So, um, well, we're very sorry that we couldn't have Carolina Figueiredo come and speak to us in person today because of the weather.
But we're really happy that we can have her come to
speak to us again, and she's going to tell us about some new results on…
How to calculate scattering amplitudes at large multiplicity. So, please, Carolina, take it away.
Well, thank you, thank you so much, Tom, and thanks.
For inviting me. Uh, so today I'm going to be telling you about some recent work that is still pretty much in progress, together with NEMA.
Alfredo, who's… who's here, Song and Chiang. Uh, which is about, uh, exactly as you said, sort of scattering amplitudes at large multiplicities, so large little n, okay, where n is the number of particles here.
So this is a regime where the sort of usual way you have to approach scattering amplitudes perturably.
becomes quite complicated because the number of diagrams grows, uh, very large, um, exponentially, in particular, for the theories that I'll be talking about.
Uh, so it seems hard to control what's happening in this regime, but what I'll show you today is that the amplitudes actually become remarkably simple at infinite n, and that we now have a way to understand that.
Okay? So, the plan of the talk's gonna be the following. So, it's gonna be divided into two parts.
In the first part, I'll start by just defining the problem, so I'll define exactly what scattering I'll be considering.
Uh, so the theory that I'll be focusing on is this simple, uh.
Scalar theory called Tr(phi^3) theory that I've talked about in the past here at one of these seminars.
And I'll define what are the sort of kinematics that we'll be considering when we go to large n, so this is going to be an important part.
Of just defining the problem. And then I'll immediately jump into just showing the sort of final results. So, what the amplitude looks like.
asymptotically in n, okay? So one very interesting feature that we'll see is that, um.
As we move around kinematic space, the answer will actually jump. So there are different asymptotic expressions, and the answer sort of jumps between these forearms as we change the kinematics.
Okay, and then in part 3, I'll tell you how we can derive these results, and all of these really comes from these new representation for trace phi cubed amplitudes via this tropical integrals, also called this global Schwinger representation. Anyway, I'll explain all of these words.
But as we'll see, these sort of new, uh, formulas for the input, you would make it easy to connect it to a sort of statistical mechanic.
problem about random walks, and that's how we can basically derive this asymptotic formulas. Okay, so all of these part one really could itself be a talk.
Uh, but we're extremely excited right now about part.
which is more recent, which is the fact that starting from these new representation, we can sort of derive.
an actual sort of dual theory for the amplitude that is valid at any n, but that in particular makes the large n limit.
Sort of very manifest and obvious. And from it, we can also derive all of the sub-leaning 1 over n corrections, so not just the asymptotics, but also.
the full expansion in 1 over n. And these dual theory is really associated to the motion of a particle on a halfline. So, the last part will be more summary, I'll just describe the main things. It's still the things that we're thinking about right now, but there were definitely.
Most excited about. Okay, so, so let me start.
So, a general introduction. So we will be studying this theory of colored scalars, the Lagrangians are following. There's a kinetic term.
A mass in this cubic interaction, so the key words to keep in mind is that they're scalars, they have cubic interactions, and very importantly, they have color, so this phi is an n by n matrix.
And in particular, at 3 level, this means that when we fix a color ordering for the external states, we only have contributions from planar graphs.
Okay, so at 4 points, that means that I would only have the S channel and the sort of T-channel diagrams.
And in particular today, I am going to be strictly focusing on the tree-level amplitude, okay? And one key assumption that I'll make throughout the talk.
is that the capital N, so the number of colors, is strictly larger than little n.
Precisely so that when I fix the ordering of the external states, I only get contributions from planar graphs, okay? So this is the.
So I will always be dealing with the color-ordered amplitudes under this assumption.
Okay. Anyway, so when we talk about amplitudes perturatively, there are two integers that play a role. There's the.
loop number and the number of external particles. Uh, and of course, the physics at large loops is interesting, as it's related with, like, what happens at strong coupling.
Uh, but, uh, these two limits are not, like, that far from each other. In particular, starting from a high loop.
process, we can sort of go down along these lines by cutting the loop propagators all the way into a sort of pure.
tree diagram, okay? This is what's called the leading singularity, and in particular, these objects form.
A significant part of the sort of full loop answer.
Okay, so this is sort of an obvious connection.
Uh, and in particular, of course, when we're talking about recoupling, the sort of.
2-2 scattering processes are the ones which are sort of dominating, but if we're interested in understanding what happens at strong coupling, the 2 to 2 is actually sort of exponentially tiny. Instead, the sort of amplitudes that will be dominating is N to N, where n is large.
Right? So… In particular, since I'm talking about a color theory, there's a naive.
hope that potentially something interesting could happen at sort of strong coupling.
some emergence of some dual theory. Of course, the best direction to see that sort of transparently would be to go to large L and large N.
But today, what we're gonna do is we're gonna sort of start going in this direction, uh, but just warming up purely within the recoupling regime, so just going at.
to large N at tree level, okay? And already see that even in this case, the amplitude sort of becomes simple.
at infinite n. Okay, so let's review the sort of standard approach we would have to make progress in these graphs, so going to larger N. Of course, we can sum over diagrams, but that becomes very complicated, because the number of diagrams.
rose exponentially with N, in particular, it grows like 4 to the n. Of course, we can bypass this exponential growth.
Using recursion relations, in particular, in this case, there are these famous Behrens-Heeler-recursor relations, which are just.
A reflection of the classical equations of motion, but basically all that we do is sort of decompose the endpoint amplitude into sums of products of lower points amplitudes, right? It's sort of a very intuitive picture.
And so, from these methods. We can, of course, make progress in this plot by building higher point amplitudes from lower point amplitudes by gluing, and then potentially go up via cuts.
Okay. Uh, but what we're gonna do today is something different. We're instead gonna try to, like, find a theory which lives directly at infinite n, okay? That does not have the sort of recursive.
character to it. Okay, and one last piece of motivation, one natural question is why are we studying this theory in particular?
Well, also from the previous talks that I gave here, I've explained that there's been, sort of, many different developments in finding new ways to formulate the amplitudes for this theory. In particular, all of this really come from this new formulation of open string amplitudes.
Uh, as a problem about curves on surfaces, and these new parameterization in the field theory limit gives us also a new way to write down.
these TraceFi cubed amplitude, okay, which goes by these global.
Schringer representation, which I'll review in a little bit.
But also, even beyond that, we've also seen that the amplitude for this theory can… are closely related with amplitude for other theories that we care a little bit more about.
In particular, for the scattering of blions and gluons, and as I'll mention at the end of the talk, it turns out that the large end results that we get for trace phi cubed can be generalized for pions.
via the connection that we know exists between the amplitudes. So I'll mention these briefly at the end.
But even beyond, like, scattering amplitudes, a lot has also been understood for the sort of flat space and cosmological wave functions for these trace phi cube theory.
Also, scenarios for which some of these results generalize.
Okay, just some general motivation. So, one last comment I would like to make before proceeding is that a question that I get many times when I talk about this, uh, is whether.
Because I'm studying sort of large and scattering whether the physics is becoming classical in some way, and there are sort of two.
points, uh, worth highlighting. So the first thing is that I will not be discussing the scattering of coherence states. Of course, we're all used to the fact that, I mean, we all know that if we have a classical way that has some large.
Amplitude, this is described by a situation where you have a large occupation number, so many particles with the same momentum, and that's what we call.
A coherent state, but for us today, that's not going to be the case. We're going to be considering a situation where.
The n particles all have distinct momentum, okay? And that's gonna be important.
And then another aspect is also we know that tree-level amplitudes are related with the.
solutions of the classical equations of motion. And of course, that's the case if I consider the scattering of N waves, say, all described by some plane wave like this.
Incoming, uh, then when I solve the equations of motion, there will indeed be a part of the solution where it will be described by an outcoming wave.
That has momentum equal to the sum of the.
incoming momentum, and indeed, the sort of coefficient of these Ps is exactly the tree-level scattering amplitude.
So, indeed, the true-level amplitude is contained in the solution of the classical equations of motion.
But at the moment, we don't… we don't see how using the classical, uh, equations of motion could give us any help understanding what this function is. So, in the early 90s, Volotion sort of studied the same problem.
But in that case, he considered the limit where all the particles were on threshold, so that the physics became effectively one-dimensional, and in that limit, the classical equations of motion only depend on time, so they become easy to solve.
And there, indeed, it plays a big role in helping you solve the problem. But precisely because here, we are interested in the sort of most generic momentum configuration.
at least at the moment, we don't know how to make use of the classical equations of motion to sort of get leverage on.
this problem. Okay, so… so let me start by defining the kinematics, uh, that we'll be considering.
