magic that I showed you. And the second
bit of related magic, although I didn't
relate them, will relate them today, I
showed you is that if you just think
about finding diagrams, you're just
supposed to add different terms. This
diagram, that diagram, that diagram, you are supposed to add all of them up together.
There's no obvious sense in which this
diagram is related to that diagram, or
this diagram is close, or far. Nothing.
They're just different things, different
numbers and you add them all up. And yet
what we did is uplift every term in the
final diagram expansion to a
differential form instead. So instead of
writing down one over propagators, we
wrote down the log of all the
propagators, and then there's this
remarkable fact that it's possible to
assign signs to every one of those terms,
in such a way that the entire
differential form is actually secretly
only a function of the ratios of
the kinematic invariants. So, that's what
we call projectivity. Now that's a
miraculous seeming fact, just
algebraically, because if you just count
equations and unknowns, there are vastly
more equations
than unknowns in order to fix all the
signs. So, it's kind of a miraculous fact
that it's possible to assign signs to
all those different pieces in the
differential form for this fact to work. And I'll sketch
quickly...and well probably flush this out
more on Thursday when we really talk
about the connection to the ideas of
cluster algebras, but at least I sketch
how just that picture of why the form is
projective suggests that you should
think about some diagrams as being close
to other diagrams, okay because
when you do this when you do the
projective variation, when you do the
projective variation of the form, you
find that you have to have the
coefficient not of all the d-logs
of all propagators but one, they have
to add up to zero, and it's just the fact
that there's precisely two diagrams that
have one fixed set of n minus four
propagators in them. And therefore if you
want the signs to add up, you have to if
you start with one sign for one of them,
then the one where you just flip, one of
the propagators to another is gonna
have the negative sign. Then you got to
flip another one, who is gonna have a
positive sign, and the negative sign, so
there's this way to walk through
everything, and that assigned signs to
everybody, and you're done. And then it's
just a miracle that it works for all the
rest of them. Okay, so this is a very
non-trivial fact that this is possible,
but it gives you an idea that you
should not be thinking about amplitudes
as numbers again, but somehow thinking
about them as forms, is exposing more
hidden facts that are not obvious in the
way they're spat out to us by the path
underground. Okay, so all of that was
motivation. All of that was motivation
that there's something going on even in
boring old planar phi cube theory. And so
today what I want to do is again go back
take the same sort of philosophy that we
had for the amplituhedron, and go to the
kinematic space associated with this
scattering process, stare at the
kinematic space for a little while, and
then ask a question in that kinematic
space whose answer is going to turn out
to be the pre scattering amplitude.That's what we will be talking about today.
An extension of these ideas, by the way
this is all work that's actually, I mean,
the connection between phi cube
theory and the associahedron goes back two
years by now. I probably won't have time
to talk about it, unless anyone wants to
come back for a super-duper bonus
informal last after lecture where I'll
talk about the connection to all these
things with a string theory, so they're
very intimately connected to a string
amplitudes, all this collection of ideas
are around the same, around the same thing. But anyway, that will be really unofficial, but
the connection between phi cube theory
and the associahedron goes back, this
explicit connection I'm talking about
goes back two years. The particular way
that will derive it is actually new, and
should hopefully come up by before the
end of the year. Okay, so, and it also
works in the simple way that I'm going
to talk about it works to one loop
order, okay, so we can do up to one loop
amplitudes for phi cube theory, and as I
mentioned last time or alluded to last
time, attempting to make sense of this to
all loop order is something that we've
been trying to do, and I think also interacts in a very
interesting way with some with novel
structures in cluster algebras which
you'll at least get some taste of next
time. Okay, but today we're just doing
very simple things. So first, let's talk,
let's remind ourselves what the
...whoops...yellow, what is the
kinematic space. The kinematic space
we're talking about, so as usual we have
n particle scattering, we have our
ordering and we defined these variables
from i to j, Xij which are pi plus dot
dot dot, plus the pj minus 1 squared. So these are chords on our favorite polygon. Okay?
So our kinematic space is just the space
of the Xij, and as I reviewed last time
the number of Xij's which is equal to
n choose 2 minus n, n n minus 3 over 2,
is also equal to the number of
independent Mandelstam invariants, so
these are not only beautiful variables
associate of the poles of cubic graphs,
they're also a basis for all the
Mandelstam invariants, ok? So, there's
something very nice about them. Ok, now, there is a lovely thing
about these Xij's, as we've said, you
stressed already, using the Xij's there
are something nice which is...there's
always slightly annoying if I wanted to
talk about, I don't know, this propagator,
I could either say that it's p1 plus p2
squared, but that's the same as p3 plus p4 plus p5 squared. Right.
So using the Xs is nice. We just call
both of them X13, right? They are just
associated the chord that cuts the
polygon in two, okay? But we still have the
little extra thing that we have to add
that Xij is the same as Xji, so it's
sort of better than before, it only
depends on the chord but the ordering of
the i and j doesn't matter, right? So, okay,
okay. So our first item of business is
then to just try to represent the
kinematic space, right? So I just want to
like represent all of these X's, but I
want to represent it in a way that makes
the cyclic symmetry obvious, right? And
the fact that Xij is the same as Xji.
And already what do you think about this
is there's it's slightly
interesting, okay? So let's try to draw a picture for a
kinematic space. There's a picture for
kinematic space for I don't know for N
equals 7, N equals 8, I want to make it
big enough so that...Okay, now just
for convenience I'm going to include in
the picture the Xii plus 1, now these
are 0, okay? Xii plus 1 is 0. That's just,
that's exactly the no momenta on the
outside, okay? But I'm just gonna find it
convenient to add the Xii plus 1, okay? So
I'll denote them with an open circle,
okay? So here's how I'm gonna do it.
Now, obviously this is gonna be a grid,
right? It's going to be a grid, because I
have to represent the ij indices. All I'm
doing here is representing ij indices, so
it's going to be a lattice of just ij
indices, but I'm gonna draw it, tilt it
over by 45 degrees, for reasons that'll
be clear later, okay? So there's
nothing deep about drawing it as 45
degrees, it's just the lattice.
The slightly interesting thing is going
to is to see how the cyclic structure on
the Xij equals Xji is reflected in the lattice.
So, but anyway, I'm going to draw it
like this. So this is going to be the
variable X12 and then X13, X14, X15, X16, X17, and X18, which is also
0, right? Ok, and then, so this is going to
be the direction of X1j, with
increasing j, right? So starting with 2
and going all the way up to 8, ok? So then
this guy here is going to be X23,
ok? And then, and so on, right? So then, so
this will be X24, X25, X26, X27, X28, oops...
Here is already screwed it up.
Okay? So is it clear what we're doing here, right? So now, so I'm
just going to keep on going like this, so
I'm just making a mesh. Oh boy! Wow!
The X34, X45, X56, X67, and so on, right?
And the picture goes in both
directions, so, and in fact the way of drawing it, so here is the
sort of it, so the whole picture sort
of lives in a strip here, whole picture
lives in a strip like that, and so in
the strip I have captured all of
the Xij's, and I've made it, so that the
cyclic symmetry is obvious, right? Cycling
everything over by one. But I get Xij
and Xji appear both in this picture, right?
Over...here, I have X14, but we'll see X41 up there somewhere,
ok? So we have both Xij and Xji n this picture. Let's say you want to just
capture one copy of all the Xij's, right?
So not Xij and Xji together. Then you
have to take a chunk out of this picture.
And there are many ways of doing it. I'll
say what the general way of doing it is
in a second, but one way of just choosing
Xij and not Xji, one way of choosing a
chunk which covers Xij versus ji just
exactly once is to for example take this.
