 Okay, thank you. So today, I was going to address the issue of expanders and shrinkers in the case of SOP cross SOQ symmetry. So we'll look at hyper surfaces M of the form of this type evolving by mean curvature flow. So for simplicity today, yesterday I discussed all the minimal surfaces that exist in this situation. And today I want to show which expanders and shrinkers you construct. And how many there are is dimension dependent. And the most interesting case is where the dimension, the total dimension P plus Q is between 7 and 4. P and Q will always be greater than or equal to 2. And so the picture for all these cases is very similar. So to simplify all the calculations, the equations that we're going to get, I'm just going to assume for today, P and Q are both equal to 2. And so that leads to the following PDs. So mean curvature flow for MT is equivalent with always this first term. And then here you get P minus 1. So that's just 1 over X times UX minus Q minus 1. That's 1 over U. So this is going to be, today is about that one particular PD. And so the boundary conditions are this. And I'm going to look for self-similar solutions. So if we look for hypersurfaces that evolve according to this rule, I always think of this as separation of variables. The situation is very analogous to what you do with ordinary differential equations. You will undergrad your PDs. If you have a linear PD, the simplest solutions that you find, you get by separating variables. You substitute a function of t that multiplied with a function of X. So here we have a function of t multiplied. And then instead of a function of X, we have the hypersurface, which is the space dependent thing. So if you look for self-similar solutions of this type, so for plus, you get expanders. For t, for minus, you get shrinkers. These correspond to solutions of the form UXT is this again. And so this kind of function is characterized by the condition that UT is equal to 1 half U minus X times UX with a plus or minus. And this plus or minus 1 half I'll abbreviate to, so I'll call that lambda. And so if the time derivative has to be equal to this, then I'll just replace that time derivative here. So the ODE for expanders shrinkers is, okay, and if I sort the terms like this, then I have the terms that have, these terms all have units 1 over length. These terms all have unit length. So they scale differently. And so we're going to, how do we analyze the solutions to this equation? To begin, first, give it a particular height A here. There should be exactly one solution coming out of here. What we did last time for the minimal surface equation were lambda zero, so you could include that here. We considered scaling invariant quantities because this term was not there, so it made sense to look at scaling invariant quantities, bless you. So let's see. Z is X over U, so that's this slope. And W is UX, that's this quantity. So in the scale, those are the two scaling invariant quantities that you can associate to a curve. And if you know these, you can reconstruct. So if you know these functions along the curve, you can reconstruct the curve up to a scaling. That's not going to work for this differential equation anymore when lambda is not zero because this equation is not scaling invariant. So what we do is we look at a three system of three differential equations. I just add the third variable is X, which is itself. And so now prime is X d dx. And if you want to, you can say we do an exponential change of variables of the independent variable, sorry, yeah, the independent variable along the curve. And if you do this, then the differential equations you get are the same system that we had yesterday plus a few extra terms. So we had, so z prime is, so d z, which is d z, d theta, that's always this. W prime now is, it's basically just these terms. So we get, right, so you get these terms. This X squared, so this term was not here yesterday. And then X prime, what is X prime? X prime is just X. Okay, so we now have a three dimensional system of differential equations. If we have solutions to these differential equations, then we can reconstruct. So first of all, we know everything is a function of X because X is one of the solutions. And just from these relations, we can then reconstruct U, right? U is just X divided by Z. So if you, if we have solutions to these, we can reconstruct the surface. Okay, so the boundary conditions that we have are, boundary conditions at X is zero are the following. So first of all, X is zero. Z is X over U. So X is zero, U is equal to A. So we're looking at, so I'm trying to construct a solution that starts at height A and does something like this. Okay, so X is zero, U is A, so this will also be zero. And W is UX. We always want this angle to be 90 degrees because then when you swing it around the axis, you get a smooth surface. So this also has to be equal to zero. Okay, so the number A does not appear in this condition. So that it will show up a little later. So the solutions that we're trying to construct start at this point and for this three dimensional system, they had, these are solutions starting at the origin. So we have to see what solutions start at the origin. So we do the same thing as yesterday. We linearize. We have one extra equation, so we'll get a three by three system. And I long ago decided never to compute a determinant or find eigenvalues in public, but this one is easy. So the system