 Okay, great, so our next talk is Cole Hookamier, and he's going to tell us about the inscribed space, sorry, square problem, I can't read. Thank you so much. It's absolutely amazing to be here and to listen to all these phenomenal talks and to celebrate Tom's birthday. I'm really excited to tell you all about this new approach to the inscribed square problem that I've been thinking a lot about. So just a little bit of background. The inscribed square problem is this 111 year old conjecture due to culplits in 1911. And it's just a statement that if I have any simple closed curve in the plane, I can find four points on that curve that form the vertices of a square. Now, this is a really hard problem, but if we restrict gamma to being in some nice regularity class, it becomes significantly easier. So we do have the following theorem, due to Schnellman back in 1929, okay, which is that if gamma is smooth, that implies there is in fact a square, right? So if gamma is smooth, we can find a square. So how do we go about proving something like this? Well, I'm going to define a configuration space, which I'm gonna call C4 tilde of R squared. What this is, is it's the set of cyclically ordered quadruples of distinct points in R2. All right, so just to clarify exactly what I mean by this, we can define this as a quotient space of sort of the standard configuration space. So this would be C4 of R2. And this is just quadruples of distinct points in R2 that are ordered. So this is just an open subset of R to the eight. And we're gonna mod out by a Z mod 4 Z action, which corresponds to just cyclically permuting the coordinates of the couple, okay? So this is a manifold. And we're interested in two specific subspaces of this manifold. We're interested in SQ, which is going to be the set of counterclockwise squares. So since the couples are cyclically ordered, if we have a square, we're getting some cyclic order on the vertices. And I'm just letting this be the set of squares where the cyclic order is consistent with the counterclockwise order of the vertices. And then we also have C4 tilde of gamma. All right, so this actually has three connected components because I'm not requiring that the four points be consistent with the cyclic order on gamma. And there are two sort of distinct ways to put a pair of cyclic orders on four points. So in particular, we're interested in the intersection of these two spaces. So I'm just gonna define inscribed squares of gamma to be the intersection square intersect C4 tilde of gamma. All right, so this is the thing that we're trying to prove is not empty. And in the case where gamma is smooth, these are both smooth sub-manifolds of this manifold and they're both dimension four, this is dimension eight. So we expect there to be some distinct set of intersection points for these things and we should be able to count them in intersection parity, more or less speaking. There are some complications with this. First of all, these are not compact manifolds. So there's that issue. And second of all, the sort of typical differential topology that you would want to use to calculate a intersection parity breaks down a little bit here. And the way it breaks down is that if we were to take the set of all possible small perturbations of a smooth Jordan curve and then see what happens to C4 tilde of gamma, we don't actually get all small perturbations of that smooth manifold. So you can't just say, let's just take a small perturbation and then apply the standard transversality theory because you don't have all small perturbations. So you actually technically have to build transversality theory sort of from the ground up for these specific situations. And that wasn't done until relatively recently. So I'm gonna, so that was done by controller Denim McLeary in 2014 and I'll just say relevant transversality theory. Obviously back in 1929, there wasn't any transversality theory at all. So he did it with just like explicit like coordinates and stuff like that. It was a really big proof. But now that we have machinery like this, it's actually really easy to just explain how to prove this fact for smooth curves. So what we can do is we can define gamma to be generic. If it has the following two properties, it's smooth and we have a transverse intersection between squares and the configuration space. All right. So this is what it means for a Jordan curve to be generic in my book. And there's a few nice facts about generic Jordan curves, which is a corollary of the transversality theorem in configuration spaces. So facts. One of them is that they're C infinity dense. So given any smooth Jordan curve, we can approximate it arbitrarily well with generic curves. And the other ones, the other fact is that they have generic isotopes between them. In particular, what's important about these generic isotopes is that they induce one manifold bordesms on inscribed squares. So if we have two Jordan curves, we'll have some squares over here and we'll have some squares over here. And then the set of inscribed squares is going to be some one manifold when we isotope between these things. So what that means is that the number of inscribed squares in a generic Jordan curve is actually invariant on the parity. So more specifically, we can say that the homology class of inscribed squares of gamma is well-defined as an element of H0, C4 tilde of gamma, with coefficients in Z mod 2Z. Also in Z, but we don't care about that. So I mentioned earlier that this manifold has three connected components. So that actually corresponds to there being three different kinds of inscribed squares that we can have in a Jordan curve. And let's just draw what they look like. There's what I call the type one squares, which are the sort of