 Ok, so, last time, we learned something crucial for the rest of the course, namely, gauss curvature, principal curvature, curvature, principal directions, and principal curvature, of course, I mean curvature, ok. So, the point was improving what kind of differentiable functions if they are differentiable and where they are differentiable, we learned also the way to compute them in terms of a local chart. So, if around the point you have a parameterization of your regular surface, now you have very simple formulae, well, some are very simple and some are less simple, but I mean, in principle, looking at your notes, you are able to compute k, h in terms of the components of the first and the second fundamental form, ok. So, if you have a local chart, if you have x, so that's kind of the strategy you have to keep in mind. You have a local parameterization of your regular surface, so around the point covered by this chart, you can compute capital E, capital F, and capital G immediately, and now I'll show you in two examples. Now, and then you compute also little, so the components of the first fundamental form, the components of the second fundamental form, out of this, you construct immediately k, remember EG minus F squared over EG minus F squared, and h, which is more complicated, I mean, it's 1 half EG plus GE minus 2 FF over EG minus F squared, and then out of this, you compute ki, k1 and k2, which are nothing but h plus or minus h squared minus k, ok. So that's the strategy, if you need to know the curvatures, all these curvatures for a surface, the only point is to be able to parameterize locally around some point with a given chart. Let's see this procedure in practice, how it works, and you will see it's very simple, ok. So, for example, so example one, let's take one of our standards, I mean, we did some examples already, the sphere, the plane, and so on, but let's, for example, suppose we take the elliptic paraboloid, one elliptic paraboloid. So, for example, the set of x, y, z in R3, such that it satisfies the quadratic equation, it came up already, we used already this example somewhere, ok. So, this is called the elliptic paraboloid. Well, maybe it's a good idea to try to imagine first how it looks like, usually the way to do it with, so actually, now I remember where we used, we proved that this was a regular surface by checking that zero is a regular value of the function x squared plus y squared minus 2z, because this is the inverse image of zero, ok. So, we know it's a regular surface, in fact it's also a graph, if you want you can put this two down here, and this shows immediately it's a graph of something, ok. So, that's also another way, and now we will use it to parameterize it, what I was saying. How does it look like? Well, the standard way, I mean the simplest way for this kind of locus to understand is to cut with planes, ok, and see what kind of curve you get. Of course, if you, generically, if you cut the surface with a plane you get a curve. So, it depends, if you are able to detect which curve is it when the plane is moving, you understand how the surface looks like, ok. Well, of course, in this case it's simple, because if I cut with planes, so I have R3, and if I pick the planes, the horizontal planes, z equal to a constant. Well, of course, for example, if z is equal to zero, the point, the origin, ok. So, the intersection with the coordinate plane z equal to zero is just the origin. Then if z moves down, becomes negative, the intersection with the plane becomes empty. So, there is nothing below the horizontal plane at height zero. What is above? Well, if z becomes one, it becomes what? It becomes a circle. If z becomes two, it's another circle. And the radius of this, I mean, the center of the circles lie on the z-axis. And the radii, of course, are proportional, I mean, besides these two are proportional to the point z. Ok, so it's a growing family of circles. Ok, so this tells us how it looks when I cut with horizontal planes. If I cut with another plane, for example, x equal to a constant, or y equal to a constant. Of course, this is symmetric in x and y. I get a parabola, which, of course, you know how to draw it. And at the end, of course, you know, you realize that the surface looks like this. Ok? Like this, if you want a better picture. Ok? This is called the elliptic paraboloid. Ok? Now, well, this is one elliptic paraboloid. So let's see our machine at work. Well, are we able to find around the point? So I fix now a point, and I would like to know, Gauss curvature, mean curvature, and principal curvatures. Well, I need to parameterize locally. I should find a local chart parameterizing a neighbor of this point. But actually, in this case, it's simple. Because this is globally. So there is one chart, which covers everything. Which is just the graph. The graph chart, in some sense, ok? X of uv, where u and v are free to move over the whole r2. So our domain u is equal to r2. And this is equal to uv u squared plus v squared, because z is equal to u squared plus v squared over 2. Ok? And this works for every point. Very well. So, let's do everything. Let's compute first the big letters. The big letters are what? Are the coefficients of the first fundamental form? Now, remember the coefficients of the first form, the entries, in some sense, they are usually called the coefficients, but it's kind of misleading in this moment, why? So the entries of the first fundamental form. What is xu? Is xu scalar product xu. And so on. So now you realize immediately, if you imagine what kind of computations that you have to do, that it's better first to compute xu, xv, xu u, xu v, xv v. Before even thinking. So you switch your brain off and you start writing xu. Xu is 1, 0, u. V is 0, 1, v. Ok, then you write. As I said, our brain now is