 So, uhhh the point I want to make is that if you compute this if you compute it and write it down you will get this inequality ok. So, let us compute it see what is uhhh so what is the expression for g. So, what is g of uhhh uhhh so g is like this g is a function of zeta and z is g of zeta alright and what is g of zeta g of zeta is just zeta plus z0 by 1 plus uhhh uhhh z0 bar zeta this is the inverse of g ok. You can check that this is the inverse to this alright and if you now now just differentiate calculate the derivative use using the quotient rule. So, you will get differentiate with respect to zeta you will get this is 1 plus z0 bar zeta uhhh whole squared uhhh 1 plus z0 bar uhhh z0 bar zeta into I differentiate this with respect to zeta I get 1 minus I keep the numerator constant I differentiate this with respect to zeta I get z0 bar and uhhh so fine. So, what I get is I get 1 plus z0 bar zeta uhhh minus uhhh zeta z0 bar uhhh minus z0 z0 bar alright divided by 1 plus z0 z0 bar zeta the whole squared and of course these 2 cancel out. So, I get 1 minus uhhh z0 z0 bar is mod z0 the whole squared by 1 plus z0 bar zeta the whole squared. So, you know if you calculate mod g dash of 0 which is what we want mod g dash of 0 you put zeta equal to 0 you will get 1 minus mod z0 squared okay that is what you will get for mod g dash of 0 that is the expression and then I will have to calculate what h dash of w0 is uhhh now. So, what is h of w0 h of w0 is uhhh this function uhhh uhhh so what is h of w h of w is neat it is this expression. So, h of w is just uhhh w minus w0 by 1 minus w0 bar w. So, if you calculate the derivative again by the quotient rule I will get the following denominator squared denominator constant derivative the numerator with respect to w it is going to be 1 minus numerator constant derivative denominator with respect to w is going to give you minus w0 bar okay and so expand it out I will get 1 minus w0 bar w minus so I will get plus w w0 bar uhhh minus w0 w0 bar divided by 1 minus w0 bar w the whole squared and these 2 will cancel I will simply get 1 minus uhhh mod w0 the whole squared because w0 w0 bar is mod w0 the whole squared by 1 minus w0 bar w the whole squared. So, mod h dash of 0 uhhh h dash of uhhh w0 and I have that is right I will I I should get this 1 minus mod w0 squared. So, you know now you plug in these 2 values in this equation in the in this equation and what you will get is I will get 1 by 1 minus mod w0 the whole squared that is the value of the derivative modulus of derivative of h at w0 and then I write mod f dash of z0 as it is I have to plug in mod g dash of 0 mod g dash of 0 is 1 minus mod z0 the whole squared this is less than or equal to 1 so that gives me what I want I get mod f dash of 0 is less than or equal to 1 minus mind you w0 is fz0 by 1 minus mod z0 whole squared okay and that is the uhhh that is the statement that is the that is the inequality of pix lemma okay. So uhhh and what you should understand is that uhhh uhhh now the of course this in when I do this calculation I have fixed uhhh I have simply taken z0 to be any point in the unit disc and I have taken w0 to be its image okay z0 is an arbitrary point. So, that that inequality holds for any z0 in delta and therefore I can I can replace instead of z0 I can put z by z belongs to delta and therefore I get the inequality of pix lemma okay. Now uhhh uhhh uhhh so the the only thing that I have to tell you is that you get equality if uhhh and only if f is a holomorphic automorphism of the unit disc. So, and that too for a single z0 uhhh if we get uhhh equality for a z0 then uhhh if you if you get equality if you get equality here that means you are actually getting equality here okay and uhhh but then you know in Schwarz lemma uhhh both the differential version and the original version of Schwarz lemma you always get equality only if uhhh the uhhh the automorphism is an automorphism only if the analytic function is an automorphism okay. So, uhhh uhhh you get mod so if I write that out h circle f circle g derivative of 0 is equal to 1 implies uhhh h circle f circle g is uhhh uhhh an automorphism of delta fixing 0 fixing the origin and this implies that because you know h and g are also automorphisms okay you will get that f which is just h inverse composition h circle f circle g composition g inverse uhhh okay is also uhhh uhhh an automorphism of uhhh uhhh of course and I say automorphism holomorphic also a holomorphic automorphism of delta okay. So, you see the uhhh what is the uhhh uhhh both the usual version of Schwarz's lemma from the unit disc to the unit disc and the differential version both are statements about inequalities okay both uhhh uhhh give you inequalities and they tell that you get an equality only in the case when the uhhh function that you are considering