 Lekšnja leksja je zelo včasna leksja. Zelo smo vse občastil občastil. Zato smo počučati, da je nekaj počet, nekaj počet, nekaj počet. A če je to občastil? Prvno. Prvno. Prvno. Prvno. Prvno. Prvno. Prvno. To approaches are important at first, but I mean principal directions. OK. This is really what we have done. The problem is, what are the geometrical meanings of these numbers, the vectors, or whatever, or forms. OK. We interrupted it in the middle of a sequence of examples. If I remember right we stopped an example. Three, just to keep on learning how to do also computations. Zelo se vseč, da egazim skojih je delalno vseč, kako sem sem izprošljala prišla. Požljam imamo dvanje obitelje. Poželjamo iz Mercedes-Selina. V sem ko všeč moraš se kaj da vseč misljami, da sem se je izroda, da je nekako poprinjen, o pa je vseč Thirty squared. nr. r squared. Kako vse to izgleda? Zelo, da so to izgleda. Vse izgleda z. z. 0 je tvoj cirk. Zelo je tvoja sekundi nr. z. Zato, da je zelo z. z. z. 0 vse poče, kaj je zelo, je na vse objev. Zelo, ki je vse objev, je na vse objev. Tako, zelo, da je prišlo, tudi nekaj zelo, in vse poče, da je vse objev. In potem je dobro, da se poče, in potem, da je prišlo, x, y. In to je tukaj, da smo tukaj što smo zelo. Zelo, da je zo zelo. Spoliv na veliko odloženje. Zelo je to ozelo? Zelo v piško. Tukaj, da se je zelo, zelo, da je zelo poznit ase in večer. Zelo, da so ko svoj tukaj? Spoliv na vsega f, oz x, yz. Zato je x vrk plus y, vsega s. In zelo s. je vsega f in verz večer. So the question is, is R squared a regular value for this function? How do we know it, as usual? We should compute, in fact, if it is a regular value, what is the tangent space? And once we know the tangent space, what is the normal vector if it's orientable or not? Does the unitary normal vector exist over everything? But we know automatically the last everything, because, I mean, so df, at a given point, will be what? Will be the linear map represented by, first we should, what is TPS? TPS is either you write it as care dfp, or the set of vectors v over 3, that the scalar product with the gradient at a given point is equal to 0. So the gradient is a normal vector. Every time it's an inverse value of a function. And what is the gradient, which is the same matrix representation as the differential? So it's take partial derivatives of the function. So it's twice x, y, 0. As a matrix, actually I should write it. Metric representation with respect to the standard basis. It's the linear map, which is represented by this matrix. The gradient is actually the vector, which has these components. So the gradient of f, the point p, is equal to twice x, y, z. And at this point, so this shows you, tells you immediately that this is a regular value. Why? How is it possible that this is the zero-momorphism? Well, it has to be x equal to y equal to zero. But p has to be on the inverse, so of course there is a point. There are many points where this is zero. Every time x is equal to y is equal to zero. But do they lie on f inverse of r squared? No, of course that's the only reasonable requirement. If r is equal to zero, this is not the cylinder. This is the line. In fact, this shows you that it's not a surface. It's a curve. But in this case, so once we know, so the tangent vector is automatically given as the normal space to this vector. So since I want the unitary normal, n of p will be just grad f at p divided by its norm. And how much is it? Well, here we wrote, of course, the two, I forget it. It's x, y, zero divided by the norm. How much is the norm? R. Again, when I say n is equal to this, of course, you could object. For you, maybe n is minus this. There is always this undeterminacy here. There is not a unique normal vector. So if I choose that, what does it mean? For example, at this point of the surface, I'm taking the vector xyz, xyz zero, which is really this vector normalized. So of course, you usually draw it pointed at the point on the surface. So that means I'm taking the normal vector going outside. You could have chosen the other one. No reason for not doing that. Sooner or later, you have to make a choice. So, as being an inverse image of a regular value, the surface is orientable, and this is a choice of a unitary normal, of an orientation. Well, now let's compute everything. I mean, second fundamental form and everything we defined last time. What do we have to do? We have to compute dn, the differential of the Gauss map at a given point p. Apply to some vector v, where v now is not any vector of r3, but it's a tangent vector to the surface. So the question is, what is this map? Well, let's give a few names. So v, suppose v is v1, v2, v3. Let's give name to the components. What do we have to do? Well, the most effective way is all. So, once you have the explicit expression of the map, compute the differential, you go back to the definition. There is no need for any subtle trick here. So, again, in your mind you take alpha, such that alpha of 0 is equal to p, and alpha prime of 0 is equal to v. And you compute what? So this is equal to dn dt at t equal to 0 of n composed alpha of t. And how much is this? Well, that means alpha of t will be x of t, y of t, z of t. So this is equal to what? This is dn dt at t equal to 0 of x of t, y of t, 0. And how much is it? Sorry, 1 over r. That's the map n. How much is it? Well, I take the derivatives and compute at t equal to 0. But these become the components of v. So this is equal to 1 over r, v1, v2, 0. So now, if you want, how much is the second fundamental form of the cylinder applied to two vectors, vw. So you see, it takes two vectors and gives me a number, without even writing. So w will be w1, w2, w3. And the definition, the scalar product of minus dn pv, scalar product with w. So I have to take minus this scalar product w. So this becomes minus 1 over r times what? v1, w1 plus v2, w2. There is no third component, because there is a 0. And now, how much are the principal curvatures? And then everything gets as a cascade. Well, either you diagonalize it by formal algebra, or you make some geometric choice. I prefer doing geometric choice. It's clear, so if it's clear. Suppose at this point here, I mean at any point, this is my point p. It's clear that I have two special directions. So