 okay so we are looking at these two versions of the monotromy theorem okay and trying to prove that they are equivalent okay. So basically so we have monotromy theorem version 1 is so the diagram is like this suppose you are having two points z0 and z1 and you have two paths gamma and eta both are both parts though both paths are defined on a closed interval AB in the real line and ticking values in the complex numbers and suppose and you assume that you are given function f analytic at z0 okay. Suppose following things hold number 1 the path gamma is homotopic to the path eta okay gamma is homotopic to eta okay which means that you can start with gamma and then you can continuously deform gamma into sequence of intermediate paths which lead from gamma to eta okay like this. So that is gamma is homotopic to eta and then number 2 f can be analytically continued not only on gamma but on each of the intermediate paths the homotopy including eta. So f can be analytically continued along any path in the homotopy in other so the homotopy is given by continuous succession of paths which start at gamma and end at eta okay and the assumption is f the function f which is analytic at z0 can be analytically continued along each of these paths that occur in the homotopy. So another way of saying this is as I had said earlier there is no obstruction to the analytic continuation of f along any intermediate path okay all these intermediate paths you can analytically continue f okay. So you must remember that first of all what we are assuming is that along each path f can be analytically continued okay in other words it means that there exist some analytic continuation of f along each path but then I already told you in the earlier lectures that you know if you have a path okay and if you have a function analytic function the starting point then the analytic continuation of f along that path is unique we have already seen this. In other words what this means is that this hypothesis mean that along each path there is only one analytic continuation of f there is only one analytic continuation possible on each path and it is there okay but the only question that the monotomy theorem answers is what happens after you analytically continue f along each path what is the function you get at the other end point z1 and the monotomy theorem says that all the functions you are going to get after analytic continuation along any path at z1 they are all going to be one and the same okay then the result of analytic continuation continuation of f along any of these any of the paths at z1 is the same. So you analytically continue this function f along any path it includes the starting path gamma the ending path beta and any path in between okay and when you come to the point z1 the function that you will get the analytic function that you will get that will be one and the same that is what the monotomy theorem says okay. So to say it in short if you analytically continue a function along two homotopic paths then the result of the analytic continuation is going to be this same at the end point you will get the whatever analytic function you get by continuing along one path it will be the same as the analytic function you will get by continuing along another path the only requirement is that these two paths should be homotopic to one another okay and of course we have assumed that through every intermediate path in the homotopy analytic continuation exists okay. So it is a kind of statement which assumes existence and gives uniqueness okay so it assumes the existence of analytic continuation on all these paths intermediate paths okay the starting function is the same the analytic function you are starting with that the starting point is the same and what it gives is the uniqueness of the analytic function that you will get at the ending point okay that is the monotomy theorem. So this is version 1 okay and then what is version 2? So what is version 2? Version 2 is it is a following statement suppose you have a suppose we have function element okay namely consisting of a pair of an analytic function and domain on which it is defined then if you take then for any such pair you have two we have defined two sets one set is called the maximal analytic extension of the given function okay it is called the region of maximal direct analytic continuation of the given function okay and then there is another set which is called the region of regularity of the given analytic function or which is also called the region of indirect analytic continuation of the given function and what monotomy and we have already seen examples that you know the region of regularity okay that is the region of indirect analytic continuation of a given analytic function can be bigger strictly bigger than its region of maximal direct analytic continuation for example the take any principle branch of the log take the principle branch of the logarithm then the region of maximal analytic continuation will be the slit plane you will have to throw out the negative real axis whereas if you take the region of regularity it will be the punctured plane so the region of regularity is bigger than the region of maximal direct analytic continuation in the sense that the region of regularity also includes points on the negative real axis and the reason is because across points along the negative real axis you can continue the log function any branch of the logarithm in an analytic way an analytic continuation exists along across any path which goes through points on the negative real axis okay so what the monotomy theorem version 2 says is that it gives you a situation when you can say that the region of maximal direct analytic continuation is equal to the region of regularity that is the region of maximal direct analytic continuation is equal to the region of the domain of maximal analytic continuation is equal to the domain of maximal indirect analytic continuation they are one and the same and the