 I want to talk about some consequences of the two theorems we have seen last time. In particular the Schwarz-Lehmann and Schwarz-Picke-Lehmann. I recall that Schwarz-Lehmann tells us something about functions which are holomorphic from the unit disk into itself and such that zero is a fixed point. So, for this class of functions of self-maps of the unit disk holomorphic self-maps of unit disk we have this important consequence. The t is that the distance from the origin of the image of any point is smaller or equal to the distance of this point which is a contraction principle. Assume that for instance you start from a point Z naught in general and you want to study what are called the orbits of Z naught. So, f of Z naught, f of f of Z naught and so on and so forth. Imagine that f is a law which applies to some set. In this case they open disk and you want to see what happens to the discrete behavior of the orbits. While this property tells you that either the function keeps the distance from the origin fixed the distance. So, it means that it has the origin fixed and it can at most rotate. So, change the argument but not the modulus and this is in fact what the theorem tells you. Or the distance of f of Z from the origin is smaller than this. So, there is a contraction and after repeating this procedure several times then you see that f of f of Z has some modulus which is smaller or equal to f of Z which is smaller. So, eventually so if you are not moving around a circle fixed distance from the origin. So, if you are not taking f to be a rotation then f contracts and maps the point up to the origin which is a fixed point. So, the limit point of the orbit is zero. This is important say in general in mathematics information when you have a contraction and you know something about the estimate how a map is pushed towards a point it is important to know. So, in some sense Schwarz-Picklema removes this hypothesis. So, consider just function holomotor from a unit disk into itself and with the use of the transitive properties of Möbius transformation it resembles the same result but with a different feature right because of the presence of the composition of with Möbius transformation. So, the result is like this for any pair of ZW and the unit disk and together with this estimate on f on the modules of f there is also an estimate in this case this is the estimate of the derivative of f at the zero which is smaller in modules smaller or equal to 1. When it is 1 then necessarily it is a rotation right the function is a rotation good and here we have the similar estimate which tells you the following. Now the question is now how can we use this to find an analog of the contraction principle in Schwarz-Picklema. So, the idea is to change the way of measuring the distance. So, this is the standard Euclidean distance right so this is the modules of a vector in R2. If we introduce a new metric right so then we obtain a new distance we will see that according to this new distance then the function which are holomotor from unit disk into itself and in fact contraction for this distance and I want to show you that this distance is not made out of crazy ideas it is somehow intrinsic with the problem okay this is the in the meanwhile let me just observe these properties in another from another point of view and assume that we want to see this in terms of an extremer problem. So, you have a disk you have the unit disk and you have two points Z and W and you want to know which functions which are homomorphic unit disk which map Z and W okay are such that the derivative at Z is maximum say Z0 and W0 just to take all right. So, you want to say okay I have this information the function maps Z0 into W0 consider F from D into D holomorphic F of Z0 is W0 consider this class of functions okay call it F of Z0 okay and consider the modulus of F at Z0 F prime of Z0 right and you want to ask which is there a supremum which kind of function to receive this supremum so we're taking the supremum in F of Z0 right well the answer is yes there is a supremum and it is actually maximum and the maximum is taken for F being maybe transformation which is maximum right and this comes from this second inequality okay well F of Z is W0 and F prime at Z0 is bounded and modulus is bounded by this ratio one okay let me write this way so we know F prime of Z0 smaller or equal to 1 minus modulus of F of Z0 of square 1 minus modulus of Z0 correct so this is in fact W0 this is true for any F in this class because we are dealing with functions which satisfy the the the hypothesis of Schwarz peak lemma they are holomorphic in the unit this into itself that's it in particular we have this information F of Z0 is W0 so that we can rewrite the second inequality given by Schwarz peak lemma in this way but we also know that when the maximum is reached at a certain Z0 then necessarily F is a maybe transformation so we can see that this problem can be solved with some function analysis problem right say give me description of function so is there any function which reaches the max the supremum max normal you say in principle this can be a supremum I can say no this is a maximum and this is taken for any holomorphic function which is an automorphism mapping Z0 into W0 okay so this is this is this solves an extramar problem if you want to see the problem in terms of with the with the point of view of functional analysis this theorem gives you important information right similarly in the case of a fixed point so go back to Schwarz lemma you want to know which of the function keeping zero fixed have the derivative of zero maximum the sense of modules okay which is maximum well this happens to be just a case of rotations only rotation can have F prime of zero of modules