 OK, good morning. We continue today the study of the automorphies of the plane, and then we extend the study to the case of the Riemann sphere to some other sets. So last time, just to call you that, we considered function from C into C. So entire functions holomorphic with inverse holomorphic. And after some consideration, quite simple consideration, with the tools we have so far developed, we concluded that the only possibilities are the following functions, with a different from 0. Correct? So we actually, just to call you what we did, actually consider entire functions and established by showing that the image of any neighborhood of the point at infinity is not tens, that the point of infinity cannot be an essential singularity for the function f. Similarly, we also said that for topological reasons, the point at infinity cannot be a removable singularity for the functional analysis. It is constant, so it cannot be invertible. And so it can be just the case that f, the entire function f has a pole at infinity, but in this case, the pole has to have order 1. So it becomes a polynomial of degree 1. So the set of automorphies of C, which will be denoted by AUT, automorphies of C, is in fact the set of function a and b are complex numbers, and a is not 0. So quite simple. Now let us observe that this transformations, we call transformations when we refer to function which represent a transformation of a set into itself with an inverse. So any transformation is an anti-transformation. Of course, includes the identity for a equal to 1 and b equal to 0. And you can invert each of these functions. So this is a group with respect to the composition of functions. This is a group. Composition is this. Small circle is a group. When we have a group of transformation, or more in general, say a group or semi-group transformation acting on a set, we say that we have a action. So the general setting is the following. We have a set E, set, and say semi-group G. And we associate to the pair G with G in G and E in E, an element E prime in E. With this property, when we consider another element of G, of the semi-group, and we consider GG prime, which is an element in G with this bullet to represent the operation in G. This is an element in G. Then we have that this element is in fact acting on E in the following way. This is like the action of G over G prime. This is the only request to have an action. Then if the semi-group has the identity, necessarily the entity acts on the set leaving everything fixed. It is expected. So this defines an action on the set E. It can be whatever you want. Very abstract and very poor algebraic structure. It's a very poor algebraic structure. So semi-group is a minimal pair request. So in particular, we can apply to our considerations this terminology and observe that given a transformation like this and it acts on C. Well, this transformation is phi. And phi of Z is an element of Z. So we can say the pair phi and Z defines an element phi of Z in C. This is an action. And it is readily seen that if F1 acts after F of Z, this is F1, which is in C, as expected. Now for actions of semi-group or groups on a set, there are some special properties which are sometimes fulfilled or not. In particular, we say that the action of G on E is transitive or G acts transitively E F for any E, E prime in E. There exists G in E such that G E is E prime. Sorry. It's mapped by the action to E prime. So the action is called transitive when you have two points on a set. And the first point is mapped onto the other, into the other, by an element of the group G. This is the general setting. So in our case, we could say that the automorphies of C acts transitively on C. In fact, given Z0 and W0 in C, I have to find transformation in C. I have to find A and B such that transformation associated with the choice of this pair of A and B, which defines the function in out C, maps, for instance, Z0 into W0. An example is the following while taking F of Z to be the map Z minus Z0 plus W0. This is an element in out C. With the property of F of Z0 is W0. Geometrically, we have here what? A linear transformation as was pointed out last time. But geometrically, this is a translation in the plane. And here, we have a dilation and a rotation so that we have all the motions, the rigid motions of the plane. That explains geometrically if you want the property we have described here analytically. Any two pair of points can be associated by, can be say, map one into the other by one of these transformations, right? So, geometrically means that, okay, we can do it. And actually here in this example, it is easily seen that we don't have even to rotate anything. See, the coefficient of Z is one. So, there is no rotation component in the transformation. But just we consider the translation, okay? So, we have two points in the plane. This is Z0, this is W0, and a transformation is this. This is exactly this vector if you wanted to play, right? Well, this is an example of transitive action. So, we continue studying now the case of the Riemann sphere and the automorphism of the Riemann sphere, okay? Let us try to describe this set, all right? So, as I said, we cannot expect to have a holomorphic function in the entire Riemann sphere to be something different from a constant function, right? Because restricted to the plane, so without the point of infinity, it will be bounded. Because it is continuous function on a complex set. Then