 Okay, so last time we learned about two crucial notions. What is a differentiable function between two regular surfaces and even more importantly, in some sense, what is the tangent space to a surface at a given point? So now, the first thing we are going to do today is a simple extension of a notion that we know from calculus, basically the differential of a map. Once we know what is a differentiable map, it's very natural to ask whether we can define the differential of a map at a given point. But that's easy. If you remember, if you go back to lecture one, well, no, lecture one on surfaces, I mean where I gave you my slightly more general definition of differential of a map, now things are very easy because if we have, so this is the definition of the differential, but it looks a bit strange because actually the symbols will be the same as we've used in the classical case. So now, suppose you have a function from S to Rn, a differentiable map. So now you know what it means. We fix a point B on the surface and we fix a vector, a tangent vector at this given point. So since V is a tangent vector, but by definition we know that there exists alpha from minus epsilon, epsilon to S differentiable such that alpha of zero is equal to P and alpha prime of zero is equal to V. This comes automatic with the definition. It's the tangent vector to a curve on the surface. So now how do we define the differential of the map F at the point P in the direction V? Well, we take the same formula we used before. This is D in the T that is the derivative at T equal to zero of the composition F composed of alpha. F composed of alpha becomes a curve, a differentiable curve with values in Rn. So we know perfectly well what it means to take the derivative at time zero. And that's it. So in another notation that we will use sometimes I just indicate it like this. Derivative at time zero. Now what is the problem? As before, as for the general, slightly more general definition I gave you from four maps between Euclidean spaces, in principle this definition depends on alpha and not just on P and V. So the first proposition we have to prove is that actually DFP of V is well defined, meaning it does not depend on alpha. And actually improving it as before we are going to get a formula which is useful for computation. So proof. How do we do it? Remember that we are always behind our shoulders. There is always the same picture. We have the surface. We have a point. We have the vector. And we have a little curve here, alpha of T, which is passing through this point with this tangent vector. But it's a regular surface. So you should add on top of this picture the other picture, which reminds you that this is a regular surface. So around this point there is a local chart. So there is an open set, a map, and some open set in R2 giving you a local parameterization of this surface. So in terms of this X, so let X from U to S be a local chart around P. Remember we proved the last time that the tangent space TPS is the image of the differential of the map X at the point X inverse of P of the whole of R2, which is the domain of the map, of the differential of the map X. So in fact just to shorten a little bit the notation, let me call the point Q X inverse of P. That means this point, there's a special point here which is the one which hits P via the map X. And of course by simple topology we can take the part of alpha whose image lies in the image of X by restricting epsilon if necessary. We don't care. So we can assume that alpha of minus epsilon, is actually contained in X of U. This is not a real restriction. Take epsilon sufficiently small. But then we do exactly what we did last time for other reasons. If we have this little piece of curve on S we can pull it back on U. We can compose with X inverse and we get some curve here. So let me call it the curve X inverse composed alpha. Now it's defined always on minus epsilon, epsilon and it has values in U. So values in R2 in particular. And of course everything is built in a way that if you compute this curve at time zero you get the point Q because you get X inverse of P. But let's do the simplest thing. Of course alpha is equal to X composed X inverse of alpha. I compose one map with this inverse I get the identity I've done nothing. But out of this stupid relationship we can take derivatives. If I differentiate this equality of maps what do I get? So taking derivative or I mean taking differential derivative at time at t equal to zero but here there is only one variable which is t so it's really one derivative. We get, let me write it here so dxq at the vector applied to the vector composed dxq applied to this vector this is equal to V. Actually I switched the equality. Alpha prime at time zero is equal to V. But if I take the derivative of this composition at time zero by definition I get this by chain rule. But the X is an isomorphism between the domain and the image. Remember dx goes from R2 to R3 so it cannot be an isomorphism just by itself but on the image it is an isomorphism it's one of the requirements of the definition of a regular surface. So since I know that of course this is in the image I can apply to it the inverse of dx. So out of this relationship I get so this implies in fact it's the same thing to say that X to the minus one composed alpha differentiated with respect to t at time zero is equal to dxq inverse of V. Let's freeze this, we'll learn this let's freeze it for a moment because actually what we really have in mind is this derivative here what is the derivative of F composed alpha. So now let's go to the how much is this? Well, what do I do? In between F and alpha I can put an X I can compose in between with X and X inverse. X composed X inverse is always the identity so if I add it or erase it nothing changes. So this