 Trami reminds me da smo počevali, da je tudi osrednji zelo tudi, in da postavimo, da smo se počevali. Tukaj včetno je, da se je očetno, da je to počevala, da počeš, da glAsječ v neko, pri tudi skupaj, početno je, pri vsej formu, očetno, kapital E, kapital F, kapital G in tudi da je tega. Tako, to je zelo vzelo tega, da je tudi, da je to, da gaustiči je niferil za izometri. To je odvršenje, da je čello, da je vse z bil dobro. Ko smo je napravili? Veseljamo, da smo v medelju prvno, ne v del, je pa več dana, pričeljamo v standardu, v standardu je to lokalne teorema, pojeljamo v lokalno čarte. Pojeljamo v standardu z kanšnjem, zgromamo vse vse. Kaj so, da je, da proč, tudi jaz evo se zdajite, z njega, da idem na čirku, pushnače nečatavno. Po infekciju se ustave. Rozumel, nr. Few, pri različen loan, razliz, večnega za pričen ljuda, s nekaj da je, večnega, večnega za pričen lenjeva do pljivna. In, nekaj je. In je je. Ko je tudi pričen ljuda, vem na preči. Zdaj musi be zelo počaje, je kako je vlastne izKN. Potom lahko vam poslite dve vrste pravi. Zobin za budem, da te vrste vrste vrste v normunji, kjer nje če vzvekj jaznj, bi ne ne ne ne ne ma odvala, ne ne ni kak стupi nekaj poten. Stupi je, že ni zelo, kak to zelo izegen z nič nič nagor, ljubek, ki to nagrožijo, tko ampak tjel je. Jel excited term, In drugi konec, čas to pravijo, je, da prijevajte, skolor nebo vse zelo, je to državno revizivno. Vse prijevajte, da je zelo vse, kako se prijevajte, in zača v vsimi štov, zača za vsimi, zača za minus. Tako što imamo je tvoje, tvoje, tvoje, izpronovate se. Tvoje vse početnje veče, da je prijevajte, prijevajte, pa je se, da je vse, način, zato vse, da je vse, tako prijevajte, Ne bo. Štete, ko sem tudi odko ci nekaj tudi pogledačila. Zato pa pizdelysne je tudi, neimi ta zeločenj viče v modlji delivati. Je tudi, kaj je neko, zeločenoji delizir usponira Poprav v zeločenji. Vespešno, tudi. Vespešno tudi zeločenje. Second why this implies this, as in then we will work on this, so now the key thing is of course what is this. But now let's freeze for a moment on this equation here, which is of course crucial because then I will argue that this expression here has the property that the Orei ki se prejmo, že je tudi mu podbavila, ešte za drejmu o vso opravljenje, tudi je to prejmo v teravljenju, za to, da bomo poživali, kako so nekaj se prijevnaš ovega tudi za pripravljenju, tako in da je K, a kako je pripravljenje. Kaj je n-u kroz n-v skalar produkt n? N, kaj je n-u, kroz n-v skalar produkt x-u kroz x-v, kaj je n-v kroz x-v. Vzivajte, da je to ovo. A n-v skalar produkt z 2-vektorim produktem, zelo, da je n-v skalar produkt z 2-vektorim produktem. Prvno, prvno, prvno, prvno, prvno, prvno, minus prvno, prvno, prvno, prvno, prvno. Lači to merce v temesih– ki je prgestak n-uv? N-v kroz z n-v. Kagop万rve skalar produkt z zelo, N-v kroz 0, میno commodities kroz zhen. kroz x v je nekaj, da je eg minus f squared 1.5. Zelo vse, da sem vsečila, da je 1.5. Zelo, vsečila vsečila vsečila. In tudi. Zelo vsečila vsečila. Kaj je tudi numerator? Vse samestvorimo, da so urbančnih šehljihing, ker je v tem, da se vsih odsnečenih šehljihti je zelo, da so da se zelo, da so da se zelo, da chci se odre hypnotizanja, da je se na понjo m區. Tako, to je tist dobalaj, to je otev. Otev, zelo možno. To je tist dobalaj, tist je tist zelo, tako, da nas zelo vse nezaj. Tist je tist dobalaj. by EG minus F squared, one-half. OK. So this is why this equation is true. And now why this equation implies star? Well, let's, well, in order to prove that it implies star, we have to substitute in this expression all these numbers, all these functions, alpha, lambda, and me, OK. So now we take this explicit expression, we plug them in, and we see what happens. So how much is N u cross N v scalar product N? Well, now in this language N, with respect to these vectors, N is very simple because it's just E1 cross E2. OK. Those are on orthonormal basis of the tangent space, so its normal vector is just the cross product. So this is N u cross N v scalar product E1 cross E2. And then we use the same algebra. And what does it happen? This is N u E1 times N v E2 minus N u E2 N v E1. And then we play the usual game. Of course, we don't know anything about N u, and then v, looking at here. But we put, we throw the derivative on the other side, because what do we know? We know that N scalar E1 is 0. N is normal, E1 is tangent. So if I take the derivative with respect to u, I get the usual equation. N u scalar E1 plus N scalar E1 u is equal to 0. So this becomes minus. For example, the first function here is minus N E1 u. Every time I switch the derivative from the first factor to the second, I get a minus. So being quadratic, nothing changes. So this becomes N E1 u scalar N E2 v. So the same derivative goes on the other side. Minus N E2 u scalar N E1 v. And now we know. Now in this form, we can take the defining equations for lambda, mi, and alpha, and plug them there. For example, how much is E1 u scalar N? E1 u scalar N is lambda 1. So now it's done, because this is lambda 1. And how much is N E2 v? E2 v is here mi2 minus lambda 2 mi1. So you see that we prove this equation by showing that actually this object here is equal to this one. Then we knew that this one is equal to this. And it's done. Very well. So nobody is really telling us why this is a good step in our direction, but now let's focus on this expression. I want to argue that this is, in fact, so basically the claim, which ends the theorem, is that alpha 1 and alpha 2 can be expressed in terms of E, F, and G, and their derivatives. Why? Well, let's prove the claim. Well, what is E1? Now let's go back to the Gram-Schmidt process. We are orthonormalizing these two vectors. Now I want to be completely explicit in terms of