 Okay so let us recall what we have been doing I mean for the benefit of continuity okay. So you know we are trying to prove the Rayman mapping theorem and the assertion of the Rayman mapping theorem is start with the simply connected domain in the complex plane which is not equal to the whole complex plane okay. So that means a domain which for which at least there is one point in the complex plane which is not in the domain but assume it is simply connected which means that any loop in the domain can be continuously shrunk to a point in the domain okay. In other words the domain does not have any holes and you can also say this as the complement of the domain is continuous to be connected it is not disconnected okay. So you take a simply connected domain which is not the whole complex plane the Rayman mapping theorem is that there is a holomorphic isomorphism of that with the unit disk okay. So you have to find a holomorphic map a holomorphic isomorphism from that simply connected domain which is not the whole plane to the unit disk. Of course you know these conditions in the hypothesis Rayman mapping theorem they are necessary because you know if you the condition that the domain should not be the whole complex plane is very important because if you take the domain to be the whole complex plane it is of course simply connected but any holomorphic function on that which maps into the unit disk will be a constant by Louis theorem. Because a holomorphic mapping from the complex plane into the unit disk will be a bounded entire function it will be bounded because the images you are saying it is taking values in the unit disk so it will be bounded and it is entire because it is holomorphic on the whole plane therefore it has to be constant by Louis theorem therefore the condition that the domain in Rayman the Rayman mapping theorem is not the whole complex plane is important okay that is a condition you cannot negate and of course the other condition is that it should it is necessary that it has to be simply connected okay. So because any domain that is holomorphically isomorphic to the unit disk is also topologically isomorphic to the unit disk and the unit disk is simply connected and anything that is topologically isomorphic that is homeomorphic to a simply connected space is again simply connected okay so if you expected domain to be holomorphically isomorphic to the unit disk then it should also be homomorphic to the unit disc ok and but the unit disc is simply connected and you know the notion of simply connectedness is preserved under homomorphism therefore your domain must be simply connected. So both the conditions in the Rayman mapping theorem namely that the domain is simply connected and that the domain is not the whole complex plane are necessary ok and Rayman the Rayman mapping theorem says that once these conditions are met then that domain is just holomorphically isomorphic to the unit disc. So we have to find a holomorphic map a holomorphic injective holomorphic map from the from that from such a domain to the unit disc ok which is which is isomorphism on to the unit disc of course injective holomorphic map is always an isomorphism on to its image. So all you have to do is find a holomorphic map which is injective and whose image is the whole unit disc ok. So what we did first the first step was we proved that we can find a holomorphic isomorphism of such a domain into a sub domain of the unit disc not the which may not be the whole unit disc ok that is the first step that we proved. And then therefore our proof of the Rayman mapping theorem reduces to studying the sub domains of the unit disc ok. So that brings us into studying the unit disc and studying analytic maps from the unit disc to itself ok and that led us to the so called short slimmer. So we had this so first we had the short slimmer so this is you know f from the unit disc to the unit disc analytic with f with the analytic map taking the origin to the origin ok. Then the modulus of the image of a point in the unit disc under f is cannot exceed the modulus of the point ok for all points in the unit disc and equality occurs if and only if f is an automorphism a holomorphic holomorphic isomorphism from delta to delta ok. And of course equality occurs when I say equality occurs equality occurs even for a single point different from the origin the unit disc if and only if this f is a holomorphic isomorphism ok. So I should say equality occurs for a z not equal to 0 in delta if and only if f is a holomorphic automorphism of delta ok. So this is a short slimmer now in the in this short slimmer you can you can actually get an infinitesimal version of the short slimmer and for that what you what you have to do is you take z not equal to 0 and divide by mod z on both sides then you will get mod fz by z is less than or equal to 1 and then you take limit as z tends to 0 limit z tends to 0 fz by z will be just f dash of 0 because f of 0 is 0. So if you divide by mod z and take limit z tends to 0 you will get the inequality mod f dash of 0 is less than or equal to 1 and that is called the differential form or the infinitesimal version of the short slimmer ok. So let me write that so infinitesimal version or differential version of the short slimmer is f from the unit disk to the unit disk analytic f0 equal to 0 then mod modless of the derivative at the origin