So, for an endpoint amplitude, of course, because of momentum conservation.
Uh, if we list the momentum of each state according to the sort of underlying color ordering.
Uh, we obtain these sort of… Yeah.
Sorry, Caroline, I was just thinking about what you were saying. So, um… Okay, so if I go… write the equations of motion for phi cubed theory.
Massive. then I can at least formally write a solution.
Yes.
Yes, you can. Mm-hmm.
In terms of an integral equation, right? Okay, so I write 5 equals 5 incoming plus.
You know, the inter… box inverse 5 squared, or something like that.
Yeah, and then you do a delta phi… Yes.
So, can't… I mean, I always had the impression that I would just.
iterate that equation, that integral equation. as a perturbation series in phi incoming.
And generate level diagrams. What's wrong with them?
Yeah, and that… Yeah, that gives you the recursion relation, exactly. That is 100%… exactly. So.
iterating the equation of motion exactly as you were saying, like, sort of generates all the diagrams by.
exactly the sort of these. Uh, it organizes the information exactly, just like in the sort of recursion relation, where you sort of get.
There's sort of two possible sub-amplitudes where you basically divide.
into all the possible subsets of momenta. That is… that is what you get when you solve the.
Okay, so I get an expansion in FI incoming, right?
And the coupling constant G.
Yeah, exactly. But you always get, sort of, exactly. This expansion is, again, recycling the info… building the, like, endpoint wave in terms of the, sort of, n-1, and the sort of lower.
sub-point that you then need to sort of continuously recurse. So that is… that's what you get. What I'm saying is that if you really want to study the situation where you have.
A very large numbers of waves. Uh, coming in, sort of like, then that is too very complicated. It's equivalent of the solving this.
Recursion relation, which you can solve on the computer, it's just… it's just not telling you if there's any.
sort of simple formula. You can't… yeah. That is the… that's the… that's exactly it.
Okay. Okay.
Okay, so… Good. So let me just… let's get back to here. So we have this momentum polygon, which is a closed polygon due to momentum conservation.
And in particular, since we'll only be dealing with planar diagrams, all of the singularities are nicely captured.
by, sort of, when sums of adjacent momenta go on shell, and these sums of adjacent momentum squared.
are related with length squared of chords of this polygon.
Okay, so, for example, here, this length squared of D squared.
And so I'll define… I'll introduce these variables, Xij, which are just the length squared of the chord from vertices i to j, shifted by M squared.
Uh, and these are the variables that, when the xijs go to 0, the amplitude has a pull. Okay, so these are just defined so that the amplitude has a pull.
When these X's go to zero. On top of that, using the momentum polygon, we can also list all the diagrams by counting triangulations of this polygon, uh, with.
seen this also before, we get the dual graph by drawing a cubic vertex inside each triangle, and from each triangulation we get the term which is 1 over the product over all of the XIJs that enter on the triangulation.
Okay? So, this is something… anyway. So, then the endpoint amplitude is a function of these XIJs, which is just a sum over all the triangulation.
And then for each triangulation, I get these 1 over the product over the xijs of the triangulation. And in particular, I'll be assuming that the momenta live in.
arbitrary D, just so that all of the XIJs are independent to begin with.
But then at the end of the day, you can always sort of specify to the finite decays and just read off what your XIJs are, and input that in the final formula.
Okay, this is just for simplicity. Okay, so now, uh, to keep track of the kinematics, we are going to use, uh, what we call this kinematic mesh, and.
There's nothing to it. Our kinematics are labeled by two indices, i and j, so we'll just keep track of them in a grid.
Which we draw in this sort of 45 degree way for.
Uh, just because that's how we've been drawing it. For now, that doesn't really matter too much.
So each point on this mesh denotes a given XIJ, and when I go along this direction, I increase i, and along this direction, I increase j.
Okay? Uh, so that, anyway, for example, this would be at 8 points, these are all the possible chords of the octagon, and I would also already like to point out that the sort of.
Points that are at the boundaries of the mesh.
Do not enter the amplitude, so they don't really play a role in our.
Uh, in the space of kinematics, okay? They're associated with the variables associated with the external edges, so they don't enter the amplitude.
Okay. Okay, so the goal, the question that we want to ask today, uh, is what happens, uh, sort of at large n when we.
evaluate the amplitude on smooth kinematics, okay? And with smooth kinematics, all that we mean.
is that there's some function x of u and v, defined on this mesh, from which we read off our xijs, right? And we just read them off by discretizing this function on this grid.
Okay? So that is the… that's the… That's what we want to study. And it turns out that in… to understand what happens on top… the variables that will play a role.
On top of this XIJs, uh… Another sort of important linear combination of kinematics that plays a role is this discrete double derivative.
That we can associate with each little square. on this mesh. Okay, so this is just gonna be another variable that will be important in the analysis.
Anyway, so to inside each of these meshes, we can associate one of these C's, and that's just a discrete double derivative.
And one important thing to point out. Uh, is that there's a difference between the seeds that live in the middle of the mesh versus those that live in the boundaries.
And that's simply because the ones that live on the boundaries sort of touch these points that are not really part of the space.
Which means… and the consequence is just that while the C's in the middle are of order 1 over N squared, because they're proportional to a sort of discrete double derivative.
The ones on the boundary, uh, since the last X doesn't enter, end up being of order 1.
Okay, so I'll give an example that will make this clear in a second, but it will be important, the difference between these differences.
So, let's start by studying the simplest possible profile that I could pick, okay? So let's.
Consider the situation where it is X of U and V is a constant, okay? Let's say it's 1.
Okay? Of course, in these limits, when the all the X's are 1, then the amplitude is just counting the number of diagrams.
Okay, which is sort of the zero external momentum limit, or if you want the matrix model limit of this theory.
But before we proceed to what this is, so in this mesh, I would give 1, assign 1 to every single point.
Right? And so, when we read off these discrete double derivatives, it's clear that every C in the middle of the mesh is 0, because the ones cancel.
But the ones on the boundary are 1, okay? So this is just the difference that I was emphasizing earlier.
Okay? So, in this limit where we're just counting diagrams, then it is well known that the number of, sort of, planar cubic trees is given by these Catalan numbers, and these Catalan numbers have also a well-known asymptotic expansion, so at large N.
They grow, like, 40DN, so this is what we had before, this exponential growth.
And then there's some corrections, like enter the three halves, the square root of pi.
And then some further, uh, corrections in 1 over n.
I have a question here. You have N momenta.
Okay.
Yes?
But, um, N squared XIJs, so are there… relation from the excite is coming from the, uh.
No, yeah. No, that's, that's a great question. Momentum conservation.
The fact that you're expressing them in terms of the momentum.
Well, naively, there could be relations because of momentum conservation, so I am allowing the momentum to live in arbitrary D so that I don't have to be concerned about.
Oh, arbitrary dimension. Oh, because if you fix the dimension, I think there have to be relations with.
Yeah.
Possible… yeah, I am learning that. For, for, you know… There are definitely relations, and they're pretty complicated. They're given by these gram-determined constraints, so… so… but I… so in general, I will let them live in arbitrary dimensions, and then at the end of the day, if you want to specify kinematics at a fixed dimension.
You just impose whatever the X's are for that dimension, but I will be.
So, let's see, I mean, to get all the XIJ is equal to 1.
Assuming that they're independent. Because…
What dimension can do that, and…
Yeah, of course. Yeah, you need to always… well, if you want, it's the zero, sort of, external momentum limit, but… but yes. You… you would need… Yeah, I mean, but the thing is that…
Oh, I see, all the PIs are zero? Then all the XIJs are just M squared.
Yeah, exactly, all the XIJs are just M squared, exactly.
Okay, so you didn't really mean one, you bet.
And so it's just 1 over M squared. Yeah, it's a constant, should have units of mass squared, obviously, so I am… yeah, so… exactly, yeah.
Okay, yeah. Okay.
I mean, it's, it's… exactly. Okay, so… Yeah, so this is just an example.
And already from here, we learned that we have these naive expectation that in a limit where we're just counting diagrams.
Uh, the amplitude sort of behaves exponentially with n for the Catalan cases, this log 4, but what we want to ask is basically what happens to this factor as we start turning on some smoothly varying function.
X of U and V. Okay, so the naive expectation is that we will instead find some.
function of this profile. Um, and that's what we want to understand, okay? So basically, what we want to learn is what does the log of the amplitude over n asymptote to, so that we learn these sort of leading piece.
