All right, guys. Sorry, I'm a little

late. Um, so today we're going to start

uh where we left off yesterday. Uh

remember at the end of yesterday

uh in the middle of yesterday we saw

there were two interesting facts um that

uh were exposed by uh thinking about the

fan of the uh of the the normal fan of

the associ. One of them uh was that uh

of course the normal fan of any polytope

gives you a bunch of cones that tile all

of space and the cones oh sorry mic's

not on.

Is that okay? All right. Of course, a

normal fan of any polytope gives you a

bunch of cones that tile all of space

and that you can think of each cone as

associated with the vertices of the

polytope.

Um, but that sort of preassumes that the

association was handed to you uh via the

construction in Carolina's lectures for

example. So the first comment uh that we

got through yesterday is how to get at

least the rays of that cone from

nowhere, right? Starting from a picture

of the surface, just recording the data

of every curve on the surface by these

words uh with left and right turns and

extracting the G vectors for the for the

curves uh from those words. Um and that

just plunks down a bunch of cones,

plunks down a bunch of rays which turn

out to bound cones that can be

interpreted in terms of triangulations

and which collectively uh cover all of

space. The second uh maybe much more

striking uh observation was that you

could generate all of these cones. you

could uh the rays and the cones um and

they're associated with these polomials

that we talked about uh and their

tropicalizations gave you a bunch of

peace-wise linear functions that uh a

few of them that when you put them all

on top of each other the domains of

linearity of these peace-wise linear

functions broke the space up into all

these cones each one of which

corresponds to a diagrams and which we

can think of as uh being put together

into the integrals that we wrote down as

giving a sort of global version of

Schwarler parameterization, some kind of

uh one single integral that as you moved

around in the whole space morphed in

pie-wise linear region by region, cone

by cone morphed into the Schwinger

parameterization for the corresponding

uh diagrams and the uh pre-tropicalized

version um uh with just the polomials

that we talked about sitting there were

instead string amplitudes. So this is

the sort of a a uh connection what I'd

like so by by the end of uh uh yesterday

we talked about where the g vectors come

from and what I want to start with today

is telling you where those u variables

come from and uh because of the time I'm

not going to be able to prove to you

that these variables do what they're

supposed to do. But I'm going to

describe what at least one motivation

for where they come from. This

motivation will still maybe seem

slightly alien if you haven't seen

anything like this before, but at least

you'll see a very simple sort of

combinatorial problem associated with

these words naturally leads you to think

about these u variables. Uh I could just

give you a formula as you'll see in 15

minutes will be a formula taking a

product of a bunch of 2x2 matrices uh

that's going to produce these u

variables. But I want to at least uh

give the the the counting problem

motivation uh for this. Not just because

it's cool, but because if you go away

and think more about the counting

problem, it'll be it'll make it clearer

why these objects do what they're

supposed to do.

Okay. So, um but uh you can really just

think of this as a recipe for producing

uh these uh new variables. Okay. So, so

um so we're going to start with uh

getting uh the U variables

and again remember all of our life our

universe is specified by this uh uh by

this uh fact. Okay.

So for for the process you draw one

representative diagram that represents

the sort of flow of color and then

everything else lives uh in this world.

And if we talk about a curve, we decided

that a curve was uh one of these tricks

uh through the back graph. And so this

curve uh 24 would have uh the word that

we described as 2 three makes a left

turn onto 13 makes a right turn onto 14

makes a left turn on to four five. And

we also discussed how we could draw it

as a mountainscape. 2 three goes up to 1

3 goes down to 1 14 goes up to four five

okay these are two different ways of

representing uh exactly the same word

okay so now I'm going to define a

counting problem associated with any

word forget forget about all this we're

going to say a counting problem

that are associated with words

and this is what we're going to do let's

say have a word that looks like a b See,

I'm going to tell you uh uh what the

counting problem is. Um you just want to

uh you want to choose

um uh every

you want to choose sort of every object

that you see uh in this word. So for

example, I could choose to uh uh I could

choose I want to pick every uh letter

that I see uh in this word in every

combination. So if you have I could

choose to pick nothing and I would write

down one. So if you had to make a

generating function that goes with that.

Okay. So I could choose to do nothing. I

write down one. I could choose to choose

A by itself. I can choose to choose C by

itself. I can choose A and C.

But if I choose B, I have to choose

everyone who's downhill from B. Right?

So I used to call this the relationships

with baggage generating function because

it's like you know it's like dating when

you date someone like dated everyone

that they've dated before. Okay. So so

uh so if you choose B you have to choose

everyone in the sort of past lyome of uh

of B. So you have to do plus b a c.

All right. Is that is that clear? So if

we do another example, let's say it's a

uh a down b up c. What is this?

>> Uh this would be again I could choose

nothing. I could choose b.

I could choose a but if I choose a I

have to choose b as well. If I choose c,

I have to choose b as well. I can choose

a and c together. Of course, then I

still have to choose B, right?

Is that clear?

All right. Now,

um now it's sort of obvious from uh high

school um how you can compute I mean you

don't have to do it this way. It's

possible to compute this uh this uh

function this generating function in an

obvious way recursively. Okay. So in

other words, let's say you give me a

word. I can kind of start from the end

and uh and and get the generating

function related to the one shorter word

where I peel off a net. Now, how does

that work? Um well, let's say that

uh let's say that so I'm going to call

these uh uh objects f associated with

the word. They're not quite going to be

uh these f polinomials that we'll talk

about later, but anyway, let me just uh

call them call them f.

So let's say I have a and it goes up and

it goes up to some b and it's going to

do something else. There's sort of rest

rest of the word there. Okay.

What we're going to do is uh is divide

this generating function into two

pieces. I'm going to call f yes and f

no. Okay. So f yes is does it choose a

or and f no is uh does it not choose a.

All right. So there's f yes and nos

according whether it does or doesn't

choose a. So what is f yes?

Okay. Well f yes

it it it it chooses a. Okay. So uh f yes

is because it chooses a it's going to uh

it's going to have a factor of uh a

there right?

Um

uh but uh uh if I choose if I choose uh

uh uh if I if I choose A then here the

question is uh do I have to choose B or

not right?

And well I can choose B or not choose B

it doesn't matter. Right? So I'm going

to call the rest of the word here is

going to have its own little F. So it

has its own little F. Yes. No, where

this uh yes or no refers to B, right?

Call this B, right?

So this is A times little F yes little F

no. It doesn't care either way. Okay,

is that clear?

On the other hand, if I don't choose A,

well, I'm not going to have anything

here, right? I'm not going to have A.

But I definitely could not have chosen B

because if I choose B, I have to have A,

right? So this is just equal to F. No.

>> And so we immediately learn something

that

uh F yes and F no

is this little 2x2 matrix A a01

times little F yes and little F no.

So when I turn it up, I get that little

uh 2x2 matrix. Okay. So I'm going to

call this M left

of A.

Right?

Well, similarly, if I go down, let's

just do it quickly here. If I go down A

down to B and the rest,

then F yes

is equal to A. But it since B is

downhill of A, I must have B here. So

this is F yes and meanwhile F no is the

thing which again I could take either

one. It could be little F yes little F

no.

So this is turning left

from A

and so we all s have a matrix for

turning right from A. Right? So we have

f yes and f no return for turning right

from a

is this other matrix

a 011

times little f yes and little f no.

So we'll call this matrix m right of a.

