 Okay, so what I am going to do now is tell you few more facts about the behaviour of a function in a neighbourhood of a critical point. So let me recall what we have seen in the last lecture, so you see we have the following situation, you have the function W equal to f of z and with well z0 is a critical point of f of order m-1 and we interpret it as z0 is a 0 of order m of f of z-w0 where w0 is f of z0 is a critical value, okay. And of course we choose rho greater than 0 such that the 0 of f of z-w0 namely z0 is isolated there in a disc centred at z0 radius rho, okay. So f of z-w0 is not equal to 0 for 0 strictly less than mod z-z0 strictly less than or equal to rho and choose or let and of course we have delta is the minimum over mod z-z0 is equal to rho of mod of f of z-w0 and then what we do is that we have this diagram if you recall where we took this disc in the complex plane the z plane centred at z0 radius rho, alright and then we are looking at this map this map is into so let me use some space above because I want to draw something above. So this is into the w plane and z0 goes to a point w0 and I am looking at this disc centred at w0 and radius delta where delta is chosen like this, okay and the fact is that so the function goes like this and what is happening is that we have so and this whole transformation is broken down in such a way so that it conformally looks like z going to z power n, okay. So how it is broken down is that you have first of all a change of variable I do not know what I called it so it is z going to zeta which is h of z if so please let me know if this notation is conflicting, okay I think it was h of z and then so this goes into the in the copy of the complex plane which is now the zeta plane and the image of this disc is something that looks like a disc, okay because h where of course we have set we have made the following definitions we write f of z minus f of z0 is z minus z0 to the power of m into g of z, g of z0 is not equal to 0, okay and then we write h of z to be z minus z0 times g of z to the power of 1 by m, okay and mind you g does not vanish on this disc and therefore there is a there is an analytic branch of the mth root of g, okay. In fact there are m of them and you choose one of them and then you call that as h, alright and then you see that if you write it like this f of z minus f of z0 becomes h of z to the power of m, okay and so what you do is that you first go by h then you z0 goes to the origin because h of z0 is 0, alright. So it z0 goes to the origin and the image of this disc is going to be well you know something that looks like a disc conformally so you know I am drawing something like this, okay but I am taking a I am taking a small enough disc there and well and from here to here I make one more I apply another function which is zeta going to zeta power m which is eta, okay and you know this will go to this will take the disc center of the origin to the same disc center of the origin except that probably the radius of the disc will get you know diminished or enlarged depending on what the radius here is, alright whether this radius here is less than 1 or greater than 1 but in any case it will go to a disc center of the origin and now this function is a function that we know very well this is the power function z going to z power m of course mind you m is greater than 1 I am at a critical point, alright m is greater than 1 and you know that how you know that we have studied this already and then to get from here to here so if you come all the way from up to here it is h and then now here it becomes h power m but h power m is f of z-w0, alright so to get f of z I just have to add w0 to it so the transformation is eta going to eta plus w0 that is a transformation is just a translation, okay so this disc will get translated to this, okay so the and well if you look at this picture now it is clear that if I so you know there are m branches of so if I take the inverse functions here the inverse function will be w going to w-w0 that will be the inverse function here, alright and which is of course eta, okay and here the inverse function will be eta going to eta to the 1 by m, okay which has several branches it has m branches, okay and then of course from here to here there is a unique inverse function and that is just h inverse, okay the derivative of h does not vanish you can check that derivative of h does not vanish and so h inverse comes because of the injectivity of h and the inverse function theorem. So you know if I compose all the way here the inverse function is unique here the inverse function is unique because it is a translation it is an isomorphism this is also a holomorphic isomorphism because it is injective so here the inverse is here the inverse is unique and here also the inverse is unique the inverse is not unique in this in here because there are m branches and each of this branches if you follow it with this forward function will get the identity here, okay. So the moral of the story is that you get functions you get several functions that so you get so you know when I have to define a fractional power, okay eta to the 1 by m is exponential of 1 by m log eta so I will have to get a fix a principle branch of log, alright and of course I have to forget the origin, okay and to fix a principle of to get a function that is analytic what I will have to do is I will have to cut out the portion of the negative real axis, okay. So in this disk minus this portion of the negative with the portion of the negative real axis removed each of these functions eta going to eta to the 1 by m are analytic therefore if I compose it what it will tell you is that you know if I take the corresponding slit disk here centered at w0 then the resulting functions