 okay so what I am going to do now is try to discuss the proof of the monodromy theorem okay of which we have seen couple of versions right and so I am of course going to prove the version of the monodromy theorem the first version of the monodromy theorem okay. So I begin with so let me put the title as proof of the monodromy theorem so this is version 1 that is what I am going to prove. So we begin with lemma okay and what this lemma says is it says the following thing it says that see suppose you have a path along which you can do analytic continuation of a given function at the starting point of the path okay then for all sufficiently nearby paths analytic continuation will exist for the same function okay and the fact is that the analytic continuation at the that you get at the end of the path will be one and the same okay. So so here is the lemma if f is analytic at z0 and can be analytically continued along gamma from z0 to z1 okay then there exists a delta greater than 0 such that f can be analytically continued along any path neta from z0 to z1 such that the distance between gamma of t and neta of t is less than delta for every t in ab moreover the analytic continuation of f along any such path neta to the same function at z1 as the one gotten via gamma. So what this lemma says is that you know so it says that if you have a path along which you can do an analytic continuation then you know sufficiently nearby paths starting at the same point and ending at the same point also will admit analytic continuation and these the result of all these analytic continuation are going to be one and the same okay so the you know so if I draw a diagram it is something like this so here is a point z0 at which function f is given and I have this path gamma which goes from z0 to z1 gamma is a function from the closed interval ab on the real line to complex plane so it is a parametrized path you know that f can be analytically continued along gamma so which means that there is an analytic continuation which is given in terms of a power series it is given in terms of a one parameter family of power series so given ft of z is equal to sigma n equal to 0 to infinity so probably okay an of t z-gamma t power of n with the mod z with this power series converging in a disk centered at gamma t radius rt where rt is radius of convergence of this power series and which is assumed positive and gamma t is the center of the power series okay. So with f0 that is fa is f okay and what is the final function you get it is so this corresponds to t equal to a and this corresponds to t equal to b okay and so this is just gamma a this is gamma b and somewhere in between you will get gamma t the point gamma t and at the point gamma t there is some there is a disk there is a disk centered at gamma t with radius equal to radius of convergence rft and you know that here well ft the power series ft lives here okay it represents it converges here this is the disk of convergence the power series ft is a power series centered at gamma t so it is an expansion in terms of powers of in positive integral powers of z-gamma t and an of t are the corresponding coefficients okay and when you put t equal to a you get gamma a and the corresponding power series is fa which is the function analytic function f you started with and it after the at the end of the path you will get a new function here and that new function is fb it is a analytic function corresponding to the analytic function with power series fb okay that corresponds to t equal to b okay and now what this lemma says is that you can find a delta such that you know if you take any other path neta which also is from a from the same closed interval ab to c and it also is from z0 to z1 so you know the neta should look something like this so your neta is like this okay so neta is also defined from the closed interval ab on the real line and the point about the what is the connection between neta and gamma for every t the distance between the point neta t and gamma t is less than delta this distance between gamma t and neta t has to be less than delta this distance has to be less than delta which means you know you are actually looking at paths this so you know it is like you can call this at delta neighbourhood of the path gamma okay so you know if one so you know if you draw a diagram let me redraw it here so here is my gamma okay and you know at every point if you give me a point then you know I take this disc with radius delta so this is my open disc with radius delta here and you know if I take it here I will get another open disc with radius delta you know and throughout it will be like this that is a that is a fixed radius delta and here also even at including the end points so you know of course you know because of the because of the compactness of the path finitely many discs are enough to cover the path of course and the point is that you what the lemma says is that you know if along this path gamma you can analytically continue f then for any other path which starts from z0 and ends at z1 which is lying in this union of all these discs okay if you have any other path like this nearby path okay so this is gamma and any other nearby path beta such that at each point the at each point t the distance between the corresponding point on beta which is beta t and the corresponding point on gamma which is gamma t is less than delta okay suppose you take a path like that then the lemma says that on that path also the analytic continuation of f is possible and not only that it says that the analytic continuation of f will lead along that path also if you do the analytic continuation you will end up with the same function as you got in the