 So, I think, given that we are already in lecture three, it would be a good idea to tell you what is it that we're doing. So you have been following everything that I've been saying, but I haven't actually told you where we want to go. So, what we want to do is to find a new formulation of possibly all quantum field theories. At least the ones that, many of the ones that we know, okay, I'm not sure we're going to be able to formulate things as crazy as the ones that Jiu-Jitsu told you about, but at least the ones that have Lagrangians and things that we understand, okay. So what is the new formulation going to look like? Well, the formulation is going to be based on, you guess it, on the scattering equations. So let me write them again. And we're concentrating on massless particles. And the reason is very simple, at least from the scattering equations point of view. If you want these equations to be invariant under SL2C transformations, you need massless particles. So we decided to look for a formula for Young-Mills amplitudes of the form. So let me write here scattering equations. And then here we have the Part Taylor formula times some integrand that depends on the momentum, the polarization vectors, and the signals. And we said we needed to interpret this integral as an integral over the modular space of an n-puncher sphere. So last time we also gave a meaning to this object as it is, it doesn't seem to make much sense because these equations are redundant. We have three degrees of freedom that we have to remove, but that matches the fact that we also have to remove three punchers or the location of three of the punchers. And so we can write a formula that is completely gauge fixed by writing, removing the punchers i, j, k, and removing the integrals or the scattering equations for particles p, q, and r. I'm putting here the Van der Monde of the particles i, j, k, and p, q, r. And once again we write our Part Taylor formula. And don't worry because this is going to be the last time I'm going to be writing this thing here because I'm going to give it a name so that we don't have to write it anymore. So I'm going to say that this is a measure for n particles, okay, is somehow a measure over the modular space of the n-puncher sphere. So this is the definition I'm giving to all this. And then we can write very nice formulas. So at this point, this is all wishful thinking because I haven't told you how to construct these integrals. Well, from the fact that we have been following these lectures, you can presume that there is such a thing that we can write down. In fact, we know that in four dimensions there is already an integrand, which is a one that comes from twisted strings, either in the formulation that Wheaton gave or the formulation that Nathan gave. But now we are looking for something that is valid in any number of dimensions, okay? So we want to write a formula for Young-Mills amplitudes, for the scattering of gluons in any number of dimensions. And guess what? If we are in 200 dimensions, can we have supersymmetry there? No? We cannot have supersymmetry. So whatever formula we're going to write is not something that relies on supersymmetry because it has to be valid in any number of dimensions. So what are the constraints on this integrand? No? There isn't? Exactly. Because I told you that anything that satisfies, remember my theorem, well, my claim really, that I made up during the last lecture, that anything that satisfies the KK relations and the cyclicity, which are properties that Young-Mills amplitudes of gluons satisfying any number of dimensions, is this algebraic transform of something that looks like partail. So I'm using the strong form of my theorem, okay? So this has nothing to do with partailer, except that it looks like a partailer. Now the magic will come into this integrand. So the integrand must be permutation invariant. That was part of the theorem, right? We needed everything else to be permutation invariant so that it would be guaranteed that whatever I write satisfies the KK relations, cyclicity and so on. However, written in this form with this measure, they will be guaranteed to satisfy BCJ relations, which are also valid in any number of dimensions. What else we need? Now we have polarization vectors. So even though Simone told you all gauge invariance is bad, so gauge invariance is something that we're having in our minds, it's not physical, so we should even start with something that doesn't have the redundancies that you see. Well, in four dimensions you can work with spin or helicity and forget about gauge invariance indeed, there is no such a thing. It's all in our minds, but in higher number of dimensions, we don't have something that works as spin or helicity, so we have to go back to our good old polarization vectors. And if you remember, every time you replace a polarization vector of a particular particle by its corresponding momentum, you should get zero. That's the way you ensure you have the right amount of redundancies. Or what