holomorphic, bounded, holomorphic entire, bounded means constant. Because of u, v of d1. So for sure the function we will be dealing with have to have at least one pole. Okay, good. So, we can just say, well, with the experience of working with transformations or linear transformation and the plane, what we expect to have is something which is the ratio of two linear transformation, right? Because this is an example of a polynomial, of a rational function whose denominator is a polynomial of degree one, and the denominator of degree one as well. Correct? We cannot have higher degrees because otherwise the order of the rational function cannot be greater than one. Because the function cannot have multiple zeros of poles of order greater than one, because they have to be injective. So I will start considering this fact. Take this M, that be the set of transformation, which are as follows, okay? With A, B, C, D complex numbers, and it is really seen that in order to have numbers, this function, this is function in the Riemann sphere, right? It can be invertible if this number is different from zero. I invite you to check, but it is, well, I think something standard for math class, okay? Well, in fact, can I associate a matrix, two by two matrix to this transformation and notice that this is nothing but the terminate, right? So if you want to invert it, what the terminate is to be different from zero. In any case, I invite you to verify that when you put W to be A, Z plus B over C, Z plus D, then you can obtain Z as a function of W if and only if A, D minus B, C is different from zero, okay? Now, what I can say is that with this assumption M, the set of the transformation, I describe this in the set, so M, the set M is in fact contained and set of half-term of C, correct? So any such transformation with these assumptions here is in fact transformation of C hat. Of course, we have to define this function apparently is defined only for Z different from minus D over C, right? And it is not defined for Z equal to infinity, but in fact we extend this when Z is minus D over C, the value of the function is infinity and when Z is infinity, the value of the function is A over C, which is exactly what you have if you replace what is inside, outside, you know, with this demographic projection, put one over Z and calculate the function is zero. Take the reciprocal and everything works fine. So with this position, this function is defined for N is A on the Riemann sphere and it is invertible. It is one to one. So it is an automorphism of C hat. So now the problem is, is this M exactly out of C hat? So other transformation beside the function we have described so far, which can be not an automorphism of C hat. Well, in this case it can be helpful to remind a very, very simple fact from algebra, you probably know, okay? Assume that you have, let me use the same notation. Well, assume that you have a subset of a group, sorry, a subgroup of the group G or transformation, G acts on E and H is a subgroup of G, right? And assume that H contains G x naught. It is indicated by G x naught. What is this? This is the stabilizer or the isotropy subgroup and then I say with respect to 2 x naught. What do I mean by this? I mean the set of all transformation which keeps x naught fixed, which maps x naught into itself, okay? So G of x naught has a set of G and G such that G x naught is mapped into x naught. I am sorry, this is not, what I am saying here, this is the pair. It is not the action. The action is indicated in this way. So, if you want, well, normally this is. But with this, we do not mean that G is a function of x naught, okay? It is, okay? The action. In the previous example, the two notation, in fact, are the same because we calculate the value of a function at a point and this was said to be the transformed of the pair transformation, okay? But this is the same most general that. I am not saying that this is the case for only for out C or for something else, but this is for the general. Well, what is the statement of this? Simple lemma. And I refer to it as the algebraic lemma, okay? The statement is that actually H, okay? If G x naught is containing H, then H is G. No, for one, it is not. I show you this. Take G and G, okay? And calculate, oh, way, I forgot to say something. Yes, wait. And H acts transitively. There are two points, we fulfilled. So, take G, any element of the group and so calculate G of x naught, which is not necessarily x naught, right? Something else. Otherwise, G would be in G x naught. So, in general, this is something else and E, okay? Now, since H acts transitively on E, there exists an H, an H such that H of 0 x naught is x naught. I can take two pairs of point and I find transformation. So, in particular, I take this point and the other point is x naught. In such a way that this is, in fact, H was G of x naught. So, that H was G as in G of x naught, okay? So, this is in G of x naught. H is sorry, G of x naught is in H. So, this is in H. In particular, H is in H. G is in H. That's it. So, for any G, in G, we prove that G is actually in H, the subgroup we consider with these two properties. And we apply this lemma for our, this is an algebraic lemma, right? For our situation. So, in fact, we have, we have that set M, the set of transformation of this form, which are called also linear fractional transformation, okay? So, these are called linear fractional transformation. And yes, I agree with you. In some