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.  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?  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.  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.  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


So, welcome back after the feed break. So, this section is a feedback session with a valedictory function. So, we will have a guest coming in later at 12.30 to conclude the workshop, but until that time this is a feedback session which is open for feedback. So, two rules here one is you are free to ask any question, but I will insist that you tell something about the workshop give us some form of feedback. So, you could ask a question, but in addition to this you should also comment on the workshop like how did you find the workshop if you had any faced any issues you could share us with us if you have any suggestions also you could share with us. So, whether or not you have a question related to the concepts or labs or any other thing we do want to hear some feedback from you. So, that is the rule number one. The rule number two is we would ideally like to give priority to those remote centers that have not had an opportunity to ask a question so far. So, if you have already asked questions in the past please refrain for some time before pressing the hand button to ask a question. We will like the ones who have not been given an opportunity so far to ask a question to take precedence. So, others please refrain the ones who did not have an opportunity so far please press the hand button. Good morning ma'am. On behalf of Siliguri Institute of Technology myself Suthapa Bhattacharya are really very thankful to IIT for organizing two week ISTE workshop on computer networking. Most of us having only theoretical concept of networking but this workshop provided us practical as well as theoretical concept. The concept of Bodhi tree to teach and reach faculties are understanding. The Bodhi tree reaches us with excellent materials. We also want to mention the talk of Professor Sridharayar and Professor Bhaskaran Raman. Their session was excellent also. We are really happy to know the IIT is very much concerned about to teach us thus how to teach different subject properly with different methods. The lab sessions were really very good but we expected some more details explanations on NS2 and socket programming. We faced some difficulties to solve last two days lab assignment apart from all these things we are mesmerized to your knowledge what are sharing to us and your presentations. Okay, thank you for your feedback. We'll definitely as I said it's an ongoing process. We do want to add on more material to make it more user friendly. Definitely more material on the labs is forthcoming. So yeah once we have the material we will again reach out to you all and share the material with you and thank you for the very nice feedback. My question is related to research topic. What is the research scope in fast TCP? Fast TCP. Yeah. See if you define fast TCP then there is no research scope. As I said TCP there are a lot of variants of TCP but as I said TCP is also another area that is more or less beaten to death. There are all kinds of work has already happened. Coming up with something new in that space I think is very difficult. So there's always some tweaking of something here something there but it's all incremental work but that said there is still work that happens. So for example in the recent sitcom the conference that I have mentioned there are some papers on TCP how to improve the performance of TCP in different settings. So you should read up on some of the recent papers that come up in some of these good conferences then you will get to know what is the research work that happens in TCP. But as I said unless somehow it so happens that you came up with a great idea in that space it's very difficult to come up with something new in that space since lot of work has already happened on TCP. I have another question. Quality of the practice quizzes was very acclaimed. Can we get some question from that practice quizzes in a single file? Yeah this I had already mentioned we do not want to give away all the quizzes like that because we do want people to use the platform otherwise there is no reason for anyone to use the platform. Yeah the MP4s are free quizzes are not free sorry about that. Very good morning ma'am this is RC 1249 Sri Ramakrishna Institute Technology Coimbatore Tamil Nadu myself Praveen Thraja and we attended so many works up under IIT Bombay DPMS computer programming but this computer networking is a wonderful works up because introduce a new like bothy tree it is a very useful and also it is a new technique to handle the theory as well as the practical part so we enjoyed a lot for the last one week so thank you very much for introducing this new technique ma'am and also on behalf of Sri Ramakrishna Institute Technology Constable Managing Trusty faculty members we show all the best for completing your assignment on the multimedia textbook thank you ma'am. Okay thanks a lot for your very nice feedback definitely I mean it feels good to know that the amount of effort you have put in people are appreciating it but that said if you have criticism also please feel absolutely free to share because one improves only via criticism you can't just feel good about the good things one has to feel a little bit bad about what you have done that's when you improve further so I mean this is an open remark for all the remote centers it's definitely feels good to get good feedback but if you have issues or something you like to criticize about it I'm not going to feel bad about it so please by all means share your criticisms as well thank you my question is for socket programming is it better to use the C or C++ or to use the java program which one is better so