one all the others have modules smaller than one okay so these are information or description of so extremal functions for some problems describing this in this geometric terms and this depends on essentially on the maximum modules right because this is what we applied for the description and for the results Schwarz and Schwarz big lemma this is okay just a small remark on functional analysis now going back to the idea of introducing a new metric okay since we want to deal with the inequality in which appear in Schwarz big lemma it is natural to consider as element of distance d a squared be d z square over one minus modules square this can be somehow surprising that we can write it explicitly d z is d x plus i d y if z is x plus i y right so this is d x square plus d y square over one minus x square minus y square square if you want to avoid this normally the element of the metric that this is the differential expression of the metric right is the square because we normally use the square but if you want the s s module d z over one minus module z which means and it is the x so this is difficult to to write right plus x square plus d y square root over one minus x square minus which of course agrees it agrees with the standard distance at zero you see there is a function here depending on the point z so at each point of the unit disk the way of measuring the length depends on the point okay so that you can imagine to introduce metric on the disk as considering a way of measuring something on the tangent right because the tangent of the disk is the plane itself and one you have well this is something which is natural to do in real differential geometry you have imaginal surface and the surface is not embedded in R3 but you want to put a way of measuring the length of tangent vectors at each point you have to define this locally and then locally and in compatible way with the charts a way of measuring the length of the tangent vectors using a function in this case this is a global way okay this is general for any z this is well defined for any z and the unit disk because this number here is positive and finite but what you see is that as soon as z tends to become closer to the boundary of the unit disk so that its modulus is getting closer to one this number here tends to zero so that d of s tends to be very big okay so two points which are close according to Euclidean distance okay can have very long very large distance using this metric good now assume that this is something artificial and I want just to say okay this is a way to measure something okay let me justify the introduction of this not as a constant or say not as a generalization of the contraction principle in Schwartz lemma for function without the assumption that zero is fixed so let me check which are the functions for this which are isometries for this metric which means they preserve the metric so take a general complex valued f okay from the unit disk and to see in general complex function and write as usual the real and imaginary part of f as u and b respectively right good so I want to say how can I characterize the isometries for this metric so I have ds squared to be modulus of dz squared over one minus modulus of z squared and I say how can I characterize isometries for this metric the answer is well and just to make calculations I have no assumption on f look f is not holomorphic as I said f is just complex value five distance and I want to see something from the geometry point of view well keep the distance keep the distance in between points mean keep the metric because then I have to integrate the distance the metric to have the distance right so my idea is to consider well all these functions are have to satisfy this so f is an isometry for the s if and only if they preserve the isometry they preserve sorry they preserve the metric so it preserves preserves the way of measuring locally so this is a differential way you know locally the distance good pardon me no I'm I'm not assuming sorry I'm not assuming that f has any other geometric problem but that the function has well it has to be at least differentiable okay so the components have to be at least c1 but not necessarily satisfying any other relation so now nothing like the cr equation or anything else so well what we'll see is surprising that if you request if your request is that this metric is preserved then necessarily Cauchy-Liemann equation are satisfied or anti-Cauchy-Liemann equation okay let me just show you how it comes from calculation there is no extra motivation so what is this numerator here here is dx square plus dy square right what is here have u okay f is u plus i v right so df is du plus i dv so df is modulus as du square plus dv square so what is du du is ux dx plus uy dy this is real analysis right and dy dv is vx dx plus vy dy therefore when I consider du squared plus dv squared I have okay ux squared dx squared plus vx squared dx squared plus plus uy squared dy squared plus vy squared dy squared plus plus so I put together the same line the elements with the same okay in the same in the same this over the same type say dx squared dx squared dy squared then I will also have two ux uy dx dy plus two vx vy dx dy okay so this is this is y squared plus vy squared vy squared right this is df modulus of df squared right so I have this as a condition right so here numerator is this plus and the denominator is this on the left hand side the numerator is like this so this implies that this and this have to be the same because here they have the same coefficient and this has to be zero because the x dy on the right hand side is not appearing so that the two condition this condition implies that ux squared plus vx squared is uy squared plus vy squared and ux uy plus