 Okay. Right. So thank you very much for the invitation to come here. It's been a long time and it's just as beautiful as before. And I'm looking forward to giving these lectures. The plan is to start slowly with an introduction to mean curvature flow, develop the equations, develop some regularity theory for mean curvature flow. And then in my part of the course, I will focus on monotonicity formula and some recent results on non-collapsing of surfaces and what you can do with it. Okay. So I start with the setup of the problem. We want to consider the emergence, smooth emergence of a hypersurface into some smooth remanian manifold. So here this is a smooth remanian manifold. And let's assume we are in the situation where there's no boundary. And here this is a hypersurface immersion. The picture you should have in mind is that there's this manifold mn and you have this map f into this remanian manifold. And here you have the image of the hypersurface, the image of the manifold sitting in here as a hypersurface. And this is nn plus one. And in here you have the image f of mn. And now it's important to consider the geometric structures that you have on this thing. The first thing you have to take into account is the induced metric on the hypersurface. This is g is just the restriction of g bar to the tangent space of the hypersurfaces. And if we have here some coordinate map x into Euclidean space, Rn, and we have some coordinate maps here on a point f of p. So here's p, here's f of p and we have some coordinate map into Rn plus one. Let's call it y. Then you can compute the coefficients of the induced metrics at the point p to be the inner product of the tangent vectors of f with respect, so these are the derivatives with respect to x. And you take the g bar metric in formula. This is to be absolutely precise, g bar alpha beta at f of p times df alpha dxi and df beta dxj at p. And here the indices by convention I use the i and j run between one and n and the alpha beta, the Greek indices, refer to the target manifold and they run from zero to n. So this is the first structure you have to take into account. The second structure that is important, you have to pick a unit normal. And a closed surface in say Euclidean space, I always pick the outward unit normal. Sometimes in the Riemannian manifold might not be clear what the outward normal is, so you have to say what unit normal you mean. And once you have the unit normal, you can compute the second fundamental form. And the second fundamental form, this is a bilinear map at each point on the tangent space which is represented by a matrix at each point, denoted by hij at p, nij between one and n. And then the hijs can be computed from the second derivatives if I have such a representation I have that the hij at p with a minus sign times the alphas component of the unit normal p is computed by the second derivatives of f. So this would be the coordinate representation. And here you have the Christoffel symbols. I assume you know what that is, the kij times df dxj dxk. And then there's an extra term coming from the target manifold, the Christoffel symbols of the target manifold alpha, gamma k, gamma rho, say, times df, gamma dxi, df rho dxj. So if you want to describe this more abstractly and if you said, say, if you have a tangent vector field EI, so in this setting this would be the df dxi. But more abstractly you can say that the second fundamental form components are given by the covariant derivative of the unit normal with respect to the ambient metric in direction Ej which alternatively is minus the normal of times the covariant derivative of Ej with respect to EI. So this is a more invariant way of writing it but sometimes it's extremely important to have this coordinate representation. And once you have the second fundamental form in this way it is important to realize that this is of course associated also with a mapping. So this is associated with the Weingarten map. Let's call it W at P which is a map from the tangent space to itself and it's defined by the fact that WP applied to some x multiplied with y with respect to the metric g is the same as A of P applied to x and y. And this Weingarten map then has components which I did not note by H upper ij at P where the H upper ij are given by the H lower ij by raising them with the metric g and these are the components of the inverse metric. Because this H ij here is symmetric this W is self-adjoint therefore it has real eigenvalues and these eigenvalues are the principal curvatures. Lambda 1 up to lambda n are the principal curvatures of the hyper surface. Now you can form all sorts of scalar invariance by taking homogeneous polynomials of these principal curvatures. For example the product of lambda 1 and lambda 2 for a two-dimensional surface in R3 is the famous Gauss curvature and we are mostly interested in the mean curvature which is the sum. So the mean curvature H is just the sum of these principal curvatures which in our setting we can now also write as the trace of W which is H ii sum over i in my notation or it is the trace with respect to g of the second fundamental form in other words we can also take the inverse of the metric g ij and sum over i in j times lower ij. And now when you look at this formula the mean curvature seems to be something like a trace of second derivative so you won't see that this is a elliptic operator but somehow depends on the choice of the unit normal so the really invariant thing is the mean curvature vector. An important object is the mean curvature vector H vector this is you see here this is supposed to be elliptic we have to take the trace of minus H so this is minus H times nu and see if I change nu the sign of this so this vector is obviously invariant of independent of the choice of nu so independent of choice of nu and it can be written as an elliptic operator g upper ij of the Hessian of f minus Christoffel symbols df dxk and then we have this nonlinear term gamma bar dot alpha beta df alpha dx i df beta dxj and this is often called the Laplace Beltrami