case of gamma namely fb okay so what this tells is it says it gives you existence and uniqueness of the analytic continuation along nearby paths okay so you can refer to this lemma compactly as existence and uniqueness of analytic continuation along nearby paths okay and how nearby that nearby is given by this delta which whose existence we have to show that is the that is part of the proof so what we do is proof is pretty easy okay so you know we just use the following fact that you know see we have already seen that we have already seen that the functions a n and r a n from close interval a b to c and r from the close interval a b to r r greater than 0 are continuous I have already proved this to you I have already proved that you know if you write a analytic continuation like this analytic continuation along a path in terms of power series then the coefficients of the power series they are continuous functions of t so each a n is a continuous function of t and the radius of convergence of the power series is also a continuous function of t so this is something that I have proved okay I am going to use I am going to use that so well so the thus if you take r of this close interval a b what you will get is some delta, capital delta okay if you mind you r is a continuous function okay it is a continuous function so you know a continuous function maps a connected set to a connected set it maps a compact set to a compact set therefore you see this close interval a b is going to be mapped to a connected set on the real line so it is going to be a real interval and it is also going to be mapped to a compact set so you are going to get a an interval on the real line which is connected and compact okay so it has to be close interval okay and of course every value that r takes as positive therefore this delta is positive so you know in other words delta is the minimum of all the radii of convergence of all these power series this small delta little delta is the minimum of all these rts so delta is equal to minimum t belonging to a b of rt you can also write this as infimum of t belonging to a b rt and of course you know infimum and minimum will be one and the same because in this case your variable is defined on a compact set okay and a compact set always contains its boundary so fine so this is the delta that I want I claim this is the delta that we need to prove the lemma okay so now so now the question is if for this delta what I have to show is suppose I give you another path neta satisfying this condition that the corresponding distance between for each t between neta t and gamma t is less than delta I have to show that on that path neta also I can define an analytic continuation and I have to show that that analytic continuation will also lead to the same function as ft as fb when you reach the point z1 for the parameter value t equal to b okay so how does one prove it is very very easy given a path neta from a b to c with distance between neta t and gamma t less than delta for all t and with neta of a is gamma of a is z1 z0 neta of b is equal to gamma of b is equal to z1 we define gt so I am going to define an analytic continuation so it is very very simple see so let me draw one more time no harm in drawing a lot of diagrams so here is z0 this is z1 this is some t and this is the path gamma so this point is gamma t I have a nearby path which is neta and the corresponding point on it is neta t and the assumption is that with respect to gamma t I mean with respect to gamma of t you draw a circle centered at gamma t radius delta then this circle contains neta t so the distance between neta t and gamma t is less than delta that is the assumption so what you do it is very simple what you do is you just define gt of z to be equal to sigma to be equal to the power series expansion of ft centered at neta t so see I wanted to understand see this if you take the point gamma t then I am having you know the at gamma t I have the power series expansion for I have the power series expansion of ft so ft is already a power series it is expanded at gamma t it is expanded at this point and it has got radius of convergence rt okay and you know this rt is always greater than delta so you know if I take this point gamma t and if I draw this circle with if I draw the disc of convergence for ft I will get a bigger disc I will get a bigger disc okay centered at gamma t okay if I consider the function ft the power series ft it will represent an analytic function the power series will represent an analytic function whose Taylor series will be this power series itself and that and where is it valid it is valid in the disc of convergence and the disc of convergence is at this is a disc centered at gamma t okay and the radius equal to r of t but you know this rt is greater than delta because delta is the minimum of all the rt's and now what you must understand is that this bigger disc I have drawn that is where ft lives the analytic function ft lives in that bigger disc okay since the analytic function ft lives in the bigger disc and this neta t is inside that bigger disc I can write the power series of this ft at neta t okay I can do this so what it will what it will give me is I will get some power I will get some coefficient sigma n equal to 0 to infinity some bn of t ez minus neta t to the power of n this is what I will get so this is I am just writing the power series expansion of the function ft at this point so it is an expansion in terms of integral positive integral powers of z minus neta t okay. Now my claim is that if you define like this my claim is that this gt is an analytic continuation along the path neta so the claim is gt is an analytic continuation along