vx vy is equal to zero all right well notice that I am not saying that this has to be one this number here comes out as one minus modulus of f of z squared over one minus z squared square but it doesn't really care I mean this has to be the same as this which implies actually this is equivalent so try to solve this problem okay so if you add or if you subtract you obtain something like ux is equal to minus one power n vy and uy is minus vx okay so it is either one and minus one here which means that Cauchy Riemann's are satisfied these are the Cauchy Riemann's equations or minus one appears here and plus one appears here so that anti Cauchy Riemann they call so the function instead of being holomorphic are anti holomorphics okay instead of solving d bar z equal to zero is d z equal to zero so somehow what I have here is that well if you want to preserve the metric and you are a holomorphic and they are complex value function as I said you are the function you are considering are holomorphic or anti holomorphic okay this is Cauchy Riemann or anti Cauchy Riemann equations so that this is proposition any function any complex valued function which is an isometry for is necessarily holomorphic or anti holomorphic these holomorphics satisfies equations yes it satisfies anti so then just minus okay appears in the reverse position anti Cauchy Riemann equations and this is something quite surprising I would say right so geometric condition you put you can you can see this entirely in real terms they have a function from r2 to r2 to define a differential metric and you can write it in terms of dx and dy right well if you want this to be this metric to be preserved that is to say that the function is an isometry then necessarily additional conditions equivalent to holomorphic to or anti holomorphic are satisfied okay now how you define a distance from a metric so as I said in say in complex in sorry in real differential geometry what you have is a way to measure the length of a vector on the tangent space so in order to measure a distance according to this way of measuring the length of the vectors tangent to the surface you consider a curve connected these two points on the surface okay since you have a way to measure the length of the vectors tangent to the surface then you can measure the length of the curve according to this way how by integrating this rectifiable curve on the surface and measuring the distance between two points using the distance so the length of the curve since you have the the measure of the modulus of the tangent you integrate this and you have the length of the curve a rectifiable curve is okay the distance the distance the the length of a curve is different in for a rectifiable curve it's defined like this so how you define in a this this is just one possibility if you change the curve of course in principle the distance changes obviously right you have several possibilities then you take the infimum because the distance has to have some infimum properties and so you define takes the one and see two two points and that is similarly the distance which will be denoted in this way as the infimum on gamma gamma connecting z1 gamma rectifiable curve in delta uh connecting z1 to z2 and well this is right the distance ds because we define ds to be what you know as the well ds means uh exactly what I said right we have a metric and we integrate of a curve so we have to put the modulus of gamma prime so passing through through the standard change with the parameter gamma is a curve this is the distance and this distance is in fact an important distance it is a name it's a one correct distance in d so can you calculate this distance explicitly just to give you an idea what happens okay so yes we can this is okay the again the infimum of gamma of what of dz modulus of dz1 minus modulus z squared this is where dz here is well what contains the information of the tangent and here is the function which defines the way of measuring the length so the closer the point is to the boundary the bigger this uh this way of measuring the counts okay but this is a distorted way it's not like the standard distance and imagine this asher pictures of the you know the same triangle and then which becomes apparently small and small but the distance preserved actually okay in this in this way you know this is kind of pictures which appears in some book and some by cornelius asher he's a famous artist he made such a game okay by repeating the same triangle okay keeping the distance and so the the triangle becomes smaller and smaller for what we see but they are in fact this with the same distance of points using this one correct distance well it doesn't really matter now i i want to show you how this if you're not familiar with asher never mind i want to show you how you can calculate okay so we'll make some very simple examples and the example is the following start from zero and take z okay so i want to measure the distance from zero to point z and using this definition so i have to take the integral over any curve connecting zero to z and take the infimum of all this distance this is the general definition but we observe that well and this metric nothing changes if you rotate z so the distance from zero to z is the same of the distance from any point on the same circle center at the origin and of modules modules of z because if you replace with z with z z star being e i theta z while nothing changes here and here so in particular i will consider the case which is easy to handle the case where z is in fact on the real axis okay i have the freedom of choosing any z and i take this z because the definition i have taken for the this for the metric is independent for rotation therefore this is as i said infimum over gamma