from the unit disc to the unit disc is an automorphism. You get equality only when it is an automorphism if it is not an automorphism uhhh by that if it is not an automorphism is supposed to be a self map okay map from a given set back to itself. So, if you have an analytic function from the unit disc to the unit disc uhhh which is not an isomorphism namely not an automorphism then you will get only a strict inequality at every point in the Schwarz's lemma statements okay and the Schwarz's lemma itself says both the differential form and the usual form of Schwarz's lemma says that if you get equality even at one point okay which is uhhh uhhh at at one point is is uhhh in the differential version uhhh if you get equality of the derivative at the origin with one modulus of the derivative at the origin with one then uhhh the function has to be an automorphism okay. So uhhh and also the earlier version uhhh the usual version of the Schwarz's lemma also says that that uhhh whenever you get an equality for uhhh point which is different from the origin okay then the analytic function has to be an automorphism. So, that condition will tell you that uhhh this h circle f circle g which is a function to which we applied the differential Schwarz's lemma that this will be an automorphism but then you can get f from this function by uhhh precomposing with h inverse on the left and post composing with g inverse on the right okay and uhhh uhhh uhhh and that is possible because h and g are of course Mobius transformations they have inverses and therefore it so this is an isomorphism uhhh this is this is an isomorphism this central thing is an isomorphism this is also an isomorphism composition of isomorphism is again an isomorphism. So, you will get f is an isomorphism so that proves Picks lemma okay. So, Picks lemma tells you that you will get equality here for a single z not if and only if you get equality there for all z for all z not for a for every z okay and that will happen if and only if the function f is an automorphism. So, if you uhhh so in other words if f is not an automorphism unit disc then there will be strict inequality here it will mod f dash z not is strictly less than uhhh this quantity on the right side okay. So, this is Picks lemma and as you can see it is just a generalization of the differential version of Schwarz's lemma because in this if I put z not equal to 0 I get the differential version of the Schwarz's lemma which says the modulus of the derivative at the origin cannot exceed uhhh uhhh uhhh uhhh uhhh uhhh you know it cannot exceed 1 okay. If I put z not equal to 0 and assume that f takes 0 to 0 that is if I put z not equal to 0 and assume that f of z not is also 0 then if you put that here you will get mod f dash of 0 is less than or equal to 1 that is the differential version of Schwarz's lemma. So, uhhh uhhh Picks lemma is just a generalization of the differential version of Schwarz's lemma okay but the point is that it is the key to uhhh so called hyperbolic geometry on the unit disc which is what we have to study okay. See so let me uhhh let me again remind you uhhh what we did earlier was that you know we were uhhh we are we are on our way to prove the Riemann mapping theorem okay and the Riemann map in the Riemann mapping theorem what we have what we are supposed to do is we are supposed to start with a simply connected domain which is not the whole complex plane and you are supposed to map it holomorphically isomorphically onto the unit disc. The first step that we achieved was to map it holomorphically isomorphically onto a sub domain of the unit disc okay. So, this was possible uhhh because uhhh it because uhhh the the domain was simply connected in the and it was not the entire complex plane okay. So, we reduced the mapping problem to a sub domain of uhhh simply connected sub domain of the unit disc okay alright. So, you have to now we are reduced to proving that given any simply connected sub domain of the unit disc you can map it conformally onto the unit disc okay. So, our problem is completely reduced to studying sub domains of the unit disc and so in other words you have to study the unit disc carefully and how we are going to do it or the way we are going to do it which will help us is study hyperbolic geometry of the unit disc and the hyperbolic geometry depends on so called hyperbolic metric and the hyperbolic metric the key to the hyperbolic metric is the Picks lemma. So, which is a uhhh nice generalization of Schwarz's lemma okay the differential form of Schwarz's lemma. So, you know that is how this enters into the discussion of the proof that we are looking at of the Riemann mapping theorem alright. So, now we go on to study hyperbolic geometry. So uhhh so let me do that. So, this is hyperbolic geometry on the unit disc delta. So, this is this is