but going to give you uh some idea of more systematically where the ideas that we're talking about the past couple of days uh come from. Oops, sorry. Just one second. There we go. Um but um uh before I do that, I want to mention uh at least some of the sort of qualitatively interesting and new facts um that this picture uh that we've been uh talking about for the past couple of lectures. Um some qualitatively new things about uh about amplitudes uh uh and the relationship the surprising relationship between uh seemingly very different theories. um uh how they've been revealed by this uh new aera picture. So I'm just going to give you an indication for what uh what some of these uh uh ideas are about. And let's begin with this side. Um I'm going to begin by talking about this phenomenon of uh hidden zeros and split factoriization. So remember uh we already emphasized the standard thing about trimplitudes is that they factoriize. Okay, they factoriize on poles. So the kind of uh normal story is um uh the uh uh the standard story is uh that the amplitudes have poles and they factorized near poles. All right. Now there is a very natural thing that you could have always wondered uh I mean since we know a lot about where the amplitudes have poles um how much do we know about uh uh about their zeros um of course uh you know if you just add up the whole amplitude put it under a big common denominator let's say at tree level you just have some giant polomial in the numerator and obviously that polomial equals zero defines some variety, you know, so there's some place where where it vanishes. Uh so of course the amplitudes do vanish somewhere. The question is whether we can predict where that somewhere is. Um or whether we can even sort of predict some uh some uh simply predict some some some parts of where what this uh uh variety of the numerator looks like where the amplitude uh vanishes. In other words, are there any predictable uh zeros of the amplitude that don't involve solving this insanely complicated uh polomial equation? And that's what we're now going to see that that that there's something completely analogous that the amplitudes have zeros. And remarkably, they also factoriize near the zeros. This is textbook. This is totally new. Okay. So, uh this we did not know about before. Now, as I also stressed this business about uh factorization is made obvious by the associ. Okay. Um but as we'll now see this the business about the zeros and the factorization is also made obvious by the socran. That's what I was uh promising is that there's this there's this new picture um uh and this new picture definitely knows all about the old things. It knows about factorization and all that stuff, but it tells us more. It tells us how all the diagrams are kind of combined into this master object. And this particular way that it is uh this particular way that it is uh uh uh combined, it tells us um new things that we didn't know before. Now let me uh illustrate what these uh zeros look like again in our in the sort of simplest example where it looks not non-trivial where the associ look like this uh uh pentagon. Okay. So recall that we get this pentagon um from this uh uh lattice wave equation picture. Okay. So we we we wrote down these formulas like x13 + x24 - x14 associated with this little mesh is equal to c13. And then also x14 + x25 - x24 is equal to c14 that's associated with this mesh. And finally this one was x24 + x35 minus x uh 25 is equal to c24 associated. Okay. with that uh pop uh uh mesh here. So this was 134 and uh 24. Okay. Now if I plot what this thing looks like in uh x13 uh x14 space, well I mean you can see from this uh uh uh uh you can see from this uh inequality that this point here is is at location C13. This point here is at location C13 plus C14. And this point up here is at location C14 plus C24. Okay. So uh so this is what and again this was the uh x14 goes to zero edge. This was the x24 goes to zero. x25 goes to zero. X35 goes to zero and x13 goes to zero for these uh uh different edges. All right. Now remember that I told you that the amplitude is going to be the canonical form of this associ. We didn't have time to talk about this in in detail but but the amplitude is that unique function uh that has uh the actually let let me uh let me talk about this uh uh for a moment. Remember the amplitude is is the unique function um or the unique form that has logarithmic singularities on and only on the boundaries of the geometry. Here we're imagining that the C's are fixed. Okay. So I so I I I I fix the C's. So it's a function just of X13 and X14. And I want to have uh I want to have a form that has logarithmic singularities on and only on the boundaries of uh of of uh this shape. And um there's many ways of building such a form. For example, I can take this shape and triangulate it. Here are some interesting way of triangulating it. I can add up the the canonical forms for these little triangles in just the same way as we discussed uh uh what forms for triangles are in general in in the uh uh introductory lecture. That'll give me a formula for the canonical form of the pentagon. Um uh but there is another formula uh which also gives you the canonical form for for the pentagon. This one uh uh uh this one needs uh uh this is something that you can actually do for uh any polygon and in fact any any simple polytope. Um so let me uh let me illustrate it maybe with this uh example. So let's say I want the canonical form for this little square. Well, what what I can do is put a line far away and uh just look at uh in uh and just look at uh the sums of the canonical forms of these pieces. This is what you could call an external triangulation. Um it's an external triangulation uh in the sense that I'm get I'm making this square uh out of this big triangle, right? And then I subtract this little this smaller triangle and I subtract this smaller triangle and I add this little triangle because I overs subtracted it. But you can see that what that looks like is just in the neighborhood if I imagine moving this line to infinity. Um it looks like in the neighborhood of every vertex I just have a form which is the product of the two uh of the two linear factors that go to zero at the vertex. Okay. Uh so so that's the that the canonical for this triangle would just be like one over L1 L2. If this L1 and L2 are the lines that pass through that uh uh vertex. Um and if this was L3 and L4, the canonical form for this triangle would be plus 1 / L2 uh L3. And the canonical form for this little triangle would be 1 over L3 L4 and so on. Okay? So plus one over L4 L1. And you can show that this always gives you the canonical form for any polytope. So long as the polytope has this properties called being simple such that at every vertex uh if you have an n-dimensional polytope at every vertex exactly n faces and no more than n faces are meeting uh at the vertex. That's a minimal number that can meet at every vertex. So every vertex locally looks like a simplex. Locally looks like there's n faces meeting. Whenever you have that uh one of many forms uh one of many formulas for the canonical form is given by taking the product of all the one over L's uh for every linear factor that that uh that meets at the vertex and summing over all of the vertices. Okay. So if we go back uh to this picture um uh we gave a few uh a few pictures. I can triangulate this pentagon. Um or I can multiply the linear factors that meet at every vertex. Okay. Um uh I can uh for this vertex I've multiplied these two linear factors and so on. Okay. But you'll notice that this last formula that looks like multiplying the linear factors that come together at every vertex has a name is known as Fineman diagrams. Okay? That's because every vertex, if you remember, is a complete triangulation of the polygon. Every vertex corresponds to a fineman diagram and uh all the linear factors that meet at the vertex are precisely the poles associated with that fineman diagram. So you can see very nicely that in this picture you the the the the associating together all of the uh all the vertices um all the finement diagrams together in a master object and even the formula that's the fineman diagram formula for summing over everything for summing over all diagrams is one of many ways of computing the canonical form of the uh of the socied but you'll see what the findment diagrams are missing uh what the findment diagrams are obscuring. You see, if you go back to this picture, every single one of these triangles uh has a has a pole in it which corresponds to this edge and this edge. Those are good poles of the amplitude, but it also has a pole at infinity corresponds to this line that we put very very far away. Okay? And that's manifested again by the fact that every single term uh every individual term again has poles at the vertex but I've just drawn it again also has poles at infinity. Those poles at infinity magically cancel out when you sum over all the uh diagrams. Well, WE DON'T EVEN TALK about those poles at infinity when we normally talk about finement diagrams because we don't have this picture of the plane on which uh we we're we're looking at this geometry. But that's what each findment that's what the findment diagrams are missing. Every term term by term has a furious pole at infinity and there's a magical cancellation that all of them uh that all of them add up to a zero. Um if we had more time I would tell you that there's a hidden symmetry associated with the uh associated with the amplitude. Um which is precisely the statement that there's