is less than or equal to 1 and equality occurs if and only if f is a holomorphic automorphism of the unit disk ok this is a differential version of or the infinitesimal version of the short slimmer right. Now peak slimmer is a generalization of this peak slimmer is just a generalization of this and what does it say so peak slimmer gives you the and estimate for the modulus of the derivative at any point of the unit disk the infinitesimal version of the short slimmer gives you an estimate for the modulus of the derivative at the origin ok but peak slimmer will give it to you at any point z the unit disk. Now what is peak slimmer so peak slimmer is so same conditions f from delta to delta unit disk to unit disk so you know in all these things of course I should mention that delta is unit disk which is set of all complex numbers such that modulus is less than 1 ok. So f from delta to delta analytic ok and see here you do not put the condition f of 0 equal to 0 you relax that condition then modulus of f dash of z modulus of derivative of f at a point z cannot exceed 1 minus mod f of z whole squared by 1 minus mod z the whole squared for all the points of delta you know in particular if you put if f takes 0 to 0 and you put z equal to 0 you will get mod f dash of 0 is less than or equal to 1 which is the infinitesimal version of the short slimmer therefore this peak slimmer is a generalization of the infinitesimal version and of course I should say that you also have this equality condition here and equality occurs for a z in delta if and only if f is a holomorphic automorphism of delta ok. So this is peak slimmer which we prove and then but the point about the important point about peak slimmer as far as the unit disk the geometry on the unit disk is concerned is the equality part ok it is the it is the so called it is what gives rise to hyperbolic geometry on the unit disk ok. So what we do we go to first defining we go to hyperbolic geometry on the unit disk ok. So what is this hyperbolic geometry so the first thing is you know we define what is meant by the hyperbolic length of a path in the unit disk ok. So you see if you have if you have a path on the complex plane so you have this interval on the real line bounded finite closed interval and you have this function gamma which is continuous piecewise differentiable piecewise continuously differentiable differentiable. In other words the image of this interval under gamma will give you a piecewise smooth contour in the complex plane so you know you get you will get something like this and in principle it need not be fully smooth it could be it could be like this it has to be piecewise smooth it need not it need not be there could be points where it need it is not smooth for example it can be a polygonal line ok. So of course you have the starting point which is gamma of A you have the ending point which is gamma of B and we all know how to calculate the length of such an arc from basic calculus ok length the Euclidean length of the arc gamma is simply given by integrating mod dz whenever you integrate mod dz you will get the usual length ok and what does it mean it means that you are just you are just putting instead of since whenever you integrate the variable of integration is on the domain of integration the domain of integration is this this oriented path ok so the z is a point on gamma ok so if you want to calculate this integral you have you can trans you have to transform it into a real Riemann integral by substituting z equal to gamma of t ok t is a variable here and gamma of t is a corresponding point here ok. So if you do that what you will get is you will get integral from t equal to A to t equal to B if I substitute d gamma of t then I will get modulus of gamma dash of t into dt this is what I will get and of course is modulus of gamma dash of t makes sense because gamma I told you is first of all gamma is piecewise continuously differentiable that means I can break this interval AB into finitely many sub intervals over which gamma becomes differentiable alright and therefore this and the derivatives are also continuous ok. So this integral can be defined as a Riemann integral and if you calculate this you will get simply the length you will get the arc length ok you will get the arc length along this along this arc or path or contour ok. Now this is Euclidean length but what we are going to do is we are going to look at very special case when your path gamma is a path not just in the complex plane but it is a path in the unit disc ok so you look at a situation where your path lands inside the unit disc this is unit disc and your path is here so here is your path it starts at a point in the unit disc it ends at a point in the unit disc and the whole path lies inside the unit disc ok and of course for this path the usual length or Euclidean length is given by this integral but then we define a new kind of length we define the hyperbolic length. So what we do is we define hyperbolic length of gamma to be you see you do not you of course you integrate our gamma but do not just integrate mod dz but you integrate mod dz by 1 minus mod z the whole square you put this factor of 1 by 1 minus mod dz the whole square you put this extra factor and when you integrate like this ok you see this integral is going to give you a positive quantity because you see mod z z you know