the first fundamental form. So of course E1. So let me write it in this form. It's a multiple of xu, so this, but which multiple? Well, of course, it's 1 over the norm. The first one, you just normalize it. So A is nothing but E to the power 1 minus 1,5. And how do you get the second vector? Well, in the Gram-Schmidt process, what do you do? You take the second vector, you remove the projection on the first, and then you normalize again, because you want something of norm 1. But that means that E2, I want to write it as kind of, is of course, it's a linear combination, meaning at every point, so there are functions for which this is true. And now the point is what are B and C. Well, if you go through the procedure I just said, you can write, I propose you this notation just to, B is E to the minus 1,5, F, let me call it delta, and C is equal to E to the 1,5, delta to the minus 1,5, where delta is our usual function. Let me write it here EG minus F squared. This is strict. I mean, just write down what are these vectors. But then what is alpha 1? Well, alpha 1, I can express it as the scalar product between E1u and E2. So E2 is of norm 1. So alpha 1 is exactly that. I don't have to divide by the norm of the second vector. So this is E1u, E2. But now let's plug in these expressions here. You see? Now let's work with this equation. I should take the derivative of E1 with respect to u. So, and then I want to express everything in terms of A and the coefficients of the first fundamental form. So what do I get? Well, if I take the derivative with respect to u, let me first do the derivative of x. So this becomes A times xu u E2 plus, now I put the derivative on the first. So this becomes the derivative of A, dA, du, xu E2. And then I use, again, the second equation. Now I can express E2 with respect to xu and xv. So how much is this? This is A, so imagine you are putting this expression here. So this is A times what? Times B xu u xu plus C xu u xv. And what then? Plus something. But is there something? E2 is orthogonal to E1. And E1 is parallel to xu. So these two objects are orthogonal. So this is 0. So there is nothing else. So this is equal to 0. So let's go on with this. Now the point is we never saw an expression like xu u scalar xu. The point is, can we express this in terms of the first fundamental form? Well, so freeze it for a second. And if you want, open a little bracket here. Of course, E is equal to xu. So that looks similar to that, except of course there is one derivative less. So how much is this? It's exactly what I want twice. So what I want is a half of that. So for example, this term becomes AB divided by 2 times the derivative of E with respect to u. That's the first part. And what can I say about the second? Well, the second is slightly more complicated, but it's as elementary. Well, my bracket is not enough. So how much is xu u xv? This time you have to make two steps, because you want to say, for example, that looks like a derivative of xu xv. This is F. But now if I take the derivative with respect to u of this, I get one term, which is the one I like. But then I get an error, because this is what I want. So on one hand, this is dF du. How do I want to write it? So my object, let me go on. What is this? I imagine that here I wrote xu u scalar xv is equal to this minus xu xu v. But now this is OK. This I can express in terms of a derivative with respect to v this time. So what is this? No, now only working on this little piece. Well, I express this as a derivative with respect to v of xu xu. Is it really the same? Well, more or less, because this is twice again. So this is equal to one half of this. So what I was getting here is equal to what? So the derivative of, so in fact let me write it now, the end of the equation. So this is plus ac times what? What did we get here? Is dF du minus one half dE dv. Give me a sign of enthusiasm. Because you see, now the moral that I was trying to convey, it's in the game. This is clearly an horrible expression. Alpha one, by no means, is a simple function of the first fundamental form. And in fact, it's not even alpha one which comes here. It's the derivative of alpha one, even worse. So you can even imagine. So if you want to write down the explicit expression of what is d alpha one in dv, it becomes a mess. Because remember that these are functions. So d alpha one in dv would be the derivative of this times everything plus the derivative of this times everything plus the second derivative of E. So it's a disaster. Nobody writes it down. I don't know anybody who wrote this equation. Because then what is E? But the point is that E is a function of the first fundamental form. And B is a function of the first fundamental form. And so on. So you see, you can imagine that this formula becomes three lines. But that's not the point. The point is not if the formula is nice or not. The point is that the formula exists. OK? Now, in order to, I mean, I've already jumped to the conclusion, but I mean, you can imagine that the same holds for alpha two. Because of course the expression is a combination of these two things. Alpha two is what? Well, alpha two, you can extract it from whatever you want, for example, you have two possibilities, of course, every time, but they end up in the same thing. So it's E1, V, E2. OK? Now, let me don't do it. The same thing. So you take both expressions. So you take the derivative of E1 from this expression, and you take E2 from this expression. You expand, and you make very, you will get the same kind of things coming up. And you manipulate them in the same way. And I can tell you that at the end you get AB over 2, DE DV, plus AC over 2, DGDU. It is slightly simpler. OK? Same proof. Now, no. Yeah, that's why I was saying it's