 Okay, so we continue with the proof of the monodromy theorem, so we have this rectangle which is product of the close interval AB on the real line with the close interval CD on the real line and we have a homotopy capital F which is a continuous function on this rectangle and it gives a homotopy between the path gamma which is the beginning path in the homotopy and the path neta which is the terminal path the homotopy and of course any intermediate path in the homotopy is given by gamma s, okay, gamma s is just F of s gamma t with s fixed and t varying, alright and what we need to show is that if we know that there is an analytic function even at the point z0 which can be analytically continued along each of these paths then analytic continuation along any path will again lead to the same function at the terminal point z1 that is what we have to prove, okay that is the monodromy theorem. So how do you prove this proof is as follows, so what we will do is we will do the following thing you know, so what is given to me is that there is an analytic continuation of F along each of these paths, okay, so let me write down some analytic continuation alright, so it is so for every s in CD there exists an analytic continuation along the path gamma s which is given by F of s, t of z sigma n equal to 0 to infinity a n of s, t into z minus gamma s of t to the power of n with radius of convergence with disc of convergence mod z minus gamma s of t is less than r sub s of t, okay which is of course positive with of course and with F of s, 0 is equal to f, okay. So what I have done is for each gamma s I have written this f of s, t which is an analytic continuation along the path gamma s starting with f, okay and it has certain disc of convergence, right. So you know so on this picture gamma s is this path which is the image under capital F of this line of this line segment, okay where s is fixed and t varies from c to d, t varies from a to b, s is fixed value between small c and small d and t varies from a to d and the image of this line segment under this continuous function f is this path gamma s and if you give me a point, if you give me a point t here a point corresponding to a certain value of t between a and b then the corresponding point here this rectangle will have coordinates s, t and the corresponding point and the image of this point under f will be f of s, t which is just gamma s of t. So it is going to be this point gamma s of t I am going to get a point here and I have an analytic continuation along this gamma s as t varies and that is this analytic continuation f of s, t, okay and at the point gamma s of t I am going to get a power series centered at gamma s of t, okay and the only thing that you have to remember is since there are two real variables or everything I mean f, a, a, n and gamma and also r they will all depend on two real variables s and t, okay. So far they were depending only on one variable if you are writing an analytic continuation along a single path then you get only one variable which is the path variable but now you are writing an analytic continuation on a family of paths, okay which means that you have also a variable for different paths which is the variable s so there are two variables involved s and t. So everything is a function of the power series in the analytic continuation the coefficients of the power series the centers of the power series the radius of convergence of the power series they are all depending on these two variables, okay. So s and t will appear in all of them that is how we write it, okay and so this is given to me there is an analytic continuation like this, okay I do not care what this analytic continuation is for the moment, alright but it is given that there is an analytic continuation. Now what I am going to do, okay so the fact I am going to use is that this function rs of t if you think of it as a function from this rectangle with st as a variable point of the rectangle then the claim is this r is a continuous function on that rectangle, okay. In fact an will also be a continuous function on that rectangle, okay so here is a claim rs of t is a continuous function of the point s, t in the rectangle av cross cd it is a continuous function. So this is the claim we use the lemma that if f is so if g is analytic in the domain d u and for z in u p g z sorry let me put z prime for z prime u p g z prime is the power series expansion of g around z prime with disc of convergence mod z-z prime lesser than r of z prime then r is continuous as a function of z prime in u. So I am just using the fact that you know if you have an analytic function on a domain and at various points you start writing its power series expansion, okay so at various points I will get various power series expansions and at corresponding to each of these power series expansion I am going to get radii of convergence so as I change the point I am going to get different radii of convergence, okay depending on which point I am expanding the function of power series about then the fact is that this radii as you change the point the radii of convergence will change continuously, okay so we proved this. So in fact what we proved is in fact we proved if you remember we proved r of z1-r of z2 mod is strictly less than mod of z1-z2 we proved the difference in the radii of