we call gauge invariance. Excellent question. No, it's not necessary, but life is good. So it must be that it's something that happens before integration and solution by solution. What else? Well, you should transform the integrant so that SL2C is also a redundancy. What do I mean by that? Well, I've been using very strongly that I need SL2C, right? In order to remove three of the punctures and in order to make sense out of this. But if the whole integrant, if this whole object is not SL2C invariant, then we're not talking about any, we're not making sense. So we want to think about this whole thing as if it was a path integral for a gauge theory and the gauge group is SL2C. So we need this object to transform correctly. Now, what do I mean by transforming correctly? Well, you take all the sigmas, you perform an SL2C transformation, and it turns out that the integrant, whatever it is, has to go to the product of gamma sigma a plus rho square times itself. So this is an exercise for you, show that this is true. In fact, note that this is a transformation that this part a low like factor. I'm going to introduce a like so that we're not confused anymore. So this object transforms exactly in the same way under SL2C transformations. So the part a low like factor and the integrant, they both transform correctly. In fact, you could say, well, just for fun, could I just remove this and put another part a low like factor in there? That would make sense as an integral over the modulate space of Riemann surfaces. Indeed, we're gonna do it, but not now. Let's keep working on junk mills. Because if we did that, we would lose all polarization vectors and therefore we would have a scalar theory. But that's not what we're looking for. We are looking for junk mills. So we need the polarization vectors. What else? Well, it has to be multilinear in the polarization vectors. From Feynman diagrams, we know that each particle has a single wave function, which is a polarization vector. So whatever we write down has to be multilinear in it. In other words, it has to have exactly one power of the polarization vector of each of the particles, okay? So I'm not gonna try to motivate it very much, but except for saying that we have basic building blocks. What are the building blocks? Well, we need transformations like this. We know part Taylor like factors transform correctly. And we need Lorentz invariant combinations of these things. So we can have things that look like this, ka.kb. This would be a natural object to use. We could also have the product of polarization vectors, but we could also have momentum and polarization vectors together. So the fact that we needed something like this for each particle motivated us to look for something that has the structure of a matrix. And this requirement would imply that new linear dependencies would appear. So you're looking for a matrix such that whenever you make this replacement, new redundancies appear. And if you construct a matrix that has that property and you take the determinant of that matrix, you're gonna get zero every time something like this happens. So using the building blocks, you can construct a matrix that looks natural. Sorry, this is K3. So I'm gonna write some of the elements. So this is one of the matrices that you would guess. You could write down using only the first kind of building blocks. Now, the next one that you can think of is the one that you get by replacing the case with the polarization vectors, okay? Now, remember what we want to do is to build something such that is a matrix that has more redundancies or linear dependencies among its columns or rows. Every time we perform a gauge transformation. So if we look at this, and we replace, say, epsilon 1, the polarization vector of the first particle by K1, we want all this thing to look like, say, the first row. Well, there is a natural thing that we can write so that this is actually satisfied. So if I write this, this becomes identical to this. If I replace epsilon 1 by K1, sorry, no, that doesn't happen. Well, if I want that to happen, then all I have to do is switch these guys, right? You see, there is no way to get it wrong. Now I repeat, you take this, you substitute epsilon 1 by K1 and you get exactly this, right? I'm simplifying my life, I'm building a matrix that has some symmetry properties. As you can see already, it's anti-symmetric. This matrix is anti-symmetric. So let me also conjecture that this part would be anti-symmetric just as that one. So I'm gonna have here something that looks like epsilon 2 dot K1, sigma 2 minus sigma 1, epsilon 3 dot K2, K1, sigma 3 minus sigma 1, and so on. So I can keep constructing my matrix in this form. So now, you see that everything here, okay. So I want everything here to contain an epsilon 1. So I can keep going. So this is gonna be K2 dot C2 minus 1. All the way to KN, epsilon 1, sigma m minus sigma 1, all the way to here, and so on. And this part is just the anti-symmetric. This part is supposed to be the minus the transpose