open unit disc centered at the origin radius 1. So, you know so you see so let us begin by uhhh uhhh recalling certain facts. You know if you have uhhh uhhh so suppose this is a complex plane and suppose you have uhhh an arc suppose you have a piecewise smooth arc or a contour which is just uhhh image of uhhh the unit interval uhhh or or uhhh any close interval on the real line by a function gamma which is piecewise a differentiable okay and which is continuous okay and such that the uhhh derivatives are also piecewise uhhh the derivative is also continuous okay. So, so if you take a contour from uhhh this point which is gamma of a starting point to gamma of b this is my path gamma or contour. Then you know how to get the length of gamma length of gamma and I will stress it I will say I will put in I will just prefix it by Euclidean okay because this is the length in the usual sense arc length. You know what is the formula for the Euclidean length of gamma or all you have to do is you simply have to integrate over gamma mod d z this will give you the length of gamma alright. And what is that integral I mean it is this integral is well you can also substitute you can this is from you know uhhh t equal to a to t equal to b uhhh uhhh modulus of d gamma t and that will be just uhhh that will be just integral from a to b uhhh mod gamma dash of t into mod dt okay and of course t is increasing. So, you do not have to put this mod here. So, this is the this is the Euclidean length of a of an arc okay. Now what we are going to do is we are going to take a special case uhhh we are going to look at this arc uhhh we are going to look at such arcs or contours inside the unit disc okay. So, you are going to have situation like this you have this uhhh closed interval a b finite closed interval on the real line and you are going to have this path or contour gamma and see the point is that this gamma lines inside the unit disc. So, you know it is something like this. So, this is the unit disc delta and uhhh this is the complex plane again alright and now I am going to again this is gamma of a this is gamma of b and this is my path gamma. And what I am going to do I am going to something uhhh new uhhh instead of defining the Euclidean length of gamma which you know is this you integrate mod dz over gamma. I am going to define the hyperbolic length of gamma okay that that is that is that is only in the special case when the path is a path in the unit disc okay. So, so here is the definition hyperbolic length of gamma is you see what you do it is also an integral over gamma okay. See if I put integral over gamma and put mod dz I will get the Euclidean length okay if I put integral over gamma and and if I integrate mod dz I will get Euclidean length but what I will do is I will integrate mod uhhh I will integrate 1 by 1 minus mod z the whole square into d mod dz. So, I am I am adding this factor 1 by 1 minus mod z the whole square which is the uhhh which is the hyperbolic factor okay. So, this is called the hyperbolic length of gamma okay. Now what is so special about this expression the special thing about this expression is you see if I now take f to be an automorphism holomorphic automorphism of the unit disc that is it is a map from the unit disc to the unit disc which is holomorphic injective bijective holomorphic. So, it is inverse is also holomorphic. So, it is a holomorphic automorphism unit disc to unit disc and then you know so what will happen is this you know this this f will map uhhh so this is my uhhh so this is the map w equal to fz okay and uhhh so this is the z plane and here I have the w plane alright and what is going to happen is that because f is an automorphism uhhh one to one onto inverse is also holomorphic what is going to happen is that the image of this path is also going to be a simple path is also going to be a path inside the unit disc. So, what I am going to get is I am going to get another path like this starting point will be f of gamma of a and uhhh the ending point will be f of gamma of b and I will get this path which is uhhh uhhh which is gamma followed by f okay. So, it is this you first apply gamma then you apply f then you get a path from a from this close interval a b into the unit disc and what is that path that is that path is just gamma circle uhhh uhhh yeah it should be f f composition gamma f f circle gamma that is right it should be f circle gamma right. So, I get this so the path gamma is mapped by f isomorphically onto the path f circle gamma now now you see the beautiful thing you calculate the hyperbolic length of f circle gamma okay what is the hyperbolic length of of f circle gamma well it is by definition integral over f circle gamma of mod dw by 1 minus mod w the whole square this is the definition of hyperbolic length where I am using the fact