again here the variable of integration is z and that is going to lie on the domain of integration domain of integration is the arc the arc is lying in the unit disc therefore the denominator never vanishes alright the integrand is a you know positive quantity alright therefore this is going to give you some positive value you are going to get some positive value and that is called the hyperbolic length of gamma. Now what is the importance of PIX lemma PIX lemma says that you see according to PIX lemma suppose I take f to be an automorphism of the unit disc suppose f is a holomorphic automorphism of the unit disc ok then what will happen is that f will map the unit disc holomorphically isomorphically on to itself so this is again unit disc to unit disc so I will get the mapping if I call it as w equal to fz what will happen is it will map this path into another path and this path will well you know you can write out the starting point will be well this point is what it is just gamma followed by f so it will be f circle gamma that is the path you will get it will be a path because it is just a path composed with an automorphism ok so it is again a path in fact a path composed by any continuous map I mean any differentiable map continuously differentiable map will also give you a path ok. So this is if this path is gamma then this path is f circle gamma it is gamma followed by f ok the composition of these 2 maps and of course it will start at the image of this point under f which is f gamma a and it will end at this point which is f gamma of b but since f is a automorphism that is a isomorphism of the unit disc on to itself the image of this is again going to be a path inside the unit disc alright and if you calculate the hyperbolic length of this path ok what you will get is you will get hyperbolic length of the image of the path f circle the image of the path gamma under f which is f circle gamma is going to be what it is by definition is going to be integral over f circle gamma of mod dw by 1 minus mod w the whole square this is the definition ok mind you this is a z plane and this is w plane so here the variable the complex variable here is w the complex variable here is z and w is fz ok and well if you if you change this integral if you make a change of variable alright by plugging in w is equal to fz you will see that you will get integral over gamma dfz mod by 1 minus mod fz the whole square and if you calculate this see this mod dfz is going to give you f dash of z mod f dash of z into mod dz. So what I will get is I will get integral over gamma mod f dash of z mod dz by 1 minus mod f of z the whole square this is what I am going to get but then what is what what does pix lemma say pix lemma says that because f is an automorphism of the unit disc you have equality in this expression which will tell you that mod f dash of z divided by 1 minus mod fz the whole square is the same as 1 by 1 minus mod z the whole square and if you plug it in there you will see that these two are equal you see that these two are equal. So the moral of the story is this of course is equal to this as I told you because of pix lemma this is because of pix lemma ok. So the moral of the story is this beautiful fact that if you define hyperbolic length by this formula then the hyperbolic length will not change under an automorphism of delta. So if I define hyperbolic length of path gamma by this formula whatever length I get will be the same if I take the image of this path and again measure the hyperbolic length of the image path under an automorphism of the unit disc. So in other words this hyperbolic length definition is adjusted in such a way that it is invariant under automorphisms holomorphic automorphisms of the unit disc ok and the reason for all that is the equality statement in pix lemma ok. So the pix lemma is the kind of you know it is a motivation to define hyperbolic length like this because if you define hyperbolic length like that then pix lemma tells you that the hyperbolic length of an arc will not change. If you map the arc by an automorphism of the unit disc you will get a new arc but it will still be this it will still have the same hyperbolic length ok. So that is the beginning of hyperbolic geometry from pix lemma which is a kind of generalization of Schwarz's lemma ok. Now and of course we are doing all this is all this because we want to study unit disc we want to study domains of the unit disc we want to study analytic functions from the unit disc into itself ok. So this hyperbolic length of a path gives rise to a metric ok called the hyperbolic metric on the unit disc ok. Now see for example let me let me take your mind back to the Euclidean situation that is usual plane for example ok. How do you define distance between two points ok. Well one is a distance formula that you know pretty well ok but suppose I ask you to define it using arcs ok then how will you define the distance between two points in Euclidean space just using arcs. One natural definition will be draw any arc from this between the two given points calculate its length and take the minimum ok you draw you so in other words the distance between two points in Euclidean space is given by the minimum arc length of all possible arcs from this point to that point and you know that minimum arc is just a straight line it is just a straight line it is a portion of