let's go and check if they are really special also for what we are doing. The two special directions are the vertical direction and the direction tangent to the circle passing through this point. These are clearly two obvious choices of vectors on the cylinder. So if v, so v could be this one, or this one, or of course in general it can be skew, but I mean let me take either this or this. So if v is vertical, v vertical, how much is dn pv? Well, vertical means it has only z component. It will be a vector of the form 0, 0, something. How much is dn? It's 0. So in particular it is an eigenvector. So you have immediately found an eigenvalue, 0. So one of the two eigenvalues is 0. Before deciding if this is k1 or k2, we have to compute the other one because I don't know who is the biggest. So how do I check what is the biggest? Let's see if the geometry is really what's behind all this computation. Let's take the other v, the one tangent to the circle. So v tangent to the circle. How much becomes this? Almost. Almost v, 1 over r. Yeah, it's immediate to check. But that means we have found already the second. The second eigenvalue. Because this means that this v is an eigenvector. This is an eigenvalue. Now, just remember, if you remember, because I keep on forgetting it, too. But in any case, so we have really to look at the eigenvalues of minus dn. So then this becomes minus the eigenvalue. So the two eigenvalues are, let's take in my notes, k1 of p, so the smallest one, this is minus 1 over r, because it's negative. Well, OK. Just for convenience, let's suppose r is positive. So we don't waste time. k1 is 1 over r, and k2 is 0. And actually, we have automatically found the principal directions. I mean, from the way we found them, we know already which are the principal directions. So the principal in the other case is the horizontal. OK. So what is the Gauss curvature? How much is the mean curvature? OK. That's the end of the story. On this example, just one thing. So this was an exercise, and with the excuse we found another interesting surface. But the output is a bit strange. OK. And this we will have to think about what this really means. The cylinder, well, if you ask my four years old baby and you say, is the cylinder flat? Well, I can tell you. I mean, he will say no. But we found it has constant zero curvature. So now as a mathematician we will say something is flat if it has zero curvature. So what's going on? This is a very important question. So we would have liked the theorem saying k equal to zero if and only if it's a plane. And now we know already that this theorem is false. Of course, if it's a plane, k is zero, but it's not the unique surface having zero curvature. So what's going on? This is a subtle question, and I want to give you the answer now. But focus on this problem. OK. Freeze it, and let's move on. Because now I really want to understand geometrically what is the second fundamental form. What is the geometric meaning of the second fundamental form? Well, let's start with the definition. We take a curve on the surface, in fact a regular curve, just to avoid trivialities. And just to fix notation, let's say it passes at time zero through some point p. Well, being a curve in R3, it's a curve on a surface, so in particular it's a curve in R3. So if I forget the surface s, I could apply what we learned in the first three lectures of our course. So I can study it as a curve in space, so I can apply to it. I mean, it will have a curvature, it will have a torsion, it will have a normal, a tangent, a binormal, and so on. Frenet, rejedron, and so on. Now, what is the interaction between all those things and the geometry of the surface? Well, in order to do that, so let's say I need to fix some notation. So k is the curvature of alpha as a curve in R3. Well, it cannot be anything else, but I just want you to... I mean, there is no conflict of notation. I mean, k... Little k is nothing else, so I mean, it's the curvature in R3. And we call cos theta the number nn. So now here there is a conflict of notation. So I have to explain to you, because remember, when we look at... In fact, let me draw a picture. We have our surface s. We have a curve. We pose this, our kind of central point p. In fact, that's a very bad place where to put p. We have a point p. Now, here there is a problem of notation, because if I study the surface, here I have a normal vector. But with the same notation, you could say I have a normal vector to the curve. And we called it capital N, unfortunately. So now we have to distinguish the two, because there is absolutely no reason why capital N... I mean, the normal to the surface and the normal to the curve should have anything to do at all. Okay? So... So in fact, presumably, if I imagine the alpha b moving... Alpha of t moving in this direction, I would say the normal looks more or less here. The normal to the curve. Okay? So, in fact, here I define so N normal to the curve to alpha in R3, okay? Meaning, take your notes of the beginning of the course and change the notation, because little N will be the normal to alpha. And capital N is the normal to the surface. Okay? And we call it the angle between the two. These are unit vectors, both by definition. So the scalar product is really the angle... Is the cost of the angle between them. Of the oriented angle, okay? Well then, if I, in some sense, kind of algebraically project the curvature of alpha to the curve, I define a new function that I call capital, sorry, k little N which is just k cos theta. I mean, what? And this is called the normal curvature of alpha. And of course, this depends not just on alpha, but on s. Okay? Because to define this function I need s, I need the surface because I need the normal vector. Okay? And this is the projection of the curvature to the surface. Okay? So pictorially, what's going on? Well, that means more or less... I mean, this is the normal line. Okay? If I imagine to take the projection, the orthogonal projection of the normal to the curve to the normal to the surface. Okay? Well, in fact, I should draw which is k N. So instead of looking at N I look at k N and then I project to the normal line to the surface and the normal curvature is exactly the length of this vector here. This is linear algebra. One