is, I think it's better if I write WZ. So I'm going to throw away all higher order terms. So for W, the only linear term that we get, ah, sorry, yeah, near the origin, this term is singular. So I can't start at zero. That would be a problem. What I do is I'm going to re-parameterize the equation and I should introduce a new variable, another theta. What I'm going to do is I'm going to multiply all these equations with Z. I'm sorry, this is not what happened. Sorry, yeah. To deal with this singularity, we have to introduce a new variable. You look at one over Z as the new variable minus that. It doesn't solve this. Yes. Yeah, so the proper way to do this is not to use Z and W, but to use these angles and then all these things don't show up, but then the equations become a bit. I don't know if I want to spend time on this. I was trying to present you a simplified version of what we actually did, but this, so the simplification is too much. Sorry about this. Okay, so if you'll bear with me for a second, I'll just have to, sorry. Yes, but I, so, okay, so what I want to do is, so if this is a regular system of differential equations, if the right-hand sides are smooth and all the, if they're polynomials, then you can linearize and there's an unstable manifold. And if you use, so that thing exists if you use instead of, so if you take the angles, if you take the angles instead of the tangents of the angles, so we would have to do a certain amount of algebra to, yeah. Yeah, so if I, right, so at X is zero, this term should not be relevant, so if I, let's see what happens. So you could add a fourth equation, U prime equals, let's do this, U prime equals X times, so U prime is X times UX, which is X times W. So if you write the equations like this, so, okay, so I've replaced this term by this and the cost is that we have to add an extra variable, X squared Z. Okay, I'm really sorry about this, X squared over Z, right, U is X over Z, so X squared over Z is X times U, but this should be just a U, yeah. Okay, so I'm, I'm going to cut this calculation short a bit, there is, yeah, so I'm sorry, I'm going to not do this because I'm not really sure how this is going to continue, I should have started with this being, with this angle calling that theta and this one phi and then work with those and then if you take the radius as a variable, you get a nice system of differential equations but the formulas are longer and I, while preparing for this, it looked like this was shorter but I must have overlooked that one over Z. Okay, so let me put it differently, so the result is, is this, so lemma, so for any positive a, there is a unique solution of that type, so to the equation, so let me call this the expander shrinker equation and now I'm still going to use that system that I have over there because once, okay, so and now we have to distinguish, so lambda and this actually, this is true for any value of lambda, so but in particular lambda, so lambda is allowed to be plus or minus one half, so for both expander and shrinkers we get a solution. Okay, so let me do expander first, so in this case the same kind of arguments that I showed you yesterday show you that this is defined for all positive values of a and again it's the, so it's the same kind of arguments, the second derivative at the origin has to be positive, at any point where the first derivative is zero, the second derivative has to be positive, so that means there can't, there can only be one of them, okay, so that you get that from this equation and the fact that lambda is positive, so they're defined for all positive, for all, sorry, for all x positive, right, so these solutions are globally defined and this is, and for all a, so we get a whole family of expander, and now what happens to these things as x goes to infinity, so as x goes to infinity, let's look at these equations, the one term, so there's one place, so x shows up here, as x goes to infinity, this becomes the most important part of the equation and so for very large values of x I'll calculate this, okay, so this is minus z prime over z squared minus w prime, and this is, now as x goes to infinity, this becomes minus one over z plus w plus, and then all those terms, one plus w squared, z minus w plus lambda x squared, okay, and now if you assume which you can prove, so you would have to, if you assume that z and one over z are bounded so that the slopes don't get close to zero or infinity and that the slope of the tangent also doesn't get close to zero or infinity, then for large values of x, the only important term is this one, you get this equation, the difference one over z minus w satisfies an almost linear equation, so there is, consider this the inhomogeneous term, this is the factor multiplying one over z minus w, as x goes to infinity, if lambda is careful now, here I have minus w prime, so all these things should have had minuses, not this one, this one needs a minus, okay, so what happens for large values of x, this becomes a really large number, this is of order one, this tells you that this quantity will decay exponentially at some rate that you can figure out, so and what is one over z minus w, well one over z is u over x and w was u x, so this tells us that if you're far away, so the interpretation is this, let's say you're on the solution of, one of the solutions of the differential equation, you're at this point, one over z minus w is the difference between, it's essentially