slightly simpler. Because here you have the difference of two derivatives, and here you get only DGDU. OK? So now, strange enough if you want that the theorem is over. OK? Now, this equation tells you that the Gauss curvature is, in fact, an expression, an horrible expression, in terms of the first fundamental form. Now, there are two situations where one simplifies a little bit this formula, which are simpler and geometrically meaningful. So let me write down at least the output of this proof in two cases. Well, in fact, in two sub cases, one of the other. In particular, out of the proof. So I will never write down the complete formula for k in terms of this function. In particular, if f is equal to 0, f equal to 0, it's a huge simplification in all these things, and you can see it y. It's simple to imagine why f equal to 0 is fantastic. Because, well, b becomes 0. Delta becomes a much simpler function. And here there is nothing like this. So every time you have to take the derivative, for example, of this and this, actually b is equal to 0. So you see this is not there. So the formal expression of alpha 1 and alpha 2 dramatically simplifies. And in this case, with a little patient, you can actually get to the final formula, which is k is equal to minus 1 over 2 square root of eg. Now remember, eg minus f squared, in this case, becomes just eg. So this is the famous eg minus f squared to the power 1 1 half, times d in du of 1 over square root of eg times d edu plus d in dv 1 over square root of eg d edv. For example, this is one situation where in finite time you get to the final formula. And as a sub case, if also, if moreover, e is equal to 1, so moreover means moreover, f is equal to 0, and e is constantly equal to 1, then actually, you see, e equal to 1 means these derivatives are not there, which also means that there is a mistake there, because I think this is dg. Well, one of these two, e was a g. And now it will take, decide which one of these two capital E is a g. Here there is a mistake in my notes. I think it's this one. I would bet it's this one, because this seems to come from here. You see, this does not have a b in front. So f equal to 0 does not kill this. So a term like this should appear. So I'm pretty sure this is a g. But double check it. Now, if moreover, e is equal to 1, then if I write this one, this appears. So that shows that this is the right one. This is the g. Double check it, but I'm sure it's this one. And then k becomes minus g to the power minus 1,5, meaning 1 over square root of g. Second derivative with respect to u of g to the power 1,5. But in this case, it's reasonably simple. Well, of course, if e is equal to 1, and f is equal to 0, there is only one function left in the first fundamental form, which is g. So everything, whatever the formula is, must be a function of g. And this is the formal expression. We will come back, because this is actually probably the only formula we will use explicitly out of the teore migratium. But you see, it holds in a very special geometric situation, because e equal to 1, and f equal to 0. It's a very special situation. In fact, how to reach it is partly what I want to do now. May I raise this? So I can. Because now we start something which looks, it's interesting by itself, and looks quite different, but then the loop will close, hopefully by the end of the lecture. So new chapter, if you want, in your notes, geodesics. Now we go back to a problem, which is very natural. Basically, the problem we are going to attack is the following. Given two points on a surface, this is something you have already seen many times in Euclidean spaces, and you used, in fact, the solution without even thinking, you take two points, and you ask, which is the curve of shortest length, which joins these two points? Well, in Euclidean space, in fact, I don't know how many of you have seen a proof of it, because, of course, everybody says it's the segment, and it's right. But I mean, would you be able to prove it? Small challenge. But now, you see, since we know how to measure lengths of curves on any surface, we can ask the same question on any surface. And in fact, there is another situation which you probably know how to handle by hand, which is the case of the sphere. If you take two points on the sphere, which is the curve joining these two points with shortest length, what your colleague called the diametrical circle is usually called a maximal circle. A maximal circle means two points on the sphere, by simple standard geometry in R3. Two points on the sphere, and the center, so three points, determine one plane. Cut the sphere with this plane. And you get, so this plane passes through the origin. So you get a circle of radius, exactly the radius of the sphere. Something like, OK. It's an equator, but it's not horizontal. It's an equator rotated somewhere. These curves are called maximal circles. And then you take the piece of the maximal circle of shortest length, because now you have two possibilities, going one way and going the other way. Of course, question mark, there is one shorter than the other, and you pick the shorter. But then, of course, was false, depending on the position of the points. If the two points are anti-podal, you make this construction, and you don't get one. If I don't even get a unique plane, if the two points are anti-podal, this elementary analytic construction does not determine one plane. You have a one parameter family of planes, because they determine the line. So each plane which contains this line would work. So this is a particular situation where you are not able to define a unique curve. And of course, any curve you construct by this construction have the same length. So there is not one shortest. There are infinitely