 Okay so this is the continuation of the previous lecture see what we saw in the previous lecture was that if you have an analytic continuation along an arc or a path then which is given by a family of power series also parameterized by the path variable okay then the coefficients a n as well as the radius of convergence r are continuous functions of the path parameter okay. Now what I am going to do now is going to tell that because of this I am going to tell that the whole analytic continuation all the functions f t is for t greater than a they are all completely determined by f a namely the initial function so I am so in other words I am saying if you have a path for which you have fixed a parameterization and if you have an analytic continuation of a function along this path then the analytic continuation the function the analytic function at the starting point of the path determines all the other analytic functions that correspond to power series later on in the path okay it is a so in so what it says is if you give me a particular parameterization of a path and if you start with an analytic function at the initial point then all the analytic functions that you will get at various points by analytic continuation they are all uniquely determined okay in other words for the same parameterization of a path you cannot find two different analytic continuations for which the power series at t are different with the initial power series being the same okay. So let me write that down so here is a given analytic continuation f t of z is equal to sigma n equal to 0 to infinity a n of t z minus gamma of t to the power of n mod z minus so this power series is valid in this disc of convergence which is mod z minus gamma of t centered at gamma t it is r t along a path gamma so the statement is f t for t greater than a is uniquely determined by f a so the starting so the power series at t equal to a namely the initial point of the path the value of the parameter t when it is equal to a it is the initial value of the parameter and the corresponding power series f a is the initial power series you are starting with and I am saying that that initial power series that analytic function that determines all the other all the other analytic functions they are going to get in terms of power series along as you continue analytically along that path okay. So in other words I am saying if you give me a path with a fixed parameterization and if I start with an analytic function at the initial point okay if I do any indirect analytic continuation along that path okay then at the end of the path I am going to get one and the same analytic function not only at the end in fact I am going to get given any other point on the path the analytic continuation of that function I am going to get at that point is going to be unique okay so the analytic continuation is completely controlled by the initial function and the path that is all that is what I am trying to say okay. So what is the proof the proof the proof essentially actually you know if you think about it is just tautological to this lemma that a n and r of t are continuous functions of t okay but let me explain that so you know so it is a proof by contradiction alright. So suppose suppose the lemma suppose the theorem where false where false then it means that you start with f a fixed you could have there could be a certain t for which you could get different you could get different families of power series okay which both start with the same f a the same function f a but at later on at some point for some value t of the parameter greater than a the corresponding power series are different analytic functions okay if that is what it means to contradict this theorem okay. So let me write that then there exists g t of z is equal to sigma n equal to 0 to infinity b n of t z minus gamma of t to the power of n mod z minus gamma of t is less than so you know I will use r tilde of t because r tilde of t will be the radius of convergence of g t of z of the power series g t of z centered at gamma of t okay. So I am using a different r okay and such that which is also an analytic continuation along gamma starting starting with g a equal to f a and with g t not equal to f t for some t greater than a okay. So the claim is that the claim is that you give me one analytic continuation like this along the path then that is the only thing you can get so long as you fixed the initial function f a okay and how do I oppose this claim I oppose this claim by saying that there is another analytic continuation okay with the property that it also begins at a with the same function so g a is f a so this power series at t equal to a starting point is the same as this power series as functions they are the same okay at the starting point but for a certain value of t greater than a the corresponding functions I am getting a different okay that is how I am contradicting the statement of the theorem and now I have to get a contradiction. Now now it is now a contradiction is very evident because you see you see you know the contradiction will just come in a very easy way let t not be the infimum of t in a in a b such that ft is not ft is not equal to gt okay so there is one t beyond a for which ft and gt are different functions okay so you can look at the smallest such t you can take the infimum you can look at the least such t then of course you know f a is equal to g a therefore this t not is greater than a clearly t not is greater than a no doubt about it t not cannot be a because f a is equal to g a alright and you see when I take infimum of all these t's okay you know the infimum is also it has also a limiting property the infimum of a set of real numbers which is bounded below is a limit of those set of a suitable sequence in that set okay so there is a whenever you study infimum and supremum in a first course in analysis you always have the so called approximation property of the infimum or supremum which says that the infimum or supremum is actually a limit okay so it is a limiting property. So this infimum is a certain limit of all t's with a certain property and it exists because all these numbers all these t's are bounded so this is you know if you recall a first course in analysis you would have come across this statement that you know that every subset of real numbers which is bounded below has an infimum is equivalent to saying that every subset of real numbers which is bounded above has a supremum and that is equivalent to the completeness of the real line okay it is a very deep statement. So it is also equivalent to the monotone