Okay.

And so this tells us how to build this

uh f. I just take the product of these

matrices, right? I take the the product

of the matrices as I go down the word.

Uh and then at the very end I have f yes

and f no. I can add them if I want to

get the uh the full hat.

Okay.

Now if you remember uh our words coming

back to our context um there are open

curves and so they're kind of special in

that they start and stop somewhere

at some boundaries.

And then I leave it as a very small

exercise. It's very very obvious, but

I'm going to leave it as a very uh small

exercise to show

that if I have a word that's like alpha

as a boundary and it goes turn, I'm

going to turn is left or right. Left or

right. Okay, alpha turn and some uh some

road A1 turn A2 turn. Okay. And it's

going to go finally uh some turn uh

beta. So alpha and beta are the ends.

Okay.

Um

okay. Then uh there is a uh in I can

look at the sort of f that I would get

from this from this big word. But in

this big word it's also natural to look

at which part of f includes both alpha

and beta. Which part includes only

alpha? Which part includes only beta?

And which part includes none of them?

All right.

So I'm going to group those things. Let

me call them to f has a yes on alpha,

yes on beta. F no on alpha, yes on beta.

F no uh yes on alpha, no on beta, and f

no on both.

Okay. So this is a this is a a 2x2

matrix.

And uh it's a trivial consequence of

what I just told you that this matrix is

in fact the product of all of these uh

matrices. Okay. So this is the product

of of the matrix for um uh uh turn at

one

turn at two.

Okay. Up to whatever it was that he had.

Okay.

Let me be a little more precise.

Um so if I take this matrix and um for

the for the so for this first matrix if

I turn left or right I'd have to put the

m for alpha right um but here you have

to put alpha equals 1 and you imagine

that this is sort of beta equals one as

well of course beta doesn't show up

there's no term that depends on beta

beta is the the last thing okay so if

you look at this matrix and set alpha to

one of course beta is passely one then

you get an entry here where this would

be uh to get the if to get the real word

for this that would have an alpha and

beta you just multiply this by alpha

beta okay so this is uh so these are the

sort of coefficients of the alpha beta

term the coefficients of the beta term

the coefficient of the alpha term the

coefficient of none of them comes simply

from taking the product of these

matrices okay so this is a matrix that's

attached to to any word

Now um

back on screen.

[Music]

Okay. Now, if we stare at these

matrices, sorry I uh uh I partially

erased it. Notice that each one of these

matrices says uh as determinant is just

given by A, right? So determinant of

this matrix A, the determinant of that

matrix is A. Um

and so one thing which is obvious is

that the determinant

of this uh matrix associated with any

word if we imagine that those variables

the ABCs are just positive the

determinant is positive. Okay. So the

determinant of this matrix associated

with the word w is just the product over

all these uh all what I call there that

little a is all the sort of a k and this

is positive if the aks are positive

so which uh we'll we'll be assuming so

if I look at this matrix so this is a

matrix that has a has a one one

two 2 1 M22 entries

then I know that M11 M22

minus M12 M21 the determinant it's a

product of all these A's is positive

and uh therefore the ratio u associated

with this word which is M12 M21 / M11

M22 is less than one and of course

Again, if all of these variables are

positive, clearly as I multiply these

matrices, they're all plus signs here

everywhere, these simple matrices. Okay,

this U is also going to be bigger than

zero, greater than equal to Z, less than

or equal to one. That's the U variable

attached to a curve.

So if you give me any word for any curve

on any surface

all you do you just build the word you

multiply the 2x2 matrices associated

with the uh with the entries in the word

uh and you take this ratio 1221 off

diagonal over diagonal and that gives

you the u variables associated with the

curve. So I'll leave it as an exercise

for you to do this for the little

fivepoint problem. uh you'll have to

multiply at most uh two 2x2 matrices um

and take the ratios and discover those u

variables that we wrote down uh

yesterday. All right.

Okay.

Now, um and that's all I'm going to uh

uh say about it. Um maybe one point to

make is that this is sort of presented

as sort of manifestly a ratio of

polomials right all of these things are

sort of ratios

>> could you say one more time what the

variables are

>> oh the variables associated with what we

call the y so let's just do let's just

write down

this example okay so back to this

example you would now go and and label

uh the internal rows not just with the

name but also the variable y. So

there'll be a y13 that goes with this

road

and a variable y14.

Okay? And then you again take your

favorite curve like 24

which is now the word would be y one two

up 1 3 down 14 up four five.

Okay.

And now the the uh the matrix associated

with this word the matrix associated

with the word M24

is this is up at one two but remember uh

the uh this I'm supposed to set to one.

So the up matrix

was a a 01 but a is one here for this

one. Okay. So I'm starting at the

boundary. So this is 1 1 0 1.

Then at 1 13 I turned right. So here I

put the matrix Y13

011.

[Music]

At 1 14 I turn left. So I put the matrix

Y14 Y14 01.

Okay.

So this gives me a 2x2 matrix

[Music]

>> and the U for 24 is going to be YZ over

XW.

which if you did it in this case you

would discover is uh 1 + y14 + y14 y13

over 1 + y4

1 + y13.

Okay, you can see how these things are

built up from products of these 2x2

matrices. See the these things look like

relationships with baggage generating

functions. Okay,

>> very quick question. Is this five or 24?

>> Sorry.

>> Is this path 25 or 24?

>> This path is 24 cuz it's 1 three to 2

three left on three. Right.

>> Then I turn uh then I turn

uh started at one two.

>> Oh, I'm sorry. I just wrote this wrong

entirely. I'm sorry.

>> So, I'm starting at 2 three. Sorry about

that. at 2 three left on three right on

14 and then left on four five. Sorry

about that.

>> Is that okay?

>> No.

>> I have a question.

>> Yes. Two questions in fact.

>> Sorry.

>> Two questions. Yes.

>> Can I ask them?

>> Yes, you can ask them.

>> But you can't ask three questions.

>> There's a sharp upper back.

>> Okay. So the first question um is um

yeah what kind of um geometrical

resolution or precision we obtain on the

two dimensional trajectories within the

formalism used.

>> Oh there's no geometry here at all. This

is pure this is this this is pure uh

there's nothing about what this curve

looks like at all. This is just a

labeling for what the curve looks like.

So it has nothing to do with that.

>> Yeah. Thank you for your answer. And um

um question number two is um directly um

related to the observation you just um

made publicly. Um well for her um where

does this formalism first arise from? I

mean physics or maths

>> both.

Uh I mean although I I should say that

these uh the the these these variables

are closely related to many other

variables that have been uh discussed.

They're very closely related to uh

xcluster variables.

uh for surfaces uh alop and gonerov

they're related to cluster variables xcl

cluster variables but there are very

special sets of them and uh this

particular way of thinking about them

and focusing on them is uh uh is uh is

new and was very much motivated by by

physics okay um the thing is that you

can talk about cluster anything you like

it's these variables that turn into the

shringer parameterization of all

diagrams and string amplitude and so on.