And the interesting thing about assuming these smooth kinematics is that it makes obvious that it's not like there's any class of diagrams that it's sort of dominating. Instead, what's happening in this limit is that the whole sum over diagrams is sort of smoothing out in some way.
Okay? Okay, so one last comment about kinematics, just to maybe keep some pictures in mind. Of course, asking for a smooth function of the XIJs on the mesh is completely equivalent to saying that you can read off your momentum polygon from some smooth curve.
For example, for the octagon that I drew earlier, we can think that I got it from this underlying curve.
And I can obtain the sort of large N limit by putting down more points along this curve.
following some distribution. Okay, so that can be… in that case, I will, of course, land on some smooth function.
on the mesh. But there are other things you can think of. One particular scenario that we also like is.
this sort of bomb situation where there's one particle that has, like, a very large momentum.
And then it sort of decays into many, many particles that have much smaller momentum, so this will be one potential curve to read off such kinematics.
And because most… a lot of the discussion that I'll have from now onwards will demand certain sort of signs for the kinematic invariance.
Uh, this situation of these sort of bomb exploding.
is actually one that will be captured by the sort of.
things that I'll be talking about in a little bit. Okay, this is just so you have a picture in mind.
Okay, so given this… I'm working in the signature mostly plus.
Sorry, what signature are you working in?
Okay, but it is Laurentian. So these are space-like momenta.
Yeah, it's Lorentzian, so yes, it is Lorentian. Yes, exactly.
Okay.
Yeah. Well, or if you want, this XIJ is already shifted by M squared, right? So there will be time-like.
But the M squared is compensating such that these XIJs are positive.
Okay, but the energy… The time component can't be too large, that's the way I'm trying to understand.
Yeah, that's exactly why here I'm saying it's restricted, for example. Here I'm saying that the delta T squared is smaller than M squared, precisely such data.
Okay. I'm here.
That's fine. Exactly. Okay, so now let me jump to tell you what the final answer looks like, okay? So, it's gonna look a little bit unmotivated, but… Just stick with me for a little bit, and then I'll show how to derive all of these results.
So, it turns out that the variables that will be important in these sort of large-end formulas.
are some interesting combinations of these X's and C's. Okay, so… Let me define AK to be all of the X1K.
So, in all of these stories, there's… you always pick some sort of underlying.
reference triangulation, which I'll get to in a second, or if you want an underlying reference index.
So in this case, one is sort of playing a special role, but we can access all the other possible choices using the sort of cyclic symmetry of the answer.
So it's just a choice. And then we also define these interesting combination BK, to be this difference between one of the Cs and X1K.
And then C-Star is this other C. Okay, anyway, so at the moment, this looks a little bit random, but it's just so that I can show you how simple the answer is, and then we'll see that these quantities naturally arise from the analysis.
So, let's look at the regime. For example, where all of the AKs are smaller than the BKs.
Okay, speak something that that's the case. Then the claim is that at large n, the amplitude asymptotes to the following simple product.
So, product over K1 over AK plus 1 over bk.
Okay, so these wiggly arrow just stands for that the log A over N convert… goes to log B over N. So this is, again, describing the leading.
piece in the exponential. Okay, so the sum over all of the diagrams sort of collapses into this single.
product. So, anyway, let me show you just some numerical evidence. So, again, since this… for the scalar theory, we can actually compute the amplitude up to.
pretty large, and using this Baronsula recursion relations on a computer.
And so, in this case, we picked the sort of AK and BK profile to be these two lines, so that AK is always smaller than the BK.
And here on the right, we have the log of the amplitude over n in blue.
And the log of A0, which is this formula over here, over n, and we see that, indeed, the amplitude's converging to it, where the offset is of order.
1 over n. And already here, I would like to remark that kind of a funny aspect of these formulas, these asymptotic formulas.
is that they seem to have poles in locations that, like, no diagram has poles.
Of course, this is not a genuine poll, precisely because the region where the formula.
uh, holds does not let BK go to 0.
Okay. But anyway, so this is one result. Another result, for example, say that all of the AK are larger than the BK, and then there's another sort of funny quantity that plays a role, which is these ratio of AK minus BK over 2C star.
Anyway, I'll motivate all of this in a second, and say this is increasing in K, so you read off the kinematics of this particular profile.
Now, the claim is that at large N, the amplitude also asymptotes to a very simple product.
Which, in this case is just 1 over… 4 over AK plus BK.
Okay, and here's the, sort of, numerical results. Okay, um… Anyway.
But there are other more interesting regimes where it's not always the case that I can give you a closed.
I mean, like a closed-form expression. For example, if you look at exactly the same regions, but now we have that this ratio is decreasing in K.
Then the claim is that the amplitude asymptotes to these also simple product, but now where the AK is shifted by something dependent on these.
Uh, sigma bar, where sigma bar is the solution to an interesting, sort of, this interesting difference.
equation, okay? Anyway, so, and here is the sort of numerical evidence of that, so we have this… a particular profile where it's decreasing with K.
This is the sigma bar that solves the equation, and here's the comparison between the amplitude and these sort of prediction.
looks to me like there are many roots of that equation for sigma bar.
Okay.
Well, the sigma bar will have to follow some particular property. It actually has to be bounded between 0 and 1. You don't know that at the moment, but you'll learn that in a second. And when you do that, then… There's an optical…
I mean… okay. So the equation… the equation for sigma bar is really a high-order polynomial equation.
And so you're saying there's a unique route between 0 and 1, and that's the one you choose?
Yeah, at large, and this can really be turned into an integral equation, right? This is… this sum can be turned into an integral.
And I'm claiming that there is a… there is a single solution, yes.
Okay.
Yeah. Okay, so now let me actually show how we can derive all of these results that I just showed you.
Okay? So… Okay, so to do this, I need to explain these sort of tropical representation of the amplitude, or this global Schwinger representation.
And so let's start with a trivial example. So at four points, we know we have two diagrams. So we have the 1 over X13 plus 1 over X24, S in the T channel.
And of course, we can Schringer parameterize, for example, the S channel by writing it as this integral, right? This is trivial.
And similarly, we can do the same thing for the T channel, write it as another sort of similar.
integral, but now it's also easy to see that we can easily combine them both into a single integral.
Uh, by basically deforming the action by a factor like this.
Okay, so we start out with the Schringer parameterization of the.
S channel, and in this Macs. is precisely such that when T13 is positive.
Then it gives me 0, so on this axis I get the Schringer parameterization of the S channel.
But when T13 is negative, it precisely fixes the exponent such that on this axis, I get the Schroinger parameterization of the T channel.
Okay? So… this seems… this… this is… trivial. And then it's also clear that at larger N, for each diagram.
We can write a term like this as an n-3 dimensional integral with the respective Schrodinger parameters TI, right? And it's also obvious that now, like, if I try to.
glue all of these different integrals together to sort of cover the full n-3 dimensional space, the amount of work that I would have to do to achieve this, it's… exponential in n, right? Because I would have to do it for every single graph, and they're exponentially many graphs.
Now, the remarkable thing is that this is exactly what the surfaceology, or these new… these.
Formulation of string amplitudes does for you at, like, low energies, okay? So these new parameterization, when you.
Sort of take the low energy limit. gives you automatically a global Schringer representation, exactly like the one that we had at 4 points.
But now, crucially, where there's only a polynomial number of these max functions.
And these polynomial number of max functions divides the whole of the integration domain into exponentially many cones, okay? Each of which corresponds to a diagram.
exponentially many cones, or regions where the action becomes linear.
Okay, so anyway, that's why this is called a tropical integrals, because it has these max functions everywhere.
Okay? So, anyway, I don't have time to derive.
Like, how to get these tropical integrals, but I'll give you an example at higher points, so at 6 points, just so… because they follow a very simple and predictive structure.
Uh, which will be important for our analysis. Okay, so at higher points, uh, so we always start out by picking some underlying base.
diagram, which we like to pick it to be the one that has the propagators X1K, that's why they play, uh.
As a special rule. And so we start out with the Schrodinger parameterization for.
this diagram. And then what we do is we extend the integration domain to the whole space, so to… from mice to plus infinity.
But we deformed the action in the appropriate way, okay? And so, uh, and this is very nicely encoded.
in terms of the mesh, what we have to do is, for the first layer of C's in the mesh.
The ones that are close to the left boundary, we add each of these multiplies a max function that only involves one of the Schwinger parameters.
So we see here each of the T's appears individually, and then we progress along the mesh.
So that then we get another set of maxes.
Multiplying the sort of next layer of C's. Now, these maxims involve intertwining two different T's.
