Differential or Infinitesimal Schwarzs Lemma,

Picks Lemma, Hyperbolic Arclengths, Metric and Geodesics on the Unit Disc So, uhhh the part I want to make is that if

you compute this, if you compute it and uhhh you will get this inequality okay. 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+z0 by

1+uhhh uhhh z0 bar zeta, this is the inverse of g okay, you can check that this is the

inverse to this alright. And if you now now just differentiate calculate the derivate

use using the quotient rule. So, you will get differentiate with respect

to zeta you will get this is 1+z0 bar zeta uhhh whole square uhhh 1+z0 bar uhhh zo bar

zeta into I will differentiate this with respect to zeta I will get 1-I keep the enumerator

constant, I differentiate this with respect to zeta, I get z0 bar. And uhhh so fine so

what I get is, I get 1+z0 bar zeta uhhh –uhhh zeta z0 bar uhhh –z0 z0 bar alright divided

by 1+z0 bar zeta the whole square. And of course these 2 cancel out, so I get 1-uhhh

z0 z0 bar is mod z0 the whole square by 1+z0 bar zeta the whole square. So, you know you calculate mod g dash of 0

which is what we want mod g dash of 0 you put zeta=0 you will get 1-mod z0 square 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 uhhh what is h of w0, h of w0 is uhhh

this function uhhh uhhh, so what is h of w, h of w is neta it is just this expression,

so h of w is just uhhh w-w0 by 1-w0 bar w. So, if you calculate the derivative again

by the quotient rule I will get the following denominator square, denominator constant derivative

of the numerator with respect to w it is going to be 1-numerator constant derivative of the

denominator with respect to w is going to give me –w0 bar okay. And uhhh so next point

out I will get 1-w0 bar w- so I will get + w w0 bar uhhh –w0 w0 bar divided by 1-w0

bar w in the whole square. And these 2 will cancel I will implicate 1-uhhh

mod w0 the whole square because w0 w0 bar is mod w0 the whole square by 1-w0 bar w the

whole square. So, mod h dash of 0 uhhh h dash of uhhh w0 (()) (04:36-04:47) uhhh and I uhhh

let us write I will I should get this 1-mod w0 square. So, you know now you plug in these

values in this equation in this in this inequation. And what you will get is I will get 1 by 1-mod

w0 the whole square that is the value of the derivative modulus of the derivative of h

at w0 and then I write mod f dash of z0 as it is

uhhh and I have to plug in mod g dash of 0, mod g of 0 is 1-mod z0 the whole square, 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-mind you w0 is fz0 by 1-mod z0 the whole square okay. And that is the uhhh that is the statement,

that is the that is the inequality of Pick’s lemma okay. So, uhhh and what you should understand

is that uhhh 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

it is image okay, z0 is an arbitrary point, so that uhhh that inequality holds for any

z0 in delta and therefore I can rem I can replace instead of z0 I can put z where z

belongs to delta. And therefore I get uhhh the inequality of

Pick’s lemma okay. Now uhhh uhhh uhhh, so uhhh uhhh 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 to for a single z0 uhhh if we get uhhh equality for a z0 then

uhhh if you get 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’s

lemma uhhh both the differential version and the original version of Schwarz’s lemma

you always get equality only if uhhh the uhhh the automorphism is an automorphism only if

the analytic function is an auotmorphism okay. So, uhhh uhhh you get mod so if I write that

out h circle f circle g derivative of the 0=1 implies uhhh h circle f circle g is uhhh

uhhh an automorphism of delta fixing 0 using 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 when I see automorphism holomorphic also a holomorphic automorphism

of delta okay. So, you see the uhhh what is the 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 can 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 uhhh an automorphism is supposed to be 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’s

statement okay and the Schwarz’s lemma itself says both the differential form and the usual

form of the Schwarz’s lemma says that if you get equality even at 1 point okay which

is uhhh uhhh uhhh at 1 point is this in the differential version uhhh if you get equality