So these things that go from 0 to 1,

they satisfy these remarkable u plus u u

uu equals 1 formula that's the uh they

they they realize in this binary way the

uh the combinatorics of uh compatibility

whether curves cross or not. Okay. So

that's uh uh that was very much

motivated by by

>> actually I'll just make just since the

the question was partially historical um

uh these variables literally these

variables

um for the context of studying string

amplitudes at tree level were discovered

in string theory before the standard

formulas for doing string calculations

okay by there's a famous formula called

the kovielson formula and these u

variables were were described by Kova

and Neielson in uh 1968. Okay. Um and

they were they were very excited about

them exactly because they made

factorization manifest. Okay. That's uh

only later that did coin Nielson

themselves realized that you could

formulate the problem instead in terms

of sort of cross ratios of points uh on

on a disk and the picture of a string

world sheet began to emerge and then the

string world sheet picture um uh kind of

took over because you could make it

systematic. You could do it for any

surfaces, think about loop corrections.

So lots of things came from that picture

and this more algebraic

uh uh and the way that's sort of

centered on curves on surfaces was

entirely forgotten. Um and uh what's

happened now is two things. First of all

um uh we understand that they also make

sense for all surfaces. Okay. So there

is no loss in going back and forth. So

all those physical things that Coen and

Neielson were excited about are actually

there and and present and uh uh and and

useful not just at tree level but for

all surfaces. But maybe more importantly

uh Coen Nielson found the U equations

and they even found how to solve them

but they only found how to solve them in

this way that I mentioned yesterday. is

most direct way solving for the U's in

terms of the U's which you can only do

in some cases and it's very artisal uh

and definitely you couldn't imagine

doing it for more complicated surfaces.

Okay. So the second novelty here is the

solution of the equations like we're

telling you how to solve them by

associating these uh counting problems

with the curves on the surface and the

product of these matrices and so on.

Okay. So, so it's both the fact that

they exist and that they're concretely

available.

And um uh and what they do what these

what the what the U variables do is they

give a sort of completely algebraic way

of thinking about what uh what not just

the the modulized space really the

tangular space of these two dimensional

surfaces but also what what the

compactification of these spaces you

know how you add back in all the

boundaries and what's again what's novel

about it is it's done an entirely

algebraic way you just write down all

these equations And you just declare the

user positive and you're done. The usual

way of thinking about the

compactification of these spaces even in

sort of a clustery way of thinking about

things or the faky boner way of thinking

about things is more local. You it's

synthetic. You go to the neighborhood of

a boundary and you figure out how to

sort of complete things in the

neighborhood of that boundary. So patch

by patch. This is not patch by batch.

And that's it's a very practical

advantage. It's conceptual conceptually

interesting but also practically

important. That's what allows us to

write down one integral. Done. Right.

And the one integral uh blows up all the

singularities and tells us how to how to

just uh go ahead.

>> Can you say a word about finitness when

you and the example are always at this

fivepoint example but as soon as we do

something more interesting everything

will be infinite. Can you comment on f

that bothers me?

>> Can you comment on finite versus

>> absolutely absolutely yes. So what what

burn is uh uh referring to is that the

moment we get to more interesting

surfaces

uh and actually uh you have to get uh it

really starts with the annulus. Okay. So

the problem has nothing to do with these

variables per se. Uh the problem has to

do something about curves on surfaces.

So if I think about this picture of an

angulus um uh I could draw a

triangulation of the angulus. This is a

slightly degenerate triangulation, but I

I hope you can see this is a triangle,

right? This is a triangle and this is a

little triangle here, right? They all

have uh uh three sides. Uh that's the

important thing. The triangles have to

have three sides. They don't have to

have uh three vertices. They have to

have three three sides in them. And

that's what lets them go along with uh

cubic graphs. So, well, what cubic graph

would go along with this picture? If

this is one and two, the cubic craft

that goes along with it would be this

little picture of an angulus. Okay, so

this line one is what's going around

here. Okay, so that would be sort of

region one or line one. This is region

two or one two. Okay.

Okay.

And now what burn is talking about is

the following. You see at this level

this triangulation is this diagram but

there's one diagram here. There's

there's a single diagram. But here I

this picture and I could just twist

everything. Okay, I could just twist

everything once. And clearly, it's going

to look a little more wound around. I'm

not even going to try drawing it. I

could twist it any number of times.

Okay, this way. That way. So, on this

side, there's an infinity. There's

there's infinitely many uh

triangulations

that just differ by winding uh or

game twist

or this is the action of the mapping

class group uh on the surface. But if

you haven't heard these words, it's just

the obvious picture of uh of winding

here.

Um and that's completely absent on this

side apparently. Okay, so it looks like

so and and the point is of course that

that precisely because all of these

things are related to each other by

winding, they're combinatorily exactly

the same, but they're all this

triangulation, but thought of as a uh in

terms of curves on surfaces, there's

infinitely many of them. Okay,

so this seems uh this seems uh um uh

problematic at first. It's actually um

it's actually uh it's actually a

blessing in disguise.

>> One of the reasons that it's a blessing

in disguise is that remember uh I

mentioned early on that um as physicists

we would like to imagine putting all the

diagrams together even at loop level and

putting them under a common loop

integration sign to sort of identify

what we mean by loop momenta from

diagram to diagram to a diagram.

And I said that when it's planer,

there's a sort of way of doing it. But

when it's not planer, there isn't there

isn't naively a consistent way of

labeling loopa that allows you to

combine all diagrams together. But in

fact, there is. And the curve on

surfaces exactly tells you how to do it.

You draw for any diagram any diagram is

some uh is some triangulation, right? Um

so you draw the curves on the surface

and then for any curve you read off the

momentum associated with that curve by

homology. Okay. So we saw it already uh

we we didn't say it in this slightly

fancy uh language but already here when

I draw you know even in in the disc when

I draw something like that I draw this

curve uh and I say the momentum

associated with that curve that I

squared to get the x associated with it

is p1 plus p2 plus p3. But you can sort

of think that I assigned momentum to

these boundary components and the

momentum of this guy is the same as the

sum of those guys because this curve is

homologous to these. I can I can

continuously deform it uh to the

boundary. Okay. So in that way you can

take any surface. Now in a case like

this you see this curve for example

cannot be this curve can't be uh uh

continuously uh uh deformed to the

boundary. So you have to give that one a

name. So we have to come up with an

element uh of a mole. So this group this

this curve I'll I'll give this momentum

a name. I'll call it L. But then this

curve is uh is is going to be L plus the

momentum associated with particle 2.

Okay. So this curve would have this

curve would have momentum L plus the

momentum particle 2 with Q with Q. If it

wind around twice it would be L + 2 Q

and so on. Okay.

And so that actually gives you a

completely consistent way of labeling

what you mean by momenta including loot

momenta for any diagram. Right?

What's the catch? The caches are that

there's now infinitely many diagrams.

Okay. Uh there's now infinitely many

diagrams. So that when you do the action

of the mapping class group, what it does

is it doesn't leave the loop momentum

invariant, right? It'll shift the loop

momentum. For example, if I wind here

once, the loop momentum will shift by L

goes to L plus Q. Now, of course, since

I'm integrating over the loop momentum,

the amplitude doesn't change. That's the

point. But we sort of found a way to

solve this essential you know kinematic

problem of how you label momenta uh

directly from homology of curves on

surfaces at the expense of having

infinitely many diagrams. Okay. So but

that's a sense in which it's a good

thing right that at least it's a first

step to letting us put everything under

the same sign. So now all we have to do

quotes all we have to do is mod out by

the mapping class group when we're done.