So they involve sort of sums of consecutive T's.
And two different T's. And then finally, there's the sort of last.
Max, or the last C, which is the sort of most complicated one that involves.
all the different T's, okay? So, that involves sort of sums, sorry, differences of adjacent T's.
And now the claim is that if we evaluate this integral over the whole of t space.
Then we get all of the different diagrams. Okay.
Okay, but now let's recall that for the large and kinematics, we learned that, uh, the sort of internal C's.
were of order 1 over n squared, while the boundary ones were of order 1, right? Provided that we specify some smooth function on the mesh, from which we read off our kinematics.
So this means that, at least a leading order in n, we should be able, like, we might as well sort of drop.
The contributions that come from the C's in the bulk of the mesh, and we can just look at the integral where we only have the boundary C's.
Of course, this is a little bit subtle, because each of them is of order 1 over n squared, but they end up being order n squared C's in the bulk of the mesh, so it's not so clear whether their effects could not build up.
Anyway, in certain cases they do, and I'll discuss that at the end, but for simplicity, let's just assume that.
The effects don't play a role at leading N.
Okay? So in that case, the integral that we get becomes fairly simple, and I can write it for general n.
So… so we have a piece that I'm gonna call S0.
Which is the part of the action where all of the Ts are decoupled.
Okay, so this part involves the X1K, and also these other boundary C's, the ones that live.
In this boundary of the mesh. And then there's a final piece, which is proportional to this corner C, which involves the sort of biggest max that intertwines all the different integration variables, okay? This is the only part.
That makes the integral non-trivial. Right?
And in particular, if we look at, like, the sort of S-naught, then we can look at each individual S-naught of K, and we see that.
For positive TK, it is sort of linear and it has a slope which is given by what we called AK before.
And for negative TK, it's also linear and has a different slope, which is exactly the variable that we called.
BK earlier. Okay, so that's where these variables are now appearing.
Okay, so to make things simpler, even to begin, let's imagine that we further set C star to 0.
Then we just get the integral over snaught, which is now a fully decoupled integral in each TK, and so from here.
We just get the product over 1 over AK plus 1 over BK.
Okay, so this is just this result. And that's… this quantity we're going to call A0.
And so, having defined this A0, it's natural to look at the ratio of AN over A0, and interpreted it as the expectation value of eta minus S1.
Under the probability distribution given by IRA minus S naught.
Okay, so that's just, I mean, just a way of.
rephrasing what the interval is. And in particular, we have in this probability distribution, we have that each TK is a random variable, uh, where, you know, at each step K, the average step size in the positive direction is 1 over AK.
And the average step in the negative direction is 1 over BK, and as we'll see in a second.
Uh, the sort of random walk will precisely emerge from these probabilities distribution for each.
When we… for the computation that determines the amplitude, this will become clear in a sec.
Okay, but so now the goal to understand what happens at large n is going to be to find a good upper or lower bound.
Uh, that the amplitude saturates at large in. Okay.
So, well, there's a trivial upper bound that we can consider.
So if we recall what S1 is. Right? Uh, then we can see already here that, uh, well, since it's a max and there's one entry which is 0, then clearly S1 is positive.
Right? If C star is positive, we're assuming. So then S1 is clearly positive, and therefore, these expectation value is small or equal than 1.
So we have that AN over A0 is bounded by 1.
So that's a simple upper bound. And also, we can trivially get a lower bound just by using Jensen's inequality, which tells us that, anyway, the expectation value of e to the minus s is greater or equal than e to the minus the expectation value of s.
Because the exponential is a convex function. Okay, so doing this, we clearly see that the amplitude is lower bounded and upper bounded in a simple way.
And so we can now start asking what happens in certain kinematical regimes.
So let's say, for example, that 8K is smaller than BK for all K.
Right, then what happens to S1? Right? Well… If all the AKs are smaller than the BKs, then clearly the expectation value of each TK is going to be greater than 0.
And if we look at the interpretation of S1, since it's the sort of max.
Of sums of consecutive Ts. Right, here's where the sensing, which is sort of.
A random walk is appearing. We care about how far did it go in, like, T space.
Clearly, since all the Ts wanna be positive, the bias random walk is gonna go in the positive direction in T space, and therefore, this max is going to be mostly dominated by the zero.
Right? Okay, so that means that the expectation value of S1 is at most of order 1.
Most importantly, it does not grow with n. Okay.
Okay, so, so, so that's great. That means that, uh, so if S1 is of order 1.
Uh, then that means that the number that sits in this exponential is also of order 1.
And since A0 itself is already exponential in N, because it's this big.
product, then we've managed to sort of lower bound the amplitude by two things which are both of.
Upper and lower bound, amplitude by 2 things which are of order 1.
So therefore, the sort of coefficient in the exponential of AN must match that of A0.
Okay, because we know that A0 already goes like E to something times n.
Okay? Anyway, so that's the… so that is the sense in which we've managed, indeed, to prove that the log of AN over n is converging to the log of A0 over N.
Okay, and this is just the result that we showed before. So we have that for these kinematics.
The amplitude is just gonna convert to this A0.
Okay, but now let's say we consider a different regime, for example, where all of the AKs are greater than the BKs.
Then now, uh, for each TK, its expectation value will want to be negative, so that this bias random walk is sort of going in the negative direction.
And therefore, we have that these max that defines S1 will be mostly dominated by these last entry.
And therefore, we have that as the expectation value of S1 is going to be something of order n.
Okay? So now we learned that in this limit.
the sort of upper and lower bound that we got are pretty bad, because here we get these e to the minus n, which clearly goes to 0.
At large ends, so we just learned that these ratios between 0 and 1, so… uh, didn't really learn anything. So clearly, we need to come up with some better way of.
upper bounding or lower bounding, the amplitude. Okay? So the strategy's gonna be, we're actually gonna find a better upper bound.
that DM would you sort of saturates at large N.
We will not be upper and lower bounding it, we'll just find a better upper bound, and I'll claim that the amplitude saturates it, we have a way to prove it, but that involves these dual theory that I won't.
be able to get doing such great detail. But let's just see how it is trivial to get a much better upper bound already. So, the original upper bound that we had was, well, we looked at these max, and we said, well, one of the entries is 0, so of course it's greater or equal than zero.
But in general, if we have a max of 3 different things.
It's always true that it is greater or equal than any weighted average of its entries, okay, where these weights, W, have to add up to 1.
Okay? So, this is a much more natural upper bound to consider, and so if we do that, we can sort of upper bound AN.
by what we get, by replacing the max. by the weighted average of its entries, okay? So that leaves you something like this.
And then what's nice about this is that after doing this, we've again fully decoupled all of the different integration variables.
Right? And so, after we combine these two exponentials, we're just left with a product.
Over K, but where now the sort of AKs and the BKs are redefined.
by some function of these weights. Okay, so… so the final result that we get is this simple product where there's a new, sort of.
AK tilde and. BK tilde, which are shifted by C star times the sigma K's.
And plus C star times sigma k, so they're shifted in a way such that there's some.
remains constant. Okay? And D sigma K's are just simply the partial sums over the weights.
Okay, um, and so… So this means that, I mean, since the weights have to be positive and add up to 1, we just learned that these sigmas also sort of need to all live.
between 0 and 1. And they must be ordered as well, because the weights are all positive.
Okay? So, anyway, so here's a new upper bound on the amplitude. We have that AN is smaller equal than this.
A0 of sigma, where A0 of sigma is a valid upper bound for any sigmas that live inside this domain.
Okay? So now the question that we want to ask is, given some kinematics that specify a profile of AK, BK, C star.
What is the choice of sigmas inside this domain that gives the best possible upper bound? Or in other words.
That minimizes this A0 of sigma. And what we'll find as a bonus is that it turns out that the amplitude actually saturates these best possible upper bound at large n.
Okay. So… so let's start by asking a simple question, which is, what is the minimum.
of this A0 of sigma. Okay, uh, this, that's a… I mean, in particular, since ANR sigma, each sigma K only appears in one factor, that's a trivial question to ask, so it depends only on the factor.
And the sigma k that makes these A0 naught.
Minimal is the one such that the A tilde K is equal to the B tilde K.
Okay, so that is one for which you have that.
Sigma star K is AK minus BK over 2C star, okay? So this is the sort of optimal, the minimum of A naught of sigma is.
Um, when sigma k is equal to this. Okay, but of course, the second question that we should ask.
is whether this minimum lives inside the allowed domain, right? Because it might be that when I specify AK in the BK, these sigmas live outside the domain.