answer is that it will happen when the domain of maximal indirect analytic continuation namely the domain of regularity is simply connected So let me state that when the domain of maximal indirect analytic continuation it equals the domain of maximal direct analytic continuation when the domain of maximal indirect analytic continuation is simply connected then that domain also becomes the domain of maximal direct analytic continuation okay that is the that is version 2 of the monodometer okay and now what I want to say is that these 2 are these 2 versions are equivalent and then I will prove this version so I will also prove that they are equivalent so it means I would also have proved this version so let us first look at the proof of well yeah let us assume version 1 and prove version 2 so what we will do is we have assumed the monoderma theorem version 1 okay and we will prove version 2 okay so what so you are trying to prove version 2 that means you have the hypothesis of version 2 which means that you are assuming that you have a function whose domain of maximal indirect analytic continuation is simply connected so let f let u,f be a function element f that is f is analytic on the domain u with domain of maximal indirect analytic continuation analytic continuation which I have also earlier called as a domain of regularity is v2 is simply connected is simply connected so that is this I do not know what notation I used probably I used v2 alright so v2 is the is the domain of regularity of f namely see remember v2 is it is the union of all those points in the complex plane such that there is a path from a point in u to that point along which f can be analytically continue so you know v2 so the diagram is something like this you have so this is if you want this is u and this is the function f it is defined on u and if I if when do I put a point z in the complex plane when do I put it in v2 if you can find a point in u z0 and you are able to find a path gamma from z0 to z and you are also able to analytically continue f along that path to z all such z you put in the set v2 okay and it is clear that you know I have already told you that whenever a function is analytic on a domain inside that domain if there is any path then you can always analytically continue f that is trivial because you will get the trivial analytic continuation alright analytic function can always be continued to any point in its domain okay and that will be just trivial analytic continuation you are actually getting back the same function okay but the whole point about indirect analytic continuation is that you might go around a loop and you might end up with a new function like what happens if you go around the origin once and start with the branch of the logarithm you will end up with the next branch of the logarithm okay so the point is it is always trivial that if you have a analytic function on a domain if there is a path inside that domain on that path it can always be analytic will continue namely it will be you will get back the same function at every point of the path alright but the question is you are if you are able to find a path that goes out of the domain along which you can still continue the function all those terminal points of all such paths you put together you get another set v2 and that turns out to be again open and connected okay and therefore it becomes a domain and this is called the domain of regularity so v2 is a domain of regularity of f okay and the point that is given to you is that v2 is simply connected alright so you know so let me write let me just write down what I said v2 is a set of all z in C such that there exists z0 belonging to you and a path gamma from z0 to z along with along which f can be continued analytically this is this is a definition of v2 and v2 by this definition itself I we checked last time it is very easy to check that v2 is both open and connected so it is a domain v2 contains you and is open connected hence domain and that is the domain of regularity and what is given is that v2 is simply connected it is given that v2 is simply connected what do I have to prove I have to prove that v2 is equal to v1 which is a domain of maximal analytic extension okay I have to prove v2 is equal to v1 see we also define v1 to be the domain of maximal direct analytic extension namely it is a largest open subset of open connected subset of C where the function f can be directly analytically continued to given analytic function okay and we if we call that as v1 we want to show v1 is equal to v2 okay so of course so let me you know it is obvious that v1 is contained in v2 right because it is obvious that v1 is contained in v2 because v1 contains a direct analytic extension of f and you know a direct analytic continuation is also an indirect analytic continuation you can treat that also as an indirect analytic continuation namely the trivial analytic continuation alright. So let v1, g1 be the pair with be the function element with v1 the maximal direct analytic extension the domain of maximal direct analytic extension or continuation of f and g1 the extended function so you know that means v1 is a domain v1 contains u, g1 is analytic on v1, g1 restricted to u is the same as f okay it is a direct analytic continuation of f alright and v1 is the largest possible domain it is the largest possible domain to which you can extend f alright. So let me write that down v1 contains u is the largest domain in the complex plane such that g1 restricted to u is equal to f okay this is the maximal domain of direct analytic extension or continuation of f okay and of course you know since g1 extends f it is very clear that v1 is contained in v2 okay clearly v1 is contained in v2 why because you see after all you know if you take any point in v1 okay then that point in v1 is connected to u by a path okay that is because you know see any two points of v1 are connected by a path why because v1 is a domain it is an open connected set