And that's what we actually do. We take

this whole formulas for any surface we

have the product of u to the x um

including when there's loops it's the

product of u to the x uh when there's

loop the x's will contain loop momentum

in them and this u to the x will have

things that are quadratic and loop

momenta upstairs again this looks

exactly like swinger parameterization uh

if you're familiar with it either in

physics or math um uh so we can do the

mmenta they're gaussian integrals that's

exactly the swinger parameterization

point so you're still left with an

integral over the y's. Now you're done

integrating the loop momentum. So

whatever you're left with and you're

integrating over all these y's is

mapping class group invariant because

you've uh integrated over the loop

momentum. Now it's totally mapping class

group invariant. So all you have to do

is now mod out by the mapping class

group. Okay. And uh you might think that

modding out by the mapping class group

would uh would uh uh involve you know

maybe identifying some fundamental

domain that then uh that then sort of

covers uh the the entire space. uh after

you do the uh action mapping class group

if you wanted to do that it would not be

such a fun thing to do but there's much

simpler ways of modeling out by the

mapping class group so what a physicist

would call the fa popoff trick what

mathematicians would call the merely

kernel okay are very straightforward

ways of just modding out by the mountain

glassass group uh to get the answer so

that's the so beyond at tree level we

don't run into this phenomenon even at

one loop we don't run into this

phenomenon but starting to annulus and

beyond

this thing that burn uh mentioned about

the infinity is ubiquitous but the

infinity is a good thing somehow across

the board that if we didn't have

infinitely many curves these U equations

would not make sense we really need

every curve on the surface for these

equations to make sense not only the

ones that wind infinitely much even the

ones that intersect themselves I mean

every conceivable curve on the surface

uh shows up when you write down these uh

>> so no way for me to look at the first 25

in a meaningful way

>> ah let's say I want to look at this 25

variable and 25.

How should I list them?

>> Yeah. Yeah, that's right. Yeah. So, um

so uh one of the I mean that this is one

of the things that makes this very very

practical. Um there are infinitely many

curves and so on. It's true. But the way

it's presented makes it clear you have

these ratios of of polomials and uh

there's a sort of concrete sense in

which you start with your parent graph,

right? You draw the simple curves.

called the simplest curves are the ones

that occur in the triangulation and then

the ones that you know where by simple I

mean literally the length of the word

you know how long does the word get okay

then as the words get longer and longer

the u variables associated with them get

closer and closer to one okay remember

they're stuck between zero and one okay

and so if the y's are so for example if

the y is just to say a super concrete

statement if you think about the u

variables as a tailor expansion in the

y's okay the ratios of the polinomials.

You can certainly tailor expand them in

y's. Of course, the y's go from 0 to

infinity, but uh somehow the the y is

close to zero means that you're in the

neighborhood of the uh of the surface uh

degenerating to the triangulation that

you started with. Okay. So, if you just

look as a tailor expansion in y at any

fixed order at the tailor expansion, a

finite number of curves have u not equal

to one. Okay. So first of all in the

sort of super precise sense you need a

finite number of curves at any order in

the tailor expansion. Secondly um even

if you let y's be anything but some

fixed number you know y have 10 y's and

the y's are 17 23 I just fixed some

fixed numbers for them as the words get

longer and longer for those fixed y's

the 's for longer and longer words

approach one exponentially quickly.

Okay, so these U's are sort of very

they're very practical. If you want to

sort of numerically evaluate these

integrals, you don't need infinitely

many of them remotely. You really need a

small handful of them because of how

exponentially quickly they go to one.

Okay.

>> Yes.

>> So these are like some type spaces.

These spaces have some dimension. Do I

know how many views I need? Are

>> there infinitely many use? In principle

there's infinitely many use. Okay. Uh

there in principle there are infinitely

many U. If you want to write down a

string amplitude, okay, uh it's uh um uh

it's your choice uh what use to put into

these formulas, okay? Because um uh

because uh um

uh uh

first of all uh there there's there's

infin there's infinitely many curves

partially because even given any one

curve when the surface is interesting

enough it can go around and self-

intersect itself lots and lots of times.

So there's all the sort of self-

intersecting uh friends of a given curve

right they're there. have u variables

associated with them. But notice if a

curve intersects itself, it's u can

never go to zero because in the u

equations it occurs in both terms,

right? So you have u plus a product of

all the curves that intersect

x uh is equal to one. If a curves

intersect itself, it shows up in both

terms of the equation. So you cannot set

it to zero. So that means that all of

the self- intersecting curves are

irrelevant for the field theory limit.

Okay? If you want to take these curves

and see what they look like in the field

theory limit, you can just throw all of

them out. That's the beginning of seeing

the sense in which we have a bigger

world of objects that uh kind of

connects to string theory in some limit

but is not necessarily string theory

because it turns out that string

amplitudes require you to put in every

single curve on the surface including

all the self-intersecting ones. Okay.

And that's why you know if someone

handed you a one loop string amplitude

and said please take the field limit

you'd be a little terrified of doing it

because of these horrible Jacobian data

functions everywhere. What are you

supposed to do? Where are those coming

from? They're coming from the product of

these infinitely many self-insected

curves. They're are literally irrelevant

baggage as far as the field theory limit

is concerned and you do not need them.

Okay. So when you ask the question what

you include it's a little bit up to you

what you want to do. Okay. Um so um but

but if I'm strictly talking about theory

limit the answer is a little bit more

interesting and it's related to

something I just want to say about these

U variables. Now

uh and then I'm just going to answer

this question but it dubtales exactly

with what I was going to say.

Um,

so

so I have these u variables that depend

on the y's

for any for any x. Let's once again um

some yk. Let's once again put yk equals

e to the minus tk just uh so I can talk

about uh uh uh talization. So again,

we're interested now what happens when

when when the sort of t uh goes to

infinity. You know what's going on

because we go to infinity. And so u is

going to go to e to the something uh e

to the uh we'll call it alpha times the

sort of tropization of uh ux. Okay,

that's uh

um sorry uh e to the trough of of ux and

this I'm going to call e to the alpha x.

Okay. So now alpha x is going to be some

peacewise linear function on tsp space.

Now the properties of the use guarantee

the following.

So sort of key point about

uh tropicalization and these alpha

variables is that these alphas are the

global swing parameters.

And so the the key point is that if I

take alpha for a curve x, it's a

function on this space where this where

all these cones live right in this t

space. If I take alpha for curve x and I

evaluate it on the g vector for the

curve y,

it's equal to zero if y is not equal to

x

and otherwise one if y equals x. Okay.

So, so the tropicalization of the U

variables lights up the G vector okay

for its own variable and a zero on on

everybody else. Okay.

And that's why if you then just write

down this integral integral you know uh

d and t whatever the dimensionality is

uh e to the minus the sum over all the

curves on the surface um the x I'm sorry

I'm doing maybe I should say some sum

over all curves c the x associated with

the c remember with every curve we just

figured out how associated momentum I

can square that momentum to get uh an x

or I could just say that there's

variables xc associated with every curve

that's my kinematics My kinematics bases

some variables XC associated with every

curve multiplied by the alpha associated

with that with that curve.

Okay, this is something which looks like

shrinker parameization in every cone.

Okay. And this is something which is

equal to the sum over all the cones

the product of the one over uh of of the

x little c for the uh uh for the c's

belonging to the cones. Okay,

which is what we call the amplitude for

right.

So now this brings me back to your to

your question. Okay. So, um uh if if

we're if we're doing this uh before

doing loop integration, we're doing this

before we're doing loop integration, all

of these axes have the loop variables in

them. We would have all these infinitely

many winding guys, right? And that's

fine. Okay, I do the loop integration.

