 Oče, zato sem izgleda, da je vsak zelo izgleda, tako če malo kursi, spaski, 3 zelo, kaj pošliš, kaj se vse prejšel, hodnje, vzelo in vse bizan, taj je odpusten, kaj je zelo odpusten, taj naprav je odpusten vzelo v 3 zelo za vsi, težko vse zelo se kaj je 3 zelo, in we have to imagine that they move around in space, as time goes on. So, the way they move, they should remember the geometry of the underlying curve, which was giving birth to them. So, now the question is, somehow have we finished the study, I mean we have introduced two functions and three vectors, tangent normal and binormal, and two functions, curvature and torsion. Is that anything else we should discover about the geometry of curves? Well, the way to say no, actually, that's it, that's the end of the story, is to prove that if you give me two functions, k that I would like to call k and tau, there is a unique curve having k and tau as curvature and torsion. If I prove this, that means these two functions, it's all there is. There is nothing else to be discovered. This is the topic of this lecture. How do we prove it? Remember, we always start with a differentiable curve and we always assume it's parametrized by arc length because if it's not, we can always do it. This is not an assumption. The only assumption really we make is this is a curve which has positive curvature everywhere at every time, so it's a regular curve. And now there is a little trick because we have learned how to associate to each time, and actually time, since it's the arc length, I will call it s. So to each parameter, I know how to associate three vectors, this famous orthonormal basis, the Frene 3a. And instead of look, so I have three vectors in R3. So just as a convention, I look at as a map from the same interval, so to each time s, I associate a vector, a column vector, t of s, n of s, b of s. So I put the components of the three vectors in column, so we get 3 plus 3 plus 3, so this is a point in R9, 3, 3, 3. And a nice way to write down the Frene formula in this matrix form is to say that the Frene formula say like t prime, n prime, b prime as a vector is equal to what? 2 k n minus k t minus tau b and tau n. So this is component by component, I just put the three equations in vector form, but I want to write it as a matrix representation in this form. So let me write it and then I comment the notation. So this big matrix 9 by 9 times the column vector t, n, b. So what does it mean, this notation? This notation means what I call 0, 3, or o, 3 if you want, it doesn't matter. This is the 3 by 3 matrix whose all components are 0. So 0, 0, 0, 0, 0, 0, 0, 0, 0. So this is a block representation. So if I want to write a 9 by 9 matrix, I can tell you 3 by 3, 3 by 3, 3 by 3, 3 by 3, and so on. And what is ID3? Well, this is just the identity matrix of rank 3. So 1, 0, 0, 0, 1, 0, 0, 0, 1. Now, in this form, this is just a notation. There is nothing new. You see, if I do the row by column multiplication, you see that here I get 0 times t. So 0 plus kn plus 0, so I get kn, which is exactly what appears here, and so on. So it's just a convenient way. We are not learning anything. But it will become very useful. So now let me state what is called the fundamental theorem of the local theory of curves. Long name. So this is the theorem, which answers the problem I was mentioning a few minutes ago. So give me two functions. So given two functions, k0 and tau0, from some interval to r, is a real value functions. The only requirement that these are smooth, because in general they will have to be smooth, and the only requirement is that k, what at the end should be the curvature, has to be a positive function. So given such two functions, there exists a curve from the same interval into space, r3, parameterized with arc length, such that k of s is equal k0 of s. So its curvature is equal to the first given function, and its torsion, and tau of s is equal tau0 of s. So this answers the first part of the problem. So give me any two functions, such that the first one is positive, such that these two functions are curvature and torsion. But I want to know more how many curves are there with this property. Moreover, so that's the second answer that this theorem is giving us, alpha is unique, and as we commented last time, unique, what is the best uniqueness theorem you can hope, up to rigid motions of r3, up to direct isometries. Remember, we just commented quickly over r3. Isometries are all two types. Basically, they are translations plus an orthogonal transformation. The orthogonal transformation could have determinant 1 or minus 1. If it's positive, so plus 1, we call it direct. So, well, that's the best you can hope. This is telling us that the problem is solvable and it's solvable in the unique way, which is meaningful. OK, let's prove it. Unfortunately, the blackboard is quite small, so this notation is useful, but let me start here, and then we have to erase proof. Let's look at the following ordinary differential equation. Look at the following. So, x prime of s is equal a0 of sx of s. Let me call this equation star, where a0 of s is formally the matrix that I constructed here. You see, this 9 by 9 matrix, but where I use, of course, my starting data are these two functions k0 and tau0. I still don't have a k and tau. OK, so formally I just define it to be 0,3 k0 of s. I write of s only here, then I drop it, because these are functions, 3 or 3 minus k0 identity or 3 minus tau0, or 3 or 3 tau0 identity or 3. OK, so I use these two functions to construct this matrix, and now what do I do? So, of course, x will be a vector in R9, because this is an equation in R9. OK, choose. Now we have to make some choice, and then at the end we will comment the effect of this choice on the final output, but choose some fixed vector a in R9. You see, if you want to solve an ordinary differential equation, you have to