 Hello, welcome to NPTEL NOC, an introductory course on point set topology part 2. Module 43, today we will present a proof of Brauer 6 point theorem which you may term as analytic proof, there are quite a few different proofs of this one and this is due to Milner. You may have come across with the following classical and famous design which we call Brauer 6 point theorem. Every continuous function from a closed disk dn to dn has a fixed point. Certainly for n equal to 1, if you take the closed interval minus 1 plus 1, it is a familiar result to you in the real analysis course. Indeed, it is an easy corollary to intimate value theorem. Maybe you have done it any closed interval AB to AB, you have a function continuous function then it has a fixed point. The general version was proved by Brauer using Lebesgue covering theory. In modern times, it is fashionable to prove this using homology theory. You may find a proof of this using simple shell approximation in my book for example or in NPTEL course on algebraic topology part 1. Milner gave a proof of this using just multivariable calculus and stone wise stress theorem. In fact, stone wise stress theorem for functions defined on a closed and bounded subset of Rn is much easier than the general stone wise stress theorem. So, in this section right now we will present a proof which is simplified version of Milner's proof and that is due to C. A. Rogers. In Hurwitz-Land-Walmans book which I am referring quite often, the authors also give a completely elementary proof quote-unquote of the Brauer fixed point theorem. However, the word elementary should not be confused to mean easy. Indeed, the concept they introduced in order to prove this theorem namely triangulation. They do not use the word triangulation at all but that is what they are doing and that is in an ad hoc manner they are doing. So that is better left to be learned properly in an algebraic topology course. So that is one reason which we are not very keen to present that proof here. On the way while learning Milner's proof, you would have witnessed you know inverse function theorem and stone wise stress theorem being applied. So that is also another motivation. So both of these things we have studied in this course itself. So the main reason to include this result which is seemingly a diversion is that it is going to play a key role in the final step of our goal in dimension theory. So that is why it is here. So let us begin with Ernest's this lemma which gives you three equivalent conditions. All of them are the statement of not a conclusion statement of Brauer fixed point theorem. Every continuous respectively C1 function from f from dn to dn has a fixed point. There is no continuous or respectively C1 function r from dn to Sn minus 1. So the codomin has changed here. Pay attention to that such that r of x equal to x for all x inside Sn minus 1 that is the boundary of dn. So such a map r is called a retraction of dn on to Sn minus 1. We do not need that word but just for your information I am telling you that. The third condition is there is no continuous respectively C1 function h from Sn minus 1 cross i to Sn minus 1 such that h of x0 is x and h of x1 is x0 a constant for all x in Sn minus 1. The third condition just says that the identity map of Sn minus 1 is not homotopic to the constant function that is no continuous means not homotopic to constant function. So in each of them I have put in the bracket with C1, C1, C1. So there are one version which is all where you have to take C1 all everywhere, other version which you have to take just continuous, continuous, continuous that is the meaning of these three are equivalent conditions. Notice that B and C are statements in negation whereas A is an assertion so that is the beauty of this approach here. So what we will do is we will prove that these three conditions are equivalent. Then in order to prove this one we can either prove this or prove this one so that is the whole idea. So let us prove that A implies B. Everywhere I am now going to take only continuous function then I will indicate what to do when you have C1 function. So assume that every continuous function from Dn to Dn has a rigid point. Now in order to prove B is not true implies A is not true that is what we will have to prove A implies B. If B is not true, B itself is in a negation that means we have a retraction namely continuous function from Dn to Sn minus one which is identity on the boundary. Take that map compose it with this alpha which sends x to minus x, antipodal map. So you are inside Sn minus one now you go back to Dn by the inclusion map that is not. What happens? What happens this function will never have any fixed points at all. See if this whole thing is f, fx is equal to x first of all means that the point x you know has to be on the boundary because fx will be on the boundary or Sn minus one. On the boundary this is identity followed by x equal to minus x. So identity x goes to minus x here. So fx cannot be minus x, fx cannot be equal to x. So that is all. So there are no fixed points for this composite function. So that is the contradiction that is denying A. So that means not B implies not A since we have A implies. Now I want to prove B implies A. Suppose there is a map F from Dn to Dn such that fx is never equal to x extend the unique line segment fx to x. See these are vectors inside the Rn. So line segment between two points makes sense. Two vectors makes sense fx to x but you take this line segment extending in the direction of fx from fx to x. So that whatever point you get it should be on the sphere. So you will get a unique point. So why this makes sense? Because the entire line segment is contained in the closed road unit disc. Maybe it is strictly inside the unit disc. So you will have to extend it. Okay as soon as you extend it because the boundary because the entire thing is a bounded set it will have to hit