these angles, except these are the tangents of angles, so it's the tangent of this angle minus the tangent of that angle, what this says is that those two are going to zero, the difference between those two are going to zero, so as you're moving along this curve, the slope of the tangent will rotate and it's going to align with the slope of this line, okay, so solutions will look like this, right, so if a solution is moving, you're on a solution and you arrived here, you're moving in this direction, the direction in which you're moving is going to align with the radius vector and so it'll just do this, right, so and this, so if you take this, it implies that the solution goes to a cone, in particular if you use the fact that this is of order one and that this is x squared, you can improve this using barriers to show that this is something of order one over x squared, so then since that is integrable, you can prove that the thing converges to a cone, right, and the order one over x squared is this order one divided by this x squared and so in this argument, so I'm going to repeat this argument for shrinkers, so instead of repeating it, let me just say what changes right now, so in a second we'll look at shrinkers, if so for shrinkers lambda would be minus one half and everything is the same except this lambda now is negative and instead of, so this differential equation for the difference between those two angles forces that difference to grow exponentially rather than shrink, so for shrinkers you would have the following picture, so if you're on a shrinker and the slope to the curve is not aligned with the radius vector, then that differential equation forces this slope to turn around, so if you're moving around it will, the solution will look like this, which if you plot solutions of this or if you try to compute solutions numerically and you have something that is sort of on a cone, this cone is unstable, if you're tangent at some point is deviated slightly from the cone then it will turn around and it will do something like this, so shrinkers look like this or they could go the other way around depending on what the instability look like, okay, so this explains why there are lots of expanders but there are very few shrinkers and so there's a, so there's a, let me at this point mention a PDE version of this, so this is not an ODE statement, this is a PDE statement and so if M1 and M2 are N, if you have two smooth self-rinkers for mean curvature flow that are asymptotic to the same cone at infinity, so they're self-rinking solutions, they're asymptotic to a cone, both of them are asymptotic to the same cone then they must be the same, okay, so what we're seeing in the ODE here is a manifestation of this fact, okay, so we have, so back to expanders, for each, for each A, so back to expanders, for each A we have an expander, it's a monotonically increasing function, it is asymptotic to some cone, there will be some asymptotic expansion that I wrote down the other day, the question is, so which cones can we achieve by an expander, so how does the slope of the asymptotic cone depend on A, so, okay, so by looking at these ODE's again and doing more analysis you can prove that they depend continuously on the shooting height A, what I want to explain is, and this is, so what are the asymptotics for this as A goes to zero, so what do the shrinkers look like when A is a small number, so you get the ODE's gone, so I'll work with this ODE, so where lambda's plus one-half, but the same arguments also work for other lambdas, and so I'm going to, so Y is X is the stationary cone, that's a special solution, we're looking at solutions that start down here, which is a singular, so U is zero is a singular point, X is zero, U is zero is a, two terms become singular at that point, so to analyze what happens here we have to rescale, so I zoom in on this neighborhood and I multiply it with the factor A, a one over A, and so what do we get, if I do that with this equation, so let me say, let me give this variable, what happens to this equation, the, let's see, so UX, all the UX's are scaling invariant, UXX becomes, gets divided by A, one over U gets divided by A, these things all get multiplied by A, so if you do, you'll allow me to do it like this, okay, this is similar again to what I did yesterday, so you get this equation where A is a small number, and now the initial conditions are V of zero is one, and VX, this becomes VY of zero is zero, and so as you let A go to zero, the right hand side goes to zero, and what we get is the differential equation for minimal surfaces, so in the limit what we'll see down here, in the limit what you see is the minimal surface equation, okay, so where this is exactly at height one, we know there's exactly one such solution, we analyzed it yesterday, it is, it oscillates, it intersects the cone infinitely often and is asymptotic to the cone, and then yesterday from the asymptotic, so what is the shape of this difference, so this is one building block of the solution, so this thing is going to show up here, and so that thing describes the solution here on a length scale A, large multiple of A, to get the rest of the solution we will have to analyze the equation in this region, but what we see is that the, so what does the solution look like here near the origin, after it has done this, come off the vertical axis, it converges to the cone and it stays close to the cone and parallel