 Tudi svoj smo zelo na teoriom kursi v spasji. Protožijo do vseho. Na vsehoj občku na vseh lektu. Vseh trafične. Tudi prejboj na vseh trafične. Vseh trafične vseh trafične. Vseh trafične je, da je to vseh treba, KoBecause the motivation for the mathematical study came from a practical problem. Tso this was the period where, of course, the human beings start learning how the word looked like and of course traveling was a key problem, going with ships to America, to India, to the Farrista. To je zelo však vsezno, da bi bilo však predniti zapunje. Tkaj, zelo je prvnega vzela v kartografiji. Zelo je predvorične se vsezno, da je neko vsezno to, da je zelo. Zela da vsezno vsezno, da je zelo vsezno. Tako je problem je da razprav chilly reproduksije počke obč成 vs. naprejven, ok. V se detak je to naša bilou, dairable. Zknutak je, da je naša bilou ca načeli v nožnosti, je, da je učetnja. Otev ni naša 1959 vsega ne dolge naša bilou policja, ok. In začetno je to dočet.ATH butf, but you would like to have up to this scale factor. Mejžering distances of your map should be the same thing as measuring real distances on the earth. Ok, so that's one thing, so distances should be reproduced faithfully. The other thing which was of crucial importance, if you imagine to being the middle of the ocean of an ocean z vsem, ki je vse boče. Vse drugi vse boče je, da je vse objevno reprezentacije v nekaj delovosti. Čekaj, da sem tukaj in na mojem mapu, vidim, da je nekaj delovosti, kako sem vse boče vse boče, da sem vse boče vse boče v nekaj delovosti v nekaj delovosti, da sem vse boče vse boče. Musim se zelo, da se čemu se boče vse boče, kako sem vse boče vse boče, da sem vse boče, ki se po vse boče vse boče, ki se po vse boče vse boče. Ko vanish, če sačeno. Da nekaj delovosti vse boče, ki sem vse boče, in jaz lepo. Zelo, ki je Gauss. Zelo, da, Prof. Gauss, to je problem, da sem zelo vas zgovoril, da je to dobro, da ga se vse ploče. In, ki je vse dobro matematik, Gauss vse vse vse ploče, in Gottingen, vse ploče vse ploče, je vse ploče, pa nekaj različ, in potem v 1827, in je to skupo, ne v boljži in vihljala u obrženju, in je bilo je zelo v povedu obrženju. Sej ta vse nabah je vzelo vzelo v povedu. Zelo, da je vzelo vzelo vzelo vzelo vzelo vzelo, če nekaj priroženje, nezelo vzelo, There is no map, which satisfies the two requirements that you want. Distances and angles. By solving this problem, he actually built the modern theory of curves. And it's quite amazing. I invite you, these are kind of night readings. But if you look at the original book, which is now available in translations in English, it was written in Latin. But now you can read it in English, if you want. It's quite amazing, really, because it's perfect. Whatever it's written there is still the best possible way to say what he wanted to say. There will be very few modifications. Of course, the theory has grown a lot since then, but the part which is there is still the status of the art. So if you want to admire one mathematician, well, Gauss for this is certainly worth your admiration. OK, so this is a little bit of history, but so what's the starting point? Remember what we did for curves? So we wanted to study geometry of curves, but we defined curves as maps, differentiable functions. Then we were really thinking of studying the image of the function. But really we had to be careful about how the parameterization was influencing our study. So what do we do? I mean, I'm making this introduction because actually today's lecture is about a definition. So today we are just going to define what is a surface, making comments and so on. Now we change a little bit perspective, because now for a surface, we are really looking to a set. So a subset S over 3 is a regular surface. So this is the keyword. If for any point in this set there exists an open set V over 3, such that this point lies in this open set. So this given point is in there. And so there exists an open set with this property. And there exists a map, which I indicate by X as a vector. X from U to V intersect S, where U is an open set of R2, X is surjective and few further requirements. First, this map X is differentiable. Again, unless explicitly stated, I will always mean by differentiable C infinity. If in some occasion we are going to study C1 or C2 surfaces, this is the only place where there is the difference. So what does it mean? X is a C infinity function. Well, you see, X is a map from an open set of R2 to some special subset, but in any case into R3. So this is a subset of R3. So really I can write it if I call UV the point in U. So this is also to set notations. The point in R2, I will denote the coordinates of R2 by U, little U, little V. Well, that means that X of UV, I can write it as a triple of functions. And now I indicate little X of UV, little Y of UV, little Z of UV. This is independent of the fact that the map is smooth or not. But now smoothness of the map is the same thing as smoothness of these maps from R2 to R. These are real valued functions, each of them. So IE, little X, little Y, and little Z are C infinity functions on U. This is another way