the boundary somewhere. Okay so hit boundary is a sphere. So this is just a logic, just a geometric way of getting this function g. In olden days that was enough for people to understand that g is continuous. Even today I can leave it as an exercise. But here because you may be seeing such things first time I will give you a full proof of why gx is continuous just because this f is continuous, x is continuous. Okay so how do you do that? Look at the entire line segment entire line from a passing to x and fx. Okay so what is the parameterization? It is t times fx plus 1 minus t times x as t raise over r. Right so this left hand side this right hand side as t raise over r they are all points inside this line. Now we want this line to be one on the sphere and from x to fx. So fx to x so it is towards x. Okay so beyond x. Maybe it is x but it is beyond x including x. Therefore I have to take the non-positive root of the quadratic equation y quadratic equation norm of the left right hand side must be equal to 1. Norm square equal to 1 will give you quadratic equation. So norm square when you write I am rewriting it we say t square times norm v square plus twice t into v this is the dot product v dot x plus norm x square minus 1 equal to 0 where v is a short form for fx minus x the vector fx minus x. You can just check that when you take norm of this function here in right hand side it means x minus x t times this and so on. What do you get in this one? Here t is the variable x is fixed x therefore fx is fixed x therefore v is fixed. So this is a quadratic with some coefficient norm v square twice v dot x and the constant term is norm x square minus 1. Okay now where does x range exceed inside the closed ball? Okay dn therefore norm x square is less than equal to 1. So norm x square minus 1 is less than equal to 0 which is a negative thing. It follows that the discriminant of this quadratic is non-negative. Okay and identically 0 if and only if this norm x square is equal to 1 okay not only that identically 0 means the other one this coefficient v dot x must be also 0. v dot x is 0 means what x and v are perpendicular to each other. So therefore what we have once is norm of fx square will become norm of x square plus norm of v square okay norm of x square plus norm of v square okay which is bigger than 1 which is observed alright. So therefore the discriminant is strictly positive and then the two roots are continuous okay. So what I am saying? Discrement is non-negative identically 0 if and only if this happens okay. If this happens there will be problem so it must be strictly positive that most we have get the two roots are continuous function. Okay whenever something is 0 when the roots are equal and so on there will be problem about continuity which root are we choosing and so on. So here the discriminant is strictly positive so there is no problem about that. So continuity of the roots follows. What we have to take? We have to take the non-positive root of it and that will give you the solution will be a continuous solution. So g will be in terms of that because put that value t okay then left hand side this right hand side will be g of x. So here the picture of and proof is completed for many people fx is here x is here and extending it towards x from fx to x and hit g. By chance if x is already on the boundary then gx is equal to x so that is the beauty of this construction okay g is like r in our statement okay that is the whole idea okay. So we have completed the proofs of a implies b and b implies a. So here again what we have proved? Not b implies not a. Now b and c are very easy look at this formula r times r of 1 minus t times x is equal to h of x t where x is on s and minus 1 t is between 0 and 1. So 1 minus t is also between 0 and 1. So this is an element of the closed disc x is an element of the boundary sphere okay r times this one will make sense for all functions defined on dn okay I am just putting h of x t if I know h of x t I could have defined r by this formula and vice versa. So this is the formula which will define either side if you know the other side. If r is given as in b when hx is continuous r will be continuous okay and if h of x h of x comma 0 is x then this just means that r of x is x that is that is what we wanted that it is a retraction okay. H is of course taking values in s and minus 1 therefore r will always take in s and minus 1 right and vice versa and finally what happens when t equal to 0 okay h of x sorry when t equal to 1 h of x t is a constant in what is that constant r of this one right. So that if you have r then h will satisfy that property and vice versa okay therefore this one formula proves both b implies c and v implies. Now comes the part where is we take everywhere c1 functions okay if this is c1 then that is c1 and vice versa so there is no problem in b implies c. Here what happens if all these functions namely fx fx is a c1 function these roots are also c1 functions in fact they are analytic functions right roots of a polynomial wherever they are you know positively defined strictly defined they are analytic functions alright. So that will take care of that wherever I have so here also alpha is also a smooth function x is going to minus right it is inclusion map so if this is c1 the composite is c1 so what we get f is with the c1 and so on okay. So simultaneously all these three statements in both c1 case as well as general continuous case are proved by the way even if you just prove c1 case here you have to argue for continuous separately we have to be done separately there is no other choice here one does not implies other you have to go through the whole thing separately but that is easy anyway. So having prepared the ground for the general theorem now the idea is to prove the smooth version of b and then use stone wise straws to prove the continuous version of a see both six point theorem we will prove like that