Lecture 26

Hello, in the last lecture, we have seen that a torus admits a Riemannian metric, which is isometric to the Euclidean metric. So on torus, so we can put the Euclidean metric and the curvature with respect to that metric at each point is zero. So using the same idea, so we will prove that, so if you take any closed orientable surface of genus greater than equals to 2, so it admits a hyperbolic metric. So you can put a Riemannian metric on closed surface of genus greater than equals to 2, such that that Riemannian metric is isometric to the hyperbolic metric. And so also, we will be proving that the fundamental group of closed orientable surface of genus greater than equals to 2 is a Gaussian group. So in fact, so what we will show that, so if you take any genus 2 surface which is closed and orientable, so then it admits a hyperbolic metric and the fundamental group of the surface is a Gaussian group. So using the same thing, you can do the same thing for higher genus surface. So this theorem, what I am going to say is called Boyk-Harris theorem. So I will be not proving the whole thing. So I will be proving that thing for only genus g equals to 2. So first, we will be proving that the fundamental group of this closed orientable surface of genus greater than equals to 2. So it, in fact, it is a subgroup of this PSL2R or PSE11. And after that, to show that it is a Fuchsian group, so that fundamental group of the surface, so this will act on the unit disk model of this hyperbolic plane properly discontinuously. So that thing we have to show. So this, that will prove that this fundamental group of the surface is a Fuchsian group. And using the same idea, if you take any closed orientable surface of genus greater than equals to 2, so then its fundamental group will be Fuchsian. And because the fundamental groups, fundamental group, it acts properly discontinuously on the unit disk model of this hyperbolic plane, and the action is also free. So the quotient will be a surface of genus greater than equals to 2. And the surface will admit a hyperbolic metric. And that metric will be coming from this hyperbolic plane and the covering map, using the covering map. So let me first begin with this genus 2 surface. So this thing already we have seen that, so there exists a regular octagon in the unit disk whose sides are geodesics. So this thing already we have seen. So this is my regular octagon. So this angle, so this angle, so this will be my pi by 4. So this angle will be my pi by 4, yeah. So this part already we have seen. So now what we are going to do here, so if we identify the sides in this octagon suitably, so the quotient space will be a genus 2 surface. So this red geodesic will be identified to this red geodesic. So this identification will be done by a hyperbolic isometry. And similarly for the other colors, like this black geodesic will be identified with black geodesic. And the blue geodesic will be identified with the blue geodesic. And the green geodesic will be identified the green geodesic. So if I make that identification, so the quotient will get to be a genus 2 surface, ok. So now the first job is to find an isometry which will be taking this, first of all this black geodesic C1 to black geodesic C2, ok. So we know that this geodesic, so this is a part of a circle. So this you can extend, so this you can extend. So this will meet the unit circle. So this is my unit circle, this is the boundary of this unit disc orthogonally. So C1, so if I call this circle to be C1, so C1 will intersect the circle S1 to be orthogonally. So similarly I can extend this geodesic and make it bi-infinite geodesic. So this will meet orthogonally and further I can extend in the complex plane. So I'll get my C2. So C2 will intersect the circle S1 orthogonally, ok. So now what I want, so I want an isometry taking C1 to C2, ok. So any isometry of this unit disc, model of this hyperbolic plane looks like this. So z plus c by c bar z plus a bar, where this mod of a square minus mod of c square is equal to 1. And let us take this a to be, yeah, so orientation preserving isometry of this unit disc, ok. So now let us consider the circle, so mod of c bar z plus a bar equals to 1. So if I take mod of c bar z plus a bar equals to 1, so this implies that, so mod of z plus a bar by c bar, so this is equal to 1 by mod c. So this is a circle in C center minus a bar by c bar and radius 1 by mod c. Now let us make another observation. So suppose, so z satisfy this thing, so z belongs to this set, z belongs to C such that mod of c bar z plus a bar is equal to 1. So suppose mod of c bar z plus a bar is equal to 1. So now if I compute this thing, so mod of minus c bar a z plus a, so this will be equal to your minus c bar times a z plus c by c bar z plus a bar plus a, so take the mod of this thing. So this will be equal to 1 by mod of c bar z plus a bar equal to 1. So we have this observation. So this tells you this thing. So if I take this c1 to be the circle, so z belongs to C such that mod of c bar z plus a bar equals to 1 and c2 to be the circle, so z belongs to C such that minus c bar z plus a is equal to 1. So c2 again it is a circle, again it is a circle whose center is a by c bar and radius is 1 by mod c. So what this equation is showing, so a of c1 is