will be analytic, okay so you know I will get functions like this these functions are going to be Zj of they are the Zj's of w, okay and they satisfy the property that this followed by this is the identity here, okay Fj, Fj, F of Zj of w is equal to w for j is equal to 1, 2, etc up to m where the Zj's are the for so what are the Zj's I mean if you start with the value w here, okay then this value w is assumed by F at m points counted with multiplicities and it is those m points which are called as Z1, W, Z2, W and so on up to Zj, W up to Zm, W, okay. So you know this so the you get points here m points counted with multiplicities and these points are the Zj of w and they depend they will change as w changes so for every point here you get m points here which go to that point so this function F is a m is to 1 mapping or outside Z0 of course Z0 goes to W0, okay and no other point goes to W0 but the point is Z0 goes to W0 with multiplicity m, okay because that is what it means to say that Z0 is a 0 of order m of Fz-0, okay F assumes a value W0 which is F of Z0 with a multiplicity m, okay. So these so for any w here you get m values of Z, okay and these values of Z are such that when you plug them back in F you will get back the w, okay. So in other words these are solutions of W equal to F of Z, you are solving from the equation W equal to F of Z you are solving for Z, okay. So you must think of this as these are inverses of F, inverses in the functional sense, okay that is the reason I put a bracket to the minus 1 on top because strictly speaking F is a many to one function it does not make sense to talk about a set theoretic inverse but these are functional inverses, alright just like you know it is the same idea for example when you have the exponential function the inverse functions are the log functions but the exponential function is not 1 to 1 it is many to 1, okay but then we but log is functionally an inverse to the exponential because it undoes the act of it the exponential function. Similarly when you take the mth root function the mth root function is the functional inverse of the function which takes everything to the power m, okay a function which take everything to the power m is not a 1 to 1 function it is a m to 1 function again except at the origin, alright. So but to undo the operation of taking something to the power m you have to take the mth root, okay. So functionally it is an inverse taking variable to the power of 1 by m is a functional inverse and how many such inverses are there are m such, okay. So in the same way if you take any analytic function if you look at neighbourhood of the critical point what happens is that this analytic function has functional inverse there are m of them, okay and each one of them is actually analytic on a slit disc, alright but if you want to actually see them as globally as an analytic function one has to go to the Riemann surface of this f inverse, okay. So you know what happens is that we have we can construct an m sheet at covering so I will draw it like this and so on. So there is an m sheet at covering with a projection and a projection here such that on this you really have a function which I can call as f inverse and again there you see you should understand this f inverse is only a one sided inverse in the sense that you know this followed by f is identity I should say this no in fact I should say this f inverse is a functional inverse which on every sheet assumes the corresponding value fj but the only what is the advantage of this f inverse, okay maybe I will even put a bracket here so that you know you do not get confused but the point I want to say is that this f this the good thing about this when compared to these guys is that these guys are only analytic on a slit domain, okay but the fellow above is analytic on the whole Riemann surface this is the this guy here is the Riemann surface for f inverse the functional inverse, okay and we have seen this already for the case of f the exponential function in which case we got the Riemann surface of the log function which is the functional inverse of the exponential function and also we have seen the Riemann surface for z to the 1 by m which is a functional inverse for f of z equal to z power m, okay we have seen two cases but more generally here is the point is that when you look at the when you look at the exponential function you see the derivative of exponential function is again itself and you know the exponential function never takes the value 0 therefore it is derivative never vanishes, alright so somehow there since the derivative never vanishes it is locally 1 to 1, alright therefore it was easy to write the inverse function itself which is a which is given by a branch of the logarithm, okay but the point because the exponential function has no critical values, alright whereas we are now in a bad situation you are what you are saying is take a function which unlike the exponential function has critical values how even if you take a critical value in the deleted neighborhood of the critical value this function also admits functional inverses the functional inverses they all can be put together as a single analytic function on a Riemann surface, okay and they are they give the solutions for the independent variable in terms of the determinant variable. So they solve z for w, okay you get z as a function of w, okay so let me write this down so the point I want to make is that the functions zj of w are analytic in the slit disc w0-delta w0 yeah I should this is what I should throw out so it is mod w-w0 less than delta-you throw out the portion parallel to the negative real axis from the to the left of and including the center w0 so it is going to be w0-delta w0 so this is the line segment from