So here I would have 12, 13, 14, 15, 16, 17
18 here, but then, okay? If I fill
in the rest of the picture this would be
23, 24, 25, there's 18. And then we go up here, we get 23 34
45, 56, 67, 78 up here,
and so you see precisely by stopping
here, if I went one over, I would have
81, right? But I already have the
18 down here in the picture. I
have a 17, if I kept going over
here, I'd have a 72, okay? So if I
want to just keep everything where the Xij versus Xji covered once, I have to
take something like this. There's one
example, I could do other, you know, I
can take many chunks out of this
infinite mesh that cover all of Xij or
Xji precisely once. This is a
particularly simple one. Okay, so, now what
would happen if we wanted to go from
eight particles to a hundred particles,
okay? Well I mean, I would just draw a
similar mesh or just get denser and
denser and denser, okay? So it's actually
convenient if I want to imagine treating
all particles in one go, just imagine
the whole picture is a strip, as a
continuous strip. This is not saying it's
actually continuous, it's just saying
that I think of kinematic space, just
abstractly is this continuous strip, and
then if I want to choose any N, I have to
just plunk down some integer
lattice on it out these 45 degree lines,
okay? So let me first give you the
description of what the continued what's the infinite N kinematic
space is, okay? So infinite N kinematic space is just
this infinite strip, okay I'm going to
call this direction t and
this direction x, later we'll interpret
theses as time and space, okay? And some
auxiliary. This is in kinematics place,
not space-time, this is some funny
kinematic space, but just for now. But all
I want you to see is what is
this, if this point here, if this point
(t,x)  is labeling ij where is the
ji point? ji is (1+t, 1-x). If this thing has, I just defined it to
have width 1, okay? So that's the ij to ji
identification. So I've been sort of a
Mobius strip here. I have an infinite
strip, but if I want to cover, but the ij
to ji identification is making this
Mobius identification on the infinite
strip together with the cyclic rotation,
right? But the cyclic rotation, here the
cyclic rotation has width 2, because
in order to just cycle everything over
all the indices over 1, if I just do it
cycling over 1, it is not taking ij back
to ij, taking ij to ji, so if I wanted to
come back again, I have to cycle it over
by 2. Is that clear? And that's... So we have a literal
identification of something of width 2
in this picture, and
there's a mobius identification, and the
cyclic rotation that cycles all the
indices over comes back to
itself when we shift everything over by
two, okay? This is the cyclic. That's a
cyclic shift. Okay. And now if I want to
cover ij and not ji in this picture, so
if I want to take out a chunk that
covers ij and not ji, then I just choose
any curve here C, I choose any curve on
the other side
I'll call, drawing it badly, any curve on
the other side I'll call C bar. The
curve C bar is what I get
take C shift one to the right and turn
upside down, so that's exactly what this
(t, x) to (1+t, 1-x) transformation is doing. It's identifying
a point here to the point there, and so
if I take anything in this region
everything in that region will cover ij
and not ji, okay? So in the continuum,
that's what I want to do, okay? So, I want
to take a curve, and in
the language of tight time and space I
want to take a space like curve and
shift it over by one in time, and turn it
upside down, and anything that is
captured between them is going to cover
ij and not ji. So we're just practicing
with what kinematic spaces, ok? So there's
many ways of picking out just that...there
isn't the way of treating of seeing all
the symmetries manifest, okay? Because of
the slight annoyance that ij and ji are
the same thing, if you want to cover
things non-redundantly, the thing which
is perfectly symmetrical is the whole
strip, but then you have to make a choice
to take a chunk out of the strip that
covers ij and not ji, ok? So far so good?
An extreme limit of this is the simple
case where C actually becomes null. So if this is C,
then C bar is just that. C bar which
is translating over by one and turning
upside down becomes this, and that's just
the picture that we talked about there.
So this now has width 2, and this is 1,
and so the inside of that region is
exactly what we're just talking about in that simple example.ok? So far so good?We are just talking about there's almost
nothing interesting going on here. It's
just the way of plucking down ij, right?
You know we just have a 2-dimensional
space of kinematic invariance ij, we
want to allow ourselves the freedom to
talk about any number of particles, so we
go to the continuum, and the only little
subtlety is how to make sure we cover
ij and not ji, okay? So already that, but
this is our canvas now. Our
canvas is this kinematic space, okay?
Pretty boring place, right? What the hell
are we supposed to do here, okay? Just
like N 4 vectors in momentum
space is a pretty boring place, and what
the hell are we supposed to do here?
Well we managed in the kinematic
space of N 4 vectors to find some
injection of the word positivity in
order to bring to life all of the
structure of planar N= 4 super
Yang-Mills, ok? So now we're going to do
something analogous here, okay? We're
gonna ask a question in this kinematic
space, just on the strip, that's all. We're
gonna ask a question on the strip, and
the answer to this question will end up
in a very similar way giving us the
amplitudes for phi cube theory. Along the
way we'll discover the associahedron,
and after we play around a little bit with what we're doing,
we'll also get a clue as to what these
generalized clusters of the associahedron.
Okay, any questions so far?
Almost nothing interesting has happened
yet, so we're just setting up, but any any
questions so far?
Okay, good. So now let's forget about all
of this,
since our canvas is just this two
dimensional space. The question that
we're going to ask in this two
dimensional space is going to end up
being, oh, that's right, let me tell you
what what the goal is. So what are we
supposed to make obvious? Sorry let me
just stress again. What we'd like to make
obvious is that, remember we have
amplitudes associated to these cubic
graph so the triangulations of the
polygon? It is not obvious that the
triangulation of the polygon are
vertices of this associahedron. I told you
it's a mathematical fact, but it's not an
obvious fact, okay? It's not obvious that
the combinatorics of the compatible chords in the triangulation
of a polygon is polytope, and it's not
obvious as that polytope factorizes on
the boundaries. What we'd like is some
picture ultimately for the amplitudes
which makes it obvious that it's polytope and then it factorizes. That's
what we want to make obvious. So we want to make all those facts just completely
manifest from some simple picture, okay?
That is our goal. Remember something
makes just factorization obvious, Feynman
diagrams make factorization obvious, the
string worldsheet makes factorization
obvious, they don't make the polytope
nature of the combinatorics obvious. We
want to make everything obvious, both the
polytope nature of the combinatorics
and factorization. So that is our goal,
okay? And that's what we're gonna do in
the next half hour. Okay, but to begin
with let's talk about the wave equation
in 1+1 dimension,
okay? So again it doesn't much matter
what I call time and space 1+1
dimensions because everything is
symmetrical anyway, but I'm gonna imagine this is time and space. So
the normal wave equation is dt squared
minus dx squared of some variable
that we'll call capital X equals 0,
right? But let me allow myself a
source as well, and I'll call the source
little c of x and t, right? So that's the
usual wave equation. And let's immediately transform to u and v,
variables, the light-cone variables which
are x plus or minus t. And we know in
terms of u and v the wave equation
becomes du dv X equals c of u and v now,
right? Ok now let me review a few things
about the wave equation. First how do we
solve the wave equation? Well what you're
used to doing, from whenever you learn it,
you're European kindergarten I guess,
if you're your American sometimes sophomore
in college... is that... but the Americans
catch up quickly, or at least they used
to, is that you're supposed to give
boundary conditions on some surface and
the time derivative there, then you can
predict the future of course,
right? It's actually going to be
convenient for us to present it in a
different way. We're gonna be interested
in boundary defined on causal diamonds.
So this is a picture of a causal
diamond, right? So here's some event, time
is going this way, so the waves go in
this direction, here is two space like
separated points, and then... so this is the...
so the causal diamond associated
these two time like separated points is this picture, right? The
intersection of the future light cone out of the point A with the past light cone
of the point B, okay? And that's quite
obvious, and I'll give you the reason for
it in a second, is that if you give me
the data on the boundary of the causal
diamond, in other words we just give me x,
nothing about x dot, you just give me x
on the boundary of this causal diamond,
so x of, if you like, x of, if you
give me x on these boundaries or these bounds, you can give
me x on any boundaries of this causal
diamond, I can fully determine what x is
on the inside. And actually let me give you the immediate general
reason for this, which is just the
following. Let's say that the value of x
here is xA , xB, xC and xD, ok? Then there is a Gauss's law
associated with the wave equation. The
integral form of the wave equation is
very simply that xA plus xB minus xC
minus xD is equal to the total current,
the integral of C on the inside of this
diamond, okay? That's very, it's very very
easy to derive. In fact you can think of this Gauss law as the
fundamental integral equation, the
fundamental fact, if you squeeze the
diamond to 0 size then the Gauss law
turns into the wave equation, right?
Because exactly xA plus xB minus xC
minus xD is the du dv derivative, so
give you du dv equals C, just like you were
used to and going from the integral to
the differential form of the Gauss's law, and I think which is a
little bit cool is that it in Gauss's law you're used to
thinking about sort of integrating
something on the boundaries, but here we
just get the contribution from the
corners, and that's because this is a
Lorentzian picture, so the the the
the the contribution from the edges are
just zero and I just get this simple simple relation, okay?
So I can use this fact to determine
x anywhere inside this diamond, right? If you if I
plunk down, if I want to know what is x
in here, I just draw any picture that I
like. For example if I draw this picture,
then I can determine x in terms of the
boundary values here and there. If I draw
it like that, I can determine this x in
terms of the boundary values there and
there, so okay? So again, time and
space are totally symmetrical here, so
you can sort of decide how you want to
interpret this fact, but in 1+1 dimensions if you give me the value of x
on any two of the boundaries of the
causal diamond, I entirely determine what
x is inside the space, okay? Alright. So
that's fact one about the wave equation.
There's another fact about the wave
equation we'll need in a second,
another simple and kind of familiar fact is
the following. So let's say that, so
what I call scrunching, so let's say
you have some solution of the wave
equation, but you take some 45-degree
strips, okay? And there's a solution out
here, let me call it x to the right, and
some solution here x to the left of the
strips. Then the claim is that from
this solution of the wave equation I can
build another solution of the wave
equation, of a different wave equation... sorry
same wave equation with different source,
just by scrunching away this region.
So what I do is just collapse this
region to zero size, I still just put xR here and that xL there, okay?
But now what do I have to do? I have
to include some source which is just the
integral of whatever the source is in
the strip as the delta function on this
juncture, okay? Now, and why is this true?