operator with respect to the metric g and the metric g bar you see this is if we adjust in Euclidean space this is just the Laplace Beltrami operator with respect to the metric g but we have to also take into account this of this of f and to see the so now we are ready to define what mean curvature flow is so now we say a smooth family of hyper surfaces so now we add a time parameter into this target manifold is a solution of mean curvature flow derivative at each point and time is exactly this mean curvature vector it's called this mean this is the problem we want to start and now because this operator here depends on the second derivatives the question is is this really like a heated question and it turns out you have to be a little bit careful because notice there's still some room yeah notice that the Christoffel symbols here these Christoffel symbols gk ij just a little check this to do the little calculation turns out this is just g upper kl times the second derivatives of f with respect to xi xj times df dxl in other words the Christoffel symbols here are the tangential parts of that vector there so this is not capturing all the second derivatives of f it's only taking the normal component so this is like g ij of the normal component of the second derivatives and this is a lower order term right plus it's called this gamma bar of the f the f so it is this is not what you would call a uniformly elliptic operator that you can immediately apply elliptic theory to it's the system of equations here it's crazy linear because it's linear in the second derivatives it's crazy linear not linear because it is nonlinear here but it's also not elliptic in the tangential directions and that's clear because it's a geometric flow which is invariant under tangential transformations of our coordinate systems x or y so there is an invariance in tangential directions so it will not be directly elliptic a physicist we have physics institute here would say there is a gauge group and that's the tangential differ morphisms invariant under tangential diffuse so you have to be careful okay I think yeah I should give you a first example of mean curvature flow right it's so far there has only been been formulas the main the main example of course that you should think of if you have a closed surface if the manifold say is sn and the embedding of the initial surface of mn is equal to a sphere of radius r0 in rn plus one in that case this equation here for the position of the surface becomes just a equation for the radius because clearly this is a radially symmetric so I just have to see what happens to this radius and if you look at this it must be true that the d dt of the radius at time t is equal to minus h times nu the outward normal so it's shrinking and the mean curvature is the sum of the principal curvatures each principal curvature is one over r for the sphere so we get this ode and therefore the surface at time t is the sphere with radius r of t and r of t can be computed explicitly it's just the square root of r0 squared minus 2n times t and therefore this exists on a time interval up to r0 squared over 2n so you have a finite time when this thing shrinks rotationally symmetric to a point and this is very different from what some of you may have seen in the harmonic map heat flow in the harmonic map heat flow you would solve a equation d dt of f equals also Laplace-Beltrami operator of f but you would do it with respect to a fixed metric whereas here our metric g is moving in time and that's why it accelerates for the harmonic map heat flow if you always use the initial metric on the say the standard sphere it would also shrink but it would shrink more and more slowly it would take infinite time until it the map reaches the central point so there's a big difference between harmonic map heat flow and inverse mean and mean curvature flow already in this example the second example is obvious from this one namely that you can take cylinders right you can take mn of t to be a sphere of dimension n minus k with a certain radius cross rk and here then r of t is equal to the initial radius squared square root of this minus 2n minus k times t and you get this picture here where you have an sn minus k and then here you have rk immediately from this compact example you get a non-compact example as well the third thing is of course that any minimal surface is a stationary point any minimal surface is a stationary point so the minimal surfaces are the fixed point of the flow and like an equator in a sphere would not move and this gives you the hope that maybe this equation generalizes the minimal surface equation just like the ordinary heat equation generalizes Laplace equation the fourth example that is important is suppose your surface is a graph suppose it's not a closed surface it's just a graph the case of graphs consider a function u on some domain in rn cross some time interval into r and set set mn of t to be the graph of u contained in rn plus one and suppose that u satisfies the following PDE d dt of u at x t is equal to the square root of one plus du squared times di du sum over i square root one plus du squared omega cross time interval now if you're in this situation that you have a graph then somewhere here is omega here you have this graph of u then turns out the so the motion here i'm describing is a motion in direction of the xn plus one direction now for such a graph uh let's take the lower unit normal nu and then this normal is equal to in terms of u you can easily compute well it has to be it has to be it has to point down to get a minus one here and it has to be normal to the tangent vector so you get the gradient of u in the first n variables and then here to make this length one you have to take one on the square root of one plus du square to normalize it so this is the lower normal to the surface and the mean curvature of course is the divergence of the unit normal and if you compare this the mean curvature is just what's coming up in this formula this is the mean curvature of a graph and then if you compare this