of what the z over 1 minus z squared right and i show you this that this is in fact the integral between zero and modulus of z of what of dt y so among all possible curves connecting starting starting zero and ending at modulus of z which is a real number i can always take the segment on the real axis our observe that the real of z is smaller or equal to modules of z right and also the square of this because modules of z is smaller than one good so the the distance can be in fact defined in this way taken the infimum that is to say this is the infimum so am i able to integrate this function real value function over an interval well this depends on our skills in calculus so assume that you are say first year student and you have to solve this problem find a primitive function say of this everything is real right can we solve this yeah okay some of some of you can solve it i want to see how do you solve this okay you know okay that this is something related to to hyperbolic tangent the inverse of the hyperbolic tangent okay but i i don't want to go into say very obscure terminology i've never introduced anything which is hyperbolic sinus sinus and cosine so it would be strange that now i start to introducing the inverse function of the hyperbolic tangent is that possible to use simply logarithms which is in fact related to this okay so the idea is as usual to to split this in okay one minus t okay times one plus t and this becomes as a one minus t plus b one plus t and i have to find a and b right all right are you so i guess that all of you ever at least once in your lives in this kind of integrals to okay and then i well from this i obtained at a minus b is equal to zero because a minus b is the coefficient of t and on the left hand side t is not appearing on the numerator whereas a plus b has to be one because it is the coefficient of t to the power zero if you want right so from this what i obtained is it to one that a is one half and b is minus one half i'm sorry where is one half as well sorry correct therefore the integral i'm looking for becomes so one half the integral of one over one minus t dt plus one half integral dt and therefore i have well this i this i is easily recognized really seen that this is this the primitive this is log of one plus t and here is minus log of one minus t right therefore i have one half log of one plus modulus of z over one minus modulus of z okay are you okay okay now what i observe here is that well this number becomes well i'm i take this notation okay this is a division it's a well subtraction of two this is the logarithm of this one half in minus the logarithm of one minus this and then i write it in and as logarithm of this well i write it this way also if you wish equivalently what i observe here is that this function is well this is a real value function right this is the real logarithm this is the real number and this is the real logarithm well this is monotone increasing with u it is to say if u one is smaller than u two log of one plus u one minus over one minus u one is smaller than log okay this is if you or either you trust me or you can see this well this is easily seen because well take you can use standard tools in real analysis so take the derivative is this okay which is positive it's one over one minus u square which is positive because u is smaller than one okay this is the primitive of this function so one over one minus u square is a derivative of this function here correct which is positive so the function is increasing and this is what we need to prove that in fact holomorphic function are contracted by the Poincare distance in fact i have this look this is number eight this is number nine first observe that we have just calculated the distance the you say the Poincare distance between zero and any point is z is it enough to conclude something yes it is because we know that the um maybe transformation are transit acts transitively on the disk and they are preserving what precisely this metric so as i put a remark here take phi and okay notom office okay okay md acts transitively in on say this it is to say for any pair of points and it is there exist phi in m of d such that phi of z is w right and we also have that d phi calculate this d phi squared okay over one minus modules of phi of z square is d z squared over one minus modules of the square because okay d phi of z is prime of z d z right and we have that the prime of z one minus since phi is an automorphism so in general we have inequality here lesser equal but since we are starting from an automorphism we have an equality here so once again the automorphies automorphism preserve the metric so they are isometries so you can see this directly okay and with the previous calculation we show that these are the only holomorphic function which preserve the Poincare metric and are impact isometries for this metric because of this I can always reduce my problem to the following so take n into points z and w in the unit disk and take phi maybe it's transformation and the unit disk such that phi of z is zero it acts transitively I map z into zero and well who cares phi of w is some other point in the unit disk then I calculate the Poincare distance between z and w keeping in mind the previous remark that is to say that the distance is the same if I consider any two pair of points mapped by a maybe transformation because maybe transformation are in fact isometries so in particular this is as omega of phi of z phi of w but phi of z is chosen such a way that this is zero and I have the freedom of choosing this because the automorphism act transitively so I can always set map one point to the origin and the other to somewhere somewhere else which I call w and this is what I can explicitly write in