observation, remember, N as usual, so the orientation of a surface is not uniquely defined. I picked one N you could have picked the opposite N. What does it happen to the normal curvature if you choose the minus one? The minus this one? It changes sign. Okay? So, in fact, when I say the length of this vector I mean the oriented length. So, for example, in my picture this would be negative. So, this is the end of the definition. Now suppose since it's a regular curve we started with a regular curve we can parameterize it by arc length and we give it the usual name S to the arc length and so we can restrict the normal vector the normal vector field to the curve alpha. So N, which will depend in general by two parameters because it's defined on the surface I restrict it to the curve. That means for every point of my curve I look at the normal to the surface restricted to alpha. Okay? Let's try to get some interesting equation. Well, what do we know? We know that N is normal to the surface and alpha is a curve on the surface. So in particular alpha prime is a tangent vector to the surface. So, the thing I know is that N of S, scalar product and this is true for any S not just at zero. N is everywhere a normal and alpha prime is everywhere a tangent. So this is identically zero. Okay? Every time I see something like this I take the derivative and see if the derivative of this equation is telling me something interesting. What do I get? I get the derivative of this. So this implies N prime and then I drop the of S. These are all functions of S. N prime alpha prime plus N alpha double prime is equal to zero. Okay? Let's freeze this equation for a second and now let's try to compute the second fundamental form at p applied twice to the vector alpha prime of zero. In fact, from now on in this time I write, so if you see a quadratic form with only one vector inside it means you apply twice to the same vector. Okay? Just to short a little bit notation. How much is this? Well, this is by definition minus dN at the point p applied to the vector alpha prime scalar product alpha prime. This is the definition dNp at alpha prime of zero by the definition of the differential so, here, what should I do to compute this? I should take alpha oh, by the way, I have alpha passing through a point p oh, I have it with some tangent vector alpha prime oh, and take the derivative of the restriction of the composition but I've given it a name to the composition so this is nothing but N prime in this notation. So this is minus N prime at zero scalar product alpha prime of zero and now I use the frozen equation N prime which actually was true for every s but now I'm using it only at zero N prime scalar product alpha prime is actually minus N alpha double prime so this is minus equal to N at zero but N at zero means N of p alpha of zero is p so N of p scalar product alpha double prime at zero ok but how much is alpha double prime? now we go back to the theory of curves this is the second derivative of alpha is a curve in R3 and this is parameterized by arc length so alpha prime would be t and alpha double prime by definition is KN it's the definition of K and N simultaneously so this is with plus or minus plus and scalar product K of p K of how do I want to call it K of alpha of zero is p so K of p N, little N and now this is the new notation not to make confusion so the normal to the curve at the point p but what is this? well of course the scalar product is linear I take the curvature out and I'm left with the scalar product between the normal to the surface and the normal to the curve which is here but we have just given it a name is the normal curvature so this is exactly KN the normal curvature to the curve at p so that's it this picture is telling us exactly what is the second fundamental form from a geometric point of view so summarizing if p p is any point on the surface but if the only thing really we have assumed alpha is parameterized by arc length meaning that alpha prime is a vector of norm one so if I want to draw a general conclusion I say if v is a tangent vector at some point with norm one then I can apply this machine and then the second fundamental form at the point p applied twice to v and now I start using the short cut notation is equal is equal to let's write it in English even though becomes long is equal to the normal curvature to the normal curvature of any and here this is why it's worth writing down any curve of any curve passing p with velocity so you see once you write it in English you realize that here there is a theorem it's not just a geometric interpretation but there is a simple corollary out of this which is interesting and quite surprising in some sense which is what if you give me a curve I go and compute its curvature its normal curvature but then the output does not depend on alpha the output depends only on p and v and the surface of course so that mean the corollary implicit in this statement is that any curve passing through this point with a given velocity has the same normal curvature of course this is not true for the standard curvature I mean a point and the velocity does not determine the acceleration otherwise the theory of curves also mechanics would be an empty subject you can have curves passing through some point with a given velocity and having all possible second derivatives you cannot reconstruct the second derivative out of the first at a given point but this is true for the normal curvature this is actually this was the way all these things were stated and it's called Euler's theorem but now we can push it a little bit further because you see now here I didn't draw v so v was some vector here tangent at that point 1 and now let's suppose to every v in the tangent space of norm 1 now how many vectors of norm 1 I have on the tangent space I have a circle because it's a plane so it's a circle it's the circle of radius 1 now remember this is something related to what we saw in 3 dimension in the example in 3 dimension but certainly you know it because of the quadratic form actually have this simple interpretation so they are the critical points of the value of the quadratic form restricted to the sphere in the vector space this is what we saw so remember when we picked a 3 by 3 symmetric matrix and we