textbooks, they are also called main transformation. But I will use the terminology for a special class, okay, of linear fractional transformation. So, I prefer to say linear fraction to say the general case, and then maybe transformation for a special subclass, okay? If you don't mind. But in some books, and also in what we'll see later, this, this set is called maybe transformation. If I say linear fractional transformation, I just refer to this and maybe transformation to something else, okay? So, what we have is that M is a subgroup of, so, out C hat plays a role of the group G in the previous lemma, and this M plays a role. So, eventually we'll play the role of the subgroup H, okay? Let me just take G in, well, M acts transitively on C hat, on this lemma sphere. While we can arrange A, B, C, D in such a way that the point, for instance, 0 is not infinity, okay? For instance. And this is what we need, because of course in M, we can also think of having the automorphism of C with, which is extended to the case of infinity, if the infinite is mapped into infinite. It is the case when C is 0, okay? Which gives you one transformation of out C. Actually, out C, the automorphism of C can be regarded as the isotropic group of infinity. In fact, the transformation of A, Z plus B, when extended to the lemma sphere with the assumption that infinity is mapping to infinity, keeps infinity fixed, okay? Good, well. Now, take G in out C hat, generic. We can, we can assume that after composing G with C of Z, which is Z minus G naught, Z minus G of infinity. G of infinity is a number, it is a number, is an element of C hat, right? Then we can assume that G of 0 is 0 and G of infinity is infinity. If not, we use this transformation and we consider G tilde to be C composed to G, okay? Well, C belongs to M, why? Because G naught and G infinity cannot be the same, right? Good. Now, with this assumption, this function G is in fact, since G of 0 is 0, in the power expansion of G in the neighborhood of 0, okay? The A naught is 0, okay? So, that I can consider G of Z over Z and this is an entire function because we don't have any problem of poles. We have no problem with poles since A naught is 0. So, when we divide, we obtain a power expansion of Z without denominators in Z, right? So, this is an entire function, all right? And since G of infinity is infinity and G is injective, the function G of Z over Z is, well, this is smaller than R. That is to say, this function here is bounded. It is bounded only at infinity. All the other values are bounded. Therefore, G of Z over Z is constant, all right? And then, G of Z is lambda C, which is still in M, okay? So, G is in M, okay? So, this proves that M is actually the set of automorphism of the Riemann sphere. So, linear fractional transformations are, in fact, the set, form the set of the automorphisms, they will say. Notice that if we imagine to associate to the transformation A Z plus B C Z plus D, this 2 by 2 matrix, matrix, okay? We are, this is in, okay? Since we have the hypothesis, linear group 2 C, right? It means that 2 by 2 matrices with entries in C and who is determinant is different from 0. But we also observe that if we take lambda, any lambda in C, okay? Lambda, of course, different from 0. Now, we multiply each coefficient here times lambda. That is to say, we consider lambda times A B C D is the matrix lambda A, lambda B, lambda C, and lambda D. So, this matrix we associate the same transformation, because we have a numerator lambda A Z plus lambda B over lambda C Z plus lambda D, which is A Z plus B over A Z plus D. So, that we can define not just one matrix associated to transformation, but a class, okay? A P 1 thing, a line of classes, a line of matrices. So, that this automorphism, this set of automorphism C hat is described, is better described by the PSL to C, that is to say the special linear, projective special linear group of matrices, 2 by 2 matrices in C, okay? So, these are matrices with entries such that A D minus B C is plus or minus 1. This is another representation if you want of the same group, which is important to know, but all right? And now, what are simple examples of these transformations? Well, the simple examples are the following. One simple example is essentially this. This is one of the cases, right? When C is 0 and B is equal to 1, right? Or a particular this. So, translations can be considered a special case. But then also, this transformation, these all belong to M, right? This transformation, the inversion, okay? One of the functions you studied in your exercises is in fact, transformation is an invertible function on the Riemann sphere, okay? Of course, this can be regarded as an element of M when we extend this function also to infinity, putting infinity is mapping to infinity. So, and if we take, so we have simple translation, inversions, and then we can have also dilation and rotation. For instance, this is also my example. Now, take this S1 to be Z plus D over C, take S2, sorry, S1 over Z, right? S2 over Z to be 1 over Z. And S3 over C, sorry, this is translation, so inversion. And here we put a BC minus AD, BC minus AD, which is different from 0 over C2 times Z, which is dilation, a complex dilation, right? Which means there is a factor which rotates and a factor which dilates in real sense. And then we also add this transformation, which is another translation. So, we have two translations. This is ordered, right? S1 is 2, S3 is 4. And