as I said you could do socket programming with either you could use java or C++ C++ gives lot of the internal details become more evident to you because java hides many of these internal details it provides just very clean APIs for you to use so it's a so I would say whatever it's more important to learn how to do socket programming so whatever you are comfortable with if you feel you are comfortable in with java just get started with it you're comfortable in C then get started with it if you're comfortable in both then I would have a slight more preference for C just because it gives lot more detail not only that if you are going to go into more advanced coding then java doesn't expose what are called raw sockets these are very useful if you are doing I mean this is an advanced concept also but certain things are not exposed via java so if you really want to use some of these things then C++ is the way to go but for a beginner either is fine whatever makes you comfortable very good morning ma'am very good morning the best thing in your workshop is your the video lectures which you have provided on the bodhi 3 but I would like to give you a bit of selection suggestion can you please embed the national anthem in the morning session of every day yeah we will give that suggestion to the people workshop staff yeah I will I'll pass that information to them because obviously you do start every lecture of appeals with a prayer so why can't with the workshop sure sure I'll pass the feedback to the concerned people thanks for that ma'am hello ma'am yes first of all I would like to thank you for organizing this workshop it has helped us in learning a lot of things just one complaint I would have is that there should have been at least one session on ns2 and socket programming so that it would have helped us in doing things better and in implementing things better and regarding today's lecture I have just one question is that is when you spoke about MPLS what information does this MPLS tag contain and how does it improve router speeds and ensure or improve quality of service okay regarding your first question about ns2 rather feedback on ns2 and socket programming the material has already been put on bodhi tree there are videos on ns2 including a demo of it as well as socket programming the expectation was that you would have watched those videos before attempting the exercises so from my side I think I did put them up so maybe majority of you or maybe some of you have not gone through it before the exercise yeah and does make it difficult if you don't go through them regarding your second question on MPLS the tag basically contains the virtual circuit identifier the label if you know the concept of virtual circuits you identify a flow with a label the same similar concept is there in MPLS except that it's not an individual flow you could identify an entire IP prefix with a label so it's more sophisticated than regular virtual circuit switching these are called forward equivalence class so you could define some such class based on some criteria and bind many flows to the same label so that is one big advantage of MPLS the speed up of MPLS that is achieved is as I said earlier typically in datagram networks you use the longest prefix match which means within your routing table which in a core router could be millions of such routing tables there may be multiple entries that match so search is the algorithmic search is lot more difficult it takes more time whereas when you use a label you could view the label as an index into your routing forwarding table so you just need to suppose if you are maintaining it as a form of an array you are basically telling this entry is at position 2032 in which case you just access it there read out the label and interface outgoing label and push it so it makes things much faster the concept of f forward equivalence class as I have mentioned is used to implement traffic engineering and thereby quality of service how it does it as I said involves lot more detail I cannot go into it in the interest of time usually the video transmission is done through VBR variable bit rates right but the nature of video is a streaming that is mostly streaming protocol or use it all why you why we use the VBR why not CBR can you explain it yeah so the nature of video is like that the so for example if you are watching a movie some scenes are very static like there is not much stuff happening whereas some action sequences for example where the hero is hitting the villains or whatever it is there is lot of motion so it is inherent in the nature of video to generate a variable bit rate stream so over time the amount of bits to capture the action on the scene are going to vary so thereby it produces a variable bit rate video but to some extent these issues are handled through this playback buffer so you do not play the video right away you buffer the content to some extent and then start to play mean the hope that some of this variability gets averaged out and you won't see this playback problems when you are playing the video because the buffer tries to smooth out some of the variability in the incoming traffic another question is is actually the we face a problem when we go for the get the contents from the site why you you didn't use the youtube as a private channel for the accessing these videos see as I said I wanted this is a multimedia textbook I wanted a structure imposed on them I wanted the quizzes to go along with the concept I want slides to be there I wanted to monitor your performance there are many things I wanted to do as part of the platform we really wanted to test it out at such a scale see what's happening so that's one of the reason in fact the reason that platform was developed is to impose such a structure in the textbook as well as to monitor the progress of the students youtube actually doesn't