to state the same thing. So this is the first requirement. The second requirement is that this map is a homeomorphism. Of course it's a monomorphism between what and what. I mean, it's a monomorphism if I write here V intersect S. If I put here R3, of course there is no homeomorphism between an open set of R2 and R3. So it's a monomorphism on its image. So now I've already, X is surjective on this set here. And now I require that it's also an homeomorphism. Of course by one, X is automatically continuous because in fact it's differentiable. It's more than continuous. I'm requiring it's surjective. So really what I'm asking with 2 is that it's also injective and the inverse should be continuous. So two things of the definition of homeomorphism are already here. One is surjectivity and the other is continuity. So the two missing are injectivity and continuity of the inverse. And then comes the third one, which is something we will have to comment on quite a lot. So we ask that for any point Q in this open set U, the differential of this map, dx at the point Q, so I'll remind you in three minutes what is the differential, but this is a map from R2 to R3, so you should know. So the differential of this map at this point is what? Well, this is always, the differential is always a linear map from R2 to R3. So this linear map here, I want it to be injective. Now we will spend half an hour commenting this definition, but let me end the definition with giving names to the objects that we just defined. So of course the first name is that s is a regular surface. The other important names are these. So x is called local parameterization around p, or local parameterization, let's say, or local chart around p. And v, this set here, v intersect s, section between v and s is again said coordinate chart, sorry, is called coordinate chart of s at p, or around p, I mean these are all equivalent ways. OK, so this ends the definition. So now we have also these names in our bag. And now comments, because this is a highly delicate definition. Well, actually before making inside the comment, I give you a reminder of what is this object here, of the differential of a map. Because actually probably I'm going to give you a slightly different definition from the usual one. So reminder of what is dxq as a map from R2 to R3. Well, it has to be a map from R2 to R3. So give me a vector in R2. So given w, a vector of R2, in fact maybe a little picture would help. So here it is our R2. And here we have our open set u. And inside this open set we have our point q. Now I take a vector of R2. I imagine this vector to be pointed here. Like as q was the origin of my R2. This is just a drawing thing. So suppose I take this w here. Well, of course any vector is the tangent vector of a curve passing through this point at time zero. So there exists some curve, alpha, from some little interval containing the origin. So let me call it minus epsilon epsilon into u such that alpha of zero is equal to q. And the tangent vector at zero is equal to w. Of course, if our domain is a subset of R2 as in this case there is an obvious choice which is probably what you have always done without even thinking. At least that's the way I was taught, the differential of a map from Rn to Rk. Basically take the straight line. You can take as a curve alpha q plus tw. That's a special choice. But of course there is no need to take that one. That's one choice. So I can take, certainly one is there, but I can take anything. But then remember in our definition, so I have a map x taking this open set u into R3. So I have a map here, x going into some R3 here, y and z. So what can I do? Of course I can compose. So if you want I can also draw minus epsilon, epsilon and here I have the curve alpha. So I have alpha and I have x. Of course I can take the composition of these two. And I call it beta. So put set beta x composed alpha and now this is a space curve because this is a curve defined from minus epsilon, epsilon into R3. I don't even write it. That's okay. So then I define dxq of the vector w. So here I gain a curve. A little piece of a curve here which is kind of beta. Okay. So if you want in the middle of this curve here I have x of q which is the one obtained at time zero. And then I define dxq of w to be nothing but beta prime so the tangent vector of this curve at time zero. Basically the velocity of this curve here. So this would be dxq of w. Okay. Well, all very nice, all very simple but there are two problems. First in some sense I knew, I wanted, I know that I want a linear map. Now written in this form it's not clear why this should be a linear map. Linear means what? If I take w1 and w2 for example I need to prove that dxq w1 plus w2 is equal dxq w1 plus dxq w2 Or if I take a real number lambda I need to prove that dxq lambda w is equal lambda dxq w. Okay. From this definition at the moment it's not clear. But there is a more subtle problem in this definition. This definition crucially depends on alpha. And there are infinitely many curves so that it was the straight line I can take this parabola, this other parabola I can