of these metrics, okay? So on this side, I have epsilon 1 dot K2, sigma 1 minus sigma 2, up to epsilon n dot K, sorry, epsilon 1 dot KN, sigma 1 minus sigma n. Okay, very good. So hopefully I got it right. So judging by the faces, probably didn't get it right. But let's check. If I compute the determinant of this thing, does it vanish whenever I want it to vanish? Well, let's have a look again. So if we replace epsilon 1 by K1, all this column becomes identical to this column. I think I got it right this time. And all this row becomes identical to this row. Now that epsilon 1 is K1 now, so it works perfectly. So in fact, it develops a null vector on the row space and a null vector on the column space. So that all looks pretty good. So this seems to be happening if I were to take the determinant of that matrix. How about this property? Is it multilinear in the epsilons? Well, it almost works because if you rescale, say, epsilon 1 by some quantity T, right? All this column will rescale by T, and all this row will also rescale by T. And therefore, the determinant will rescale as T squared. So we got something that is not multilinear in epsilon 1, but it has two powers of epsilon 1. So we almost got it right. And now you know why I chose this to be anti-symmetric, right? Because if we could only take the square root of this determinant, then we would get it right. We have two powers of each polarization vector. So if we could take the square root, we would get only one power. But we just don't want to take the square root of any determinant because that would give us something horrendous. But if we have an anti-symmetric matrix, we know that the determinant of an anti-symmetric matrix is a perfect square. In fact, it's a square of what is called the Fafian. So if we call this matrix psi, the determinant of psi is equal to the square of the Fafian of the matrix. So it seems that we got it. So we got a matrix. We compute its Fafian. The Fafian clearly vanishes every time something like this happens. For each one of the particles independently, you can check that it transforms correctly under SL2C, is multilinear in the polarization vectors, and is made out of these building blocks. So could this object be the Fafian of our matrix? Well, sadly, the Fafian of that matrix vanishes. So oh, OK. We are this close. But when have we stopped because the determinant of the Fafian of something vanishes? We didn't stop last time, right? We also found the same problem. We had that the Jacobian, the determinant of the Jacobian matrix was 0. This time, you can check that you have two null vectors. So this matrix has two null vectors. They are, these are the null vectors. This matrix is 2n by 2n. That's why I divided it in blocks. Each block is n by n. And you can check that these are null vectors. And they are the ones responsible for the vanishing of the Fafian. But following from what we did this morning, maybe we can use the same trick and define some special Fafian, which you can get by removing two rows and two columns of that matrix and then computing the Fafian. But of course, the rows and the columns, they have to be in the place where the action happens. That means that you want to take them in the first n dimensional part of the matrix. And likewise, as we did in the morning, now the pay, the price you pay for doing this is also the Van der Monde determinant. But now of the 2 by 2 matrix made out of the i and j rows of this object. So you get 1 over sigma i minus sigma j. And this object happens to be permutation invariant. So it doesn't matter which rows and which columns you delete as long as they are in the first n or the first n and the first n. And this object is permutation invariant. So the proposal is that this object computes Young-Mills amplitudes for any number of particles and in any number of dimensions. So this proposal has now been proven by showing that this satisfies recursion relations that will build the amplitudes anyhow. And the recursion relations are satisfied by this structure. Now what I want to do is to actually, if you probably notice, maybe some of you notice, that I wrote here a calligraphic a and not my standard a for the partial amplitudes. So what I want to do is to write down the formula for everything. So I'm going to squeeze this a little bit here. And I want to put here the sum over permutations. And I want to put here the trace. And this will be the formula for the complete amplitude with the color structures and everything included. OK? Very good. Let me rewrite this slightly. I'm going to rewrite this as follows. Let me say that, let me introduce this notation and say that if you have a, b, c, and so on, the vertices correspond to interactions of the structure constants. Any two vertices connected by a line mean that we are summing over all the values of the label e and so on. OK? So that's what this diagram means. It turns out that you can replace this by sum over permutations belonging to Sn minus 2 of the items where you fix say 1 and n. And