give any of this that said finally when we face these problems we could share the videos instead of a google drive I could put it up on youtube I just didn't do it just I mean I don't know youtube I have never uploaded videos there is some learning curve it just was easy for me to share with google drive I didn't give it much thought this is Vidya from remote center 1303 Srinayana Guruvillam College of Engineering Kerala and we have a question like in the transport layer we have different versions of TCP like TCP Renault Tahoe QBQ etc what would be the general TCP version commonly used in real time systems so typically the real time systems use TCP stack the selective acknowledgement option of TCP most of the real time systems use that regarding the feedback actually ma'am your talk about future trends in computer network was really good and some of our faculties I hope they would be able to identify areas in computer networks for their research and think fair share by Sridhar Iyer was really helpful and the flipped classroom concept I think we don't really able to implement because we are having a predefined method of teaching but as far as from your perspective the classroom was really good and the lab session I hope we got a really wonderful hands-on experience with the NS2 wire shark as well as TCP dump your writing tips by Bhaskaran Raman was also good we wish you and your technical team best of luck for your future projects thank you ma'am okay thanks a lot for your feedback first of all I want to thank you briefly for your wonderful programming training scheme and I would like to thank Mr. Iyer for the team fair and fair scheme and Mr. Bhaskaran for the team on perfect handling system and this training program for the good thing that the teachers can gain some knowledge to express their knowledge and everything to get a new thing to their life so and I would like to all the staffs for participating this thing and my question is that in during the serial data transmission for transmitting the different numbers of data if one packet will be lost and it will be transmitting through the different numbers of half node and what will happen in the receiver side and what is the role of the half node during this transmission so when you're thank you for your feedback and secondly when you're sending a bunch of packets yeah if a packet gets lost and if the packets are taking different paths what this manifests at the receiver is you could get out of order packets so for example you send one two three four you may get one two four three or some packets internally I mean some packets like two could be lost in which case you will get one four three something like that can happen at the receiver so most of the depends upon the context so if you are doing some link layer link layer protocols like ARQ at that level which normally Ethernet doesn't support so let's talk about the internet within the internet often this is handled at the transport layer so the transport layer will reorder the packets as well as recover from the lost packets through retransmissions put them all in order and then pass it to the application layer so that is being handled by the protocols within the internet to ensure that even though many such things can happen you still have mechanisms in place to overcome them at the appropriate layer which is often the transport layer I'm very much happy to attend this workshop and after attending this workshop I have a great interest to research on this topic after listening to you and different professors I'm somewhat excited to attend this program my question is ma'am how the five layers are installed in the system whenever we are connecting internet then that is transmitted to these five layers so how these layers are installed in the system so as I said all these layers except application level is a user level thing you run it as processes so you can run any application you could for example you wrote that send udp.c you executed it and you ran it so that's all being done at the application layer that is something which you could do that is what we call runs in the user space applications but anything below it which is the transport network link drivers all of them are implemented at the kernel space so these are implemented as different modules which get loaded on you could for example if you change your network card and you install some other card it can come with another driver so you load that module so that you can control so all this is in the kernel space so unless you do some kind of kernel programming it's not easy for you to get access to this code so if you want to know internally what is happening you should read some books on kernel programming then you will know what is happening internally within a host I would like to thank you ma'am for patiently hearing the questions and answering the same and I would like to give the feedback regarding the lab we are experiencing here very new aspects of computer networking and we are enjoying the packet delivery packet transformation and dropping out of the package so thanks a lot for showing us and making us aware of new aspects of this computer networking thank you ma'am okay thank you for your feedback hello yes go ahead thank you and your team for the excellent material that you have provided online and I'll show also the clarity and the quality of the video that you have provided is very good so whatever input you have given during these things actually really it has helped us to sharpen the knowledge we have and so that we can also distribute the same knowledge to the students which will help them to understand it better the concepts and even in the practicals but at the same time so expecting maybe in the future some more practical materials and also some concepts in the advanced computer networking if you can have so that will be great for us it will