Um, and so if they live outside the domain, like, what is… what is the best possible upper bound, so what's the best value of sigmas that I can pick that is still within the domain, right? It's gonna live now in somewhere in the boundary of this domain.
Okay, so… so let's see how from this… this we can, uh, like, do some analysis to get already the old case, so let's look at the case where all of the AKs are smaller than the BKs.
Then clearly, the sort of best possible value is negative for all.
Okay, right? So some… so the kinematics tells me that this.
Sigma star K is some curve that lives beneath.
Uh, the axes, right, beneath 0. And so that means that the best possible upper bound that I can get is the situation where all of my sigma k's are equal to 0.
Right, within the domain, that's the closest I can get.
Okay, and so, indeed, in that case, we have that A0 of sigma precisely turns into our previous product.
Okay? Anyway, so this is just restating what we had learned before.
But now we can ask, for example, what happens when AK is smaller than… sorry, it's greater than BK.
But also, let's say that the sigma star K is greater than 1, so we have some curve like this, then the best possible upper bound is the other limit, where all of the sigma k's are equal to 1.
So we have this other expression. And anyway, so here, just a, again, a numerical result, sort of showing a situation like this, where the sigma star K all live above 1.
And this is the comparison between the upper bound and the actual amplitude.
Anyway, then there's a sort of, uh, what we call the tracking case, which is a situation where the sort of sigma star K.
All leave between 0 and 1, and are increasing with k, which means that this curve takes.
some form like this as a function of k. So, of course, in this case, the minimum will happen when we pick the sigma such that they precisely agree with the sigma star K.
And in this case, the amplitude just turn this A0 sigma turns into this 4 over AK plus BK.
This is precisely the regime that lets us access the sort of zero-dimensional kinematics.
That corresponds to setting all of the AKs to 1, or M squared, say, and the BKs to 0, and precisely, then, this gives us this.
40DN growth, as we expect from the Catalan asymptotics for just counting diagrams.
Could you translate that back into what the particles are doing?
Okay.
Uh, what do you… like, well, here, this is just a case of, like.
I mean, you know, you started with a problem with momentum of particles scattering.
Okay, so now you've got some interesting regime where you could say something about the large N, but what does that mean? How do I think about this in terms of the particles?
The momentum of the various particles.
Yeah, exactly. So, so you're asking, like, I mean, given a particular sort of, uh, curve that specifies my momentum, say, uh, like, which regimes will I be landing on, right? That's, that's, that's what you're asking.
Yeah, I mean, I understand that what could translate back, but it… I mean, there are a lot of equations, so… Um, I mean, so you have the exact… you have some result, symptotic result for the amplitude.
Now I want to… okay, so that's the answer to a question.
But I want to know, what is the question that… to which that is the answer?
Yeah, that is now going to depend. Yeah, that will depend, like, on the… on the particular, sort of, curve that you pick to read off your kinematics. Like, that, I mean, uh, so, I mean.
given a curve, for example, that, like, sort of massless, uh, that, sorry, massless, no, the sort of case where I describe as the bomb.
Where I had something like this. Uh, where, you know, there was a particle with a big momentum that was then decaying into a bunch of small momentum.
Oh, okay, the bomb is a goodie, too. Yeah.
Yeah, that is a situation where they would live… but depending on the exact shape of this curve, you might be in these different, like, regimes.
So that's something that you need to sort of.
I see, so you can't say for the case of the bomb, this is the amplitude.
Uh, defined from the.
Because there's still too many choices, and the choices of the outgoing Momenta PI.
Of the… of the things coming out of the bomb.
Yeah, I would need to double, like, because… Yeah, there's still some choices about… Yeah, like, of exactly what… Exactly, exactly how I put this curve, like how I draw this curve, uh, it might be, uh, that.
Yeah, I don't… I'm not sure if it's gonna… it depends which regime I'll land on, exactly.
But this is a regime that is described by Diskyn Mars that I'll be talking about, like, that's… Okay.
Anyway, so… so this is a… this is just the comparison of, like, for example, in this case, I pick my sigma star K as this curve.
Which is increasing with K, it starts off. Underneath zero, so clearly the best thing that I can do there is put all my sigma k's to 0.
And when he goes above 1, I need to stay at 1, obviously.
And so here's again the comparison between the amplitude and this upper bound.
So, let me discuss one final thing, and then we're ready to say what happens in general.
So, there's another case where I could have, which is the situation where my curve lives between 0 and 1, but now it decreases with K.
Right? Now, clearly, I can't have my sigma k struck, because it's decreasing, so… so since they have to be increasing, the closest that I can get to be decreasing is the case where they're all equal to a constant.
Right? Anyway, and so this is where we obtained back the sort of equation that we got before, which is that, well, the best thing that I can do is put them all equal to some constant, sigma bar.
And now I have to basically vary. sigma bar, uh, and basically find the value of sigma bar that makes this minimum.
Right, and just the minimizing this function, where all the sigma k's are equal to some sigma bar, gives me this simple.
Uh, equation, right, which turns into an integral equation for very large n.
Anyway, so here's, for example, again, the example I showed before.
But so therefore, this means that if I have some.
Uh, interesting sort of profile where many different things happen to find a sort of best possible upper bound, all that I need to do is look at each piece, right? And for example, at the beginning.
If it starts out negative, then the best thing that I can do is put all the sigmas.
to 0, then while the function is growing, I can track for a while.
Then eventually, since this curve. turns around and it will decrease at some point.
I clearly, I can't track forever, so I will need to switch into some constant.
Uh, and then eventually it will track again, and.
stay at 1, right? And now, the best possible upper bound is determined by asking what… I mean, so I vary over these K star, so the point from which I switch from tracking to, sort of, constant.
And find this value of k star for which this function is minimized.
Okay, so anyway, here's an example of, like, some numerical situation where it's like that, so… This is the best possible upper bound corresponds from this profile of sigma k.
And here's again the comparison between the amplitude. And these upper bound.
Okay, so anyway, so this finishes this part of the discussion. Uh, the main point to emphasize is that, well, we learned that the asymptotic behavior of the amplitude is.
given by this interesting optimization problem, so it's this minimization of this function.
Uh, but what's interesting about it is that it makes manifest that the complexity of this optimization problem.
is not really determined by N itself. Instead, it's determined by the number of bumps in the kinematic profile that we pick.
So in other words, if we say that we're reading off our kinematics from some smooth.
kinemax or some smooth curve, then these underlying curve that I draw here is already predetermined.
Uh, so increasing N is only going to discretize this more and more, but the complexity of the optimization problem will be exactly the same.
Okay. Anyway, so I know I have only, like, 7 minutes left, so I'm gonna be fairly quick on these.
final part, but I definitely would like to at least describe to you this sort of, uh.
Second part, which is about these sort of dual theory that we have for the amplitude at.
NEN. So… The idea is the following. You start with the previous integral that I wrote before, with these interesting max.
And you can ask, well, can you think of it as some sort of path integral?
And something which is very… quite unsatisfactory is that it has a very non-local character.
in tea space, right? Because to evaluate these max, you need to know some global information about all Ts to know what it turns into, right?
And so the idea to get these dual theory is that we can actually sort of.
massage this integral and turn it into a one higher dimensional integral, where we completely destroy this max and turn it into a purely local integral in a new space.
Okay, so let me show you what the integral looks like.
So these new integral is now given in tau space, where differences of tau, adjacent tau's are what we previously called ts.
Okay? And the integral that we get is sort of describing the motion.
of a particle on a halfline. Okay, where this piece of the action, you should recognize as sort of all, like, like the… The full length of some path.
In tau space, there is discretized into n steps.
Right, since here we have just the sort of differences between the steps.
But it has an interesting metric, if you want, in the sense that.
You know, each length is weighted by some factor, which depends on the kinematics.
Okay. Anyway, so that part is just the main action, and then on top of that, there's an extra factor that sort of tells you that the path.
cannot be very deep in tau space. It should be the center of mass of the path should be sort of close to zero.
This factor, of course, is very important, because since these P's.
is fully translational invariant in tau, this integral would be divergent if this exponential suppression did not exist.
Okay, so this is the integral that we have over in tau space.
And I mean, we say it's a particle on a half line because since the tau's all range from 0 to plus infinity.
This corresponds to some one-dimensional path. That always stays above.
the origin in tau space. Where n now is clearly the sort of number of time evolutions.
Okay, so… Let me start by describing to you what the usual sort of diagram picture.
turns into, in this dual theory. Uh, so… Anyway, we're integrating over all such paths in tau space, so all such pictures like this, so a very simple way to, you know, dividing the paths into different classes.
is by listing the location at which they reach a minimum.
