Lecture 04

Hello everyone. In the last lecture, we have seen that PSL2R, it is a subgroup of isometric group of this upper half plane. So, using that, so we will be proving that. So, Geodesics in upper half plane with respect to hyperbolic metric. So, they are the semicircles whose centers lie in real axis and the vertical lines. To prove that thing, we need Morbius transformations. So, let us first go through the Morbius transformation. So, let us first define what you mean by Morbius transformation. So, a mapping P from this complex number plane to complex number plane of the form dz equals to az plus b by cz plus d. So, where this a, b, c, d are my complex numbers and this ad minus bc is not equal to 0. So, this mapping will be called a Morbius transformation. This Morbius transformation, so what we will prove in this lecture that this Morbius transformation, it takes circles or straight lines to either circles or straight lines. So, first let us see so why this Morbius transformation, it is a bijection. So, let us define. So, let s z to be equal to your dz minus b divided by minus cz plus a. So, this s again it is a Morbius transformation because ad minus bc is not equal to 0. So, s is again it is a Morbius transformation and now if I do this thing, so t compose s z. So, this you can check. So, this is nothing but just the identity map. So, identity map is, so identity of z is equal to z and also s compose t. So, this is again an identity map. So, this proves that this s is, so this implies that s is the inverse of t. So, because this t is invertible, so this proves that so t is a bijection and also note that, so this already I have said, so if s, t are Morbius transformations, then s compose t is again a Morbius transformation. So, this proves that, this proves that the set of all Morbius transformations form a group under composition of maps. So, this definition of Morbius transformation can be extended to C union infinity. So, first of all, what is C union infinity? It is the one point compactification of this C. So, let me write first this thing. So, C infinity, it is C union infinity. So, this is just the one point compactification of complex plane C. So, what are the open sets in this one point compactification? Open sets of C together with the set V equals to C minus K union infinity, where this K is closed and compacted in C. So, these are my open sets in C infinity. So, one can extend the definition. So, one can extend the definition of Morbius transformation to C infinity. So, that is your C union infinity. So, how do you define this? So, so Tz is equal to Az plus b by Cz plus d, if I take z belongs to your C and where this a, b, c, d, so these are my complex numbers and ad minus bc is not equal to zero and T of this infinity, so this is defined to be nothing but your a by c. So, T of infinity is defined to be a by c. So, that is why, so this T is a, now it is a map from C infinity to C infinity. So, in fact, also you can check that. So, T of this minus d by c, so this will be equal to your infinity, ok. So, now what are the examples of this Morbius transformation? So, first of all, translation. So, Tz equals to z plus a. So, this is a Morbius transformation. So, a belongs to your C. Second is dilation. So, T lambda z is equal to your lambda z, where this lambda is not equal to zero. Third, rotation. So, it is R theta z is EI theta z. And the fourth is inversion. So, Iz is equal to your 1 by z. So, one can check that. So, this Morbius transformations are composition of translations, dilations and inversion. So, let me prove this thing. So, proposition, a Morbius transformation is the composition of translations, dilations and inversions. So, what is the proof of this? So, let us take a Morbius transformation. So, let Sz equal to az plus b by cz plus d. So, this is my Morbius transformation and where this ad minus bc is not equal to zero, ok. So, first suppose that c is equal to zero. So, then Sz is equal to your a by Tz plus b by d. And if I write this S2z to be my a by Tz and if I define S1z to be your this translation z by z plus b by d. So, then Sz is nothing but your S2 compose S1z. And now assume that c is not equal to zero. So, then put S1z to be equal to your z plus d by c, S2z to be equal to this inversion 1 by z and S3z to be equal to your bc minus ad by c square z. So, this is this dilation, S3z dilation. So, then one can show that, so this S is nothing but composition of S4, S4 composed with S3, S3 composed with S2, S2 composed with S1. Yeah, so I forgot to write the S4z. So, what is my S4z? So, this is S4z is nothing but z plus a by c. So, next we will learn about the fixed points of this Mobius transformation. So, let S be a Mobius transformation and suppose Sz is equal to your az plus b by cz plus d. So, in order to find the fixed points of this S, so we need to consider these equations Sz equals to z. So, this will imply that az plus b by cz plus d. So, this is equal to z. So, that is c z square plus d minus az minus b is equal to 0. So, this is a quadratic equation, equation in z and so therefore, so this equation, so star has at most two solutions. So, therefore, so this implies that Sz can have at most two fixed points. So, next we will see this thing. So, if I take three distinct points in C infinity, so let us take the C infinity. So, let us take three distinct points. So, let a, b, c be three distinct points in C infinity and suppose S is a Mobius transformation such that S a equals to alpha, S b equals to beta and S of c is equal to your gamma. So, now if T is another Mobius transformation taking these three