help us to a great extent in later point of time and also at the same point of time I'd like to ask a very general question ma'am so basically when you talk about computer networking there are two approaches one is a top-down approach what exactly in the IIT you follow for teaching the students and also what you think the best approach to go for the students to teach so that they can learn better okay thank you for your feedback definitely I had mentioned this before also we do want to add on more material to the current concepts add more labs also to the current concepts so this is an ongoing thing as and when we add we will share it with all of you that said coming to your second question related whether top down or bottom up is better I personally prefer bottom up because of two things one is application layer I feel I I mean I have this generic feeling that it doesn't have any major concepts like anything related to computer networking the concepts are at the link layer network layer and the transport layer so application layer kind of doesn't have much concepts for so when you start from the top down you spend a lot of time on teaching the different applications but it doesn't give you much concepts the other reason why I prefer to go bottom up is it also gives this very building block layer approach to teaching so you start with how to communicate just between two hosts then you build how do you communicate between tens of hosts then you tell how do you communicate between thousands of hosts which is the extended line then you talk about tens and thousands of hosts which becomes the network layer so it gives that kind of a very structure to it but that said so when we teach our own CSC undergrad students we do bottom up just because we can get to spend lot more time so when you start with the application layer what happens is the first three four weeks go there so the amount of time left for the bottom layer gets kind of squeezed when you do bottom up you spend lot more time on the relevant concepts and then you can squeeze in the application you could decide to skip some applications if you there is no time or add on if there is time so it gives some flexibility like that but when I teach so we do teach computer networks to other department students also for example metallurgy is the institute level course we offer whenever I teach them I do top down because when it comes to other department people they do not really need a lot of in-depth concepts they are coming more as a users of computer network they are more interested in how the applications work and provide just an overview of the network transport and link layer so there we spend lot more time on the application layer and little bit less time on the lower layers so that is how I do it it is a personal preference as I said up to you whatever logic appeals to you yeah mom actually if you go for a bottom-up approach so in that case basically as far as the lab is concerned so from our side from my experience I am telling so sometimes we face difficult in the lab so because if you go start from physical layer then what are the sort of lab we can conduct as the beginning for the students okay so even for that so again the way so I do it bottom-up and there is an associated lab for it so this is typically what I do the very first so as you saw even in my lectures there is an introduction that talks about OSI protocol stack the interlayer communication this is fundamental so the very first lab is more or less identical to your lab one and lab two so the ones that you have done multiplexing demultiplexing encapsulation all that is part of our lab one here the second lab again I do that as you saw I also measure I also have a concept in introduction on goals and metrics which talks about what are the goals of computer networks how do you measure delay latency packet loss blah blah so my second lab is on NS2 that deals with those concepts so which is basically they get a grounding in the introductory material they understand what delay is what losses are what is packet stream so on so forth and through the first lab which is on encapsulation they get to know the OSI protocol stack well and from that point on so the third exercise which I often give is this torchlight based communication which is basically dealing with framing error correction error detection where they have to design things on your own you could also do for example NS2 based on CSMA CD type of exercises so you could measure what the throughput is as users are increasing so on so forth so that's where link layer some of the ARQ mechanisms through simulation all of them can be incorporated as part of NS2 or as implementation the torchlight based communication then you cut to the network layer again network layer you have seen some of the labs ARQ based DHCP based we also have labs using a tool called BNU ML which lets us do bridging which lets us do IP configuration misconfiguration of IP addresses subnet mask assignment so on so forth so we do all this as a network layer and then once we get to the transport layer again we use some simulator NS2 to do the TCP level analysis TCP Reno versus new Reno or TCP Tahoe versus Reno that kind of TCP level and when we come to the application layer we basically ask them to do some project based on socket programming so that's how we go bottom up actually so if you can give a list of experiments and do some details to which to be followed as far as your TCP I mean bottom up approaches tension that will be really help us see whatever exercises I could do I already shared with you the other exercises dealing with network configuration as I said involves this VNU ML tool which is a nightmare for us itself to install so I I mean I absolutely cannot provide any support to if you somehow figure out how to install it on your own I can give you additional exercises but