restore the equation, so how we have to give an initial data, and this will be the thing. And, sorry, choose a vector in R9, but of course we are interested, so such that we have to split it in the first three components, the second three components, and the third three components. These are the geometrically interesting bits, so such that if I call t0, as a vector, the first three components, a1, a2, a3, and not a4, a5, a6, and b0, b equal a7, a8, a9, OK, sorry, of course, you're right, OK. So the requirement I make on the initial data, a, is that if I take the first, I split into three vectors, t and b, formally, they are still not a frenetriadron of anything, but t and b, I would like them to form a positively oriented, so such that these three vectors form an oriented orthonormal basis of R3. That's a very natural thing to ask because at the end I would like this to be true at every time. So this is kind, I want to start from a configuration which looks like an orthonormal basis. Then I want to evolve following this differential equation, and I want to prove that it stays an orthonormal basis. OK. So now, formally, how do we do, how do we do, set f to be a function from i to R9 solution of star? OK. Solution exists because this is a linear ordinary differential equation. OK. But then if we take a solution in R9, again we split it in the first three components, second three components and third three components. OK. We define t, sorry, t, which now is a function, a vector valued function. OK. To be f1, f2, f3, n will be f4, f5, f6, and b will be f7, f9. OK. And now these are functions of s. Sorry, of course I didn't write it, but it goes without saying. So set f to be a solution of star with initial data a. OK. The unique solution with initial data a. OK. So what do we want to prove? We want to prove, first, that actually these three vectors are presented orthonormal basis for any s. OK. We know it for s equal to zero. We know it at the beginning. And we want to know that this property is preserved. So we want to prove that if I put now t and b as three distinct vectors, this is and orthonormal positive orthonormal basis of r3 for any s. So this is the point. How do we do it? Simple trick. Look at this matrix here. M of s. How do I know if something forms an orthonormal basis? I can consider the matrix of all possible scalar products. So I take the 3 by 3 matrix given by tn, tb and so on. tn, it will be symmetric, of course, nn, nb and then tb, nb, bb. So let's look at this matrix here. Another way to say to state our claim is that this matrix is constantly equal to the identity. So the point is that this matrix automatically satisfies a nice differential equation because x was satisfying star. It's a simple computation to prove the following. If I take twice the derivative I get exactly this. A of s M of s minus M of s A of s where now this is a 3 by 3 matrix. This is a 3 by 3, 3 by 3 and A will be a 3 by 3 matrix. A of s is the matrix constructed. Formally, you see, this form here but instead of imagining this of b 3 by 3 blocks I just put them a number. So 0, 1, 0, so this is just a matrix. 0, k0, 0, minus k0, 0, k0, tau 0, 0. Of course, of s. The function. k Now this is very simple. I'm not cheating you. In three lines you prove that star implies this equation here. And the fact that we started from an initial data with the property that if I split it I get an orthonormal basis I also know that M of 0 you see, at time 0 matrix is the identity because of the property of t0 and not b0. But now again this equation here has a unique solution given an initial datum. So the only thing I need to observe is that M of s constantly equal to the identity is actually a solution. Put M equal to the identity of course the derivative of the identity is 0 because it's all made of constants. So you get 0 is equal a minus a, 0. So it's a solution. But the system has a unique solution given the initial datum and this satisfies the initial datum of course. So this must be the solution. But that's exactly the claim. That's exactly the claim. This matrix here is constantly the identity but that means that this is constantly equal to 1 this is constantly equal to 0 this is constantly equal to 0 and so on. So exactly the claim. This is an orthonormal basis. There is actually a little comment to be made here. It was a positive basis at time 0. It stays orthonormal for every time and that should be a positive orthonormal for every time. Please speak much louder because of properties of the determinant. What does it mean positive? It means that if I write the matrix transformation which takes one basis to the other for two different times positive means it has determinant 1. But now how is it possible? Suppose that for some s this has become negative. This is positive or negative because it's a basis. But by continuity of the determinant it means in between there has to be a time where it was 0. The transformation between the two basis had to have determinant 0. But this is impossible because it has to be an isomorphin for every time. So if it starts positive it stays positive for all time. So also this property is easily checked. Ok. Up to now what have we done? Given two functions k0 and tau0 we have constructed the only and now we will comment why it's the only frene basis t and b which a curve, a possible curve could have if it had curvature k0 but So now the problem is where is the curve? So now we have t and b how do I construct alpha? But that's not the problem. Now define alpha alpha from the same interval to r3 to be what? Alpha over s if I know the tangent vector it's enough to know t because if I know t alpha of s will be just the integral between s0 and s of t of u in the u. What is t? T is the first derivative vector. So if I integrate I get alpha. There is only one alpha who could have t as a tangent vector. So the first part of the theorem