no poles at infinity. uh that's just trivialized by the canonical form by the fact that the only singularities are on and only on the boundaries of the geometry. Term by term fineman diagrams destroy that symmetry and so you don't see it and only when you sum over all the diagrams do you discover that it's there. Okay. So uh so that's the that's the uh that's the picture for uh uh where finding diagrams sit in this story. Okay. Um but now let's uh now let's uh uh uh uh go back uh to our um uh what I want to tell you about uh about about hidden zeros and factorization and factoriizations. So uh so we've just seen again that the amplitude is a canonical form of the uh is this canonical form of the associ. Um but this fact makes uh something else obvious. How can we make the amplitude vanish? There's a very simple way that the amplitude could vanish. It's if the if thisahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahedan collapses to something lower dimensional. This is after all supposed to be like a two form in this example a two form that lives on this space that has logarithmic singularities on and only on the boundaries of this pentagon. But if the pentagon collapses uh to one lower dimension, this canonical form obviously has to go to zero. Okay, so that's a trivial way to see that there's going to be some uh some some uh zeros associated with the amplitude. And remember that uh that but this this this picture that that we just have just makes it obvious that there are some limits where we can make this happen. For example, uh uh associated with shutting off some of these C's. You see for instance here, let's say I send C13 to zero. If I send C13 to 0, it's like I'm dragging this point to the origin. So if I send just C13 to zero, I'll be left with a shape that looks like this. Okay. If on top of that, I now send C14 to zero. All right. If I send uh uh C14 to zero, this would be now uh uh C14 after I set C13 to zero. If I send C14 to zero, I'm just going to collapse the entire thing so that it'll just look like a little interval. Okay? So if I send C13 to 0 and C14 to 0, the full associed collapses to that shape. All right? If I send C14 to 0 and C24 to zero, then it collapses to this shape. And uh but it doesn't always collapse. If I send C14 uh to zero and um uh and C sorry, if I send C13 to 0 and C24 to zero, then it just collapses to this triangle. Okay. So, uh in fact uh this is related to a point I was uh uh uh stressing yesterday that the associ sum of simple pieces. For example, in this case it's C13 multiplied by this interval plus C14 multiplied by this interval plus uh C24 uh sorry C24 plus C14 multiplied by this triangle. So because the assoc is a sum of simple pieces many of which are lower dimensional uh then it's obvious that if you shut off enough of these C's it's going to collapse in dimension and therefore the canonical form is going to vanish. Okay. So uh so um uh so we've learned that there are some very interesting pattern of zeros that are just obviously there because uh of the uh because because the amplitude is a canonical form of the uh uh so in fact if we go back to our picture of the mesh let's see what these uh look like. Um uh we see that we could send C13 to zero for example and C14 to zero. that corresponds to setting these guys to zero. Or we could send these guys to zero. Okay? Or we could send these guys to zero. If I send these C's to zero, the amplitude vanishes. Now, it's very easy to see the what the what the general story is. Okay? So, uh if you uh I'll just tell you what the general story. So, so if you have any kind of picture at any end, so you draw this sort of uh kinematic mesh, okay? Then you pick sort of any point on the boundary here and you fire out you fire out uh the 45 uh degree lines in this way. So it defines some uh rectangle uh in the interior. If you put all the C's in the rectangle to zero, the amplitude vanishes. Okay, so that's some interesting uh now uh uh uh these C's remember are not individual X's. So there's no poles here. Okay. Um so that's why you can send them to zero without worrying that the C's are X plus X - X - X. uh the amplitude is poles only when the x's go to zero. So of course there's no poles but not only are there no poles when you send these pattern of zeros for any such rectangle you actually get uh a zero. Now, we can actually go back to this picture. Um, we can we can uh go go back to this picture and uh let's say that I let's say that I I shut off uh just to give you an example, let's say I'm in the neighborhood of one of these zeros. If I'm in the neighborhood of one of these zeros, um uh let's say I have that the zero where I send C14 to zero and C24 to zero. Um then what things? Oh, sorry. Uh so so that that that gives me that gives me uh uh that gives me uh a zero. But uh let's say that I just u uh uh let's say that I just turn off um uh C14. Okay. Uh so let's say I just send uh C14 to zero. If I send C14 to zero, then you'll notice that that the is associed is still topdimensional, but it looks like this. It has uh C13 in this direction and C24 in this direction. Okay, so the pentagon collapses with this little rectangle. But that little rectangle is the direct product of two intervals. So what we've learned is that close to this zero uh the amplitude actually factorizes into the product of two fourpoint amplitudes. Okay. Um so that's also uh surprising. In other words, in the penultimate step right before it it collapses entirely. If I turn on one more C uh it actually factorizes into the product of uh into the product of two lower point uh amplitudes. So not only do we have new zeros but we have a new pattern of factorization of the amplitude near those zeros. That's uh that that that we call splits. So if we go back to this picture the claim is that if I if I take the amplitude and now I just turn on any one of these C's. I turn on I turn back on any one of these C's like this one. Okay. that now the amplitude factorizes into a left amplitude which is up here, a right amplitude which is down there. Okay, so uh so so the amplitude doesn't go to zero anymore. It goes to uh a left amplitude times a right amplitude times a little four-point factor that's associated with these two ends. like a little four-point amplitude uh uh if this is x bottom and x top times a little fourpoint amplitude which is uh uh which is basically 1 /x bottom plus 1 /x top and uh now uh each one of these amplitudes the kinematics for these amplitudes is slightly shifted I don't have time to explain how it's slightly shifted depending on precisely where you turn this guy back on, but it's shifted in a very simple and uh and canonical way. But the important point is that in the neighborhood of the zero, we discover that the amplitudes factoriize again. Right? So this is a completely sort of interesting qualitative behavior. It's very analogous to what happens near poles. But again, near poles, we've known it for 80 years. Uh this is the the fact that the amplitude has zeros and that it factorizes near the zeros is something new. um and is something which is made obvious by this connection between trace cube theory and the uh associ okay and um now this is some location in kinematic space you turn on uh you you take a bunch of x's c's or just some linear combination of x's so you just take these c's and you set them to zero and you observe that the amplitude vanishes okay and uh once you see this you can wonder how general this is Does this happen in other theories? And well, there's two close cousin theories of trace 5 cube. Trace 5 cubed uh is a is the simplest uh uh theory uh for n byn matrices. Okay. Uh uh simplest theory where the where the degrees of freedom are have color. They're uh n byn matrices. There are two other obvious theories that uh have the similar feature. One of them is the nonlinear sigma model. Uh just the where here the uh the uh the grantian is like du dagger du where u is e to the i pi but pi is also n byn matrices. Um and the other one is of course yang mills. Okay. So uh uh of course here naively things are are are different because we have polarization vectors and so on. But I explained at least briefly yesterday how we can uh represent the degrees of freedom of polarization vectors in purely scalar terms. Okay. Um once you do all of that, you just experimentally discover that these things have exactly the same zeros and splits just observationally. Okay. you experimentally take the amplitudes and you you observe that they have zeros in the same spot and they split in exactly the same way as this uh trace 5 cube theory does. Now in the particular case of the uh NLSM, in the particular case of the NLSM, of course you know that pion amplitudes have a famous zero have the famous Adler zero. When uh when uh a pion uh some particular momentum uh goes soft, the amplitude goes to zero and that reflects the uh shift symmetry that reflects spontaneous symmetry breaking, the gold stone phenomenon and the shift symmetry for goldstones. And uh so you can wonder what the relationship is between uh this and the new uh and the new zeros that we're talking about. And the point is that these new zeros are more fundamental. The new zeros are more fundamental. There's many more of them. Uh and you can actually show that they imply the Adler zero but not the other way around. Okay? So the adl zero does not imply the new zeros but these new zeros uh uh imply the u adl zero. Now all of this uh uh all of this turns out to be explained by the following fact that I don't have time to uh uh uh derive for you or even