that installation is a very tricky installation it doesn't work very easily ma'am we are much beneficial from IIT workshop that avails at our doorstep this remote center so I was beneficial through the first workshop organized in 2009 by IIT Mumbai that is SDS since our January was started from that workshop and today's test will continue and in future going to be continued so ma'am at this moment I wish to share my my output that here in this workshop this workshop excellent in terms of different concepts we having like in fair and fair flip the platform and ma'am but the most expected from your and that that is our workshop should be majorly focused on issues in different technology like in computer network networking programming and other ma'am my question is that what are the current development in 128 bit encryption of SSL protocol and second question is that now what are the utility of RU pay payment gateway gateway recently launched by the government of India ma'am so your first question is on SSL 128 bit okay that's a security thing as I said I have very minimal background in security so I don't want to handle any security questions now your second question I did not quite get government has initiated what are you pay payment gateway payment gateway like visa and government of India launched are you pay payment gateway actually I don't know about it so I have to I cannot comment because I'm not aware of what this gateway is ma'am I would like to know that if two different wireless technology can be converged or not means if I need to send one packet from Wi-Fi and I am getting the same packet from another terminal which is having a Bluetooth then this convergence is possible or not and if it is possible then we charge the tools necessary to successfully this convergence so currently each technology for example Wi-Fi defines its own physical layer its own media access control Bluetooth has its own physical layer its own media access control so it's like two people talking in different languages so you cannot understand what the other person is telling though you could detect their energy so Bluetooth can detect Wi-Fi's energy if it is transmitting in the same spectrum but you cannot interpret any of the bits because it's following something different modulation scheme which you don't know how to demodulate so currently definitely it is not feasible for them to interact in the fashion you suggested so for any modulation technique that you use the receiver has to use an equivalent modulation technique for it to interpret the data to some extent with the use of these things called software defined radios where you configure the radios based on parameters so at the receiver for example you know you have the sender has used modulation 1 it could potentially in software load that modulation and decode it but that's slightly I mean there is work happening in that space but the radios are pretty expensive so anything currently let me just conclude that you cannot make them both talk with each other in the fashion you have discussed you try to make any mediator device which can be convert from Wi-Fi to Bluetooth from the same server and then configure both the device with the same server then it is possible or not so it depends on what you are doing for example you want to send a packet from Wi-Fi to some other user who is sending a Bluetooth so you could send this Wi-Fi packet to another intermediate host which has a Wi-Fi interface based on which it will decode the packet it will send up the protocol stack and then you are again using this intermediate host has two interfaces one is Wi-Fi another is Bluetooth then it will again come down and go out via the Bluetooth interface that is feasible so it's something like in the common language you have a translator who understands both languages who is sitting listening translating it and passing it on the other end for that which tool is useful there is no tool you have to write you have to do the implementation you have to set up a host that can have two interfaces there is nothing readily available like that you have to set it up so we have here professor Patak as well as professor Gaitonde so I will request them to please come here and give some concluding remarks of the workshop good afternoon and I am very happy to welcome professor Gaitonde who has been my colleague for almost four decades and with me he has been a great participant and supporter of this T10KT program and with me he is also embarking on a new journey of offering some of our courses as massive open online courses first through EDX and maybe later on an Indian platform so welcome professor Gaitonde please have a seat before handing over to professor Gaitonde and requesting him to say a few words as a valid address I thought I would like to go over to a few remote centers myself but please don't ask me any questions on networking that is the domain of professor Kameshwari but I would like to have any general feedback about the workshop and how you have benefited you can go to two or three first of all I would like to say that I am removing all the hand raises that have been raised so far and I would like to request only those remote centers which have never interacted with us at all during this workshop never interact they did not get a chance to interact at all and I would like only those remote centers to raise hands and then we will go on first come first serve basis so I am requesting my colleague to remove all raised hands now and I request only those remote centers which are never interacted to raise hands can you can you clear up Reenaatji Institute over to you for some comments good afternoon please go ahead good afternoon sir first of all I want to thank IIT Mumbai this workshop was very fruitful as well as the students we got a lot of knowledge about networking especially the board triple platform