obvious type where we just like go through all four vertices in a circle. There's the type two squares. Oh, I should say these, these are also called gracing squares. Gracing squares. That terminology is due to our rich shorts. So there's also the type two squares where we go through the vertices in a zigzag. All right. Like this. And there's the type three squares where they go through the vertices in the opposite cyclic order from what we're supposed to. It goes like this. So these are the three different kinds of inscribed squares and each of these individual types of inscribed squares has its own parity for generic Jordan curves, which is invariant for all generic Jordan curves. So let's calculate those parities. We need an example generic Jordan curve and any ellipse, which is not a circle, is generic. And this has exactly one inscribed square, which is grace. So what that tells us is that this is always odd. This is always even. And this is always even in any generic Jordan curve. All right, so we understand generic Jordan curves. That's nice. But what about non-generic? So let's let gamma be non-generic. There's this attempted argument that you can make for non-generic Jordan curves, which goes something like this. We're gonna take generic Jordan curves, gamma one, gamma two, dot, dot, dot, and they're going to approach gamma in some sense. And we know that all of these generic Jordan curves have inscribed squares in particularly. They have gracing squares. So we can let S one, S two, dot, dot, dot, be gracing squares inscribed in these Jordan curves. And although these might not approach anything, they're all within some bounded region in space, right? So we can take a convergent subsequence. So we can take a subsequence that converges to some square S. And then we have an answer of squares in gamma, and then we're done and we solve the inscribed square problem. Sorry to interrupt. That's exactly what I'm about to talk about. So you can, of course, approximate any continuous curve with smooth curves in the C zero topology, right? So if I can approximate smooth curves with generic curves, then I can approximate anything with generic curves in C zero, right? So you can do this, but the issue is that when you take this convergent subsequence, the squares might shrink down to a point. And that might sound like a trivial sort of technicality with this argument, but it's actually super important. And the thing that makes the inscribed square problem so hard is the fact that this convergent subsequence of squares can shrink down to a point rather than a actually non-trivial square. So you can get around this issue for certain regularity classes. And the way you do that is by selecting these generic approximants in such a way that there's an epsilon greater than zero so that no squares of size epsilon are smaller appear in any of those generic approximants. So if we were to have, say, a corner in gamma here, so it's not generic, of course, it's not smooth, we can take these sort of smooth approximations that look like a hyperbola locally, and they won't have any inscribed squares. Like there's no inscribed squares within this part of the picture. So there's no small inscribed squares in a smoothing of a piecewise linear curve. And you can do this sort of argument for all sorts of regularity classes. As I just pointed out, you can do it for piecewise linear. You can do it for C1, and you can do it for piecewise C1 with no cusps. And some other classes as well. Piecewise C1 with no cusps, that means that when I'm connecting two pieces of two C1 curves at the spot where they're connected, I'm not allowed to have the derivatives be like negatives of each other. So I'm not allowed to, in C1, I can have lots of little loops like this, so I'm not allowed to have something like this in a no cusps piecewise C1, because that would be able to have small squares. So yeah, so what's important about all of these classes is A, that we've solved the inscribed square problem for them, and B, that they have no small squares. So I call this the small square barrier. And it's like the really hard, it's the separating wall between the regularity classes that we can solve the problem for and the regularity classes where we can't solve the problem for. There are some things that technically break the small square barrier, but they don't break the gracing small square barrier. So an example of that would be curves that locally look like the graph of some continuous function for some choice of coordinates, right? So we could have lots of zigzags like this that go all the way around, and they could be like fractals, zigzags. And as long as we can choose coordinates at any point so that it passes the vertical line test locally, you can't have gracing squares. You can have inscribed type two squares locally, but you can't have gracing squares. So that would still, you'd be able to apply this argument. And breaking the small square barrier is really hard. And when we're like facing such a difficult problem, sometimes the right way to deal with it is to look at the sort of simplest case where everything breaks down horribly. So I have a regularity class that I've been focusing on, which I call ACS for arbitrarily complicated singularities, but it's a fairly simple regularity class. This is just curves. This is just curves which are smooth except finitely many points, right? So that's ACS, curves which are smooth except are finitely many points. And crucially, the reason I'm calling it arbitrarily complicated singularities is we have no restriction on the regularity for these curves near the bad points. So I want you to imagine just like a spiral where the two strands are coming in to each other, and then there's like smaller spirals