take a curve. There are clearly infinitely many choices of alpha. If I just ask alpha to pass through this point at time zero with a given velocity. Okay. So in fact even before proving that this is a linear map I need to prove it's well defined. So that this definition does not depend on alpha but only on its velocity at time zero. Okay. You see the problems because actually we solve the problems at the same time. With the same computation we will prove that it's well defined and it's linear. Okay. How do we do it? Just a little bit of notation. We have already fixed the notation that u, v are the coordinates in R2 and x, y, z are the coordinate in R3. And this is the coordinate u and this is the coordinate v. Now, another little bit, another piece, I mean in fact not, yeah. Another little notation. Let me call E1, E2 the standard basis of R2. Okay. What does it mean? It means E1 is equal to 1, 0 and E2 is equal to 0, 1. Okay. And let me also denote by R3 the standard basis of R3 and I don't repeat what it means. So F1 is 1, 0, 0, F2 is 0, 1, 0 and F3 is 0, 0, 1. Okay. So with this notation let's do the explicit computation of how this map looks in coordinates. So let me, so of course alpha of t is a curve in R2 so it's given by the two coordinate functions u of t v of t for some given functions u of v. What does it mean? Alpha prime at time 0 is equal to what? I get it by chain rule. Okay. So this is u prime at 0 times the vector 1 plus v prime at 0 times the vector E2. Well, sorry, this is not even chain rule. This is a triviality. So the vector alpha prime is the vector u prime v prime. So then if I write it in terms of the standard basis this is just u prime E1 plus v prime E2. So this is just notation. And how does beta look like? Well, beta of t will be what? Will be x of u of t v of t y of u of t v of t and z of u of t v of t for the functions x, y and z which parameterize the map x, capital X. So let's try to compute beta prime. So here's where I need chain rule. Okay. Beta prime of t is what? And I want to express it directly in terms of f1, f2 and f3. Okay. So of course beta prime of t will be the derivative of this function in the first component the derivative of this function with respect to t on the second component the derivative of this function on the third component. Means is the component along f1 f2 and f3 because these are the standard vectors. So what is the component along f1 of beta prime? Is the derivative of the first function. And what is the derivative with respect to t of the first function? Well, by chain rule it's given by two things because I can take the derivative of x with respect to u times du dt plus the derivative of x with respect to v times dv dt times f1 on the vector on the first component meaning times f1. Okay. The second component will be the same in the sense that now we are differentiating the function y. So this is dy plus du dy du times du dt plus dy dv times dv dt f2 plus the third component which is the same thing with z. So dz du du dt plus dz dv dv dt f3 But now if I look at it I can write it down easily in matrix form with respect to the two bases. You see now I have a basis of the domain and the basis of the code of the target. So I can write it simply in this form here. So this is dx du dx dv dy du dy dv dz du dz du dv So this matrix times times du dt with the straight d and dv dt. If you make the computation row by columns this is exactly what you get here. Okay. This times this plus this times this on the first component Okay. So this is just a way to Now this is the expression at every time t. The differential of the map x is exactly this vector at time zero by definition. Beta prime of zero. So so this proves everything. This expression here tells you at the same time the two things we were looking for. Okay. Because if I evaluate this object at t equal to zero what does it happen? You see this object here at t equal to zero is what? It's a vector w No? It's w written with respect to this basis. So it's independent of alpha. So any alpha will have the same object here. Okay. And here evaluated at t equal to zero I get what? Well I get partial derivatives of the functions given which form the map x evaluated at where? At alpha of zero. Alpha of zero is q. So any other alpha would have the same q. Okay. So again this is telling me automatically that this object evaluated at t equal to zero does not depend on alpha but only on q and w. Okay. And moreover it tells me it's a linear map. Well it's written as a linear map. In fact I already found I automatically found the matrix representation of this map with respect to the standard basis. If you want this representation is given exactly by the Jacobian the Jacobian matrix. Okay. So I don't know how you learned the differential of a map in calculus. The way I learned it was always to pick the straight line. So it was exactly this but alpha was always the straight line. So here the point I'm making besides a reminder is to make you observe that this definition actually doesn't care about which curve you take. The only important thing is where it pass from at time zero