as flip class flip classroom is very active we will definitely implement this concept from this session thank you sir thank you go to the other centers good afternoon sir good afternoon on the behalf of Maman IIT Bhupal I am Sanjeev Sharma we are very thankful to you sir for organizing such a very great kind of workshop for us especially I am thankful to you to provide the material that was very good quality material and the approach that you are provided a practical approach for the capital networks because before this workshop we are not aware of such a kind of practical approach in the capital network so your quality of the material was very good and the course was very very good the assignment that you have provided was very good and especially this workshop provides us to interact with the experts and all of the remote centers are interacting and they are raising the questions so it is also clearing our various doubts regarding the subject so we are very thankful for organizing such a kind of workshop sir thank you thank you I will just comment briefly while he goes to the next remote center you mentioned that the approach was very practical well first of all Professor Kameshwari is to be complimented for making a very rigorous set of tools available to all the remote centers where you can do experiments but I would like to mention that there cannot be good engineering education unless there is a lot of hands on and what you mentioned raises some concern in my mind from old times that in many of our colleges most subjects the teaching concentrates on theory teaching whereas corresponding hands on experience in the lab is absolutely vital for students to understand proper problem solving so I am glad that you mentioned that and I hope you and all other colleagues across the country would actually adopt this approach in their teaching let us go to the next remote center please this Thakur College of Engineering the sessions were very good especially the lab sessions we got a very good knowledge and the videos which were uploaded were very nice and the quizzes were a little bit difficult but if you try to attend you can go for it and thank you very much so the observation by the lady that things were a little bit difficult I will tell you the approach and philosophy of IIT education and I feel no reason why the same cannot be adopted everywhere else our students are routinely given only a chance to answer simple questions our examinations also contain simple questions these fail to challenge the minds of students it is wrong to assume that if the students do not have adequate preparation they will not benefit from attempting to solve difficult questions this is another advice I would like to give to all the colleagues that whatever question papers that you said whatever quizzes that you said please ensure that there are at least some very challenging questions including some which even you cannot solve easily that is the way to challenge the minds of the students and you will find unbelievable improvement in the general learning of all students thank you for that comment let us go over to center 1271 Jawaharlal Nehru national college yeah over to you good morning sir regarding the tps implementation is somewhat tough to us implement as per our syllabus I govern by vtu and the technical education board tps implementation is not implementation is somewhat tough as well as coming to a classroom not going to work as we have but outcome based education so all the things are we need to have another option for these kind of things and apart everything was fine we have worked too much good too much well in the labs as well as I missed on all thank you sir thank you very much my observation is that independent of which method of education your university follows or independent of the scheme of teaching learning useful material in any subject can always be used innovatively by the teachers to benefit their students so please do not tag any of our efforts or any other efforts in the world as being compulsorily attached to a specific model of teaching and learning it is what is important is all the basic notions are discussed practical implementations are discussed students are challenged with the problems and then no matter what kind of education system you follow you should be able to use these basic building blocks in your own teaching thank you so much let's go over to another remote center sir this is Manoj Yadav from Dronachar College of Engineering we are thankful to IIT Mumbai who given us such type of important things about the computer networks and our participants they have learned a lot from this workshop thank you thank you one and all sir thank you so much remote center one three four two good afternoon sir am I audible yes you are please go ahead sir first of all I would like to thank the team IIT for conducting such a very nice good workshop although there is very good quality of the content moreover I would like to ask that there is one scope about the improvement here there is the video lectures that you had played the audio quality was not appropriate second thing all the live sessions the quality of all the live session is very good and I would like to request you kindly provide us the video of all the live sessions that held during this five days workshop if it is possible kindly upload on the google drive most important a very good huge round of applause for all the experts especially Dr. Deepak Fatak, Madam Kameshwari, Dr. Ayur, Dr. Bhaskaran and a very very special thanks to the technical team of Mr. Sushant so thank you very much sir thank you huge I would like to ask all my participants to have a huge round of applause thank you so much we will be uploading these videos not on google drive but there is a standard site on which all these videos are uploaded I will request Sushant to make note of the