Well it's simply if I only care about x to the right
and on the left, back in this
sort of full geometry, I would just draw
one of these causal diamonds, Gauss's law
tells me x plus x minus x minus x is
that total current, and so it doesn't
know that it's finite size thing, right?
As far as this is concerned, all I have
to do is reproduce the total amount of
charge that's in that region that I scrunched away, okay?
So these are two simple facts about
the wave equation. You have some
complicated solution of the wave
equation some source, you can just
scrunch away some region and replace it
and the rest of the wave
equation is unaffected, okay? So, okay? Is
that clear? Okay. Those are the only
things we're going to need to know about
the wave equation. Okay. So now comes the,
so still nothing too exciting happening.
So now comes the idea is that we're
going to demand positivity. So we're
going to solve the wave equation du dv
capital X equal C. We're going to demand
two things. We're going to demand that
the source is positive, and we're going
to demand that on the time like boundaries of the space
that we're going to look at, on the
boundaries x is... sorry we get a demand
that C is positive, we're going to
demand that x is positive. And that's
actually what we're really interested in,
okay? We're interested in
characterizing the space of all
solutions of the wave equation, which
have the property that for positive
source it stays positive, everywhere
inside the sort of domain that we're
interested in. So now we're going to talk
about asking that this is true in a
chunk, and in one of the chunks that we
talked about before, okay? So the kind of
simplest thing we're going to do is to
ask that this is true. We're going to ask
that this is true. And a spacetime that
looks like this, right? So I'm going to
try to solve the wave equation in this
spacetime, okay? And you can quickly
convince yourself that trying to find
these solutions, the solutions to all the
x's that have this property, you first
have to solve a slightly more restricted
problem where x is actually 0 at the
boundaries. So it's 0 here and 0 here,
okay? So I'm demanding that x equals 0
here, x equals 0 here, and then we just want to ask that x is
greater than or equal to 0 in there with
the source is greater than or equal to 0.
okay? So that's what we'd like to
understand. We'd like to characterize
solutions of the wave equation with this
property.
Know that we can think, again since this
is time, we can think of this as having
specified the state on this initial kind
of null surface, and I'm evolving forward
in time, right? Or I can think about it
the other way because time and space
are symmetrical, I can also think
that I'm evolving forward in time in
this way, but the way I've drawn t you would think of this some
initials, so a null surface that you
defined the data on and that were
evolving forward in time. So in that way of saying things
what entirely determines the solution of
the wave equation inside here is giving
us boundary values here, okay? So let's
try to understand why that is
just explicitly for a second. So let me
put this in here so I want x greater
than or equal to zero here, and C greater
than equal to zero here, okay? But
we want to ask in the space of possible
sort of x(u) on this slice, what
allowed boundary conditions? Can I have
for x(u) such that it's compatible
with x staying positive everywhere
inside when the sources are positive,
okay? Is it clear the question? Sorry, yeah. With
that, so x(u) will have to satisfy the
boundary conditions 0 at this end and
0 at that end, and it satisfies these
boundary conditions that we go on forever
in both directions. We could have
have drawn another...I mean
again we... this is for any chunk, I could
have also drawn something that just
looks maybe slightly more familiar, it
looks like that, this is another chunk.
This is another chunk of the spacetime,
and it's exactly the same question, right?
So now I'm imposing that x equals 0
at this entire top, x equals 0 at the
entire bottom, and I want to ask what
constraints do I put on x on this
boundary which has the property that we
want, okay? (A question was asked by a student.) No, no, we're not. And in
fact we can uniquely solve for
any x inside by the Gauss's law constraint,
okay? So let me just quickly
convince you of that, back we're gonna
come to this in a second, but first of
all you see that given any x on this
boundary, fully expressed the x on this
boundary, because this Gauss's law
tells me this plus this minus 0 minus 0
is that constant which I know, right? So
given the x's here, I fully specify the
x's there, then I'm done. Then I know the
x's anywhere. Give me x's anywhere. If I want to know
the x here, this plus this minus that
minus zero is equal to a constant, okay?
And I just determine what that guy is
in terms of that guy. So I fully
specified the solution of the wave
equation inside, just given this
boundary condition of x on here, and
these boundary conditions that x equals
zero at the top and the bottom, okay?
But we'll be seeing is quite explicitly in a second. All right, so
just to say again we want to see in
the space of all functions x(u),
these functions x(u) look something
like, you know, so here's a function x(u),
u equals 0, it has to be 0, u equals 1, it has to be 0,
and it has to be positive everywhere
there, but I want to know somehow in this
space there's some function x(u)
that's positive everywhere that goes to
0 at the boundaries, is that function
compatible with x being positive
everywhere inside or not? So I want to
put some constraints on what x(u)
looks like. I want to figure out what the
constraints are on what X of U looks
like in order for it to be positive
everywhere on the inside. Now this
problem is hard because we are asking
some infinite dimensional question, right?
In the space of all functions, right? So
let's make it a little bit easier, and
let's just ask about the values of x on a
few select points on the boundary, ok? So
in fact the simplest thing we can do is
just ask what can we say about x on a
single point? So what can I say about x
here? So let me call this point, I don't
know, A, what can I say about the value
of x.. let's call it x1, what can I say
about the value of at x1? Well let's draw
the obvious light cones in this picture,
right? So here's x1, so here's some
friend on the other side, I don't know
let's call it Y1, ok? And let's write down
what the Gauss's law condition is. The Gauss's
law condition says that X1 plus Y1 is
equal to the integral of the current in
there, right? C, let me just call that C,
ok?
Alright, so what do I know? I know that I
want X1 to be positive, but I also want Y1
to be positive, right? At least I
want those to be positive. So I
want X1 and Y1 to be positive, with this condition. And so that means
that X1 has to lie in this region
between 0 and C, right? So in X1 space we
already learned something interesting
that X1 cannot be arbitrary, it has to
lie in this little interval, ok? On this
end X1 goes to 0, on this end, Y1 goes
to 0. C is the fix. We're imagining
that we're imagining that this
positive source is fixed, so the only
thing that we're varying is the boundary
data, ok? Ok? All right, that's a little
interesting. Let's move on to sampling at
two points. Ok so let's say that we
sample at two points X1 and X2. Let me
draw the light cones, ok? Now this is
slightly more interesting, so I have X1
and Y1, and X2 and Y2, and this third
point here Z because on the inside, okay?
And now I want to somehow say something about all of these things being positive,
right? So I could write down, I could
literally write down the Gauss law
condition through this mesh, that mesh,
and that mesh, right? That's three
equations. So let's write them down. So
this mesh, it says that X1 plus Z minus
X2 is some constant C. This mesh so...
let's put some C, C', is C".
This mesh says that X2
plus Y1 minus Z equals C,' and this
one says that Y2 plus Z minus Y1 is C",
okay? And now note that I
want all of these five variables X1, 2, Y
1, 2 and Z to be positive. My data is X1
and X2, so I want to see in X1 X2
space there's gonna be some nice region
where this is true, right? But here are
three equations that I can use to solve
for Y1 Y2 and Z in terms of X1 and X2,
okay? Now I could actually sit here and
take linear combinations of these
equations to do it, but we can actually
go back to the picture and just let the
picture do it for us. Let's say I want to
solve for Z in terms of... I want to solve
for everything in terms of X1 and X2,
right? Let's solve to Z in terms of X1 and
X2. Well, that's easy. That's just this
equation right away, so we don't have to
do anything. So Z equals C plus X2
minus X1, and we want this to be
positive, right? Well let's solve for Y1. See I can solve for Y1 just by looking
at the Gauss's law on this picture. X1 plus
Y minus 0 minus 0 equals C plus C'.
Of course that's a linear combination of
two of these equations. This case is
not so hard to do, to take the linear
combinations, but I want you to see that
in a second going to be drawing more
complicated things that you can just
sort of read it off immediately, right? So I want to get an equation for Y. So
this is also the answer to your question.
So we're explicitly seeing now we're
solving for every one in terms of the
internal guy. So Y1 plus X1 equals C plus
C'. Therefore Y1 equals C plus C'
minus X1. Similarly Y2,  Y2 plus X2 minus
0 minus 0 is C' plus C",
so Y2 is equal to C' plus C "minus
X2. And I want all of these guys to be
bigger than zero, and of course I also
want X1,2 to be bigger than zero. So let's draw what this space looks like.
In X1 and X2 space, here's X1 and X2. First of all, X1 and X2
have to be positive, so I am in the
upper quadrant, right? But I have to
demand that X2 minus X1 is bigger than
negative C, ok? So that means that there
is a line coming out like this, and I
have to be above this line, right? X2
minus X1 has got to be bigger than
negative C. The length of this line is C.
But I also have to have that X1 is less
than C plus C'.
And now notice how interesting it is,
that C plus C', because the C's are
positive, is guaranteed to be further
over from where this line started, right?
That was not...if I didn't say
anything about the signs of the C's, I
wouldn't know what this shape looks like.