switched off. So, xu u, xu u is 0, 0, 1. Xu v is 0, 0, 0. Xv v is 0, 0, 1. And then what? We know. I already imagine what kind of computations I have to do, I need to compute a unitary normal vector field. So I pick the standard one. Remember there are two. Ok, so this is a moment where there is no one answer. I pick n to be xu, wedge, xv divided by its norm. Ok, so let's compute it. These are the vectors 1, 0, u and 0, 1, v. So what kind of vector is this? Ok, so this is the wedge product and then I have to divide by its norm. So this is minus u, minus v, 1 divided by its norm, which is square root of 1 plus u squared plus v squared. Ok, now we can switch on the brain. We have on the blackboard every ingredient. How do we mix them? Well, capital E is xu, scalar product xu by definition. So it's 1 plus u squared. F is xu, scalar xv. So it's uv. And g is 1 plus v squared. Ok, now how much is little e? Little e is the entry of the second fundamental form. Ok, so in this case let me rewrite what it was, the definition. The definition was it's minus dn xu, scalar product xu. Ok, but now instead of this remember we proved a little, we have a little observation that this is equal to n xu, and now you see why this formula looks a little bit better not to have to take derivatives of this. Ok, this is just a technical thing, of course they are the same, but here it's less likely to make mistakes. Ok, so how much is it? Well, I have to take n, scalar product xu, so it becomes 1 over the square root. Ok, so what is little f? Little f I use the same trick and I find out that this is nxuv. Ok, so n is that one, xuv is here 0. And what is little g? Little g is n scalar xvv. So it's the same as before, it's 1 over the square root. Ok, so now you have everything and you pick the formula and you find what? That k, now if you want just to be careful, k at which point? Well, k at the point x of uv, so if I fix uv on the domain I'm looking at the point x of uv on the surface and I'm computing the Gauss curvature at this point. Ok, so k of xuv is given by this formula. Ok, so it's eg, so that becomes 1 over this without the square root. Ok. Minus f squared, but this is 0. So this becomes the numerator actually now becomes a denominator, but it's divided times 1 over eg minus f squared. Ok. Eg minus f squared is this one. It's 1, so let's compute here eg minus f squared which is something which appears everywhere. Ok, so let's compute it on a side. So this becomes 1 plus u squared times 1 plus v squared minus u squared v squared. And this is what? This becomes 1 plus v squared plus u squared plus u squared v squared minus u squared v squared. So that's it. So that's again the same thing. So I write it just for once, but then of course this becomes 1 over 1 plus u squared plus v squared. Everything squared. Ok. And that's for the Gauss curvature. Well, now you pick h, ok, let's not waste 10 minutes. You have everything here. You substitute and you get something. I can tell you, I mean h, h at the point of x, u, v at the end of the substitution will be just one half. And then it's 2 plus u squared plus v squared over 1 1 plus u squared plus v squared at the power 3 over 2. Ok. Is there anything we can draw out of this computation? Any conclusion we can draw out of this computation? Well, for example, these functions are known constants, but, for example, k, and in fact even h, both are strictly positive, for example, if you want just to observe something. So you can say in the language we are building that the elliptic paraboloid is all made of elliptic points. Fortunately, otherwise the name was kind of very unlucky. Ok. If the elliptic paraboloid had hyperbolic points it would have been a language crisis. Ok. Every point is elliptic, for example. Just one conclusion. Ok. Let's make another example. Much more difficult is to draw any conclusion out of the sign of the mean curvature. In fact, the sign of the mean curvature is not a geometric thing, because we have already observed that if I had picked the opposite n, this would have changed sign. So the fact that this is positive is just an accident of my choice. Ok. If you had picked minus n, you would have gotten minus h, so it would have been negative. So maybe the only thing which is geometrically interesting is that this is never zero. Because that's the only thing which is invariant by the freedom of choosing n. Ok. Let's build another important example. Well, I mean, this is not particularly important, it's just one example. But now the next one is a famous surface and sooner or later exposed to that. In any way, the point of this computer is to show you how easy they are. So now do ten exercises and then everything will become quick and automatic. Ok. Example two. Well, this is a famous surface which is called, it has a name, it's called the helicoid. What kind of surface is it? Well, the idea, actually you have seen it many times. Some of you might even live with one of these in their house. So this is R3. The idea is to take at a given point here a line orthogonal to the z-axis. Ok. I take an orthogonal vector for example of norm one, just to fix. I mean, it's irrelevant, which norm it is. It's a line in this direction. So in particular it has to be a surface covered by lines. And then I take the object. So then as I move this point I would like this vector to rotate at uniform speed. Ok. So if I take as a parameter the height of the point I would like at the same time, so if I move this height vector to change at the uniform speed always in the orthogonal direction to the z-axis. Ok. How can I get an explicit formula for an explicit parameterization for something like this? Well, that's not too difficult. Look at this. So you take X of UV let me remember now. So the height so it will have third one is kind of the parameter of