points a, b, c to alpha, beta, gamma, so this will imply that, so then, so this T inverse compose S, so it fixes a, T inverse compose S, it fixes b and T inverse compose S, it fixes c also. So, T inverse compose S, it is again a Mobius transformation, it fixes three points. So, this will imply that and if you take a non-identity Mobius transformation, it will have at most two fixed points. So, therefore, so this T inverse compose S, it has to be identity map. So, this implies that T is equal to S. So, the action of this Mobius transformation on these three distinct points, so it determines that Mobius transformation uniquely. So, here in the previous thing, so let S be a Mobius transformation and we will also assume S to be not equal to identity. So, now let us take, let us take three points z2, z3, z4 belongs to your C infinity and they are distinct. So, then define S from C infinity to C infinity. So, Sz is defined to be this Mobius transformation, so z minus z3 divided by z minus z4 divided by z2 minus z3, z2 minus z4. So, if the z2, z3, z4 belongs to your C and you define Sz equals to your z minus z3 by z minus z4. So, if z2 is equal to your infinity, Sz is defined to be z2 minus z4 by z minus z4. So, if z3 is equal to your infinity and this is equal to z minus z3 divided by z2 minus z3, so if z4 is equal to your infinity. So, this is my definition of this Sz. So, then we show that, so S of z2, so this is equal to your 1, S of z3 is equal to your 0, S of z4 is equal to your infinity and by the previous observation, so there exist a unique S with the above property. So, given any three points in complex plane, you can have a Mobius transformation which takes those three points to 1, 0 and infinity and that Mobius transformation, it is unique. So, now let us move to the definition of cross ratio. So, if z1 belongs to C infinity, so then define this bracket z1, z2, z3, z4. So, this is defined to be S of z1. So, where this S, the definition of S is this thing. So, this is my definition of S. So, if I take S of z1, so that is defined to be the z1, z2, z3, z4, so this is called the cross ratio. So, if you put z1 in this definition of S, so S of z1, it is just the cross ratio, that is defined to be the cross ratio and it is denoted by bracket z1, z2, z3, z4. So, next proposition is Mobius transformation preserves the cross ratio. Mobius transformations preserve cross ratios. So, that is if I take, so if I take this four distinct points, that is if z1, z2, z3, z4 are four distinct points in C infinity, so then, so this cross ratio, so this is equal to T of z1, T of z2, T of z3, T of z4. So, these two cross ratios are same. So, what is the proof of this? So, let S z equals to this cross ratio, so z, z2, z3, z4. So, then the cross ratios are preserved for any Mobius transformation. So, if I take any Mobius transformation T, so then the cross ratio z1, z2, z3, z4 is equal to the cross ratio T z1, T z2, T z3, T z4. So, now how to prove this thing? So, let us take this Mobius transformation S z, which is just the cross ratio z, z2, z3, z4 and now what you do? So, let M is equal to S compose T inverse. So, M is again a Mobius transformation and M of T z2, so this is nothing but, so this is equal to your 1. So, this is, yeah, so this you can do in the following way. So, S compose T inverse of your T z2. So, if I apply this, so S of z2, so this is basically S of z2. So, yeah, so S of z2, so this is equal to your 1. So, similarly M of T z3, because S of z3 is your 0, so this will be equal to 0. And M of T z4, so because S of z4 is infinity, so this will be equal to your infinity. So, therefore we have this thing, so this will imply that, so M of z, so this will be equal to your z, T z2, T of z3, T of z4. So, M takes this T z2 to your 1, M takes T z3 to equal to 0 and M takes T z4 is equal to infinity. So, therefore by this previous thing, M will be of this form. So, M will be of this form here, so which is basically, so M will be of this form here. So, you replace z1 by z, z2 by T z2, z3 by T z3 and z4 by T z4. So, this is my, this M and now, so again, so M is, so M is again equal to S compose T inverse z. So, therefore what we have here, so S compose T inverse z is equal to this cross ratio z, T z2, T z3, T z4. So, this imply that, now if we put T z1 in place of z, so what we have, so the right hand side it will be equal to T z1, then T z2, then T of z3 and then T of z4 and the left hand side is equal to S z1, so which is nothing but the cross ratio z1, z2, z3, z4. So, this will be equal to T of z1, T z2, T z3, T z4. So, this proves the proposition and next thing is very crucial. So, what will prove that, so if the cross ratio, if it is a real number, so then this four points corresponding to the cross ratio, it lies either on a circle or a line and the converse is also true. So, let me write this thing first. So, proposition, so let z1, z2, z3, z4 be four distinct points in C infinity. So, then, so the cross ratio is a real number if and only if, so this four points z1, z2, z3, z4 lie on a circle or straight line. So, how to prove this thing? So, again let us take, so let S z to be equal to your this Mobius transformation. So, then by definition of the inverse image, so S inverse R, so this will be equal to your, the set, so set of all complex number belonging to the C infinity such that this cross ratio z, z2, z3, z4 is real. So, S inverse, again it is a Mobius transformation. So, if you can prove that this S inverse R, it is either a circle or a line, so then we will be done. So, what we will prove that, so we will show that, so we will show that, so