and which velocity does it have. Okay. And this is crucial. We will appreciate this extension often in our course because in general on our objects there won't be any straight line. So in this case since our domain was still R2 it was okay. But for example if you want to extend this definition to a surface, to a sphere you are not allowed to take a straight line because there are no straight lines on a sphere. But you can still make sense of this definition here. Well we will come to that. So of course unfortunately I had to erase the definition of a regular surface so this was the reminder in the definition of regular surface on this map here. Well we were asking it to be injective. It cannot be better than injective because it's R2, R3. You can never hope this to be an isomorphism or anything. So what does it mean? Now that we have the matrix representation we can rephrase the injectivity condition by saying in equivalent way injective from R2 to R3 means this matrix here as rank 2 at every point. So since it's given by two vectors so injectivity is the same thing as requiring that the two vectors the vector given by the partial derivative with respect to u and the partial derivative with respect to v which form these two columns here. So if this matrix has to have rank 2 at every point these two vectors have to be linearly independent at every point. And now how to check linear dependence it's up to you, you are grown up mathematicians but a simple way if you have two vectors to detect if they are linearly dependent or not for example is to make product. So since there are only two vectors that's... Ok, so this was to extrapolate some mathematical content to one of the bits of the definition but other comments in fact the first comment must be a picture. So we are asking we are taking some subset s inside R3 so really I mean pictorially how do you see if something is a regular surface or not well what we are asking is that this set has to have the property that for every point there is an open set of R3 v ok and a map from some domain of R2 u which takes u into the intersection between v and s ok and it has to do it in a differentiable way it has to be an homeomorphism you see the gray areas must be homeomorphic via the map X and moreover the differential must be injective at every point of the domain ok so this is kind of the picture every time you think of a regular surface but for example regular surface regular surfaces contrary to regular curves with this definition regular surfaces do not have self-intersections see our definition of curves allowed nodes ok this was a beautifully parametrized smooth curve a two-dimensional object like this is not allowed as a regular surface ok so now let's build families of examples of the objects which are regular surfaces well the first observation first proposition if you want is that if you give me any function f from some domain of R2 so now from now on if I don't say anything u will always be an open subset if you take a function from an open set of R2 into R so a real valued function well there is of course and this is smooth a smooth function then if I define I look at the graph of this function and I indicate it by capital gamma of f so the graph is what is the set of points u v f of u v if I look at the graph the graph is always a regular surface this is a simple observation how do we prove it so you see I have my think of a possibility picture so here you have in fact for example this would be gamma of f for some function f defined on some possibly the whole R2 so how do I produce think of this picture and prove that the graph has this property so for every point I can find an open set and so on and the map X so how do I do it I take a point look at this what is the simplest thing you can do of course it's to project you can project it back to the domain of the function if it's a graph you can go up and you can go down this is not good as X because X must go from u to s not from s to u but this looks pretty nice so you can invert it you can take the inverse of this function which is actually the graph function again so take as a function X of u v to be exactly the function which takes one point u v on the domain corresponding point on the graph but it's easy to write it down and notice that this is a simple situation in what sense that you see a priori in the definition of a surface and we will see it's actually crucial for every point I need to produce a domain and a map here the nice accident is that there exists a map which works for every point and it's easy to prove that this map here satisfies all the properties that we want for any q on s so this maps works for everything I don't have to change it depending on the point which in principle I should in the definition it's a degree of freedom I have so let's check everything well X is certainly differentiable that's the same thing as saying that little f is differentiable X is clearly one to one so injective and surjective on the image that's the definition of a graph the graph is exact so the surjectivity if you want is the definition of a graph because the graph