This is in Schwinger form. So, I know

what I get. I get uh I get the uh

semantic polinomials except these you

would call surface semantic polinomials.

So they're not semantic polinomials for

a fixed graph. They're the semantic

polinomials associated with the entire

surface. Okay. So there's an analog

semantic polinomials that we put all the

curves on the surface together. They're

the same old polinomials very similar to

the polinomials that you had before

except with these alpha C's showing up.

Okay.

Um and now you have something which is

mapping class group invariant. So you

have to mod up by mapping class group.

You model out by the math class by

putting a little extra factor of a

kernel in this measure that uh does

again to a physicist the fa popoff trick

for modeling out by uh by by by

symmetries. Okay, there's a very

standard and simple way of producing

this kernel. You can do it for all

surfaces at once. Uh it's algorithmic.

You can put it on the computer and do it

to 10 loops. It's it's all it's all

fine. Okay. So So these formulas are

absolutely concrete. There's nothing

formal about them. you can put them on

the computer and integrate uh okay so

Julia Salvator has put them on computers

and integrated them into 10 loops okay

so there's no issue

you know say there's an epsilon here a

rotate there just uh of course if the

kinematics is positive ukidian etc etc

once the kinematics gets interesting all

the usual physical issues about

thresholds and Ion those of course don't

disappear but I just want to stress this

is not some sort of formal object in in

the regime where the integral looks well

defined. It's 100% well well well well

defined. But now coming back to your

question you see the purpose so what

what I'm uh uh supposed to do in general

after I loop integrate is put this

kernel there. The purpose of the kernel

is to sort of kill uh the the far away

windings. Okay. So that's effectively

what it does.

>> Can you say a little bit more about the

kernel?

>> Yes.

>> You don't wake up every morning and

think about me.

>> No, I know. I'm sorry. Yeah. So let me

instead of saying uh let me let me say

it in a simpler let me say what it is in

a simpler example.

So um

and this is really all that's going on

but we can see it already in this

example. Let's say you have a a function

of of just one variable. Let's say have

an f ofx which is translationally

invariant by some amount a. So it's

equal to f of x plus a.

Okay.

But what you want to do is uh make sense

of integral minus infinity to infinity

dx f ofx.

Okay, this is of course infinite exactly

because it's uh translationally. Okay,

so there are these uh translations t

uh so this is equal to infinity.

And so what's our usual attitude about

this? Our usual attitude if you're a uh

you know if you're in high school

student or undergrad or something is to

find a fundamental domain right so in

this case fundamental domain would be a

little interval that starts at b's any

old b and goes to b plus a okay and so

you say really what I should do is I

want to make sense of integral dx fx mod

translations whatever this means mod

translations okay but what this should

mean is just the integral from b to b

plus a dx of f ofx. Okay. So this is the

fundamental domain idea.

All right.

And you can try to do that for surfaces

too. Just a mapping class gets more and

more complicated. This is not a fun

problem to try to identify a fundamental

domain. You can do it for a taurus, you

know, but things get uh I mean already

for a taurus slz is not, you know, a

walk in the park. It's not that hard,

but it's not the most trivial thing you

do. And then it gets more and more

complicated from there. Okay.

All right. But the the sort of fa popoff

or but physicists run into this issue

all the time, right? When you define any

path integral in the gauge theory, it's

infinite because it's gauge invariance

and the volume of the gauge group is

infinite. Um so we have to figure out

how to mod out by these uh by these kind

of symmetries all the time. And when we

do the path of gauge theory, we

absolutely do not do this. You don't

find the analog in the fundamental

domain which is insanely complicated.

That's a sort of space of all possible

engaging variant states. Ridiculously

complicated. You would never think of

doing that. Instead you uh judiciously

insert one. Okay. So what you do is you

say um I'm going to just pick any old

function. This case I'm going to pick

any old function g of x. Okay.

Let me pick some g of x. I'm gonna draw

g of x. Here's g of x. And g of x is

going to look like this. Any old random

g of x you like. Okay.

Okay. But I'm now going to insert one is

equal to the sum over all k of g of x +

k a

just translating g divided by the sum

over all k g of x + k a. So clearly I

haven't done anything. Okay. And I'm

just taking the function and all of its

translates. Okay.

So far so good, right?

So I'm going to insert that into my

integral. My integral is integral minus

infinity to infinity dx to begin with

one. But I want to somehow mod out by

the translations. Okay. 1 * f ofx. But

the one I'm going to write as a sum over

k g of x + k a over I'm going to call

this big sum g. I'm call this big sum

capital g. Okay, capital g. The only

thing I need is that capital G does not

vanish anywhere. Okay, so I want this

formula to be meaningful. So capital G

better not vanish anywhere. So G of X

had better be kind of like enough non

zero so that when I translate it, it it

can itself even vanish in a lot of

places like G of X could even look like

this. And G of X could look like this.

So long as this length is bigger than A.

Okay. So that when you translate it, the

translates sum together never go to zero

anywhere. Okay, so I just want this

denominator to be multiplied

and then it's sort of sorry times f ofx

and then it's sort of obvious what's

happening because precisely by

translational invariance in every term

of this sum I could translate back to k

equals z if you like okay so modding out

by the so if there was a fundamental

domain even without finding it I know

that uh that that this uh is going to be

the same as the integral minus infinity

to infinity just one term here for

example just g of x / capital g * fx

okay

if you like all the other ones are just

copies of this one by the action of the

mapping class and modeling precisely

throws them out now this is very cool of

course if you make a choice for g of x

to be a a a step right with size length

x a. This goes back to the fundamental

domain picture. Okay, you can put g

anywhere you want and this goes back to

the fundamental domain picture because

the sum of the g's is just equal to one.

But you don't have to be smart to do

that. Just choose a random g, any g that

you like. Um uh it's a little fun to do

it, you know, with a Gaussian or

something. It's slightly surprising that

this integral over everywhere without

thinking gives you the right answer, but

of course it's designed to do that.

Okay. So, this isn't what people what

physicists call the fa pop-up trick.

Perhaps it's not being dignified with

that. I'm sure did it in the 18 in the

1700s.

>> This is your way of taking a quotion.

>> This is a way of taking a quotion. This

is a practical way of taking a quotient

without manifestally identifying a

fundamental domain. Okay, that's the

sort of key key point.

>> Can I follow up? But absolutely one more

only one not two.

>> Okay. So, I'm going to I'm going to ask

the what's in it for me question and

it's as follows. So in this institute

there's a geometry group.

>> Yes.

>> And there's an algebra group.

>> Yes.

>> So now you moved you know everything

into the geometry sense of timma space

and the algebra.

Yeah.

>> So I would like to take something back

to the study of algebraic curves.

>> So the moduliz space mg or mgn

>> and I'm interested in algebraic curves.

>> Yes. Yes.

>> Not in remon surfaces. So what can how

can I make use of this technology to

study algebraic curves?