I mean analytic continuation along the path neta and why is that true see what is the condition for our what is the condition for an expression like this to give rise to an analytic continuation so the first condition is that for each t you must get a power series of positive radius of convergence that is the first condition and the second condition for nearby t's the functions that are given by the power series they should be one in the same analytic function okay so you see now for each t it is very clear that this function has a this power series has a positive radius of convergence because what is the radius of convergence of this the radius of convergence of this is at least this distance from neta t to the end of this disk okay because that is where the ft lives okay so what you must understand is so let me write this the radius of convergence is you know so I should say at least it is at least this distance rt minus the distance between gamma neta t and gamma t it is at least this much okay so you know if I draw this line like this this whole length is going to be rt alright and this length is the distance between gamma t and neta t and at least this much should be the you know at least here this power series will live because it is the power series of ft so the power series of ft will at least this power series will have at least radius of convergence this much which is the distance which is the difference of rt and this distance between neta t and gamma t okay and that is positive so for each t the radius of convergence is positive okay that is the first condition so for each t I am really giving you a proper power series so I should not give you a power series with 0 radius of convergence that is not allowed okay so for each t I should first ensure that I get a power series with positive radius of convergence that is that is the first condition second condition is for near by t is the power series should represent the same analytic function that is what that is the second condition for an expression like this to give to define an analytic continuation so how does one see that it is very simple one see so we have so let me write this fact we have so it is a I think so this is the radius of convergence of gt so let me write this r sub gt of t is positive for all t so this is the radius of convergence of gt this radius of convergence okay so that is the first condition second condition is so you know let me again draw another little diagram so it is like this so here is my gamma here is my neta this is z0 this is z1 so you know well so I have this point gamma t and I have this point neta t and of course I have taken a disc centered at gamma t radius delta so this is how I have chosen my neta okay now you see if you take t prime close to t okay of course neta t prime will be close to neta t and gamma t prime will be close to gamma t and that is just because of continuity. So you know if you take a t prime very close to either on the left or on the right let me take it on the right so this is gamma t prime which is for t prime close to t and here is neta t prime for t prime close to t okay so you see if t prime is close to t we need to check that ft that is gt is equal to gt prime as analytic functions you have to check this okay so one of the conditions of the analytic continuation is that it is given by power series but the power series are all expansions of analytic functions at different points okay the points are varying along the path but the condition is if you go to nearby points then the power series will be different but they should give you the same analytic function okay the power series will of course be different because you have changed the center of the power series okay so what I will have to check is that this gt for t prime close to t the analytic function defined by the power series gt and the analytic function defined by the power series gt prime are one and the same I have to check that only then this will this be truly an analytic continuation and that is very easy because you see if you check carefully you see so it is just a matter of so you know what will happen you will see that gt so you know gt prime of z if you calculate this will be ft prime of z okay as an analytic function because that is how I have defined it because gt prime is defined to be the power series of ft prime so it will be equal to the analytic function ft prime that is how I have defined it here gt is the power series of ft therefore the function represented by gt is just ft okay so gt prime of z is ft prime of z but you know f is an analytic continuation so for t prime close to t this will be the same as the analytic function ft of z so this step is because f is ft is an analytic continuation this is because ft is an analytic continuation and t prime is close to t okay but then ft of z as analytic function is the same as gt of z because that is how gt is defined gt is defined to be the power series of ft centred at eta t so therefore so this implies that gt prime gt is indeed an analytic continuation of ga to gb okay okay so you have proved existence of an analytic continuation along the path meter along any path meter which is in a delta neighbourhood of the path gamma okay now what is ga but you see what is ga ga is fa because gt is after all the power series expansion of ft okay so ga is just fa and you see but ga is fa and gb is also equal to fb by definition so the analytic continuation of fa so that is an analytic continuation along the path meter of f equal to fa and leading to fb okay now use the fact that if you give me a parameterized path okay then the analytic continuation is unique if you fix the starting function so because along