z. Ok. So it's what I would like to call the speed of this special point. Ok. So let me call it AU for some given number A non-zero. Ok. Pick a non-zero number and U is exactly the movement of this point there. As here cos U sin U. So what is this point? Which point is it? So the point given by, let's say suppose A is equal to 1. I take the point at height U and then I take the point the vector cos U sin U. Ok. Which is exactly a vector orthogonal to the z-axis of norm 1. Why do I like this expression? Because when U moves this is actually describing a rotation at uniform speed in the circle. Ok. So this becomes the angular momentum of this vector. Ok. But now this would be a curve. In fact this is the helix, one helix. Ok. Now basically for every point of this helix I want to take the whole line in the second parameter, which is V. I mean this is not a surface of course it depends only on U. It's a curve. But I want to which line do I want to take? The line passing through this point with this velocity. Let's see. Ok. So for V equal to 1 is exactly this point here. For V equal to 0 is exactly this point here. And then it's linear. I'm picking all points on this line. So why do I say that you should be very familiar with this surface? So now if you want to draw the surface this becomes a bit more complicated. So I've printed well, this is a helicoid. No. So you imagine you have this line straight line which rotates when you go up or down. I mean here there is no problem in going down. Ok. So this keeps on rotating or if you want a nicer picture but it's less clear it's something like this. Ok. So this is, you should be familiar with this I mean certainly you have seen it many times because for example stairs, I mean circular sphere, stairs are built out of this. You discretize this and you make stairs going up. Ok. So now, so I don't draw it keep those pictures in mind. Well, the first problem is is it a regular surface or not? Because now I've given it as the image of this map but now it's not clear in principle that this map satisfies all the properties required by the definition of a regular surface. Ok. So you do it. Ok. So it is a regular surface. It is automatically, so actually where U and V lie. Well V is the parameter on the line. So V lies in R. The whole R. U lies in R too. Ok. Because you want Z to be free to go up and down over the whole Z axis. Now. So, you can check in 5 minutes that this is a regular surface. It has the nice feature of being covered by one chart. Ok. By construction. So, let's see the machine. What is the machine telling us now? Again, now you switch off the brain. I have X, X U. X U is minus V, V cos U A X V is equal cos U sin U 0. Then X U U is minus V cos U minus V sin U 0. X U V X U V, for example. Let's take this one. So it's minus sin U cos U X V V is 0. Which is ok, because it's linear in V. If I take two derivatives in V. Ok. Then what else? N, normal vector. N, I pick again the standard choice. X U cross X V divided by its norm. So instead of writing N let me write immediately minus, let me make the wedge product. Minus V sin U cos U A And on the other line I have to put cos U sin U 0. So how much is this wedge product? Ok, so this becomes 0 minus A sin U. Then here 0 minus A cos U with the minus, so plus A cos U. And on the other line minus V sin squared minus V cos squared so minus V. So then I can write N. The normal with my choice is exactly this one. I don't repeat it, but it's divided by its norm. And the norm is what? D squared plus D squared it's A squared. A squared plus V squared Ok? And now let's start. Ok, now we are back in business E. E is the scalar product of this with itself. So it's V squared sin squared plus V squared cos squared so it's V squared plus A squared. V squared plus A squared. F, F is the scalar product of these two. So it's minus V sin cos plus V sin cos 0 plus 0. F is 0. G is what? Is this times itself, so it's 1. Little letters N scalar X U U. So besides the square root we will remember about this at the end sin U so it's plus A V sin cos Ok? Minus so plus, sorry, plus A V sin cos minus A V sin cos. So they disappear plus 0. So E is 0. Very good. Little f is this times this. So it's plus V squared plus A cos squared so A plus 0. So it's A, but now I don't have to forget the denominator. So it's A divided by the square root of A squared plus V squared. And then little g. Little g is, well, it's 0 times something. Very nice. So Modulo mistakes let's see. K at the point X of U V is how much? E g 0 minus f squared, so it's minus minus A squared divided by A squared plus V squared divided so times 1 over E g minus f squared but E g minus f squared is another A squared plus V squared. OK? So this becomes just minus. Choose the way you like it most. OK? For example. H I can tell you, you substitute all these numbers. In this case it's easy because there are plenty of 0s. Of course when you get the 0 you are very happy because formula becomes much easier. Then in this case XU little miracle is constant equal to 0. OK? We will talk about these kind of surfaces at some point. This is a first well, it's not the first example because the plane is another surface like this. But it's the first meaningful example of what is called the minimal surface. In this case well, the only thing, the only conclusion again we can draw is what? Now, this surface all the points of the surface have negative gauss curvature. So the helicoid is completely covered by hyperbolic points. OK? Now, the point of this lecture is to be able to say this kind of conclusion just looking at the picture. OK? So you could have told you that the helicoid was made of hyperbolic points just by looking at the picture and also for the elliptic paraboloid we did before that there were elliptic points. There is a simple way just by so what