this way logarithm of one plus modules of phi of w over one minus modules of phi of w and since I have that phi of z is zero I can also say well this is our function so I have w minus z over one minus z bar w to be modules of phi of w so I have explicitly written the distance of any two points maybe well it can be misleading that z appears to be as a constant so let us put z naught here z naught and I have z naught here so I have this z naught z naught bar here all right so this is the distance and since we know that look f of z minus f of w over one minus f of w bar f of z is smaller or equal of z minus w over one minus w bar z for any z and w in delta and f holomorphic from the unit disk into itself from here you can see that what I am doing here is that I am considering I am considering this I have phi of f of w of f of z okay which is on the other hand I have phi of w z modules which is z minus w bar w sorry over one minus w bar z correct this is another way to rewrite the statement of Schwartz-Beklema but then I have that this number is smaller than this okay so that I have that the distance bonk rate distance between f of z and f of w which is expressed as one half logarithm of one plus phi of f of w f of z because this function as a real value function is monotone increasing and this number is greater than or equal to this this is smaller equal to one half the logarithm of one plus modules of phi w z over one minus modules of phi w z which is precisely omega of z w so here we finally have the contraction property of holomorphic functions for this metric which is important okay this is a model and as we will see it is not just by chance a model to be to be chosen it is a model well if you want to investigate a bit more in detail this geometry associated to the metric we have introduced so far you can prove that this is a model of hyperbolic geometry in fact the curvature for those who you know what it means the curvature is constant negative so even though we are looking at what is a disc well with this metric it would be like something which is hyperbolic okay so this is a model of non-euclidean geometry and what we have here is that well this in this non-euclidean geometry holomorphic functions are contractions and isometries are automorphies so finally let me just give you what it is a simple geometric descriptions of things in this geometry so and this is 12 right good so I want to consider and the unit disc a portion of a circle which is orthogonal to the boundary of the disc well two curves are said to be orthogonal if the tangents okay at this point at this point at the at the intersecting point are in fact orthogonal okay so what I'm considering here is something which is like this where okay this is so the tangent to the curve I'm considering up to the circle I'm considering is parallel to the ray okay so take complex number C whose modulus is greater than one and let us describe this circle in terms of the standard equation say this is say that's a distance from the point C okay this is C and I want to describe the circle C circle orthogonal to the boundary of the unit disc this is 0 and this ray has length 1 so I take this what I have here is that well I can describe this circle in this way the distance distance from C is this right distance square and this is modulus of C minus 1 modulus of C square minus 1 okay you see this all modulus of C is this this is 1 modulus of C square this is this minus 1 which is this okay good so I define C in terms of this equation and this describes the set of points which are in C and in D okay so that well I can also make this condition more explicit I write it as modulus of C square plus modulus C square minus 2 real part of C bar Z equal to modulus of C square minus 1 or modulus of C squared minus 2 real part of C bar Z plus 1 equal to 0 good so I can describe all such circles using these equations so this is the equation of the portion of the circle orthogonal to the boundary as of course C as modulus greater than 1 and in particular among all other cases we can also consider the case where C tends to infinity which is well affected this as a diameter diameter is also somehow a limit position of disc of this form good now what I want to see is that that if f is an automorphism of the unit disc then f of C is C prime so an automorph is so an isometry for the metric keeps circle orthogonal to the boundary of the unit disc so it keeps map circle onto circle orthogonal to the boundary and possibly being also the case of diameters let us see how so take well first if f is a rotation which represents a sub group of the group of automorphism rotation center at the origin well this is obvious okay so obvious if f is a rotation of center so we can consider f of z to be a transformation like this okay without the the rotational term because if it is a rotation it's obvious if you rotate everything and you have something which is orthogonal and you have something which is autonomous somewhere else but it is okay and let me make this calculation instead of considering the image I consider f minus one of z and I assume that this is on the circle z okay and then I want to prove that this is possible if and only if z is on a circle okay equivalently just to make me less calculation as as less calculation as I can okay that's why I prefer this expression because f minus one of z is z minus z naught over one minus z naught bar z which is more familiar so and I use this here good and what is this well we cannot avoid mean not avoid calculation we have to make these calculations okay so just okay this is this over one minus z naught bar z minus two real part of z hat then I have z minus z naught bar sorry z minus z naught over one minus z naught bar z plus one okay then I