many, all of the same length. Now this is a key difference, of course, between spherical geometry and Euclidean geometry. We will come back to this maybe on a special lecture, give you a special lecture on the birth of non-Euclidean geometries. Probably some of you know the history of the fifth postulate, and this situation is critical for that. Now, the point is, how to make, so these are probably the two cases that you know, presumably. Certainly, all the airplanes know this. If you want to go from Trieste to New York, unfortunately, you have to change three airplanes. But I mean, if there was a direct flight, you will end up very close to the North Pole. So Trieste and New York, they are more or less at the same height, but you don't go this way. This is much longer than this on the sphere. So every time you take an airplane, you solve this problem, more or less, modular wind. Now, what I want to do today is to attack this problem on a general surface. This problem makes perfectly sense on a general surface, and I would like to see what is the mathematical problem coming out of it. So the story starts a couple of steps back. It's very similar to what we did in the definition of minimal surface with respect to the area. In some sense, we have done something which is already a bit more difficult in some sense, because there we were taking a piece of a surface, measuring the area, moving it, and see how the area was changing. So now we are going to do in the slightly simpler situation of a curve. We move it, we measure the length of this one parameter family of curves, and we want to see how the length of the functional works, behaves under this deformation. So as we did in the definition to define minimal surfaces looking at the area functional, we need to define what is a one parameter family of curves. As we did for a one parameter, remember that it was the graphical image of a function over a surface. We are doing something similar now, but with curves. So of course, from now on, as usual, is a regular surface. So definition, a family, or differentiable family, or one parameter family, doesn't matter in this situation. A family of smooth curves on S is a map, let me call it capital gamma, from some fixed, well, let me say, first is a little interval, which is parameterizing the variation, how I'm moving, and then some fixed interval, AB, into S. And that's it. So gamma, it's a smooth map. It's a smooth map. You see, this is not as innocent as it looks, because, of course, a curve, at least in our language, has always been a differentiable function. But this will be the parameter on the curve. Now I'm requiring that also the parameter of the variation is smooth. So for such thing, notations, you write, so gamma will be a function of two real parameters. So one is called S, and one is called T, just to give names. And if I fix S, so fixing S in the minus epsilon, epsilon interval, I can look at gamma of S, T. So now this becomes S, imagine it's fixed, and then only T is moving. So I prefer to give it another name to remind me that this is kind of a curve. So this is the curve at parameter S. So this is a smooth curve as a function of T. Now, so definition one. Definition two is exactly a smooth curve, is a geodesic. So this is the key word, of course. If, well, this smooth curve, let's say it's gamma defined on some interval AB into S. A smooth curve gamma from AB into S is a geodesic if for every smooth family of curves, for any family of curves, for every family of smooth curves, such that gamma naught is equal to gamma. So for those families, which at S equal to zero give me the old curve, the starting curve. And they all have the same starting point and final point. And gamma S of A is equal gamma of A, and gamma S of B is equal gamma of B for any S in minus epsilon epsilon. Then I'll draw you a picture just to have a geometry. But I mean, I hope it's clear. So these curves, so you are taking one fixed curve gamma. OK, let's draw it now. I don't draw the surface, let's even draw the surface. S, we have somehow two special points, which are the initial points and the final point of this curve gamma. So this is gamma of B, this is gamma of A. So this curve is a geodesic if every time I change it with a family of smooth curves moving from this one in some way, but in a way that, of course, it starts with this and the family always starts from this point and then in this point. So imagine that you are taking variations of this curve but always starting here and ending here. So under these conditions, something is a geodesic if the length, the derivative with respect to S. So you see, for any S, so these are the gamma S. For each S, this is a curve from A to B. So I can measure its length. So I can ask, what is the variation of the length of gamma S with respect to S? And in particular, I would like to know what does it mean that something is a critical point for this function. So D in the S, a test equal to 0 of the length of gamma S and actually here, sooner or later, we will drop all these things, the length between A and B, because that's the whole curve. And we say something, is geodesic if it's a critical point. So for any variation which respects the end points and starts from gamma, this is equal to 0. This is completely analogous to what we did for minimal surface with the area. OK? There is a subtle difference, but it's subtle. In that case, we were taking normal variations. So we were taking one surface and we were pushing around in R3, using the normal vector. In this case, the family of curves must lie on the surface. OK? So just to tell you honestly, which are the differences. OK? So what is the, as we did for the minimal surface equation, remember here we had the area, a type of variation, we