>> Uh I don't know um uh um uh what what

what what I would say is that um uh uh

is that what makes it useful for us

>> um uh is is in fact it's distinctly

algebraic flavor. I mean that's the

point. You just have these algebraic

equations you solve as ratios of

polinomials and this does something. it

uh it gives a gives an algebraic way of

characterizing

>> but the algebra is on the on the

geometric analytic side right

>> sorry

>> it's we have to somehow cross the

transcendental divide I mean you spoke

about ugly phobi theta functions they

seem nicer

>> oh yes well

>> we have to somehow go from the from the

analytic side to the algebraic side

maybe this I'm willing to do numerical

computations even with Julio

>> right well um um

Maybe one one one interesting point uh

uh to make is also related to your to

your earlier questions about the uh

infinitely many use. So you can ask I

mean there are we know that uh we're

talking about string theory at loop

level. There are jacobi data functions

everywhere. Where do the jacobi data

functions come from in this world?

>> For example,

>> the jacobi data come functions come from

the product over infinitely many curves.

Okay. So that's so that of course you

know the product representation for

dedicant theta or jacobi theta is like

product of 1 - y the k right what are

those product 1 - y the k in this

language the 1 - y the k is coming from

these f polinomials and there's

infinitely many terms of the product

because we have infinitely many

self-intersecting curves so that's uh uh

and uh maybe at least one thing that I

find uh interesting is that a lot of

this story uh can be abstracted away

from curves on surfaces. Uh there's even

more general settings where you can have

uh associated both with cluster algebra

and quiver representation theory and so

on and so forth. There's more general

settings where you can associate all of

these things. There's sets of variables

and u variables that go along with them.

They satisfy U equations. There's

infinitely many of them and so on. And

so it's natural to wonder whether

whether the things like jacobi theta

functions that are associated with

surfaces have similar generalizations in

these bigger sets that um uh it's

somewhat for example uh many just to say

a negative if I give you a jacobian

theta function the most exciting thing

about it is it modular properties and so

you can ask are modular properties

obvious in this way of writing things

absolutely not as far as I can see you

know what this tells you is that it's

good to write the the formula product of

1 - y to the k that comes out of this

sort of way of thinking

>> but we should see this and then using

the u equation formulism we should see

remon theta functions and all that

>> yes absolutely absolutely yes

>> and it could be a tool it could be a

tool to study and evaluate

>> it could be and I think one really a key

thing which is uh which has come up in

the question I also haven't uh answered

yet is it gives a natural regularization

of these things okay so that's kind of

kind of the point um if you're a if

you're a street theorist talking about

uh calculations at high loop order.

There are some greens function on a

surface that you have to compute. You

compute it roughly by the method of

images. It involves infinite products

that have to be regulated. They're very

subtle things. When they come out in the

uniform, they don't have to be regulated

because they're manifestly finite

products of things that are bounded

between zero and one. So uh uh and it

and it gives you a picture of what's

regulating it. The sort of degree of

complexity of the curves that you're

talking about is what is regulating it.

So I think the that connection what what

the probably the most uh the most useful

thing about it is a way to sneak up on

these very infinite objects in a well-

definfined finite way that has geometric

meaning right in terms of putting these

cuto offs on the complexity of the word

and on uh on how complicated the curve

looks on the surface. And finally coming

back to your question um you see this

now means so what was the purpose of the

g of x the purpose is kind of mostly to

shut off a little bit away from a right

so now of course uh when we when we do

this in our setting with the mapping

class we don't choose random g of x's we

choose the g of x's to be made out

exactly out of these alphas right so

it's very easy to build you know sums of

alphas whose translates uh are never

zero anywhere that you can put in the

denominator, right? But that means that

in this formula in the numerator, you're

just going to have the alphas for some

finite set of curves because there's a

finite set of curves upstairs. It's

there's literally a finite region in

this fan in which this kernel is non

zero. Okay? And so you don't need to

keep infiltur.

Okay? So that makes it uh uh that makes

it that's that's what really allows it

to be completely practical because uh uh

uh I told burn that these u variables

mostly go to one or if you tropicalize

them only in very narrow cones far away

uh do they matter but in a precise sense

you can throw them out if you're doing

uh the field theory uh computations when

you mod by the mapping class because

they are literally you don't go there

okay it's just that a fundamental domain

is a low brow way of saying it is

choosing one representative set of finer

diagrams. So if you choose one

representative set of diagrams to cover

the space. If you do that that's going

to look very ugly. That means you have a

weird collection of cones uh that you

say okay this collection and their and

their translates are going to cover the

space. But you have to artisally choose

the collection of cones. Instead of

doing that you just sort of light up

some region by this method. Okay. say

here's some sort of wedge and that's

what I shove upstairs and then I'm done

by by this uh by this uh by this idea.

So long as I just do the integral over

that cone that's going to it's it's

going to pick a third of a vinement

diagram from from this cone and 2/3 of

one from the other cone I'm going to add

them up in some nice way but without

thinking I cover the uh entire space and

then it's finite then it's really a

finite number. Yes,

>> I know you stated the third commandment

that is forbidden to ask the third

question. Yes,

>> I will cross this forbidden.

>> Yes,

>> things. Uh, okay. So, um it seems that

you are trying to um go back and forth

um in this crosspollination between

mathematics and physics to this problem.

Um it's normal you got this the kicks

for with abstraction.

Um yeah um one natural question um uh

yeah which arises I think um are there

any toy models in three dimensions with

the same formalism uh of course without

the monstrosity focus on the right

topic. Um

>> I I will stop you and say this is

already a a toy bottle. It's already

it's already a toy model uh that that

actually uh uh is dimension agnostic. It

could be in three dimensions if you

wanted. That's uh I I never said what

the dimensionality was in the story.

>> Yeah. So you're you're saying that

there's a natural generalization between

>> generalization or specialization, you

know, that's uh there's a there's a

there's the number D. It doesn't have to

be an integer. It could be complex, you

know, it could be anything you want. Um

uh but um that's one of the nice things

about working with the the scalers. As I

mentioned yesterday, although I won't

have uh uh I won't have time to explain

it in detail, it is surprising that

starting with this formula just for

scalers secretly knows about gluons as

well. Okay, so there's a way in which it

knows about gluons and pons also in any

number of dimensions. So that adds a lot

of uh physical excite and novelty about

these ideas that uh that you're saying

but uh but nowhere here do we say

anything about the number of space

professor you're saying that the

generalization is completely obvious and

we have only three three matrices for

example for three.

>> Oh no no no that that three is nothing

new in any number of uh spacetime

dimensions you'd always have the same

2x2 matrices. If you're asking if there

it's a generalization where you have

instead of words I don't know some

threedimensional words and you'd

multiply 3x3 matrices and 2x2 matrices I

have no idea probably there is because

mathematicians are very interesting

people who invent all kinds of things

but uh but uh anyway uh I think we have

to stop now for the break right for the

tea break anyway so but so what what

what I what I can do in the time that I

have is give you what I mentioned in the

beginning is the sort of a very new

application of uh these uh of the ideas

that we've just been talking about. Um

uh

that gives us access to a very

interesting region of uh uh physical

processes involving uh scattering with

asmtoically large number of particles.

Um so

so um so we're going to again talk about

this uh trace by cube theory but I want

to imagine I have amplitudes

um oh this a nice truck in trace 5 cub

um but where I have sort of n particles

um and n goes to infinity. Okay.