is the difference between these two surfaces? Well, they have nothing in common of course. But now this is this one. On this surface imagine that you pick a point and you draw the tangent plane at this point. Where is the surface? On one side. OK? It's all on one side. So there is only one point where the two things touch which is of course the point where you are taking the tangent plane and the whole surface lies on one side. OK? Well, the key properties of course it's locally but I mean it is true. Now what if you do the same game here? Pick a point and draw the tangent plane at this point. Where does the surface lie? On both sides it crosses. OK, that's exactly the geometric property that characterizes elliptic points well, almost characterizes we will see. There is one delicacy that we have to be careful about but I mean this is the key property which tells us just by looking at the picture which are possible usually elliptic or hyperbolic points. OK? How do we do that? Step further in our kind of analytic machinery. Given a function on a surface we learn how to compute the differential of this function and we spoke about critical points and regular values. This was all about first derivatives of a function at a given point. What about second derivatives? After all the function on R2 and you ask me to look for minimum and maximum I need second derivatives to distinguish between a minimum and a maximum point. I would like to do the same thing on a surface. Well that's now S is a regular surface now we pick a function with real values OK? A differentiable function I don't even say because otherwise it would be impossible to do derivatives. Now I pick a critical point on S. So suppose that there is a critical point and I call it P so P critical point for F. OK? Now I would like to define the derivative of the function but of course the surface is two dimensional so there is not anything like one second derivative even if the surface was the plane you don't have one second derivative you have a second derivative in one direction second derivative in another direction mixed second derivative in fact you have a second derivative in some sense in any direction exactly that's exactly what I want to define this symbol so this squared at the point P it has to take one direction and give me a number so the second derivative in that direction OK? So it should be a map from the tangent space to R doing what? Well it has to take a tangent vector hence I have at least one alpha such that alpha of T alpha of zero is equal to P and alpha prime of zero is equal to V but then it's obvious what I I take F compose alpha now it's a function of one variable T and I take the second derivative now this object whatever it is at the moment we don't even know it's well defined so in any case it's called definition of F at P so now immediately we have to prove that this is a good definition and actually the philosophy is always the same so what is the problem is that in this formula I have alpha exactly as for the differential now it's slightly more complicated but it's exactly the same problem and we are going to solve it exactly in the same way so we produce a formula for this object quite complicated but we don't care in which alpha is not there ok how do we do it so remember that was exactly the philosophy for the differential when we proved we picked a local chart and we said the differential of a map is the linear map which is associated with respect to the standard basis to the Jacobian so we play the same game it's just that we have to take two derivatives instead of one so proposition and in fact let me list few remember that we have this condition here we are not taking any point this is a definition given at a critical point so but let me repeat let me stress it just for if p is critical some sense proposition zero is that the action d squared f at p is well defined and then we have also some properties it is also a quadratic form on tps this is like an ambiguous statement but it's clear what it means second if p because after all it was a critical point so now let's distinguish if p is a local maximum and in brackets the other case respectively minimum then the action is in the maximum case is semi definite negative in the maximum case and positive in the minimum case and then exactly as we did in calculus for functions of more than one variable is the converse so true not exactly so three you have to put a stronger assumption to so if d2 squared is strictly negative is negative definite negative and of course in the other case positive definite then the converse holds then p is a local maximum or in the other case minimum this is exactly the same statement and in fact you will see it's exactly the same proof as for functions of R2 so how do we prove it basically we prove it at the same time everything we pick a chart so take a chart around p and give it a name for example to the point there will be one point on the domain u which is sent to p via x suppose that this q for example q will be a pair of numbers corresponds to some pair of numbers ab now we have taken a curve alpha on the surface s as usual by picking its domain sufficiently small we can assume that the image of alpha lies in the image of x ok so let me just say all these things by saying I pick alphas and epsilon such that the image of alpha is contained in the image of x ok and then exactly as we did for the differential define beta to be the curve on u by pulling back alpha ok but then beta is a curve on a domain of r2 so beta will be given by a pair of functions u of t, v of t ok and the fantastic idea we used already for the differential was to write f composed alpha which is the one we have to differentiate twice of t as f composed x inverse composed alpha ok that's exactly what we did last time, few