convergence theorem that any sequence of real numbers that is decreasing and bounded below converges or equivalently any sequence of real numbers that is increasing and bounded above also converges okay all these statements are equivalent to the completeness of the real line which is a deep fact okay so the infimum exists because this all these t's are bounded below by a okay and the infimum has the infimum if you call it as t0 it has a limiting property so it is a limit of a sequence and since the closed interval a b is closed the infimum also belongs to that interval okay and the infimum cannot be a it has to be greater than a right now I will get easily a contradiction for any t close to t0 and you know lesser than t0 if you compare see because I have to use the fact that you know ft is a ft is a is an analytic continuation therefore you know for t prime for every t for t prime sufficiently close to t ft prime has to be equal to ft similarly for all t prime close to t gt prime also has to be equal to gt so what you will get is for all t prime very close to t0 and lesser than t0 okay you will get ft prime is equal to gt and ft prime is also so you will get ft prime so ft prime is equal to ft0 and you also get ft prime is equal to gt0 and it will contradict the fact that ft not is not equal to gt not okay so let me write that down we have ft prime is the same as we have t0-epsilon okay and you will have and that will be equal to f that will be equal to g t0-epsilon which is g equal to gt prime okay you will have this see you will have this because ft is an analytic continuation so for t prime close to t0-epsilon ft prime has to be equal to ft not-epsilon and but ft not-epsilon has to be equal to gt not-epsilon because t0 is the smallest of those values for which ft is not equal to gt so anything lesser so the infimum is called the greatest lower bound anything less than that is not a lower bound okay so t0-epsilon is lesser than the infimum so it is not going to be I mean so this condition is not going to be true for t0-epsilon so which means that ft not-epsilon should be the same as gt not-epsilon right because this t0 is the least if you go to the left of t0 the ft is and gt is will be the same right so these two are equal because the definition of t0 and these two are equal because gt is an analytic continuation okay but also ft not-epsilon is the same as ft not and gt not-epsilon is also equal to gt not for epsilon small enough this is also true alright probably this is the maybe even the previous statement is not important maybe this is the statement that is important see if you choose epsilon small enough then ft not-epsilon should be the same as ft not and gt not-epsilon should be equal to gt not okay alright and so you will get so this will tell you that ft not is the same as gt not okay and you see the same argument will also apply instead of minus epsilon if I put plus epsilon also it works see because here I can put instead of minus I can put plus or minus okay here also I can put plus or minus that is because ft is an analytic continuation if you take t not and you take a small neighbourhood around t not okay then all the ft prime should be the same as ft not okay and similarly since gt is an analytic continuation if you take gt not and you take a small neighbourhood surrounding t not all the gt prime should be equal to gt not okay and so what this will tell you is for epsilon small enough so I can put plus or minus and then you will of course if you take it with the plus you get a contradiction if you compare these two alright and if you compare these two alright what you will get is that ft not plus epsilon and gt not plus epsilon will be the same for epsilon small enough so this implies ft not plus epsilon is equal to gt not plus epsilon for epsilon small enough a contradiction to the definition of t not see I am just looking at a neighbourhood of t not okay if I look at values before t not then all the ft primes in the gt primes are the same alright and they should also represent the same function to the right of t not also okay but then to the right of t not ft not and g the ft is in gt are supposed to be there are supposed to be points to the right of t not as close to t not as I want where ft and gt are different because of the approximation property of the infimum and that is a contradiction okay so let me repeat this if you look at the point if you look at the point t not at the point t not if you took at the power series ft not that ft not has to be equal to the power series ft not plus or minus epsilon for epsilon sufficiently small and gt not also similarly has to be equal to g sub t not plus or minus epsilon for epsilon sufficiently small but if you take minus if you look at t not minus epsilon okay then the ft's and the gt's are the same because the definition of t not so what this will tell you is that ft's and gt's are also the same for values greater than t not which is not for all values in a neighbourhood to the right of t not but that is not supposed to be because given however small a neighbourhood to the right of t not there is a value t for which ft is not equal to gt that is the approximation property of the infimum okay so so this contradiction it is a settled contradiction uses some very basic analysis so this contradiction tells you that your theorem is true okay so this is that is the end of the proof so the moral of the story is the following the moral of the story is you know if you give me a parameter parametric path if you give me a path and you have parametrized it and you start with the function at the starting point of the path then that then any analytic continuation along the path if it exists then it is uniquely determined okay so in particular so everything depends on the parametrization of the path and this initial function okay everything depends on that fine so anyway the moral of the story is that if you give me a if you give me a parametric path and if you give me an initial function then the then the then there is only one analytic continuation which starts with that function there cannot be more than one okay but of course it may happen that there is no analytic continuation at all there may be a path along which you cannot continue a function analytically that could happen so what I want to do next is I want to tell you about two things first of all you could have a path along which you cannot have an analytic continuation at all okay and that will happen for example even with a function which has a non-removable singularity at a point okay so you know for example so let me make