of the derivative at the origin with 1 modulus of the derivative at the origin with 1. Then uhhh the function has to be an auto 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 the function which we applied the

differential version the Schwarz’s lemma. That this will be an automorphism but then

you can get f from this function by uhhh pre-composing 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 Moebius transformations they have inverses and therefore uhhh so this is

an isomorphism uhhh this is this is an isomorphism this central thing is an isomorphism, this

also an isomorphism and composition of isomorphism is again an isomorphism, so you will get f

is an isomorphism. So, that proves Pick’s lemma okay, so prick

pick’s lemma tells you that you will get equality here for a single z0 if and only

if you get equality there for all z for all z0, for for every z okay. And that will happen

if and only if the function is an automorphism, so if you uhhh so in other words if f is not

an automorphism of unit disc then there will be strict inequality here, it will mod f dash

z0 is strictly less than uhhh this quantity on the right side okay. So, this is Pick’s 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 z0=0 I get the differential version of the Schwarz’s lemma which says that modulus

of the derivative at the origin cannot exceed uhhh uhhh uhhh uhhh you know it cannot exceed

1 okay. If I put z0=0 and assume that f takes 0 to 0 that is if I put z0=0 and assume that

f of z0 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 Pick’s 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 see uhhh. 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 mapping the Riemann mapping theorem what we

have what we have suppose to do is we are suppose to start with a simply connected domain

which is not the whole complex plane. And you are suppose to map it holomorphically

isomorphically on to 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 uhhh the the domain was simply connected uhhh and and it was not the entire

complex plane okay. So, we reduced uhhh the mapping problem to a sub-domain of uhhh a

simply connected sub-domain of the unit disc okay alright. So, you have to uhhh now we

are reduce to proving that given any simply connected sub-domain of the unit disc, you

can map it conformally on to 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 on 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 Pick’s lemma. So, which is a uhhh nice generalisations of

Schwarz’s lemma okay, the differential form of Schwarz’s lemma. So, you know that is

how it is enters into the discussion of the proof that we are looking at the Riemann mapping

theorem alright. So, now we go onto 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 centred at the origin radius 1. So, you know

so you see so let us begin by uhhh 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 uhhh uhhh

suppose you have a piece wise smooth arc or a contour to which is just uhhh image of uhhh

uhhh the unit interval uhhh or or a any closed interval on the real line by a function gamma

which is piece wise differentiable okay and which is continuous okay. And such that the uhhh derivatives are also

piece wise uhhh the derivatives are 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

see it I will put in bra I wil 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 uhhh all

you have to do is you simply have to integrate over gamma mod dz, 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 t=a to t=b uhhh uhhh modulus of d

gamma t. And that will be just uhhh that will be just

integral from a to b uhhh gam mod gamma dash of t into mod dt okay and of course t is increasing

so you do not have 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 a 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

lands 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 is

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 the Euclidean length but I will do is I will integrate mod

uhhh I will integrate 1 by 1-mod z the whole square into d mod dz. So, I am I am adding

this factor 1 by 1-mod z the whole square which is the uhhh which is the uhhh hyperbolic

factor okay. So, this is called the hyperbolic length of gamma okay and what is so special

about this expression. This 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 inverses also holomorphic. So, it is a holomor holomorphic

automorphism of unit disc to unit disc and then you know so what will happen is this

you know this this f will map uhhh uhhh. So, this is my uhhh so this is a map w=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 1 to 1 onto inverses 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

closed interval a, b into the unit disc and what is that path that is that path is just

gamma circle f 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 uhhh path gamma is map