that uhhh my my variable here is w and the variable here is z so I am using correct variable alright. But you see uhhh watch carefully what is uhhh uhhh so now comes now comes the you know importance of pix lemma okay now comes the importance of pix lemma you see what does pix lemma say see by pix lemma what you will get is mod f dash of z is is equal to 1 minus mod f z the whole square by 1 minus mod z the whole square for all z in the unit you get this you get equality in pix lemma because f is an automorphism of unit disc pix lemma says that you get the inequality of pix lemma will become an automa will become inequality if the map is an if and only if the map f is a automorphism okay you get this. But then you see see in this the in this integral you know I can make change of variable by putting w equal to f z if I make a change of variable then this integral is the same as integral over gamma okay mod d f z by 1 minus mod f z the whole square okay. But what is this this is integral over gamma uhhh d f z is f dash of z d z so I will get mod f dash of z mod d z by 1 minus mod f z the whole square this is what I get okay if I make a simplification if I make change of variable from w to z using w equal to f z I will get this. But what is this equal to by by pix lemma mod f dash of z by 1 minus f z the whole square is simply 1 by 1 minus mod z the whole square therefore what you get is this is the same as this these 2 are equal these 2 are this this this expression is the same as this expression because of pix lemma because of the equality in pix lemma which comes because f is an automorphism unit disc alright. So what is the moral of the story the moral of the story is the following the moral of the story is if you define the hyperbolic length of an arc or contour in the unit disc the hyperbolic length will not change if you apply an automorphism of the unit disc whether you take the hyperbolic length of gamma or whether you take the hyperbolic length of its image and an automorphism unit disc you will continue to get the same length the same hyperbolic length therefore the so we expresses by saying that pix lemma the equality in pix lemma actually asserts that for the automorphisms of the unit disc preserve the hyperbolic length the equality in pix lemma assert that you know automorphisms unit disc preserve the hyperbolic length ok. So let me write that down the automorphisms the holomorphic automorphisms of delta preserve the hyperbolic length so this is the this is a geometric you know statement concerning the equality in pix lemma ok and that is the importance of this expression instead of just integrating over mod dz which will give you the Euclidean length you integrate over mod dz by 1 minus you integrate mod dz by 1 minus mod z the whole square ok that is the that gives the hyperbolic length right. Now you know you know the hyperbolic length is I mean the unit disc is anyway as far as Euclidean space is concerned or the that is the plane is concerned the unit disc is bounded and you know if you take the ordinary length the ordinary length between any 2 points is going to be finite of course it cannot exceed 2 which is the diameter of the unit disc right if you take a straight line segment the length is less than 2 ok but what about hyperbolic distance so you know you can we can it is rather curious we can make a computation you know if you take the unit disc and you know take a point 0 take the point z ok take this straight line joining 0 and z ok and then try to calculate what this length is what is Euclidean length Euclidean length so you know to calculate Euclidean length first of all I need a parameterization I must think of this as a path so you know what I do is I just map 0, so if z suppose I call this point is z0 and z0 is r0 into e power i theta0 ok where r0 is this length which is actually the Euclidean length and theta not is this angle so this angle is theta not and this length is r0 from here to here ok then you know you know how to parameterize this path so this path can be parameterized as 0, r0 map to you just map 0, r0 to t well to the unit disc by simply sending t to t e power i theta not ok when you put t equal to 0 you will get the origin when you put t equal to r0 you will get z0 and you will get this straight line segment joining the origin to z0. So this is your path gamma now the straight line segment from 0 to z0 where z0 is a point of the unit disc what is the Euclidean distance Euclidean length of gamma is going to be integral over gamma mod dz by 1 minus mod z the whole square this is the sorry this is I should not put 1 by this it is just mod dz so what I will get is well if I substitute for the expression for gamma of t so z is gamma of t so I will get t equal to 0 to r0 and I will get here mod d of gamma of t but gamma of t is t e power i theta