you have a1 or aw2 up to awm minus 1. And you put the corresponding part Taylor like factor here. And this formula is valid for an engaged group. Now, was this just luck? Maybe. One observation that we can make already is that I told you that these things transforms under SL2C exactly in the same way as this guy did. OK? And we saw that the problem we had at the beginning with computing the determinant of the matrix psi was that it transformed or it had two powers of each polarization vector instead of one as we needed. But if you have two powers of each polarization vector, you can make a polarization tensor. So what theory could you construct that has in the state of polarization vectors, it has massless particles with polarization tensors with a rank 2 tensor? Well, yes, you're guessing it correctly. This is gravity, just Einstein gravity. OK? So here each particle, so if we have an amplitude, each particle is defined by a momentum vector and a polarization tensor. And you can show that using the gauge redundancies, this can always be chosen to be the product of two polarization vectors. Now the natural guess is that this object could be computed by the same measure that we had here. Remember, this was my definition so that I wouldn't have to write it over and over again. Times the determinant, but now not the regular determinant because it also vanishes. It would be the determinant prime. So in order not to get confused, let me write it as Fafian of psi prime squared. So that's a proposal that this object now computes the complete three-level S matrix of Einstein gravity in any number of dimensions. OK. So that sounds like a little miracle, but let's try to understand if that's actually possible. Let's try to compute something and see if that works. So let's try to check. One of the standard checks that you should always run on any formula when somebody gives it to you that declares is a formula for gravity or for young males is what happens with soft limits. Soft limits are something that you can control. And Weinberg told us what the behavior of any amplitude of gravitons or gluons is when you take one of the particles and make it soft. So when we study the soft limits of this object, we're going to find that they are all wrong. And the same thing with this one. They said, what? This is impossible. This matrix satisfies everything that we needed. How can it be that they are wrong? Well, it turns out that there is something else we could have done here. There is another way we can modify this matrix so that all the properties we mentioned are still satisfied. And the only place where you can modify the matrix is actually in these entries. So let me call this C11 up to CNN. So all the diagonal entries could be more general than what I wrote. Now, what can they possibly be in order to satisfy, say, gauge invariance? Well, we know that if we substitute epsilon 1 by k1, this thing has to be equal to 0. Well, but we know also that whatever the object is, it's going to be integrated on the support of the scattering equations. So here is something that does vanish when we need it to vanish. Why does it vanish? Well, note that C11 will have the polarization vector C1, epsilon 1. So let me write C11. We'll have epsilon 1.kb, sigma 1 minus sigma b. And when you substitute epsilon 1 by k1, what do you get? You get one of the scattering equations, so it vanishes. So miraculously, this thing can also be added. And all the properties we mentioned are still satisfied. Now let me show you why these things were important, because we're going to show how the soft limits work. So let's go back to this formula. And you're all experts in soft limits, because we did it this morning. So what do you do? You take, say, particle n, and you replace it by something that looks like this, and you take tau to be very small. Well, let's look at our metrics and see what it does. Is there color? Maybe something bright. Let's try this one. So we look at these terms. They all depend on the particle momentum kn. So they all go like tau. So if you take tau to 0, all these components, and all these components, they all vanish in the limit. So now you see that if I had left here a 0 and a 0, I would get that in the soft limit, I would get strictly 0. But now, cnn doesn't vanish. In fact, the momentum kn doesn't even enter in the definition cnn. And I'm computing the determinant of these metrics. Now I want to go back to the definition, where this thing is 1 over sigma i minus sigma j square, now times the determinant of the metrics where I remove i and j and i and j. And I'm going to take i and j to be anything, but the one that I want to take soft. This is just my special choice so that I can prove to you that this has the correct soft limit. So I'm going to choose i and j to be anything, but the one that I'm removing. Now if I'm computing the determinant of these metrics after I remove say 1 and 2 or any i and j here and i and j there, well, you wouldn't be very smart if you didn't choose to expand the