is d in dt at t equal to zero of what? d of F composed X composed X inverse composed alpha but then do you agree? Sorry d in dt at t equal to zero I put it here but then how much is this? Again I use the chain rule and now apply to this compose with this forget that each of them is a composition. I look at it as a composition of these two things one and two but if I do that what do I get? I get that this is equal to the differential of the map F composed X of the first one at the point q applied to the derivative vector of the second one at the corresponding point which is X inverse composed alpha at time zero here there is a derivative sign of these two minutes before these were not lost minutes because we have a nice expression for this vector this is the part which contains F and I keep it but this one I can manipulate so this becomes the differential of F composed X at the point q and then instead of this I rewrite this formula applied let's be a bit more inverse of V so now now this seems in a strangely complicated formula but here what do we have here on this side on the left we have exactly the definition of the differential so what is really written here is that so written in another form this is the differential of the map F at the point p evaluated on the vector V is equal to that expression there let me rewrite it I don't change it I have composed the Xq dxq inverse V so in fact I can write it in terms of maps so this is for any vector V I get this equality that means as linear maps this map is equal to this map here and that's it now it doesn't matter how this expression here does not contain alpha anymore so the objection of your colleague is an interesting objection he's saying well now I removed alpha from the formula but I'm paying the price of introducing X no problem because depending on what you want to prove we are trying to prove that this is well defined the only problem coming from the definition is alpha is not X so if I give you a formula for the differential which does not contain alpha I'm done of course this expression depends on X but if I take Y I take another chart Y from another open set to a neighborhood of P on S of course this expression will change I mean this will change and this will change of course the two together no because of this but each of these piece will change but this is irrelevant for our problem so for any chart I have an expression that's okay the important thing is that this expression does not depend on alpha but only of the value on alpha of alpha and it's derivative at time zero it's a good question but the logic of the problem makes it irrelevant if this at the end depends on X or not so this expression does not depend on alpha and this ends the proof of course even without raising everything here I gave you the definition of the differential of a map when the map has a domain a regular surface and a target a Euclidean space remember last time we learned not just what it means for a function of this type to be differentiable but also what it means for a function among between two surfaces S and S1 so you might say is there another definition of differential so suppose you take now a function if your old function was defined on S but you knew the target was S1 well, but you see the point is that this definition covers also this case because another regular surface is in particular the subset of R3 so the differential of a map from S to S1 now is well defined it's a special case of this moreover and here I need at least few lines of blackboard I keep it for a second the definition so if F was a function from S to another regular surface let me say F1 same definition and I have a well defined map DFP for any P in S and for any V in TPS I know what it means this symbol here using the same thing you forget that the values are on S1 it's just something in R3 but now if you look at me closely what it's where actually is this any vector of R3 because of course the different you see here the differential of the map now goes from the tangent space to the surface into Rn in general okay this is in principle this is any vector of Rn now in this specific particular case is this vector special yes because it's what it is the velocity at time 0 of a curve of this form in particular a curve of this form is a curve on S1 alpha goes from an interval to S F goes from S to S1 the composition is a curve on S1 so then this vector where does it belong to the tangent space at which point at the point F of P of S1 okay so in general the differential of the map goes from the tangent space to the domain to the tangent space at the corresponding point of the co-domain okay in this particular case if the co-domain is a surface we have given it a name the symbol and so on now I won't prove because actually now that you have seen the definition and you have recognized that the definition is exactly the one the classical one there is only the restriction that the curve lies on S that's the only so it's a special case in some sense of the classical definition of maps between Euclidean spaces all the algebraic properties that held for Euclidean spaces holds also now what do I mean chain rule for example so if I have a picture like this I have S1 and S2 I have a map here F and now suppose you have another map from S2 to S3 you have another map G okay both are differentiable of course otherwise we cannot even talk about the differential then first observation the composition is differentiable that's something you can check immediately and you have the same formula for the differential the differential of the composition at a given point here you pick a point here okay now you look at it as a map from S1 to S3 so this object here takes a tangent vector to S1 at the point P to the tangent vector to S3 at the point G composed F of P okay but