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-mod w the whole square is this 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 the correct variable alright

but you see uhhh watch carefully what is uhhh uhhh so now comes if now comes the you know

importance of Pick’s lemma okay, now comes the importance of Pick’s lemma, you see

what does Pick’s lemma say. See by Pick’s lemma what you will get is

mod f dash of z is less than or eq is equal to 1-mod fz the whole square by 1-mod z the

whole square for all z in the unit disc, you get this, you get equality in Pick’s lemma

because f is an automorphism of the unit disc, Pick’s lemma says that you get the inequality

of Pick’s lemma will become an automo will become an equality if they map is if and only

if the map f is a automorphism okay you get this. But then you see see in this d in this integral

you know I can make change of variable by putting w=fz. If I make a change of variable

then this integral is the same as integral over gamma okay mod d fz by 1-mod fz the whole

square okay. But what is this, this is the integral over gamma uhhh d of z is f dash

of z dz, so I will get mod f dash of z mod dz by 1-mod fz the whole square, this is what

I will get okay. If I make a simplification if I make change of variable from w to z using

w=fz I will get this. But what is this equal to by by Pick’s lemma

mod fz of z by 1-fz te whole square is simply 1 by 1-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 same as this expression because of Pick’s lemma because of the equality in Pick’s

lemma which comes because f is an automorphism of the 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, you whether you take the hyperbolic

length of gamma or whether you take the hyperbolic length of it is image under automorphism of

the unit disc, you will continue to get the same uhhh the same hyp uhhh uhhh the same

hyperbolic length. Therefore the so we expresses by saying that

Pick’s lemma the equality in Pick’s lemma actually asserts that for uhhh uhhh the the

automorphisms of the unit disc preserve the hyperbolic length. The equality in Pick’s

lemma assert that you know automorphisms of unit disc preserve the hyperbolic length okay.

So, let me write that down the automorphisms, the holomorphic automorphisms of delta reserve

the hyperbolic length of an arc. So, this is the uhhh this is the geometric

uhhh this is the geometric uhhh you know statement uhhh concerning the equality in Pick’s lemma

okay 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-uhhh uhhh you integrate mod

dz by 1-mod z the whole square okay, that is the uhhh that gives the hyperbolic length

right. Now you know uhhh it for the hyperbolic length

is uhhh uhhh I mean the unit disc uhhh is anyway as for as Euclidean uhhh at spaces

concern are the that is the plane is concerned 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 alright uhhh if you take a straight

line segment uhhh the length is less than 2 okay. But what about hyperbolic distance, so you

can it is it is it is rather curious we can make a computation, you know if you take the

unit disc and you know uhhh take a point 0 take the point z okay, take this straight

line to 0 and z okay. And then try to calculate what this uhhh length is what is the Euclidean

length, Euclidean length, so you know uhhh to calculate uhhh the Euclidean length uhhh

first of all I need a parameterisation I must think of this is apath. So, you know what I do is I just map 0, uhhh

so if if z suppose I call this point as z0 and zo is uhhh r0*e power i theta 0 okay where

r0 is this length which is actually the Euclidean length. And theta0 is this angle, so this

angle is theta0 and this length is r0 from here to here okay, then you know you know

how to parameterize this path. So, this path can be parameterized as 0, uhhh r0 map 2 uhhh

which is map 0, r0 to uhhh uhhh t uhhh uhhh uhhh to the unit disc by simply sending t

to t e power I theta0 okay when you put t=0 you will get the origin, when you put t=r0

you will get z0. And you will get this straight line segment

joining the origin z0, so this is your this your uhhh 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 uhhh gag a gauhhh uhhh of a of gamma is going to be integral

over gamma mod dz by 1-mod z the whole square, this is the this is the defi uhhh sorry this

is uhhh I should not 1 by I should not put this it just mod dz. So, what I will get is well if I substitute

for the x-for gamma gamma of t it is a z is gamma of t, so I will get uhhh t=0 to r0 uhhh

and I will get here mod d uhhh d of gamma of t but gamma of t is t t power i theta0

okay. And if I if I simplify this I will get integral 0 to r0 uhhh I have I have to differentiate

this I take differentiate and the uhhh variable of integration is t, so I will get uhhh I

differentiate this respect to t which will give me e power i theta0 uhhh. And then I will get dt and then I will have