uh or even really motivate. But um uh so let me first uh uh uh tell it to you just purely as a field theory statement uh for particle amplitudes to begin with. Um, say you have the amplitudes for the trace 5 cub theory which depends on these xigs. And what I'm going to do is now do the following peculiar thing. I'm going to shift xig goes to xig plus delta j where delta j is equal to delta if i and j negative delta if i and j are both odd positive delta if i and j are both even and zero otherwise. Okay. So now I have the the this amplitude for the trace 5 cub theory with the xig plus delta j and now I'm going to expand for when the x's are much smaller than the delta. So I've introduced a new scale delta. I'm going to expand when the x's are much smaller than the delta. Then I then I'm going to get first of all uh then uh so so this this shift only makes sense uh both odd and both even is something that only makes sense when n is even. Okay. So I'm just going to look at the trace 5 cubic amplitudes when n is even. I'm going to look at it when x is much much smaller than delta and I'm going to get one over delta to I'll see lots of uh uh cancellations. Then I'm going to get one over delta to a leading power. I forget what what the power is uh uh off by heart. But what multiplies it is the amplitude for the nonlinear sigma model thought of as a function of the x's. Okay, so this trivial shift in kinematics uh takes you from the trace cube theory to the nonlinear sigma model. Now these theories again seem to be utterly unrelated to each other. Okay, so one of them has cubic interactions and no gold zone phenomenon. The other one is only even point amplitudes and uh and uh u has you know an infinite uh lrangeian. It has u shift symmetries atler zeros and so on and so forth. But in fact they're secretly identical. Okay, the amplitudes are secretly identical. There's just one amplitude, the trace 5 cubed amplitude and I just expanded around different points in kinematic space and that moves me around between the uh uh the trace cube theory and the uh NLSM. Let me just illustrate this with a simple example uh just at four points. So at four points your amplitude was 1 /x13 + 1 / x24. Okay. So this is uh for n= 4. This was the trace 5 cubed amplitude. Now I'm going to shift this is 1x13 minus delta because 1 and 3 are both odd plus 1 / x24 plus delta. And now I'm going to look at this when the x is much smaller than delta. Very nicely. You see that the leading term cancels and the subleading term gives me 1 over delta^ 2 minus x13 - x24. Um if you remember your uh pion amplitudes this is minus s minus t which is in fact exactly the amplitude for the four point non nonlinear sigma model. Okay. Um, in fact, this connection is really by far the fastest way to calculate the amplitudes for the nonlinear sigma model since it's so fast to calculate them for the trace 5 cube theory. Just this little shift gives you uh the amplitude for the nonlinear sigma model. And in fact, this is a special case of a more general phenomenon. Um, uh, uh, I mentioned uh, at the end of last time that there's this stringy amplitude which now depends also on on an alpha prime for the trace cube theory. Um this I wrote as this integral uh associated with these uh with these uh u variables d y13 over y13 over d y1 n -1 over y1 nus1 let's say uh the product of these uj to the alpha prime xig I didn't really explain too much about where this uh uh formula came from but um uh uh and just by convention there's a power of alpha prime in front here um so that the alpha prime goes to zero limit uh the alpha primes uh uh uh disappear. Alpha prime goes to zero is the field theory limit. Okay, that's obviously the limit where the string length gets arbitrarily tiny. But now we're going to do exactly the same thing. We're going to shift a of x goes to xig goes to xig plus delta j exactly the same way. Maybe I should put n uh well just plus uh uh plus uh uh uh delta j. Um and the claim is in general that now there's a new scale uh alpha prime. So if I look at what this delta is um uh here's zero. Of course at at zero I I go back to the trace cube theory. We just saw that when delta is small compared to alpha prime. So if I ignore the stringiness like out here I get the NLSM in here the moment I turned it on I get the NLSM at low energies uh but that actually turns out to be true everywhere here I get the NLSM at low energies everywhere until I hit delta alpha prime delta equals 1. So when delta equals 1 over alpha prime at this point I actually get Yangmill's theory. Okay. So as I move uh so all these amplitudes are in fact identical. Okay. So there's one function which is this stringy trace 5 cubed amplitude. And as I shift the kinematics by this way by this funny peculiar way um uh I see that um uh that it uh uh it interpolates between the trace cube theory the nonlinear sigma model and the theory of uh uh of of the glor. Now what is this funny peculiar shift? Well, this is something that you could actually go away and really work out and discover yourself. Remember that our zeros associate are involve putting a bunch of C's to zero. So now you can ask the the question is there a shift of kinematics that preserves all of these C's? Is there some way of shifting kinematics such that all possible C's are unaffected? If such a thing existed, then it would leave unaffected all of the zeros that we're we're talking about. And it turns out that this funny shift by shifting delta with even even and odd odd is in fact the unique shift of the kinematic variables. Just a linear shift to kinematic variables. It's the unique linear shift of kinematic variables that actually preserves all of the C's. Okay, so that's the sort of motivation for this strange shift. Um and uh uh and uh and this final comment is also now what lets you understand why the zeros and splits are inherited by all of these theories. These theories are all are all related by the simple shift of kinematic variables. Uh and since that shift is precisely the one that preserves all of the C's, uh it preserves the entire pattern of zeros and splits uh that we saw before as well. All right. So that's just kind of uh that's a little a little tour of some of the things that we've learned about uh now real world amplitudes. There's no super symmetry here. These are non-superymmetric gluons. Uh non-supmmetric pions obviously. Um uh some of the things that we've learned that have been made obvious by this uh by this uh uh uh geometric picture. uh and so maybe I'll pause now to see if there's any questions and then if not I'll spend the last part of the lecture at least giving you some indication of the systematic place that these ideas come from. So maybe we can stop to see if there are some questions now maybe I have a question. So um so as you said the other zeros are related to a to a symmetry to spontaneous breaking. I didn't understand. So these more general zeros are also related to some other symmetry. >> No the the these these that's that's the point I was trying to make. Um these are uh these are uh these are zeros that have no conventional symmetry explanation. They have no conventional explanation at all. Uh I mean that's why they they they they could have been noticed at any point in the last 60 years. Uh they were not noticed till a couple of years ago. There there's no absolutely no reason they they they couldn't have been dis uh discovered in in the 60s when people first started computing these uh uh even pine amplitudes. Um the reason is exactly they're just not obvious in the lrangeian. Of course, by now there's lots of papers where once you know it's true, you can go and say, "Oh, this collection of diagrams adds up and cancels that collection of diagrams." And you can do a lot of work to see it like that. Of course, then it's no idea why it's true in the stringy case. It's very difficult to see why it's true in the uh you can do that for the trace 5 cube, not for any of the other examples. Um uh but the point is that there's a sort of picture of what the amplitudes are as associated with volumes or canonical forms of these shapes. Uh just makes it obvious. You you don't have to think about it. It just sort of comes out out of the box that uh that they have to have these properties. So if you're a particle physicist, you know, as uh uh or model builder as as I spent a lot of my life being, um your antenna should really pick up on this uh on this point because they're exactly a phenomenon of some zero uh with no conventional Wilsonian naturalness symmetry explanation. Okay. and uh sorry Wilsonian symmetry explanation and um and you know our conventional picture for symmetry is this very diagrammatic one diagram at a time or small collections of diagrams cancelling together. This is something else. This is something that you cannot see in diagrams. You cannot see from the lrangeian um but is made obvious from this other other picture. Um uh uh you know uh it's definitely inspirational to the idea to the thought that maybe we mechanisms might like this might be relevant for uh finding new ways to attack the hierarchy problem or the cosmological constant problem that are precisely examples where some quantity is tiny or zero uh if not zero tiny um with no conventional symmetry explanation. Well I had never seen examples of this phenomenon before. Turns out they're under our noses, you know, everywhere in uh in uh in amplitudes for realistic theories. Um uh but indeed it's good to highlight that that point. They do not have a conventional symmetry explanation. What makes them