looked for the eigenvalues that's what we essentially did because we picked minimum and the maximum of the associated quadratic form remember we had a and we looked at the function aavv this was our f and we looked for critical points of this that's exactly what the general picture eigenvalues are the critical points of this is the quadratic form associated to a so if I use this standard interpretation of the eigenvalues of the quadratic form and I apply to the second fundamental form what do I get I get the principle curvature so now in some sense what we are doing is simpler than in this example because this is dimension 2 we did it 3 by 3 now if we apply that we know that k1 at p and k2 at p have to be the minimum and the maximum ok this is smaller so this is the minimum and this is the maximum are the minimum and maximum of the quadratic form of which they are eigenvalues so that's the second fundamental form restricted to the set of vectors of norm 1 if you don't restrict it to the sphere in this case it's a circle because it's one dimension of course there is no minimum certainly there is no maximum because if you keep on multiplying by lambda you get lambda squared so you need to restrict it to a compact set so it has a minimum and a maximum and this is the interpretation but now we know that the value of the second fundamental form in one direction is exactly the normal curvature of any curve passing we re-read this statement and that means that means that if I take any curve passing through this point on the surface it's normal curvature it's bounded between k1 and k2 ok that's automatic interpretation but now now I need to so let's make another picture just of the tangent space we have this point I don't draw the surface otherwise the picture becomes too complicated I have the tangent space at the surface I'm restricting myself now I'm looking at the set of vectors of norm 1 so I'm looking at the circle in this plane in this circle there are two special vectors the eigenvectors to the second fundamental form so second fundamental form has two eigenvalues and two eigenvectors so let's suppose you have an E1 let me call it E1 and E2 and they have to form an orthonormal basis because it's a symmetric operator so let me fix just as fixing the notation 2 orthonormal basis of TPS of eigenvectors well since I prefer linear maps to quadratic forms I mean this is of minus dmp so that's the linear map which has eigenvectors ok now let me use this basis so now if I pick any vector V in this of norm 1 so given any V it will be what? it will be something like cos theta E1 plus sin theta E2 for some angle theta ok it's the basis of a plane now we have just learned that the normal curvature is equal to the second fundamental form applied to V well I mean this is the normal curvature somehow in the direction V this is by definition minus dnPV scalar product V but V is this one so let's substitute here this is what? this is minus dnP cos theta E1 plus sin theta E2 scalar product cos theta E1 plus sin theta E2 let's expand it remembering of course and E1 and E2 are not just any basis but it's a basis of eigenvectors ok so so for example of course dn is a linear map so dn of cos E1 is equal to cos dn E1 but how much is dn E1 dn E1 is by definition dn E1 I didn't say E1 is of course associated to K1 and E2 is associated to K2 ok so this becomes cos theta E1 cos theta K1 E1 plus sin theta K2 E2 scalar product cos theta E1 plus sin theta E2 but then E1, E2 are orthogonal and unit so of course this times this becomes 1 but I get a cos squared so this is minus cos squared theta K1 and this one scalar product E2 is 0 they are orthogonal this one times this 0 and this one times this one is minus because there is a minus in front minus sin theta squared K2 ok so this is telling me look do you want to compute the normal curvature in one direction the only thing you need to know is the angle of this direction that you are interested in so give me the angle with respect to E1 and then without doing any computation you know how much is Kn actually here probably I mean this is Kn of V ok in every direction there is a normal curvature and this is called actually that somehow the way Euler got it so this is called Euler's beautiful interpretation of the second fundamental form and we are pretty satisfied now now let's give another few names to some of two objects let's start attempting to understand what is the Gauss curvature because this is critical but we won't really be able to give a completely satisfactory I mean complete answer to that so we have to be a bit more humble now a point on the surface S is called so let's give names to special points on the surfaces so it's called elliptic there is a 1 elliptic if K of P is positive hyperbolic if K of P is negative now you would expect parabolic if K is equal to 0 but we have to be a bit slightly more careful because that's true parabolic if K of P is equal to 0 but we have two ways of getting 0 curvature to make a difference between the plane and the cylinder that's kind of the example you need to have in your mind the plane has 0 curvature because so the curvature is a product of two numbers the Gauss curvature is the product of two numbers so how it can be 0 it can be 0 in two ways they are both 0 or one is 0 and the other is not and this is exactly the difference between the cylinder so we want to give two different names parabolic is somehow the cylinder in your mind so K of P is 0 but one way to say not both K1 and K2 are 0 D and P is not the 0 map K1 is equal to K2 is equal to 0 means both eigenvalues are 0 that means the map is the 0 map and then of course we call it planar and this you don't have to make any effort to have a model in your mind if Kp is 0 because D and P is actually the 0 map but let me first go on because now natural question is at least do these points characterize the plane I mean the answer is no it's still no the surfaces which have planar points and the surfaces are not the plane but let's go back let's come back to this later because the list is not over if the two so there is another algebraic and in some sense already geometric for this Euler interpretation of the second fundamental form special situation but when you diagonalize a matrix special situation if the