if you consider S4, S3, S2, S1 of Z, this gives you precisely, so that any transformation, any linear factor of transformation is in fact a combination of translation, inversion, and dilation, okay? So, I invite you to verify this with this. I checked once, okay, that this choice of coefficient is correct. So, you can arrange in case in order to make it correct. But the idea is clear, right? So, in fact, now I want to show you, show to movie about this fact, okay? Movie, yes. Okay, these are translations. As you can see, this is dilation, right? The square is dilated and then it will be rotated. What is difficult to see is the inversion, right? To imagine. So, what is in, out? Good. But in general, the complete KT situation is that the combination, okay? Of this is an inversion, rotation, and rotation, and then probably also, so now Riemann sphere comes in spinning, the rotation, and then inversion is this. Okay, so we have in some sense that any linear factor of transformation is a combination of simple, or simpler linear factor transformation is a translation, dilation, and inversions. And this is important to remember. Now, I prefer not to use the terminology which is somehow adopted in other cases, that is, maybe it's transformation for linear factor transformation, because I refer to maybe it's transformation just for automorphism of the disk. So, we just consider two cases of automorphism sets of unbounded sets, right? So, consider transformation of the plane and of the Riemann sphere. Now, I want to consider just one case, and it will be not the one case, but the case, as we will see, of a bounded domain. And the bounded domain we choose is denoted, will be denoted by D, okay? Blackboard D is the set of Z and C such that the modules of Z is smaller than one. This is called the unit, okay? Now, assume that you have, now, comes another lemma, quite interesting, it is called Schwarz lemma, which is not, which is a tool for many other things, but it will be fundamental for the description of the automorphism of the unit disk. Assume that you have a function f, holomorphic, from the disk into itself and assume that this function maps zero into zero. So, then we have the following fact, then f of Z as modules smaller or equal to modules of Z and f prime of zero as modules smaller or equal to one. Now, furthermore, f, we have that f prime of zero is in modules equal to one, or there is a Z naught in D, but not zero, such that modules of f of Z naught is equal to Z naught. So, we have an equality here or an equality here, right? Then necessarily, there exists theta in R such that f of Z is EI theta times Z. So, it is a rotation keeping zero fixed and EI theta is precisely f prime of zero. So, all this information are in this very simple, apparently, Schwarz lemma, but it is considered a lemma, this is a fundamental theorem in the theme. As I will see, the proof is extremely simple compared to the importance of this lemma and I don't know precisely that the history of the evolution, you know that in the history of mathematics, there are some kind of approximation of results. Normally, there are some first steps, which describes something else, then some other else comes in and says, okay, but maybe we can remove this hypothesis, okay, you can adjust the conclusion. I think that this version is due to Poincaré. Okay, but in any case, this is known as Schwarz lemma. So, the proof, the proof goes like this and well, I didn't put it here. This is 8 and this is 9. So, since we have a automorphic function keeping zero fixed as already observed in the previous case of the automorphism of C hat, then we can consider this function and this is holomorphic. Since f of zero is zero, it's holomorphic in D. Once again, I repeat the idea, take the expansion to zero of this function and A naught, so the first coefficient is not bearing. So, it's not in business. So, that we can divide by Z and the result is an analytic function. Good. So, this is holomorphic. We take now, this is the unit disk, this is zero. We take this disk, R smaller than one and greater than zero, okay, and we consider the modulus of f of z over z. If you want, call this function g of z. This is the modulus of g of z. Well, we can say the following. Since the maximum of the modulus is taken for sure on the boundary, we can recycle our consideration on the boundary, right, because this is a holomorphic function. So, we apply maximum principle. So, that, well, this number here is smaller than what? The maximum on the circle of f of z and modulus of z is r, right, but the maximum of f since f of d is contained in d can be at most one, because f of z is in d. So, modulus of f of z is smaller than one. So, this is one of r, right. Remember that f of z is smaller than one as modulus smaller than one and the maximum is taken at the boundary of the disk. Maximum modulus principle tells us this, right. So, this is valid for any r. So, when I take the limit as r tends to 1 from the left of modulus of g of z, this is independent, right, and so we obtain that this smaller is tends to 1, but this means then f of z over modulus is smaller or equal to 1, right, or f of z is smaller or equal to, as modulus smaller or equal to modulus of z. Similarly, this number here, this, sorry, this function here can be regarded as the incremental ratio of f at 0, because it is f of z minus f of 0, this is also f of z minus f of 0, which is 0 over z minus 0. So, that this, the same