is proved. Because now we got alpha. What is the tree addron associated to alpha? By construction t so what is the tangent vector to this alpha is precisely this t by construction. What is the normal vector to this alpha? Well it's the derivative of t times k but this is exactly the derivative of t has been constructed in a way that it is exactly n k normal of alpha again I take the derivative of the normal and so on. So it's all constructed in the way that the tree addron of this is exactly t and b and the curvature and the torsion of alpha are exactly k0 and tau0. There is nothing to be proved. So one comment about uniqueness because we said this curve now we have a curve with the desired properties in which sense it is unique. Well I had one degree of freedom here because nobody has told me where to start to integrate passing from the tangent vector to the curve itself. What is the effect of changing this base point? It's just a translation in our tree. It means I'm just starting the curve from another point translation. And then what else? You see every time you see the word choose that means you had the freedom. So let's go back to see if I had picked another vector a of course the properties I wanted so another vector if you want a prime whose first three components second three components third components formed basis. And then repeat the argument. Of course f would be different because it's a different initial data. But then m would be m is always the identity so f will be different and the three vectors t and b will be different because are the three components again splitting of this function. But how do this change? Well, if a and a prime they have this property they are related by an orthogonal transformation of r3 and then pick this orthogonal positive with determinant 1 pick this orthogonal transformation of r3 and that means that t if you want you have t and b and t prime and prime v prime everything is related by this transformation. The final output is related by the same transformation because t and t prime are related by an orthogonal transformation so also alpha and alpha prime would be related by an orthogonal transformation. So there is really nothing to write except identifying the places where you are making choices and see that the output is translation plus orthogonal transformation of r3 of positive determinant. Ok, so that sends the proof of the theorem. Well as of course the name of the theorem says that's really the end of the story for curves in r3. It means we have discovered everything that was to be discovered so a curve is completely determined by curvature and torsion and so that's it. So besides making some nice exercises there is this is really the end of the first part of the course. This kind of differential geometry in dimension one it's over. But I would like to use this last part of this lecture to tell you about a classical problem in the theory of curves and how it's inspired so much research that I think you should be aware of it. So the problem I would like to describe you in the solution is about planar curve. Simple closed planar curves. What does that mean? Well clearly it means we have a function our curve now goes in r2 automatically because it's planar. What does it mean it's closed? Well suppose we give names to this interval so suppose that this interval is given by some a, b I'm slightly stretching our definition because usually the interval is open that doesn't really matter now suppose it's defined on a closed set and differentiable means the obvious extension to the probability so it means you have a limit of the derivative going to the left extreme coming from the right and the limit of the derivative coming to the right extreme from the left. So closed means that alpha of a is equal to alpha of b so it's really a loop. Simple means no self-intersection if you want alpha is relative. And when I say differentiability I also mean that the tangent vector coincides also when you take the limit on the two extremes. So really we are talking about the simplest thing. So it doesn't really matter where I put alpha of a is equal to alpha of b. So you start from here you go around, you go back here at the other extreme and the two tangent vector coming from this way and this way it must coincide. It has to be smooth also in this direction. So there is a beautiful problem coming at least from the third century before Christ so this is the first time at least we have a written record and actually you have to apologize your lecturer now because I'm full of western civilization and of course we tend to think that Greeks invented everything and all our mathematical culture come from Greece from ancient Greece. It's very likely that in India or in China or somewhere in Africa similar problems were studied couple of thousands years before but at least that's the way we came to know it. What the problem I would like to describe you is called the isoperimetric problem. So for what we know it appeared in a mathematical treaty about that time so what does this problem say? Question. Among among all planner simple closed curves of given length of length l suppose you give me a positive number so among all planner curves of that length which one bounds the region of largest area May I ask you how many of you have seen this problem? Please raise your hand. One one Good enough to repeat it. You will be a bit bored. You understand the problem. So for example suppose this curve as length, I don't know one meter more or less so here is another one with probably the same length and so on. So actually this question is hiding a very delicate problem. So a simple closed curve in R2 separates the plane in two parts. One which is bounded and another one which is unbounded. This is very intuitive but it's actually not easy to prove. It's called the Jordan separation theorem. It was one of the early success