wrote down what it meant on a chart, so the Euler Lagrange equation for this, for this function. I want to do exactly the same thing now. So what does it mean? Assume now that this picture is inside a patch. OK? So I have my usual map, some domain U in R2. OK? If this is the situation, gamma S of t, every time I have this curve, I can pull it back here. So here I had two special points, maybe these two, which were the two initial and final points of this picture. I had one special curve, the inverse image of gamma here. And then I can take the inverse image of every gamma S. OK? So in some sense I can draw the same picture now on the plane here. OK? So there is a variation of this planar curve. OK? Well, that means gamma S of t, I imagine it to be X of U of S t, V of S t. OK? That means I just transplant this family of smooth curves from the surface to the chart. OK, to the domain of the chart. OK, notations, because now things are going to be quite complicated. So notation, because you see, we have two parameters. So S and t. So the derivative with respect to U, with respect to t of whatever function actually, now I write it for, I'll call it U dot, OK? Or, of course, the same thing for V, or for any function. OK, the derivative with respect to t, I will use the notation dot, OK? While the derivative with respect to S, I leave it explicit. OK? Just to avoid to forming of which are too bad. And then second part of the notation, let me isolate this function that I call capital R, which is nothing but the norm squared of the tangent vector. OK? So this is E U dot squared plus 2F U dot V dot plus G V dot squared, OK? This is the norm squared of the tangent vector to the curve gamma, OK? Remember, these are all functions of S and t, OK? So with this notation at hand formula looks a bit better, because what is the length? I will drop also the AB. Every curve is defined from A to B, so the length is between A and B. So what is the length of gamma of S? Well, the length of a gamma of S will be the integral between A and B of the norm of the tangent vector. So R to the power 1 half in the t. T is the parameter giving the curve. So I fix S, and I move t, OK? And now the question is, what is this equation? So I have to take of this object here, of this functional, the derivative with respect to S. OK? So how much is the derivative with respect to S? Well, of course, I mean, since everything is smooth, I don't have regularity problem. I can plug the derivative inside the integral, OK? And this function is a function of S and t, and of course, I take the derivative with respect to S. So this is, but of course, I have a 1 half coming from here, the integral between A and B R to the minus 1 half dr dS in the t. OK? And now, how much is dr dS? Let's focus. OK, so this formula will be there for a while, and let's work on how much is dr dS? Being a function. OK? So, of course, these are functions of u, which is a function of S t, v, which is a function of S t, and these, which are directly a function of S t. So how do I take the derivative with respect to S by chain rule? OK? So dr dS becomes what? If I take the derivative, for example, of e with respect to u times du dS. And then I will take the derivative with respect to v times dv dS, OK? While when I differentiate this, of course, I take the derivative with respect to S, because they are already the functions of S, OK? So if you do that, then becomes, so let me already group the things. For example, I mean, I want to take a group. Let's hope this space is enough, times du dS, OK? So what it will be inside the bracket du dS? Well, there will be this one with respect to u. So there will be dE du u dot squared. And then similarly, this with respect to u plus 2 dF du u dot v dot plus dG du v dot squared, OK? So this will be the coefficient of du dS, OK? Similarly, there will be something times dV dS. And what will be the something? The same thing replacing u and v, OK? So this will be this with respect to v dE dv u dot squared times 2 plus 2 du dF dv u dot v dot plus dG dv v dot squared dv dS. Am I done? No. There are still the derivatives. So this hands up the story of the derivatives of these, of E, F, and G. But then I have the derivative of these with respect to S. They are honest functions of S. So plus, what do I get there? Well, let's see. For example, what would be the derivative of this with respect to S? It will be twice E u dot, the derivative of, not say, u double dot, because u dot is the derivative with respect to t, and now we're taking the derivative with respect to S. So it's twice E u dot du dot in dS, OK? Excellent. So plus, and then doing the same thing for everything, OK? It's twice times E u dot plus F v dot, OK? So because now I group it with respect to du dot in dS, OK? Plus, similar thing, plus 2 times F u dot, yeah? U dot plus G v dot dV dot in dS, OK? Not really nice, eh? We have to work. Now, in order to get the Euler-Raglandz equation, what do we have to observe, for example? How can we simplify this? Well, at the initial and final points, we know that gamma S of A and then B are fixed, are independent of S. What does it mean analytically? That U of S A, so if I transport it here, of course, this will hold also for the curves in the plane. They will all start from the same point, and that will be because this is 1 to 1, OK? So the functions U of S A always take the same value. It's the first coordinate of this point. And the same is for V of S A. And the same is for U of S B and V of S B, OK? Because they are the U and V coordinates of these two points here, and they have to be fixed, independent of S, OK? Analytically, how can I use this information? I can use them because that is telling me that at A and B, the U, the S, and the V, the S