to put a mod uhhh and of course mod e power i theta0 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 uhhh straight

lines even from 0 to z0 is r0 which is Euclidean length okay. Now let us calculate the hyperbolic

length what is a hyperbolic length and of course I should tell you that uhhh uhhh 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 the hyperbolic length of gamma,

hyperbolic length of gamma well it is integral over gamma mod dz by 1-mod z the whole square,

see this is the uhhh this is the formula for the hyperbolic length okay. So, if you calculate

that so you will get et is that, so well I will get uhhh integral from 0 to r0 uhhh so

again substitute this I will get uhhh mod d uhhh d of d e power i theta0 by 1-mod d

e power I theta0 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 uhhh and and the denominator I am going to just

get 1-t square okay and you know how to integrate this, you split as partial fractions uhhh

you know it is 1 by 1 uhhh –t-1 by 1t if I am not wrong

uhhh maybe I should a + here let me I will get 1+t+1-t is 2 by 1-t square, so I will

have to divide by 2, so this is what I will get uhhh dt. And you know what this is going to be this

is going to just give me uhhh uhhh uhhh, so this is this is uhhh half uhhh 1 by 1-t is

log 1-t uhhh so this is lan if you want lan of course t is less uhhh t is positive, so

this lan 1-t uhhh into -1 and here I am going to get +lan of 1+t and I am going to take

limits from 0 to r0. So, this is just lan 1+t by 1-t alright, so I am going to get half

uhhh lan uhhh 1+r0 by 1-r0, this is the hyperbolic length. So, the hyperbolic length, the Euclidean length

is r0 okay which is the modulus of z0 whereas hyperbolic length is you have a fine funny

expression it is half lan 1+r0 by 1-r0 and you can see something the Euclidean length

is finite okay it is bounded by 1. But you know if z0 tends close to the border of the

unit disc okay 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 infinitely okay uhhh 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 okay, so this is going to approach uhhh

0+. So, this quantity is going to approach infinity +infinity and lan of that is going

to go into +infinity, so the moral of the story is the hyperbolic length will tend to

infinitely as the point z0 moves to the edge of the unit disc as it goes to the units circle

okay. So, you know the so this is the beautiful

fact about the hyperbolic length, the hyperbolic length uhhh makes this in as far as the hyperbolic

distances concerned this is not uhhh this is this is not uhhh this is not a bounded

thing the unit the unit disc is not bounded okay, the disc even the straight even the

even a segment if you take uhhh uhhh straight line segment if you compute radial segment

it is 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 it makes the the

hyperbolic distance it makes this a uhhh in in the in the sense of the in the sense of

the hyperbolic distance it makes unit disc unbounded okay. So, that is one fact that

you have to notice alright, so well uhhh now you know I have to uhhh uhhh, so I will tell

you uhhh that uhhh what is what is 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 of the unit disc then it acts like a contraction okay. So, uhhh

what I have defined uhhh so so you know to define the notion of a contraction I have

to tell you what a metric what a metric is because you know a contraction map is defined

between metric spaces, it is defined from 1 metric space to another. And a mapping set to be a contraction map

if you know it decreases distances you take 2 points in the in the source space they have

certain distance but you take their 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 okay, what I have defined so for is just hyperbolic length

of a of an arc in the unit disc I am going to define the hyperbolic distance between

2 points in the unit disc okay, so let let me do that next. So, uhhh 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 okay okay and uhhh of course here also I should have mark this is 1, here

I should have marked it as 1 as well, this is the origin, this is the origin alright.