eigenvalues are all different it's kind of the generic situation if multiplicities of the eigenvalues jump you say well this is kind of a special thing well let's give it a name if at some point you have the two eigenvalues are the same so you have one eigenvalue with multiplicity 2 P is called umbilical ok so I don't know how much English umbilical means it's this one if you want a model for an umbilical point you have it looking down ok well especially if like me you have kind of a round situation that makes it even more umbilical ok now ok we will come back to that the point is let's start with this one this seems to be the most special case so in some sense if it's most special it should be easier to understand so what is a standard the simplest example of a surface where you can find umbilical points inside your body well if you start with your usual list this is the plane are there umbilical points on the plane all of them they are both two curvature is equal to zero ok but this is kind of stupid let's go to example one the sphere are there umbilical points on the sphere every point on the sphere we have already so the sphere whatever center whatever center you have and whatever radius you have in the second fundamental form and it turned out to be one over r the identity minus dmp minus dmp is one over r the identity that means that the two eigenvalues are the same and they are one over r so is made of umbilical points and you can ask well are these two situations so not that if a surface has an umbilical point then it is a plane or a sphere that would be really too much one point cannot tell you the global geometry of a surface but I mean here what's happening is something more stronger I mean if every point is umbilical on a surface is the surface the plane or a sphere and the answer is yes let's prove it if s is connected otherwise you have to argue on each connected component in principle your surface could be a plane and a sphere as far as long as they don't touch a plane union a sphere it's a regular surface ok it's all made of umbilical points and it's not a sphere so if f is connected and every p in s is umbilical then I cannot say s is the plane or the sphere because maybe it's just a piece of ok so then s is contained a plane or in a sphere ok so that's the only reason possible statement and that's true so let's prove it immediately it's going to be quick how do we prove it well suppose we have such a surface and suppose we take a local chart around some point I mean doesn't matter so let's first prove that the image of a chart as far as the proposition then we will argue that then the whole surface ok so let's take a tangent vector so let's take a w in tps being a tangent vector I can decompose into the standard basis if I have a local chart so w will be some kind of a x u plus b x v for some choice of a and b for some cases ok and how much is dnp of w dnp of w now we use the fact that every point is umbilical if every point is umbilical the two eigenvalues of dnp are the same at every point but what is it this is a transformation between a two space to a two space the two eigenvalues equal that multiple of the identity ok so this is equal to but this multiple, the factor which is k1 or k2 which are the same in principle depends on the point nobody here is telling me at one point they are the same at another point they are the same but who tells me how much are the two same I have no reason, I don't know so let me write it lambda p w there is a function which may be k1, k2 now, so in fact really the heart of the proposition is that because what does it happen on the sphere or on the plane is that this function does not depend on p in one case is 0 in the other case is 1 over r it doesn't depend on p but in principle in this statement I don't believe this free ok so in fact now in some sense I want to prove that lambda does not depend on p that's the line of the proof how much is d and p w well if w is this vector d and p w is nothing but a and u in our notation because I apply so a goes out and d and p x u is the partial derivative of n with respect to u d and v and here it's written that this is exactly lambda of p times so of course everything here is evaluated at p lambda p times a x u plus b x v but that means what you see that means that the partial derivative of n with respect to u ok so just one little this is true for any a and b so for any tangent vector I have this so I can take for example if I take a equal to 1 and b equal to 0 this is true but for a equal to 1 I get what a n u is equal lambda x u ok and if I take a equal to 0 and b equal to 1 this equation holds so n v is equal lambda x v in particular so these are interesting for two reasons in particular they imply immediately that lambda is a differentiable function because the principle was nothing and no regularity whatsoever because it's the well maybe you can argue automatically that it should be a continuous function because it's the eigenvalue of a differentiable family of maps but eigenvalues of a differentiable family of maps are just continuous not more than continuous because remember what do you have to do to compute the eigenvalues ok exercise simple in general in principle they are nothing more than continuous in fact it's simple to write down examples where they are just continuous so in this case thanks to these equations lambda is a differentiable function you see it I mean it's the factor of proportionality between differentiable vectors I mean if lambda was c0 this vector would be c0 and not c1 but this is c infinity now this is important because the way I want to argue that lambda in fact does not depend on p here I drop the lambda of p no remember here there is a dependence on p the way I want to argue that lambda does not depend on p is by taking its derivatives but first I need to know that I can I am allowed so I am allowed and now let's take the derivatives how do I take derivatives as usual here everything is differentiated with respect to u let's take the derivative of this equation with respect to v here everything is differentiated with respect to v let's take the derivative with respect to u why this is always the same trick because of course in both cases on the left I get the same thing so this is a partial derivative