are zero, because these are constants, OK? But now you see, I can play some game here, because this is the R, the S, it's this expression. On these two terms, I can try to use some integration by parts, OK? And then, of course, I cannot just try. I can. But when I do that, I introduce an error term, which depends on the value of the function at the end points and at the boundary. But what I'm telling you is that these terms are not there at the boundary, OK? So the moral is that integration by parts tell me the following thing, is that I can rewrite dL gamma S in the S at S equal to zero in this form. So there will be some two functions. So it will be the integral of A and B of some function p, the U, the S, plus q, dV, the S, in the T. But now the functions p and q will be quite horrible, where? So let's take our time. And p will be 1,5 R over minus 1,5 times dE du u dot square plus twice dF du u dot v dot plus dG du v dot square minus, that's coming from, d in dt R to the minus 1,5 dE du dot plus F v dot, OK? And q, similarly, it's a similar expression. Now everything is replaced with respect to derivative, with respect to v. And then it's actually the same expression. R to the minus 1,5 dE du dot square plus 2 dF du dot v dot plus dG du v v dot square minus, the same thing, minus d in dt R to the minus 1,5 F u dot, well, not the same thing, because it's this other part, F u dot plus gv dot. OK. I'm not cheating you as usual. When there is something subtle, I warn you, OK? Here, this is just brute force integration by parts using this information. We have to use the fact that these two functions at the boundary disappear, OK? Now, of course, these two functions, when I write, because since I've already written at s equal to 0, it means that these two functions for here are already evaluated that, in principle, they were functions of s and t. But now I already evaluated s equal to 0, OK? So now these are just functions of t, OK? Well, now depends how fresh and optimistic you are, because we have written, in some sense, the Euler Lagrange equation for the length function. Because we said, well, so gamma is a geodesic. If this is 0, for any possible variation, gamma s, OK? So you see, any variation means that these vectors here are free to do whatever they want. Because these are the way you are moving each of these points in the family, OK? So if I have a family, if I have a curve, the derivative with respect to t is the tangent vector. If I have a family, the derivative with respect to s at the given t is telling me in which way I am pushing one point in my family, OK? So now I propose, because I would really like to. So you see, we have written, remember, for example, what happened for the mean curvature equation, for the minimal surface problem? We got the variation of the area in our variation at s equal to 0 was the mean curvature times something, times the variation, little h, OK? And then we said, well, a minimal surface is something for which this is 0 for any variation. We argued something like this. We said, this is the Euler Lagrange equation. And this has to be 0. So this is something defined on the surface at s equal to 0. And this has to be 0 for any function h, so that this is equivalent to h equal to 0. So the only way that this is 0 for any little h is that capital H is 0. Now, I just want to convince you that we are exactly in the same situation. Don't be disturbed by the fact that the functions are horrible. Mathematically, it's exactly the same problem. Now, it's slightly more complicated, because it's a combination of two functions, but multiplied by something which is free to be whatever they want. So how is it possible that this is 0 for any du, ds, dv, ds? Dv, ds, dv, ds are exactly in this language h, because h was the function that was moving the surface. And these are exactly the vectors which are moving the curve. So philosophically, it's exactly the same thing. So guess, how is it possible that something is a geodesic? Something is a geodesic if and only if p is equal to q is equal to 0. Identically, these are functions of t. So for any time, these functions are 0. You can write down the precise proof as we did for this equation here. Well, of course, in one way, it's obvious. If p is equal to q is equal to 0, this is 0, and that's it. But for any variation, the other way around is always slightly more tricky. Because you say now, suppose if it's a geodesic, this is 0 for any du, ds, dv, ds. So then you argue by contradiction. If p was not 0, you have to construct a family which violates the equation, this is equal to 0. Exactly as we did for the mean curvature equation. It's simple. It's very simple. Now, these are almost useless, I would say in general. There is one way to simplify them significantly, which is assuming that the initial curve gamma was in fact, so remember, we have the parameter on the initial curve gamma was t, any parameter. You see, if t was arc length, at least there are great simplifications here. What does it mean, it's arc length? It means the norm of the tangent vector is constantly equal to 1. So velocity is constant norm equal to 1. But the norm of the tangent vector, this is actually the norm squared, but it's the same. So that means if we start with a curve, which is parametrized by arc length, r is equal to 1. And you can see that at the end, this is nice, because that means that this derivative is not there. I mean, this goes out. This goes out, but there is the same thing here, so you can cancel because you are saying this is equal to 0. So at least this nasty r to the minus 1 half disappears. So let me write the final moral of this. After all, requiring that something is parametrized by arc length, we understood that it's almost an empty, but if