and an open connected set is also path connected okay an open connected set is path connected. So v1 is a open connected set therefore any two points of v1 are connected by a path therefore if you take a point of v1 and a point of u they are of course connected by a path inside v1 okay. Now along that path at the starting at the point of u you have the function f which is g1 restricted to u and out at the but throughout that path g1 is defined. So you can think of g1 as a direct analytic continuation of f along that path okay so what it means is that you can analytically continue along any path from a point in u to a point in v1 and analytic continuation of f is just given by this function g1 therefore all these points in v1 will also get in will also be points in v2 okay the only problem the only point with v2 is that it might contain points at which you will not be able to directly analytically continue f okay but at which you may be able to only indirectly analytically continue f okay that is why v2 could be bigger than v1 okay. So certainly v1 contains v2 alright now what we are what the monotromy theorem version 2 says is that if v2 is simply connected then v1 has to be equal to v2 that is what it says we are given that v2 is simply connected and you have to show v2 is equal to v1 so you have to show that v2 is containing v1 okay we need to show v2 is equal to v1 you have to show that alright that is what you have to show. So how do you do that it is pretty easy so what you do is you know what you do is so what we do is see v2 is going to be equal to v1 then at every point of v2 also you can directly analytically extend f and that extension is going to be g1 because g1 is a maximal analytic direct analytic extension of f to the subset to this largest possible domain v1 okay. So in other words you know what is happening what is happening is that v2 in v2 what is happening is you are getting various points to which you can add directly you can indirectly analytically continue f along paths but you do not know what the function is that you are going to get at the end of the path okay but trying to show v2 equal to v1 is the same as trying to show that even after you analytically continue it the function end function you are going to get is actually g1 which is just a direct extension of f okay you are trying to show that for every point z in v2 okay whenever you have point z in v2 whenever you which means that you have path from a point in u to that point whenever you analytically continue f what are you going to get you are going to get just g1 you are not going to get anything else that is what you are trying to show. So what we will do is we will define a new function so you know the picture is something like this you have u of course you know the way I am drawing the pictures I am always drawing bounded domains I am drawing simply connected bounded domains okay but they need not be like that what is so this picture is only to help you to think alright but it is you should not take it to be accurate alright so the sequence of things is that you have u the set on which f is defined domain on which f is defined that is contained inside v1 that is the you know maximal domain to which you can extend f and the maximal extension is g1 okay and then there is v2 and v2 consist of all those points to which you can analytically continue f you know that you can analytically continue f but you do not know what is the function you are going to get and the claim is that if v2 is simply connected then v1 is the same as v2 that is the claim. So what you do is you do the following thing take a point z in v2 by definition there is a point z0 in u and there is a path let me call this path as gamma sub z alright and there exist an analytic continuation of f so I start with f here and I get a function f sub z okay so now define g2 from v2 to c by g2 of z by g2 of z is equal to fz of z so look at this definition look at this definition it is a very clever definition but anyway very simple definition the definition is very simple you give me a point z of v2 small z in v2 then by definition there is a path which nzz starts at a point of u say z0 and along that path your f can be analytically continued and you know once you know we already know that once you can analytically continue f along a path then the analytic continuation is unique as far as that part fixed path is concerned or along the same path you cannot get two different analytic continuation there is a uniqueness of analytic continuation along a given path okay that we have already seen. So the ending function that you are going to get is going to be some function I am calling that function as f sub z because it is at the end point z alright and my what is g2 of z it is f sub z of z I am defining a function like this okay this function is well defined function there is no problem mode it is well defined as why so here is where the simply connected hypothesis will come see the point is this z is connected to z0 by a path gamma sub z okay. Now what you must understand is you know I could have taken instead of taking z0 I could have taken z1 okay so let me draw it so that I can draw so instead of z0 I could have taken z1 so this is z0 I could have taken z1 you know and I could have gotten another path and this path well I can call it as neat as z right so try to understand what is happening when I so this is the this is the this is where I am trying to say that this function g2 is well defined okay I am trying to say that g2 is well defined. See what I said earlier was f sub z is well defined f sub z is well defined because it is analytic continuation of f along the fixed path gamma gamma sub z okay which starts at z0 which is a point of view but then for the same z I might have another point z1 in u I may have another path starting neta z starting from z1 and ending at the same z along which also f can be analytically continued it can happen after all it can happen and as far as