equal to c2. So a of c1 is, so what we have shown that a of c1 is subset of c2 and a it is a Mobius transformation, so in fact it will take a of c1 to c2. So what we have assumed that, so this circle, so this circle is c1 and this circle is c2. So we need to find an isometry which takes this c1 to c2. So we have some, so there exists some Mobius transformation which takes c1 to c2, and now what we are going to find this coefficients a and c. So this a and c that we are going to find, ok. So first let us use this notation, so o means center of this unit disk which is nothing but this thing. Let us take p to be minus a bar by c bar which is center of c1. So now what you do, so join this op. So suppose this is my center b, so what you do, so you first join this. And now let us take this point to be my b, this point to be my b, and now you take this line segment. I will just take a Euclidean line segment here. So we have this triangle and this point is my o. So now note that, so what will be this angle. So this angle is pi by 4. So why this angle is pi by 4, so let us see this thing. So now what you do, so you draw a tangent at b to the circle c1. So you draw a tangent here. So this is my tangent. So this angle, so this angle it will be 90 degree, so it is pi by 2. And also note that, so this angle, so this will be half of pi by 4. So this is pi by 8. And now also note that we have taken this regular octagon such that the internal angles are pi by 4. So this angle is pi by 4. So all these angles are my pi by 4. So therefore, if I take this angle, if I take this angle, so this will be half of pi by 4. So this is my pi by 8. So this is pi by 8. So therefore, this angle, so this angle, so this will be my 3 pi by 4. So therefore, this angle it will be my pi by 4. And let us call this point to be my Q. So if I take this triangle PBQ, so we have, so this is a right angle triangle at P. So one angle is pi by 4. So the other angle, so this angle QPB, so that will be equal to your pi by 4. Ok. So let me write down. So we have written joint OP, so that thing we have done. So BQ is tangent to C1 at P. So internal angles are, of the regular octagon, are pi by 4. So this will imply that, so the angle OBQ, so this will be equal to pi by 8. Yeah, angle OBQ, so this will be pi by 8. And now consider triangle OBP. So angle this BPO, so this is equal to your pi by 2 minus pi by 4. So that is equal to your pi by 4. So this will imply that angle PQB, so this is equal to your pi by 4. So we have this picture. So because this angle BQB is equal to angle PPQ, so therefore the sides PB and BQ will be same. So here now we are doing this Euclidean geometry. So therefore, so this PB, so equal to BQ. So these are my Euclidean lengths. So PB will be equal to your BQ, and this BQ will be also equal to your OQ. So if I take the triangle OBQ, so we have two equal angles. So therefore, OQ will be equal to your BQ. Also we have this thing, okay. So now, so what is radius of the circle C1? So this is, so this is 1 by mod C. So radius of the circle is 1 by mod C, which is nothing but the length of PB. So PB is the radius of the circle C1. So radius of C1, so that is equal to your PB, so that is equal to your 1 by mod C. So therefore, OQ, so this will be equal to 1 by mod C. And the center is P, so OP is equal to your minus A bar by C bar. So this is equal to your A by C, mod of A by C. And also note that, so this PB equal to sin pi by 4 times BQ. So PB, so we have this right angled triangle at B. So PB equal to sin pi by 4 times BQ. So that will imply, so this will imply that PQ, so this will be equal to PB by sin pi by 4. So this is equal to OQ root 2, which is equal to 1 by mod C root 2. And now also see that this OP is equal to your OQ plus QP. Yeah, OP is equal to OQ plus QP. So that is equal to your OQ plus OQ root 2. So OQ times 1 plus root 2. And now mod of A by C, so this is equal to your 1 by mod C times 1 plus root 2. So this will imply that mod of A is equal to 1 plus root 2. So now consider the center of this C1 and the center of C2. So if I go back to this picture, so center of C1 is P and center of C2 what I want is, so this to be my, so this is my A by C bar. Yeah, center of C2 is A by C bar. So from here, ok. So we have this relation here between these two centers of C1 and C2. So A by C bar is e to the power minus i pi by 2 A bar by C bar, yeah. So if I draw a picture here, both this point lie at a same distance from this origin. So this angle is pi by 2, so therefore we have this thing. So A by C bar is equal to e to the power minus i pi by 2 A bar by C bar, ok. So that is, so this A is equal to minus i times A bar. So let us write this equation to be star equation. And now suppose if A is equal to mod A ei theta, so from this equation, star, so what we will get? So ei theta, so will be equal to minus i times e to the power minus i theta. So this will imply that this theta is equal to 3 pi by 4. So from the earlier thing, so mod of A we have already computed, it is 1 plus root 2. So therefore what we have, so this, so therefore this A is equal to 1 plus root 2 and e to the power i 3 pi by 4, so which is nothing but minus 1, 1 by root 2 plus i by root 2. So we got this A and mod of C is, so will be equal to root over mod A square minus 1 and also we know this