as we zoom in, and those spirals wrap around each other worse and worse and worse, just like exponentially fast as we zoom into that point. So I'm not gonna attempt to draw something that terrible, but I want you to, I'll just draw it like this. So this is a curve in ACS, all right? So we'd have some point where everything gets terrible and maybe a few other such points, but finitely many. So if you can solve the inscribed square problem for these curves, you can break the small square barrier. So it would be a really big deal if we could solve this inscribed square problem for ACS. And I have an idea of how to do this, and I've somehow convinced myself that the key to solving the inscribed square problem for this class lies in sort of a, as far as I know, unexplored area of mathematics. And that's essentially the connection I wanna show you guys in this talk today, which is relation avoiding paths. So in particular, we're gonna have a vector space, V, which is gonna be a vector space over the complex numbers. Find a dimensional, and we're gonna have relations, R1, R2, dot, dot, dot, and these are all, finally many of them, Rn, and these are all going to be linear sub-spaces of V times V, okay? And what is a relation avoiding path? A relation avoiding path is just a continuous path in V, which has the property that no two points of that path are related. So in particular, I'm interested in relation avoiding paths that sort of approach the origin of V. So I'm gonna define a relation avoiding origin path to be some function P, which goes from zero infinity to V continuously, and it's gonna have the following two properties. It's going to have the property that if I limit as T goes to infinity of P of T, that gives me zero. It's gonna have the property that there does not exist any pair of times T1, T2, so that P of T1 is related by any of my relations to P of T2. All right, so this is a relation avoiding origin path. All right, now I really wanna show you guys how this is related to the inscribed square problem, but this is a new concept to me, let alone you guys, so I think it's worthwhile to go over a couple examples of these paths to sort of get a feel for how these objects behave. So we're gonna do example one, and for all these examples, we're just gonna be looking at the one-dimensional case. So V is just C, right? So what is our relation that we're gonna be avoiding? Let's start with something simple. X plus Y equals zero, this is a relation. So two things are related if they're negatives of each other, and let's try to figure out what it means for a path to avoid this relation. Well, if I have a path that sort of approaches the origin, I want none of these points to be the negative of any other point, so that's the same as if we take this path and we rotate it 180 degrees around the origin, we want this path and this path to be disjoint from one another, and that's the same as avoiding this relation. All right, so let's take a step back and suppose we're trying to make a relation avoiding path and we're gonna start at one in the complex plane, and our eventual goal is to reach zero. From here, all we really know is that we're never allowed to touch the point negative one, right? We immediately see something interesting here, which is that there's infinitely many homotopy types of path from one to zero that avoid negative one, right? We've already exhibited one, we could just go straight there and then that would give us a relation avoiding path, but then you could ask the question, are there relation avoiding paths from one to zero that wrap around negative one? And if you try to find one, right, we'll just do this, you immediately see that the answer's probably no, right? This crosses. So it's rather hard to prove that it's impossible to find such a relation avoiding path, just a sketch of the idea here as to why it's impossible. Here's one, here's negative one. If we try to wrap around negative one, then this path is gonna go over here, and now we have to wrap around this whole thing because we're not allowed to touch it. So we have to go all the way over here, and now this is gonna wrap all the way over here, and you can see that you just get stuck in this infinite loop of spiraling farther and farther away from the origin. So in this case, there's only one homotopy class of path from one to zero for the relation avoiding paths. Okay, so we've solved this example, let's look at another example. I'm just gonna erase all of this here so that we have some space. So we have example two. Here we're gonna just look at X plus two Y equals zero. So in this case, we have our plane, and we have one, we're just gonna start at one again. Now there's two points we have to avoid. It's gonna be negative one half and negative two, all right? But for the same reason as that we can't wrap around negative one, we can't wrap around negative two either because we'd spiral farther and farther away. So we can just ignore that. So what types of homotopy paths can we get wrapping around negative one half? And you can see that if you sort of wrap around and get a little bit closer, then this will get smaller and you can actually spiral into the origin with both of these paths. And you can wrap around as many times as you want using these spirals. So you can actually get all homotopy types of path if you're avoiding this relation. So clearly something interesting is happening because what homotopy types of path we can get depend on what relations we avoid. So this is already a little bit interesting, but in my definition I had multiple relations. So let's see what happens when I have multiple relations. We have the relation X plus two Y equals zero and we have the relation X plus three Y equals zero. So in this