we state and prove these two things separately no confusion okay. So there is no c1 function r dn to Sn minus 1 such that r is equal to x for all x belongs to Sn minus 1. Now this is an assertion earlier this was just a statement equivalent to two other statements now we are proving this one okay so what is the proof assuming that there is a function let us just tentatively put Sx equal to rx minus x and for each t in 0 1 let us put rt of x equal to 1 minus t times x plus t times rx joining identity map and rx this can be rewritten as x minus tx you combine with rx and write it as t times Sx because Sx is rx minus x okay just look at this formula rt is from dn to dn because rx is from dn to Sn minus 1 x is identity and this is like joining the two points you can see so it will be convexity of dn we will tell you that this is always inside dn however this difference may go out also you do not know where it is so this S is from dn to rn because I started with r as a smooth function all these are smooth functions now let us say notation here bn do you know so open this namely set of all points where in norm x is less than 1 for each fixed x inside bn take the total derivative of rt okay there is a linear map from rn to rn okay there is a total derivative okay note that the total derivative of rt at any point is the identity map of rn plus t times the derivative of S operative on it okay he is also a linear map from rn to rn so d of tr d of rt is identity plus t times d of S these derivatives are taken with respect to x okay t is also very well you may wonder what is happening rt where t is frozen here you see rt is one single map for each fixed x now you look at the function t going to the determinant of this linear map see this is a linear map from rn to rn right so you can talk about determinant d of determinant of d of rt at x okay x is fixed now one linear map is there look at a determinant but t is there so determinant of this identity plus some matrix it will be a polynomial in t of may be possibly of degree n right t occurs in each entry here okay so this is a polynomial function therefore if I define the function f from closed interval 0 1 to r by taking the integral of this function determinant of d rt okay integrated on the entire of this open ball bn okay remember remember that this function is a continuous function okay because we started with c1 function and then we have taken one derivative so derivative is a continuous function so this is just a ordinary Riemann integration okay n fold in iterated integral you may say okay so there is a variable t involved here therefore it will become a polynomial in polynomial function t because integrated integrand is a polynomial okay this polynomial we are going to show that it is a constant function what is that integral ft is a constant function then we will compute f0 and f1 separately and show that they are unequal that is a contradiction we started with something in the contrary there is no function in the statement assume there is a function so we are getting a contradiction that will complete the proof so what does it remain we have to compute f0 and f1 okay so first claim is there exist 0 less than t0 less than 1 okay such that this rt is injective for all t between 0 and t0 okay look at 0 you have no control what is r0 r0 is just x there you are lucky it is injective okay so we now have we now say that there is a some positive there is an interval of positive here in which rt is injective okay look at the function s from dn to rn is a given map right therefore there is a constant c positive such that this mischievous condition holds sx1-sx2 the norm is less than some constant times x1-x2 norm okay this is true because dn is compared you know closed and bounded and c1 so you look at the derivative take the maximum of the derivative that can be taken as t that is the general statement here this is some calculus now suppose x1 is not equal to x2 and rt of x1 equal to rt of x2 remember we have this formula right rt is x plus t times sx now that will imply that x2-x1 is equal to t times sx1-sx2 okay because rt of x1 is equal to rt of x2 so you put equality you get this one but then norm of x2-x1 will be less than t times norm of this okay t is some positive constant 0 to 1 right norm of this norm of that is less than c times norm of x1-x2 alright but then these two numbers are the same here and they are x1-x1 is not equal to x2 this is a non-zero number the non-zero number is less than equal to some number times the same number means this must be bigger than 1 okay once it is bigger than 1 I can choose 1 by c as t0 and then for t less than t0 we will never have this that is all okay so this can happen only beyond this t0 that is why we take something less than equal to t0 it will never happen that means that s is injective the rt is injective alright alright now let us go ahead okay as before bn denotes the open unit bar inside rn we claim that there exists a t1 now another constant between 0 and open 0 this time again positive constant around less than or equal to t0 we do not want to go out of t0 such that this function rt from bn to bn on the open disk is a difumorism for all t belong to 0 t1 so this is a claim now there is such a t it may not happen for all of them okay if this t1 is larger than t0 we will take a smaller than that once it is true for all that then you can take let t0 you can take t1 equal to t0 itself that is all that is not a big problem so you have to find a such a t1 that is the whole idea which is positive now look at dr of x which we have seen that identity of rn plus t than ds right therefore it follows that there is a t1 belong to 0 to and positive t1 is what I want to say then I will choose it less than t0 such that the determinant of tr of x is positive why because when you put t equal to 0 what is the determinant it is determinant of identity map which is 1 by continuity of the