determinant around the row or the column that has the most zeros. That's what we learned in while it depends on where you are. I guess high school. This one is definitely high school. So if you have a determinant, and you want to calculate the determinant, and you have something that has zeros everywhere, except somewhere here, that's the row you want to choose to expand. So you get these times the determinant that is left over, after removing these and these. So you get the factor c and n. So let me write the determinant of i, j, i, j for m particles is equal to c and n times the determinant of the matrix that is left over. But the matrix that is left over also has another, now it has a column that has lots of zeros. So once again, you expand around this one, and you get 0, 0, 0, and c and n again times something a smaller jet. So you get another power of c and n times the determinant of the matrix that you would have constructed if you had n minus 1 particles, and you had removed i and j and i and j. So this is looking very, very good. In fact, all these entries, they have this form. I'm removing the particle n from these entries. But this piece is multiplied by a momentum that is vanishing. So this is order tau times q. So this is also going to 0. So literally, if I take the limit, this determinant splits clearly into something times the object that you need to c d n minus 1 particle amplitude. And we're getting close, because that's what Weinberg told us. Weinberg said that in the soft limit, you should get something that only depends on the particle you are removing on the particle that is soft times the n minus 1 particle amplitude. So this is looking very good. What's next? The measure. So what do we do with the measure? The measure, remember, it was something of the form product over sigma a's with a different from i, j, k. Again, I'm not going to choose n. So that's part of the trick times the product over the delta functions imposing the scattering equations. Again, I'm not going to choose n as part of this times the van der Mondes. So this was a measure. So let me just write it like this. Now what I want to do is to separate particle n from here and from here. And I'm going to put the integration over particle n here and the scattering equation for particle n. I'm going to put it here as well. All I'm doing is trying to separate everything that has particle n in it. Now the scattering equations, as we saw in the morning when we counted the number of solutions, for any particle that is not particle n in the strict soft limit, they become the scattering equations for n minus 1 particle. So this piece is looking very, very good. This thing is the measure for a n minus 1 particle system. And I have this piece left over. So now we're ready to put everything we have learned together and see what happens in the soft limit. So our amplitude looks like the measure. It has a measure for n minus 1 particle amplitudes. It has the determinant prime of n minus 1 particles. This is all looking great times the integral over the nth particle puncture or the nth particle location of what? Of c and n square, but c and n is sum over n kb over sigma m minus sigma b. All that is square, sum from b1 to n minus 1. Now I also told you in the morning that every time you see a delta function, I'm lying to you. It's not really a delta function. What I mean by that is a pole. And all these integrals are really contour integrals done on contours that pick residues on the poles that I selected when I wrote down a delta function. So now I really need to use the truth. So I don't want to write a delta function of this. I want to put it as a pole. So I'm going to put it as a pole. And the contour gamma is a contour that encloses all the zeros of this thing. So in the sigma m plane, this object has a bunch of zeros. How many of them? Well, you should know. This morning we found how many zeros this thing had. It had m minus 3. So we just have to sum over computer residue of this object, just solve a polynomial equation of degree m minus 3 with coefficients that are crazy variables and depend on the particular solutions of the m minus 1 system. That's all what you have to do. Well, clearly, that sounds almost impossible. And indeed, I think that's impossible. I mean, actually doing that is going to be incredibly complicated. But if we don't do that, this object cannot be pulled out of the amplitude of this object. We want to get this with something that can be pulled out. So by some magic, this object better be independent of the other punctured locations. How can that possibly happen? This thing has all the other punctured locations all messed up everywhere. Well, where are the other punctures? The other punctures are here. Say sigma 1, sigma 2, sigma 3, and so on. Well, I drew this in a particular way so that you'd be inspired to think about something. This is a contour integral. And you are told to do the calculation over a very complicated set of poles. But you also see that your function outside of