thing. So minus A bar by C bar, so this will be equal to your mod of this minus A bar by C bar. So this is equal to A by C and so this is making an angle pi by 8, yeah, so this angle is pi by 8 with the x-axis. So therefore we have this equation. So minus A bar by C bar, it is just the point P. It is the point P in the complex plane and it is making an angle pi by 8 with this real axis. So therefore we have this thing. So from here, so what we will get, the C is equal to minus root over mod A square minus 1 by mod A times A times e to the power i pi by 8. So we know the value of A and we know the value of mod of A, so we can compute the C. So also check this thing. So we got hold of this A and C. So Az is Az plus C by C bar Z plus A bar. So you can check this thing. So trace of this matrix, so if I take the mod of this thing, mod of trace of A, so this will be greater than 2. So check this thing. So this will imply that A is a hyperbolic isometry and let us take, so let A2 be this isometry A. So A2 of this C1 is C2. So this we got and we already computed the matrix of this A, ok. And now let us take this isometry. So phi z equals to e to the power i pi times z. So this again it is a isometry of this unit disk model of this hyperbolic plane. So this in fact it is an elliptic isometry, it is just rotation by angle pi. So it will fix the origin. And now let us take, so A1 to be defined to be this conjugate of this A, A2. So phi A2 phi inverse, ok. So if I take this isometry A1, so what this A1 will do? So A1 will take this geodesic, this blue geodesic B1 to this blue geodesic B2. So if I call this to be my C1 prime and if I draw the same circle here in corresponding to this B2, call it to be C2 prime. So this, so this A1 it will take this B1 to B2. So this A1 will take B1 to B2. So check this thing. And now if I take psi z, so let psi z to be this rotation by angle pi by 4 around origin and let us denote this A1 prime to be this conjugate psi inverse A2 of psi. And this, ok, so one notation mistake. So let us call it to be A2 prime and let us call this A1 prime to be this conjugate. So one mistake. So this should be A1 prime should be this conjugate phi A2 prime phi inverse. So now what you can check that. So this trace of each of this Ai, take the mod, so this will be greater than 2. And trace of this Ai prime, take the mod, so it will be greater than 2. So all this, this A1, A2, A1 prime, A2 prime, so are hyperbolic isometrics. And now if I take this product of this commutator, so A1 inverse, then take A1 prime inverse, then A1, A1 prime, then let us take A2 inverse and A2 prime inverse, then A2, A2 prime. So let us take this product of this commutators. So this is nothing but this commutator A1, A1 prime and A2, A2 prime. So now if I take this vertex, so let us go back to this diagram. So if I take this vertex, so let us call it to be my V. So if I apply this isometry, this product of this commutator, so this is again it is an isometry. So if I apply this isometry on that vertex, so you can check that. So it is again equal to V. So this will imply that. So this will imply that. So this isometry, because it fixes a point of this, interior point of this unit disk, and it cannot be an elliptic isometry, so it has to be identity element. So therefore, so if I take this group generated by A1, A1 prime, A2, A2 prime, and we have this relation, so we have this relation, so this groups, so this groups is in fact it is a subgroup of the isometry of this unit disk. So in fact what we have shown that, so this in fact it is a subgroup of this PSU 1 1, which is again a subgroup of this isometry of this unit disk model of this hyperbolic plane. And this group is again nothing but the fundamental group of this genus 2 surface. So, so this is my genus 2 surface, yeah. So what we have done so far, so we got 4 isometries. So one isometry, so that is my namely by A1, it will take the black geodesic with my black geodesic. And the other isometry, so A2, it will take this B1 to B2. And one of this A1 prime or A2 prime, it will take this green geodesic to green geodesic. And the other one, it will take the red geodesic to red geodesic. So we got hold of 4 geodesics, which will do the side pairings. And we have a relation, only one relation. So the product of the commutators is equal to 1, yeah. So therefore, so what we have seen, so the fundamental group of this genus 2 surface, so this we got, this genus 2 surface we got from this regular octagon by side pairing. So this fundamental group of genus 2 surface, it acts by isometries on the unit disk model of this hyperbolic plane. So in fact, so one can show that, so this action, so acts as a covering space action or properly discontinuous action without any fixed points on the unit disk model of this hyperbolic plane. So this will prove that, so this will prove that this pi1 of sigma2 is a Fuschian group. So what we have done, so if I take this disk with hyperbolic metric, so this pi1 of sigma2, it acts properly discontinuously on this unit disk. So therefore, this quotient space, so this is is homeomorphic to sigma2 and in fact, so in this space, so we can put a hyperbolic metric, okay. So maybe I will stop here and in the next class, so I will show that how to give this hyperbolic metric on this genus 2 surface.