fact that not just the lecture videos but the interaction videos should also be uploaded this will be done in due course of time as you are aware we always release all contents in our T10KT workshops in open source so that not only the participants such as you but even other students and teachers can benefit thank you very much for that suggestion we will go to one or two more institutions J.S. College of Engineering Kalyani a very good afternoon sir this is Soham Sengupta from J.S. College of Engineering remote center 1335 so first of all I would like to thank on behalf of all the participants from the center to IIT boy and especially the speaker professor commissary for holding such a big event and which was really I felt it was needed because for the last 10 years of my teaching I have been digging into networking and everything so I have always felt the need to make the students feel the necessity of the what should I say that the TCP IP stack they feel the stacks are just the figures on the textbooks so what I do is quite matched with madam's you know frequency so what I make my students do is to explore themselves the stack by using worship tool and also I use them to get on to hands-on with a wind PC cap uprooted by Intel on which they can use a socket or a half web socket programming which makes the underlying things so thank you so I'll brief up things things that could have made this event a bigger one and more prosperous one first had we got chances to work hands-on with worship not only worship but also packet generation tools like it's cap and second which had been better if we could get some more time to know get in the into the programming of a kernel of a Linux to make a custom TCP stack and third was I felt this AV software is sometimes making a lack of synchronization because I think the software is using web socket technology which is not properly synchronized to use the user interface and that's all from my side as I hand the mic over to my colleague. Hello I want to stress that this workshop was extremely well organized and we all congratulate you and thank you for that. My next request would be to continue a line of communication between the centers and the coordinating body so that if we have any other queries in future we can always raise that in a forum. Secondly we would like to request if there is any scope for collaborative work between the centers and IIT Bombay under your mentorship. Very good points all I just briefly commend you see if you are not able to work on wireshark sufficiently it would be true about many other tools and that's the whole point the whole point is that each one of you when you go back to your own respective colleges don't stop that is where you start working actually get a group of students who are interested introduce new lab practical sessions introduce new BTEC projects or BE projects and continue to work with them in fact working in isolation you will discover many facets on your own which will be both challenging and more rewarding of course you can always interact and to answer to the question from the point from our other colleague we do propose to keep the moodle on for quite some time however you must appreciate that the moment this workshop is over like you professor Kameshwari also has about 1000 other things to do don't forget that she was awarded the excellence in teaching award of the institute which is a very rare award and which means that once the semester starts 24 by 7 she will be working on that I am proposing to the institute and to the ministry that we could keep a set of teaching assistants regularly attached to the moodle over coming semester in a year who could perhaps answer these questions who could perhaps classify them and some unanswered questions could be forced up to professor Kameshwari who could then release a single video lecture for anybody to watch where she might want to answer those questions do you think that makes sense yeah I do agree that this communication should go on further because I am also interested in the space if there is anything you could potentially use so we will look into seeing maybe moodle is the way forward so you could post things there they'll be brought to my notice and I will get back to you based on what the question is and what the requirements are I'm personally interested in it I would like to add more concepts to the video add more lab exercises to it and if as you as a remote center we are also collecting this post homework where you're supposed to create resources so during that process if you have certain ideas and you want to create additional labs and come up with that we could collaborate on that aspect as well so the communication line definitely from my side is open though it may not be a direct communication line but I'm interested in the space so any questions I will definitely pay attention to them whether I will act on it is will be a function of my life at that point what priorities I have but I'm definitely interested so if it's something that is interesting I'll definitely interact and collaborate we are going to Don Bosco Gohati first of all I would like to thank the entire team of IIT Mumbai for you know conducting this tremendous practical oriented workshop and also I would like to thank our remote center Don Bosco and thank to Kameshwari for you know designing such a good and very helpful course for the teachers thank you all thank you so much we just go to one more remote center and then we'll close this over to you. Hello sir. I'm Ajita from Moms & Queens Tamil Nadu and we are very thankful to you and your teams especially ma'am professor Kameshwari ma'am she introduced it some of the it was very well and and she today morning she went on with some modern trends in networking also that was too good and we would like to have something sir we'd like to have an open source tool for