Alright, so say, anyway, this path has a minimum at k star.
And so we can divide the sort of integration domain.
Uh, into the sort of places where tau of k star is the minimum.
And so if we do that, we're clearly left with a path on the right and a path of the left, which are of the same nature, such that this precisely… anyway, decomposing the paths in this form.
leaves us with the sum over k star of what looks like the sort of amplitude on the left.
times the amplitude on the right, because we get all possible such paths on the left and on the right.
Okay, so this is precisely the analog of this Behren's Euler recursion relations.
Okay? And it's clear that if we keep iterating, so for the path on the left and the path on the right, we pick the sort of place where they reach a minimum.
Then eventually, we'll just land on what a diagram corresponds to. Okay, so diagrams correspond to regions in tau space, which can be easily read off… anyway, I'm going to skip this part, directly from the graph, so… They're just basically given by some particular region in tau space.
Okay, so the point that I want you to keep in mind is that.
Diagrams are fairly global. in tau, in the sense that we need to know something about an ordering of all tau's, and they build up the full answer in an additive way, right? Each chunk.
of tau space corresponds to a diagram, we have to add them all up to get the large… to get the full answer. And that's why these sort of large n limit is not really.
very obvious from the Feynman diagram picture. But instead, once we have this purely, like, local integral in tau space, a much more natural way to approach it is instead to ask what happens when we integrate out the last tau.
Right? That's sort of the analog of what time evolution would look like. It's an operation which is sort of purely local in tau space.
And if we do that, if we integrate out the last tau, we're left with some function of tau n-1, right? And if we keep on going, then we're clearly building up the full answer in a multiplicative way, right? Because we're carrying out the multiple integrations one after the other.
And this is the reason why this new picture is very useful to make the sort of large N limit obvious.
Okay, um… Anyway, some interesting sort of things that are… that make it… that are extremely counterintuitive from the standard picture that we learn.
from these, sort of, as a consequence of this locality in this tau space, is that, for example, in the regime where all of the AKs are smaller than the BKs.
Using this picture, we can derive the direct sort of prefactor that corrects the A0, and it depends exclusively on the kinematics of the sort of last part.
Of the or. Near the end of tau, so AN-1 and BN-1.
And all the corrections that come from changing the kinematics at the beginning are non-perturbative in N.
Okay, so this is, uh, quite remarkable. Anyway, I don't… I don't… since I don't have much time, let me just skip this part, or just a comment.
But now let me tell you how we actually use this to sort of compute the answers. So, uh, since we have these.
Uh, integral one natural way to think about it is through the transfer matrix description.
Where we think about this sort of one-time step as being this integral transform, or this operator.
Which takes some function of tau prime. and integrates it against this kernel to give a function of tau.
Okay, so this is… anyway, this… Turns out to be equivalent to this acting with these operators, so this is the analog of these.
time evolution, and you can ask, what are the eigenfunctions of these operators? Turns out that they are simple linear combinations of.
Cosines and sines with this particular set of. eigenvalues.
And so, uh, using this, then we can now fully just solve our integral.
Uh, just for simplicity, let me consider an example where all of the AKs are equal and the BKs are equal.
Okay, just for simplicity. Then, really, the amplitude just turns into this integral, where we have some.
Function that depends on the end time and the beginning time, and then we're basically acting with these.
time evolution and times. So that, it really is just the Euclidean transition amplitude for this particle on a halfline that we evolved for a time.
N. Okay? Anyway, and so, of course, knowing the space of eigenfunctions, we can just decompose the in-states.
and projected into the eigenfunctions, and then just pass it through the integral, and when doing that.
Uh, we find, and we get precisely the sort of correct.
Uh, asymptotic factor out front. And then we're left with an integral over the sort of frequency space.
Uh, from which, if we perform a trivial saddle point analysis, we can extract all the subleeding.
1 over n correction. So, for example, here you see.
The appearance of these end to the three halves.
Which famously appears also in the. Catalan numbers, okay?
Anyway, but so, let me just finish with some sort of general comments of what this picture also makes manifest, is that even if we go to other regimes where we have some effective at K and B tilde K, and they're different.
We can just ask, what is the spectrum of.
eigenvalues of this operator. It's clear that since we're acting with this operator many, many, many times, uh… Then we're mostly gonna be dominated by the eigen… function that has the biggest eigenvalue, right? It's like we're converging to the fixed point, if you want, which just comes from the sort of biggest eigenvalue.
And we see that depending on the values that we choose for… depending on the kinematics, these biggest eigenvalue is going to change. In particular, in the regime where they are equal to each other, then the spectrum is such that there's.
No gap between the biggest eigenvalue and all the other eigenvalues, and precisely because there's no gap there, then power law corrections to the sort of leading behavior in N.
But as we change A tilde K and B tilde k, then the spectrum actually develops a gap.
Okay, and so the biggest eigenvalue. is the sort of, uh, is gapped with the continuum.
And therefore, we see that in this regime. Uh, we… every sort of sub-leading correction is exponentially small.
Okay, so that is… anyway, so we can just analyze what's happening by looking at the spectrum of these operators.
Okay, so I'm gonna finish now, just give you my outlook.
Um, and so… So, for future things and that we're still understanding, of course, if we want to understand the full 1 over n expansion.
Clearly, we do need to account for what happens when we add back the internal C's, because even though they're order 1 over n squared, they're n squared of them.
But it turns out that for the leading order in n, there are certain regimes where they just don't matter.
Uh, but then there are other cases where they do matter, and we already know how to account for them, so the formulas that we… that I wrote before are slightly deformed by being shifted by some sums of some C's inside the mesh.
But this is all just for the leading behavior in N, but of course, to know the full systematic 1 over n expansion.
Uh, we need sort of more than this. And in particular, these dual theory that I described only really holds for the case where I don't have the internal C's.
Uh, but, uh, this is really what we're thinking about at the moment, and while… so I told you that the nice thing about this integral is that it only involves, sort of, interactions between the two nearest taus, or terms that involve the two.
Adjacent tiles, but it seems that when you turn on all the internal C's.
The generalization of these integrals is just something where.
We have some kinematic factor, and then now, instead of having just 2 adjacent tails.
The towels are compared between certain integrals, or sort of strings of indices that are determined by this kinematical indices.
Okay, so it's a remarkably simple formula for the amplitude. We're just trying to understand exactly how.
How we can use it to extract large and behavior as we did for.
This previous case. Okay, just very quickly, the connection between large and trace phi cube theory also gives us large and pions via the simple kinematic shift that exists at the level of the answer.
Uh, so if we apply that, we learn that the large N pi on amplitude is also sort of.
Remarkably simple, and we checked all of this also numerically.
Another interesting generalization is actually to string amplitudes. At large multiplicity, and also at high energies, so the claim is that.
For the field theory one, you have this optimization problem.
Uh, then to get the, the. String theory one, all that you have to do is you uplift these local 4-point amplitude into the full.
Beneziano 4 point, and. From here, when you extract the high energy limit.
Uh, it turns into a factor like this. Uh, and then you, again, you.
optimize this new function, and that's, uh, what. what strings do at high energies and large N. So this is related with the fact that.
Turns out that the solutions of the scattering equations, like, we have a new form of writing down the scattering equations that makes manifest at large n.
The solution sort of dominated by a piece which is much simpler, and that's related with this term.
Okay, and that's it. So, of course, the main goal would be to understand what happens for completely general kinematics, even beyond what I described before.
I told you about the connection with pions, we also know that there are ways to relate it with gluons, so it would be interesting if we could say something in there.
And of course, beyond tree level, understanding what happens at higher loops.
is something that we'll definitely think about. Okay, thank you so much.
I'm… I'm done.
Okay.
All right, let's thank Carolina.
Are there, uh, more questions?
Yes.
Can I ask a question? Okay, so, do I understand correctly that if I would like to calculate a cross-section in the three approximation.
You don't have control over all kinematics. That would allow you to do the cross-section calculation.
Or can you estimate. Actually, what the cross-section is, when I integrate of all of phase space.
2N.
Yeah, no, like… Yeah, one thing you would need also, on top of that, is that you would need to sum over color ordering.
So that would be actually, like, that is sort of another aspect that I didn't discuss that would be important. And that… so, if, you know, if you have some smooth kinematics in a particular color ordering, in a different color ordering, that might not look.
So smooth anymore, right? So, to compute that, so I would not know, uh, how to do.
the full cross-section at the moment. Even just because of the step.
Okay, in a sec… And the second question, then, so this was massless, right? Phi cubed.