not ok and if I simplify this I will get integral 0 to r0 I have to differentiate this I take differential and the variable of integration is t so I will get I differentiate this with respect to t which will give me e power i theta not and then I will get dt and then I will have to put a mod and of course mod e power i theta not is 1 so I will simply get integral dt from 0 to r0 and I am going to get r0 which is what I expect the length of this straight line segment from 0 to z0 is r0 which is Euclidean length ok. Now let us calculate the hyperbolic length what is the hyperbolic length and of course I should tell you that of course r0 is strictly less than 1 because z0 is a point of the unit disc r0 is strictly less than 1 alright but what is a hyperbolic length of gamma hyperbolic length of gamma well it is integral over gamma mod tz by 1 minus mod z the whole square so this is the this is the formula for the hyperbolic length ok. So if you calculate that so you will get what is that so well I will get integral from 0 to r0 so I again substitute this I will get mod d d of t e power i theta not by 1 minus mod t e power i theta not the whole square this is what I will get so I will get integral from 0 to r0 as usual this is going to give me dt alright so and the denominator I am going to just get 1 minus t square ok and you know how to integrate this you split it as partial fractions you know it is 1 by 1 minus t minus 1 by 1 plus t if I am not wrong maybe I should put a plus here let me I will get 1 plus t plus 1 minus t it is 2 by 1 minus t square so I will have to divide by 2 so this is what I will get dt and you know what this is going to give me this is going to just give me so this is this is half 1 by 1 minus t is log 1 minus t so this is ln if you want ln of course t is less t is positive so this is ln 1 minus t into minus 1 and here I am going to get plus ln of 1 plus t and I am going to take limits from 0 to r0 so this is just ln 1 plus t by 1 minus t alright so I am going to get half ln 1 plus r0 by 1 minus r0 this is the hyperbolic length so the hyperbolic length the Euclidean length is r0 ok this is the modulus of z0 whereas the hyperbolic length is you have a fine expression it is half ln 1 plus r0 by 1 minus r0 and now you can see something the Euclidean length is finite ok it is bounded by 1 but you know if z0 tends close to the border of the unit disc ok if z0 gets close to the boundary of the unit disc then r0 gets close to 1 and as r0 gets close to 1 this approaches infinity ok I mean this denominator approach is 0 alright and therefore this quantity approach is infinity because r0 is going to tend to 1 from the left ok so this is going to approach 0 plus so this quantity is going to approach infinity plus infinity and ln of that is going to go to plus infinity so the moral of the story is the hyperbolic length will tend to infinity as the point z0 moves to the edge of the unit disc it as it goes to the unit circle ok so you know the so this is the beautiful fact about the hyperbolic length. The hyperbolic length makes this in as far as the hyperbolic distance is concerned this is not this is not this is not a bounded thing the unit disc is not bounded ok that is even the straight even the even a segment if you take straight line segment if you compute radial segment its length tends to infinity in the hyperbolic distance if the if the end point goes closer and closer to the unit circle so this is the point about the hyperbolic metric it makes the hyperbolic distance it makes this in the in the sense of in the sense of the hyperbolic distance it makes unit disc unbounded ok so that is one fact that you have to notice alright so well now you know I have to so I will tell you that what is what is it that we need actually we are looking for a statement like this we are looking for a statement which says that if you take any analytic function from the unit disc to the unit disc then if it is not an automorphism unit disc then it acts like a contraction ok so what I have defined so you know to define the notion of a contraction I will tell you what a metric what a metric is because you know a contraction map is defined between metric spaces it is defined from one metric space to another and a mapping is said to be a contraction map if you know it decreases distances you take two points in the in the source space they have certain distance but you take the images and then you measure the distance the distance becomes smaller so if this happens for a map it is called a contraction map and essentially the statement that I need for the proof of the Riemann mapping theorem is a statement that if you take a analytic map of unit disc to itself which is not an automorphism then it will necessarily be a contraction map but contraction with respect to what metric it is with