to the cone, so now we have to analyze what solutions in this region look like, and the boundary conditions we have is that they are close to the cone here, parallel to the cone, okay, so let me not do this, the calculation that I'm showing you is, it's a typical applied math calculation, it's a formal match asymptotic expansion for the solution, and so the formal match asymptotic expansion by itself doesn't give proofs, but these are ODE's and the, so the calculations can be justified afterwards, okay, so I want to show you the results, okay, so we assume that the phi of x is much smaller than x, we substitute this, so now here we go back to the original equation again, and the only difference with what we had before is that we've given up the boundary conditions, so the boundary conditions are, no, we're not looking at these boundary conditions, but we're assuming that the solution is going to match with whatever comes out of this small corner here, okay, so, okay, so if you said u is equal to x plus a small function phi of x, you end up linearizing this equation, and so let me tell you the result of the linearization, actually most terms, so this term and those terms are all linear, so those will just stay like this, this, if you linearize around x becomes plus one over x squared times phi, and here u of x is roughly one, so, okay, so the linearized equation that we get is this, and then of course because you're linearizing there are terms of order phi squared and phi x squared, and so part of the formal part of the calculation, this being a formal calculation, means that we're going to ignore those quadratic terms for now, so this is the linear equation, so the solution to this linear equation, this is a linear differential equation, it's one of the standard equations, these, you will find them in all textbooks, it has two, there are two solutions, there's two dimensional space of solutions, if you look at, so it's not constant coefficient, but if you look at the coefficients near x is zero, it's a regular singular point, and so this means that you can find power series expansions, so phi one and two x, both are of the form x to the power r times a polynomial in x, the times a power series in x, and the coefficients are that you find by substituting here, so the characteristic equation is this, you can find the roots, and let me not calculate them, but the roots that you get are r one and two, they are complex, so this is the one place where in this story that is dimension dependent, so depending on p and q you will get different roots, if the dimension p plus q is less than eight, you get complex roots, so here you get minus alpha plus or minus i beta, so alpha is a positive number, so these are the complex roots, so both solutions, so near zero phi of x is x to the power of minus alpha plus or minus i beta, and so you would have to take, these are complex valued solutions, so you take the real part, so you get something like this, you get x to the minus alpha, that's from this factor, this constant, complex constant, you take the absolute value out, what remains is a phase theta, and then you get the real part of x to the i beta, which is this expression, so near zero, what do these solutions look like near zero, they oscillate, which is good because they have to match with the oscillating minimal surface, so near zero, what phi looks like is something that oscillates like this, and as x goes to zero, log x goes to minus infinity, so this thing oscillates infinitely often as you go into zero, on the other hand, this x to the minus alpha as x goes to zero becomes unbounded, so these solutions, they oscillate infinitely often, but the oscillations also grow, they become really large, which means that this phi, this approximation that we made, that u is equal to x plus phi, can't be valid for all values of x, as soon as this function phi becomes a little bit large, we have to switch to the other minimal surface expansion, so match with the inner expansion, how do you do that, so what was the inner expansion, that is u of x is a times v of x over a, and v is this, I keep pointing at this picture because it's almost the same picture, the minimal surface looked like this, we have to rescale this thing down, and that will give us a function, that's the function v that we have here, and we have to match that with this expression, so this leads to a certain amount of algebra, and let me tell you the result, so first of all, as you let a go to zero in this expression, this goes to infinity, right, so to match this for small a with these functions, we need to know the behavior of v for large values of a, and so v of y is, well for large values of y, the thing is asymptotic to the cone, it is y, plus, and now you need to know the deviation of v from the cone, and you can do a calculation similar to this one, and what you'll find is the same linearization, except here you're dealing with minimal surfaces, so that means that this term is gone, lambda will be zero, so the linearization is the same, so what we find is that v of y has an asymptotic expansion, and it is, so for y going to infinity, it is y plus, and then the next term is something like this, in particular the exponent alpha and beta is the same, because they come from the same differential equation, and then there are extra terms which are small compared to this, okay, so now here I have to replace y by x over a, so we get two, okay, so