It would depend on whether it cuts
here or here, but I'm telling those C's
are positive, so I know for a fact that
X1 which is bounded by C plus C' is
out here somewhere. So there's like,
there's a line that's going to look like
that, where this point is, so this point
is C, this is C plus C', and
similarly on the other side I know that
Y2 is got to be less than it C'
plus C", and so that's what
the picture looks like. So this is C' plus C", OK? So I have to lie on the inside of
this interesting region what you notice
is the pentagon. Yeah. (A student was asking a question.)
Because it's bigger than C'.
So because this intersection point is
exactly at C' okay, because...sorry,
thank you! This point here is exactly at
C', right? Because it's a 45 degree line. So the next one is above that, and
so it gets intersected above it, okay? All
right, so you see that already we're
getting some kind of cool stuff, but for
one point it had to lie the inside of
an interval. At two, just for two points
that's the line inside this Pentagon. As
you remember from last time, the interval
was the one-dimensional associahedron,
and the pentagon is the two-dimensional associahedron. And let's
keep going. So, what if we have three points on
the boundary? Okay? If I have write down
all the equations, but we'll just do some
counting. So if I have three points on
the boundary,
so I have X1, X2, X3, and then I have
how many other points, 1, 2, 3, 4, 5, 6 other
points. I have a total of 9 variables.
How many mesh relations do I have? 1, 2, 3, 4, 5, 6 mesh relations, and so we're going
to get a three-dimensional polytope, so
it's manifested as a polytope, I hope
that's clear, right? It's manifest we're
just cutting things out with the
inequalities. So we're getting a 3d
polytope with 9 faces which are
associated with these demanding the
positivity of all those variables, just
like here we got 5 faces because we
demanded the positivity of all 5
variables, so we got, okay? And you can
just, if any of you have Mathematica
open, you can just like plunk down these
equations right now. Just do region plot
3d of all the equations, and you'll get a
beautiful picture of a three-dimensional
associahedron, okay? So these guys are associahedron in general. The X's have
got to lie inside an associahedron. So
in this case the associahedron is
answer to this extremely simple question,
but again the word positivity was
important, right? So we have to ask for
positive solutions of the wave equation.
But positive solutions of the wave
equation in kinematics space force the
X's to lie inside and associahedron. And..
So this is a very particular realization
of an associahedron. It's not just some
random combinatorial associahedron. It's a very particular way of cutting out the
associahedron with linear equations
that's associated with the wave equation
in this way. But instead of going through
the example and showing the associahedron,
I now, if you want to prove that it's an
associahedron. In other words, what I want
to now show is why this picture does
what we set out to do. This picture makes
it obvious that we have objects that are
polytope. That's clear, again, just by
construction. We're asking
once we sample it finitely many points,
we're manifestly talking about the
interior of some region that's cut out
by a bunch of linear equations, okay? So
it's going to be manifestly polytope,
and now comes the magical part -- it's
going to manifestly factorize on its
boundaries, okay? So I want to show you
how we see both the polytopality
and factorization, okay? So let me now
explain why we have
factorization. And let me again just go
back to this continuum picture. I draw a little
bigger.
Okay, now I'm interested in solutions of
the wave equation which stay positive
everywhere inside, right? So what does it
mean to go to a boundary of this space?
What it means to go to a boundary is to
put X->0 somewhere inside, OK? ll right so
let's say you put X->0 here,
what I'm now interested in doing is
figuring out what does the space look
like. That's compatible with x equals
0 there. And what I'm currently going
to show you is that the space is a
direct product of two triangles that
look like exactly the same thing, okay?
So let's see how that emerges.
Well, the first natural question to ask
is if I put X to 0 here where else can I
put X to 0, right? And what I'm about to
show is the key point of this
entire story, and it's basically one
extremely simple factor we'll talk about
in a few ways. But the point is that if I
put x equals 0 there, then there's
this entire region if I draw the natural
light cones in this way, I cannot put x
equals 0 in any of these other places,
okay? In other words, said another way, if
I put x equals 0 here, there're all the
points that are time like separated from
this one, either in the future or in the
past, but which are causally connected to
this one, so that there's a causal
diamond that has this one on one end and
the other one on the other.
None of those points can go to 0. In
other words, if I put, so let me say the
more general statement, if you draw any
causal diamond that fits in the
space-time, any causal diamond at all, if
I put X equals 0 here, on one corner
of the diamond, remember this is time, and
put X equals 0 in the past, I cannot
put X equals 0 in the future. Why not?
Because remember if this is A B C D. I
have to have Aplus B minus C minus D is
something positive, because I assume that
the current is positive. And so if I put
A to zero and B to zero, I get a
contradiction, the sum of two negative
things is positive,
okay? So that's the sort of key point
here that causality whether or not you
have points that are causally connected in the sense that
if you think of these as observers, that
the future observers and the causal...
is that the past in the future
observer a part of a causal diamond in
the space-time, if one of them goes to
zero the other one cannot go to a
zero, okay? Yeah. it's zero at the top and
the bottom. You put a zero here,
you can't put a zero anywhere in here,
right? All of these things there's
someone who fits up to that point, right?
And now, but you can put it to zero there,
here, or there, or there, okay?
Furthermore, so I can't put X equal to
zero in anywhere in this shaded region,
but furthermore, I don't care about the...
Yes. (A student was asking a question.) Yes. Sorry? Well this is some charge
density here. So X is zero at the top, and
X is zero at the bottom. That which we're
trying are we talking about.  (A student was asking a question.) Like this
one? This one? What are you talking? Sorry
this one? Yes. No, no the charge inside
does not have to be...
Yeah. That's not a causal diamond in 
space-time, okay? So we don't have an
equation for that guy, right? Okay? So it
has to be an equation that involves this
plus this minus that minus that. Any other
questions? Okay? All right. So, but notice
that everything about what's inside
these shaded regions, I can reconstruct
just by knowledge of boundary values, for
example here. So not only do I, not only
can't I said anything to zero there, but
everything if I want to know what the
X's are in there, I can reconstruct
them by knowing what these
boundary values are. Therefore, as far as
the kind of degrees
of freedom that I actually have, the
degrees of freedom that I actually have
involved the X's here and the X's there,
because I can certainly reconstruct all
the degrees of freedom in there just
from all the other
boundary values that I have, OK? And by
scrunching, this is exactly the same as
the following picture. So I'm now just
going to scrunch away this region and
that region. And what does the new
picture that I get? The new picture that
I get is a triangle on top of another
triangle. I have to of course add some
delta function positive source on this
bottom triangle, to like make up for the
sources that are there, so I add some
extra source there, okay? But, that's now
what I get, after I scrunch away those
regions, and this point here is the point
that I set to zero.
Okay, so what do we conclude? We conclude that all the solutions of the wave
equations that are positive, for this
problem, when I set X to zero, are exactly
the same as the solutions are completely
determined by the solutions of the
wave equation for this problem where I
set X equal to zero here as well as X
equal to zero there. This is just a
direct product of two triangles. We see
factorization. And therefore it's
completely manifest that once I sample
this guy at finitely many points, I get a
finite polytope, but the finite polytope
is a property that on the boundary it
factorizes into lower polytopes of
exactly the same time, okay? So. (A student was asking a question.) Yeah, well
what I'm... there's no momentum
here, right? So what I'm saying
is that I'm trying to characterize all
of the X's here which, all the X's here
that are compatible with the conditions
that it's positive everywhere and zero
there. I'm telling you that all of those
X's are in one-to-one correspondence,
so all the X's that satisfy exactly
the same rules on this picture. So, and
that is exactly one problem, and another
problem, okay? And it factorizes, this
literally factorizes. Now, of course there
are more variables here, so there's some
concrete formula if you want to know
what all the X's are here, they're
entirely determined by giving the X's
here and the X is here, okay? So if you
give me, so because this
boundary is nothing other than this part
plus that part, okay? So if you give me
all the data on this part and all the
data on that part, if you give me all
the data here and there, from which I can
reconstruct everything,
okay? So I'm giving you a sort of a map
between this problem and this one, and
that just tells you that everything here
on this boundary looks like the direct
product of two lower
problems of the same sort, okay? Okay, so
this is really the sort of key
conceptual point is that the
causal dependence in this one
plus one dimensional space-time, this
auxilary one plus one dimensional
space-time which is how we're
interpreting kinematical space, so this
is not, has nothing to do with ordinary
space-time, there's some funny auxiliary
one plus one dimensional space-time,
they were how we're interpreting
kinematical space, that causality in this
one plus one dimensional kinematic space
tells you that tells you a notion of
whether two events are compatible or not.