w0-delta to w0 with the right end point w0 included and the left end point w0-delta not included this is what you are throwing out from this disc to make it a slit disc and on this slit disc each of these functions are analytic, okay and of course you know but all of them glue together to give a global analytic function, global single valued analytic function f to the minus 1 on the Riemann surface for f to the minus 1, okay and the Riemann surface is you know is obtained by pasting carefully the upper you have to take m copies of this disc the slit disc, okay but do not throw out you take the cut disc not the slit disc, okay which means you do not throw out in the slit disc you throw out the this piece but now you do not throw it out you just cut it and you do a cut and pasting process where the portions above the line segment in one disc are joined to the line segment and the portion below in another disc and you do this repetitively, okay and then the last disc you pasted with the first one, okay this is something that is very hard to visualize in three space, okay but then you can do this and you get a surface Riemann surface that is the Riemann surface on that Riemann surface this you get a function which is on each piece that you have pasted equal to the corresponding Zj so on the first piece it will be Z1 and as you move to the second piece which has been pasted to the first piece the function becomes Z2 then as you move to the third piece it becomes Z3 and so on, okay and of course the last piece is connected to the first one, alright. So you finally get one single function which I call as if you want one can call it as Z equal to f minus 1 w, w hat, okay and why I am calling this as w hat is because this w hat is a point above the point w below, okay so you get a function on it is not a function here it is a function there, alright it leaves as a function on a Riemann surface, right. So it is nice to see that you are able to even at a critical point you are able to get inverse functions that is the whole that is the, that is the beautiful thing that you have to notice, okay even at a critical point you are able to get inverse functions functional inverses for your function you are able to get, okay. Now what I want to say is that I want to say more I want to actually say that these functions are actually solutions of a polynomial equation, okay the solutions of a polynomial equation whose coefficients are analytic functions, okay and it is a that is a very deep fact and so let me explain that, so let me do the following thing consider the functions the symmetric functions functions in the Zjw, okay consider this, so you know if you give me a set of functions then for example if you give me a set of variables then you know how to construct the symmetric polynomials in those variables, okay namely what you do is you the first symmetric polynomial is simply the sum of the variables. The second symmetric polynomial is a sum of variables taken two at a time, okay and then the third symmetric polynomial is sum of variables, sum of products of variables taken three at a time, okay the second symmetric polynomial is sum of products taken two at a time and similarly the jth symmetric polynomial is you take a product of the j variables and take all possible such products and take sum, okay and of course in this when I say product the product you do not worry about the order, okay the variables are assumed to commute, right. So in the same way if you give me any set of functions I can construct symmetric functions from the given set of functions by looking at the corresponding I use the same method, I use the same definitions that I use to construct symmetric polynomials in a given set of variables only now that I am thinking of the functions as variables. So what will be S1, S1 of Z1, S1 of Z1, W etc. Zmw the first symmetric polynomial will be just the sum of the ziw or zjw, okay of course j equal to 1 to m. So the first symmetric polynomial in m variables is simply the sum of the variables, alright. The second symmetric polynomial in this will be now you have to take sum of products of variables to a time. So it will be in the form sigma zj1w zj2w where j1 is strictly less than j2, okay this is second symmetric polynomial and so on you can define the earth symmetric polynomial which is now you take the sum of r of them, sum of the products of r of them so which means that you write zj1w into zj2w and so on you go on up to zjrw you take r of them take the product and then you take the summation over all possible uhh such uhh r indices, okay and if you go on like this the last one you will get is the mth symmetric polynomial which will simply be the this should be the sum of products taken m at a time but there is only one product taken m at a time because there are only m of them so this will be only a single uhh expression not a sum of expressions it will be a single product this will be just product of it will be just the product of all the zj of m over j, okay. So you get these are the m symmetric polynomials in all the zj's, okay and what is the property of the symmetric polynomial this property of the symmetric polynomial is that property of a symmetric polynomial in these m variables is that if you change the variables by permutation the value of the polynomial does not change, okay. So that is what symmetric polynomial means, okay it is invariant under the action of the symmetric group, alright for every sigma in Sn so S so you know I will okay. So let me use uhh let me use let me use small s because it is better so that I can use capital S to denote the symmetric group okay which is just set of group of permutations permutations of n symbols m symbols it is a set of all bijective maps uhh from a set of