convergence of the power series expansion at z1 and z2 cannot exceed is smaller than the distance between the two points, okay. So this is the fact we need to use so let me draw another diagram so you know the situation is like this so here is my rectangle so this is when this is A this is B this is C this is D and this is the variable T this is the variable S, okay and here is my particular value S well if you want you know I can take S0 and then I can take certain T0 so I will get this point which corresponds to S0, T0 and I have this F and what this F is going to give me is well it is going to map this segment point of course gamma which is gamma A rather gamma C this is the path from z0 to z1 and then I have this other path below which is neta and neta is gamma D, okay and corresponding to this line segment with Y coordinate S0 I am going to get the intermediate path gamma S0 this is gamma sub S0, okay and this is the point that corresponds to gamma sub S0 of T0, okay this is what I am going to get. Now see I have to show that I am trying to show that RS of T is a continuous function of ST, okay to show that a function is continuous it is enough to show it is locally continuous, okay. So it is enough to show that RS of T is continuous in SNT in a neighbourhood of each point S0, T0 in the rectangle, alright. So you know so you know fix so let me write that down it is enough to show RST is continuous in a neighbourhood of S0, T0 for every point S0, T0 in that rectangle it is enough to show this, alright because continuity is a local property to show that a function is continuous it is enough to show that on an open cover, okay. So you know so I have frozen this S0 and T0, okay now for the moment what you do is you see you just look at this path gamma S0, okay along this path gamma S0 there is this analytic continuation which is given by F S0, T there is an there is this analytic continuation F S0, T and it starts with F S0, 0 which is F it is an analytic continuation of F along this gamma S0, alright. Now you see by the previous lemma, okay for all T for all paths close enough to this path the analytic continuation are the same, okay. We have see the previous lemma that we proved in the in the in the preceding lecture was that if you have analytic continuation along a path then along sufficiently close paths the analytic continuation is going to exist and is going to be and is going to lead to the same function at the end, okay. So what you must understand is on nearby paths, okay, nearby means for S close to S0, okay if you take nearby paths then the analytic continuation are going to be the same as the analytic continuation along the path gamma S0, okay. So by the previous by the lemma of the previous lecture, namely the lemma that I proved at the end of the previous lecture there exists delta of S0 such that for every S with mod gamma S of T minus gamma S0 of T is less than delta of S0 the analytic continuation F S, T is going to be the same as analytic continuation along F S, T is going to be the same as along F S0, T. This is what we so you know I am saying that you know if you choose any S which is very close to S0, okay. So if you choose S close to S0, alright then of course the gamma S will become very close to gamma S0. So this is gamma S, this is gamma S and this is gamma S0. If S is close to S0 then gamma S is close to gamma S0 that is simply because F is continuous and gamma S is the image of this line segment that corresponds to S and gamma S0 is the image of this line segment that corresponds to S0, okay and nearby a continuous function maps nearby objects to nearby objects is just continuity. So as you make S close to S0 the gamma S comes closer to gamma S0 but then if you have chosen S so that the distance between the point gamma S, T and gamma S0 of T for each T is always less than this delta S0, okay then the analytic continuation along gamma S is the same as analytic continuation along gamma S0. This is what we prove in the in a lemma in the previous lecture. In words to state that we prove to state what we proved was is that if analytic continuation exists along a path then analytic continuation will also exist along sufficiently close paths and all these analytic continuation will really lead to the same function at the ending point I am just using that lemma, okay. The only thing is that this delta will now this depends on that path S0, gamma S0, okay. So if I use that so what I get from this if I use that lemma is that so in fact you know in fact what we if you go back to the if you go back to the proof of that lemma what we proved was that the analytic continuation at gamma S of T the analytic function you get at gamma S of T is the same as the analytic function you get at gamma S at a gamma S0 of T, okay because what we did was we actually defined on a close sufficiently close path we defined an analytic continuation by simply writing out the power series expansion at that point, okay. So in fact the analytic continuation the function you get at ST is literally the same function that you get at S S0 T for S sufficiently close to S0. So let me write this in fact going back to the proof of that lemma of that lemma the function f ST is the same as the function f S0 T as functions S such that mod gamma S of T minus gamma S0 of T is less than delta S0 for