fact tells us that the derivative of 0 of f is a modulus smaller or equal to 1, and this is the first part of the statement. Let us come back to the same, okay, description, because everything is on this inequality. Assume that you have an equality at one point. Assume that f of z0 over z0 is modulus equal to 1, which means that, okay, we have equality here, right, at a point inside the disk, different from 0, of course. Then we have that the function g is in fact achieving its maximum inside the disk, so it is constant. It is constantly equal to z0 in modulus. So, after rotation f of z is z0, okay, which tells us exactly what we want. So, this has very important consequences. One of these is that we can describe the automorphies of the unit disk in the following way. We will describe it to be the set, which we will call the Mabuse transformation, the set of Mabuse transformation, defined as follows, where theta is a real number and z0 is an element in the disk, which should be somehow familiar to you because in one of the exercises without this rotation factor, we have to study this and say, okay, this function, okay, z0 is given in D, okay, this, all this family is formed by functions which are holomorphic in the entire disk up to the boundary, actually. So, it is holomorphic in the closure of the, the reason is that, well, this is polynomial and this is also polynomial, this is a ratio polynomial. The only problem could be that this polynomial vanishes. In fact, it vanishes. It has a pole, but it is outside the unit disk. So, inside the unit disk, this function is holomorphic, but this doesn't mean anything else except that first we have to prove that it maps the disk into itself. So, we consider this difference. This difference, if we prove this difference is positive, then we are done. So, we prove that the image and the z is in D, of course. Taken any z in D, if this number is positive, we prove that the image of any z and for any z0 in D is in fact in D, okay. So, the modulus of EI theta is 1, so we don't really care about, but this becomes 1 minus z0 bar z square minus z minus z0 square over 1 minus z0 bar z square, which is a positive real number when z is and z0 are taken in D, sorry. So, to prove that this is positive, it suffices to consider this difference on the numerator, right. So, we calculate this separately and prove that this number is in fact positive, okay. This is, okay. The module square is the number times its conjugate, which gives us the following. So, we have 1 minus z0 z bar minus z0 bar z plus and then we have modules of z0 square times modules of z squared. Then minus modules of z squared, then minus modules of z0 squared and then I have plus z z0 bar plus z0 z bar, which luckily cancels this to other z0 z bar and minus z0 bar z cancels z0 bar z here. So, at the end of the day, we have 1 plus module of z0 square module of z squared minus z square minus module of z0 square, which can be also written as 1 minus module of z0 square times 1 minus module of z squared, which is positive. Since we are taking z0 and z in a unit disk, okay. So, this is the claim. So, the first step is to prove that these functions are in fact mapping the unit disk into itself and this is done. They are holomorphic and this is important. Are they invertible? Yes, they are because there are special cases, there are special cases of the linear fractional transformation. The coefficient is e i theta. We can see, oh, sorry, sorry, sorry. We cannot see, sorry, sorry. We can prove it directly, but if you want, you can regard this automorphism of the unit, well, this set and this may be transformation as a subset of the linear fractional transformation, right, with specific choices of the coefficients. They are not the freedom of choosing a, b, c, d as you want, but you have just that you choose b and a and then you have c and d given automatically and b and a is to be taken, well, a modules 1, b with modules smaller than 1, okay. So, verifying that this is the case, so we have e i theta times d minus and then the modules of Zeno square, right, which is smaller than 1. It cannot be that a, b minus c, d, a, d minus b, c is equal to 0. We are on the safe side. However, if you don't want to use this such an argument, you can prove it directly and say, well, if w is e i theta Z minus Z naught over 1 minus Z naught bar Z, what is if I want to express Z in terms of w? Is it always possible? Yes, it is, as we will see, and it is instructive to make it once in your life so that you can see how is the inverse of the function, maybe its transformation given. So, in fact, this is w minus w Z naught bar Z. So, if I collect everything, say on the right hand side, for instance, I have the z e i theta plus w Z naught bar is w plus e i Z naught 2. That is this number here. Remember that w is in d, Z naught bar is in d as well. So, this can never have modules equal to 1, this product. So, you can never cancel this, which has modules 1. So, we can divide. This is never 0. So, we have this is, which looks like memory transformation, but then we have to adjust and put this way e minus i theta and I have 1 here plus, which is sorry, minus minus e i theta with a minus in front. So, minus i i theta Z naught bar and then w and here I have w minus e i theta minus e i theta Z naught. So, call c to be minus e i theta Z naught and observe that this is e minus e i theta w is minus c over 1 minus c bar w and this belongs to d. So, it is as