remember that one minus z naught bar z module square is nothing but one minus z naught bar z times its conjugate so it is one minus z naught z bar so that if I multiply everything well this is zero right if I multiply everything times modules of one minus z naught z squared I cancel this cancel some cancel this and I and it will appear here only this term inside the real part right and I have to then multiply one times one minus z naught c module squared so I have finally z minus z naught module squared minus two real part of c bar and then I have z minus z naught and one minus z naught z bar okay as I said okay plus one minus z naught bar z square equal to zero okay so I multiplied everything times one minus the module squared of one minus z naught bar z which is not zero as we all know or should know okay good so I make these calculations and I remember that this is modules of this squared plus modules of z naught squared minus two real part of z naught bar z okay then I keep the other term here fixing some sense so but I multiply what is inside so I have z minus z naught modules of z squared then I have minus z naught right then minus plus z naught squared z bar and then I have finally plus one plus modules of z naught squared plus modules of z squared well I can omit the bar because okay the module is the same and minus two real part of z naught bar z equal to zero so this is like this okay so I have modules of z squared plus modules z naught squared plus one plus z naught squared times the modules of z squared then I can put this real part inside of this right then I have plus minus actually minus two real part of then c hat z right minus c hat z naught modules of z squared minus c hat z naught minus c hat c bar sorry z naught squared z bar and then I have minus two z naught bar z plus yes because the minus in front of us force two because I have two right and two right this is plus thank you all right okay so I notice that here I can also have what this is c z naught c bar z naught times minus okay plus z squared this is correct right and here I can this is equal to zero right and this part here can be written as one plus modules of z squared times let me see one plus z naught z one plus z naught squared plus z bar squared right so that I factor out this term from here and from here in such a way that I continue on this and you piece of paper and I have one plus modules of z squared times one plus modules z naught squared which is this and then I have minus then I say plus right two real part of z hat z naught is it because this is a real number then I can take from the real part so the coefficient is minus c z naught minus two is in front two c bar z naught times one plus z squared then plus minus sorry minus two real part of the rest which is c bar z then I have plus two c naught bar c let me see c bar z sorry I forgot plus c hat z naught bar c naught squared and that's it right one two three one two three yeah is it correct c bar z then I have two z naught bar z and then I have z bar z naught squared this is what is left after after putting this term in this part here now what can what can we have here well first case assume that one plus z naught squared plus two real part z bar z naught is different from zero this is a real number and it can be different from zero or equal to zero if it is different from zero then we can rewrite this condition and this way I have we can divide right so I have one plus modules of z square plus minus two real part of what is here yes it is something like this c z naught bar is it appearing here sorry z bar c z naught bar so what is missing is this right you probably wanted to point it out is that I forgot a z bar here so I have well that this is the situation at c bar z plus two z naught bar z plus z bar z naught squared c bar over one plus module z naught squared plus two real part z hat z naught and this is equal to zero which has the same aspect as one plus real part minus two sorry real part of c tilde z plus modules of z squared equal to zero c tilde comes out from this that is to say that is c is in c prime with a c tilde as a center the second possibility left is the following so I have well if one plus modules of z naught squared minus two real part of c hat z naught is zero this is equivalent to saying that well this number well this is a plus z naught minus z naught is in c you see because it satisfies the condition or since f of z is z plus z naught one plus z naught bar z that is to say f of minus minus one of zero is z naught right because zero is mapping to z naught by f f minus one of zero is z naught or sorry f of z sorry as I said f of minus one of z naught is zero or f of f of sorry minus z naught is zero so f of f minus one of zero is minus z naught that's more convincing which means that well f minus one of zero belongs to c and so if this is the case what is left is that we have from the previous calculation we have that the real part of c hat plus c z naught bar squared plus two z naught bar c is equal to zero and the previous equation we had this left the first part is zero because the term one plus modulus of z squared is zero the term the coefficient of one plus and so what is left is here and this is what this is a line in the passing through the origin which means it is a diameter so in both cases what we conclude is that any portion of circle orthogonal to the boundary is mapped into portion of the circle orthogonal to the boundary is mapped by an an automorphism or a diameter which represents again a circle but this is the geometric property which leads to this conditions this condition to this observations this 14 this is sorry 14 15 this is 16 right so for this metric the geodesics are so conclusions circles in the orthogonal so this