And I want to say something about what

these uh uh amplitudes look like. Now

again, uh I'll just say this uh uh I

said it already before as part of the

motivation. Um what I like about this

this question is it forces you to think

uh it makes it very clear what part of

what we're doing before is continuously

connected to standard ways of doing

physics and what could be really new

because none of the ways you think about

computing uh amplitudes whether we

interpret amplitudes as canonical

forearms or BCFW or other kinds of

recursion relations every one of these

pictures has something recursive built

into it right a canonical form is an

object

that recurses to a canonical form,

right? Um so uh you build up higher

point amplitudes by gluing together

lower on points, right? So it's all a

picture that the simplicity is in few

particles and complexity is in many.

Okay? And so we're looking for something

essentially new where the simplicity

when the partic number of particles is

huge. So that's what what we're looking

for. Okay.

And um the clue that that such a thing

is possible is from the tropical

representation of the trace 5 cube

amplitudes that I now want to write on

this board. Okay. So if I look at the

tree amplitudes for trace 5 cube for n

then I can write it as this integral dn

minus 3t

e to the minus some s

okay that depends on t

and s of t uh this is now associated

with this picture of the mesh okay

so s oft T is the sum of X1 J T1J.

Okay. So these T's are associated with

this bottom uh boundary. So this is this

is the mesh.

Um we mentioned it I mentioned it before

but that's associated with this kind of

uh this kind of triangulation where

that's one 2 3 4. Okay. So um

um and

uh so again this is just a picture uh as

a pneummonic for thinking about all all

the variables in the problem that makes

it easy to write this expression. Um and

then you have for every internal for

every internal C you have a sum of these

internal C's. So the internal J is CI

times something. It's a max of zero and

a bunch of things, right? And um so just

let me just give it concretely in this

example. Um just I I hate it when people

write formulas with bounds on indices

that you can barely read on the side.

Okay? So I'm not going to do that. So

I'll just sort of show what it looks

like uh in a big enough example where

you can see how it generalizes. Okay. So

if I have this example that's 131415.

Okay. So my action is t13

is uh t13 x13 plus t14 x14

is t15 x15.

Okay. And then uh associated with this

C. So this is C13. I have plus C13 max

of zero and minus C13.

Okay. So in Karolina's language remember

this was associated with this little

that little uh uh mic in that direction.

This guy's the other direction. This

guy's uh uh uh the other direction. So

there's C13. This would be 2 4 5 2 36 35

26 and 46. Okay. So then I'd have plus

uh so I'm going up this way also because

organized by how complicated these uh

these things are. So plus c24 max of 0

and minus t14 plus c35 max

0 and minus uh t15

and then uh here and here I have plus uh

c14

max of zero and you see I've gone up to

14 here. Okay, so it's max of minus P14

and minus P14 minus P13.

Okay, so I go as far up as I go and I go

down from there to see it's a max of uh

all of those guys. Okay, so this guy

instead would be plus uh C25

max of zero and minus T15 minus T15 -

T14.

Okay. And finally this one would be plus

C15

max of 0 minus T1 5

- T15 - T14 - T15 - T14 - T2.

By the way, I apologize if these things

have minus signs on them. You might have

thought it would be convenient to uh uh

you know just reverse all sign

conventions and call all these things uh

plus signs. If you thought that you

would agree with all my collaborators uh

and so in in the papers you'll find it

in that uh officially intelligent way

but I have my own personal reasons for

preferring it this way. So screw you

all.

All right. Um, and that's a technical

term for my collaborators.

Um, okay. So anyway, so what again the

the and so I hope that the the pattern

for any n is is is clear, right? You're

getting you got a max of zero and these

strings of sums of uh of uh of negative

t's. Okay.

So once again in this example already we

see we have you know uh 1 2 3 4 5 6 we

have six little uh tropical functions

that are going to turn this

three-dimensional space into 14 cones.

Right? So here we have 14 diagrams at

six points as catal numbers we have 14

diagrams and so we're starting to see

the point right that as you build a

large n there's n squared roughly of

these uh of these maxes they have order

n terms each and magically they they

they uh their domains of linearity turn

into all the roughly four to the n

diagrams at larger. Okay.

So that's why there are some sort of

vague hope that uh maybe this integral

it's a single integral. I'm not talking

about summing diagram. It's a single

integral. It's a single integral made

out of like simple objects that if you

sort of squint even maybe have some kind

of seeming nice continuum limit like

these sums of consecutive t's maybe look

like an integral. You go to large n you

replace sums with integrals or that

integral kind of turns into a path

integral. There's all sorts of sort of

vague words that you could say about why

you expect something nice to happen here

in the large end limit. Okay, that's

exactly what what we're going to see.

Um, but I first want to uh uh I want to

do two things here. Um uh first I want

to be a little bit more precise about um

what a large end limit could mean

because you take a limit you always have

to say what you're holding fixed. Mhm.

>> Okay. So, we have to have fold something

fixed as we go to infinity.

And then I want to do the super simplest

case of the large end limit just to give

us an idea of what what we're looking

for. Okay.

So, the first comment is uh what are we

holding fixed in the large end limit.

Okay. Uh what's fixed? So large and

kinematics essentially I want to talk

about

and if you're a physicist we could

actually spend a fair amount of time on

this topic. think it's actually uh uh

interesting to talk about but let me

sort of give you a quick impression.

I'll make a very very precise statement.

Um so if you're a physicist you'll care

about the physical implications of all

these things but we can just sort of

make the precise statement quickly. So

remember we said that uh we started with

this picture where we imagined you know

we have momentum of particles that we

put uh end to end uh to make this

momentum polygon to give us uh this

picture of a momentum conservation.

Okay, so this picture immediately so as

n goes to infinity, you have to somehow

give me a large number of momentum,

right? And but in some way that

something is held fixed. So the obvious

thing to do here is to say that what's

held fixed is just this curve. I draw

some curve. Okay, I draw some curve C

and then for any finite end, well I

just, you know, point down end points on

this curve.

Of course, maybe I plonked them down

with some density. You know, there's

more details we could talk about about

how I plunk them down, but at zero

order, I'm just going to plunk down a

bunch of points on this curve, and I'm

going to use that to define my momentum

polygon. Okay, so this defines my

momentum polygon

for any finite.

Okay, so at least here we've defined

something, right? Here we've uh here

we've defined uh uh here we've defined

something uh that is going to uh stay

fixed as n goes to infinity. And so

that's what we want to know. We're going

to have the amplitude is going to depend

on these n momenta. But the hope is

again this is sort of vague that at

large n the amplitude will only depend

on this curve c and it won't depend on

the particular way that you put uh

points on it. Okay, that's the that's

what that's what uh we want to see if

it's true. something like that is true.

Now that picture

immediately translates to a kind of an

obvious state the language of this mesh.

If we think about this mesh as defining

our kinematics space, don't worry, I've

got to write down a simpler version of

this equation again if you're missing

it.

um is that uh so another way so so so so

kinematic limit to another way of

talking about it slightly more general

way of talking about it is draw this

mesh

okay

but now just imagine that the x so so my

kinematic variables are some xigs in

here right so and I'm just going to

imagine this is a very fine mesh okay so

I'm going to imagine that the xig's on

the inside are really secretly some

smooth function I'll call it x of u and

v but when for example u is i n and v is

j overn

okay so it's like here's u it goes from

0 to 1 here's v it goes from 0 to one

the other way I don't know right so I

have a smooth function of x of u and v

inside this triangle and I'm just

discretizing I'm just plotting down a

mesh and reading off the x objs from the

value of that smooth function evaluated

on uh the discrete points. It doesn't

have to be a uniform mesh if we do it in

uh in in many ways.