the straight line segment namely the portion of the line that passes unique line that passes through those two points of course if those assuming those two points are distinct ok. If they are one of the same point of course the distance is always defined as 0 right. So this is a this is a even if you do not know the distance formula ok you can define Euclidean distance by as an infimum ok you can you can define it as a minimum of arc lengths of all possible arcs between the two points. Now you do the same thing adopt the same procedure to define distance function on the unit disc so what you do is given any two points in unit disc you take the minimum possible arc length hyperbolic arc length of all possible arcs between these two points in the unit disc and that gives you a metric it is called the hyperbolic metric on the unit disc ok. So this defines hyperbolic metric on the unit disc so let me write that down this is what I explained yesterday I mean in the last lecture. So hyperbolic metric distance rho h of z0, z1 is equal to infimum over gamma of hyperbolic length of a path gamma from z0 to z1 in the unit disc ok you take so in other words what you do is that given any two points unit disc you draw any arc piecewise smooth arc ok and you calculate its hyperbolic length as a form using this formula right and then you minimize over all possible such arcs then the fact is that you get a distance function on the unit disc and so then you can ask this question you see in the usual Euclidean geometry if you give me two points what is the minimum distance it is a straight line distance ok. So it is a straight line segment which gives you the arc of smallest possible length ok these arcs in any geometry the arcs of smallest possible length are given special names they are called geodesics ok they are called geodesics. So in the Euclidean geometry the geodesics are straight lines ok segments along straight lines segments of straight lines ok. Now you can ask the same question if you take the hyperbolic geometry namely the unit disc with this new metric this new distance function given any two points in the unit disc you can ask what is the what is the geodesic ok so what is that minimum what is that arc of minimum hyperbolic length connecting these two points and last lecture I ended last lecture by stating a theorem which tells you what that is it is very simple the geodesics in the case of the hyperbolic metric are circles and they are actually circles perpendicular to the unit circle which is the boundary of the unit disc circles orthogonal to the unit circle are the geodesics and they are the analog of straight lines in Euclidean geometry ok. So here is the theorem which is what I ended with yesterday stating so the theorem was if z0, z1 are in delta then the unique then the unique path of shortest hyperbolic length from z0 to z1 ok is the arc of the circle through z0 and z1 which is orthogonal to the unit circle which is the boundary of the unit disc which is z1 ok so the so this theorem answers what this hyperbolic distance is so you know if I draw a diagram it is going to be like this so this is the unit disc and you know if you give me well if you give me two points z0 and z1 what you have to do is you have to draw a circle passing through z0 and z1 and which is perpendicular to the unit disc so you have to draw something like this you know you will get something like this. So at this point the angle between the tangents of this circle and the unit circle will be 90 degrees and at this point also the angle between the tangents will be 90 degrees ok and this it is this length it is this arc of this circle which is orthogonal to the unit disc which is the hyperbolic geodesic it is the shortest distance in the hyperbolic matrix. So in particular you know if you join this if you take the straight line segment from z1 to z0 to the unit disc and measure its hyperbolic length you will get a bigger value so it is rather so you see it is a very funny geometry so geometry in which the straight line distance is not the smallest there is some there is a curve along which the distance becomes smallest and it is hard to imagine with respect to usual geometry this can never happen ok with respect to the usual Euclidean geometry that we are familiar with the shortest distance between the two between two given points will be the straight line distance but here is a strange geometry by the distance the shortest distance between two points is by not by the straight line segment joining the two parts the two points but through you know a curve joining the two points ok that is a and actually it is a reflection of it is something that you cannot imagine right always a curve distance should be more than the usual distance that is because your usual definition that is because you are used to the usual Euclidean distance ok but here you know our distance function is not you have twisted the length by this factor 1 by 1 minus mod z the whole square that is what is creating this strange thing and let me tell you one more thing this is also this is also indication of another fact in differential geometry according to which you know the hyperbolic disc that is a disc with the hyperbolic metric has negative curvature ok. So you know it is very difficult to imagine an object with negative curvature alright so this has negative curvature alright and