if I take this with respect to v and this with respect to u I get the same thing so that means that this with respect to v is equal to this with respect to u let's write it down what does it mean this with respect to v lambda v x u v sorry lambda v x u v so here of course these are vectors this with respect to u lambda u x v plus lambda x u v so there is another nice accident and I get so let's rewrite it just to see it better lambda v x u is equal to lambda u x v very nice because x u and x v are bases so how is it possible that this is equal to this it's impossible unless lambda u is equal, lambda v is equal to 0 which is exactly what I want this is at every point so lambda if you want to think back of the principle, the only principle curvature in this game because they are both the same is constant is constant on the connected component if the surface was at the beginning connected is constant everywhere now we have to argue that depending on this constant the surface is a plane or a sphere so now lambda how do we distinguish the two cases well of course we know the answer and we go back lambda is equal to 0 so now I know that lambda was a function and now it's a number so if lambda is 0 of course I have to prove it's a plane it's a piece of a plane well how is it possible that lambda is equal to 0 if lambda is equal to 0 the gauss map is the 0 sorry the differential of the gauss map is the 0 map everywhere but then this implies that n is constant it's a function whose derivative is 0 everywhere so being connected so it's constant on the connected component in particular if it's connected it's constant so lambda equal to 0 then n is constant well n constant it's a plane it's a piece of a plane that's a trivial case now suppose that lambda is non-zero let's look at here and now with the extra information that this is not depending on p what do I get out of this information that d and p minus lambda the identity is 0 but then again this implies or actually where can I get it even clearer wait a second because it's right look at these two equations this is the best place where to look now look at these two equations which of course come from there but here it's transparent x and n are two functions in two are three which have the same partial derivatives in respect to the same coordinates up to a constant that means that x and n differ by a constant well x and the suitable sample of n differ by the constant that means that x is equal to 1 over lambda which is good because lambda is non-zero 1 over lambda n more or less plus a constant vector because of course when I take derivatives I don't see constant vectors so this situation I can say that in this situation x of uv so they depend on u and v minus 1 over lambda n of uv and here remember the little speech I made about the ambiguity of n being defined on the domain of the chart or n being defined on the target of the chart but it's ok n of uv is equal to some constant vector constant vector but then what is the sphere which is the right sphere this implies that x of uv which is the point on the surface minus c which norm how much is its norm n is of norm 1 so this is just 1 over absolute value of lambda because I don't know if lambda is positive or negative so in fact it's more usual to write down the square so this is exactly the equation of the sphere of radius c and sorry of center c and of radius 1 over 1 over absolute value of lambda ok just one thing how come in our computation lambda which is the k1 or k2 in our computation it was plus or minus 1 over r plus why you should not be surprised so you could have said if this lambda is negative it's a sphere but no this is clearly a wrong theorem because if I switch n in this proposition there is no n so lambda on the same surface can be positive or negative because I pick one choice of n I get positive for example in the way we compute we got negative if we do the other choice we got positive so you see there is no k2 lambda positive k3 lambda negative it would be completely wrong 0 and non 0 these are the only two geometric distinctions but positive or negative here it would be crazy ok we have proved up to now we have proved that the image of a chart is contained either in a plane or in the sphere but then the whole since every point is contained in a chart every point is contained in a chart so there is only one thing to be worried about would it be possible on a surface to have two charts one contains on a sphere of some center and some radius and the other one contained either in another sphere of another center and another radius or even worse on a plane so I mean why these two cases these two cases were built with the local chart so if you change chart in principle you could jump from one to two and in fact two you could jump from different radius but no by continuity lambda this function lambda is defined over the whole surface and it's continuous over the whole surface ok so I use the trick the local chart trick to compute to get equations to whatever but I mean lambda cannot be one here and zero here well I mean no in principle it could but since I could join I can join on the surface I can go from one point to the other with a sequence of charts intersecting each other of course by connectedness but then see on the overlap lambda has to be constant so it has to be the same on the two sides so lambda is equal to lambda here and for the same reason lambda here is equal to lambda here lambda here is equal to lambda here so certainly I cannot pass from a sphere to a plane certainly if I pass from a sphere to another sphere the two spheres must have the same radius because lambda so now is it possible that passing from one sphere to the other sphere of the same radius I change center in principle it's possible from the way we have built things because C which is the center of the sphere is just kind of a constant of integration so think of it think about this very well now just one technical thing and actually improving the technical you have seen how important it was in this kind of things to realize that our geometric functions are regularity because usually the way to get interesting equation is to take derivatives so now that you understand because the first time you see this come on this will be certainly