all T, okay. So this is again going back to the proof of that lemma, right and so you know so what this tells you is that you know if you if you take if you take S sufficiently close to S0, alright then you are getting all for any value of T the analytic function you are going to get is the same for a fixed value of T, okay all for all these S which is close enough to S0 and for any fixed T the analytic function f ST is the same as f S0 T, alright. Now what I want you to understand is that but you see if T prime is close to T of course f ST prime is the same as f ST and f S0 T prime is the same as f S0 T that is also true that is because it is a analytic continuation, analytic continuation is required that as the T variable comes close to a particular value then the analytic functions given by the power series also coincide, okay. So what all this tells you is it tells you that you know it tells you that there is a neighbourhood around S0 T0 where all the functions f ST are a single analytic function, okay. So in other words there exists a neighbourhood around S0, T0 where all f ST represent the same analytic function and call that function as G, call that function as G S0, T0, okay then R of RG S0, T0 of that is radius of convergence of the power series of G S0, T0 at the point S, T is just RST, RS of T in our notation is a continuous function of S, T in that neighbourhood by dilemma, by dilemma so you know I will call this is I will say lemma star so I will label this lemma star, this is this lemma star, okay not to be confused with the lemma of the previous lecture, okay this lemma star is the lemma that if you take an analytic function in a domain then if you expand the analytic function as a power series at each point then the corresponding radii of convergence will be continuous will be a continuous function of the point. So you know there is a small neighbourhood here where all the f STs here all the f STs represent the same function G S0, T0 what it means is if you take the image of this neighbourhood here, okay you will because of the continuity of f I can find a small enough neighbourhood here into which the image of this neighbourhood goes, okay and for all points in that neighbourhood you are actually expanding the same function G S0, T0 in this neighbourhood which contains that neighbourhood, okay. See you take this point gamma S0 of T0 which is the image of the point S0, T0 which is the image of the point S0, T0 under f, okay then you take a sufficiently small neighbourhood of gamma S0 of T0 where this G S0, T0 lives. See after all G S0, T0 is an analytic function which is it is not defined here, G S0, T0 lives here so it is this so this is the point at which G S0, T0 is analytic, okay G S0, T0 is analytic at this point which is the point gamma S0 of T0, okay it is analytic there and there G S if you in that neighbourhood if you take any point and if you write the power series expansion of this G S0, T0 at that point and look at its radius of convergence then the radius of convergence is a continuous function of the point, okay. So the radius of convergence of G S0, T0 at each point in this neighbourhood surrounding gamma S0, T0 is a continuous function of the point gamma S of T but gamma S of T is a continuous function of S of T because it is actually f. So R of R S of T becomes a continuous function of S and T, okay. So in fact so let me write that properly so to in fact there exist a neighbourhood of gamma S0 of T0 where G S0, T0 is analytic and the radius of convergence of the power series expansion T G S0, T0 at gamma S of T is a continuous function of gamma S of T in that neighbourhood of gamma S of T0, okay this is what I am saying, right. And R S T is actually and R S of T is actually the radius of convergence of the power series expansion of G S0, T0 at the point gamma S of T which is continuous function. So and mind you this is just R composed with I am here now I am thinking of R as composed with f of S of T because f of S of T is gamma S of T, okay. So it is a composition of f, f is continuous and R is continuous therefore composition of continuous function so it is continuous. So you know R is indirectly a function of S of T so I wrote it directly there but if you want it more explicitly I have written it here, okay this is the reason why R is a continuous function of S and T, okay. So what I have proved is R is a continuous function of S and T locally, okay but that is enough to say that it is continuous globally because continuity is a local property, right it is a property that can be verified at each point in a neighborhood of each point. So I have proved this claim, okay so I have this claim that R S T is a continuous function of S, T in this rectangle, okay. Now how do I proceed? I proceed in the same way I simply take the image of that rectangle and under R and I notice that the image will again be a compact interval and it will have a minimum and I am going to call that minimum as delta, okay. So you know so R S of T is so this settles the claim that R S T is continuous in S T varying in this rectangle and take the image under R of rectangle, okay. See this rectangle, this rectangle is anyway compact and connected, the rectangle on the plane is a compact and connected set of course it is a connected set because it is actually path connected, any