because I have proved that along the path meter there is one analytic continuation which starts with fa and ends with fb uniqueness will tell me any analytic continuation along the path meter which starts with fa will always end with fb so I will get the same analytic continuation okay so any so you know here I am using this fact that you know if you have a path and you give me if you have a path on which you have parameterized the path and along that path if you have an analytic continuation of a starting function then that is unique the analytic continuation is unique you cannot you cannot get different analytic continuation in fact analytic continuation along each point is uniquely determined for every point other than the starting point okay so it is uniqueness of the analytic continuation along a fixed parametric path that is I am using okay so any analytic continuation continuation of fa along meter leads to fb so that finishes the proof of this lemma okay it is a it is a it is a pretty simple lemma nothing complicated about it but the power of this lemma is that it tells you that if you have analytic continuation along the path then along nearby paths also analytic continuation will exist and they will be the same they will give rise to the same analytic continuation okay and of course I should again let me again repeat the risk of being overtly repetitive that we always keep using this fact that if you give me a parametric path and if you give me a starting function that is an analytic function of the starting point then there is only one analytic continuation of that function along that path this is essentially because of the identity theorem okay so that is something that we keep using alright so this is so this is a lemma that says that sufficiently close paths along which you have an analytic continuation will also lead we also have analytic continuation but you will get the same result okay now I am going to use this and proof or use these ideas and proof the first version of the monotromy theorem so let me write it down so given so let me draw the diagram so you know you are given point z0 and z1 of course the way I am drawing it z0 and z1 seem to be distinct points but there is no there is no restriction they could be one and the same point okay so it includes that case also okay so z0 and z1 are two points okay and you have two paths there is a path gamma from z0 to z1 and another path neta from z0 to z1 which are homotopic so there is a homotopy from gamma to neta okay so there is a homotopy from gamma to neta by intermediate paths which means I can continuously deform the path gamma to the path neta so I have this of course because the complex plane is simply connected any path like this can be deformed to a path like that okay and how do we write this homotopy so we are so given a homotopy of paths so I am drawing the diagram the other way maybe I should have drawn it to the right side but anyway it does not matter so let me do it like this so you know you have you have this you have this real line I mean you have the real plane r2 and I have the t parameter and I have here the s parameter and this is the unit square okay and I have a I have a continuous function f from here to here okay and what this continuous function does is so this continuous function is written as f of s, t okay and what it does is that if you freeze an s if you freeze s then it gives rise to the path gamma s so f of s, t is gamma s of t okay. So for fixed s of course I have put I have you know of course we can use instead of using the unit square you could have used a, b, c, d you could have used a product of two closed intervals but I am not doing that or perhaps I could even do that I mean it is alright our gammas are all or paths are all defined on a, b so let me also use that so let me do this maybe I can take some a here and b there and my homotopy could be defined on c, d could be like this I mean the argument is going to be no different so I can as well have it like this. So I will get this so rather I will get this rectangle which is given by a, b cross c, d okay so t lies from a to b and s lies from c to d so this should be c okay so the picture could have been like this so what is happening is that gamma s is for every s in for every s if you give me a certain s value which lies from c to d I have this gamma s which is defined from a, b to the complex plane giving so you know if I fix a value of s from c to d and if I take the and if I let t to vary from a to b I am getting going to get this line okay and the and as I move as t moves along this line this is going to trace the curve gamma s okay gamma is a path alright and when t equal to you know when s equal to 0 I have gamma when s equal to d I mean sorry when s is equal to c I have gamma and when s is equal to d I have neither so you know so all these intermediate paths they all started time t equal to 0 so this is time t equal to 0 this is sorry not t equal to 0 t is equal to a and this is time t equal to b okay you think of it as a time parameter and you think of the interval a, b as time it is time as a parameter as you are moving along the path okay so what is happening here is all these paths gamma s of a is always z0 gamma s of b is always z1 so this is what is called a fixed end point homotopy that is all the paths involved they start at the same point and end at the same point okay and gamma a, gamma c is your gamma gamma d is neta okay so this gamma is gamma sub c this neta is gamma sub d and you have in the intermediate for intermediate values of s between c and d you get the various intermediate paths so actually what is happening is that you know you are having you can think of all