the path neta sub z is concerned if I continue the same f along neta sub z I might end up with another function okay that tells me that this definition seemingly is not well defined because this is f sub z depending if we take the path gamma sub z and this is also equal to well if you if you go along the path neta sub z you might get some let me call give some other name to it g sub z of z if we go if we take the path neta sub z okay it can happen and if you want g2 of z to be well defined these two have to be the same otherwise it is not well defined okay these two have to be the same otherwise it is not well defined. Now the point is that you know well the point is you know z0 and z1 they belong to u anyway okay so what you can do is you can connect z0 to z1 by a path delta okay you can connect z0 to z1 I think delta is not a very good let me use some other symbol let me use lambda okay so you connect z0 to z1 by a path lambda okay and now watch I start with f along lambda take the trivial analytic continuation because after all lambda is a path inside u and along a path inside u you can always take the trivial analytic continuation which means you your analytic continuation is the same function f along each point of the path that means along each point of the path you are simply writing the power series of the same function f at centered at that point you have the analytic you have the trivial analytic continuation of f along lambda okay followed by the analytic continuation starting with f at z1 and leading to gz along neta z okay so if you put them together you will get the analytic continuation of f to gz along the path which is gotten by lambda followed by neta subset alright on the other hand you also have the analytic continuation fz which of f starting at the point z0 via the path gamma subset alright now you see the path lambda followed by neta z is homotopic to the path followed to the path gamma z that is because both are because both point end points are inside v2 which is simply connected and both paths start at z0 and end at z1 see what is the property of a simply connected region if you have two points any two paths starting at these two points starting at a fixed point and ending at a fixed point okay any two paths are homotopic to each other okay therefore what will happen is that the path lambda followed by neta z is homotopic to gamma z but now I have assumed monoramy theorem version 1 which says that whenever you have two paths which are homotopic to each other and along all of the intermediate paths the there is no obstruction to analytic continuation along see both these paths the region between the paths is a region inside v2 and inside v2 there is no obstruction to analytic continuation because v2 consist of all those points where you can analytically continue f okay so both conditions of the first monoramy theorem first version are satisfied okay so that will tell you that fz is equal to gz you will get fz equal to gz just because of version 1 okay so that will tell you that g2 is well defined okay. So let me write this connect lambda here connect z0 to z1 to sorry z0 to z1 by a path lambda in u then gamma z is homotopic to lambda neta z by lambda neta z I mean lambda followed by neta z concatenation of two paths to give a new path okay also the homotopy also any also a homotopy can be found from gamma z to lambda neta z via such that each intermediate path is in v2 along which analytic continuation of f exist so you see I can find I want a homotopy between gamma followed by neta z and f so I can find intermediate paths like this and all the intermediate paths are lying in v2 okay why am I getting this homotopy because v2 is simply connected okay since so here is where v2 is simply connected is used so this is where I am using simply connectedness of v2 any two paths starting at the same point and ending at the same point are homotopic to one another for points and paths lying in a simply connected domain okay so because a simple connectedness of v2 these two are homotopic and the homotopy can be chosen in such a way that every intermediate path is also lying inside v2 but what is the property of v2 any along any point for all points in v2 you can always you have indirect analytic continuation so along all the intermediate paths of this homotopy okay also you will have indirect analytic continuation so there is no obstruction to analytic continuation of f along each of these paths so you know all the conditions of version monotromy theorem version 1 are satisfied and therefore what this will tell you is that fz is equal to gz so let me write that down so maybe I will draw a line like this thus all conditions all hypothesis of version 1 of the monotromy theorem hold so gz will be equal to fz these two functions are the same okay so what this tells you is that g2 is well defined so I have managed to define a function g2 okay so let me again emphasize you give me a point in v2 I am going to define a function I am going to define a value at that point and what is the value I define I do the following things since it is a point in v2 I can choose a point of v1 I mean I can choose a point of u and a path along all of starting from that point and continuing and I continue f analytically I can continue f analytically along that path and I will get a new function at z I take that function take its value at z I call the new function as f sub z I take its value at z that is how I am defining g2 the only problem is that I could have gotten analytic continuation along some other path starting from some other point in u and it is because of simply connectedness and of v2 and it is because of the assumption of the first monotromy theorem the first version of the monotromy theorem that I am able to show that g2 is well defined okay now g2 is well defined g2 is analytic is very easy because g2 in a neighborhood of the point z okay g2 is simply the analytic function fz fz is of course an analytic function okay