of modern topology. But of course everybody assumed it was true. You see, things can get very, very complicated because this is a simple closed curve and so on. Now, if I tell you which is the interior of this curve, is this point interior or exterior? So of course it's a matter of enlarging. Of course, there are topological techniques which are actually quite easy. A little bit what is called degree theory gives you a simple answer to this problem. But nevertheless it's true. So let's assume I'm not going to make any topological consideration up to now. So this question is about of course the area of the bounded, so every curve splits the plane and of course you can measure the area only of the bounded part because the other one will have infinite area, whatever the curve is. Actually the first Greek contribution, the place where this problem was born, the only contribution it was made to this question was to prove that if you give me a number, a positive number, the area enclosed by the regular polygon of that perimeter grows as the number of edges grows. See what I mean? So the theorem that was proved here, for example if you take an equilateral triangle of a given perimeter the area inside is less than the square of the area inside the square of the same perimeter and the area of the square is less of the area of the pentagon and the hexagon and so on. Actually this is a guessing the solution but the solution escaped the author of the problem. The solution, not much mystery is this one. So if you give me a length of course you can construct the circle of that perimeter and the area inside the circle is maximal among all simple close curves with that length as perimeter. Actually this has been popularized at the beginning of, I mean around in mathematical language it was really formulated precisely about I think 200 after Christ by a mathematician called Papus in a very intriguing treaty about the life of bees so why somebody studying bees should come up with this problem that's a curiosity of course. If you look at what is called the hive then of course you would like to give us an explanation why bees live in hexagon and inside they form circular tunnels and somehow this is related exactly to this question they want to maximize the space where to live of course building their home costs energy because it costs wax so you want to make the biggest space out of the least possible wax so in some sense bees know the answer to this solution ok and the other historical comment before putting our hands in the proof is that this was actually became a very famous problem in mathematics because it's part of a famous history by Virgilius so this is about 50 after Christ where he's telling us story of the queen Dido in fact it's always it's often referred to Dido's problem so Dido was a Phoenician queen which was for some reason had to live Middle East and with some people escaped to Cartago which is in modern Tunisia ok when they get to Cartago they asked the king of Cartago in some land where to build a new town of course the local people didn't want to give a land ok but there was they were obliged in ancient times of course hospitality was supposed to be a holy thing so instead of saying no the king said ok you can take all the land that you can put inside a cow of course this seems like a trick ok because you kill a cow you take the skin and it's probably more or less like this ok so what kind of town can you build but of course Dido knew mathematics so what has she done she has taken a cow she killed, she took the skin and she made a little wire with the skin and then with this wire she embraced a huge piece of land and to maximize the land she actually put the wire into a circular configuration so she knew mathematics and she guessed the solution and of course she was able to build a pretty big town by solving this problem and cheating the king who was actually trying to cheat her very noble story ok that's why you will always see it referred in books as Dido's problem but now end of story let's go to mathematics actually the first complete mathematical solution to this problem came by wire stress was given by wire stress in 1870 around 1870 well there was probably another proof but it's not clear whether it was really complete 30 years before so wire stress wrote the complete proof the one I'm going to tell you is given by Schmid about 1914 and since then we now have many solutions many proofs of this of this fact the reason I'm telling you first because I think it's good you know a little bit of history actually if you like I mean this and other problems of this type are beautifully explained in a book that I strongly recommend you is called the parsimonious universe by Anthony Tromba by Tromba and I think also Ildebrand probably there are two authors but I remember this is a set of mathematical problems all coming from the same philosophical point it's that nature minimize effort in doing whatever it wants to do and this is one place where we can see it now so the proof I'm going to give you heavily relies on Green's theorem so let me remind you the version of Green's theorem of Green's formula actually it's a classical theorem in calculus it's a corollary of stocks theorem for what it comes so suppose you take two functions of two real variables f and g at least of class c1 over rt and suppose r is the interior region or the bounded region given boundary sorry bounded region inside a curve c in r2 ok now by a curve I really mean the image not just a parameterized curve so Green's formula tells me this so then if I take the integral over r of the function dg dx minus df minus df dy in dx dy this is equal to the integral over the curve of f dx dt plus g dy dt in dt this is a linear integral so it depends on one parameter t ok so this is a general theorem actually the way I'm going to use it f is equal to minus y and g is equal to