two points can be joined by a path in fact even by a straight line if you want, okay and so it is connected certainly and it is compact because it is closed and bounded because I have taken the closed rectangle. So it is compact and I have a continuous function R defined on this compact set, okay so the result will be the image under R of this compact set will again be a compact subset compact connected subset of the real line. So it will again be a closed interval in the real line, okay and of course R is always positive so I am going to get a closed interval with minimum with left hand point greater than 0, okay. So which will be delta, delta with delta positive, okay so this is where I use the continuity of R, I need the continuity of R to say that the image under R of this rectangle is you know compact and connected and a compact connected subset of R is just a closed interval and of course this R refers to various radii of convergence, they are all positive radii of convergence. So the R values are always positive number 0 therefore there is going to be a minimum value of R that is going to be small delta and there is also going to be a maximum value of R that is capital delta, okay. Of course you know in all these situations I am really not worried about the case when at some point you know you get a power series whose radius of convergence is infinite, okay. So you must always remember that see you always see me writing the small delta capital delta, this capital delta tells you that you know the R is finite, the radii of convergence of finite and you know radii of convergence of finite means that at the circle of convergence there is a singularity for that analytic function because if there were no singularities the radius of convergence would have become infinite, okay. So always when the radius of convergence is finite on the circle of convergence there is a singular point for that function there is a point beyond which you cannot extend that function, okay. So there is a point at which you cannot extend that function so there is a singularity. So you always see me writing this delta here capital delta and I just wanted to make this remark that you know I am never looking at the case when radius of convergence is infinite because if radius of convergence is infinite it means that one of the functions you are that occur in the analytic continuation is entire. If one of the functions is entire then there is nothing to continue because an entire function can be continued everywhere. So you know you are not going to you are not going to get you are not going to get anything you are only going to get that function no matter how you analytically continue it, okay it is going to be just direct analytic continuation just extension of that entire function to the whole complex point. So there is nothing to prove, okay so all these things become interesting only when the radius of convergence are all finite, okay. The radius of convergence become infinite even for one point all these results they become trivial there is nothing there is really no real question there to answer, okay. So that is the reason I am always thinking of r positive and finite, okay fine. So now comes the now that I have that I have this delta see now I am in very good shape so you see how I use this delta is as follows what I do is I have this you see this rectangle that I have A B cross C D you see I can actually divide this rectangle into by a series of lines parallel lines you know S0, S1, S2 and so on some S k and so on so that you know well maybe I rather call this line as S0 that corresponds to C then I have S1 I call this as S2 call this as S3 so maybe instead of writing it here I will write it here this is S1 this is S2 well and this is S3 and so on then I end up with S k and finally I end up with S well some n capital N which is D alright and of course this value will correspond to S n minus 1. So I can find these S's in such a way that you know if you take the image of each of these rectangular strips okay you will get a piece of this homotopy leaf there is a piece of this leaf like this region in between these 2 paths such that you know the distance of the points corresponding to given T is less than delta okay so let me write this so let me draw this diagram first so here is how the diagram is going to look like so you know so this is gamma S0 which is just gamma C this is gamma S1 then I will have gamma S2 and so on then finally I have this is gamma S sub N which is just gamma sub D and this guy here is gamma sub S n minus 1 okay I can find these values starting from S0 to S n for sufficiently large n such that you know you give me any value of T give me any value of T then of course the image of something like this will be something like this well if I draw it will be something like this okay that this will correspond to a given T alright it will be a it will be this point this first n point Z0 when T is A and this thing will collapse to the terminal point Z1 when T equal to B but in between the image of this line segment will be something like this that will also be a path connecting a point connecting this the point corresponding to T in the first path with the point corresponding to T in the last path okay and but the point is that you know if you take any 2 successive points the distance