Two events are compatible...are
incompatible if they fit on corners of a
causal diamond. They're compatible if
they don't, okay? And compatibility or
incompatibility, from this causality
point of view, maps precisely in the
mapping that takes a point inside here
to the chord of a polygon, it maps
precisely to what the chords cross or
don't cross, so that's the 
entire little point here. It is that there
is a mapping between causal
diamonds in this one plus one
dimensional picture and just the simple
question of other chords crossed or don't
cross in the identification. If
you like, all we've done in this entire
lecture is, when we draw
that picture of the polygon and we say
all the chords are associated Xij, it's
very it's so easy to do, we
do it right away, but we kind of in our
mind see a one dimensional picture just
of the polygon, but if we really want to
see all the variables, we have to draw a
two dimensional picture that's what we
did, and we made a map between the sort
of two-dimensional picture of the
variables, we drew a pic, we just you know
characterized the space of the
two-dimensional
the variables, but now we can ask in that
picture what does whether chords cross or
not cross look like. And if chords cross,
it means that the points are inside a
causal diamond that fits in this
space-time; if they don't cross, it means
that they're not, okay? And just that
primitive fact then tied with this
simple picture of the wave equation
gives us both manifestly an object
that's polytope and that factorizes
on the boundary. Yeah. (A student was asking a question.) That's right, yeah.
Yes. Well the ??? also factorizes,
the ??? also factorizes the
two copies of itself, but there's
a lot more going on, there's
K, there are different kinds of
boundaries, so on different boundaries it
factorizing two copies of itself into
different K, and so on, right? Right. Yes,
you can also work into the wall take the
limit as the number of particles goes to
infinity for that, but here yeah, that's right, you can
do the yeah. Okay. So now what are we
doing for a fixed number of particles, so
finally this is the geometry, so just
emphasize, just to emphasize
the again, what we're doing, so let's say
I do this, I look at this this trunk
again. So this is 12, then 13, 14 ,up to
1n at the bottom, and so on, right? So we
fill in this mesh, then I'm asking for
all the Xij to be positive. I'm asking
for the X's of satisfies the wave
equation. Now the wave equation is the
statement that I can associate with any
ij here. So with some ij, the wave
equation just says like Xij plus Xi+1 j+1, minus Xi+1 j
minus Xij equals Cij, okay? So I'm
associated with a given point inside
here, I'm going to, I'm gonna just call
that source in here Cij, okay? And so you
see that the number of these C's is smaller than the number of
X's, precisely because all the X's at the,
I mean, I'm associating a mesh with every
variable, right? I'm associating a mesh to
the right of every variable, except
there's no right to the variables right
at the end,
okay? So I have precisely n(n-2) over
2 minus (n-3) of these
Cji''s, just the ones that I see in this
mesh, okay? And the claim is that
these equations are cut out the
associahedron, the n-3 dimensional
associahedron.
And I've given you the sort of
conceptual reason for it because they
clearly cut out a polytope and the
polytope manifestly factorizes on its
boundaries, OK? Some polytopes factorize on its boundary. (A student was asking a question.) Does it matter? Yeah,
it doesn't matter no that's right it
it does matter. No, that's right. It does matter. Okay, so okay, so this is yeah. (A student was asking a question.) No, they don't have to be
zero. I mean, yeah, so if you literally
set the chart in fact let me just
interpret this in a second, if you set
some of these Cij's to zero, then it's
possible to kind of shrink the edges and
maybe even some of the phases of this
polytope. Yeah, you can certainly do that.
So, yeah, so if you want to see the
kind of generic behavior of the polytope, you want all
these things to be possible. Okay? Yeah.
Well, then first of all, none of it
would work, but the sort of the ultimate
sort of combinatorial fact is the very
simple fact what I told you. If you
look at, if you plot Xij's in this two-dimensional mesh, Euclidean or Lorentzian
you don't care, just some two-dimensional
mesh, even if you draw it right side up,
so it doesn't look like light cones, 
draw it however you like, the
invariant fact is that two chords
crossing corresponds to two points on
opposite sides of a diamond, okay? And
that's the, so that's what brings
causalities to life, okay? Because you
want to have something care about
whether the points whether the causal
diamond does or doesn't fit inside the
space-time. The ordinary wave equation
cares about causal diamond, the Laplace
equation does not, okay, so that's that is
the only reason,
okay? Okay. But now let me say, so
this is the geometry, so if I kind
of, uh okay, so now my kinematic space
is X space. In X space, I have a positive
region, but I won't even drawn a curvy
way because it's not curvy. I have a
positive region which is just where all
the X's are positive, so it's like a
simplex, right? So this is P, okay? But I
have found in kinematics space, I have
found a subspace. What is the subspace? The subspace is the
solution of the wave equation, okay?
Here's the subspace S, right?
And the subspace intersects the positive
part, so remember this P is just where
all the excerpt Xij's are positive. The
subspace is solves wave equation, okay?
Sorry. (A student was asking a question.) Well, the subspace is labeled...the subspace has some positivity constraints
on it. The C's have to be positive,
okay? So there's a subspace for some
positivity condition, but for any C that you choose that's positive, I get
a subspace. The subspace intersects the
positive part in a positive geometry. In
this case, it's just a polytope,
it's the associahedron, okay? And now
what is the amplitude? The amplitude is
an n minus... so this is an (n-3)
dimensional subspace. The kinematic space is n(n-3)/2 dimensional. And so
the scattering amplitude is an (n-3)
form, with the property that when you
pull it back to this subspace
you get the canonical form of the
associahedron.
So this story is I hope you see
qualitatively identical, but much more
straightforward than the story we saw
for the amplituhedron, okay?
Everything is polytope. Now, and but
but the amplitude is thought of as a
differential form on this whole space,
but when you pull it back to this
subspace, you strip off the overall
volume form on the subspace, all those
relative signs become plus signs and you
literally get the amplitude, okay? So
that's how the amplitude is recovered as
well. The subspace depend on Z, so just
in the analogy with the amplituhedron,
here we have the space of
all X's. In the amplituhedron we had
momentum twister space. Here the in the
amplituhedron, the positive part was
this fancy thing where all the i plus
1j j plus 1 wind around the same way
and there is some net winding number,
okay? So it has positivity and some
topological thing on top of it, but
remember it's not dissimilar. All the ii+1
jj+1 is positive. We're just
the poles are positive, right? That's all
the poles are positive. It's just needed
this extra topological information too
about the actual winding number. Here
it's simpler, just the poles are positive,
ok? In the amplituhedron, you have to find
a subspace that was 4K dimensional that
was this linear subspace where you took
some initial Z star and then you
translate in some direction delta that give you a 4K plane,
and we have to ask that that Z star and
delta gave us a positive matrix. We
thought it was that Z star and delta as
the ??? data of the usual Y space
picture of the amplituhedron. Here it's
the C's, so you give me positive C's,
they specify some hyperplane. In both
cases when that hyperfine intersects the
positive region, we get a positive
geometry. In the amplituhedron case,
the positive geometry is the amplituhedron.
In this case, the positive geometry
is the associahedron, okay?
In both cases, the
scattering amplitude of the canonical
form that lives on the whole space. When
you pull it back to the subspace, it
gives you the... its some form, lower
form that lives on the whole space that
when you pull back to the subspace gives
you the canonical form for the positive
geometry. And in this case the canonical
form is just the, the canonical form is just associahedron. Actually this is the last thing
that I want to mention.  Associahedron are
examples of things called simple
polytopes. A simple polytope is one which
has the property that at every vertex, if
it's an n-dimensional polytope, there's
only n faces that meet at a vertex, ok? So
that means that locally in the
neighbor of every vertex that looks like
a simplex. It's so for example obviously a tetrahedron is the
simple polytope, but a square pyramid is
not, because there are so locally, the
top vertex is being like there's four
faces meeting on it, not 3. Ok, now
if there's a very simple, for simple
polytopes, there's a
very simple formula for the canonical
form for the polytope. The canonical
form is just the sum over all the
vertices of the dlog of all of the
linear equations that associated with the faces
that meet on V.
So I think we we talked about this, and
what's the overall sign given by the
orientate... but in an obvious way by the
by the orientation of that
simplex locally around the
vertex V. One simple way of
understanding this formula is, if the,
remember for polytopes the canonical
form is also the volume of the dual, okay?
And if the polytope is simple, that means
that at every vertex you have, you know,
like it's a three-dimensional polytope,
at every vertex there's three faces
meeting. What does it mean in the dual?
That means that the dual is what's
called simplicial. That means that in the
dual polytope, all the vertices become
faces, that means that every face looks
like a triangle, every face is bounded by
exactly... every face looks like the
triangle, so if the polytope is simple,
the dual polytope is simplicial, the faces
don't look complicated, every face this
looks like a triangle. And therefore
there's a very simple way to compute the
volume if every face looks like a
triangle, just put a point on the inside,
draw a tetrahedron out to all of the
triangles and add them all up, okay? So
that's why there's obviously a
formula that just sums over all the
faces of the a dual polytope, and that
formula interpreted back in terms of the
original one is just taking the wedge of
the dlog of all of the linear variables
that vanish on the faces that touch the
vertex. So when you have a simple polytope, this is a simple way of getting the
canonical form for it, okay? Now what are
the vertices of this associahedron?