So, uhhh so you take 2 points in the uhhh uhhh in the unit disc alright take 2 points

of unit disc and what you do is the following for z0, z1 it delta define the hyperbolic

metric, rho uhhh uhhh. So, hyperbolic metric or hyperbolic distance,

distance function, the distance from z0 to z1 I am I am using the single rho okay to

be uhhh you know maybe I will I will I will I will maybe I will use rhos of h uhhh to

insist that this hyperbolic and you know what I do, I do the following thind simply join

uhhh z from from z0 to z1 you choose any uhhh path contour gamma measure it is hyperbolic

length and then minimise over all such possible paths. So, the hyperbolic distance from z0 to z1

is the least is the is the least of hyperbolic lengths of paths from z0 to z1 various paths

from z0 to z1 inside the unit disc you measure they are hyperbolic lengths and then you take

the minimum, you take the infimum okay. So, here is the definition this is equal to infimum

of uhhh hyperbolic length of uhhh paths from z0 to z1 in the unit disc okay. So, this is the this is the uhhh this is the

hyperbolic uhhh 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-mod z the whole square,

this is the hyperbolic metric where of course gamma of uhhh uhhh such that gamma uhhh is

a path in the unit disc from z0 to z1. So, this is the hyperbolic length okay okay, so

uhhh you know I am taking uhhh of course all all lens all these integrals are non negative. And uhhh uhhh you know uhhh so this infimum

does exist alright but what is it what is this infimum and uhhh and you know does the

infimum uhhh the infimum value does it correspond to actually length of a particular path that

is the question and the answer is yes even any to so here is the important statement

about hyperbolic geometry. So, even any 2 points in the unit disc there is a special

path from passing through from this between these 2 points which is called a hyperbolic

geodesicik it is the path of shortest hyperbolic length from uhhh uhhh. I mean between the 2 given points and what

is that path the answer to that is the the that path is a circle passing through those

2 points which is orthogonal to the unit circle okay, so that is a theorem. So, the theorem

is here is the very important theorem, the theorem is for any 2 for any 2 points z0 not

equal to z1 in delta uhhh the uhhh the the arc of the circle through z0 and z1 and orthogonal

to the unit circle okay 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 minimisation it is the minimum you are you are suppose to take all possible uhhh

path inside the unit disc from z0 to z1 measure their hyperbolic lengths and take the minimum

okay which seems a very uhhh it is not a 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 uhhh clearer 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 2 points

and which is orthogonal to the unit circle namely it is uhhh where it hits the unit circle

it will hit at 90 degrees okay, you know 2 you can also talk about the angle between

2 curves at a point uhhh at an intersecting point, it is bit of by definition the angle

between their tangents at that point. So, uhhh we circle are orthogonal if they

intersect uhhh at uhhh say 2 points and at each point of intersection the tangent to

the 2 circles are perpendicular to each other alright. So, you know so the 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

uhhh if I take a point if I if I take a 2 2 points like this okay then you know my uhhh

my hyperbolic uhhh uhhh geodesic will be something like this, it will it will be this will be

the geodesic from here to here. And uhhh that is because this is the circle

which passes through these 2 points okay and which hits the unit circle at 90 degrees.

So, you know so this is z0, this is z1 and uhhh at this intersec at this point if I draw

the tangent to tangent to the given uhhh 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 uhhh hyperbolic uhhh paths of uhhh paths of numeral minimal hyperbolic

uhhh lengths which are called geodesic you will get things like this see you know if

you you know if you take if you take 2 points uhhh along the along a diameter okay. Then

uhhh the hyperbolic geo the geodesic will be the diameter itself any diameter is a geodesic. Because if you take 2 points if you try to

find the circle passing through 2 points which lying on a line in principle there is no circle

but you think of it as a circle with uhhh you think also straight lines as circles with

the third point at infinity okay. So, if so the point is that the geodesic will look like

this you know this will be 1 geodesic any diameter will be a geo geodesic then you know

if you go the little to the left the geodesic will become like this. And you know if you if you get smaller the

geodesic will become smaller, this is how the geodesic will look like okay and the fact

is that any diameter will be a geodesic okay and the theorem says that these are uhhh the

paths of shortest uhhh length, that is how you get the path of shortest length. So, this

the theorem that we will have to uhhh we will have to give a proof of uhhh and we will we