these poles only has poles that look very simple when the puncture n hits any of the other punctures. So it's very natural to say, well, what I have to do is to use a residue theorem. And instead of calculating over these guys, I'm going to calculate the integral over the other poles, over the green poles. So that's what I'm going to do. This is now going to be the sum of contour integrals of the same object, but now on the contours that circle the other punctures. This is an identity. Well, but near any of the punctures, only one of the terms here contributes. And only one of the terms here contributes. Precisely the term KA here contributes. This one gives you a 0. But this one gives you a double pole. And therefore, you pick up exactly one contribution. And the contribution you pick when you do this contour integral is epsilon n dot KA square divided by KN dot KA. And now you have to sum over A. And this object has become independent of the sigmas. So you can pull it out and conclude that in the soft limit, your amplitude has become the m particle amplitude, or the m minus 1 particle amplitude, which is defined by this, times a factor that contains this combination. But this combination, if you remember, from the times you were studying quantum field theory, is exactly Weinberg's soft factor. So we have succeeded in proving that our object reproduces the correct soft behavior. But note something interesting. The way we did it was by deforming the true solutions of the scattering equations, where we are evaluating this object on random, or genetic, sorry, not random, but genetic points in the modulate space. These are genetic points in the modulate space of the n-pointure sphere. So the amplitude is being evaluated on fat Riemann surfaces. But in order to prove the soft theorem, we use a residue theorem that ended up localizing us on the boundary of the modulate space. So we are somehow seeing how Feynman diagrams are emerging from this fat Riemann sphere. In fact, each of these terms, we correspond to a three particle factorization, and we see the propagators coming out. These are the propagators of the three particle factorizations. So how much time we have? We started at 2.30, right? We have 10 more minutes. OK, very good. Now let's try Young Mills. Well, that would be an exercise for you. Now take a partial amplitude without all this, just the standard part Taylor-like factor. Do the same calculation and see what happens. Note that now you have only the fafian. You don't have the full determinant. So how many factors would you get of c and n? When we computed the determinant, we got c and n square, but the fafian is a square root. So you're going to get only one factor. Now you would say, well, but there are no poles. This is crazy, except that now you have the part Taylor-like factor. So in Young Mills, you're not going to have this, but you're going to have the pieces of the part Taylor that contain c and n. Sorry, that contains sigma n. And sigma n talks to the neighbors. So those are the only two poles you're going to get. Exactly what you would expect if you were taking the soft limit of a gluon. The gluon only talks to the neighbors. And that's encoded in the fact that the only poles that you would get are the ones coming from the part Taylor. Very good. So in the final minutes, I want to use the power of analogy again. I want to say, well, look what we have done here. We have this thing. Let me call it cn part Taylor. And in order to get gravity, all we did was to replace this object by another copy of the fafian. In fact, somehow they are exchangeable because they transform in the same way under sl2c. Now, if I did it in that direction, what would happen if I did it in the opposite direction? Imagine that I replace the fafian by another copy of this thing. So I want to delete this and put a second copy of this. What do you think I'm going to get? Well, we're going to get something that has no polarization vectors. This theory had one copy of polarization vectors per particle. This one had two copies. And the one we're going to write down has zero copies of polarization vectors. So a theory with polarization vectors can only be one thing. It's a scalar theory. But at least we would expect that this is going to be some sort of a scalar theory. So we're going to write our measure again. And we're going to write these factors. So I'm going to say that this was the factor for the group un. Those are the traces that we have there. So I want to be bold and propose that here we can have a different one. Maybe n tilde. So what could this possibly be? It should be the theory of a scalar particle, phi, which is massless. And somehow it has a flavor group, which is a product of two un groups. And it has a double-colored order decomposition. You can decompose it, color decompose it in terms of one un and the other un tilde that you have. And the color decomposition happens to be exactly in this form as if they were glooms. So they have to be in the adjoined representation. So