The vertices of these associahedron are
just when you set as many X's as you can
to 0. But we know what all those are,
sending it as many X's you can to 0 or
exactly triangulations of the
polygon, that tells us that it's exactly
simple because I can't set more than (n-3) of them to 0, so we see that the
associahedron is a simple polytope, and we
see every vertex of the associahedron is
associated with a triangulation of the
polygon or a dually Feynman diagram. So I
just told you that it's an associahedron, but now I just proven to
there's an associahedron in detail, right?
Just by seeing that we can
label every one of the facets
exactly by some chain of these X's that
go to zero that end up at a vertex
giving me what looks like Feynman
diagrams. Therefore one formula for the
canonical form of the associahedron
is the sum over Feynman diagrams, so
that's one very nice
thing about this story, which is easier
in the case of the amplituhedron,
is that we can see very easily the
connection to Feynman diagrams, okay? So
Feynman diagrams represent one
triangulation of these associahedron,
sort of one of the easy ones which you can associate with any old simple polytope just taking the wedge
of the dlog of the variables associated
with every vertex, okay? So we discover
associahedron in kinematic space.
But from there one simple formula for the canonical form is just
Feynman diagrams. However we know
something, we know that Feynman diagrams
are hiding a fact. They're hiding that
this form has no pole at infinity.
It's obvious this polytope has no form at
infinity because it only has
singularities on the boundaries of these
associahedron. However term by term that's
invisible, and that's related to the fact that this
form is projectively invariant. This
form that lives on the big space is
projectively invariant. Well that's not
obvious term by term in the Feynman
diagrams, ok? So that's again, we've said
this a couple times, what we're seeing it
in more more detail now. The object, this
is an intrinsic definition of
what the object is. It has various, we see the polytope nature of the combinatoric, so
we see factorization, we see this hidden
projective invariants, all as the
features just sort of built into the
problem. We also see Feynman diagrams is
one quite nice way of triangulating. The
answer which however breaks some of the
obvious symmetries, ok?. So it doesn't manifest this hidden projectively
invariance. And this hidden projectively
invariance is literally a prediction
about the amplitude which is non obvious,
and the prediction is if you take the
amplitude, you pull back to one of these
subspaces, so you literally have a
function of X's and these C's. The
prediction is that if you hold these
funny C's fixed, and you send the X's to
infinity, it goes to zero faster than you
would naively think from each Feynman
diagram alone,
okay? That's literally the absence of a
pole at infinity,
just like in BCFW, right? Just like in BCFW, when we went off to infinity,
naively it went with some power law, but
in fact it went faster to 0, faster than
that. It's a very specific prediction
which is not true of any diagram term by
term, and it's not true of any sum of a
subset of diagrams, it's only the sum of
all of them together, which has a sort of
remarkable feature that it vanishes
faster infinity than you would
naively predict from any one of these
terms. So this is the sort of hidden symmetry that I've alluded to
many times that we see even in this
simple by a joint theory. Ok? Now, let me
just, almost out of time, let me say a
couple of things. First, something we
probably won't have time to talk about,
but since there actually brought out
by a person who has gone, but since this
question was asked already, it is natural
to ask what happens when you take these C's and you should have shut
them off, right? So it's actually
especially natural to ask what happens
when you like, obviously if we shut off
all the C's, we shrink the entire
polytope to zero, right? But it's sort of
natural to ask what happens when you
turn on one C at a time. You turn on one C, 
another C, another C, another C, okay?
If I shut off all but one of the
C's, I get a much simpler polytope.In
fact, it's not even generically top
dimensional. If we take some of
these simple examples like the pentagon,
let's say you take the pentagon.
I'll just tell you what the answer
ends up being. If you take the sort of
pentagon case, and you shut off C and C', I forget which ones I
called them. But if you shut off two of
them, what you'll be left with an
interval. If you shut off another two,
what you'll be left with is another
interval, and the last set actually gives
you something it looks like a triangle,
okay? So in these limits, when you shut
off all but one of the C's, you get very
simple objects. And there's then a very
pretty interpretation for what the whole
polytope is. There's a notion of a
Minkowski sum of two polytopes. It's the
most trivial thing you can imagine as
the sum could be. If you have a space A and
B, the Minkowski sum of A and B. If the
vectors in A are V and the vectors in B
are W, there are just all V plus W's, okay?
Instead of all V plus W. So V in A and
W in B. But if you take the Minkowski
sum of, so if the variable that's
left here is C1, C2 and C3, the
Minkowski sum of C1 times this plus C2
times that plus C3 times that gives you
the entire polytope, basically by
definition. So I got the polytope by
shrinking some of the C's, and I'm just
adding them back. So this is interesting,
that we're not just getting the
associahedron in some random
realization. We're getting the
associahedron naturally each mesh, each
little piece of the mesh, is giving us
some little building block of the
associahedron. Each little mesh
giving some little building block, and if
I take the Minkowski sum of the
building blocks that I get from all the
meshes, then I get the entire associahedron,
right? Now, I might have, I might allude to
this at the very end of the next lecture,
but this is a very natural fact about this way of building up the
associahedron. And that fact is a hop skip
and a jump from string theory.
So string amplitude, the famous Koba-Nielson formula, the beginning of string
theory from the Beta function to the
sort of general N-point formula of the
Koba-Nielson amplitude has a direct
interpretation, you've never heard of
strings, you've never heard of anything
else, was just playing this
game and you want to find a way to write
this canonical form. Well there's a
general strategy for writing canonical
forms of any polytope, associahedron, any
polytope at all, which involves certain
integrals, has nothing to do with string
theory, just it works for any old
polytope, right? It involves certain
integrals. In fact these integrals are
written down by in various guys by
mathematicians decades ago. It's nothing
to do with the strings or a world sheet
or anything, but you write down
certain integrals, and these integrals
give you naturally the canonical forms
for polytopes, especially naturally when
the polytopes are presented as a
Minkowski sums. So if you have a set of
polytopes that are naturally given to
you as a Minkowski sum, you shove it into
this machine, you'll get some object that
will depend on a parameter alpha Prime,
and that ends up being the
string amplitude. So there is a kind of a
very direct way without any mention of a
world sheet to go from this picture of
the associahedron in kinematics space to
string amplitude, so I probably won't
have too much time to talk about that,
but I might say a little something about
it. But now you want to say something
else that we will talk about next time.
So here we just sort of talked about the
solution of the wave equation is a
global thing. We should have you know
plunked it down, and we found what it
looks like, and factorize it, and so on.
Well, there's a very natural question
that you could ask. How can I solve? I
mean here we gave various, we
solved these mess relations in various
ways to determine all the X's on the
inside given these X's on the boundary.
But there's a very natural question to a
physicist. How do we do this gradually in
steps? In other words, I want to do, I want
to determine all the X's on the inside
by time evolution, right? I want to
imagine this with some initial surface
and then from this initial data I'm next
going to determine what that point is. So
that's my first step in the time
evolution. The next step of my time
evolution will be I'm going to determine
what that point is. The next step I'll
determine what that point is, and so on.
Then I get to this one. Now I have a new
surface. Now I'll determine where that
one is, and so on. We're going to talk about what a
natural picture of just evolving in time
is one step at a time,
okay? It's a very natural thing to ask.
And as we'll see, asking that question
motivates drawing a certain quiver
associated with this diagram. We're going
to draw a certain quiver associated with
this diagram, and the rule of time
evolution will end up being an extremely
simple one at the level of the quiver,
flipping a bunch of arrows of a certain
type on sources of things. So that is
going to give us a rule for starting
from a quiver and walking just by
everywhere we see sources or, let's say
we decide they're their sources,
everywhere we see sources we flip the
arrows on the sources, we solved for a new
variable associated with that node by
the mesh relations by the wave equation
and we keep on going, okay? So this won't
if it's not obvious you will see it next
time, but we're going to start next
time with trying to interpret the
solution of this wave equation is time
evolution. When we term it as a time
evolution, we'll abstract that away, 
just so you have an idea where
we're going. You know, this picture, the fact that
they're all on a null cone is going to
be on the null line, it's going to be
represented by the following quiver. The
fact that they're all 45 degree lines in
one direction is the fact that all the
arrows are oriented like that. But at the
next step in the time evolution, you see
I can solve for this variable, given
everything that I had here I can solve
it for this variable first next, and so
this now looks exactly the same except
at the end I'm flipping to go the other
way. So at the next step
we're gonna have a quiver that looks
exactly the same, but at the end that
arrow will be reversed, okay? And so on,
okay? So we're going to have a picture of
time evolution, and we'll start with a
bunch of variables X1, X2, blahblah, X(n-3), and
we're gonna have some new variable here,
some X',
so I change one variable to another
reversing the direction of all the
arrows, and with a very simple formula
from the mesh equation that tells me
with the new X's in terms of the old one,
okay? So we'll just sort of practice
interpreting time evolution in this
diamond in this slightly more abstract
way in terms of starting with a quiver
and mutating, which means going to a
source that's only a sink in flipping
the direction of the arrows and writing
down a new variable at that node with a
simple rule and continuing, okay? In the
case of these linear quivers, we're
just gonna reproduce what we saw with
the wave equation. But you can then
abstract this rule away to a general
quiver, okay? If you abstract this rule away to a
general quiver, something miraculous
happens. If you do it for a totally
arbitrary quiver, you find that while in
this case the process sort of ended in a
natural place and the polytope so that
closes up and we get this finite object,
if you do it for a totally random
quiver, the process never ends, the
polytope never closes up and it just is
some infinite crappy things. But for very
very special quivers, it ends up giving
you a finite polytope. Those special
quivers are associated with Dynkin
diagrams. The quivers are nothing but
orientations of Dynkin diagrams. This is
a miraculous fact, the fact that Dynkin
diagrams show up as the answer to this
question,
finding these sort of quiver like
generalizations of associahedron is not
obviously related to the classification
of Lie groups question, okay? So the fact
that Dynkin diagrams make a second
appearance here is quite remarkable. But
it's not coincidental. And the objects
that we're going to define in this way
are actually the polytopes associated
the cluster algebras, even though I won't
have told you what a cluster algebra is,
okay? But we'll have done a lot of the
stuff about cluster algebras will that
actually tell you what a cluster algebra
is, okay? And in particular it's known the
cluster algebra are finite,
precisely when they're associated with
quivers that are Dynkin, okay? So this
sort of the fact that Dynkin appears in
both cases is something that you can
actually see, from this much more pedestrian point of view, ok? Having seen all of that, there is something else
that we'll abstract the way,
I'm just giving you just a sort of
teaser so that some of you show up on
Thursday even though you have exams and
stuff like that. So, that, in this case
when it's a chain, all this business, it's
a wave equation and factorization and
all the sort of beautiful stuff is
abstracted away to a very simple thing.