this definition definition we say f analytic function function f at z0 is analytically continuable along a path gamma starting at z0 if there exists an analytic continuation ft t belonging to ab where gamma is the path from ab gamma is a path parameterized with a parameter variable in this interval such that fa is f and of course and of course gamma fa is z0 so I am saying that if you have a point z0 and you have a path starting from z0 and you have function which is analytic in a neighbourhood of z0 if you we say that it is analytically continuable along the path if you can find a analytic continuation given given by a family of power series parameterized by the path parameterized by the interval which also parameterizes the path such that the beginning function is your given function okay. So the fact is so the question is so our aim is he starts the analytic function at a point try to look at all possible paths along which you can continue it and try to find out what is the analytic function you are going to get at the end of the path okay so that is the aim and the philosophy is that if you do this then you will get all possible branches of that analytic function okay so this is by so this analytic continuation will give you all possible branches of a given analytic function right. So there can be analytic functions cannot be continued along the path so for example you know so here is an example if f is analytic on a domain D then of course f is analytically continuable along any path in D then f is analytically continuable along any path in D gamma in D it is very simple right because you see so you know if you have some if you have some D here in the complex plane and it is a domain and you know if you have a path if you have a path like this gamma then you know so f is defined here values in C any analytic function within its domain of analyticity can be continued along any path in the domain of analyticity it is very very simple what you do is you just define at each point of the path just define the power series to be the power series just the Taylor series expansion of f at that point that is all okay. So it is it is trivial that if a function is analytic in a domain then it can be analytically continued along every path in that domain and and all these all these analytic continuations are not going to produce anything they are going to give you back the function f okay. So so you know if you have this situation you just define f t of z is equal to power series that is Taylor series of f centered at gamma that is all if you make this definition then f t will give you here automatically it will give you a analytic continuation of f along that path and what is the analytic continuation it is a trivial analytic continuation it is simply you are going you are not going to get anything you are going to just get back the same function along the whole path okay all the f t is okay all the f t is the power series will be different because the power series will change as the center of the power series changes the power series will change the coefficients will change okay but finally what function do they converge to they all converge to the same function f. So all the f t's are as a function they are the same f okay but only thing is if you write them as power series you get different different power series and the power you get different power series because you have changed the centers okay. So this is this is just a statement that if a function is analytic on a domain then I mean you have it is trivially analytically continuable on any path on the domain okay then the other important example is of a function which you cannot analytically continue and that is the case when you your path ends at a point which is a bad point for the function it is a singular point for the function. So example you know if you take f of z is equal to 1 by z cannot be continued analytically continue continue any path passing through the origin that is obvious because the origin is a you know it is a simple pole okay. So you cannot analytically continue it along a path which crosses the origin because at the origin you cannot find after all the function 1 by z is defined its domain of analyticity is a punctured plane it is a whole complex plane minus origin. So only point where it is not defined is the origin where at that point it has a simple pole alright now at that point you cannot extend it okay so if you have any path passing through the origin you cannot analytically continue it. So you know this gives you a hint that you know somehow if you keep track of all possible paths along which your function can be analytically continued then you will get points which are good points reasonably good points for the function okay. So this leads to a definition so let me make a statement so here is one more important example the analytic function you know log z which is principal branch of the logarithm of the logarithm which is given by log z is ln mod z plus i times principal argument of z where principal argument of z is taken from minus pi to pi this has maximal domain of analyticity the slit plane namely complex plane minus you remove the line segment from minus infinity to 0 okay. So it is a slit plane you just throw out this 0 the origin and the negative real axis on the rest of it this is analytic okay that is an analytic function you cannot extend it to cannot extend this analytic function to any point on the negative real axis okay simply because at any point in the negative real axis the argument function is discontinuous and the argument function is the imaginary part of your log function okay. So you cannot do anything so you take this analytic function now the amazing thing is you know you take a path that does not cross the origin even if it crosses this negative real axis you can still analytically continue it okay. So even though this whole negative real axis is full of bad points its points they are all non isolated singularities you see they are all this is a whole continuous ray it is a continuous line they are full of singularities for the function okay in spite of that you know if you draw an arc which crosses the negative real axis you and then you can really analytically continue log and you know what it will be you are going to continue log to the next branch which is actually a continuation which is happening on the Riemann surface of log z which is being reflected here okay. So log z can be analytically continue on any path in C minus 0 it is amazing even if that so even if that path crosses the negative real axis even if the path crosses