cloud implementation very interesting that you ask this question IIT Bombay as we speak is involved in implementing an open source cloud it's called the open stack model there are a few implementations already around the world as I said as we speak we are trying to implement it in fact the MOOCs platform for Indian usage that we are adopting from EDX is likely to be installed and deployed on an open stack model so that is the open open source cloud for you as we learn more about it we'll share that information also thank you very much for bringing it up in fact I would like in next six months to one year each college that you represent could have a small set of servers where you implement a mini or micro cloud and study all utilities on that micro cloud which are available in open source and use it extensively for your own activities thank you so much we'll now close this interaction thank you for your very positive feedback you have taken note of the points that you have raised first of all let me introduce professor Gayatunde I briefly mentioned that he has been working with us for a long time and more importantly he has been participating in all of our teacher training workshops in fact it was thanks to his initiative that a large number of other colleagues from mechanical engineering and other engineering departments started participating making this whole T10KT effort a very inclusive effort across the engineering departments and spectrum I'll now request professor Gayatunde to say a few words thank you professor Fatak as professor Fatak said I have been rather very willing guinea pig for his experiments particularly when they extend beyond his core competence of computer science and engineering and he knew this because we have been associated with each other since maybe 1985 or so when we really started working and both of us have a common reputation is that when asked to do something we will initially will not say no we will do it at least once very faithfully very sincerely and as properly as is possible within the given resources and time constraint second time when asked for we may say no saying I have done it once I have shown how it is to be done we will not do it again however when it came to the NMA ICT or Ecolovia project I have not said no to a second time I think professor Fatak is doing something third or fourth time because this has been a really enjoyable experience particularly because it was the empowerment of teachers that was the main objective and although the main objective was the empowerment of teachers the derived or derivative objectives are maybe better students come out hence better engineers come out and finally those students who graduate from better empowered teachers will be better engineers better postgraduate students and finally maybe the next generation of better faculty members in many engineering colleges so there is some sort of a loopy purpose here one of the comments which was made directly or indirectly during the interaction took me back to my thermodynamics courses there the complaint was in the feedback that sir we know now what thermodynamics is all about we understand it very well we know how to learn it and how to teach it very well but this doesn't really help in our students clearing the exams which they have to write for getting the degree from the university I think here we have to do a bit of re-meaning of the empowerment the teachers are empowered not only to do better teaching this teacher should also be empowered to put pressure on that subgroup of teachers who are members of boards of studies and various university committees see the university syllabus and the examination system is pretty old and by today's standard pretty old means maybe even 5 or 10 years old is not just out of fashion things are changing so much 10 years ago the power of ICT was not really appreciated today it is appreciated so use this empowerment to not only do good teaching but also to canvass and pressurize your colleagues who are on boards of studies and other university committees to get the syllabus and the examination system changed appropriately because it is only you can do it just by listening to these courses and studying some subjects in a modern way and a better way is not the end of it since it is expected that you will use this for betterment of your teaching learning processes and even educational administration I am told a survey is being planned and all of you will be partners in that survey not only the teachers who have registered for this particular course but even maybe about 1.5 lakh teachers so far who have registered 85000 teachers so far who have registered for some course or the other in this series of TKT and T10KT programs thank you thank you so much I really appreciate a large number of colleague teachers from various colleges who have participated in this workshop to make it successful I take this opportunity to thank all of them take this opportunity to thank all the remote center workshop coordinators and the RC coordinators and the technical staff there needless to add all of this will have been absolutely impossible unless our own teams working almost round the clock had made it possible to conduct this workshop in a seamless fashion so I would like to take this opportunity to thank my workshop team my video team and the other backend teams I am very sure that you will not let the momentum slip and you will complete the post workshop assignment which is supposed to be a team assignment kindly do as good a job as is possible please use this opportunity to learn something more about collaboration perhaps long distance perhaps through email but it's a very exciting experience and you'll all benefit from that thank you so much I declare the closure of this workshop now please do fill up the online survey and the other assignments in the afternoon thank you