it is something that you can physically not very easily imagine and all that is all because of this definition of hyperbolic length ok so it has negative curvature and it is because of negative curvature that the hyperbolic length through this arc of the circle is smaller than the hyperbolic length along the straight line segment travelling along this curved path gives you shorter length than travelling along the straight line path because your space is curved the space is negatively curved ok that is what is happening geometrically right. So this is the theorem so in fact you know if you draw all these geodesics you know they will look like this if you take two points like this it will be like this on the other hand you know if you take two points which lie on a diameter ok the geodesic will be just any diameter will be actually a geodesic because you know if you take two points along a diameter ok if you take z0 here and z1 here this will be a geodesic because you see what does the theorem say the theorem says that if you want a geodesic you have to pass a circle through the given two points which is orthogonal which is perpendicular to the unit circle but if two points lie on a diameter through those two points if you try to draw a circle which is perpendicular to the unit disk you will end up as a limiting case you will get only the diameter ok. So you know basically though of course the diameter is not a circle it is part of a straight line but you should think of it as a circle with the third point at infinity ok for determining a circle you need three points ok. So these are two points z0 z1 there is a point at infinity so the view point is if you think of even straight lines as circles and this is something that you should have already come across when you studied the Riemann sphere ok the Riemanns in the straight lines and circles on the complex plane on the they correspond to just circles on the Riemanns sphere ok. So it is natural to think of a straight line also as a circle but if you want you can even think of this as you know the limiting case of circles which are orthogonal to which pass through these two points and which are orthogonal to the unit circle if you take the limiting case you will get the straight line. So every diameter will be a geodesic ok alright every diameter will be a geodesic these are the only geodesics which are straight lines but if you go away from the diameter your geodesics will become you know they will start getting curved they will become like this your geodesics will get curved ok. So this is how these are how the geodesics look like on the hyperbolic disc and the geodesics the beautiful thing is that the geodesics play the same role in hyperbolic geometry as straight lines in Euclidean geometry. So in fact if you take all of Euclidean axiom except the parallel axiom ok the parallel axiom which says that given any point and a line not passing through that point there is a unique line through this point which is parallel to the given line that is called the parallel postulate or parallel axiom you throw that out all the remaining Euclidean axioms Euclidean axioms are satisfied by these geodesics. So they give rise to a new geometry called hyperbolic geometry ok and you got lot of nice things happening for example what happens is that in Euclidean geometry you know that you know you can form a triangle by 3 lines ok and the sum of 3 angles of the triangle is 180 degrees you can show that in hyperbolic geometry you can form a triangle like this again using 3 geodesics so you know basically you can form a triangle like this so this is a hyperbolic triangle and of course you know you can also have a triangle like this you can have this as one geodesic this is another geodesic alright and then you can have a third one like this. So these are 2 types of hyperbolic triangles but the beautiful thing about hyperbolic triangles is that the sum of the 3 angles of hyperbolic triangle will be less than 180 degrees ok the sum of the 3 angles will not be 180 it will be less than 180 ok so this is hyperbolic geometry. So anyway so there is a new kind of geometry on the unit disc and beautiful thing is that it is connected with analytic isomorphisms and analytic mappings of the unit disc onto itself ok and so I will have to we will have to prove this theorem first that is what I am going to do next. So what I am going to do is I am going to first prove so I am going to give you or rather let me yeah so lemma lemma 1 so the proof of the theorem will be in 2 or 3 steps and all the steps are very easy ok. So here is a here is a first lemma so it so what it tells is that if the hyperbolic distance from 0 of a point z0 is actually the hyperbolic length of the line segment line segment from 0 to z0 and it is actually equal to we calculated it in the previous lecture it is just 1 of 1 plus half long 1 plus mod z0 by 1 minus mod z0 ok. So here is a lemma so you know you take 0 you take a point z0 in the unit disc ok and then what is the hyperbolic distance between 0 and this point I told you that as a I mean if you believe the theorem in retrospect I told you that all the diameters are all geodesics ok so you know I can find a diameter that passes through 0 and z0 that is just this radial line from 0 to z0 and the claim is that this is the geodesic lemma lemma this lemma 1 says that the geodesic from the