to our characterization so you take fx of t to be um x comma u of x comma t in rn plus one and then you check here what is d dt of f then you see that d dt of f when you take this vector and you multiply it with the unit normal that you get exactly multiply so you so you get the vector zero um square root one plus du squared uh times h multiplied with the vector du and then here minus one over square root one plus du squared and you see this there's a cancellation this is exactly minus h so the normal component of d dt f is equal to the mean curvature so up to tangential diffeomorphism this the solution here up to the graph of u up to differential up to tangential diffeomorphisms is a solution of mean curvature flow solves mean curvature flow up to tangential diffeomorphism so that's an important equation and it shows you that in some sense since you can write any surface locally like a graph at least for a short time if it is smooth this complicated system of equations can at least locally be seen as a solution of a scalar partial differential equation which has this typical quasi linear type. I have the examples now of course we want to know what happens to the flow more generally we want to see how the shape of a given initial surface in this very general setting how does the shape of the surface evolve as time goes on now you may ask uh what about can you can you solve it at all so there's the question of short time existence and there's the question of what happens in the long run let us first concentrate for a moment on getting some more information on this equation before I talk about short time existence so let's see what we can infer from this basic evolution equation here this is the start this is the equation what more information can I get so in particular here here I only get told how the position changes let's compute from that how does the metric change how does the unit normal change how does the second fundamental form change then we have more information and this will help us also to solve the existence problem there any questions up to here concerning my notation okay so the next step is to compute evolution equations first one let me let me just state them first and then prove some of them let's follow this order here the structure here the d dt of the induced metric turns out to be minus two times the mean curvature times the second fundamental form then turns out d dt of the volume element turns out to be minus mean curvature times the volume element thirdly we can compute the d dt of the unit normal is just the tangential gradient of the mean curvature then fourthly we compute that d dt of the second fundamental form turns out to be the Hessian of the mean curvature minus the mean curvature times the square of the second fundamental form plus a term that comes from the ambient curvature so the Riemann tensor in the target manifold and the zero here means the normal direction to the hypersurface and this of course corresponds to an equation for the vine garden map so we can lift one index and since the inverse metric also depends on time this will lead to something like this and so right here so that you can sort of write in your notebook that you have all the important equations sort of in one view now turns out that one can change write equation four in a slightly different form namely you get the d dt of h ij can be written as in the Euclidean case at least as Laplacian of h ij i spare you the formula in a Riemannian manifold which is very complicated but in the Euclidean setting you get d dt h ij plus minus two h h i l h lj plus a squared times the second fundamental form and the a squared is the sum of the squares of the principal curvatures also five bar in so this is in rn plus one these last two formulas are not in the Riemannian manifold but in rn plus one and you get d dt of h upper ij is Laplacian for h upper ij plus a squared times h upper ij okay this this is a sort of the first the crucial set of this this is our going to be our toolbox because this this is these equations now tell us how all the interesting geometric quantities along the flow change right the metric the normal the second fundamental form the vine garden map once you know these formulas you can compute everything else right so let me sketch the proof of a few of them so let's for example do the first one d dt of g ij well you go into this equation you have to solve this equation now here's an important trick how to do these computations a very very important trick i said the equations are all invariant under diffeomorphisms so we are completely free at any given point in time where we do the computation to choose our coordinate system cleverly and you see there's all these christoffel symbols floating around so it's clever to choose at a point where you do the computation coordinate system normal coordinate system where these christoffel symbols vanish then you can ignore all the terms where these christoffel symbols appear except when you have to differentiate them the different derivatives of the christoffel symbols of course do not vanish but the christoffel symbols systems symbols themselves vanish so and also the first derivatives of the metric so when i do the computation here i'm allowed to ignore the first derivative of g bar if i'm sitting in normal coordinates okay so i choose a coordinate system where the first derivatives of g bar are zero and then i get here d dt of g ij is just in such a coordinate system it's just d dx i of d dt of f and d dt of f i insert my equation h times nu df dx j and the other term is um df dx i times d dx j of minus h times nu in a product with respect to g bar and now if i differentiate h i get a normal vector times the tangent vector is zero so i only have to differentiate the normal and here i get minus h times d f dx j d dx i the normal and now but this term here is just the second fundamental form and similarly here in fact it's the same term because the second fundamental form is symmetric so i get this equation and now i argue okay i only computed this in one