these lines you can think of various lines like this and the images of all these lines are going to give rise to this so the image of this whole square I mean this whole rectangle is this it is just a deformation of this except that you have collapse this whole end to the point is it not and you have collapse that end to the point z1 okay so you know if I take this thing and collapse it I will get something like this okay this capital f is just a continuous function now so this is a this is so modern theorem says that you are given paths gamma and neta which are homotopic and what else are you given you are given an analytic function here which can be continued along gamma and not only along gamma it can be continued along each of these paths okay I am given a starting analytic function at this point f which I can analytically continue along any of these paths that is given to me and the question that the modern theorem answers is about what is the function you will get at this end when you go along different paths the modern theorem says you will get the same function you will not get a different function you could expect that if you go along gamma you got one function but maybe if you go along neta you may get another function and if you go along some intermediate path you may get yet another function okay so you would expect that if I start with f and analytically continue along gamma s I will end up with the function fs and you would expect that fs could change but the modern theorem says no it says that would not happen it says always you will get back the same fb which is one and the same function that you would get analytically continuing f along any of these okay so the modern theorem says that if you have a homotopy like this and there is no obstruction that means a given analytic function at the starting point can be analytically continued along all these paths then the function you get at the end along any of these paths is they are all one and the same you are going to get only one analytic function right so that is what we are going to prove so let me write that down if the analytic function f and z0 can be analytically continued along any gamma s then the result of any such analytic continuation d the same analytic function at z1 so this is the modern theorem if you have if at the starting point z0 I am given analytic function and suppose I can analytically continue it along any of these intermediate paths intermediate paths of course you also include the starting path gamma and the ending path eta okay then no matter along which path you continue analytically the final function that you will get at the other end point at the terminal point z1 it will be the same that is what the modern theorem says. So if you want to state it in compact language modern theorem says well analytic continuation of an analytic function along homotopic paths will lead to the same function if paths are homotopic then analytic the result of analytic continuation of a given function will be the same that is what it says okay but of course the technical point is that you must make sure that the analytic continuation is possible along every path in the homotopy that is a that should not be some there should not be some hole here that should not be for example it should not be it cannot be like this you know that for example the origin is here and here you are starting with the branch of the logarithm and certainly a path which crosses the origin that is a path along with the branch any branch of the logarithm cannot be continued analytically. So such a thing should not happen there should not be a point in this leaf like region that I have drawn there should never be a point where there is obstruction to analytic continuation there should not be a point where I cannot continue certain the function that I am worried about that should not be there so there should be no obstruction to analytic continuation so well let me say a few words about how we are going to prove it the method is very very simple what we are going to prove is it we already seen this lemma that you know you can find certain delta rather we have seen this lemma which says that nearby paths have give rise to the same analytic continuation okay. So what we are going to do is we are going to show that you know you can and how and the nearby there was given by a delta so we are also going to find a delta here okay it is a different delta okay but that delta will give you you know distances so that will break up these homotopies into you know strips like this and along each strip the analytic continuation are going to be the same okay so by breaking this whole thing just like I have done in this diagram into finitely many strips of thickness delta okay and by using the fact that along each of these successive pairs of paths the analytic continuation along the top path is the same as the analytic continuation along the bottom path and then you do this finitely many times you will get the analytic continuation along the first path will be the same as the analytic continuation along the second path and then the analytic continuation along the second path will be the same as the analytic continuation along the third path and then you go by induction and finally you end up saying the analytic continuation they will all be equal to the analytic continuation along the last path okay. So this is the idea of the proof so the idea of the proof is to be able to to exactly draw a diagram I mean diagrammatically to get something like this breaking the leaf into smaller leaves of thickness delta okay that is the idea of the proof. So idea of the proof is pretty easy so I will explain that in the next lecture.