case, we start at one, right? And we have to avoid negative one half and negative one third, right? So what I'm gonna do is I'm going to draw four of these homotopy types. And you guys can just guess to yourselves which ones you think are going to be relation avoiding and which ones you think will have no path in that homotopy class which is relation avoiding. And you guys can quiz yourself to see how many you get right, just for fun. So here's one homotopy class. We go between them and then we go to the origin. Here's another one. We go around but we go between them up and then we go back to the origin. Another thing we can do is we could wrap around once and then go between them. And another thing we can do is we can wrap around once and then go between them twice. All right, so I'll give you all a second or so to make your conjectures about these paths. So here's the answer. This one can be made relation avoiding, this one cannot, this one can and this one cannot. So obviously something non-trivial is happening here. And there's actually just an amazing fact about how to completely classify which curves are able to be relation avoiding and which ones are not. So here is the theorem. And this is only in one dimension. So only for one dimensional V. And here it is. The possible homotopy classes are exactly those of logarithmic spirals. So as you can see here, obviously we can make a logarithmic spiral with this homotopy class, not with this one, because we have to double backwards with this one also. And the reason we can't do it with this one is that one half and one third are actually a little bit close together. So there's actually no logarithmic spiral that will go through them twice here while only going around once here if it has to start at one. So yeah, so this is the classification of exactly which relation avoiding passive one dimension we can find. Something of the form V, I'll just say x times e to the at where x is some starting point and then a is some complex number with negative real part. Yeah, there you go. That's the solution. Okay, I'm gonna give a sketch of proof of this just really quickly. So here's the sketch of proof. So without loss of generality, just work inside of the unit disk and we can assume that our starting path goes from one to zero. And all of the other paths, we're gonna have some points we have to avoid. And all of the other paths are just gonna be scale rotations of our starting path. So what I'm gonna do is I'm going to take one over two pi i log of this. And this is gonna take me to the upper half. So my starting path is going to become some path L. L and then these are all gonna be translations of L that are sort of Z periodic in here. And I know they're all disjoint from L because it's relation avoiding. And what I wanna prove is that there's some straight line that separates all the starting points on the left from the starting points on the right. The straight line doesn't actually have to be disjoint from these curves. It just has to separate the starting points. Okay, so it's suffices to show that any starting point on the left has argument angle in the complex plane greater than any starting point on the right because then such a line will exist. Right, so the way we're gonna prove that is we're going to define PQ to be what's called a split pair if the following is the case. P is topologically to the left of L meaning it has a path disjoint from L to negative real infinity. Q is on the right of L and P plus L, Q plus L and L are disjoint. So the cool thing about these split pairs is that we can define an operation on them which transforms one split pair to another. And I should mention that to prove the theorem we wanna prove, it suffices to show that every split pair has a property that the argument of P is greater than the argument of Q. So we have this operation that takes a PQ and what we're gonna do is if the imaginary part of P is less than the imaginary part of Q we're going to subtract P from Q. So then it's gonna become P comma Q minus P and then otherwise it becomes P minus QP. And what does this do? Well if we have our split pair, so here's L, here's P, here's Q, Q and P. What we're doing is we're drawing a parallelogram like this and we're adding this one. So we're switching Q to this one and we can see that this path is gonna be on the right of this path because the relationship between these two paths is the same as the relationship between these two paths. So that's why it worked. And you'll notice that if you keep applying this operation, so we keep applying, drawing these parallelograms, what's gonna happen is the paths, the points are gonna get farther and farther away from each other and one of them's gonna go to negative real infinity and one of them's gonna go to positive real infinity, except in a few exceptional circumstances such as the imaginary parts having rational ratio or P and Q having real ratio which those cases can be dealt with separately. And what that means is that says one of the things is going to negative infinity, one of the things is going to positive infinity. Why does this tell us anything interesting? Well, what's interesting is that this operation preserves the invariant which is the order of the arguments of P and Q. So you'll notice that like here's my parallelogram. If P and Q, if P is angularly to the left of Q, then P will angularly be to the left of Q minus P. So this preserves the order of the arguments and then it splits them across plus minus infinity which means that at the beginning they already had arguments in the right direction. So that tells us that every split pair has the argument of P greater than the argument of Q and that tells us that all of these points are to the left of some straight