times ago ok so I want to see it as a composition of these two maps and now I start taking derivatives I mean let me just under this notation of course this becomes f composed alpha f composed x composed beta but beta I've given names ok so this is f composed x at the point u of t v of t ok I've given names to this to the components of this so I use that ok so in fact we should pick exactly the first part after all we have to compute the second derivative to compute the second derivative and then to take another one so pick your notes if you so how much is f composed alpha d in dt but now I don't have to evaluate it at t equal to 0 because I'm going to take the second derivative how much is this well once I write it in that form of course that becomes u prime since I'm not evaluating at 0 so this will be functions of t now ok but I don't write it otherwise I need 3 blackboards u prime times f composed x differentiated with respect to u plus v prime at the corresponding point by chain rule plus v prime f composed x differentiated with respect to v at every point now I take the second derivative and now if I want I can already evaluate at t equal to 0 this is what the definition of the action is asking so I need to take another derivative of this with respect to t let's do it this is a function of t so this becomes u double prime now if you want you can put 0 just to I'm hoping to stay in one line so it's u double prime f composed x with respect to u plus u prime times times the derivative with respect to t of this but then I have to apply again the chain rule because this will be a function of u of t v of t so this is u prime times u prime of f composed x u u plus v prime f composed x in fact there was no hope to stay in one line u v and this that's it for this piece then I have to play the same game here so plus plus v double prime f composed x v plus v prime times what in this same game u prime f composed x u v plus v prime f composed x v v there is something I can group not much but you see well actually in fact here which is exactly this one but that's it, that's the only thing that I can group you see it's u prime v prime times f composed x u v so this will become if you want drop one and put here two but it's not a great improvement now if you understand the problem you would say that the proof is over what is the problem we wanted to find a formula for the action where alpha did not appear where is alpha here well everywhere but the point is what of alpha is here now remember everything here now is evaluated as zero ok when you evaluate this as zero u prime at zero but that's ok because this does not really depend on alpha it's contained here in this bit of information so when I see first derivative of the functions u and v I'm ok these objects here do they depend on alpha no these are kind of intrinsic of the chart and the function ok so the partial derivatives of f composed x are ok in my spirit so where is the problem the problem so this this observation tells me look this piece and this piece are ok this piece would be ok and this piece would be ok but here there is u prime u double prime and here there is v double prime now two different alpha passing through the same point and with the same velocity do they have to do they need to have the same second derivatives no the velocity at the given point cannot determine its acceleration at that point the second derivative is independent of the first at the given point ok so this is the problem I can easily construct two different alpha with the same p and the same v and completely different these so why the theorem is true well notice that up to now we haven't used anything that's exactly the moment where this comes in critical and the two given curves of the same thing so the only way the theorem is true is that it's true because this is 0 ok so I need to put but that's exactly I know that the critical point f composed x at u is equal f composed x v is equal to 0 at p at the point corresponding to p at everything here is evaluated I've given names so at the point a b because I'm translating everything back to the domain of R2 ok so if p is critical this first derivative is vanish so then these subjects are not there and then that's it now so now what is the output of this output it's good to so we found also an interesting formula so the action at the point p at v if p is a critical point we have a local expression in terms of a chart which is just u prime squared it's also quite easy to remember u prime squared f composed x u u at the point a b it's interesting about where I'm computing what ok and this is this plus 2 u prime 0 v prime 0 f composed x at a b but this is differentiated with respect to u v plus sorry u prime squared u prime squared times this plus twice this plus v prime 0 squared v v at the point a b so you see it's easy to remember no? because basically it's a quadrat now in fact this formula proves essentially everything why? is it a quadratic form well yes it is a quadratic form ok in the entry v well it proves it's well defined for everything we said up to now it is a quadratic form if I write this quadratic form in terms of the standard basis I've already given you a matrix representation of this it's the matrix given by f composed x u u f so on the first line f composed x u v then on the second line f composed x u v again symmetric v v but now this is the matrix representation with respect to the standard basis and I'm writing in terms of the second derivative so the standard dash this is the old dash of the function f composed x do you agree? so f was on a surface but f composed x is on the domain of R2 so you can apply all the theorems you know the quadratic forms are the same basically I'm telling you the quadratic form associated for the function on the surface is the same thing