is exactly the set of points written in this form so this map has to be surjective but this map has also to be injective of course so we have two points if you had u and v going and u prime v prime going to the same point but this is impossible because if this is equal to u prime v prime f of u prime v prime well of course here there's enough on the first two components to conclude that u is equal to u prime and v is equal to v prime so one to one is okay so myomorphism on the image so now I know that at least an inverse exists so now I have to argue that the inverse is continuous so but why the inverse is continuous well the inverse not it exists so I am allowed to write it because of the previous observation but now this is continuous because it's the restriction if you want to the graph of the map is the rest so because it is the restriction to gamma of f of a continuous map define over the whole r3 of the map which takes xyz and goes to xy which is clearly continuous so it's the restriction of a continuous map to our set and then what else I so we have done with 0.1, 0.2 of the definition, 0.3 the differential of the map x has to be injective at every point well is it true or not so what is dxq well dxq if you want we have already, I mean this condition is now you have many ways either you compute it or you use the little observation we did 5 minutes ago so injective if and only if the two vectors dxdu and dxdv are linearly independent but what is dxdu dxdu is nothing but 1, 0 dfdu and what is dxdv is 0, 1 dfdv ok so the first two components are enough whatever f is to conclude that these are linearly independent ok well this is already a good class of examples of regular surfaces but definitely not satisfactory ok so this ends the proof one example of a surface of a regular surface which is not the graph well the prototype of such object is our little universe ok take the sphere the sphere is not a graph of a function it cannot be the graph of a function so is it a regular surface or not yes, otherwise our theory would be really bad ok now just s2 if I don't put any other symbols s2 will mean always the sphere of radius 1 and center the origin ok, if there is some need to specify center or radius I would call it something like this ok this is just one notation for so if q is equal to 0 the vector 0 and r is equal to 1 I just indicated by s2 so what is s2 s2 in coordinates x, y, z such that x2 r3 x2 plus y2 plus z2 is equal to 1 ok so draw a big picture and z so, what does the definition ask us well, can you find for any point an open set of r2 in the map which locally parameterizes the sphere well, there are of course many ways to do it the simplest way also to go back to the proposition we just proved if the point for example that we are looking at lies in the upper hemisphere we can say something like this we can construct x1 x1 of xy to be equal to the graph so, xy 1 square root of 1 minus x2 minus y2 where sorry, x1 has to be a point of r3 ok xy lies in u and I have to tell you, what is u u will be the disk in the z equal to 0 plane ok so u is the set of points xy such that x2 plus y2 is less than 1 ok, in this way this function is fine and it is certainly differentiable because I am avoiding I am avoiding 1 here ok, I am staying away from 1 so x1 is a differentiable function so by the previous proposition the set covered the image of x1 is a regular surface what is the image of x1 as the third coordinate positive ok so it is really the upper hemisphere ok so this is really x1 of u I have to play this game for every point of the sphere so now I say clearly, if I take a point which lies in the lower hemisphere I play the same game just by defining x2 of xy is equal xy minus square root of 1 minus x2 minus y2 with u to be the same as before ok, the same u and now for the same reason the image of this is a regular surface and the image of this will be the lower hemisphere so are we done? no, because there is an equator missing ok we need to cover also z equal to 0 plane how do we do it? so now we are here well, but again z was just one coordinate ok, one coordinate now let's look at y so if this point on the equator as positive y what can I do? I construct x3 of xz to be equal x square root of 1 minus z now where I want to look at this function now I will change u if you want this will be a u prime u prime will be the set of points xz in R2 where x squared plus z squared well, actually it's the same thing as before but it's always the disk it's just I'm changing name to one of the variables if you want graphically I'm making now this disk here and I'm looking at the image in this way ok but actually this domain has always to be a domain in R2 the way I put it in R3 in my picture it's my problem nobody has asked me ok and then again if the point on the equator had negative y meaning it was here we construct x4 of xz same thing with the minus minus square root ok are we done? not really we are mathematicians so we have to be pedantic we are missing we are missing two points ok because it is possible of course for a point on this sphere to have both the ny equal to