derivative continuity of the determinant function here actually as a function of t okay for t some positive number determinant of the left hand side here determinant of rt must be positive okay now assume that t belongs to 0 to t1 okay now t1 as we chosen is right so I say this is good enough then put gt equal to rt of bn see remember rt is just some map from bn to rn but we want to say that the interior he said dn to rn but take only the open ball the open ball goes inside the open ball is what we have to show so look at gt equal to rt of bn okay we want to show that this gt is inside bn first of all okay why gt is inside bn because t is less than 1 I have chosen to be something okay here 0, t1, t1 this one is less than 1 okay 1 minus t times x plus that is the formula you have to use that by inverse function theorem rt from bn to rn is an open mapping this is very important here it is an open mapping into the whole of rn as a map into the whole of rn okay by step 1 rt is injective also because now we are taking t1 to be less than t0 so both of them will be true here okay so it remains to prove that gt is equal to whole of bn that is only thing which we need what so far we have observed is it is injective okay it is an open mapping and so it is a homeomorphism diffeomorphism on to gt so if we show gt is actually bn you are done okay suppose gt is not equal to the whole of bn you have already observed that it is inside bn okay so if some open subset it is not the whole space the whole space is what bn another open disk that is all then there is a point y which is in the boundary of gt but is inside bn okay bn is after all convex thing so you can take a boundary point you take a point inside y inside gt join them it will intersect boundary of t somewhere and that somewhere must be inside bn already therefore you can have a sequence xn xk okay inside bn such that rt of xk tend to y okay because they are elements of gt rt of xk will be element of gt so they convert to y because y is on the boundary of this set passing to a subsequence if necessary we can assume that xk itself tends to some point x inside dn because dn is compact right so but then rt of x is limit of rt of xk right because rt is a smooth function actually so it is continuous also but rt of xk is equal to y belongs to gt because rt of xk we have assumed they convert this to y the limit is y so that is element of gt because it is rt of something okay gt is what gt is that image of rt so which is the open set but which was a point y inside the boundary that is absurd therefore gt is equal to bn so in two steps we have proved that we have located a positive number between 0 and 1 such that for all points inside this one rt is a homeomorphism on to the entire bn to bn okay now we can conclude the lemma step 2 combined with the change of variable formula for integration implies wherever it is diffeomorphism for all those this function ft must be the volume of rt volume of the whole thing volume of bn or volume of bn they are the same okay so why look at this one this is determinant of dt is what we have got what happens to dt if t is inside that interval this will be a invertible map okay because determinant is positive there so whenever you have invertible function diffeomorphism you have the change of variable formula okay so some deep not very deep but some really good analysis have been used here so volume of ft inside 0 to t1 is volume of the dn is given by ft but that is a constant so what is the meaning of this one in this non-trivial interval okay positive interval a polynomial function is a constant therefore this ft must be equal to volume of this dn okay on the whole of t1 in fact if it makes sense on the beyond observability 0 a polynomial which is a constant on a some non-empty open interval will be constant everywhere okay so we have computed in particular f0 okay f0 is a volume now we shall conclude f1 then see that it is a different value and that is a contradiction okay now look at r1 of x that is equal to rx for all points inside sn-1 okay for all x inside bn therefore if you look at rx dot rx equal to 1 that is the same thing as saying rx is of unit length right it is a dot product therefore for any v inside rn if you take the derivative of this equation that is 0 it just means that in any direction the direction derivative dv of r okay dn of direction derivative dv of r dot rx must be 0 the derivative of this one in the direction of v is nothing but dv of r dot rx plus r dot rx dot dv of rx that is the Leibniz rule okay and that is 0 so twice that means 0 means this is 0 okay what does this mean the entire image under this linear map see v is a vector dr of x operating upon v is the direction derivative so direction derivatives are all perpendicular to rx therefore this entire linear space is perpendicular to rx it is contained in the hyper plane perpendicular to rx but that is of dimension less than n this implies the linear map okay associated that as determinant 0 okay because its rank is n-1 at most it may be smaller or I do not care so this happens for all x inside here okay therefore integral f1 will be 0 because determinant is 0 the right hand side the integrand itself is 0 okay therefore f1 is here okay f is a constant f1 must be f0 but on the other side we have volume volume of the closed decor open disk is not 0 that completes the proof of the statement b for what smooth function given any topological space x and homeomorphism c from x to dn the conclusion of lemma 9.31 is valid for any continuous function c from x or before we take f equal to c composite g from c inverse dn to dn and apply theorem 9.31 so what I have done is I have just proved the c1 c1 version here now you can apply Weisstrass theorem to get the continuous version as soon as I prove any one of them what all the three statements get proved in particular Braves point theorem gets proved okay so that is all so thank you.