this is exactly like okay and the proof is exactly like the one you know if this was Rn, Rk, Rp write down the proof and that's it okay now we will refer to this as kind of the chain rule for maps between surfaces okay it's a kind of an exercise about this definition let me give you two examples of computations about the differential a little bit of spice to it I will draw also some conclusions so example one suppose you take a symmetric so A a symmetric 3 by 3 matrix so what does it mean? Symmetric it means I'm looking at this as a transformation of R3 R3 with the standard scalar Euclidean scalar product so symmetric means the scalar product between Avw is equal to Vaw for any Vw in R3 okay this means symmetric now one of the basic properties that you know about symmetric matrices probably you should know that they are diagonalizable with an orthonormal basis okay let's see if we can prove it at least in dimension 3 using the notion of the differential of a map it's nice it's kind of fun okay so let me define if I want to use the differential of a map I need to better define a function so what kind of function I look at I take f from the two sphere I mean if you want we center the origin and radius one okay from the standard this sphere to the real numbers taking a point P and computing AP so in particular P as a point on the sphere it's a vector of R3 so A acts on P okay and I take AP scalar product okay okay this seems like a strange definition now in fact this gives me an excuse to give you a general definition so what is a critical point of a function defined on a surface whatever wherever okay so critical point a critical point for a function I repeat it's not related just to this function okay in general function what does it mean a critical point is a point on the domain so in particular in this case on the two sphere such that the differential of the map at this point is zero zero meaning the zero homomorphism is the zero map okay now problem in this specific case are we able to characterize critical points so which are the critical points of this function well we have to compute the differential of this map okay so we take V in TP as two and we compute DFP V how do we do it well we have only the definition so take the definition we imagine we don't even write it down since this is a tangent vector there is a curve on the two sphere such alpha such that alpha of zero is P and alpha prime of zero is V okay and this is D in DT by definition of F composed alpha we have the explicit expression for F so this means this is A so this is D in DT at equal to zero of the function A applied to the vector alpha of T scalar product alpha of T okay how much is it scalar product gets differentiated like the usual product so derivative of the first times the second plus first times derivative of the second so let's do it A is a constant matrix so A doesn't matter so this becomes A alpha prime at zero times alpha of zero plus A alpha of zero alpha prime at zero and now these objects have an A this is what this is A V scalar product P plus A P scalar product V okay now A was a symmetric matrix so I'm free to choose which one I like most I can take A on the place I prefer and nothing changes I prefer this one so I put this A here and this becomes exactly the same thing because the scalar product is symmetric again so this I write it as 2 A P scalar product V okay so this was the computation of the differential of the map now how can P be a point be a critical point so now suppose that this map is identically zero so that means what does it mean it's identically zero it's identically zero means that whatever V in the tangent space at P not for any V in R3 so now here becomes the only delicate point for any V in the tangent space to the sphere at this point this object is zero but that means A P is orthogonal to the tangent space say it geometrically but what is the tangent space to the sphere we have checked it last time just to be sure that we were on the right track the tangent space to the sphere at a given point P is the orthogonal to P so put these two informations together and how is it possible that P satisfies this equation for any V so the moral is that P is critical if and only if you know if you put V is free to move in the whole tangent space so which is the only vector which is orthogonal to the tangent space at the point P zero is orthogonal to everything P now of course I said it in a slightly non-precise because of course there is not only one vector there is only one direction so P is critical if and only if this vector here is that one up to a multiple so it lies on the line I cannot be sure which multiple on the other hand multiple is not free to be anything in this case because if it is true that AP is equal to lambda P take the scalar product with P and I get what AP is equal lambda one because I am on the sphere of radius one so lambda is not free to be anything it has to be exactly this number and this number is what in our language f of P so if and only if AP is equal f of P times P well up to now algebra and this computation you have to take it as an exercise about computing differentials now to prove all this was behind everything now the claim is there is an orthonormal basis of eigenvectors for A okay why well first case if f is constant if f for some accident is the constant function well the constant A has to be a multiple of the identity in that case it's already diagonalized I mean the standard basis is the basis of eigenvectors okay so in this case A is equal if you want me identity and in this case the existence of the orthonormal basis is stupid okay F is not constant it will be kind of the general case so how do I argue well if it's not constant I just observe that the two sphere is I would like to say oh observe here maybe what is