at z so g2 is just f in a neighborhood of z okay so g2 is locally analytic and therefore it is analytic because analyticity is a local property to check that a function is analytic you have to just check that at every point it is analytic okay so since g2 coincides with an analytic function locally it is analytic okay so let me write that since g2 coincides with the analytic function fz in a neighborhood of z g2 will be analytic okay so what you must understand is probably this also requires a little bit of little bit of more thought namely what you must understand is you know so let me draw one more diagram so that you know you really understand what is going on so you know this is so this is my u then I have this b1 on b1 I have this f which is defined here extending to the maximal extension g1 and then I have this v2 which is assumed to be simply connected you see I take a point z at this point I define g2 to be fz g2 of z to be f of fz of z this is my definition and how did I get that fz I start at the point z0 here I chose a path lamp I mean gamma sub z along which f can be analytically continued and at this point when I reach this point z I get a new function analytic function locally there I call that as f sub z and take I take its value at z okay now what I want you to understand is well if you take a small if you take this neighborhood where this f sub small z is defined that will be some disc okay surrounding the point z so suppose this is a disc where you know f sub z z is defined alright suppose this is certainly f sub z is analytic function at z so it is going to live in it is going to be analytic and disc surrounding z now what I am trying to tell you is you choose any other point z prime in this disc okay then the g2 at z prime will be the same as fz at z prime okay g2 at z prime will be the same as fz at z prime okay and why is that so so I mean that is the claim I am making here g2 coincides with fz in a neighborhood of z okay so what is going to happen so the situation is going to be that you know I am going to have another point say z2 this is a point of view okay and z prime is here alright and you know by definition z prime is in this disc where fz is defined but the point is z prime is in v2 so which means that there is a point of view z2 and then there is a path like this from z2 along which f can be analytically continued and you know this path can be called something well let me use something so let me use capital gamma of gamma sub z prime so capital gamma sub z prime is a path starting from a point z2 in u along which I continue f and then I get a certain function f sub z prime whose value at z prime is what I have defined g2 of z to be z prime to be so g2 of z prime is just f sub z prime of at z prime where f sub z where f sub z prime is the analytic continuation of f along capital gamma sub z prime from z2 in u to z prime this is how I have defined my g2 my claim is this fz prime is same as fz in other words I am saying g2 is always the analytic function fz it is only one function that is the claim and what is the reason for that the reason for that is very very simple see you see the reason is you know I do the following thing you see I connect if you want I am I am in a disc so you know with z is a center z prime is another point of the disc I can actually take a radial line like this I can take a line from z to z prime ok along z to z prime I can analytically continue fz trivially ok and now if I take this point from z0 if I take this path starting with z0 along gamma sub z and follow it by this line from z to z prime I will get a path from z0 to z prime ok and along that path if I analytically continue I will only get fz prime because after all I mean I will only get fz because after all I started with f at z0 I analytically continue it along gamma z I ended up I ended up at z when I ended up at z I got fz ok when I move along this radial line and come to z prime I am still keeping the same I am simply trivially analytically extending the fz at z to fz at z prime because that radial line is in that disc where fz is well defined and wherever a function is well defined I can always I always have trivial analytic extension trivial analytic integration. So from z0 I have another path to z prime along which my analytic continuation gives rise to fz but I have already proved that if you take two different paths starting from two different points of view and you analytically continue f and you end up at the particular point the function you get is the same ok. So the moral of the story is that by connecting z to z prime by a trivial path and by taking a trivial analytic continuation of f sub z along that along that radial line ok and putting these two things together you can see that fz prime is the same as fz ok but fz prime is equal to fz they are the same ok. And therefore so this implies that g2 is equal to fz in a neighbourhood of z in a disc surrounding z ok that is the statement I made here since g2 coincides with the analytic function fz in a neighbourhood of z g2 is analytic it is locally analytic so it is analytic at every point z g2 is analytic that is all. So I have managed to produce a global function on v2 which is analytic and what is that function restricted to you that function restricted to you is just f that function restricted to you is f so what you have proved is that you have this global analytic function v2 the g2 which is defined on v2 and which is directly extending your analytic function f on you so it proves that v2 also is contained in v1 it tells you that v2, g2 is a direct analytic continuation of u, f so it means that v2 and v1 is supposed to be the domain of maximal direct analytic continuation so v2 has to be contained in v1 we have already proved v1 is contained in v2 so you get v1 equal to v2 and we get the external analytic function to be g2 but g2 but g2 is what g2 has to then be equal to g1 ok so that finishes the proof that version 1 implies version 2.