expected another maybe transformation with different coefficient Z naught and different theta minus theta. So, it is important to remember this. And finally, what I will not prove and invite you not to waste your time in this kind of calculation is that when you compose to such maybe transformation, you obtain a maybe transformation, which apparently is the easy part to be proven, but it is not. So, take call this this one e i theta 1 Z minus Z 1 over 1 minus Z 1 bar Z, the first transformation and the other is theta 2 Z minus Z 2 over 1 minus Z 2 bar Z call it phi 1 over Z phi 2 over Z both phi 1 and phi 2 are maybe transformation, maybe transformation. So, what is the situation when I compose these two maybe transformation? Well, in fact, this is the odd part, a lot of calculations we done, but luckily some someone else have already done it before of us. So, we can just copy the results and trust him. But in any case, just for the sake of completing, I show you that this is e i theta 1 plus theta 2 then along this is e minus i theta 2 Z 1 Z 2 bar over 1 plus e i theta 2 Z 2 Z 1 bar, which of course, has modulus 1. This is the this is a complex number and here we have the conjugate of this complex number. So, this has modulus 1 and will play the role of the constant in front e i something times Z minus I write this way H over Z 1 over 1 minus H of Z 1 bar of Z, where H of Z 1 H of Z sorry is e minus i theta 2 Z plus e i theta 2 Z 2 over 1 plus e i theta 2 Z 2 Z and then put a bar here, which has the aspect of a maybe transformation. So, what we have time being is the following, we do not we cannot describe anything else, but a subset of the automorphies of the unit disk. So, I call M. So, this this M is different from the other ones as the set of maybe transformation either you write this way or you put this double dots umlaut in German on the letter some maybe transformations and I can say that M is in fact a subset of the the automorphies of the unit disk because I prove that well it is stable I proved I claimed that it is stable with respect to composition and I proved that actually each of this is a transformation of the unit disk or it maps the unit disk into itself it is holomorphic with the numbers which is also holomorphic. What is left to prove is that actually we have an equality here and this can be done well theoretically this is very quite difficult to prove there are subsets and anything else, but this can be done putting together the algebraic lemma and the false lemma well notice that I take f from the unit disk into itself and assume that it fixes zero all right then in general we have that f of z as modules module equal to modules of z right. So, in particular take f in the subset of the automorphies of the keeping zero fixed so take the isotropic groups of the identity those identity of the origin right so we restrict our k our consideration to the those automorphies of the unit disk which keep zero fixed so we know in general this is valid not for the automorph for any holomorphic function that when it fixes the origin then we have this inequality now this means that in particular f has an inverse because it is an automorphism and we indicate this inverse f minus one now we have that if w is f of z the modules of w which is the modules of f of z is smaller equal to modules of z right, but the modules of z is also the modules of f of minus one of w right because we are dealing with an automorphism and in particular since f is in this isotropic subgroup also f minus one belongs to this isotropic because of course is zero is mapping to zero the inverse maps zero into zero so also for f minus one we can apply schwarz lemma and we conclude the following this is so that this f is e i theta c because we have inequality right so any element of the automorphies is in fact sorry any element of the automorphies group is contained all right so because of the description of those function which map the unit disk into itself keeping zero fixed in general in particular for the elements of the isotropy subgroup at the origin we have this and this quality right which this w is in d which implies using schwarz lemma this that the function is the rotation right good that is starting from here we can conclude that it is maybe transformation it is maybe transformation it is for z not equal to zero therefore we have the set of maybe transformation contains the isotropy subgroup of zero which is the first ingredient the secret ingredient is that well it is easily seen that starting from any z with then sorry with starting from any z naught in the disk this map map this written here this transformation written here map z naught into zero and this suffices to prove to say that m acts transitively we have a point any point this point is mapping to zero take another point z one this is mapping to zero by the corresponding maybe with the choice of z one here takes take it its inverse so we have the composition of the two maybe transformation which is still a maybe transformation maps z naught into zero take the inverse zero to z one so maps z not to z one for any pair of points in not and z one right are you okay okay so we have the two condition m acts transitively maybe i write it on this final slide m acts transitively on the sense for any z naught m d this function call it field z naught even without the rotation coefficient is such that f of z naught calculate the z naught is zero therefore if i take z naught and