Forget even about the orientations, you
have a Dynkin diagram that looks like
this,
just An for this case. And the way it
factorizes is simply that you remove a
node, so that's why I factorize them into
two smaller Dynkin diagrams of exactly
the same type, okay? So that's why An
factorize into An cross An, and one
cross An too. That's exactly the kind of
factorization that we expect for tree
amplitudes, okay? But of course this ends
up being exactly the story in general. So
if we look at the other classical type
Dynkin diagrams, let's say for B and C
we have Dynkin diagrams that look like
this. So if I call this B, how should B
factorize? B should factorize by removing
a node into on one side that looks
like an A, but on the other side it still
looks like a B. Now what kind of
amplitudes have this property? That their
amplitude of something factorizes into
itself times something that looked like
an A which was this like phi cubed,
planar of phi cubed amplitudes. These are
exactly amplitudes for imagine you have
some sort of colored particles, but also
they met a graviton or a dilation, or
something that's uncoloured so it can go
everywhere. So let's say you look at all
amplitude in that form, then how does
this factorize? Well, however you
factorize, the guy that's emitted is on
one side, and it's not on the other, so
therefore it has
look like the boundary equals B times A.
What about the final classical type,
which is Dn? So the Dn Dynkin diagram
looks like this. And so what do we get
for the boundary of D? Well, once again
if I remove one of the nodes here, the
boundary of D looks like D cross A, but
there's one other kind, I can remove this
guy, this is just another A, right? Now
where have we seen this kind of
factorization before? This is one loop,
because at one loop you can have one
loop diagrams that factorized into a
loop times tree on one side that's the D
cross A, or we have our friend the
forward limit terms where I just take
this leg and I open it up, I cut the
internal propagator, then I just get a
tree, a different tree with two more legs,
and that's exactly the second kind of
factorization, okay? So, and this is going
to allow us to attach polytopes just
like we add for phi cube theory. First
it captures all the common torques
of the Feynman diagrams, but also the
canonical form of the polytope, exactly
the same story actually computes the
amplitude, okay? At least the integrand
for the amplitude. And that will happen
for these generalizations of the
associahedron for these classical cluster
polytopes for B, C and D type, okay? So all
of that will be able to sort of
understand simply by kind of following
our nose. There is this sort of step of
abstraction to it's very very natural to
think about this time evolution. I didn't
mention that there even as you time
evolve along, there's a notion of entropy
that will discover, as time evolve
along there's a certain quantity that
monotonically increases, and it's
important to keep track of this
monotonically increasing in quantity as
we go along which tells us
where to stop, okay? But anyway, you have this idea of
time evolution more abstractly
associated with the quiver. See in this
picture you know where to stop. If I just
tell you to do this some rule on quivers
and mutate on on sources or thinks, you
could in principle just keep going
forever. So some things I ought to tell
you where to stop, and there is a natural
notion, there are some monotonically
increasing quantity that hits a maximum
value, then you have to stop when it hit that maximum value. So, that takes some
inspiration to kind of think that you're
supposed to have stacked that away from
this picture. But once you abstract that
away from this picture, you can ask the
exactly the same questions for any
quiver, and then in this way discover all a
lot of the non-trivial polytope will
juice of all of these and this finite type of cluster algebra, and the classical
ones are attached to scattering
amplitudes in exactly the same way as
the associahedron is attached to phi cube
the tree amplitudes. What is the EA
polytope? Who knows, right? What if
physical interpretation doesn't have? Who
knows? That we don't know obviously
something that doesn't make sense for
arbitrarily large N. And the last thing I
will say which we won't talk about next
time because probably we won't even
really get to talk about the string
amplitudes next time, but just as I told
you that this picture of the polytope is
a Minkowski sum, immediately suggests the string amplitude, so there's an analog of
a string amplitude associated with all
these other cluster algebras as well.
They have all the usual magical
properties of string amplitudes. They're
exponentially soft, they have Regge
behavior, they have channel duality, they
have all these facts that string
amplitudes have, with no worldsheet
inside, okay? So there's no, they're not
canonical string amplitudes in any way,
so I think it's an interesting question
to figure out if they have any purpose
in life, and of course as alpha prime
goes to zero in all these classical type
cases they reduced the field theory amplitude, in a way very analogy to ordinary string amplitude.
Anyway, that's our plan for next
time, there's a bunch of ancillary things.
I'm just telling you many facts, but
a plan for next time is to slowly
go back to this picture, think about in
terms of time evolution, abstract it away
to these quivers, so I can at least
tell you something about general cluster
algebras and
we claim about amplitudes through at once. Okay! Yes. (A student was asking a question.)  Is it there? Not yet. Well, I mean
there are many many connections, so there's a connection between
the simple roots of the Lie algebra, and the normal facets of these
polytopes for example, okay? So, there are
lots of connections, but there's no
obvious sort of purpose in life to the...
There is actually. For AN,  there is. For
AN, there is
because there's a naturally a Lie
algebra, there's Lie
structure associated with the moduli space
of endpoints on P one, and that's
not unrelated to the fact that
the associahedron has an AN.
But the meaning of the, for B, C, D and the
other ones, yeah,  there's no
obvious that, at least not
obviously to me at the moment.
And as I said, the fact that I think I
mean, this is of course a very
famous fact in math. I'm not adding
any, I'm not, this is not surprising news.
I mean, it's an amazing thing, but
it's been well appreciated for decades.
It's very mysterious that Dynkin
diagrams show up as the answer to very
seemingly really radically different
questions, so it's not like, it's not,
I mean, the your the question that you
presume here in order to figure out that
the objects, you have finite
polytopes with this funny
construction, looks absolutely nothing
like you know Dynkin's original question
about the conservation of Lie groups at
all, so it's kind of very surprising
that Dynkin diagrams keep up
showing up. There's kind of three, I
know of kind of three different kinds of
questions where Dynkin diagrams show
up the answer, and it's not obvious,
certainly not obvious to me, and I've
talked to a few mathematicians. We're
sort of obsessed with this question, and
at least according to them it's not
really conceptually obvious to them why
keeps showing up, so that's
why precisely because it's not
obviously the same question there's no
obvious, there's no obvious
correspondence between the Lie
theoretic aspects of the Dynkin diagram
and the polytope ones, other than some
correspondence as I said between the
normal aspect of Lie and the polytope and the root system. Yes, right. (A student was asking a question.) Very much so. It's kind of obvious what
what they're gonna be, right? If you think
about this AN, kind of is like
descritization of this, what should the other ones be? The other
ones, the N is basically the same for
everyone, right? So as you make N
big, the only difference between A, B, C,
and D is just at one end, right? So I mean
what the Dynkin diagram looking at
just a long chain with a little antenna at
the end, or with a little two arrows at
the end, or with nothing at the end. So
what are the space times? There are strip,
with a boundary condition
at the top, it's like you put a brain at
the top, or you do a folding.