the philosophy of this exercise if I know that the function has a critical point I'm happy you see here there is written exactly that the point p is an eigenvector is a multiple of p A of p is a multiple of p hence p is an eigenvector so my question is do I know that the function f has critical points in some sense what I'm saying is that to prove that every matrix as an orthonormal basis of eigenvectors I need to show that there are at least three critical points okay so if f is constant you can check I mean if f is constant essentially everything is a critical every point is a critical point and I can do whatever I want now suppose f is non-constant who is going to tell me that there exists at least one let's start with one in fact once you get one two it's impossible to get one that's if you study algebra okay solving the eigenvalue and eigenvector equation is what you have been taught in algebra as two is compact the sphere is compact so every function has as a maximum and a minimum that's why I mean you get automatically two and you need to observe the maximum and the minimum is a critical point okay in this sense actually the proof is exactly the one you know okay there is nothing new to be proved so let's give them a name so s2 s201 is compact f is of course continuous I mean it's differentiable so let me call them and p2 on the sphere maximum and minimum for f okay so in particular critical points so if I look them as vectors of r3 p1 and p2 p1 and p2 are automatically eigenvectors of norm because they are on the sphere okay so they are automatically normalized to be of norm 1 okay so what do I need to prove I need to check whether they are orthogonal or not because remember the theorem is saying there is an orthonormal basis of eigenvectors so claim p1 let me write it in symbols p1 orthogonal to p2 well of course I should compute p1 scalar product p2 okay but let me compute this instead this little variation of this so I multiply what I really interested in p1 scalar product p2 with this number you will see now it's just computational trick this number is non-zero f is non-constant and this is the maximum and this is the minimum so the maximum and the minimum cannot be the same value so this number is non-zero okay this is important because now I will argue that this is on the contrary that this object altogether is zero so I need to conclude that this is zero okay well how do I manipulate this well I put so this is equal to p1 scalar product p2 minus p1 fp2 p2 take the first one I put it here the second one I put it there but then I go up here if I have a critical point I have this equation okay so f1p1 f1p1 times p1 is actually ap1 and this one is ap2 I know that p1 and p2 are critical points so this expression is equal to the scalar product ap1 p2 minus p1 ap2 and that's it so choose the one you like most for example put a here and you are done this difference is zero because they are the same as I said before this is not zero and so this is zero which is my claim well it's part of my claim because now what do I know I know that this matrix A has two orthogonal eigenvectors of norm 1 I miss one there should be a third one and that's what but actually you see up to now R3 was irrelevant you could have taken a symmetric N by N object and these two would be there now that's the only moment where the final part of the proof works only because you are in R3 because if you have two eigenvectors orthogonal where is the only possible third one so you just take the wedge of those two and you check that it works and it's the only possibility there is nothing else so sorry Hilbert Schmidt well I don't know how you I mean but compact and symmetric are not automatically the same thing if you want to give them a name I mean it's the beginning of spectral theory if you want but this is really the very beginning of spectral theory now another example it was fun we introduced a few concepts which are nice to know critical points and so on but really the excuse was to compute the differential of a map and to make some consideration about it let's make another example of this type with a more geometric function so now instead of taking a problem which is really a natural geometric meaning and we want to compute the differential of this map so in this case we take the distance square remember we have a little list of interesting functions so example two we have a surface in R3 a regular surface we have a point independently whether this is on the surface or not we don't care at the beginning the function f of p is the distance square from the point so f I think of f as a function from s to R in particular to the I mean non-negative R but doesn't matter now I would like to understand whether this is I mean what are the critical points of this map well if I want to know what are the critical points I need to compute the differential and to analyze the expression that I get so what is the differential of the map f again goes without saying if v means I have a tangent vector at a given point okay so it goes without saying but I say it because we are at the beginning of the story I take a curve alpha on s such that alpha of 0 is equal to p and alpha prime of 0 is equal to v okay and this is by definition d in dt at t equal to 0 of f composed alpha now I take the expression for f but the norm square of course is the scalar product between the point and itself the vector and itself so this becomes alpha of t minus p0 scalar product alpha of t minus p0 sorry this is equal to the derivative at time 0 okay as usual the scalar product gets differentiated in usual way and I get what? I get twice I know already that it's becoming the same thing v scalar product p minus p0 okay so let's try to draw let's draw a picture what's going on we have a surface I mean let's suppose the point