z one in d and i define f of z one to be this maybe transformation i compose f of z one minus one with f of z naught this is transformation so it has an inverse and this is still a maybe transformation and this is the maybe transformation which maps z naught into zero from zero to z one that is z naught to z one good and also i have shown using the schwarzlema that m contains the esotope group of the origin therefore m s sorry sorry this is because of the algebraic lemma where m plays the role of the former h and this is esotope so group in the d is the set e right in the general stuff in the general setting now we have a description of the automorphism of the unit disk and this is important so we have also description of of set of transformation acting on a bounded set so just to conclude let me point out that we can have some extra result from the schwarzlema several results actually and in particular what is somehow easy to remove is what apparently is essentially in the proof is to say that f of zero is mapping to zero so take any homomorphic function f from the unit disk into itself f of zero is called w zero which lies in d well in general so consider the generic situation called w naught f of z naught all right so z naught is somewhere here and w naught is somewhere else so what i'm doing is the following i compose the function f with two maybe transformations such a way that the composition maps zero into zero all right so i have this preferred pair of points z naught and w naught two two points if z naught is zero and w naught is zero so we are exactly in the situation where the assumption of schwarzlema but if not so consider f of z naught to be z well the inverse so i want this okay this is a maybe transformation which is the inverse of the one with the minus okay and then take those so phi of f of z naught to be z minus f of z naught over one minus f of z naught bar z okay i could put also this as w naught okay then i take phi of w naught composed f composed phi of z naught minus one observe that this function here maps zero into z naught so it's inverse map z naught into zero but no i take sorry i take this function precisely so they maps it maps zero into z naught correct then this maps z naught into f of z naught is w naught and then good f of z naught here cancel the numerator so this is a function which has is holomorphic and the it's a composition of holomorphic functions in the and such that of this function g g of zero is zero so let us repeat again so zero is mapped by phi of z naught into z naught f maps z naught into f of z naught or w naught and phi of f of z naught maps f of z naught into zero all right so what we know is from schwarzlema then modulus of g of z is smaller or equal to modulus of z and then we have the other two conditions with the quality at a quality one point implies the quality at any point then and the function is up to a rotation value and here as precisely again f g prime of zero is modulus equal to one then it is that g of z is e i theta z so the statement i'm going to summarize as known as schwarzlema and tells us the following take f from d into d holomorphic then for any z and z naught and here we have that f of z minus f of z naught over one minus f of z naught bar f of z is smaller or equal to z minus z naught over one minus z naught bar z and and also f prime how is it not as modulus if a quality holds then f a quality holds means then to be more precise if there exists a z and z naught a pair of points such that an equality holds or we have this equality for one z naught okay then necessarily f is a maybe it's not formation okay if a quality in one of the two inequalities so either you find a pair of complex numbers more smaller than one such that this is the case with equality or you'll think oh i have one point z naught in the disk such that this the derivative of f to z naught has modules satisfying this equality then necessarily f is an automorphism and this comes essentially from schwarzlema because here you can recognize that we have all the ingredients to conclude because this inequality written in terms of g okay becomes something else in terms of f just conclude with this very quick observation then we have remember that we have the g of z as module smaller or equal to z but g was was phi of f of z naught composed of f composed of phi of z naught right now call f of z naught z to be w so i can also say that what this is well g of sorry this is phi of f of z naught f of w smaller or equal instead of z i put phi of minus one of z naught at w right and this concludes everything because because this means that on the left hand side we have well i copy it and this is the modules of w minus z naught over one minus z naught bar w remember that phi of z naught was a function z minus z naught plus z naught one plus z naught bar z and this is the inverse the inverse changes the sign right we have no so the f a i theta equal to one so everything simplifies and this is the inverse and here we have f of z f of w sorry minus f of z naught over one minus f of z naught bar f of w which gives us the first inequality the second equality comes from the fact that g prime of zero smaller or equal to one then g was phi of f of f of z naught composed f composed phi of z naught then we apply the chain rule make the calculation and we obtain the other inequality but i think that this can be done not in two minutes so we stop here and then we continue on Monday thank you