No, actually, so… so yeah, actually, no.
Because just to make the… if it's massive. then you have to have an interplay that for a given energy.
You don't… you can't produce more than a certain number of particles.
So N would be bounded by the mass and the total energy.
So, I didn't see this consideration in what you were doing.
Yeah, exactly. So, like, I mean, that is just sort of all captured in my XIJs, like, so, for example, I had that description, sorry, of a situation where I had a particle with a large momentum, and then everything else had a, like, very small.
Momenta, right? So everything else, I drew some curve that looked like this. What I was reading off.
Anyway, so if you… this was the picture where I had a particle with a big mass, which was sort of decaying into particles all with, like, a very small mass.
So, of course, I am always assuming that the masses are such that this decay is possible.
Right, and that's just kept through by my XIJs, if you want, they're being precisely.
But the kinematics is far off shell, isn't it?
built such that the process can happen. But one actual… The kinematics is off-shelf, it's true. Yes, the kinematics here is actually off-shell, because… precisely because… yeah, I mean, so in other words, really, like, I'm more precisely dealing with, uh, the correlator, but for the scalar theory, a correlator and the amplitude, I mean, they're related just by a factor out front, so that's why it doesn't.
It doesn't make a big difference. But let me make a comment which is actually related with what you're saying. So, for massless.
trace phi cubed. Literally. We actually know a little bit, so I didn't talk about it, that's something that we're still understanding, but in that case.
We actually know something that even persists at the level of the cross-section, and the reason is because if you literally are dealing with a massless.
cubic interaction, right? Then, uh, you are gonna prefer the diagrams, obviously, this is the standard picture of jets, right?
prefer the types of diagrams which are these sort of forever branching trees, just so that you sort of maximize the dot products that are of the form pi dot piplus1.
Right? Because in these cases. You know, well, we have energy, and then 1 minus cos theta.
Right? Enter squared. And so… so this piece, right, is giving you water 1 over n squared.
But then if they're very close to each other, you know, this is the standard collinear thing from these being also order 1 over n squared.
These dot products are enhanced, I mean, I order 1 over n to the 4th.
Right? Um… While, uh, if you take any PI dot J, which is not very close to it, then you're just going to get 1 over n squared, just because the angle's not gonna be small anymore.
So, in this case, it turns out that the leading behavior of the amplitude.
So, anyway, from a diagram that contains mostly these type of propagators, we get 1 over.
1 over n to the 4th. Right? And then there's order and such propagators.
Right, so that's a product of N such propagators. So a leading order gives you something like e to the root of 4 and.
log in. Okay, so this is the sort of standard thing that we see, uh, because of the collinear.
Uh, divergence here. Anyway, so this is a different regime than the one that I've been describing so far. The one I was describing so far precisely because it has these masses, and it is off-shell.
Uh, then we have instead this exponential growth. And in this case.
Since there's a class of diagrams that's actually dominating, then you can actually estimate the cross-section at leading order, because you can show that basically these diagrams.
Just dominate even when you sort of average over, like, orderings and all of that.
Anyway, but this is a different regime that I didn't really talk about.
So, are there more questions?
Go ahead, Greg.
I have some questions.
let's say I got a couple, actually. So, uh, let's start with the pions on the nonlinear model.
Did you say a little more about how you… By pions, do you mean?
Yes.
Do you mean the non-linear Sigma model for… Um, gauge group.
Any non-aboolean, like, yeah, S-U-N. Yes, yes.
S-U-N or something like that? Okay. So, um…
The two derivative… the two derivative theory, right? Just the… trace D-U-D-U dagger, that's what I mean.
Right.
Right. Well… Okay, yes, do you need that?
Yeah, okay, so how do you get from, um, the trace by queue to that?
Yeah, so this is via this interesting… this… this… this interesting shift.
Uh, which, uh… Yeah, so… Well, so I can ex… so I… Well, the simplest way, if this depends on how long I can talk to you about, but this could take just 10 minutes, and it will be a proof.
Well, I got a couple of questions, so… so maybe if you keep the answer, sort of… From a purchase point.
Okay, uh, anyway, this I, this, this is what I like. So, in general, so there's this funny shift that I wrote there, where you shift your XIJs.
with plus or minus delta, depending on whether i and j are even or odd.
Okay? If I and J are both even, then it's plus delta. If they're both odd.
is minus delta, and otherwise is 0. Okay? So… so, and anyway, this shift, then the claim is that then when you expand for delta much larger than X, so you can think of this as giving it.
a mass, like, so it's like you're giving a mass square to some of the propagators, and then you look at low energies compared to these mass squared.
Of course, some of the particles will be tachyons, others will be normal particles because of these different.
sign, right? But then when you look at low energies compared to this scale, the claim is that at leading order, you get the.
pi on amplitude, okay? You can understand this purely from the amplitude, because it's easy to see that this shift.
is destroying all the poles that are not allowed.
Okay, because any. odd, odd propagator.
Or even, even, is gonna basically correspond to a factorization channel where I have an odd point times odd point.
amplitude, and we know that pi on amplitudes only have even point interactions.
Precisely because of the expansion. Okay, so you learn this as the correct pulls, and then you can show that it also has the Adler zero, so the.
the result of the shift symmetry as a consequence of the fact that both of these amplitudes share a common set of zeros.
Okay, so this is… and therefore that is crucial that these signs work well.
So this is the way we understand it from the amplitudes and how we discover this shift. Now I can ask, how can I see this as a spontaneous symmetry breaking process? And this can also be seen.
As the spontaneous symmetry breaking of the theory with the cubic potential, which is a little bit not the usual thing, right? Usually we do spontaneous imagery breaking for a quartic potential.
But if you have a theory that has some cubic potential, then you can break down, sort of pick up a VEV.
Where, you know, your fuel… so, say you have some fire that has a cubic potential, which is also an N by n matrix, so there's some underlying SUN.
And then the SUN is going to be broken to SUK times.
N minus k, but where basically you pick up.
some… your VF to be such that, you know, there's a… half, whatever, K of your… Entries are in the sort of situation where you get a tachyon, or in the maximum, and.
N minus. Are in the sort of minimum. And if you ask what happens, uh, you know, if we expand around this wave.
You get exactly, anyway, cubic theory whose amplitudes are these delta-shifted ones.
Anyway, so…
So, you know, I guess I don't understand something. I mean, the cubic theory has SUN global invaliance.
Yeah, so the theory would be an SUM.
Right? The global symmetry. Right, where I conjugate the phi field by, uh… S-U-N matrix, right? The 5 field is a Hermitian matrix.
Yeah, joint of the SUN, exactly, yes.
Okay, so, but then the… But the pion theory has an SUN times SUN global symptom.
That's for the, the, the… the standard symmetry.
breaking. I mean, usually in the standard symmetry rink, we have SUN times SUN breaks down to the diagonal, right?
Um, that's the one that we have most naturally. Here, the symmetry breaking would be this one.
Where the V breaks down the SUN down to the SUK.
Okay. Okay, well. Okay, well… Just don't clicker.
times issue n minus k. So this is if you want to… yeah, anyway, but… But…
So, yeah, let me move on. Um, yeah, okay, that's pretty interesting. So, um… So what about this comment you made about string theory? So I was wondering during your talk, okay, so what about.
Presumably, you're… the particle in this case is the tachyon and a boson obstring theory.
So…
How did…
Yeah, so it's really, like, yeah, we're… so… We're leading… we're talking about these integrals that we shift the tachyon away. So, so, like, we're at low energies, it gives you trace phi cubed. Anyway, it's not an official string theory, but you can do that by redefining the XIJs.
Well, when you said you had the strength there, I mean, were you actually… are you actually analyzing.
amplitudes over the modulized space of G is 0 with n punctures.
Exactly, it's exactly that, exactly that, yes. That's exactly what I meant. And there, if you ask, so, uh, so what we… so, it's actually interesting, so I even wrote it here, just so that… you know, for the tropical, I wrote it here already, uh, well, not really here, but for the…
For the drop, for the filtering amplitude, the whole thing of deriving what happens at large n was a way of finding an upper bound.
Right? That is, that is what, that's what we did. Turns out that you can… Also, write an upper bound for the full string theory.
sort of the integrand, if you want, over the modularized space.
Okay, so… because anyway, we write it in terms of these polynomials, you know, for us.
Anyway, the string amplitude is in this new parameterization are some.
Anyways, a function of these variables, and then you have.
Sort of polynomials. Anyway.
Sorry, Lena, these are open string amplitudes, right?
They're all open strings, sorry, they're all open strings, yes.