very nice functions and so on no in principle not so in this specific case we have argued in this way what about the regularity of Gauss curvature and mean curvature and in general k2 because here turned out to be a smooth function only under this unbelievably strong assumption every point was big I mean we used it so generally on a surface k1 and k2 which are really continuous because you compute them as eigenvalues or a family of differentiable maps are they more than continuous and Gauss curvature is it more than continuous and so on so this is kind of a technical thing but you have seen already one example where it was the crucial technical thing so let's spend 10 minutes on this problem sorry? you don't need to put complicated machinery we will prove immediately the best possible result by hand proposition k and h are always smooth functions so no hypothesis but k1 and k2 are continuous functions which are smooth on the open set smooth on the open set of non umbilical points now in some times this is kind of a dual of what we have just proved in the case of if every point is umbilical of course the open set of non umbilical point is empty so this theorem proposition is quite sad is smooth on the empty set but actually in that case we found an adopt trick to prove it's smooth ok so this is saying some kind of the dual of what we have seen before on your surface you have few points, particular points which are umbilical points they naturally form an open set sorry a closed set the set of umbilical points is always closed because it's the equation k1 is equal to k2 they are continuous functions so the set where two continuous functions coincide so if I remove it I get an open set and this proposition is just telling me well on this open set functions are also the principal curvatures are differentiable proof well since it's a local statement I mean differentiability or not depends on the behavior of the function near one point let's take a chart and let's do local computation local chart with the standard notation now we have n so we look at n which actually probably in the when it was born we call it nx but now I am getting tired of this ok I confuse the function the normal vector defined on u with the normal vector defined on s because they are naturally the same thing as long as you change the domain ok and we have the standard basis of the tangent space at every point ok so this is what we have in our hands but now since we have a basis so let's just to fix a choice of n so I take this choice of n and not minus it ok I do the computation with this choice of course the statements are orientations invariant I mean if for some choice of n this theorem is true for the opposite choice of n this theorem is true because the worst thing that can happen is that things change sign if something was see infinity the other one is see infinity so I have a standard basis of the tangent space and first fundamental form second fundamental form and the differential of the Gauss maps are all operators on the tangent space so I can represent them with matrices with respect to this basis both on the domain and on the target so let me say that the first fundamental form is represented so I use this symbol to say with respect to this standard basis is represented by the matrix m the second fundamental form will be represented by the matrix sigma and plus or minus I mean we will find it let's say minus d and p is represented by the matrix A just to fix notation well what does it mean actually since I am going to write it down anyway what is m so what is the quadratic form what is the matrix representation of the quadratic form with respect to the basis of the vector space m will be the 2 by 2 matrix where here you put the quadratic form evaluated to first vector first vector here you put quadratic form evaluated to first vector second vector here second vector first vector but it's symmetric so you get the same thing quadratic form applied to second vector in the case of the first fundamental form that means just scalar product of the first fundamental form is just the scalar product so scalar product of x u with itself so norm of x u squared here I put scalar product x u x v here I put the same thing doesn't matter the order and here I put the norm of x v squared ok now historically again Gauss for some reason gave name to these functions there was really no particular need for that but in every book you will find that e capital E is the function x u squared f is the function x u x v and g capital N is equal to x v squared so basically the matrix representation becomes e f f g now clearly these functions are smooth x u and x v are smooth vector fields so the scalar product does not disturb differentiability very well what is sigma well again Gauss decided that sigma was the matrix e f f g ok just to check that if students have enough symbols in their e f f g but what is literally for example literally will be the second fundamental form applied to x u x u ok so the definition would be this but we have already done some computations ok that tells us but we can redo it immediately if you want but this becomes equal and here is where I hope I have not forgot the minus should be with a plus N x u u so remember this by definition is minus dN x u x u N u ok and now you take the only information that you know is that N scalar x u is equal to 0 and you take the derivative in respect to u and this tells you that N u x u plus N x u u is equal to 0 so it's correct so minus N u x u is equal to plus N x u u that was hidden in a proof we did a couple of times ago ok well for the same with the same trick f which would be by definition x u x v it turns out to be N x u v and little g which would be by definition x v x v is in computation this is because this is very easy to compute ok of course you can also use the definition but I mean this formula I mean you will see why in a second but I mean it's clear you don't have to take the derivative so what is the difference I prefer not to take the derivative of N it's just psychology if you want but N is born as a vector divided by its norm division by the norm means that here there is a square root of the sum of the squares so since I'm lazy I don't want to take the derivative of the square root to the minus 1 ok