obvious is this other picture of the world which is not about lrangeians and path integrals and diagrams but is about these uh shapes in in kinematic space. >> Thanks. >> Any other questions? I don't see other questions. >> Okay, great. All right. So, on this final part of the talk, I want to give you an idea of uh a systematic picture of where these things come from. And again, I have like 45 minutes, so I won't be able to like uh I won't be able to systematically tell you where the uh where the uh where the uh systematic origins are, but I at least want to give you one aspect of it that you can go away and uh and uh and play with. Okay, let's come back to this picture of the pentagon. Okay. And already here what I want to do is is focus on something that when you study polygons and polyopes and so on, it's a very natural thing to do. Let's look at these normal vectors to every let's look at the normal vectors inward pointing normal vectors to every uh sort of facet. Okay, so that's what these inward uh uh pointing normal vectors look like. And if I just plot them, I get a picture like uh uh if if I plot them all sort of uh uh starting at the origin, I get a picture that's something like this. Okay. Um this picture is sometimes called the normal fan. The normal fan associated with the uh uh uh geometry um uh with with with a polytope in general. If I label again the things that vanished here x14 vanished here x24 vanished x25 x35 and x13. So if I label these things here this is this is the vector x13 this is x14 x24 x25 and x35. Okay. So let me uh let me plot this again. Let me just draw it a little bigger. Okay. So, we got this whole picture. We got this picture remember from the wave equation and and positivity all that stuff, right? All that stuff gave rise to this uh picture. And now I'm capturing one important feature of this picture just by looking at the uh at these uh uh uh this this normal fan. Now uh look at look at what the so what what is this what is this fan doing? You see that just plunking down these vectors has divided the space up. This whole this sort of two-dimensional space for example labeled by these variables x13 and x14. Just blocking down these vectors has divided the space up into a bunch of cones. Right? There's this cone here. There's this cone here. There's this cone here. This cone here and this cone cone here. And if you look carefully at the cones, it's cool that uh you know what are the rays that bound the cone. So this one is x13 and x14 and x13 and x14. If I just draw them on my momentum polygon, well, this is one of the triangulations of the polygon. That's one of the fineman diagrams. Similarly, this cone here, it's one over x2 x14 x24. This is another one of the fineman diagrams and so on. Okay. So this is the um 1424. This is another one of the uh uh uh diagrams. Now this is of course obvious by the fact that it came from the associ. Uh that's one of the things that the associ does. If I go go go back uh to this picture for example in this exactly this neighborhood of uh x14 and uh and uh xx24 these are exactly so maybe I can draw it like maybe I can draw it like like this here um uh this this this one way of drawing this uh normal fan is that uh everywhere here like in this region uh I'm just going to put uh a vertex uh a vertex here uh and I'm just going to connect things so that it looks so that they look normal so that the uh so that the rays coming out look normal. Okay, you see this is exactly the same picture as this pentagon. It's just flipped uh it's just flipped because I drew here I've drawn outward uh pointing normals instead of inward pointing uh normals but it's exactly the same pentagon. And so it makes sense. It makes sense why this region uh this region is associated with this vertex. Okay. Um and that vertex all the vertices of the pentagon all the vertices of the associated correspond to individual finement diagrams. So this is a general fact. If you're given any polytope and you draw the normals um uh and you get this normal fan, every cone in this normal fan corresponds to one of the vertices of the uh polytope. So it's kind of a duel of the polytope. So every cone in the fan uh corresponds to one of the vertices of the polytope. And so in our present context uh what we see is that this this normal fan is an interesting way of of dividing this space up into regions into cones such that every single cone by itself um uh has exactly the uh that the rays associated with that cone or exactly uh uh are exactly the the x's that correspond to individual fineman diagrams. Okay. So this is again the sort of magic that it's possible to combine all the fineman diagrams together in a single object. This is like a dual manifestation of that magic. The associed combines them into diagrams as vertices and the normal fan combines all the diagrams together each one as comp. Okay. Now so far this is all coming from the ultimately from this lattice wave equation positivity picture that I pulled out of nowhere uh a couple of lectures ago. Um, and now I'd like to explain to you at least where this fan comes from. Okay? Uh, where this uh where this fan comes from in an utterly elementary way. Okay? All right. So, let's uh so uh so this is called the uh the fineman fan. What we call the fineman fan. So I want to now tell you about this uh uh finan and to do this we're going to go right back to the to the beginning. Okay. So uh so uh we know that we know that at uh we know that uh we've just said that we can think about all the tree amplitudes. Um we draw this sort of momentum polygon associated with them. Sometimes I draw it like this uh like this uh for example at five points I draw it like uh uh a surface in this way and well that's that's the statement that all of the diagrams at tree level are planer. Okay so and we can think of them all as triangulations of this uh of this of this surface of this uh of this surface which is a disc with uh with five mark points on the outside. Well, this is the general fact that at any order in the topological expansion as any order in uh in the loop expansion the genus expansion at large n the diagrams are associated with some topology the diagrams are associated with some surface um but now we're going to take a slightly different point of view uh so I want to characterize what the what the surface is okay well um in order to give you a surface uh it suffices to give one triangulation of the surface. Okay, so that's what what we're going to do. I'm going to think of this not as a fineman diagram, but I'm going to think of it just as a single triangulation uh to specify what I mean by the surface at tree level is sort of easy. But if I have some ridicul if a very complicated surface and any loops just even tell you what the surface is, I'm going to have to sort of define some triangulation to tell you what uh uh what uh uh what the surface is. and the the then the triangulation is literally the same as giving you uh a particular diagram. This is not a fineman diagram if you like. It's like just like a color diagram. Okay, it is a diagram uh which is telling you how the color indices are flowing. And you'll notice that I'm putting the 1 2 3 4 5 in the in between regions here. So I'm putting it like like like like this. This is this is this is the color one which is flowing here. this is the color too which is flowing along this line and so on. Okay. So one way of reading this picture of specifying the the the surface is that every sort of vertex every uh uh every triangle in this picture every vertex in this picture corresponds to a triangle and every propagator in this picture every propagator in this picture corresponds to some way of uh uh of two triangles that meet on a common uh edge uh in in the triangulation. Okay. So this is a very standard thing. A given uh uh what we what you can call fat graph or double line notation graph or mathematicians sometimes calls ribbon graph. A given fat graph I'll call it a fat graph. A given fat graph just specifies a particular surface. Okay. Once you give the fat graph you've specified the surface that we're we're talking about. So if we're going to talk about the kinematics space before even getting started we have to at least give one fact graph to specify uh what the surface is that we're uh talking about. Okay. So, um, by the way, there's another way of thinking about this uh picture which is useful, which is that, you know, when I draw this picture here, I should not think about the points 1 2 3 4 5 as sort of belonging to the space. They're kind of like uh marked points. Um, so something that uh so I can if I take this picture here on the right and I just shrink the uh the regions 1 2 3 4. or if I just sort of shrink them to zero, then if I shrink them to zero size, I go from this picture here to this picture here. So that's another way of thinking about um uh how the triangulation is uh how the triangulation turns into uh the uh sorry how the fat graph turns into the surface. If I take these sort of red regions and I shrink them to zero size, then I'll go from the picture on the right to this picture on the left and maybe I should color them in red. Okay. Okay. So good. So now we've specified our surface. And now so now so this surface is going to be our world for the less for the rest of this lecture. I can have this surface uh for this fivepoint example that we're talking about. If I wanted to do something at one loop, let's say at four points, I could draw a surface like this. This is a surface. This specifies a surface and so on. Okay? So you draw anything you like. uh