this massless scalar field is in the bia joint. So what's interaction? Well, you can try to write down the simplest possible interaction. And the simplest possible interaction is an interaction with the structure constants of un and with the structure constants of un tilde with our bia joint scalar field. So that's a proposal that this object computes a complete three-level-less matrix of a massless bia joint scalar with this cubic interaction. Can we check that? Well, I think it would take more than five minutes. But let's do something simpler. Let's go back to four particles and extract from that object some particular pieces of it. Remember, each of these factors contains all possible combinations, so we can start to multiply them and get things. So I want to take as a first object, I have to choose two orderings now. I'm going to choose this order. For the first trace and for the second trace, so this is going to be the integral over mu 4, 1 over the part Taylor. So I'm going to use my shorthand notation for part Taylor like factors. And remember, this morning, we solved the scattering equations for four particles. There was only one solution. So it's actually trivial to perform this calculation. So I would like you to try and discover that what you get is 1 over s plus 1 over t. Now you do 1, 2, 3, 4. And you change something here, say 2 and 3. And you repeat the calculation. Any guesses on what you can get? This is what you get. Remember, these things are cyclic. So this is basically all you can do. Modular relabelings, but these are all the calculations for four particles that you have to do. And they indeed agree with the theory that we proposed over there. But can you see anything happening here? Well, if you draw the particles on a circle, so we have this ordering 1, 2, 3, 4. I'm always going to keep the standard ordering here. And now the next order, I can draw it. I'm going to draw it here in another circle. So what I found here is the sum over all cubic diagrams, say lambda phi cube, all cubic diagrams that can be drawn on a plane respecting this order and this order. Well, here is trivial because I have the same order. So you see that I drew the diagram 1 over s. And I also drew, well, that one can also be drawn here very nicely. But I also had the diagram 1 over t. The diagram that produces 1 over u cannot possibly be drawn on the plane respecting this ordering. Now let me keep this order and change this order. So this is 1, 3, 2, 4. And ask again, what are the possible trivalent diagrams that I can draw that respect both orderings? As a trivalent graph, you can move it in a space, in the three-dimensional space. But then I want you to take it and paste it on the blackboard so that it's planar and it respects this order. And it should also respect this order. If you take it out, move it and paste it again. It should also respect this order. Well, clearly the diagram that produces 1 over s fits here, but it wouldn't fit in here. The only diagram that fits in both is this diagram. And that's why you get 1 over t. I knew this was going to happen. So you see, once again, there is no way to get it wrong. You just have to do it. So the proposal is that take these same pattern follows for any number of particles. Take any number of particles, any orderings. And these interiors compute for you the sum over Feynman diagrams in a trivalent theory, all possible Feynman diagrams that are compatible with the two orderings that you have. So these objects will make a matrix where the indices are permutations of labels. That matrix, if you build it, turns out to have a very surprising property. The property that it has is that when you find the inverse of that matrix, you get exactly what kawaii, lewelland, and thaii, or KLT found in the 80s as the object that you have to use to merge Yang-Mills amplitudes to produce gravity amplitudes. So you probably have heard the statement that gravity is a square of Yang-Mills. Well, we more or less saw that here in the sense that Yang-Mills was the Fafian, and gravity was the determinant. But it's not actually true that one is the square of the other, because you would also have to square this. So modulo that. But you probably have seen or you probably have heard the fact that you can take closed string amplitudes, deform the contours, and separate the complex integrals in terms of two sets of real integrals. Each set of real integrals are now integrals over a disk. And therefore, they become integrals that compute for you these amplitudes that gives you Yang-Mills amplitudes. So you get gravity as the sum of a linear combination of products of Yang-Mills amplitudes. And there is something in between that KLT computed, which is a horrendous object that somehow merges these two things together. That horrendous object is so horrendous because it's the inverse of a nice matrix. And the nice matrix is the matrix made out of these objects. So if you're interested in that discussion, we can have it at some point. But I think I'll stop here.