zero ok meaning here and here so these two points in principle could be a problem but they are not because of course and now I use the coordinate x one will have positive x and the other will have negative x so it's clear I can construct x5 and x6 if you want now I write it as a shortcut in this form plus minus square root of 1 minus y squared minus z squared yz define on the set yz y squared plus z squared less than 1 ok and now I'm done so I also cover these two points and the whole sphere is covered now this was highly non canonical ok keep this in mind the way you prove that an object regular surface is up to you so the number of charts so in for example just to learn how to use the words that we have just defined we have proved that the sphere is a regular surface by covering it with six charts ok so for example you could ask was there a possibility to cover it with what one in principle you would ask which is the least possible number of charts well one is always the least possible is it possible to cover a sphere with one chart no why no finish your argument and then I will repeat it loudly so why this is impossible perfect that's it perfect argument the definition of a regular surface requires x to be an homeomorphism between the domain and the target so if your surface is covered by one chart in particular that means that your surface is homeomorphic to a domain of R2 an open subset of R2 by definition the sphere is compact an open domain of R2 will never be compact ok so one is impossible so you could ask was it possible to cover the sphere with two charts well we will see it it's a classical example we will see for example that it is possible ok so there is nothing magic about 6 that's what I'm trying to say and if you for some reason came up with a proof with 26 ok doesn't matter there is no a better solution as long as you prove that your object is a regular surface ok there is a beautiful theorem calculus which simplifies a lot usually the proof that something is a regular surface which is the implicit function theorem I will assume that you know it but I'll give you the statement which is more convenient for us also to fix notations again so this is a calculus theorem nothing to do with surfaces up to at the beginning kind of calligraphic o be an open set over 3 and take a point p x0 y0 x0 y0 z0 in o pick also a real number a and the function f define on this open set o into r differentiable and let's say as usual I exaggerate with the regularity because suppose it seems suppose this number was chosen to be the value of f at this given point for example and suppose that there exist at least one partial derivative f is a function of three variables x, y and z so suppose that the value is a and suppose that df dz for example but it could work any partial derivative you have to change the final output accordingly but suppose that there exist one derivative which is non-zero at this point so then there exist an open set such that x0 so which contains the point given by the first two coordinates of the point p in r2 so there exist an open set u this is automatically in r2 and an open set and an open set v in r with z0 so the third coordinate lies in this other interval v there exist an open set and an open set and so three things and a function and a function g from u to v such that u cross v so the rectangle if you want I mean this is not I never really said this with shape it has so the product open set u cross v lies in o g of the point x0 y0 is equal z0 and let me write it here the set of p in u in u cross v such that f of p is equal to a this is equal to the set xy where xy lies in u of course here I didn't say a function c infinity function now this is really a key theorem in calculus probably you have already commented it many times when you were doing when you were studying it so this is basically saying suppose you want to parameterize the set of points with a given value and you want to understand how does the set f of p equal to this value looks like so the level set of this function well if you have one partial derivative non-zero at a given point then locally basically this is saying that locally this level set is the graph of a function where you put the function you substitute the coordinate which was giving you non-zero derivative so if here you had df dx different from zero you can change the statement accordingly so this would be x naught so it will be x the variable which becomes the graph a function of the other two so the mechanism here is clear so z which is the one which is giving this property becomes the graph so if here you had df dy it would have been y who had become a function of x and z just by changing names or if you want to call it this p naught so I need one point in the level set where something happens if you have this special point then locally around this point, not globally notice that this is not telling you that the whole thing, the whole level set is the graph but around this point that you have chosen locally it's a graph you see, we have just done explicitly this procedure with the equation of the sphere if f was x