we have the inner expansion is u of x is a times v of y of x over a, you get that y becomes an x, everything gets multiplied with a, so we get c0 times a to the power, this y minus alpha becomes, here we get a multiplied with a to the alpha, one plus alpha, x to the minus alpha, and then we get this cosine, cosine beta log, and then here I have to write x over a, and then using the property of the cosine that's beta minus beta log a plus theta not, okay, so a large formula, and the outer expansion, okay, so there's one more ingredient that I have to tell you, I said, so for the outer expansion I'm going to use this expression, and we will compare with this one, but now here I wrote this is true for every fee for every solution to this linearized equation, but there are two of them, so what angles do we get here, and so how I'm going to select those two solutions, so before I do that I have to, so the way we select the solutions here is by looking at their behavior at infinity, so I'll come back to this here, I'm going to write u, and so we're going to look at the behavior as x goes to infinity of this, this equation, so it's a classical equation, so you could just, you could look it up, even Wikipedia will tell you, or the internet will tell you what this thing does, a quick way to remember what happens is that there are, as x goes to infinity, you could assume that either the first, the function phi and phi x, that they are the dominant term, so then this would have to be zero, and so if this is zero and this is negligible, then this, the solution to this is that phi is approximately a linear function. The other option is that the second derivative is actually the largest, and then the largest terms are this phi xx and x times phi x, and this tells you, so you integrate this, this tells you that phi x is roughly like e to the minus lambda x squared, and this, so the sign, so again lambda can be plus or negative, positive or minus, sorry, positive or negative here, you get two very different behaviors here if you change the sign of lambda, so for expanders, lambda is positive and this means that there is a solution that decays to zero really fast. If you're looking at shrinkers, then what you get is this minus lambda becomes positive and this means that most solutions blow up really fast and this is the same phenomenon as what I showed Drew before. If you follow solutions of the shrinker equation and you go out along a cone, the cone is unstable and the solutions will tend to turn around either going up or down, so it's the same phenomenon. Okay, so at this point we don't need this equation anymore. So let me stick to the case where lambda is positive, so for expanders, in that case, so what this shows is that there, most solutions will grow linearly at infinity and there's one special solution that decays exponentially, so I'll let, so I'll let phi one be the solution that has this behavior at infinity, for a solution that has this behavior at infinity, it is x plus little o of x and I'll let phi two be the solution that is e to the minus lambda x squared plus little o of x squared. Okay, so this choice determines, this choice uniquely determines phi two, this one determines phi one up to adding multiples of phi two because if you add a multiple of this to that, you still get something that is x plus little o of x, so you just pick one of these. Okay, so now we have, so this thing is like x at infinity and this one is actually, yeah, so phi two of x is like, and this one is like, has this expansion, this one. Okay, so this phi two is uniquely determined by this asymptotic expansion, so that means that the coefficients that I have here, this c, these are uniquely determined by that choice, these are specific numbers, so this phase angle theta two and this number c two are completely determined by that choice and now I can, this other one will have, so this phi one, where we have a lot of freedom, we can choose the asymptotic expansion here at infinity, it'll be c two times x to the minus alpha cosine beta log, sorry it'll be cosine beta log x minus some other phase angle and what I'm going to do is I'm going to choose that phase angle so that it's, that this becomes a sign, I'll do that. So then the general solution is, this will be x plus c one times phi one x plus c two times phi two x and what I want to do is I want to find c one and c two by writing this out and comparing what I have with what I have up here, let me do that. So near x is zero, what you get is x plus and here you get c one, so both these phi one and phi two contain, contain this same constant c two and the only difference is that one of them has a cosine and the other has a sine, x to the minus alpha and one of them has cosine, okay so that's the expansion we have near zero so now we have to compare this with that and unfortunately that tells you, right this is a cosine and you have to compare it with this thing, we use the addition formula so you get, I'm going to write this as beta log x minus theta two plus beta log a plus theta not minus theta two and I'll apply the addition formula for cosine to these two things and what I get is one thing, one term with a sine and one term with a cosine and so you can imagine that that's going to be a long formula instead of writing all that out. Let's see which term we want. Once we know the expansion of u to be this, then we can look at, so what, what do we have? We have the