They have some exponents, they have these sort of polynomials over these parameters, which are parameterizing the moduli space.
Um, and these are the things which, basically, when you tropicalize them, gives you the maxes in the field theory limit, but anyway.
But already at the level of these polynomials, you can use.
the sort of untropicalized version of the inequality which bounds the maxes.
to sort of write, you know, 1 plus Y1 plus Y1, Y2.
So either this is greater or equal, then. 1 over W1. Anyway, with some weights, just like before.
But you just do that. Anyway, and so…
So, if I'm understanding what you're saying, you could have… You could have started with the… open bosonic string.
Endpoint tachyon amplitudes. derived it… derived your formulas there, and then specialized to field theory, is that correct?
No, that is not correct, and that's what I'll explain. So, but we have… anyway, so using this, you get an upper bound on the integrand.
Mm-hmm.
Right? But now, this is an upper bound which depends on the Y's, right? So it depends on exactly what place you are in the integration domain, right? It's on the integrand.
So, it's not very… I mean, when we… when I find the best weights.
Right? I mean, it will, of course… it's clear that there's not a best possible weight for all Y's.
Right? Like, that's sort of obvious. But the thing is that if you go to high energies.
Okay.
Right? You know that the answer is going to be localized by a point in the modularized space.
Right? That's what's given by the scattering equation, right? The saddle point.
Like, you're gonna localize on some saddle. And that's exactly the thing… and so, basically, then, at high energies.
Indeed, the sort of best possible weight is going to give you the sort of exactly the point where you localize.
Um, it's gonna match with what you get from the scattering equations. The interesting thing about writing it with this upper bound is that at large n, it makes clear that there's a piece of the scattering equations which is completely subleading.
And so, what you want to solve is a much simpler part, because in general, it would be very hard to solve the scattering equations at very large n.
Right, because they become… they become very complicated. But what these upper bound makes obvious is that.
Anyway, at large end, there's a piece which is so bleeding, and so you want to find a solution of something which is much simpler, which is basically the minimum of what I wrote before.
So, in other words, I don't know what the large N string scattering is, but I know what it is at very high energies in large N.
Just because it sort of localized to this point, uh, in the modularized space.
Yeah, except, Carolina, there's something…
physically, that I don't understand about this, because there's…
There's a large region
of real kinematic space and string theory.
Where at large energy, you know that what happens is you produce black holes.
And then… and then the final… the final state is lots and lots of particles.
Yeah, so this… this is still an is much smaller than…
And so, it's large energy and large N, but you're not going to get that from these formulas.
Yeah, no, no, definitely not. So here, I am considering large energies compared to alpha prime.
I see, okay.
But it's still much more than M Plunk. So… so that's… so that… so that is the… so that's the regime that I'm talking about. It's high energies compared to alpha prime.
But much smaller than Mplunk, so I'm never…
Oh, wait, you're fixing the energy and taking the number of particles to infinity, right? All your results are about asymptotics, and little n goes to infinity.
Yes, exactly. But yeah, but here I'm assuming that the X's can also, I mean, that X times alpha prime is very large.
And then back to Eliezer's point, you have to go more and more off-shell in order to do that.
Uh, right, so I can fix a curve.
Yes, yes, that is true. Yeah, so this is a pure… exactly, that is true, yes.
Okay. Yes, I had two other questions.
Um, early in your talk, you introduced this X of UV.
Mm-hmm.
Yeah, so I thought, oh, if she's going to write some kind of… she's going to view X of UV as some field on a Riemann surface.
I mean, it was that… Were you meaning to suggest anything like that?
All right, a related question.
Ah. No, not… no. It was really just a field on this particular plane. The UNV… Or just because they were sort of… these directions are 45 degree directions, that was all.
I see. Okay. A related question is that the XIJs look to me like cluster variables.
cluster coordinates for a flat bundle on a Riemann surface. The Riemann surface would be a momentum.
the interior of your momentum polygon. And the fly connection would have a, uh, irregular singularity and infinity on the XIJs would be.
Yeah, so it's…
Flat cord, that's… parameterizing the Stokes data of that connection. Is there anything like that going on here?
Well, the only thing which, I mean, at the level of the.
string amplitudes, right, the XIJs are sort of related… I mean, they're the things that sit in the exponents of the cross ratios of the… You know, the sort of SL2R invariant cross ratios, right? These, uh, UIJ, what we call these UIJ, which is just this ZI-1JZ.
Anyway, I, J-1. Z-I-J.
So… so… Anyway, these themselves are not, some of these are cluster variables.
Okay, some of these. Not all of them are, I mean, uh, yeah, so that is the, the, the, the… the closest… but this has come… I mean, but only here at the level of now thinking about the world sheet itself.
Do I see there's something which could be. Uh, that is related. The XIJs themselves directly as purely just the kinematics.
Um, I mean, I… it's just a sort of a nice way.
to keep track of, it's just the fact that the similarities of this theory are associated to curves on the surface, right? Then the curves on the surface, when you think about the marginalized space associated to the surface itself, the curves on the surface.
are also naturally associated to these cross ratios. And these cross ratios themselves are then.
Um… Okay, it was cluster variable.
Okay, I mean, I'm thinking about a… I'm thinking about a flat connection for SL2 on a disk with, um… with, uh, end Stokes raise of infinity.
And then I, um… I can get the different cluster.
Um, uh, cluster regions. by the different triangulations, exactly the picture you drew.
So there are different, um… There's an atlas, a cluster atlas, right?
And those are labeled by the triangulations, just what you drew.
Yeah, so different triangulations correspond to different regions.
And, um… So you get a Catalan number of those.
Um, and then the analogs of your chord coordinates.
are like the Stokes coordinates. Uh, sorry, cluster coordinates on the moduli space of these.
flat connections with specified irregular singularity. So I was just wondering if there was some connection to that mathematics.
Yeah. I think the only connection to that mathematics that I see is sort of, uh… Well, it's also in the same way. I mean, so the… these UIJs are all associated to a given curve.
And then given a triangulation. I can pick different ways to basically parametrize these UIJs.
And each of them give me, if you want, an atlas or a sort of a way to parametrize the whole moduleized space that gives me, you know, where the different triangulations are located in different regions.
Yeah, but you're talking about the triangulation of a world sheet, right?
is… Okay, good, then.
Yeah, and we'll talk about translation of a disk with n mark points on the boundary, that's all, yeah, so… That is…
Sorry, sorry, there is an analog in kinematic space of the whole story, which I think is closer to what Greg is asking.
So, these new IEAs are, as you know, dual to XIAs.
Oh, I see.
And from the mesh, you can see that every mutation you do in
Muaya is essentially just…
transposing, uh, particular,
time-like curve in the match. So if you try this… if you write this timeline curve in the match,
That's a payment item. You can do one mutation of that Feynman item, and that would be a transposition of the time-like curve, essentially a move where you shift one
point in the curve to.
another point in there.
In the time. It's kind of the way that the mesh was originally discovered, especially from the cluster Polytool.
But this, like, completely kinematic…
analog of what's happening, what's happening in the… in the modai.
Yeah, in other words, but that's sort of equivalent to the fact that, I mean, when you.
I mean, when you tropicalize the sort of modulized space, right? You also… you get the sort of… anyway, the polytope that you get at infinity is.
Right. It's a cluster polytope, so that's how…
sort of pair request. is a cluster polytopia. The associon is a cluster polytope, yeah.
Now, I guess to say, what you say is happening in this world sheet, but…
As you said, if you tropicalize,
Then the story ports to kinematic space, which I think is closer to…
Yeah.
Okay, that's what I was, yeah, I was asking about kinematic space.
Yeah, in that sense, XIAs are in the cluster
cluster variables. These are… these are each…
Oh, okay, so then… okay.
Their compatibility rules, the compatibility rules.
Yep.
But they don't satisfy, like, a cluster, like, the cluster, uh… They are compatibility rules, but they don't satisfy the cluster algebra equations, so that's.
Right, but…
Well, are they parametrizing the moduli space of some flat connection?
That's all.
With irregular singularity.
Yeah, maybe in the worksheet there is a description like that, but…
I don't think so, as far as I know, but uh…
Okay. Okay.
Yeah, in the world sheet, I think that that is… that is the case, but, like, anyway, then when you tropicalize, you sort of get the… this cluster polytopes. They all live in the… kinematic space, but anyway, they satisfied its compatibility relations, but they don't.
Okay.
Of the cluster algebra structure that you see for the world sheet variables.
All right, thank you, thank you.
Okay, let's thank Carolina again for a really nice talk.
Thank you.