centre of the unit disc to any point on the unit disc is just the radial line segment ok and so how does one prove this right. So what one does is you know we will do the following thing you put so let z0 be r0 e power i theta0 ok so z0 let z0 be this and what you do is I will first rotate this so that I bring z0 to the real axis ok. So what I will do is I will give this map w is equal to z into e power minus i theta0 ok this is just so it is e power minus i theta0 times z it is just rotation ok. So if I do this what will happen is of course this is a rotation of the unit disc so it is an automorphism of the unit disc which takes the origin to the origin and you know we have already seen as a corollary of Schwarz's lemma that any automorphism of the unit disc which fixes the origin is a rotation ok so this is one of those but it has been adjusted in such a way that you know finally I get this situation like this so you know so z0 will go to r0 ok and this line segment from 0 to z0 is going to get mapped to the line this line segment on the real axis ok. And and I and the basic point about this first lemma is to show that you know this is the geodesic right mind you the hyperbolic length of an arc does not change under an automorphism of the unit disc ok. In fact that is the motivation for defining the hyperbolic length with this formula which comes from big slima. So under the hyperbolic under the under any automorphism unit disc hyperbolic length will not change but how is the hyperbolic metric defined hyperbolic metric is also defined using the hyperbolic length. Therefore the moral of the story is under any automorphism of the unit disc the hyperbolic metric will be preserved ok. And in other words for the hyperbolic metric any automorphism unit disc will be an isometric ok any automorphism of the unit disc will be an isometric which means the distance between two points will be the hyperbolic distance between two points will be the same as the hyperbolic distance between the images under an automorphism unit disc ok that also just comes because of this reason that the hyperbolic length is invariant under an automorphism of the unit disc ok. So you know to show that this is the to show that the straight line segment from 0 to Z naught this is the this is the geodesic in the hyperbolic sense it is enough to show that this is the geodesic. See what you must understand is that an automorphism unit disc will also preserve geodesics ok that is because it will preserve distance ok and you know because it is an automorphism of the unit disc and you know the automorphism of unit disc are just mobius transformations ok and they are conformal. So if the geodesic is a circle passing through the given points is orthogonal to the unit circle then its image under the mobius transformation will again be a circle or a straight line which is orthogonal to the unit circle because the mobius transformations you are considering are going to take the unit disc to the unit disc therefore they will take the boundary of the unit disc to the boundary of the unit disc and conformality will make sure that you know whenever two curves intersect at an angle the image curves will also maintain the same angle. So if I take a circle which hits the unit circle orthogonally its image under a mobius transformation which takes the unit disc to the unit disc will again give you another circle or a straight line which is perpendicular to the unit circle. So a geodesic will go to a geodesic ok under an automorphism of the unit disc a geodesic has to go to a geodesic ok just because of the fact that any holomorphic map which is with non-zero derivative is conformal so it will preserve orthogonality it will preserve angles between curves ok. So to prove that this is a instead of proving that this is the geodesic from 0 to Z0 its enough to show that this is a geodesic ok. So how do I show that this is a geodesic I have to show that this is a path of shortest distance. So what I do is I just compare the length of this path the hyperbolic length of this path with any other path from 0 to R0 ok but that path could be curved and I show that this path which is a straight line segment from 0 to R0 has a minimum distance that is how I prove this lemma ok. Once I prove this lemma I am more or less I am done with I am more or less proved this theorem ok because I know how the automorphisms of the unit disc look like right. So let me say let me write these two statements note that if f is an automorphism of delta holomorphic then for Z0, Z1 in delta we have through hyperbolic distance between f Z0 and f Z1 is the hyperbolic distance between Z0 and Z1 and further any circle or straight line orthogonal to S1 is mapped by f again onto a circle or straight line orthogonal to the unit circle due to conformality ok. So this is the statement that I said that any automorphism unit disc will have to preserve the hyperbolic metric and it has to preserve geodesics it will map geodesics to geodesics right. So to show that this is the geodesic from 0 to Z0 it is enough to show that this is the geodesic from 0 to R0 because this map is an automorphism unit disc ok and to do this I will have to compare this distance with the length of this hyperbolic distance along this straight line segment with the hyperbolic length along some other path gamma from 0 to Z0 to R0 ok. So let me stop here.