is the point where I'm measuring distances from just to draw a picture I take a point p somewhere on the surface and I'm asking is this point a critical point well critical point as before means dfp is equal to 0 as a linear map but we have this expression so I can phrase it geometrically quite easily what does it mean that the point p is a critical point it means that for any tangent vector if I draw if I take the vector p minus p0 p minus p0 is orthogonal to the tangent space at p so for example the point I've picked here is not or it doesn't look like okay where would be a critical point more or less somewhere here there is a chance of having a critical point because if I draw there the tangent space more I mean if my picture is not okay the normal line to the surface at this point meaning the normal direction to the tangent space hits the point p0 okay so in fact let me phrase it geometrically in this way the point p is critical for f if and only if the normal line and the explanation of what is the normal line is just the picture okay is the line passing through the point p with direction the orthogonal direction to the tangent space that's the normal line if and only if the normal line at p so if you want to be a bit too precise the normal line to s to the surface at p passes through but now again simple okay the exercises is over I mean we have already computed the differential but let's play the same game we did before if s is compact so again how do I know that there are critical points I don't know in general but if s is compact I have a minimum and a maximum again there are at least two critical points and now depending of what are you thinking which fixed and what are you thinking moving because in some sense there was a parameter in this discussion which was the place where you put the point p0 so for example if you reverse our imagination now you can say the following if s is compact and if I look at the normal lines at every point you see to draw the normal line I don't need p0 I have a surface and at every point I think of the normal line then I take another point I have a normal line so for example you could ask how big is the set by the normal lines I mean is every point hit by one of these lines is every point of R3 hit by one of these lines natural quest the answer if s is compact is yes in general I don't know but if s is compact pick the point that you want to hit you want to hit this point here okay take this point construct the distance with this function and take one of these two critical points you have reversed the process so you want to hit this take it construct the function take one of the two critical points that exist and by what we just observed here the normal line that the critical point hit this point p0 okay so every point of R3 lies on one normal line at least one maybe more okay which is kind of nice I'm sorry but the piece of paper is it the whole plane or a piece a piece of the plane how can it be compact it's not a surface it's a good exercise a disc with its boundary for example if you want a rectangle with its boundary it's not a regular surface in fact and it's crucial because you see after all what are you measuring here with this function the distance so if you take a disc and you take a point here of course the point of minimal distance will be on the boundary and the point of maximal distance will be on the boundary and you cannot take them because they are exactly where the definition of regular surface fails so they will not be critical points so they are critical points only because you have taken a small piece of something but if you complete it they are not okay you can say well we have just observed if S is compact then every point of R3 is hit by a normal line other extremes somehow inspired by the sphere so what happens in the case of the sphere well in the case of the sphere if you look at the normal the set of normal lines well of course basically you are taking polar coordinates okay I mean you take lines because the usual identification not the normal vector to the tangent space is the point itself so the normal line is exactly the position vector okay that's okay so it's clear that every point is hit okay now but here there is another accident it's not only true that every point is hit but there is one point through which every normal line pass through this okay this is an accident this seems an accident is it an accident or not so now forget that this is the sphere call it S suppose if there exists a point P0 in R3 such that each normal line pass or to S so you have your surface and each normal line to S passes through the same point P0 then it is a sphere of course I don't know which sphere I don't know the center well in fact the center is P0 I don't know the radius so S is equal to in our language P0 R for some R and how do you prove it if there is a point with this property take this point build the function distance squared and what is what's going on every point on the surface is a critical point because of this every point is a critical point but then what is a function which functions for which functions every point is a critical point must be a constant okay but what does it mean it's a constant it means S is exactly the locus norm P minus P not squared is equal to something C to a constant C but that's exactly the equation of the sphere in fact here I've been a bit sloppy because maybe it's not the whole sphere maybe it's an open subset of the sphere okay let me leave it like that meaning open subset so you can phrase it as a little theorem if all the normals to a connected surface pass through the same point then the surface it's an open set of the sphere it's a simple characterization well I guess this is a good moment to stop today we have a shorter shorter lecture okay and we stop here