So this is the this is the sort of

slightly more general picture. I mean

obviously everything that looks like

that will give me something smooth in

here. Okay, but now we're just going to

say it in here. I'm just going to have a

smooth X inside this uh have a smooth X

inside this triangle.

Okay, is that clear? So, and so once

again, the hope now is that the

amplitude as n goes to infinity is now

just going to become a function of this

smooth x of u and v.

Okay.

Okay. Now, now let's get a

>> can you say one more time how I compute

the x from the

>> I'm sorry.

No, but suppose I have a continuous

curve

>> and I want to compute the continuous x.

willing to do a continuous computation.

>> Yes, a a a a a continuous computation.

You know, I pick uh you know, a point

here and call uh uh zero. Okay. So, my

my my curve is labeled by some little x

that depends on a parameter, let's call

it u that goes from 0 to one. So, here

is u equals z, u equ= a half,

>> u comes back to one here. Okay. Mhm.

Uh and so my x of u and v the sort of

first guess for x of u and v would be

little x of u minus little x of v squ.

Okay. And if the particle has a mass

plus m

okay

so that's what we're that's what we're

talking like in the smooth curve or

>> Yeah. Yeah. That's that's that's how I'm

defining the uh the uh the Okay.

Is there some notion of some shellness

in the in the curve like the E square?

>> You could you could you could ask for

that or not. Okay. Uh and actually so

this this gets into the longer physics

uh uh discussion. You could ask for that

or not. The simplest thing to do is just

say these are whatever they are. You

would call this an offshell correlator.

Uh the formulas are exactly the same.

You could interpret as an onshell

amplitude with a little bit of

scaffolding on top of these guys if you

wanted. Okay. So um and and that that's

where there's a there's about a 45minut

discussion of all the different ways we

could draw this curve that could be

space-like. They could be timelike. They

have an interpretation of a bomb going

off. They have the interpretation of

lots of particles going in, lots of

particles going out. So there's lots of

different sort of physics words.

>> But then there was also the C. So the C

is the H. You know that

>> uh that that's we're going to come to

the C's in a moment. That's going to be

the sort of key thing that makes that

that that that gives some hope that

something simple is just going on. Okay.

Okay. But in fact, so so so before uh

before before getting there, I want to

do the very simplest version of a smooth

X, which is all x's equal. Okay, so

that's the very simplest thing you can

do. What what if all the x's are equal?

Put it here.

So if all x's are equal,

then the amplitude is very simple. What

is the amplitude? Well, every single

finding diagram, every diagram is the

same. It's like 1 /x to the power of the

number of propagators, right? So all the

x's are the same. So the amplitude is

just 1x the n minus 3 multiplied by the

number of diagrams.

And number of diagrams are the catalan

numbers.

And it's very easy to see that the

catalan numbers grow like four to the n

at large x. So we said said that

already. So already here we see that

this goes like 4x

to the n at large x.

Okay.

So that's our sort of first more precise

hint for what we're looking for. And if

you're a physicist, you're very excited

to see that adding the exponent. Okay,

because the amplitude looks like

the amplitude looks like e to the n

times some little some function a of x.

In this case, a of x is just uh you know

log 4x.

Okay.

or I put a minus sign there and it would

be log of x over 4

either way

but it's very common in physics when you

have a small coupling in a theory that

the leading amplitude or partition

function or whatever looks like e to the

minus some coupling normally we call it

1 / g ^2 times the thing that sits here

is controlled by some classical

physics

that's the leading thing this is called

the leading classical behavior and then

you do a semiclass expansion around it

and so on and so forth but the sort for

example in the path integral uh in the

finest path integral we have e to the 1

/ h bar it's pine's constant that goes

downstairs in the action okay so

>> so in some sense this equal one is a

complicated case because the

dimensionality of your curve also goes

up

>> should I think about your polygon

staying in a so your curve being a fixed

dimension.

>> You can think of it as saying in a fix.

>> This is not a good model.

>> That's right. This is not a good model.

Exactly. It's a bad model for that.

Exactly. Exactly. So, this is the

slightly longer uh uh discussion about

exactly what does or doesn't work for

the story that I'm telling you. Just to

to to skip to the end, there's a the

story I'm telling you is valid for a

fixed curve and fixed dimension. But

you're right, x equals 1 is not a model

for that because we cannot realize x

equals 1 from any curve in a fixed

number of dimensions. Absolutely. Okay.

But somehow uh I mean uh it's kind of

interesting in this entire story.

Everything is about the x's in our

stories. Everything is about the x's.

That's the whole point. The moment are

somehow far away somewhere, right?

Everything is about the ex. So it's

natural to ask this question in our

story just to get some sort of first

handle on what's going on. But I want to

stress that this is now looking exciting

because it looks like at large n we

should be looking for some kind of

theory whose weak coupling is 1 / n.

Exactly. Right. One over g^2 to be n. So

n is 1 over g^2. So it looks like

there's a dual theory at large n. Okay.

So if you're a physicist those were in

the beginning of any talk on the subject

you know you would say this is evidence

for some kind of dual theory at large n.

We're going to be finding this dual

theory at large n. That's exactly what

we're looking for.

Okay. So, but of course, as Burn was

saying, that's a very very special case.

Um,

let's imagine more general X of U and B.

And now comes a second uh uh interesting

point. Maybe before we get into this uh

a second point, um let's go back to this

picture. This is actually essentially

repeating uh slightly more slowly uh

something uh uh Kolina was saying at the

end of her talk yesterday. Let's say we

have this uh this example at six points.

Again, sorry for the bad drawing. Uh

and it's just kind of interest. So the

whole associated derivative, the whole

amplitude language involves turning on

all of these C's. Okay, but let's just

see what happens if we turn some of them

off. Okay, Carolina mentioned that there

are some limits where we turn so many

off that this association collapses the

dimension that the amplitude gives us a

zero. But let's just back up a little

and see what some other sort of simple

things look like. One simple thing you

can do, remember these guys were

intervals, right? Whatever direction

that was, right? So, one kind of obvious

thing to do is just turn off all of

these guys and just have these

intervals.

Then the object is really simple. It's a

hyper cube just a cube in this case,

right?

And also the amplitude is extremely

simple in that limit. Okay, so if you

remember I erased the action but let me

write it again. You know these

contributions. So my action was uh t13

x13 plus t14 x14 plus t15 x15.

Remember I'm integrating the amplitude

is integral uh dt e minus s. So in this

example, this is S has these linear

pieces and then these guys were the

simple uh things whose whose maxes uh

only involve one variable at a time. So

C13 max of 0 and - X13 plus uh C14 max

of 0 - X14 sorry C24 and C35

max of 0 and - X15

T. Thank you.

Okay. Um, so these integrals are really

simple too. It's just, you know, a

single integral, right? Uh, so what is

that that integral? What is this action?

This s like when t is positive, uh, when

t13 is positive. Uh, it's equal to t13

x13.

And when t13 is negative, uh, that's the

max. So here it's equal to this slope uh

so this is c13 minus x13

uh sorry this is the times minus p13

okay so just as a slope of different

slopes I made them look the same but

different slopes on this side and that

side and so when I do the integral from

that I get 1 /x13

from the integral from 0 to infinity and

from this side I get + 1 / c13 - x13

Okay.

And so I just get a product of those

factors for each one of these things. So

times 1 x24 plus 1 / c 24 - x14 1 /x uh