line compared to all of these points. So then there's a straight line and that straight line becomes a logarithmic spiral that is in the same homotopy class as this line and spirals in past all those points. Voila, all right. So this is a beautiful theorem but what does it have to do with the inscribed square problem? Well, the inscribed square problem is actually can be solved if you have a higher dimensional generalization of this fact. So I have a few conjectures about how this theorem generalizes to higher dimensions. And I'm just gonna call that the spiral conjecture and the spiral conjecture implies the inscribed square problem for this regularity class here. So here's the spiral conjecture. All right, so the spiral conjecture is if P is a relation avoiding origin path then there exists some H which is a homotopy from zero one times zero infinity to V. And this H has the following properties. Limit as T goes to infinity, H of S T is zero. I'll just make this one, two. H of zero T is P of T. This is where it's a generalized logarithmic spiral. H of one T is going to be equal to sum. I goes from one to N, X I, E to the A I T where X one dot dot dot X N are linearly independent and A one dot dot dot A N are complex numbers. All right, and then there's a fourth condition which is sort of that it respects the homotopy class. So what this is, is that there does not exist any pair S T such that H of S T is related by any of the relations to H of zero T. So it's only having to avoid the relations of the start point of the path as we homotop and that start point is allowed to move over time. So I have slightly stronger versions of this conjecture as well. You'll notice that it doesn't actually have to be a relation avoiding origin path for all time. So the strong spiral conjecture is the statement, one through three is the same and four is that H of S at S is a relation avoiding origin path, right? So this is the strong spiral conjecture and this in turn is implied by an even stronger version which is what I believe to be like the real fact underlying all of this, which is what I call the monotonic spiral conjecture. Now I should mention I have not proven the strong or monotonic spiral conjectures even in the one dimensional case. All we have shown is the spiral conjecture one dimension and the monotonic spiral conjecture, again one through three is the same. I wanna point out only co-dimension one relations matter because in this homotopy a small perturbation can be made to avoid anything of co-dimension more than one, complex co-dimension one, so that's real co-dimension two. So in four what this is is that the set of co-dimension one relations we avoid is increasing in S. So that means that if we avoid a relation at time S then for all future, for all S greater than that we also avoid that relation. So I think that we can actually not just find a homotopy that avoids the relations but find one that just strictly increases the relations we avoid until we reach one of these spirals. And essentially the reason these spirals are special is that they are sort of within this upper set of this ordering where we add an ordering on all the paths based on the set of relations we avoid and then it seems to me like these spirals are sort of, they're not quite the maximizers. It's more like if we're only allowed to go like up in our two with like 45 degree angles then it's sort of like this where there's this upper set and then once we reach a certain point we can always find a path to one of these spiral things. That's sort of a vague picture. I don't know if that made any sense at all but this is kind of nice and we have the following sequence of implications with these conjectures. So we have the monotonic spiral conjecture implies the strong spiral conjecture, implies the spiral conjecture which in turn implies the inscribed square problem for ACS. All right, so this is great. So I've already explained these two implications but let's talk about this implication for the remainder of the talk. So I'm going to prove, assuming the spiral conjecture that every ACS dording curve has an inscribed square. And the key theorem here, the key theorem that we're going to use is what I call the square envelope theorem. So I'll just state that and then explain what I'm talking about later. It's every bad dording curve has a square envelope. All right, so what's a bad dording curve with the square envelope? A bad dording curve is just a dording curve that we're trying to prove doesn't exist. It's just a dording curve with no inscribed squares at all. So what's a square envelope? So a square envelope is a square depending on time, A of t and then we have B, C, D, square, these are all depending on time and it has the following properties. A and B are inside gamma, gamma being our dording curve and C and D are outside. Actually I want to flip these. We have that the limit as t goes to plus or minus infinity of the side length is zero and we have that A and B wrap around gamma as time progresses. So what do I, let's sort of draw this. They shouldn't exist, but I'm gonna draw it anyway. So here we have a dording curve and we start with a square over here with A and B. Here's A, B, C, D. So A and B are outside and C and D are inside and then somehow A and B wrap around the whole curve. And then we have a small square. We have a small square over here with A and B outside, C and D inside and then somehow while keeping it a square the whole time C and D have to go over here which is obviously impossible, right? Obviously if we try to move A and B around the dording curve C and D are gonna bump into the curve. So you'll lose property one. So it seems completely obvious that these don't exist. And I mean if you can prove they don't