Alright, and that's it. So we,
these one dimensional spaces, we
do these few things, and we discover B, C,
and D, like that. (A student was asking a question.) The wave equations. It's
a wave equation. It's a wave equation
with, but you have to be careful about
what the boundary conditions are. So now
when we put boundary, we have to think
about what the boundary conditions are, and then... So that's a very
poor man's way of discovering all these
things, even without any of this
quivering nonsense, okay? You see there's
something cool about the wave equation,
great. You want to play with the wave
equation. Now you want to say it's a
strip, but now there's something at the
top, okay? I want to put some degrees of
freedom at the top, like some extra
particle at the top, good. So I put this
there, but now I have to write an
equation for that guy at the top, and now
I'm guided by something, and the guide is something that happened automatically
and was quite not was magical, 
it happened automatically, but which I
stressed, which is that the shape that
you got did not depend on these C's,
right? They're all positive, and the
positivity guaranteed that the kind of
new wall that you put in, you never have
to worry about whether or cut a wall
in front or behind of a wall you had
before, so you could call this property
sort of C independence, or sort of
shape independence. That the independence
of the of the polytope of on the
parameters. If you ask that the equations
that you write down, preserve this
property, it turns out to be almost
impossible to do, and you get
you get roped into very very specific
kinds of boundary conditions to put,
let's say at the top of the strip, those
boundary conditions don't have to be DN.
If you now take the strip and
instead just fold it in half, demand some
symmetry around some plain and fold it
in half, then that gives you again
something else,
that turns out to be the B and the C types. Of course from here you
won't discovery ???,  those are not
on the large N limit,
there are no good continuum description.
So there is something conceptually more
correct about starting from these
quivers, so that really tells
you everything. But if you just want to
think that there are some, but they're
not obviously connected to amplitude, so
if you want to think that you're asking
questions in kinematic space, then then
the kinematic space gets sort
of further, you know you put some
strips you'd make some identification as
you play, and you find the other of the
other examples. Yeah. (A student was asking a question.)
No, no, that's an excellent question.
What is the interpretation of the
differential forms in this case? They're
not Fermionic variables. It's just,
here they're just, it's a complete Bosonic theory, but what
I can tell you there's something kind of,
there are some interpretation of them,
but it's a rather remarkable interpretation which you don't see
the full force of, until you consider not
just one ordering, but all orderings, the
completely non planer theory, ok? So
here's the
interesting fact. So let's say you
imagine drawing any cubic diagram, a tree
tree diagram, of course any tree diagram
can be made planar some way, but I don't
know anything about the ordering of the
legs, right? This might be 1, 1, you
know 6, 4, 3, 2, 5, ok? Now with any cubic
diagram, if your computing
in the full phi cube theory,
this by joint phi cube theory, if you
weren't doing the color trace
decomposition, with any diagram you would
just associate a bunch of fabc factors,
right? So associate with every diagram is
naturally some color factor
given just by multiplying fabc's on all
the vertices, okay? So with any
diagram gamma there's some color factor
of C_gamma, which is just given by fff,
okay?
But there's another things you can
associate it with a given diagram. You
can associate given diagram gamma,
there's a certain form omega sub gamma,
they can just think of as the product of
the Ds's, the d of the propagators that
that occur. I'm not even using the X
variables anymore, because it's not just
one ordering, I have all orderings, okay?
So it's not even convenient to use
those X's anymore. I'm just going to use
the s's in one, okay? So also associated
with gamma is something else as a
differential form, which is the product
of, so this is the product in the f, this
is the product of the dlog s's. You
see these are kind of dual, that they use
dual information. C involves the vertices
of the graph, omega involves the
edges of the graph, okay? Now the C's
satisfy Jacobi relations because there
are color factors, and in particular the
easiest way to think about the Jacobi
relations is as relating three graphs to
each other. We're out here, who knows
what's happening to the rest of the
graph, but inside you look at this plus
this plus the cross equals zero, okay?
That's exactly what the Jacobi relations says, because it says whatever else
is happening outside, ff plus ff plus ff is
equal to zero, okay? So it's a statement
about C_gamma_s plus C_gamma_t plus C_gamma_u is equal to zero. The remarkable
claim is that the omegas satisfy exactly
the same identity. So something made out
of kinematic invariants satisfies
precisely color Jacobi relations, and the
reason for this is really simple, it's
the thing that you learned early on in
field theory. When you think about I mean just
kinematics, if you have one p1, p2, p3, p4,
you learned that s plus t Plus u
is actually the sum of the
masses, so it's p1 squared plus p2
squared plus p3 squared plus p4 squared,
but this means that ds plus dt plus du
is the d of all of these things, and
therefore if you wedge ds
plus dt plus du into the dp1
squared dp2 squared dp3 squared dp4
squared, which is what you do if this
went out to the rest of the diagram, you
get 0, okay? So this sort of basic fact,
this basic seven term relation, actually
explains that color factor is satisfied,
that these
kinematic forms satisfy exactly the same
algebra as color factors, and so in fact
we can interpret these forms as actually
giving us color information. So it's kind
of fascinating, that you can think about
for example in Yang-Mills theory, you can
think of that the Yang-Mills
amplitude is actually a differential
form on kinematic space, directly a
differential form on kinematic space.
Of course it has some polarization
vectors in momenta, but there's some
polarization vectors even in old-fashioned
language, you have polarization vectors
in momenta and you have some dlogs of
kinematic variables, Ok? The claim is
that on kinematic space there is a
unique projectively invariant
differential form, no color factors
anywhere, which is gauge invariant, on
shell gauge invariant, involving
polarization vectors in momentum,
ok? And and that form is actually the
Yang-Mills, the full colored Yang-Mills
amplitudes. And now two claims, one how do I get the Yang-Mills amplitude from it?
Well everywhere I spend it out of the
form, everywhere I get ds ds ds ds,
everywhere I get one of those guys I
replace it by the corresponding ffff, and
that's the way I go from the form
to the full amplitude, or B let's say
want to get a particular trace. What I do
is take this differential form on the
full kinematic space, and I pull it back
to exactly the subspace we defined for
the associahedron, but that involves
choosing an ordering. As you choose some
ordering, these X + X - X - X equation,
you like to choose this ordering, you
pull back this form that lives
everywhere onto that subspace, and you
get the Yang-Mills amplitude for that
ordering. So there's this very
intimate connection between kinematics
and color, which you know then has many
other consequences as well. It's related
to the fact that gravity is gauge
squared and so on and so forth, but at
this most basic level it appears that
these forms are actually about color, and
so the reason why we don't particularly,
I didn't see a real role for them, so far
that I only told you about the story in
a single order, but there's a natural
extension of the story beyond the planar
limit for all orderings, and this way of
thinking about forms is kind of
formizing or geometrizing color. The
amplituhedron way of thinking about
forms in momentum twister space is
geometrizing
helicity. And so it's very strange, I
don't know what to tell you, it's very
strange that we have two very very
different ways of uplifting amplitudes
two forms that have very different
purposes in life, and we don't know, we
don't yet know how to combine the two of
them, and we don't really know what
the general essence are.
You know more generally I hope it's
clear that, we don't have, we don't have any, you know we're
like that famous Groucho Marx joke, right?
We don't have any. You don't like our
principles, we have other ones, right? You
know, so we don't know, we don't have
any principles in this system. We're
not, we don't yet have any principles,
we're just starting to get guesses, right?
For what the nature of the
questions are. But I still
think it's rather striking that it's a
very similar kind of question in very
very different settings with a very
similar flavor to them, and you know it's
not just pushing around equations, and
you know interpreting something from a
pretty language, blah blah, it has beef, it
tells you things you didn't know before.
Formulas that are true
formulas that we're not known before,
that are exposed by this way.
There's a third example. So I've told you
guys about almost all the examples. An
example we won't have time to talk about
at all is in cosmology. So there's a
question about calculating the
wave functions of universe,  to calculate
wave function of universe in perturbation
theory. You can think of this as the kind
of geometry underlying old-fashioned
perturbation theory in the quantum
mechanics, ok? And that ends up being
associated with a polytope as well. So
that's another question where there's something polytope, now
it's something there's nothing to it
scattering amplitudes, that's just for
something about the old-fashioned perturbation theory of
quantum mechanics or computation of
cosmological correlators in de Sitter
space or the flat space wave function in
a quantum field theory. So those are the
examples so far that we've seen this
kind of connection, but it's the same
thing every time. The object was in
kinematic space, there's a positive
question in kinematic space, there's a
subspace, and there's a form,
so that's the uniform thing in all the examples. (A student was asking a question.)
I don't know. I mean I would, in
in previous years I would swear
up and down that all of these things are
very very special to gauge theories of
gravity and so on, but since we've seen
this business for something as done in
phi cubed theory, I'm less less sure, so but
I still somehow feel that something
that that Lorentzian physics is crucial
for all this stuff,
I really doubt anything like, certainly
there's no what not a scintilla, a reason
to expect for some, you know generic
condensed matter systems, phonon, blah blah,
anything nice is happening, I would be
shocked if anything nice is happening. I
think it has everything to do with
relativity and the quantum mechanics, but
that's pure speculation. Okay, all
right. Gosh!
Oh my! That's what happened.
That's what happened. I was wondering why,
why it seems so free and easy.
Yeah so matter's done. That is done. Okay, great!