in this formula N is left as it is and x gets another derivative so you need to compare this formula with minus N u x u now depending on the situation it would be easier anyway but now also these little functions e f g are smooth functions because it's again scalar products of smooth vector fields ok what is the matrix A so now A depends I mean we can write A in terms of M and sigma A I'm pretty sure sooner or later there will be a minus jumping minus M inverse sigma I think this is a plus I think this is a plus double check it next time I double check it too I'm pretty sure this is a plus in any case let's go on for a second with the minus otherwise we change sign everything at the end doesn't matter ok how much is this well this would be minus if that's true M is the capital 1 e f f g inverse times the matrix e f f g little independently whether there is the minus how much is the inverse of this well this is easy fortunately because it's a 2 by 2 ok I don't have to write down very complicated formula so this would be so minus I have to keep it 1 over the determinant the determinant is just e g minus f square times g e minus f minus f ok and then times always the same thing e f f g well but then now do the product and so if as usual A is the matrix which has entries a i j ok with i and j going from 1 to 2 you should find this formula for example a11 so what is the first entry of this matrix well means just take this times this ok and this is where no my computations are consistent with this minus ok so this will be let me first write this because it gets minus f f but there is a minus in front so this becomes f f minus g divided by e g minus f square then how much is a12 well a12 you find it by 1 2 so for the same reason it's f g f g minus f g divided by e g minus f square and a21 you have to end up with the same no ok a12 and a21 ok this will be probably must be correct sometimes a21 should be this times this so it will be f e f e well you have to do it because of course when you read it it sounds stupid it's f e minus f e what is capital and what is not so minus so it's f e capital f e minus capital e f ok divided by e g minus f square and then you get a22 which is 22 ok plus f f minus e g divided by e g minus f square ok because of course double check everything because here mistakes are very easy to do and now so what is the for example the gauss curvature the gauss curvature is the determinant of this matrix because it's the product of the eigenvalues I don't need to know the eigenvalues I don't need to compute first the eigenvalues then compute the gauss curvature I take the determinant of the matrix and that's it so k which is just that a well do the computation now my possible mistakes are irrelevant because my possible mistakes are a minus somewhere and switching these two I can tell you ok in some but in any case the determinant doesn't care so that a turns out to be e f minus f e divided by e g minus f square well maybe actually we should have done a comment sooner or later dividing by e g minus f square it's ok it's the determinant of this matrix of the matrix m now m is certainly an invertible matrix because it's the scalar problem ok cannot have the determinant zero must be ok must be and how much is h also the trace of a matrix so h is the trace minus trace of a over 2 because actually it's the average of the eigenvalues no well also this one you don't need to compute the eigenvalues you can actually compute automatically given the matrix and turns out to be 1 half e g plus g minus 2 f f divided by the usual thing e g minus f square ok just one comment everybody's doing a little bit of geometry of surfaces should remember by heart this formula I don't know anybody who remembers by heart this formula don't worry it actually takes 5 minutes to recover it and that's the best way for me to do mathematics this and actually I've never studied mathematics trying to learn something by heart by the end of this course you will have to have done this computation so many times that you will remember it it's a way to check that if you have studied enough don't try to remember if you have made 150 exercises this formula will be automatic you will learn it by brute force in any case in any case what are this formula telling us with respect to the actually this formula is more important in actual proposition but first corollary the first line of the proposition because once we have argued that e f g capital and e f g non-capital are smooth functions k and h are smooth functions ok and they don't care about umbilical they don't care about anything so why the principle curvature care well once you know the determinant and the trace of a 2 by 2 matrix you actually know the eigenvalue so you can go the other way around in general in the exercises what you really do is you compute the eigenvalues, you make the product and you make the average but here we can do it on a conceptual level we can play the opposite game we have k, we have h what are k1 and k2 so ki meaning 1 or 2 is what the trace plus or minus the square root of h squared minus k I mean meaning it's completely general this formula between the trace, the determinant and the eigenvalues of a function of a 2 by 2 matrix ok so how do I prove the second part of the proposition well these functions of course the problem is that there is a square root sooner or later a square root had to come out because in this line of proof comes out here now the square root of course is smooth as long as what is inside is different from 0 so how is it possible that the thing inside is equal to 0 well how is it possible that there is a factor the trace squared is equal to the determinant I mean the average is equal to the determinant you check it in 3 seconds this is if and only if the 2 eigenvalues are the same actually no sorry the fact that it type as if and only if the 2 eigenvalues are the same it's here I mean if this is 0 k1 is equal to k2 so you are at an umbilical point so outside the umbilical points this is smooth that's for sure now if you want to know more right sorry now so the point is that you have to convince yourself that this equal to 0 happens only at umbilical points ok so where the 2 eigenvalues coincide but standard 2 by 2 linear algebra ok and we are done for today