uh um uh it doesn't have to be planer uh you know uh if I'm doing uh uh uh two loops and the leading one over end corrections I would draw uh I would draw something with the topology of a Taurus um so anything goes okay so any fact graph of any sort uh goes but um but we're now uh we've been talking about tree level these this this whole lecture so let me deal deal with this picture all right so this is my uh this is my uh picture Um and this is my world now. Okay. So, uh this is specified my my surface. Okay. Now, um we we keep saying that kinematic space is very important. So, we need to label our kinematic variables. Remember all of the magic uh before started by drawing the x's on this grid and doing all that stuff. But again, we made some choices. We drew this grid. We anyway, we made various uh little choices in that in that picture. Now, we're going to make no choices at all. Okay? So, this is my world. Now, I have to somehow talk. Let's just go back to the picture uh for for for a second. Um uh you know, I want to talk what does it mean to talk about X13 or X24 in this uh in this picture? How would I represent X13 and X24 uh in in in in this picture? Well um uh one one little tiny point is that uh exactly for the reason I said for example if I wanted to represent x13 you know one here as we said is now this region here three is this region here. So the most naive way of representing x13 would be to draw some kind of uh chord like this right that that that would be what what what would correspond to x x13 but that's a little arbitrary. How do I decide where to draw it and so on? I mean um there's something a little bit more more natural once you draw this uh once you draw this uh uh picture um which uh in in the literature on surfaces is called instead of working with the chords working with what they call uh laminations I'm going to just call them curves instead of the fancy word uh lamination. And the idea is is very simple that uh that you see here 1 and three don't really belong to the surface. So if I want to talk about the curve x13, I'm going to label it instead by moving a little bit to the right of one and a little bit to the right of three and draw this curve instead. Okay. Now, this is completely canonical because moving to the little bit of a right is something that only requires an orientation to specify and we're we're dealing of course at tree level it's trivial for the disc but for all the surfaces that we're dealing with are orientable. So this notion of moving the uh end points uh uh clockwise is totally well defined. Okay. So, but therefore what I when I when I want to represent the 13, I will represent a curve here that starts uh on this road here between one and two just a little bit to the right of one here and it ends a little bit to the right of three. Okay, you see this is now natural on the the surface. Uh x13 is kind of like a path that I take through this. If this is my map of the world, if this is my my my this is this is my my world. The x's are just uh are are are paths that are uh paths uh on this map. Okay, paths on this map of the world. So that's what x13 would look like. Um x14 would look like let's just do Oops. So x14 would look like this. X24 would look like this and so on. Okay. All right. So, uh let's draw a picture fresh again. Okay. So, every path here is uh let me draw this one. This is x24. That's that's the path uh that corresponds to uh uh x x uh x24 as an example. All right. So, this is our kinematic space. Now, we draw a surface one triangulation. one triangulation to specify the surface and then all the curves all the kinematic variables are associated with a walk through this map that uh that maybe starts on a boundary road and ends on a boundary road okay so now comes the question how do you specify such a path right let's say you're blind and you couldn't see what I just drew how would I specify this path completely completely systematically. Well, the way to specify the path uh systematically is to do um uh imagine it was like uh uh it was uh GPS navigation system like Google navigate or something like that. Um how would you specify to take this green path that corresponds to the line 24? Well, you just say what what what you do. So, you know, here you I'm going to label um I'm going to label every one of the roads in this picture. I mean, I I I could label them. Uh, A, this is road A, road B, road C, D, E, F, G. Okay. So, I'm just going to label every one of the roads. Um, uh I uh here because it's a little bit more uh convenient and it's not uh ambiguous to do it. I'm just going to label them by the the regions that that that they touch. Okay. So this road is going to be labeled by 2 three because it has two on one side and three on the other. So this would be 1 2 2 3. The boundary roads are like this four five and the uh intermediate roads are 13 and 14. Okay. All right. So um Okay. So uh so how would I specify now that I've done this? How would I specify this road uh uh 24? Well, I just have to see what what what I see. So I start on the road 23 and then what I have to do is make a left turn on the road 13. I keep on going here, right? Start here and make a left turn. Now as I keep on going uh when I reach reach the the next intersection I have to make a right turn onto the road 14 and I keep going and uh then finally I make a left turn again on the road 45. Okay. So this is the road that corresponds to x24. Let's do another example. What's the road that what's the uh so uh you can think of every curve as associated with a word like this which just involves the roads and a left right choice uh as I encounter every uh uh vertex. The only thing I'm not allowed to do in this picture is I can't like uh travel along uh I can't uh travel along uh I can't do something like this. I can't go on 13 and say that I uh and say that I changed my mind and uh halfway down down the road 13 decide to come back and do uh uh and go to one two. I can't double back. Well, once I go down a road, I have to keep going down the road until I hit the next intersection and then decide if I'm going uh left or right. That's th those are the only choices. Okay, so let's do another example. I'm going to have the road uh so actually let's let's let's do all of them. Um so I'll draw this picture again. Okay. So let's do uh the the the interesting roads like 13. So 13 is like one two right uh uh on 13 left on 34. Now I'm going to do one more thing just by convention just so I don't keep writing uh right and left. Um uh uh I'm going to write this as uh one two uh right I'm going to uh denote by a down uh by by a down step and left I'm going to denote by an upstep. Okay. So that's the that's for 13. What do I have for uh 14? Uh well 14 looks like this. Uh so that is one two right on one three right on 14 left on four five. So this is one two right on three right on 14 left on four five. All right. And let's just I'll just finish doing uh uh the other ones. So 24 is 23 left on 13 right on 14 left on four five. Uh 25 I'll do up here. 25 is uh 2 three uh left on 13 right on 14 right on five. And finally 35 is uh 34 uh left on 14 right on five. Okay. So these are the uh these are the words for every every curve. Okay. I hope it's clear that I'm not uh I'm just recording every curve. That's the I mean I choose there's many of course it's not unique which graph I specify the surface with. I can choose any fact graph representative to represent the surface. But once I make my choice uh I'm done. Okay. My my world is set. Um, and just even labeling the variables, just even labeling the variables involves giving these words that just tell you how you walk through the uh how you walk through the uh surface. Okay. And now something uh something remarkable happens. Um uh so so I want to record the information about these vectors. I want to uh uh if I want to record the information about these words. Oh, sorry. Uh let me let me do one more set of uh let me do one more set. Um so let's look at the kind of more trivial curves like what is a curve uh that are just the boundary curves, right? So where I just go from 1 2 to 2 3 for example, right? So that would be the curve one two. These are the kind of uh these are the trivial ones. Let me put them in a different color. These are the the boundary curves. So this is one two and one two just turns left onto 2 three. Similarly 2 3 is something that uh that goes uh starts on 23, it turns left onto 1 13 and then it turns left onto 3 four. Okay. Uh and so on. Uh 34 is 34 goes left onto 14, left onto four five and so on. Okay. All right. So, uh now um uh if we had uh 15 more minutes, I would even tell you where this step comes from um uh in a completely canonical way, zero thinking way, but uh but uh just for the sake of time uh uh I will take a little shortcut um uh u for the following point. Now, now I'd like to record the information about these words. Okay, I want to be able to, you know, not just write out these big words every time. I want to just record the non-trivial information uh in these words. And one thing that you can see is that there's an interesting difference between the non-trivial words and the trivial words, the words that correspond to the trivial boundaries. The words that correspond to the trivial boundaries just go up. They just go either left, left, left. If you read them the other way, they'll go right right. Okay. Uh while the words that correspond to the uh non-trivial uh paths on the inside have peaks and valleys, right? They can go up, they can go down, they have peaks and valleys. Okay? So, I want to