is euclidean euclidean orthogonal to the boundary of d are preserved by isometries of the Poincare so for those of you familiar with this terminology the portion of circles of so on to the boundary are in fact the geodesics for this so if you want to find the minimal the curve which realize the minimal distance connecting two points using this metric then you have to take the circle orthogonal to the boundary passing through these two points and it's not surprising that when you restrict the case to the real axis when we have already done this we didn't make any mistake because in fact on the real axis what we have is that well the connecting points are in fact on the diameter which is the connecting path minimizing the distance is in fact the portion of diameter which is okay so in this in this geometry if you want a triangle becomes well three portions of like this three portions of part of circles orthogonal to the boundaries and this is a triangle this is you see the hyperbolic shape of a triangle if you want okay normally it is seen on what is more familiar probably yeah any parabola and you rotate it and you have a model of hyperbola using the say surface but here we are in the flat stuff and this is what you have okay thank you good so in the last five minutes I just want to to make a short very short observation and remark so this is 16 14 15 so we have applied several times the properties of the automorphies of the unit desk okay but we have also the class of automorphies in the plane and in the Riemann sphere let me just point out some effect about fixed points of automorphies so if you remember we have this class of automorphies in the plane with a different from zero right so the question is are this transformation always without fixed points or sometimes without expanse so we know for instance that we have translations and translations are without fixed points but we also have here dilation and rotations so for instance b zero necessarily while the origin is fixed so the question in general is what are the fixed point sets for out c c hat and out d okay so in other words I want to describe which subclasses of automorphies have or do not have fixed point and if I can characterize them so in the case of automorphies of the plane well we have to solve such an equation which obviously has a solution if a is equal it's different from one because you have that this is z a minus 1 minus b and if a is different from one then we have that this c is b over 1 minus a so a fixed point exists f and only f a is not one well if b is zero and sorry yes if b is zero and a is one well necessarily all points are fixed because we have then the identity but if b is different from zero we have just one fixed point and of course a is to be different from one if a is equal to one there is no fixed points for any b except yeah for any b now what about the case of linear fractional transformation right so we have to solve an equation like this which becomes quadratic equation right so it is a quadratic equation if c is not zero now it becomes a c z square plus t minus a c minus b equal to zero right d z minus z okay so this is quadratic unless c is different from zero unless z sorry unless c is equal to zero but if c is equal to zero we go back to the previous case right because with them we have automorphism of the plane so we know okay everything but if c is different from zero then there is a fixed point a pair of fixed points okay so in other words there is no automorphism of the Riemann sphere without fixed points and finally we have the case of the maybeson formation and also in this case since this maybeson formation can be considered as special cases of linear fractional right so we have a quadratic condition but surprisingly not any maybeson formation I have a fixed point because we are strictly set right so this is the condition which becomes okay becomes okay z naught bar z squared plus z e i theta minus 1 minus z naught e i theta equal to zero square quadratic unless z naught bar is zero but if z naught bar is zero look if z naught bar is zero then we have a rotation and the rotation is obviously zero as a fixed point so we can assume z naught bar z naught bar to be different from zero therefore this is the last things I'm going to tell you today therefore we want to look for the zeros of this polynomial so the roots right and so this can be approached in this way so the modules of p of z is of course greater equal to zero precisely when z is a root right so this is smaller or equal to what to the sum of what of z naught and well if I find a z such that its modules satisfy this equation then I say this is zero and this is zero in fact I'm interested more in the modules than in what is precisely the zero and I observe the coefficient of the quadratic part and this of the zero zero zero degree term are the same so that the two roots have modules this alpha is modules of z 1 and beta is modules of z 2 the two roots which of course exist because we are in c are such that the they are one the reciprocal of the other because the modules of z 1 times the modules of z 2 are such that well it corresponds to the quotient z naught modules of z naught time over modules you know which is one which means that either one is in the p so one or a solution as modules smaller than one and the other is greater than one or both as modules equal to one I'm not saying that they are always distinct okay if they are distinct and of modules different modules one is inside so there can be one fixed point and the other is outside all there are none and the unit desk and they are on the boundary and they can be distinct or coincident okay and this will be used later on in our discussion so I stop here and I thank you for your attention