squared plus y squared plus z squared minus 1 we have picked we have picked points and written the sphere as locally as graph it was impossible to write it globally but around every point we found the function g so that's exactly the theorem which is generalizing the example of the sphere no, g is unique around the point p if you fix the point your your object can be the graph only of one function so this function is uniquely determined so this is a general analytical statement let's phrase the useful corollary in our theory and in fact inside the corollary I would like to give a definition so let a a number let b be a regular value of c infinity function f so I stop a second what does it mean a regular value in case you haven't given this definition i.e. so a value a value in the image of a function it's called a regular value if p is different from 0 for any point in the inverse image of this value so to decide whether a value is regular or not I need to look at the inverse image of this value and at every point in this inverse image I compute the differential of the function and I see if it's 0 or not not is that in principle I never said that a was in the image of f so f inverse of a in principle could be empty in that case mathematicians have decided that if you cannot check a property the property is true so if something is not in the image it's a regular value in any case given a regular value of a function the corollary of the inverse function theorem of the implicit function theorem that we want to use then f inverse of a is either empty just in the case I just said it could be empty or a regular surface how do we prove it let's take a point in the inverse image where f is f inverse of a so let p x0 y0 z0 d a point in in the inverse image and by this I automatically exclude the case that this is empty so either it's empty or there is at least a point suppose I take a point but this was a regular value by assumption so I know what I know that df at the point p is different from 0 because it's a regular value but what is dfp we have done the computation in general at the beginning of this lecture for the differential of a function from r2 to r3 this is a function actually from r3 to r but the matrix representation if you want of the differential in the same way with the same definition that I gave you before and the matrix representation is given by the vector of the partial derivatives df dx df dy df dz so meaning that this vector is non-zero it means that there exists at least one of them which is non-zero so let's assume for example it's the z1 otherwise I change name to the coordinates and I rename the right one z so assume df dz at p is different from 0 otherwise I change names but then the implicit function theorem is telling us what it's telling exactly the implicit function theorem implies that s is f inverse of a f inverse of a so the intersection of s and u cross v meaning I take the u and v given by the implicit function theorem and I intersect it with s so you see this is my open set of r3 which was necessary in the definition of regular surface v the open set of r3 now it's called u cross v we don't have enough names enough letters but the implicit function theorem is telling me exactly that this piece of this set s is the graph of s infinity function it's written here but so by the first proposition we proved this part is a regular surface this part take another point maybe it will be another one the coordinate which makes the derivative known 0 but there has to be 1 because it's a regular value so also this other part will be covered also this other point will be covered by another chart and so on so it's really automatic out of the statement let's make an exercise together to extend a little bit our tank of examples quadrix for example quadrix in r3 what does it mean quadrix quadrix means you take a 4 by 4 actually so let a b a 4 by 4 symmetric matrix and define the associated quadrix the affine quadrix to be and let s to be the vectors of r3 such that if you take well and now I will use a vector here it will come automatically in column so the transpose of a vector will be a row it's just one notation so this will be a 4 line vector times a times 1 v is equal to 0 this is 4 by 4 so if I multiply on the left and on the right by a 4 vector provided it's put in the right position I get a number so why this is an example of the implicit function theorem if you want or when it is an example of the implicit function theorem this is a function f so from r3 to r so now I take any vector and I look at the function giving exactly this 1 v transpose a1v so now this is a function f from r3 to r and my plausible I mean conjecturally surface is what so s is the inverse image of 0 every time you have a locus given by an equation equal to 0 of course if you take the equation as a function you can write the set as the inverse image of that so this falls exactly in the category of things we were studying before I mean we were studying before so the question is if you want to apply the corollary the question is is 0 a regular value for this function or not because