they are smooth, it's not even almost. It's an empty assumption, so we are not losing any generality. Basically, the corollary of this computation, of this long computation, is the following. If gamma on s is parametrized by arc length, then it is a geodesic, if and only if. So let me write down for once the system d in dt. I'm just, but now I'm really just taking p equal to 0. I take this on the left, and I rewrite. d u dot plus f v dot, the derivative with respect to t of this is equal to 1 half d e d u dot squared plus twice d f d u dot v dot plus d g d u v dot squared. And then the other equation is d in dt here. f u dot plus g v dot is equal to 1 half, the same thing with respect to v d e d v u dot squared plus twice d f d v u dot v dot plus d g d v v dot squared. So, comments out of this computation, geodesics depend, I mean, the fact that the curve is a geodesic or not, depends only on the first fundamental form. See, these equations are equations on, which are how the values of the coefficients of the first fundamental form behave on the curve gamma, OK? So again, simple corollary, if you have an isometry, geodesic goes to geodesic, because, OK? So they are intrinsic of the surface, OK? They do not depend on the second fundamental form, but only on the first, OK? But there is an important geometric interpretation of this, because one can say, OK, but what is this measuring, OK? Well, interpretation of this system. You see, this is a system of equations. So gamma is a geodesic, if you write it in a local chart, because this is written with respect to the local chart functions of the curve gamma, u and v, OK? So the piece covered by this chart must satisfy this system of differential equations. Now, what is the geometric interpretation? Gamma is a geodesic. This is extremely important, is a geodesic, if and only if its acceleration is normal to s at every point. First let me prove it, and then let me comment on this, OK? Well, if gamma, so we said gamma of t, it's, again, this is a local property, so I can take the piece of gamma, which is in a patch, and work with the piece covered by the map x, OK? So gamma will be x of u of t, v of t, OK? As usual in our picture. What is the tangent vector? Well, the tangent vector is by chain rule, OK? So the tangent vector gives me u dot, u dot, sorry, u dot derivative with respect to t, OK? U dot x u plus v dot x v, OK? This is, we used it hundreds of times now. So what is the acceleration? Well, of course, the acceleration is second derivative. So this is first derivative, this is gamma prime. So the second derivative is the acceleration, but so the acceleration being normal means what? It's orthogonal to every tangent vector, so I can say, if it's orthogonal to x u and x v, it's normal, because they are a basis, OK? So acceleration normal means the derivative of this object, u dot x u plus v dot x v, the scalar product between this, for example, and x u is equal to zero. And the same thing with x v, OK? Well, now, let's see what this is taking us, because now let's expand this derivative, of course, OK? If we expand this derivative, maybe I should leave it as an exercise, because now it will take 10 minutes, but I mean, OK, because here, because you see, if I expand the derivative immediately, I would get, well, no, in fact, let's do it, OK? This would be u double dot x u x u, which means e, OK? Plus u dot d in dt x u, OK? Which I can express again by chain rule. Is x u u, x u u, sorry, it's u dot x u u plus v dot x u v, OK? And then I've done the derivative of the first one. So it's this times x u, OK? And so on. Yeah, no, no, it will take 10 minutes. So, exercise, that this equal to zero, actually mixed with the other one, is exactly this system of equation. Because also here you can expand. You see, here it looks like first order, because I always said u dot v dot, but here there is a derivative with respect to t. So this is a second order system, OK? So you can also expand this. And you get exactly these things. So it's just a matter of expanding and checking that you are getting exactly the same thing, OK? But morally, this is extremely important. Because what is saying? Suppose that you are a bug, OK? This is a typical game in differential geometry. I don't know why differential geometry is right from the beginning of the history thought to be bugs. Zero dimension, you are a point living, and your universe is a surface, OK? What can you do? What do we do? Because actually, besides the fact that our universe is not a surface, it's not two-dimensional, we are exactly in this situation, OK? But now suppose that for some reason you know that your universe is two-dimensional. Now, what can we do? We can do, we can measure angles between directions. We can measure lengths of curves, because, of course, I know the distance between me and you. I can measure it, OK? Leaving on my surface. I don't have to go out. So this is the key conceptual game that you have to play, OK? So what are the things that you can do? Suppose you are a bug on the Earth. Which are the things that you can do because you are a bug on the Earth and because you have a rocket going out and taking a picture, OK? Leaving on the surface means measuring length of curves, measuring norm of vectors, tangent vectors, because if I can measure length of curves, it means I can measure the first fundamental form. So everything which depends on the first fundamental form is something you can actually know just leaving on the surface. Everything which depends on the second fundamental form, we know it, because we are three-dimensional objects, so we know what is a normal vector and we look from outside. The normal vector is by definition extrinsic, as we say. So there are the difference between intrinsic