solution, the expander that starts at height a, small value of a near the origin. We can follow it along the minimal surface that has been rescaled, then it fall, then near the cone it follows, it stays near the stationary cone for a while, behaving like this and then we want to know it's asymptotic behavior as x goes to infinity, so we have to look at what happens to this expression as x goes to infinity and that's, okay, so as x goes to infinity the term with phi two just goes to zero really fast, okay, so this term we can forget about. This one grows linearly, so that indicates that the solution is going to deviate from a cone and the coefficient multiplying phi one, this c one here tells us how much it will deviate from a cone, so what I want to get out of this calculation is c one, okay, so that means I have to get the term multiplying, I have to get the term with sine of beta log x, so of the addition formula, so cosine alpha plus beta is cosine alpha beta, it's minus sine, so we get minus sine of this, okay, so the addition formula gives you this and the thing to look at is what is c one, this c one x to the minus alpha matches this and almost that, so the conclusion is the coefficient multiplying phi one is c zero a to the one plus alpha divided by c one, absolute value, and then this sine, so we get a minus sine and again this is a formal calculation, so the actual value is a small perturbation of this and part of making this rigorous is to show that the perturbation is much smaller than this, but all that can be done, okay, so this implies, so this, let's, these are just other constants, so this is my, this is a constant times a to the power one plus alpha times sine beta log a plus some other phase angle, okay, so it has this form and the conclusion of this is, so now we can erase phi two, so the shape of the expander with u zero is equal to a is the following, on the scale a here, it's a minimal surface, so that's the thing that we started with, then, so it intersects the minimal surface many times, then for a large length it follows the cone and at some point it is, so when you're far away it is x plus c one times x and c one is given by that formula, okay, so in particular the asymptotic slope of this cone one plus c one of a, I'm a little over time, but let me just write down the, let me graph this, this is a, okay, what is the graph of this function, the graph of this plus a, well it goes to, as a goes to zero it goes to zero like a to the one plus alpha and as a goes to infinity this log goes to zero, this log a goes to minus infinity of this, so this sign will oscillate up and down very often and what you get is, get a curve, so the slope as a function of a oscillates up and down infinitely often and so one conclusion, so one conclusion is, so what was the question, which cones can we, for which cones can we find expanders and how many expanders can you find, so if you pick a certain value of a here then these are all the expanders that you find, right, so the conclusion is that the closer the cone is to the stationary cone the more expanders there are, so in particular the stationary cone itself has an infinite sequence of expanders, so smooth, so similar expanders coming out of the cone, the, so this is the story for expanders, now what about shrinkers and so it's the same story, so it could in principle take as long, but instead let me tell you what happens, so very quickly, for shrinkers you do the exact same thing, except at this point here what are c1 and c2, you determine c1 and c2 in the exact same way, it's when you look at the behavior at infinity you have to do different things with the c2, so as x goes to infinity for shrinkers remember this phi 2 of x doesn't decay like either the minus x squared, it blows up like either the plus x squared, so the only, the only solutions of this type that correspond that will actually give you shrinkers that are asymptotic to cones are the ones where c2 is 0, so to get shrinkers you have to do this calculation but you have to find c2 as well, right, which is this, which will be this other term here and that thing has to be 0 because otherwise you will get a solution that blows up differential equation that is not asymptotic to a cone, so for shrinkers what you find is that you don't get a shrinker for each value of a but there's an infinite sequence of a's because this thing has the same argument, it's beta log a plus that phase, so as a goes to 0 there will be an infinite number of times where this hits a multiple of pi plus pi over 2 and making this cosine 0, so you get an infinite sequence of shrinkers and the asymptotic slope of those shrinkers are given by this constant which is the exact same formula that we have here, okay, so for shrinkers what we find is that there are not, there's not a whole family of them but there's an infinite sequence of slopes, well there's a sequence of a's going to 0 and for each a in that sequence there is a shrinker with that asymptotic slope and the asymptotic slope converges to 1, so for large enough and you get a smooth, you get a smooth shrinker whose slope will be somewhere over here, so let's say it would be this one and then for that slope there are many many smooth expanders, so that's how you get the non-uniqueness, okay, so I'm sorry about the mess in the beginning, I would have preferred to spend more time on this but if you have questions I would be happy to explain further detail.