make some distinction between the words that have peaks and valleys. Um and in fact, the the distinction is going to end up being precisely that. I'm just going to characterize where the peaks and valleys of the words are. Okay. Um I'm I'm going to just tell you how to do that in a moment, but I'm just going to look at the the the peaks and valleys for the words. And so that's automatically going to assign and I'm going to make a vector out of the peaks and valleys of the uh of the of of of the words. Okay. So um so this is called this is going to be called a uh in in in the literature this is called a g vector um uh associated with with every word and it's simply the following. First I imagine assigning uh to all of the internal uh to imagine assigning some unit vector E13 and E14 in general some unit vector to all of the internal boundaries of the all the internal propagators of your uh back graph. Then once I do that uh the the uh the G vector for any word is just equal to the sum of the E in all the valleys minus the sum of the E in all the peaks. That's it. Okay. So that's going to be some vector if I have uh uh if I have e internal uh propagators, it's an e- dimensional vector. So in this case, it's going to be a two-dimensional vector that I'm going to associate with every word. Okay, you see uh with this uh with this definition the the the G vector for the boundary uh curves are all equal to zero um uh just because they're they're equal to and in this case it's a two-dimensional vector 0 0 okay just because they don't have any peaks and valleys but now let's look at what the G vectors are for the non non non-trivial words okay so um so let me uh let's uh copy these over Okay. So let's see what are the what are the uh uh G vectors here. So what is the G vector for let's start here. What's the G vector for X13? The G vector for X13 is equal to well it only has uh so I'm going to write these in the in in this basis E13 E14. So it's going to be a two-dimensional vector. Okay. So G13 it has 1 13 as a value and uh um uh and nothing else. So G13 would be the vector 1 0. Okay. Let's look at X14. G14 would be it only has 14 as a value. So it's 01. Meanwhile, let's look at X24. X24 uh has X13 as a peak and X14 as a valley. So the g for 24 is minus1 uh uh for 13 because 13 is a peak and plus one for 14 since 14 is a valley. Okay. So finally let's look up here. G for 25. The g vector for 25 is equal to 1 13 is a peak. So it's min - 1 0 and that's the only peak it has. And 35 is 0 - 1. Okay. All right. So, is it clear what what what we've done? We specified uh we we've specified the surface by giving a fact graph and now every curve on the surface has this word associated with it and we're capturing the data of this curve uh just by giving the G vectors. It's very easy to see that once you've given the G vectors, you can fully specify the rest of the word. Basically, because you know uh given where the peaks and the valleys are, you know, locally in the neighborhood of a uh of a valley that you're going left, right. Locally in the neighborhood, sorry, of the peak, you're going left to the peak and then right. Locally in the valley, you're going right and then left. And so that's it. So just giving the location of the peaks and the values is in fact obviously enough to fully specify the where all the rest of the lefts and the rights are and specify the rest of the word. So that's why this information suffices to re to uh reconstruct the uh uh entire word um and so it faithfully captures uh the information about uh what the curve is. Okay. So to specify a curve I have to just give you these g vectors. I hope it's clear that nowhere here have I asked whether curves cross each other, whether curves, you know, whether they come together to make fineman diagrams, nothing like that, right? All we've done is specify the surface and and uh uh uh specify what what curves are and find this uh simple rule that captures all the information about curves in these uh uh little vectors. Now comes a wonderful surprise. Okay, the first uh uh the first uh surprise it actually turns out to be quite simple to uh to prove, but let's just uh see what it looks like here. I'm just going to take these curves and draw them. Okay, so uh so uh so I'm going to uh so here's my my E13 E14 space. My basis is E13 and E14. So I'm just going to draw the vectors. This was the vector for X13. Okay, here's a vector for x14. Here is a vector for x24 -1 1. Here is a vector for x25 was - 1 0. And for 35 was 0 - one. And so now you see the first little miracle simply listing the curves. Nothing else. Okay? simply listing the curves uh drawing their G vectors automatically breaks up the space into cones and the cones are finement diagrams. Okay, so this is exactly the picture of the fan that we saw before. Okay. Um except no lattice wave equation, no things pulled out of the blue. I just take the surface, I record the I record the information about uh uh the curves by making this G vector. I draw the G vectors and that automatically divides a space into cones uh that are bounded precisely by X's that can come together that give me a triangulation of the surface that correspond to finement diagrams. This works for every single surface. This works for every single surface. Um and when you do this it um uh when you do this it uh uh it always divides the space into cones. Okay. Um, one of the fascinating things that's related to the question that was asked yesterday about uh about uh about uh closed strings uh or uncolored particles is if you do this for more complicated surfaces, you find that always the sort of topdimensional I mean the the whole big region is divided into cones like this. The cones correspond to diagrams done. Um but you find that there are lower dimensional sort of halfdimensional regions in the space or lower dimensional regions in the space that furthermore are divided if you include the G vectors uh by the same peak value rule. If you include the G vectors not only for the open curves but also the closed curves on the surface then you find that those the ones that correspond the closed curves all live on lower co-dimension regions of this uh space of the fan. And if you look carefully, YOU SEE THAT THOSE GUYS themselves make cones that correspond to the diagrams that you would associate with the uncolored particles. And the uncolored particles put together with the colored ones as well. OKAY. SO ALL OF THE finding diagrams for this extended theory that has both the colored uh uh particle as well as the uncolored particles and as well of them interacting together, they're all sort of given to you for free. You don't have to think about finding diagrams. You just have to list the damn curves on the surface. The open curves, the closed curves. You just list them. You draw their G vectors and automatically they divide the space up into regions that uh cover uh the whole space. Okay. So that's the beginning of a systematic story. That's the beginning of of of the story that uh uh that uh that that does not involve making any choices and which extends to all loop order. Okay. So having told you this, let me just take the last five minutes to tell you where the U variables come from in this picture. Okay. So where the U variables uh come from is that uh is uh is uh associated with uh one of these words. Um so uh and I and I keep uh I've mentioned a few times that the essential ideas uh in this uh subject are ultimately uh uh combinatorial. Um, so there's many interpretations of what the U variables are. There's more conventional interpretations in terms of cross ratios of points on the world sheet and so on. But the but the the the deepest way of thinking about them that makes their properties uh totally obvious um or as as as obvious as possible is not that is not a is not this directly geometric way of thinking about them but is a somewhat more abstract combinatorial way of uh thinking about them. And that's associated with a certain counting problem uh that you attach to every one of these words. Okay. So, every one of these words is associated with an interesting uh uh uh uh counting problem. Let me just tell you what this counting problem is. Okay. So, um uh what uh so let's say I have this uh uh let's say I have any any any word like a up b. Okay. Um uh and I'm just going to associate some abstract variable with every single uh with every single element of this of this word. Um uh the counter problem is is is what I sometimes call the relationships with baggage uh generating function. So what you have to do is choose one. Let me let me do a a simple example. Uh you have to choose you can choose any one of these guys that you want. For example, you can choose nothing and you'd write down that with one. I can just choose A if I like. I can just choose C I like. I can choose A and C if I like. But if I choose anyone, I have to choose everyone in their past like if I draw this sort of picture. If I think of them as a little sort of mountainscape, if I choose anyone, I have to choose everyone uh downhill from that person. So in this case I cannot if I choose B then I have to choose A and C as well. Okay. So I have to have plus b a c. Okay. So that's uh that's a little counting problem that I can associate with uh with uh uh every word. Instead if I did a let's say down b up c I can choose nothing. I can choose b. If I choose a I have to choose a and b because b is downhill of a. If I choose c I have to choose c and b. Finally, I can choose A and C. If of course, if I choose them, I also have to choose B. Okay, so associated with