quantities and objects and extrinsic things. Intrinsic are the ones which depends only on the first fundamental form, because these are the things we can do leaving there. Extrinsic are the things that we can do because we know that there is something outside. So no matter what, if you think the game is stupid, don't think so, because now we are playing this game in a situation where we have the possibility of taking a rocket, go outside, take a picture and go back. But if I ask you the same question, in your universe, we are not able to say anything, OK? So if we substitute to a surface, a n-dimensional manifold, the output is that these type of questions are unsolved, OK? And that's why we are studying so carefully surfaces, because in this case where we can actually take the rocket and take a picture, we want to know what we can do, OK? So the... Now, a geodesic is something that a bug leaving on a surface knows. So he knows, he it knows, I don't know, it knows, if he is moving on a geodesic or not, because he, in principle, he could write down this system of equations, OK? Geometrically, so in principle, he could write down the system of equations and seeing whether he is solving it or not. And that's one way. But the other way, so suppose the answer is yes. Now take the rocket, go outside, and you say, looking from outside, how do I see that he is moving, it is moving on a geodesic? I see it because x acceleration is normal to its universe, OK? So that means for the bug, this movement is without acceleration. If I ask the bug, are you accelerating? He must say no, because the acceleration is normal and he doesn't know what it is normal. He doesn't know even normal exists, OK? If it's only normal, the answer the bug will say is no. I'm moving without acceleration. So his definition of uniform motion will not be moving on a straight line as we did for a few thousand years, because the straight line doesn't exist in his universe. His answer will be I'm moving on a geodesic. Uniform motion is movement on geodesics, OK? So if this bug lives, so now next time I will show you a few examples of geodesics. For example, we will check on the sphere these great circles are indeed geodesics. So if you move on these great circles and you live there and you think there is no normal vector, so if looking from outside, this is not uniform motion. Even if the norm of the speed is constantly equal to 1, we see a rotation, OK? But he will think that it's a uniform motion, OK? This is actually a point, I sometimes point to something historical and says we have three minutes left. It's just comments. If you take the most influential thinker in Western civilization up to 200 or 300 years ago, which was Aristoteles, he wrote a treaty which was called On Physics, and he starts exactly with this mistake, OK? So I would have been killed 300 years ago, 400 years ago for having said something like this. But the book On Physics by Aristoteles starts exactly by saying uniform motion is a motion on straight lines or great circles, OK? So now you see why it's so important to determine whether geodesic exists. Because in principle, you see, this is a system of equation and in principle it could have no solutions. So from the physical point of view, this is crucial now. Do geodesic exist? Do they always exist? Is it possible to go with a geodec? You see, there are so many accidents in the Euclidean geometry that we don't even realize. If geodesics are straight lines, for example, in Euclidean geometry, in fact, it's an axiom, it's not a theorem, given two points, there is a line joining them. Is this true on a surface with geodesics or not? But this is true in Euclidean geometry by definition of Euclidean geometry. So now the question is, give me any surface. The most complicated surface you can think of, in fact, even non-compact if you want, OK? You can give me any two points. Is there a geodesic starting here and ending up here? Connected, suppose it's connected, otherwise, yeah. Well, OK, on a connected surface. These are highly nontrivial problems, OK? I can tell you, it's also impossible to determine explicit, even if you are able to solve them, I mean, it's completely impossible to write them down. There is no hope, unless very simple symmetric examples, like the plane, the cylinder, the sphere, but I mean, in general, even on a torus you have to, which is kind of the simplest nontrivial situation if you want, even on the torus of revolution, writing them down, you can do it. We will do it probably as an exercise, OK? You can realize that this is already quite complicated. And again, final warning about the words. I started by asking you, is it possible to connect two points with the curve of shortest length? And we ended up with the definition of geodesic, which is a critical point of the length functional. So, in principle, they are not minima, even when they exist. Nobody's telling me that the second derivative of the length is positive. This is the same as what happened for minimal surface. This, fortunately here, they didn't say minimizing curves. Here they gave a different name, geodesic, and that's it. For minimal surface there was this problem. They called them minimal, and then maybe they are not minimal at all. They are just critical points, OK? Because, again, in general, there is no reason why the geodesic should be a minimum. Even on the sphere, in R2, in R3, it is true, the only geodesic is a minimum, the segment. There is only one segment, and this segment minimizes. But on the sphere it's not true. Take two points, and there are two arcs of joining them on a maximum circle. One is minimal, and the other is not. The long one is a critical point, but it's not a minimum. So next time we will speculate on geodesics.