coordinate system but what i got in the end is a tensor is a tensor so since it's true in one coordinate system this tensor equation must be true in any coordinate system that justifies this first equation okay you see here this is a coordinate derivative with respect to nu this is not quite uh the covariant but this the difference would just be a christoffel symbol which is zero so i saved myself in with this trick writing down all the christoffels this is well established how to how to do this um i just wanted to show you this trick once you have this um you can compute of course d dt of the determinant of the metric since d mu is a square root of the determinant of g times dx so this evolution equation implies immediately this to compute three this was one to compute three the trick you have to use is that the d dt of the normal times the normal because it has length one this is of course zero so you can write d dt of the normal um just as the it will only have tangential components right the normal component will be zero so you can write this as d dt of nu times the tangential component of the metric so you can write it like this because d dt of the normal times the normal is zero so you just have this guy and then using the fact that the normal is vertical to the tangent director you can flip this to the other side with a minus sign and then you bring in again the time derivative of f and that brings in h and that gives you this formula it's easier yes easy exercise but if i differentiate h if i differentiate h then i'm left with the normal vector times the tangent vector so right and so you use also that d dt of nu df dx i it's identically zero and then you get the result so that is uh that is three now to compute four you go into this formula here that's why i wrote it down all right so so to compute four compute d dt of h ij is equal to d dt of the inner product of minus d2f dx i dx j times nu and this term here when you when you multiply it with nu is zero so i can ignore this term but i cannot ignore this term so you get minus so so you get this term here plus gamma bar of df df times nu and when you do this computation here you again do the trick i told you about for example it looks horrible that you have to compute this but here you don't have to differentiate df because this is zero don't have to differentiate that don't have to differentiate that you just differentiate gamma bar and the gamma bar term leads to the curvature term so this is that leads to this term and the equation and in here you use the evolution equation for f and you use the evolution equation for nu and then somewhere in there when you do the computation you also have to use the other Weingarten equation which i didn't write so you have to use that d nu dx i can be written as at places where the Christoffel symbols vanish as h i l g l m sum of l and m df dx m plus terms yeah okay it's a term because you have to be careful here plus there's a term gamma bar alpha alpha alpha beta gamma d f beta dx i and there is a nu gamma something like this so if you use this then you get two terms from differentiating these Christoffel symbols they combine to give that and then you use once more this formula here i think it's a good exercise to see in any ways explained in the literature exercise to check that you end up with this formula and to get five you just note that the time derivative of the inverse of the second fundamental form the inverse of the metric is minus d dt of one over u is minus one over u squared d dt of u so it's minus the inverse of the metric times the inverse of the metric times the derivative of the metric this is a general formula for the inverse of a metric if you use that and you combine it with this evolution equation to raise the index then you can easily pass from here to there and the minus sign switches because of this term here all right so this is equal to plus two h h about ij and explains this equation now to to move from equation four to equation four bar um one uses an important identity of differential geometry it's a so-called simons identity which occurs in many other contexts so it's a crucial tool that's i think it's important i mentioned it here so for four bar we have to use simons identity simons identity is a commutator identity it tells you that you can compute the laplacian of the second fundamental form in other words the trace of the second derivatives can be computed from the second derivative of the trace all right the trace of the second derivatives is the same as the second derivatives of the trace except for some commutator terms and these commutator terms you have to compute using the gauss equations and the kodatsu equations and they look like this and if you just state in euclidean space this is it's much more complicated in a general ambient manifold in euclidean space it's that and then you just insert this here and you get and you get this equation and then uh raising the indices you also get five bar okay this uh completes the proof as an immediate corollary we get an evolution equation for the mean curvature which is important because mean curvature is the speed we want to know how the g speed changes and uh well we just take the trace of equation five why we just have to take the trace of equation five and we see that the evolution equation of the mean curvature is the laplacian of the mean curvature plus the mean curvature times the second fundamental form squared plus the trace of that riemann term but the trace of the riemann term is exactly the ritchie tensor you may say the ritchie tensor would be the trace over all indices but if i they're only in this i'm missing if i take the g trace is the one with zero but if you have two zeros next to each other because of the anti symmetries in the riemann tensor it doesn't contribute so you get here the ritchie tensor in direction of the normal of the free manifold of the manifold and this is uh not surprising let's stare now at our equations this first equation here is just what you know from the first variation formula right this is