the negative real axis you can analytically continue it. So you are able to so here is an example of an analytic function which can be analytically continue across a point along a path which passes through a very bad point and here is one that cannot be continue along a path which passes through a bad point. So you see so this is so you know analytic continuation is one more tool of trying to distinguish between analytic functions in terms of properties okay. So I will try to explain this in more detail later but I will like you to check that this is the case okay. So in particular I would like you to try it as an exercise you take a path which goes like this you start with log z here okay start with log z here and at this point you take log z so let me write it so here I will take log z plus 2 pi i okay and this log z plus 2 pi i is certainly analytic continuation of log z along that path you can make it okay. So I want you to try that as an exercise show that along this path which crosses the negative real axis log z can be analytically continue to log z plus 2 pi i. So you know the beautiful thing is even though you are crossing this this ray there is another avatar of there is another branch of the of the log function which continues it which leaves okay across across that negative real axis is full of bad points okay. So I wanted to prove this as an exercise so okay so show that if you take a arc like this going from the upper half plane to the lower half plane crossing the negative real axis you start with log z above and you end up with log z plus 2 pi i two different successive branches of the logarithm show that you can actually you can find analytic continuation that goes from here to here okay. So our aim is you know you have a function which is analytically continueable and you want to know what is the final function that you get at the end of the analytic continuation okay that is the aim. So the monotrom there is something called the monotromy theorem which says when the final analytic continuation is going to be the same even if you change the path okay. So there are many versions of the so called monotromy theorem so I will give you one I will give you the first version of the monotromy theorem monotromy theorem so here is the monotromy theorem. So the idea is like this you see I start with a point z0 I start with a point z0 and I end with a point z1 okay and I have path okay gamma okay and along the path gamma I have an analytic continuation given by a power series ft of z sigma n equal to 0 to infinity a nt z minus gamma t the power of n for mod z minus r of t less than sorry z minus gamma t less than rt. So this is one analytic continuation of fa to fb okay where a is this is just z0 is gamma a and z1 is gamma b you know that fb is completely determined by fa okay we have seen that for this particular path. Now suppose so as far as I fixed you see if this path is fixed parameter is fixed and starting function is fixed then I know that every analytic continuation I am going to get later on each of those functions is fixed that we have already proved that is that was this theorem okay. The question is how will the ending function change if you change the path so you know instead of gamma suppose I give you another path eta so eta is another path which also starts at z0 and ends at z1 okay and suppose you have another analytic continuation along eta okay such that eta t of z is given by something sigma n equal to 0 to infinity a so some b nt z minus eta t to the power of n mod z minus eta t is less than r so here may be for clarity I will put r of gamma of t and here I will put r of eta of t. So suppose I have another analytic continuation okay and assume that eta a is the same as f a okay with the initial functions are the same f a is the same as sorry I should have used not eta, eta is the path I should have used g so let me use g so f a is g a okay. So I am having on this all the f t's it starts with f a here and ends with f b here and along this path I have g t's which starts with g a which is same as f a and I am going to get a g b what is the relationship between f b and g b so the question is the monodermy theorem 1 version of the monodermy theorem says when you can guarantee that f b and g b are the same and the answer is it is a topological answer it is when gamma can be continuously deformed to eta okay. If you can continuously deform gamma to eta then f b and g b will be the same in other words if you start with a given analytic function and you analytically continue it along a path that analytic continuation you get at the end of the path is going to be independent of the path so long as the path the paths are the same up to a deformation of one another which is what in topology is called as a homotopy okay. So that is monodermy theorem version 1 there are other versions which also I will explain so given analytic continuation this and this with f a is equal to g a if the path gamma can be continuously can be continuously deformed is homotopic eta then f b is equal to g b okay. So if you if you start with the function and so long as you deform it along so long as you analytically continue it along a path the final function that you get is not going to be is not going to change if your path is going to change only up to a continuous deformation namely up to a homotopy okay. So so this is this has a very nice statement in terms of covering space theory also it can be rephrased as the action of the fundamental group the action of the fundamental group on the germs of analytic functions on the covering space okay so it can be rephrased I will try to explain that also in a later lecture. So the point I want to make is that if you have a path gamma and if you have another path eta which are homotopic this is called fixed endpoint homotopy which means that you know you can find you can find paths like this continuously you can find a continuous sequence of paths a continuous family of paths which start with gamma and end with eta like this so I deform this path gamma to the path eta if I can do that then the analytic continuation along a eta and the analytic continuation along gamma and the analytic continuation on eta along eta will be the same in fact it will be the same for every you take any anything in between which is a deformation for all the cases the final along any of these paths you know the function you get there upon analytic continuation it will be the same as this FB you will get only one okay so that is the content of the monodermity okay. So I will explain the proof of that in the next lecture okay.