respect to the hyperbolic metric which I am going to define now ok what I have defined so far is just hyperbolic length of an arc in the unit disc I am going to define the hyperbolic distance between two points in the unit disc ok so let me do that next so so hyperbolic metric on the unit disc so here is the so here is the definition of hyperbolic metric so here is my unit disc ok and of course here also I should have marked this as 1 here I should have marked it as 1 as well this is the origin alright so you know take two points in there in the unit disc alright take two points in unit disc and what you do is the following for z0 z1 delta define the hyperbolic metric so hyperbolic metric or hyperbolic distance function the distance from z0 to z1 and I am using the symbol rho ok to be you know maybe I will use rho sub h to insist that this hyperbolic and you know what I do I do the following thing simply join from z0 to z1 you choose any path contour gamma measure its hyperbolic length and then minimize over all such possible paths so the hyperbolic distance from z0 to z1 is the least is the least of hyperbolic length of paths from z0 to z1 various paths from z0 to z1 inside the unit disc you measure their hyperbolic lengths and then you take the minimum you take the infimum in fact ok so here is the definition this is equal to infimum of hyperbolic length of from z0 to z1 in the unit disc ok so this is the this is the this is the hyperbolic metric so in other words this is infimum over all gamma such that the hyperbolic length is given by integral over gamma mod dz by 1 minus mod dz the whole square this is the hyperbolic metric where of course gamma of such that gamma is a path in the unit disc from z0 to z1 so this is the hyperbolic length ok ok so you know I am taking of course all all lengths all these integrals are non-negative and you know so this infimum does exist right but what is it what is this infimum and you know does the infimum the infimum value does it correspond to actually length of a particular path that is the question and the answer is yes given any two so here is the important statement about hyperbolic geometry so given any two points in the unit disc there is a special path from passing through from this between these two points which is called hyperbolic geodesic it is the path of shortest hyperbolic length from I mean between the two given points and what is that path the answer to that is the path is a circle passing through those two points which is orthogonal to the unit circle ok so that is the theorem so the theorem is here is a very important theorem the theorem is for any two for any two points z0 not equal to z1 in delta the arc of the circle through z0 and z1 and orthogonal to the unit circle ok is the unique path of minimal hyperbolic length from z0 to z1 so this is the theorem so the theorem is that what is this hyperbolic distance it gives you see the hyperbolic distance is defined by some minimization it is the minimum you are supposed to take all possible paths inside the unit disc from z0 to z1 measure their hyperbolic lengths and take the minimum ok which seems a very it is not a it is not a definition that will help you to make calculations because you have to find minimum but the theorem makes it clear it tells you what is that path which will give you the minimum hyperbolic length that path of minimum hyperbolic length is nothing but the arc of a circle passing through these two points and which is orthogonal to the unit circle namely it is where it hits the unit circle it will hit at 90 degrees ok you know two you can also talk about the angle between two curves at a point at an intersecting point it is by definition the angle between their tangents at that point so we say two circles are orthogonal if they intersect at say two points and at each point of intersection the tangents to the two circles are perpendicular to each other alright so you know so the picture is like this you know if I take the unit circle if I take the unit disc you know if I take a point if I take a two points like this ok then you know my hyperbolic geodesic will be something like this it will be this will be the geodesic from here to here and that is because this is the circle which passes through these two points ok and which hits the unit circle at 90 degrees so you know so this is z0 this is z1 and at this point if I draw the tangent to the given to this circle and the unit circle this will be 90 degrees similarly here if I draw the tangent from here and here this will be another 90 degrees so this will be the hyperbolic path and the beautiful thing is that you know if you draw all these hyperbolic paths of paths of minimal hyperbolic lengths which are called geodesics you will get things like this see so you know if you you know if you take if you take two