as the old dash for the function f composed x on the domain of R2 but then you know all these theorems and these theorems are exactly the old ones of course you need to translate p to q but then of course maximum minimum who cares something is a maximum on s if and only if q is the maximum of f composed x and so on so everything is done in one shot so this formula tells you immediately everything now is this making us closer to understand the geometric interpretation of all the of all the curvatures that we introduced well it's not clear yet it will be clear after we make a couple of examples of computations of the action for a couple of interesting functions height function remember this was one of the examples listed also for the differential what are we doing we have our surface we pick a point p0 and we pick one plane actually for making my picture a bit ok and we take a plane passing through p0 ok sorry a plane this is capital p ok a plane now what was the height function as a function on s with respect to this plane well the idea was simple just take a unit a unit normal to this plane call it a ok so p the plane capital p is nothing but the orthogonal complement to a vector and now I assume that I already picked a vector of norm 1 ok there is no so the definition of the height function was h h we call it h of p was nothing but p minus p0 scalar product a ok now what was which were critical points for these functions we gave a geometric characterization of those so p is critical if essentially the normal remember we wrote it in lang in english ok saying if the normal line through the point p passes through p0 but that's the same thing as saying that if n of p since now I've picked the unitary normal to the plane n of p has to be a or minus a suppose it's a we don't care ok if it was minus a change a with minus a ok minus a is again a unit normal to the plane p so up to change of sign this is ok ok do you remember this? now compute the action so in fact geometrically in my picture it's a guessing name it's true it's somewhere here ok so the normal at this point should be exactly a ok so this would be my point p now this is the new the new ingredient so how much is this well it's the second derivative at t equal to 0 of h composed alpha I have an explicit expression for h so alpha of t minus p0 scalar product with a ok so how much is the first derivative constant vector p is a fixed point the only thing which is moving is here so the first derivative would be alpha prime at t scalar a so the second derivative is alpha double prime now I evaluated 0 scalar a ok but a is the normal vector because p is a critical I mean this would be true at every point but now I know that a is n of p so this is alpha double prime scalar product n of p why this reminds me of something this is exactly what we found in Euler's theorem ok this is exactly what this is by Euler's theorem the second fundamental form at the point p evaluated twice in the direction v ok so the action at the critical point of the height function gives you a quadratic form exactly the second fundamental form now in particular there is always so this is true for any plane p0 and so on but in particular there is always one special plane to look at if I now I reverse the process I was fixing here p0 and p and I was looking for a special point on p now suppose I take a point here and I want to construct all this picture based on this point basically what I'm saying this is my p now and suppose I take p0 equal to p and I take as a plane in R3 the tangent plane to this point of course I can construct a height function on this why this is interesting well because of course p is automatically p is always a critical point for this height functional so I have automatically a nice critical point for the height function with p0 and a equal n of p if you want you can rely on by construction p is critical if in fact this was an if and only if if I take the tangent plane of course the unitary normal will be the normal by definition of the normal vector ok so in this case this is obvious and so it p is a critical point ok today I see you a bit maybe I'm not particularly clear today but I see you a bit confused is it hard I mean is there some point you want to go through again ok so if I reverse somehow the process I first fix the point on the surface and I construct the plane that the point the original point is a critical point for this height function but this property was true for every height function ok so in particular it is true for this one ok but then this implies without proof that's what I hope you agree the following theorem one if the Gauss curvature at some point is positive meaning IE the same thing IEP is elliptic sorry IE is a Latin form to say the same ok you know Latin more than your Italian colleagues very good so if you take an elliptic point then there exist it's better to write it in English there exists a neighbor a neighbor of P in S which lies on the same side of D and let me say now in quotation marks affine tangent plane to S at P P is the only point of intersection I stop here but I mean of intersection between these affine tangent plane and the surface itself so just one comment before commenting why this is true what is the affine tangent plane remember when we define the tangent plane I mean the tangent plane to a surface at any point so really if we want to make some precise pictures all these pictures are wrong ok so the vector space it's really this one parallel transport to the origin I mean a vector space something a two plane which does not pass through the origin cannot be a vector space ok on the other hand it's very convenient at P ok that's exactly the affine tangent plane so it's the translation of the tangent plane to the point P ok ok so let me write also part two and then we