m elements back to itself okay it is a group under composition is a symmetric group and you and the point is that Sr of uhh if you take sigma x1 etc sigma xm is the same as Sr x1 etc xn, okay. You apply the value of the symmetric polynomial does not change if you substitute for a variable for the variables a permutation of the variables and that is the reason it is called symmetric polynomial, okay and find and what is the advantage of looking at the symmetric polynomials you see the way I have defined them these are also functions of W, okay these are also functions of W and of course they are all they are all built by taking products and then taking sums so all these fellows they will all be certainly analytic functions on the they will all be analytic functions on the slit disk but the big deal is that they are not only analytic functions on the slit disk they are actually analytic functions on the whole disk that is the amazing thing, okay. So the so here is a fact all each or each Sr of z1w etc zmw is analytic in the whole disk this is the this is the amazing fact, okay and so let us let us grant this fact for the moment then what does it say it says that it says the following thing. So consider the polynomial z-z1m z1w into z-z2w and so on z-zmw consider this polynomial this is a polynomial of degree m, okay whose roots are these m functions I mean whose zeros are these m functions this polynomial becomes 0 whenever z is one of the zi's, okay. So the zi's are roots of this polynomial, alright of this zeros of this polynomial, alright but then if you expand it out the questions will come out to be as you would have seen in any course in algebra the symmetric functions up to a sign, alright. So what you will get is this will become z power m plus minus 1 to the power of m minus 1 s s1 of z1 of w etc zmw plus minus 1 to the power of m minus 2 s2 of z1 of w and so on zm of w and so on it will end with the constant term which will be it is going to be minus I have to worry about my my signs in this case is going to be I choose I think let me adjust the signs this is going to be finally I take minus 1 to the power of m sm of z1w and so on zmw only things I will have to take the sign for example and I have to add the questions also say here I want z to the m minus 1 so which means I how do I get the question of z to the m minus 1 I take I choose m minus 1 of these to be just z and the remaining one I choose it to be minus of certain zjw and then add it. So I will get a minus here so it will just be minus ok and here I will get a plus probably so this will be a minus this will be a plus and it will keep on alternating like this right so this is what I get so and here I should of course put z to the m minus 2 alright I should put z to the m minus 2 and so on and this is the constant term this is the constant term which is simply the product of all the ziw minus ziw ok so this is what you get ok. Now so what did I just say I said that you see the ziw are zeros of this polynomial but what is this polynomial it is the same as this polynomial but what are the coefficients now the coefficients are s1 s2 etc sm up to sin but what I say about s1 s2 etc up to sm they are all analytic functions so the m these m functional inverses for f are actually zeros of a polynomial with coefficients which are actually analytic on the disk on the target disk that is a fact that I want to say ok so and it is thus the solutions to z for w equal to f of z are zeros of a polynomial with analytic coefficients this is the fact this is a algebraic fact ok it is an algebraic fact it says that you have you have found you have found bunch of functions which are which are zeros of a polynomial with analytic coefficients ok and in fact this is of great significance in Raymond surface theory alright probably at some point later at some later point of this course I will try to explain that but now for so you know so the moral of the story is that all these functional inverses are zeros of a nice polynomial ok the only the only fact that needs to be checked is that each of these is analytic alright now to prove that it is analytic there are let me at least first try to begin by convincing you that they are at least continuous leave alone analyticity see the problem is you know if you go back these functions they will you know if you if instead of throwing out this this ray ok if you keep it but you cut the portion of the disk on the ray and below it and separate it from the portion of the disk above it then on that they will become continuous they will become continuous it is just like the log function you take a branch of the logarithm the branch of the logarithm if you take the principle branch of the logarithm for which the imaginary part is a principle argument which varies from values starting from minus pi and going all the way up to values strictly less than plus pi if you take that as a function the principle branch of the logarithm then you know the real part is always continuous because it is the natural logarithm of the modulus of the complex number which is which is a continuous function the problem is only the imaginary part which is argument and what is the problem the problem is that you know if for all points on the negative real axis and below the negative real axis the principle argument is you know is greater than or equal to minus pi ok whereas for a point just above it above the negative real axis the argument is close to plus pi and lesser than plus pi so there is a jump of almost 2 pi on the negative real axis and therefore you have discontinuity on the at every point on the negative real axis. So suppose I separate the points of the points of the negative real axis above the points of the plane above the negative real axis from