of those points is less than delta okay so you can find such a finite collection of points okay so there exists there exists S0 equal to C strictly less than S1 and so on lesser than S n minus 1 is equal to D such that for every T for every T in AB the distance between gamma S of T and gamma S prime of T is less than delta for S, S prime belonging to any sub interval Si, Si plus 1 i equal to 0 and so on up to n minus okay you can divide you can divide this rectangle into small thin rectangular strips with this property this is purely by compactness okay okay so it is a compactness argument that you can further expand and try to write down but it is intuitively obvious and easy to write down so you can do this now once you have done this once you have realized that you can do this the proof of the theorem is over because you see what will happen is you see because the distance between because the distance between all the paths in between gamma Si and gamma Si plus 1 is less than delta they will all define the same analytic continuation okay that again the proof of that is again following the proof of the lemma of the previous lecture. So what will tell you is that on each for each of these pieces the analytic continuation along the upper path is the same as the analytic continuation on the lower path and then you go by induction okay so the analytic continuation along gamma S0 is the same as analytic continuation along gamma S1 the analytic continuation along gamma S1 is the same as the analytic continuation along gamma S2 and by induction finally you will get that the analytic continuation along gamma S0 is the same as analytic continuation along gamma SN. In other words analytic continuation on gamma gamma Si which is gamma is the same as the analytic continuation along gamma D which is eta okay and that proves the monodermithia ok. So what you must understand is that uhhh it is a kind of uhhh cleverly playing upon the ideas of the lemma that we proved in the previous theorem and also critically using the fact that the radius of convergence is a is a continuous function that is a very critical fact that you keep using and also uhhh let me again uhhh repeat the uhhh the main idea in the proof of the lemma of the previous lecture was that you know if you take sufficiently close paths then there is only there is a there is a unique penalty continuation on that path and it is simply defined by expanding the the relevant function on the given path into a power series ok. So uhhh if you if you take 2 nearby paths and if I have this analytic continuation along the path gamma S0 on along a nearby nearby path gamma is the analytic continuation how is it defined it is very simple what you do is you simply define the analytic continuation by simply expanding this function at gamma S0 of t0 at gamma S of t0 and you do this for every t0 ok. So uhhh the fact the the whole idea is you know if a function is analytic at a point it lives in a neighbourhood and therefore that function itself can be used to define power series in paths in that neighbourhood ok. So it so in other words if you have a path and you give me a point on the path and you give me an analytic function at that point then there is a disk by definition of analytic there is a disk where the function is analytic and whenever there is any other path which passes through the disk along the portion of the path which passes through the disk I can simply define the analytic continuation to be the power series expansion of this function that lives ok that and that is a crucial this is a very very crucial idea ok. So that is a crucial idea that is being used and also the idea that the radius of convergence is uhhh uhhh is a continuous function of the of the point ok. So so let me write down uhhh by the uhhh uhhh proof of the lemma of the previous lecture uhhh analytic continuation continuation along gamma si is the same as that along analytic continuation of f along gamma si is the same as that along gamma si plus 1 uhhh for every i starting from 0 to n minus 1 thus analytic continuation continuation uhhh of f along any gamma s leads to 1 and only 1 function at z 1 and that is the proof of the monotomism. So let me again at the risk of repetition let me again stress the whole idea is if you have a path and at a point uhhh gamma of t if you if you have if you are given an analytic function f s uhhh I mean f t analytic function here then if you have any other path which hits this disc where f t lives along this path along the portion of the path from here to here leaving out the end points there is this f itself has a trivial analytic continuation along this that is the whole idea that uhhh is being used again and again and you are of course crucially using very very crucially the fact that the analytic continuation along a path is unique once you fix a parameterization of the path the analytic continuation is unique for a given starting function okay you cannot have two different analytic continuation with the same starting function for the same parameterized path that is one important fact the other important fact is the radius of convergence that varies continuously uhhh as as the uhhh it is a continuous variable of the point where you are expanding or writing the power series about okay so these are crucial points so I will stop here.