will do that in the next lecture. So, uhhh so let me write let me add here uhhh

uhhh paths of shortest hyperbolic length are called geodesic for the hyperbolic metric

and uhhh so let me uhhh finish with one uhhh important statement you see if you are looking

at the Euclidean distance Euclidean metric okay. Then the geodesic are all straight lines

okay, if you take any 2 points in Euclidean space what is the shortest uhhh what is the

path of shortest lengths it will just be the line segment joining those 2 paths, straight

line segment. So, the geodesic 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 uhhh geodesic that we get for uhhh the hyperbolic metric

okay they will play the same role as straight lines play for Euclidean geometry okay. So,

all the axioms all the Eucli all the Euclid’s axioms except the parallel axiom that hold

for straight lines, I mean parallel axiom also holds for uhhh it is it is also taken

for Euclidean geometry. But this all those axioms will work with the

geodesic for the hyperbolic metric except that the parallel axiom uhhh you you have

to throughout the parallel axiom okay, all other axioms that you have for straight lines

okay uhhh. The same axioms will hold good for the hyperbolic geodesic, so the so all

these curves on the unit disc they are the analogues of straight lines on the Euclidean

plane. The analogues of the straight line on on on

the on the on the Euclidean plane which are important for Euclidean geometry, the analogues

here are these are these curves the hyperboloic geodesic and you know you can check a lot

of uhhh statements like you know if you have 3 lines which uhhh uhhh you know if you take

any 2 lines which are uhhh uhhh uhhh if if if you take 2 lines if they are not 1 in the

same uhhh uhhh. And of course you know if they are not parallel

then they will intersect at uhhh you know 1 point, 1 point in the finite plane and of

course if you think of infinity as a point then they will also intersect the infinity

alright. And the same way here also you can check that uhhh if you take 2 geodesic which

are not uhhh you know one in the same they will uhhh they will hit at one point okay. And if you take 3 lines you can you can use

3 lines to form a triangle and every triangle is formed by 3 lines okay uhhh which are the

lines uhhh passing through the lines the sides of the triangle. In the same way you can also

define uhhh hyperbolic triangle, a hyperbolic triangle will be will be something like this

you know uhhh, so you know I can, so if I draw a hyperbolic triangle it will be like

this uhhh it is given by 3 hyperbolic D geodesic. So, 1 like this, 1 like this and 1 like this,

so this is the hyperbolic triangle okay and uhhh you know that in the Euclidean geometry

the sum of uhhh 3 angles of the triangle is equal to uhhh 180 uhhh degrees what will happen

in hyperbolic geometry is that the is the sum of 3 the 3 angles of a hyperbolic triangle

will be less than 180 degrees okay and so you will have all these uhhh nice things happening

uhhh uhhh differently from uhhh what you know in Euclidean geometry. So, hyperbolic geometry uhhh will give you

uhhh uhhh gives set of properties okay and the basis for all this is the so called hyperbolic

length which is defined uhhh for a for an arc but the key to the fact that hyperbolic

length does not change under an automorphism of the unit disc is the equality in Pick’s

lemma okay. So, Pick’s lemma we called in Pick’s lemma is a is a beginning point for

a whole geometry okay. And uhhh now I will now let me tell you if

you take this hyperbolic uhhh metric then if you take any automorphism of from the unit

disc to the unit disc any automorphisms of unit disc to the unit disc will be an isometric

with respect to the hyperbolic metric okay, that is what that is what you will get okay.

So, uhhh so the whole beautiful point about Pick’s lemma will be that you know any automorphism

of unit disc will uhhh will actually be an isometry of the hyperbolic distance of the

hyperbolic metric. And any map which is not an automorphism of

the unit disc will be a contraction with respect to this hyperbolic metric and that is the

statement that we need uhhh to proceed with the proof of Riemann Riemann mapping theorem

okay which we will do in the next uhhh lecture.

Nice lectures Sir thank you!

These videos are gold. Thank you so much for the thoroughness.