x if I substitute there what do I get well if f is equal to minus y of course here I get minus 1 with a minus it's a plus and dg dx is another plus ok so here I get 2 the integral of 2 but that's exactly twice the area of r twice the area of r is equal exactly to well if c is our simple closed curve parameterized by alpha from ab so this is the integral between a and b and then f well let me first write g which comes with a plus so g is x d dy dt then plus f so minus y dx dt in dt ok so this is actually the corollary of the Green's theorem that I'm going to use ok and so now I can state the theorem which goes under the name of isoperimetric inequality because actually so under all the notations that we are using I'm not going to repeat it l squared is greater than or equal 4 pi area of r and moreover equality l squared equal to 4 pi area of r if and only if c is a circle or if you want alpha parameterizes so this is the complete solution to the ancient problem in mathematical language give you the proof I need to make a little picture so so the picture I want to something like this so suppose I have my simple closed curve ok this is my c in some I take I decide to bound it by two vertical lines so the first one tangent will be this one the second tangent will be something like this and then of course my picture won't fit but I mean now here so this gives me a possible a candidate diameter for a circle so below here I try to draw somehow the corresponding circle with this diameter ok but more or less ok so here there will be more or less the center of this and this I call x and this I call y so this is my curve alpha of s suppose I am going in this direction this is alpha of s and suppose I choose time and suppose this is alpha of zero after all changing names of the point doesn't matter and suppose that this is alpha of another point and now I am not free to choose another name I cannot say this is one so suppose this is alpha of s1 so these are the two points where there is a vertical tangent which is bounding the curve itself ok so this is the region inside r and this is our curve ok so this is kind of a background background picture to understand what's going on now so in my picture these are called l and l prime so referring to this picture I call this the center the center of this it's clear how it's constructed so these are some lines which are the first one on both sides which intersect the curve c so this gives me some kind of an amplitude and below my curve I draw a circle with this diameter I am not going to write in language the explanation so if alpha if c is parameterized by arc length as alpha of s so this is a planar problem so it has two x of s y of s it has two components again I choose the interval in such a way that these points corresponds to alpha of zero and these points correspond to an unknown number s1 ok if such under this condition we can choose a parameterization of the circle we can choose a parameterization of the circle below in fact let me call it c prime ok of c prime as again beta of s so it will be another curve beta but I can fix this first coordinate to be x of s but now of course the second component will be another function that I call y bar ok y bar is just another name for another function ok if you want to call it another name it's ok now remember that s was arc length so actually the interval where the curve c is defined is zero because I'm assuming I'm starting from zero there to L L is my given perimeter in the isoperimetric problem ok arc length measures exactly the length so the final point will be parametrized at time L now let's try to give an estimate about the area the area of r just one second ok remember the formula I just erased let's comment one second on this formula remember we have just observed actually first I put the two here so this is one half so the formula we wrote before was the integral between the two extremes now the extremes are zero and L ok of what of x y prime minus x prime y of t of s in ds ok so this was the formula we took from green's theorem ok but now the point is that these two reintegrals are the same because the curve is closed you see now the only little observation is this one if you take the integral between zero and L of x prime y for example in ds by integration by paths how much is it you see I can imagine this to be x y no between zero and L minus y x y prime x y prime ok but this is zero because it's closed it's a closed curve so if I evaluate the functions x and y at L I have to get the same thing as zero but then if these two things are the same you see these are the two things which come here I can erase one of them and erase the two ok so let me keep it to the positive one so this is zero L x y prime in ds so if I also give if this radius is r how much is the area inside the circle below there but that's a circle of radius r ok but by the same reason it is what so this is by geometric consideration remember in fact we still don't have any estimate on r so this is just an almost useless at the moment consideration but I can use the same trick I was using up here for the curve beta ok so this is equal to minus the integral between zero and L of x prime y bar you see now I choose the other one the same property but I choose the one with the minus ok ds so what do I get out of this the area of r plus pi r squared plus this but this is equal to the integral between zero and L of the sum of these two integrals so this is x y prime minus y bar x prime in ds but now this is less than or equal to the integral between zero and L of the square root of the square is a simple fact x y prime minus y bar x prime squared and then other manipulation this is less than or equal to the integral between zero and L x squared plus y bar squared times x prime squared plus y prime squared in ds ok, I leave you to check this but these are arithmetic nonsense I'm not