every word is this little um with this little counting problem. Um and uh and uh it's kind of an an exercise in uh in in high school combinatorics um to uh to uh to to to figure out that if you have something like this and and you have the the the the the rest of the word here that you can work out what the word is for the bigger guy from the knowledge of the word for the smaller guy just by just by asking the the uh the question of uh first if going up or down. Um uh you just have to ask the uh uh uh the question uh uh of whether uh whether the the piece of the word that you're looking for contains a or doesn't contain a. So there's a there's a yes, it does contain a and no, it doesn't contain a part of this word. And if I'm going left here, it's easy to see that uh an and a is the element that that I want to have in the word. It's easy to see that uh it has to look like this. Okay? Where uh where now these little fs are whether yes or no they contain b. Right? So for example, this is saying that yes, if it did, if it does contain A, then it might have not contain B or it might have contained B. Okay, but if it does not contain A, that's F no here. If it does not contain A, well then it definitely could not contain B. Okay, it could not contain B because if it contain B, it would have to contain A by going uh downhill. Okay. So you see that that if I'm going upwards from a uh then the then the parts of this word the the the parts of this counting problem that contain a either yes or no are given by this little 2x2 matrix times the lower vector of the same sort and I have a similar matrix if it goes down uh to some B and the rest uh uh uh happens um uh in in in that case if it goes down. In that case, I I would again have F yes and F no. But now instead, it's going to look like uh A011 on little F yes and little F no. And now it's for a a similar reason. If it does include A, then it must contain B. Okay? Because B is downhill from A. So that's why I have an A F yes and zero F no. Meanwhile, if it does not contain A, well, it could either uh uh contain B or not not contain B. And that's where these one ones uh come from. Okay, so you see that there's this little matrix M for turning left which depends on A. And this other matrix M for turning right that depends on A. And these little 2x2 matrices are what you need to now multiply one after the other um in order to figure out what the full uh what the full word is. Okay. Now uh all of our words have the property that they start from some boundary and they they do something and they end up in another boundary. And so it's actually natural to divide these in turn into that part of the word that has uh alpha and beta. The part that only has alpha, the part that has only alpha not beta, the part that has uh uh beta not alpha, and the part that has uh neither of them. Okay, that contains neither. Um that's just I mean the alpha and beta are boundaries. So it's obviously just natural to keep track of of uh of which uh of which parts of this counting problem uh uh contain the boundaries and this turns out to be uh calculated literally by taking the product of those matrices as you see them uh uh go along. So, so if you have some word that begins with alpha and then goes to some a1, a2, a3 and so on and uh and does whatever it does and ends up at some uh beta. Uh I associate variables. I I can call them y a1 y a2 y a3 um associated with each one of these uh uh of these internal uh of these uh uh internal parts of the word and I literally take the product of the of the left right matrices. For example, if I have this word 2 3 up 1 13 uh down 1 14 up four five, I write down the matrix here M left that starts at Y23. Then uh at right uh at 13 I turn right. So it's M right of Y uh 13 m left at Y14. Um uh and that's it. Uh then then I stop. Now remember that 2 three is a boundary. So uh the the y's for the starting ones they're not really variables. So I set their y23 to one. Okay. So and this is what turns out to calculate exactly that 2x2 matrix that I was telling you about. So with every word there's a certain 2x2 matrix. With every word w there's a certain 2x2 matrix m of w that's associated with this counting problem. Okay. So with every word there's a little uh uh 2x2 matrix. I can just read it off by taking the product of these uh little uh 2x2 matrices. And well one one thing which is obvious is that if you take the uh determinant of any one of these m left or m right this determinant is equal to y. Okay. And in particular if I say that all the y's are positive I discover that that this determinant of every one of these matrices is positive. uh and therefore the determinant of any word the matrix associated with any word is also positive. So the determinant of the matrix associated with any word is also positive. So if I write the matrix for this word as matrix 11 one matrix one2 uh uh in terms of its uh uh matrix elements then I learned that uh well uh m11 m22 is bigger than m21 m12 and so this gives a motivation uh to associate with every word w with every word w there's a a u variable uw which is exactly this off diagonal m12 of w m21 of w divided by m11 of w m22 of w. Okay. Uh uh I'm not expecting you to to uh understand uh uh every one of the steps here in real time. I just want you to see how concrete it is. Um all we're doing is we take the surface. Uh we have these words that associate something to every curve. And um uh so first with with every curve associated with the word the G vectors uh do this magic of dividing the space up into into a pieces uh each cone corresponding to finement diagrams uh and that ends up connecting to the uh uh the systematic completion of the story of the associated but also the story uh that connects to string amplitudes is that these words also uniquely specify the U variables. Okay. And I'm just showing you the steps. So given the given the word you just multiply out these 2x2 matrices associated with the word. The crucial point is that everything is local to this word. I don't need to know about any other curves on the surface. You give me this one curve and I multiply these 2x2 matrices. I get this uh associated with this counting problem. Uh and then I just form this ratio uh associated with the with the matrix that I get that is guaranteed to be between zero and one. Okay. So this uw is is guaranteed to be positive and is guaranteed to be less than one. Okay. The miracle is that having done this and again I stress that every u is just defined one word at a time right no knowledge of any of the other words on the surface the miracle is that the 's so defined uh satisfy the following equation. So there's a U for any curve C and it's a U for the curve C plus the product of all the curves C prime of the U of the curve C prime to the number of times the curves C and C prime intersect is equal to one. Okay, this turns out to be an almost immediate consequence of thinking about this counting problem. Okay. So, uh this is a part that I don't have time to uh explain, but this very simple counting problem, this very simple uh uh relationships with baggage uh uh uh counting problem and this definition for the use ends up to make it relatively obvious that in this uh uh that uh the second term in this product has uh well anyway that that uh that uh uh every u is the product. every u is is is is uh uh is 12 2 1 over one22. Um but you can think of every element in this product as being associated with the little subp part of the word. Uh and so in this big product over all curves on the surface uh there's there's a there's uh for a simple reason the telescopic cancellation in this big product that makes this identity obvious. Okay. Now this identity uh this formula generalizes what we saw in the case at tree level where either curves cross so their intersection number is one or they don't cross and their intersection number is zero. If they're intersection number zero, of course, they don't occur in this formula. I didn't write them down before, but this is the general way of writing it for any surface. And for any surface, the intersection numbers can be complicated, but they're just some positive or zero integer. Okay? But this still has the magical property that if a given u goes to zero, uh the uh all of the 's for the curves that cross it have to go to one. Okay? So and so that's why uh uh that's why once you have the uh once you have these u variables for any surface you can define an amplitude to be the product of the integral over the dy over y's from 0 to infinity. These are the y's that are associated with the internal edges that we just talked about and then the product over every curve on the surface u to the alpha prime of the kinematic variable that you associated with every curve on the surface. And just like at tree level we saw that the kinematic variables were associated with curves on surfaces. So at loop level including the loop propagators every kinematic variable is associated with the curve on the surface. And so that's why this gives you uh a formula for the uh amplitude. It gives you a formula for the amplitude that's guaranteed to factoriize as any x goes to zero. As any x goes to zero um I get 1 /x c. I get that propagator. But precisely because all of the curves that cross C uh have their U's go to one. Precisely the same logic as at uh as at tree level uh the amplitude factorizes into what I get from the surface that where you cut along that propagator. And that's precisely the factoriization properties that we need not just at tree level but at all loop order and not just for field theory