make comments part two the opposite if K of P is negative then P is hyperbolic then in every neighbor of P in S there are points which lie on both sides of the affine tangent plane well why this is true and in fact it's true now without a proof you have already the proof because pick this point construct this special height function the action of this height function is the second fundamental form so what does it mean that the Gauss curvature is positive is telling you exactly that the second fundamental form is strictly definite now here there is the obvious indeterminacy between geometry and analysis because depending on the choice of N whether it is positive or negative well no no it doesn't matter the point is you fall exactly in the cases of the theorem we stated before about the action of a function where you know so it was the last one so the action of this function is strictly negative or positive depending on N and that means the height function has a maximum or a minimum depending on orientation but that's exactly the analytic way to say that the surface lies of course how much is the height function at the point P if P is equal to P naught is zero so at this point is zero and suppose it has a maximum that means it's always negative it's negative in a neighbor I mean of course you cannot draw global conclusions this is a local analysis but there is a neighbor where the function has to be negative but being the height function negative means exactly that if you draw the graph it lies below but that's exactly what's written here it lies on one side and it's impossible that there exist of course if I go sufficiently away at this point I don't know but certainly there is a small neighbor where there is no other point where the height function is zero because it's a strict maximum ok you see why there is nothing to prove and the same thing here and now exactly as in the case exactly as in the case of functions of two variables so remember there is this kind of annoying between the two theorems about the Hessian of a function if it's strictly positive then you have a local minimum if it's strictly negative you have a local maximum if you have a maximum or a minimum it's not true that they are strictly something they are semi-definite so in all cases where there is semi-definite you cannot draw analytic conclusions and in fact you cannot draw now I'm going to show you examples that in fact you cannot even draw geometric conclusions so the point is is there a line 3 in this theorem starting if k of p is equal to zero then the surface does something the answer is no you see k of p is equal to zero is exactly for all this for all we said today the case where the determinant of fundamental form is zero so it's semi-definite examples to convince you there is no no such theorem with k with zero Gauss curvature so all the following examples are interesting because they are examples but the point of this observation is there is no such theorem for k equal to p k of p is equal to zero ok how do I convince you that there is no such theorem explicit examples well, remember that there are two types of points with zero Gauss curvature because essentially the second fundamental form could be identically zero in every direction or it could be one eigenvalue equal to zero and the other non-zero ok, so planar and parabolic so planar so I want to show you that in fact you cannot even if you distinguish these two cases there is no theorem well planar of course there is example one which is the plane so give me a surface with the planar point well the plane any point what happens to this theorem well in some sense it's through the ok, this is kind of delicate I mean if I take the plane the height function is of course identically zero the height function based at the point of the plane is identically zero ok but there is also an example like this take x of u v equal to u v in fact it's a graph u v and then you pick as a function u cube minus 3 v squared u I should have practiced before coming to this lecture to draw you a picture and so I didn't so now let me try it's something like this ok not too bad and so on, so here you have curves going now this is a famous name it's called the monkey saddle what is the point that we are really interested in x of p is equal x of zero zero which actually goes to zero zero zero it's exactly this central point of the saddle ok let's do the exercise there is no time to move on to other things let's make these examples complete let's compute the Gauss curvature of this surface well first I should prove I should argue that this is a surface it's ok, it's a graph of a differentiable function it's a polynomial so there is no problem it's covered by one chart so whatever I'm going to say will hold at every point at the corresponding point I hit every point of the surface so x u 3 u squared minus 3 v squared x v 0 1 minus 6 v u x u u 0 0 6 u x u v 0 0 x u x u v minus 6 v x v v 0 respect to v 6 u normal vector so in fact you start learning tricks about being quick so write x u and x v one on top of the other order in a reasonable way so that you can make the wedge product automatically so what is n n will be 0 so minus this thing so 3 v squared minus 3 u squared then so this is 0 so I'm glad this is 6 v u and then 1 divided by its norm that I don't care it will be some function I will always indicate it by square root and then ok then e capital e capital e is this times this well actually and again now for you it's the second one so it's a bit too early but for a graph how much is e g minus f squared we know it let's see again and check the previous example so e is equal to 1 plus 3 u squared minus 3 v squared squared this is e f f is equal to this mess ok there is really no point e f and g whatever functions they are ok little e little e will be n