image of R union infinity under a Mobius transformation is a circle or a line. So, if we can show this thing, so then we are done. So, now let us take, so let S z to be equal to your A z plus B by C z plus D, where this A D minus B C is not equal to 0. So, this is my Mobius transformation. Let z equals to x belongs to your R and w to be, let us take it to be S inverse x. So, x is your Sw. So, S of w, it is a real number. So, this means that, so Sw is equal to the conjugate of this Sw. So, this implies that A w plus B by C w plus D is equal to your, this conjugate. So, by cross multiplying, so what we will get, so A C bar minus A bar C times mod of w square plus A D bar minus B bar C times w plus B C bar minus D A bar times w bar plus B D bar minus B bar T. So, this is equal to your 0. So, we will get this equation. So, now in this equation, so if this A C bar is a real number, so then, so A C bar, so this will be equal to the conjugate of this A C bar. So, this will be, so this complex number, so this number A C bar minus A bar C, so this will be equal to your 0. So, now let us, if A C bar is real, so in the above equation, the coefficient of mod w square, it vanishes. And now let us write, so that alpha equals to twice A D bar minus B bar C and beta to be I times your B D bar minus B bar T. So, one can check that, so check this thing. So, imaginary part of alpha w minus beta, so this will be equal to 0. So, this implies beta is equal to your imaginary part of alpha w. So, this is a real number. So, beta it is a real number. So, again this will imply that this imaginary part of this alpha w minus beta, so this is equal to 0, beta is real. So, now let us write, so this w to be equals to capital X plus I capital Y and alpha to be equal to your alpha 1 plus I alpha 2. So, then what is my imaginary part of this alpha w? So, this will be equal to your alpha 2 x plus alpha 1 y and we know that imaginary part of alpha w minus beta is equal to 0. So, beta is real number. So, we have this equation alpha 2 x plus alpha 1 y minus beta is equal to 0. So, this is an equation of straight line. We are done if AC bar is real. Now suppose, so now next if AC bar is not real, so then the equation star becomes of the form mod w square plus gamma bar w plus gamma w bar minus delta equals to 0 for some gamma belongs to C and some real delta. So, delta it will be a real number. So, that is obvious because, so this middle term, so if I take this middle term, suppose it is equal to p, so p is equal to your p bar, ok. So, the middle term, so this will be your real. So, this will imply that delta equals to mod of w square plus a real number and mod of w it is a real number, so delta will be a real, ok. And now again, so if I put, so w to be equal to your x plus iy. So, the above equation, so what we will get to be x square plus y square plus 2 gamma 1 x plus 2 gamma 2 y minus delta is equal to 0, where this gamma is equal to gamma 1 plus gamma i2 is an equation of circle, ok. So, we have proved the proposition that if I take any four distinct points and if the cross ratio z1, z2, z3, z4 is a real number, then this four points it lie in a circle or a straight line and also the converse is also true, ok. So, the next proposition is also very important. So, it is, so it says that a Mobius transformation, it takes circles to either a straight line or circle or if you take a straight line, the image of that thing under a Mobius transformation, it will be either a circle or a straight line. So, let me write this as a theorem. So, a Mobius transformation takes circles and straight lines to circles or straight lines. So, what is the proof of this? So, let gamma be any circle or a straight line. So, let us take three distinct points. So, let z2, z3, z4 belongs to your this gamma and they are distinct. Now, put wj to be equal to your S of zj, so where S is a Mobius transformation, ok. So, again, so this wj, so these are my distinct points because S it is a bijection. So, we have three points, three distinct points. So, then these three distinct points, it, so from these three distinct points if it is collinear, so then there exists a straight line passing through these three distinct points. If it is not collinear, so there exists a unique circle passing through these three distinct points w2, w3, w4. So, wj lies on a unique circle or a unique straight line and let us call it to be gamma prime. So, let gamma prime be the circle or straight line where this all these three points wj, w1, w2, w3, w4 lies. So, now the claim is, so S of gamma is equal to your gamma prime. So, we know that this Mobius transformation, it preserves the cross ratio. So, if I take this cross ratio z, z2, z3, z4, so this will be equal to your Sz, Sz2, Sz3, Sz4 and this will be equal to your S of z comma w2, w3, w4. So, if z belongs to your gamma, so in the, so then, so the left hand side of the above equation, so this is a real number, is a real number. So, by the above equation, so this will imply that Sz w2, w3, w4, so again it is a real number. So, again by the previous equation, so this tells that, so this Sz w2, w3, w4, it lies on a circle or a straight line and because there is a unique circle or a straight line passing through this w2, w3, w4, so this implies that, so Sz belongs to your gamma prime. So, therefore, we have proved that, so S of gamma is equal to your gamma prime. So, we are done. So, in the next class, we will be proving that this Mobius transformations are conformal, means that it preserves angles and then we will move to the geodesics of the upper half plane. So, I will stop here.