amplitudes but for full string amplitudes as well. So these u variables are the are the are uh one more level of magic beyond just the fact that the g vectors give you this fan that covered the entire space. Um unfortunately I don't have time to tell you how the uh how the u variable story and the g vector story are uh are are related to each other but they're related to each other in a very simple and natural way that arises when you think about what happens uh when you take the field theory limit as alpha prime goes to zero. Um uh so this is alpha prime to the number of edges uh minus one outside as alpha prime goes to zero uh these integrals become dominated by going far away in y space by going to extremes either the y's go to zero or infinity um these u's are ratios of polomials um but it but if the integral becomes dominated by when you go far out along uh uh in in the space uh all you need to do is keep track of which one of the monomials in these polomials are the most important and that process of keeping track of which mon of monomials in a polinomial are the most important is known as tropicalization uh in mathematics. It's again something you could explain to a high school student. But the tropicalization of these polomials associated with the u variables give rise precisely to the picture of these fans and cones and the g vectors and all of the uh rest of it. So the U's are like a nonlinear version of the of the of the of the story involving the the G vectors and those little simplicies that you add up together to give the uh uh the uh associ um are are similarly again directly tropicalizations of uh the polomials that are associated with the um uh uh with these uh counting problems that define the U variables. So this last bit was really meant to be very impressionistic. I've gone uh over time. Um uh but I just wanted you to get get an idea. Again, the the crucial thing there's there's a magic in both in both parts. Um that you just ask for something one curve at a time. You see it seems like all of the drama of the of the space-time processes and quantum mechanics adding everything up together. Um our usual picture is that we manually draw all fineman diagrams. Then we have to sum them all together. Okay. Um we draw the diagrams. That's all possible space-time processes. We manually sum them together because Fineman told us that's how we we're compatible with quantum mechanics. So we're thinking globally about all the curves, how they come together, how we produce them to make diagrams. Instead, in this other picture, it's it's it it's totally the opposite. We draw the surface. We specify a single curve. A single curve through all the other curves. We haven't we don't care about anyone else. We just look at a single curve. We we either get a G vector out of that curve, we record it. We can also get a U variable out of that curve. We can we uh record it. But then if I just take the picture of all the G vectors and just plunk them all down, they magically automatically divide the entire space up uh into cones, each one of which corresponds to a diagram. So we didn't need to think about all the diagrams. They're just handed to us. We didn't think that we had to put them all together. they're automatically all put together covering the entire space. Okay. Um this local to global phenomenon is the sort of magic here. And exactly the same happens with uh u u variables. Um one curve at a time, this counting problem associated one curve at a time produces these objects that then magically satisfies these global equations that the u plus a product of all the other u's is uh to appropriate intersection number powers is equal to one. And that's what again uh uh uh guarantees that uh that that that curves that cross uh can't uh can't occur together uh and uh and gives us the stringy completion of the uh story. Um, in all of these cases, the the the structure emerges by just thinking slowly about the kinematic space and super explicitly about how you would label things, how you would uh describe things if you couldn't, you know, see uh but just systematically labeling the kinematics and the uh uh associated with the objects. And um as I stressed uh uh that this has given us uh uh new insights into what uh amplitudes do that as I hope I emphasize a few time there's no super symmetry in sight here. Uh the connection to pons and gluons is there at all loop order for totally non-supmmetric uh theories in any number of space-time dimensions. Uh and there are qualitative facts about the amplitudes. the fact they have these hidden zeros, the factoriizations near near zeros. Of course, dramatically more efficient ways to compute them. I haven't stressed that at all. I don't care about that myself so much. Uh but of course, it comes along for the ride when you have uh uh uh a a new conceptual viewpoint on things. Um uh and uh uh so uh there's there's a lot of uh things to uh continue to develop in this picture. Um I didn't have any chance unfortunately to tell you about the kind of beginnings of the steps of these ideas in uh cosmology. Uh but um uh but that's something which has been happening uh been happening recently. Um but I will say that I do think that kind of if just stepping back I alluded to it in the answer to one of the questions uh that uh what I'm personally most fascinated about is is is the fact that this sort of new picture this new dual picture for what uh the amplitudes are make these properties of the amplitudes including mysterious symmetries that we didn't know about uh zeros that we didn't know about uh obvious. Uh and those are exactly the kinds of words that if you're a model builder, you would love to hear applied to the hierarchy problem or the cosmological concept problem. Some hidden symmetry, something that we can't see from lrangeians that the usual local picture of physics uh hides, but this other way of thinking about things makes obvious. It would be amazing if some of these mechanisms that we're uh uh that that that we've been finding uh might shed some new light on these uh on these sort of grand old old mysteries of particle physics and cosmology. All right. And with that I will end and apologize for going uh so far over time. Thank you very much. Uh so thank thanks very much Nemo. Um can you mention just for the students that want to follow up is is there some kind of review article or just point to some papers or something that they >> Yeah. So so so uh the the um uh uh so the the formalism that I was uh uh talking about was uh was put out um uh in in this paper from 2023 called uh scattering amplitudes as a counting problem. uh and then uh I guess early 2024 a paper called uh hidden zeros for and the unity of pions and gluons something like that colored scalers pions and gluons. Um uh there is also uh for a succinct uh presentation of just how you go about computing these things as quickly as possible just fits on a page. Um there's an appendix of a paper uh I wrote um uh whose uh with with with friends all with young friends of course uh uh that that whose title was something like uh tropical langians for colored amplitudes or colored particles or something like that. The appendix of that paper uh just gives you know minus all of the bells and whistles just gives the the super practical way for uh computing u variables g vectors all of that stuff. So the things that I talked about in the last 20 minutes are are uh are are reviewed there. Hopefully we're going to have a long delayed paper um uh coming out where we give a completely systematic exposition of the U variables where they come from. Um but it's it's alluded to in the accounting problem paper, but that paper still was more of a users's guide for how to do computations. Um, so hopefully soon we'll have a paper coming out that uh is maximally pedagogical and uh explains where everything comes from from from from the bottom up. >> Great. Thanks for that. Um so uh I'm going to um first of all say uh thank you very much to all of not just to Nema but uh to all of the lecturers uh Rajes, Marcos, Lara and Roberto um all five of whom gave I think you would agree um incredible wellprepared um lectures um and so uh let's thank them for all of their efforts and making this school a really wonderful experience this time. Um and then um I want to thank um uh my fellow co-organizers Anzer Pavel and Francesco for um all of their efforts. Thank you very much guys. Um and then just quickly um uh thanks to uh uh INF for um their co-sponsorship um of the activity. Um and in addition uh to Victoria who was the administrator for the activity. um the uh ICTP housing office, finance office, the security guards, the guest house reception staff, uh the people that work in the catering, the people that do the cleaning, the receptionists, um all of whom have in one way or another made this uh possible. So, please let's have a thanks for the ICTP staff. And uh yeah, thanks everyone for coming. Um and I hope uh you enjoyed it uh got something out of it. Um please keep in touch with each other with uh whatever you know develop on the relationships you've built here and the things you've learned. Um, and I wish you all a pleasant journey onwards and see some of you next year. Um, Nema, I hope >> I really hope you can make it next year. Bobby, if you invite me again, I'll come. If you're not sick of inviting me. >> No, it'd be great to see you here in person. I know it's hard for you to travel sometimes, so that's fine. We're this this we really enjoyed this, so thank you very much. >> Thank you guys. Bye-bye now. >> Bye. Bye.