scalar product this ok so it's 0 0 6 u over the square root f is minus 6 v over the square root and g is minus 6 u over the square root that's actually the only thing I care I mean I'm not trying to get the complete expression so how much is k well k is e g minus f squared so it's minus 36 it's not really important the function because we have not computed the other one so in any case e g minus f squared minus 36 u squared plus 36 v squared divided by e g minus f squared that I don't know so what was the only point I was interested in was x so I was looking for a planar point did I find it? well for u equal to v equal to 0 of course the corresponding point as curvature 0 so it's either planar or parabolic and then I go back to the single coefficients for u equal to v equal to 0 they are all 0 so p is a planar point that actually is the only conclusion I wanted to draw otherwise you write down the complete expressions ok it's a planar point but if I draw the in this case actually it's since the point is the origin the affine tangent space and the tangent space are the same thing ok if I draw the affine tangent space where does the surface lie on both sides in every neighbor whatever small neighbor I pick there will be points where it's positive and points where it's negative you can actually check it immediately here ok whatever small disk in uv you pick this function is somewhere positive and somewhere negative so this seems to tell you look the theorem could be if you have a planar point the surface cuts the tangent plane but you have the plane and the plane does not do that it's not true that the plane stays on both sides so for planar points end of story well what about parabolic points well what is the prototype you have of a parabolic point give me give me a hint so here we had the plane and then we had to construct something strange what is kind of the model of a parabolic point that we have seen well on one hand we have cylinder every point of the cylinder is parabolic because the Gauss curvature is zero but it's not true that both principal curvatures are zero ok so if you look at this example you might say so the theorem could be the same as for elliptic points because what happens here on the cylinder is that you throw the tangent plane at a given point well the surface actually lies on one side now it's not true anymore that p is the only point of intersection there is a whole line but maybe that's the only part of the theorem which fails no here is an example in this case as an exercise so take the yz plane in the yz plane take the curve z equal so it's a graph of a function z is equal to y cube ok where z so the part of this which lies for example z in minus one one ok how does it look like well of course the graph of the function y cube is something like this ok and I'm taking exactly this part the part I drew is the one I am interested in now take this curve and rotate so of course for y z equal to one corresponds also to y equal to one so this is the point one one so draw this line here so this is the z equal to one line in this plane and rotate this curve around this axis and now I made a mistake in my blackboard because I have no space to rotate it so I need to cancel this so that means if I pick a point on this curve I have to add the whole circle just with this center now if I pick this point this is this would be the center and this would be the radius so more or less something like this and so on how do I draw it in one shot well I draw the symmetric curve if I if I am able to do that which looks more or less like this and then every point is rotated every point I join now the interesting part is this I don't like drawing with colors but let's so points on this curve here are more interesting than the others so first observation this surface actually my picture is lightly misleading because here there is a problem because actually the curve itself looks a bit more like this I mean so when I rotate I really get a casp a nondifferentiable point so in fact that's why I'm taking the open part of this curve so this point is not on my surface every other point since it doesn't touch the axis of rotation produce a regular surface so it is a regular surface and now the point is that all these points so yellow points are parabolic this is something you need to check you need to find the parameterization of this of this subject here exercise but that's the only way you know how to compute if something is parabolic or not you pick a point find a chart around this point and compute e f g e f g e v metal form so you can detect whether the gauss curvature is zero and you will find zero but you have also to argue that it is zero not because everything vanishes but only one of the two curvature principle curvature vanishes do that and these two objects cannot go together in a theorem in this case the surface was on one side and in this case clearly the surface cuts the tangent plane sorry after one hour and a half you start getting a bit myself but it's okay imagine the tangent plane of the picture it will cut the surface in every neighborhood no surprise I mean it's easy to construct examples of functions where the action is semi definite and the function the original function had not a minimum and a maximum so it would have been very strange that the height function so the only thing that we are doing is that we are taking instead of any function the height function of a surface on the tangent plane so it's not particularly surprising but in any case the moral of this lecture is clear show me a picture and I can tell you with minor possibilities of mistake okay minor there is some but I'm almost able to tell you immediately which are the elliptic points hyperbolic points I must be very unlucky to get I can make mistakes only at flat points okay, that's it