using anything now now I use something this is the fact that alpha was parameterized by arc length if it's parameterized by arc length that means that it's tangent vector as norm one but what is the norm of the tangent vector to alpha is the square root if you want of x prime squared plus y prime squared if it's that's equal to one also the square of it is equal to one which is exactly this object here is exactly equal to the integral between zero and L of the square root of x squared plus y bar squared in ds but what are the functions x and y bar parameterizing note note that I'm using the fact that s is the arc length for alpha I have no idea if s is the arc length for beta in general it won't but here I'm using its arc length for alpha but now I'm left with this but xy bar is a parameterization of a circle of radius r so x squared plus y bar squared is r squared of course under square root it has to be positive so it's the integral of the constant r between zero and L so this is equal to L r so we are done we are done because well at least we are done improving improving the inequality because now remember I write it down here so the simple of course arithmetic that I'm using is this I actually a little hint to prove this I already used that property here now I use it again it's Cauchy inequality as you want you are free to choose your favorite proof now let me use again this property this simple property here and now I tell you the square root of the area of r times the square root of pi r squared so that tells you what is a and what is b this is less than or equal to 1 half times area of r plus pi r squared by that nonsense but what we have just proved we have just proved that this is less than or equal to L r divided by 2 so this is less than or equal L r over 2 but that's exactly the statement of the inequality because now divide by this and square it and you get exactly the isoperimetric inequality now I leave you with an exercise because the theorem was made of two parts you had the inequality which is here but you had also the characterization of the equality so if in this equality I get equal I have to prove alpha itself was a circle how is it possible that here I get a quality what it means that here I had an equality it means here and here I had a quality so exercise for you if these are a qualities x and y parametra is a circle ok completely trivial I remember when I was a student it took me a bit of time to prove this that's why I keep on giving it as a nice exercise because a little bit of sadism ok now now we are really done but let me tell you so this is all very classical and you can see we have not used very sophisticated mathematics 19th century mathematics and yet this kind of question is still an active area of research once you manipulate the question in a clever way there are two possible clever manipulation of this type of problem first of all well of course Greeks and I think I can be sure also Indians and Chinese and Africans or whatever nobody has thought about geometry on something different than a plane before 200 years ago ok it has been since the birth of the topic that we are going to study starting next lecture I mean differential geometry that people has realized that the same question we have been studying in Euclidean geometry should have been studied on a different ambient space now what does it mean in ambient space instead of taking a simple close curve in the plane what does it happens to this problem if your universe is for example a sphere so draw a simple close curve on a sphere well again this encloses well now you have first you have a problem because you have two regions of bounded area so which one you choose the other one should be the exterior but still is there an isoperimetric inequality so among all curves with a given length on a sphere is there one which maximizes the area inside and if so which one is it and then why the sphere why not asking the same question on any surface that you can think of why you should rule out your universe to be something like this then start drawing simple close curves you can certainly measure length you can measure areas and you can ask the same question this is much more difficult and actually we don't know in general the solution okay in some in the two cases I drew we know because these are surfaces of revolution okay we have this kind of rotational symmetry which makes the problem easier but in general we don't know and again but then there is a more sophisticated upgrade of the problem so what we are really asking is the following think of the question in another form we are fixing a one dimensional I mean we are fixing a length okay and we are trying to maximize the area inside a curve of that length well length is a one dimensional measure curves are one dimensional objects surfaces so the inside it's a two dimensional thing well we should be ready to ask the same question in any dimension meaning what fix a number but then prescribe it to be the volume of an n minus one dimensional object can you feel this n minus one dimensional object in a maximal way with an n dimensional object you see in how many ways you can ask the same question the question is always the same it's just an upgrade up to modern mathematics okay but in all these forms there are open problems and these are kind of fundamental problems because again you can also mix the two problems I was telling you together you can ask the higher dimensional question in a space which is not Euclidean okay this is a fascinating part of active research in basic differential geometry and analysis in some sense okay I mean this is ground zero okay this is the first issue of a series of problems which are open up to now and I wanted you to have a little feeling about this okay so I think for today that's it and this ends our study of curves okay so on Thursday we will start with surfaces