just the first variation formula of a minimal first variation of the area if you move the surface with speed f then the change of the volume this surface volume is f times mean curvature now our speed is mean curvature so you get mean curvature squared and if you have a minimal surface and you compute the second variation then you get exactly this operator and that's because now our speed is the mean curvature and we compute the change of the speed it's just like the second variation formula applied to h second variation operator applied to the speed h if you are on a what else can we learn let's see what we can learn in the compact case from from this equation sort of quick quick conclusions suppose our hypersurface mn is a closed hypersurface then a quick conclusion is of course that d dt of the total area of the surface is a minus h squared just from this equation here so it's always decreasing and in fact this is by the Helders inequality this is decreasing faster the decrease under n for for some other speed f the decrease would be like this and if the f has the same l2 norm than our speed then by Helders inequality this one would be larger so it's this fastest decrease of area that you can arrange with respect to the l2 norm so this why sometimes mean curvature flow is called the gradient flow with respect to the l2 norm the other thing you can see immediately so the first observation is this the second observation is by the maximum principle for scalar functions if you start with positive mean curvature it remains positive mean curvature so if mean curvature is greater equal to zero at t equals zero this remains the strong maximum principle says it has to be in fact strictly positive later unless it was already zero must have h greater zero for positive t unless h identically zero at t equals zero that's the maximum principle and carlo will tell you more about the maximum principle and how you can invariant properties of the flow so but that's something easy you can just read off from the equation okay one thing I'm not going to prove is the following I refer to PDE theory that you can solve this equation for short time if you have smooth initial data so detailed proof zero mn into nn plus one g bar is a closed smooth hypersurface immersion closed means no boundary compact in say c2 alpha there exists a unique solution of mean curvature flow on some short time interval it might be long but I only claim there's a solution on some short time yes no no no no this is if we this equation I wrote down if our speed is h if the speed is h and then of course the second variation on the minimal surface is also zero because it doesn't move the general one it would be would be applied to some f right if you have a minimal surface and you do the second variation in f you get d dt of the mean curvature is plus in f plus f times a squared yes if you flip the sign of your unit normal then h would be zero everywhere right if you you the sign of h itself is is depends on your choice of unit normal so when I say h is greater equal to zero this is equivalent to h lesser equal to zero but with respect to the other normal depends on your choice yeah so the the real question is does it have a sign yeah I'm only talking about orientable surfaces here yeah so so the difficult the more difficult surfaces are we will find out later the more most difficult surfaces are where h changes its sign inside the surface yeah but particularly nice class of surfaces is the surfaces that have positive mean curvature because then we have much more information they always flow in the same direction that makes them much more stable allows much fewer singularities as we will see so in later I will prove a lot of nice theorems for surfaces of positive mean curvature that I cannot prove for general surfaces yeah not necessarily no not necessarily no could be our n plus one in particular that's completely different story I said at the beginning the ambient manifold is remanian if the ambient manifold is a lorenzian manifold then it's a completely different story because then this sum of what I said would apply if you have a space like hypersurface but even then some of the signs will be different from the remanian case yeah it's completely different story so I have to concentrate on some things in this course yeah so guess I'm running out of time I should I think I should stop at 1130 is that right yeah so let me just state the just make make yeah maybe maybe take two comments on this proof the proof is using there's two ways to prove this either abstractly you go for the system and then you have these tangential diffeomorphisms so when you linearize you have a kernel and you have to find the integrability condition to deal with this kernel there is an abstract way of doing that very elegant but you know you use this huge amount of machinery more down to earth would be to write the surface as a graph over the initial surface and then you can reduce it to a scalar equation just like I showed you you can reduce it to a scalar equation in r and plus one then you have a scalar quasi linear PDE and that's what you can solve by linearization using the implicit function so it's not that hard to get this theorem and it uses standard PDE techniques so that's why I don't give this proof what I want to prove next time and that's related to regularity estimates is I want to show that you can extend this short time solution until the curvature blows up so the next time I prove the following theorem and I close here with this for each smooth initial data the solution exists on a maximal time interval 0 t max where this maximal time interval is certainly non-empty it may go to all the way to infinity but if it does not extend to infinity if the we cannot extend the solution further then soup of the second fundamental form on the surface m and t becomes unbounded as t approaches t max in other words we can detect when there's a problem by just looking at the curvature that's what I'm going to prove next time together with regularity estimates for the flow thank you