the negative real axis and the points below ok by separating it I am saying that you know the points above the negative real axis are not close to the points on the negative real axis and the points below that is what this cutting means ok and when points are not close I do not worry about continuity, continuity I worry about only when points are close ok so once I cut it if I try to check continuity at a point on the negative real axis I am only going to check in a neighbourhood in a disk half disk which lies in the negative real axis and below it I am not going to worry about points in the half disk that lies above it alright. So you know if I cut this instead of deleting it if I cut it ok then these fellows are going to be continuous ok but then what happens as you go across the negative real across this line segment if you go from below to above what happens is that the zj the values of zj will change from zj to zj plus 1 ok depending on whatever you are ordering is either they will go from zj to zj plus 1 or zj to zj minus 1 depending on the way you wrote it ok and that is because the function itself changes and what you do is you cut and paste these various copies m copies of this of this cut domains not the slit domains the cut domains to produce this remain on surface in such a way that as you cross the this ray on one piece ok the function zj changes to zj plus 1 which is the function on the next piece ok and that is reason why all these functions agree and they become a continuous function on top ok. So the moral of the story is that as you the problem is with the problem is with points on this ray for each of these functions ok what is the problem as you move across that ray the zj will become some other zj plus 1 or zj minus 1 alright but if you look at the symmetric functions ok and you do the same business you see a symmetric functions will not be affected ok for example you know if I take the sum of all the zj's and if I take if I take the w and move it across if I move it across this ray ok that I if I move it across this ray what will happen is each zj will be replaced by the corresponding zj plus 1 ok each zj will become the corresponding zj plus 1 ok and but then the sum will remain the same ok. So what I will get is instead of getting z so you know if I start with so let me if you want let me even draw it so that you understand what is going on so here is my ray this is my w not and you know suppose I start with the w here w1 and I move across this ray and go to w2 ok then what will happen is that you know zj limit w tends to so I will put keep this as w and let me call this as w prime w tends to w prime of zj of w what you will get is you will get zj plus 1 of w you will get this this is what will happen ok or you know it may be zj minus 1 of w prime depending on how you rotate alright this is what you will get but it will always be plus 1 or always be minus 1 depending on how you rotate. So what happens is so you know if I take limit w tends to w prime of for example the first symmetric polynomial of z1w etc so on zmw what I will get is I will get it back I will simply get I will get the first symmetric polynomial evaluated at if I of course the symmetric polynomial is a polynomial in the variables so it is continuous in the variables so I can push the limit inside and apply the limit to each of the variables ok and if I do that I will get well I will get z2w and so on and you know and the point is that zmw will go to z1 because it is cyclic alright. So it will go on at some point I will get zmw and then I will get zm plus 1w or I will get the other way round the index may come down it may be zmw z1w and so on zm minus 1w this is what I will get but both these are the same as this in any case both are equal to s1 of z1w and so on zmw they are the same because after all the s1 is supposed to just add all of them and whether I add from z2 to zm plus 1 including zm or zm to z1 to zm minus 1 and then zm I am going to get the same result ok. So what this demonstration tells you is that the value of s1 of the first symmetric function of these m functions that value does not change ok. So it means that this first symmetric function is a continuous function of w and of course you know I will have to put I have to put w prime since there is a here these are all primes these are all primes and so these are also primes ok these are primes. So the moral of the story is the symmetric function will not be affected and I have written it down for the first symmetric function but you know for all the symmetric functions this is going to happen ok what is going to happen is when you when you cross the troublesome ray ok the symmetric functions are only going to be permuted by a permutation which is given by shift in the index alright and but when you make a permutation the symmetric function is not going to change the value of the symmetric function is not going to change because it is even in terms of permutations therefore what is going to happen is each of these si's each of these sr's they are all going to be continuous even on the even on this even on this ray ok and therefore you can now believe it is very clear that each sr is therefore certainly continuous on this on this disc ok. The only issue is now that is left to be fixed is that it is actually analytic ok and the fact that it is analytic can be for example seen by what is called the principle of analytic continuation ok which is what I am going to discuss in the forthcoming lectures ok. So I am going to next go on to discuss analytic continuation and the so called monodermy theorem alright and it will follow from that discussion that each sr is not only continuous as we demonstrated it is actually even analytic ok so with that I will stop now.