exist you've solved the inside square problem. And the reason that I'll prove this theorem in a moment but first I wanna show how the fact that every bad dording curve has a square envelope tells you that ACS dording curves have inscribed squares. And this is actually very close to being a relation avoiding path. In particular the set of squares in this sense is a vector space where C and D depend linearly on A and B. You can just write down linear functions that create C and D out of A and B. So set of squares is isomorphic as a vector space. So C squared. And we have a relation which is the sort of the right corners of one square bumping into the left corners of another square. So if the A and B of one square bump into the C and D of a different square that's gonna be our relation. So it's technically eight different linear relations corresponding to all the ways that that could happen. So we have a system of linear relations telling us when the A B curves hit the C D curves and we have this square envelope which is then a relation avoiding path in this system. And you're gonna notice that on one side you might have a normal looking square but on the other side you're gonna have a square where the curve wraps around it in a weird way. And that can never happen for a smooth curve. Like if you zoom in on a smooth curve there's only one way to put a square here which is like this, right? With A B inside and C D outside. So that means that these have to be close to your singularities. So you actually get that on one of the ends of your square envelope you have a limit to a point. And without loss of generality you can choose that point to be the origin and you've constructed a relation avoiding origin path. And in particular you can show that this home will be tight from the curve wrapping around the square in a weird way cannot be represented by a generalized logarithmic spiral. So that's why the spiral conjecture implies that ACS-Jordan curves have inscribed squares if we have the square envelope there. So in the last five minutes I'm going to attempt to prove to you guys the square envelope there. So we're gonna let gamma, gamma be a bad Jordan curve. And what we're gonna do is we're gonna choose a function f which is positive inside and negative outside and zero on the curve. We can of course do this and we can make f continuous. And what I'm gonna do is I'm gonna define C1 and C2 of AB and C2 of AB to be the linear function that constructs C and D out of A and B, all right? So if I have A and B, this is C1 and this is C2 of these two, they're just linear functions. And what I'm gonna do is I'm going to create a map which I'm gonna call phi and this map goes from distinct points on the Jordan curve. So this is just points on the Jordan curve minus the diagonal. And it's gonna map to R squared minus the origin. And what the map does is it takes the point x comma y and it maps it to f of C1 of xy, f of C2. And you'll notice what's special about this map is that if it were to be zero, we would have an inscribed square. Yeah, what is f? Oh, sorry, f is a function that's positive inside the Jordan curve and negative outside and zero on the curve. So if we were to have a point where it's zero where this function phi is zero, that would correspond to an inscribed square. That would correspond to an inscribed square. All right, so this is a map from a cylinder to a cylinder topologically. So it actually has some sort of homotopy type that we can calculate. So I'm gonna call that the bad wrapping number. And you can calculate the bad wrapping number. You can just do some calculations using generic approximants. And the wrapping number is actually equal to the parity of type one squares plus the parity of type three squares in any generic Jordan curve, which is then equal to one. So we know that this map between cylinders wraps around an odd number of times. So how is this at all useful? Well, what I'm gonna do is I'm gonna look in R squared minus the origin at this quadrant Q, the upper right quadrant, right? And what I'm gonna do is I'm gonna look at the inverse image. I'm gonna look at phi inverse of Q, all right? So here we have our cylinder, gamma squared minus the diagonal. And we're gonna have some sort of set Q here. And what I know about Q, since I know that this map is essential, I know that there cannot be any loop which wraps around the cylinder that is this right from Q, right? Because such a loop would then wrap around the origin that I would touch Q, right? So what I can say by Poincare duality is that within this open set, there was a function from R to this open set that goes between the end points of the cylinder. That's a square envelope. Well, it's almost a square envelope. What is it really? We have our Jordan curve and it's a square with two corners on the outside and the other two corners inside because F has to be positive on both of those corners. And for it to go from the bottom to the top of the cylinder, these points have to wrap around the curve. So we can just push these points off the curve slightly and then that gives us a square envelope. So there you go, that completes the proof. I think I'm done. Oh yeah, so here's why you need more than one. There's actually some examples of specific systems where you need more than one. And there's a notion of taking a product of one of these relation systems. So you basically just take a Cartesian product and then it's just two relation systems independently. And you can choose homotopy classes so that those relations independently have to have different exponents for their logarithmic spirals, right? So in that case, you can make, yeah. Oh, well, it can't be bigger because XI have to be linearly independent. Yeah, any other questions?