points along the along a diameter ok then the hyperbolic geodesic will be the diameter itself any diameter is a geodesic because if you take two points if you try to find a circle passing through two points which lie on a line in principle there is no circle but you think of it as a circle with you think of also straight lines as circles with the third point at infinity ok so if so the point is that the geodesics will look like this you know this will be one geodesic any diameter will be geodesic then you know if you go a little to the left the geodesics will become like this and you know if you go get smaller the geodesic will be smaller this is how the geodesic will look like ok. And the fact is that you need a diameter will be a geodesic ok and the theorem says that these are the paths of shortest length that is how you get the paths of shortest length so this is a theorem that we will have to give a proof of and we will do that in the next lecture. So let me write let me add here paths of shortest hyperbolic length are called geodesics for the hyperbolic metric and so let me finish with one important statement you see if you are looking at the Euclidean distance Euclidean metric okay then the geodesics are all straight lines okay if you take any two points in Euclidean space what is the shortest what is the path of shortest length it will just be the line segment joining those two paths straight line segment. So the geodesics in the Euclidean metric they are just straight lines okay alright and straight lines are important for Euclidean geometry right in the same way these geodesics that we get for the hyperbolic metric okay they will play the same role as straight lines play for Euclidean geometry okay so all the axioms except the parallel axiom that hold for straight lines I mean parallel axiom also holds for it is also taken for Euclidean geometry but this all those axioms will work with the geodesics for the hyperbolic metric except that the parallel axiom you have to throw out the parallel axiom okay all other axioms that you have for straight lines okay the same axioms will hold good for the hyperbolic geodesics so the so all these curves on the unit disc they are the analogs of straight lines on the Euclidean plane the analogs of the straight line on the Euclidean plane which are important for Euclidean geometry the analogs here are these curves the hyperbolic geodesics and you know you can check a lot of statements like you know if you take two lines if they are not one and the same and of course you know if they are not parallel then they will intersect at you know one point one point in the finite plane and of course if you think of infinity as a point then they will also intersect infinity alright and the same way here also you can check that if you take two geodesics which are not you know one and the same they will they will hit at one point okay and if you take three lines you can use three lines to form a triangle and every triangle is formed by three lines okay which are the lines are passing through the lines the sides of the triangle in the same way you can also define hyperbolic triangle a hyperbolic triangle will be something like this you know so you know I can so if I draw a hyperbolic triangle it will be like this it is given by three hyperbolic geodesics so one like this one like this and one like this so this is a hyperbolic triangle okay and you know that in Euclidean geometry the sum of three angles of a triangle is equal to 180 degrees what will happen in hyperbolic geometry is that the sum of three angles of a hyperbolic triangle will be less than 180 degrees okay and so you will have all these nice things happening differently from what you know in Euclidean geometry so hyperbolic geometry will give you new set of properties okay and the basis for all this is the so called hyperbolic length which is defined for a for an arc but the key to the fact that the hyperbolic length does not change under an automorphism of the unit disc is the equality in pix lemma okay so pix lemma the equality in pix lemma is a bearing point for a whole geometry okay and what and now I will now let me tell you if you take this hyperbolic metric then if you take any automorphism from the unit disc to the unit disc any automorphism unit disc to the unit disc will be an isometric with respect to the hyperbolic metric okay that is what you will get okay so the whole beautiful point about pix lemma will be that you know any automorphism unit disc will actually be an isometric of the hyperbolic distance of the hyperbolic metric and any map which is not an automorphism unit disc will be a contraction with respect to this hyperbolic metric and that is a statement that we need to proceed with the proof of Riemann Riemann mapping theorem okay which we will do in the next lecture.