if it's a regular value automatically we have a theorem ok but what does it mean so is 0 a regular value well this is the same thing as take v in the inverse image of 0 ok and compute the differential of the function at this point and compute dfv and the question is this 0 or not f is a function from r3 to r so df is a function from r3 to r ok so to compute df that means I have to take a vector in r3 v is a vector of r3 but don't be confused I am thinking of this as the base point of my function and the differential takes a vector of r3 and gives me a number ok by definition how do I do it I need to take a curve alpha which so take alpha of t parameterized by minus epsilon epsilon such that alpha of 0 is equal to v the point and velocity at 0 is equal to w and compute what so this is by definition d in the t of f composed alpha at 0 we can make this computation explicitly because we have the explicit expression for f so that means what what is f composed alpha f composed alpha means just of t this is the function 1, now not be confused between the transpose and the time t stands for infact I should put it like this just to confuse you a bit more it's alpha of t as a vector of r3 transpose so it becomes a line times a we got a in between and then I get 1 alpha of t what do I have to do I have to take the derivative at time 0 so what does it mean of this well this is a product matrix products get differentiated like usual products so derivative of the first times everything else plus derivative of the second times everything else and so on so what is the derivative of this evaluated at t equal to 0 well of course 1 becomes 0 alpha becomes alpha prime evaluated at 0 is w so this is w transpose a and then a is a constant matrix so times this evaluated at t equal to 0 so times u1 v because alpha of 0 is the vector v plus what plus now I evaluate this which is 1 v transpose a is always a it doesn't depend on t so I can take the derivative of this I have to take the derivative of this which is again alpha prime evaluated at 0 is w so this becomes 1 sorry 0 w but this is the same thing because it's symmetric a so in fact this is equal to 2 1 v transpose a 0 w or the other one there is no reason to choose one or the other but the question was is the differential injective because we wanted to know if 0 was a regular value so how can be df v w equal to 0 now we have the expression it has to be this is equal to 0 and how is it possible this is if and only if oh sorry of course given v you expect to find many w's for which this is 0 but the question of the regular value is what given v is it possible that df v is the 0 homomorphism you see it's enough to have 1 w different from 0 1 w for which this is different from 0 that I'm happy ok so the question is is it possible that this is equal to 0 for any w given v is it possible that this is equal to 0 for any w you are being a bit too optimistic because the answer is yes it is possible so this happens if and only if by accident this object here 1 v transpose a is a multiple of the first so it's all it's lambda 0 for some lambda you see if this happens multiply by 0 w you get 0 and this is of course the only possibility because if this vector has something in the last 3 components remember that this is the 0 vector of r3 well of course I can find the w for which this is non-zero so this is the only possibility but remember that we need to check this property or whatever so df v 0 or not but not for any v but for v in f inverse of 0 and we never used it up to now so is it possible that this happens with v equal in the inverse image of 0 well if sorry if f is equal if f lies in this level set you see if f lies here in this equation here for some lambda go back to the definition of f 1 v transpose a is equal lambda 0 times 1 v this is what it's lambda and we are saying that this has to be 0 because v was a solution of the equation f so this implies that lambda is equal to 0 so putting the two things together the only possibility for which 0 is not a regular value is that you have solutions so 0 is not a regular value if and only if the equation 1 v transpose a 1 v transpose times a equal to 0 has solutions well in general this could have depending on a depending on a this could happen this could not happen there will be why do we call them quadrics because these are if you call v xyz and you expand this equation so this is the defining equation of the locus this is clearly a quadratic equation it will depend quadratically on x, y and z so it's a polynomial of degree 2 in three variables so like the sphere the sphere clearly is one of these paraboloids cylinders hyperboloids we will make a special lecture on quadrics just if you haven't if you have never seen them it's worth so for those quadrics for which this equation has no solutions we have a regular surface so exercise you will find it in the next homework but I mean I can anticipate you try to produce examples of quadrics for which this has solutions or this has no solutions especially if you can get them by some kind of geometric intuition it's even better ok I think we can stop here