# The Fundamental Region in the Upper Half-Plane for the Unimodular Group

ok so ah so let me ah begin by trying to ah place in perspective where we are at this point of our discussion so ah so we have so we have the ah we have the fallowing of a half plane ok and these up[per] the upper half plane the ah on the complex plane and then we have the quotient of the upper half plane by the unimodular group and ah we have defined a function ah j telda ah which is holomorphic and takes complex values and we are verified that this functions j telda this holomorphic function j telda goes down to this quotient which we know is a riemann surface ok and it goes down to a function j so this diagram commutes in other words j telda is ah invariant for the action of the unimodular group ok and ah so we we[re] we are trying to show ah we are trying to show to show that j is a holomorphic isomorphism in other words we are trying to show that ah the natural riemann surface structure on u mod psl two z namely the set of orbits of the unimodular group on the upper half plane is the same as the usual complex plane with the natural riemann surface structure on it ok and of course you know that the interpretation for this is this is also the set of holomorphic isomorphism complex tori so you think of ah if you give me if you give if you take a tau in the upper half plane then you take the corresponding complex torus define by tau ok and you take its isomorphism class that corresponds to unique orbit half psl two z and ah this is the map that takes tau to its orbit can also be thought of as a map that takes tau to the holomorphic isomorphism class of the complex torus defined by tau ok and of course the whole point is to say that the on the set of ah holomorphic isomorphism classes of complex tori ah that is ah the classification is achieved of the isomorphism by single invariant and this is call the j invariant ok and ah so you see we have to we have to show the j of course j is j is holomorphic ok ah that can be seen by ah looking at how this ah how the riemann surface structure on u mod psl two z is gotten ah in fact we ah ah we showed that psl two z as properly discontinuously on u and ah how this map is ah a ramified covering ok and if you analyze ah that construction you can show that because j telda is holomorphic j is also holomorphic ah so the other thing that one has to show is the j is both injective as well as surjective because you know bijective holomorphic map is is an isomorphism ok which means its inverse is also a holomorphic map so i have to prove j is injective and surjective ah as i told you the easier part is to show that ah j is surjective and we did that last time by proving the j telda as surjective ok ah we have proved that j telda is surjective which implies j is surjective ok that’s because is already surjective map this is just it is just the quotient map ok

so sending every point to its orbit this is just the space of orbits ok so ah usually we we ah ah put a double arrow head to denote a surjective map and i am saying that since j telda is surjective so j is also surjective ok and of course in the proof of j telda is surjective we essentially made usage of fundamental theorem of algebra ah and also namely that the complex numbers are algebraically close that is every complex po any every polynomial in one variable it complex quotients has all complex roots all its roots in complex numbers and of course we also used the fact that the function ah lambda the partially modular function lambda which is not invariant enter the whole psl two z but invariant under the congruence mod two subgroup of psl two z that function doesn’t take the values zero one one we use that also in the proof of ah the fact the j telda is surjective ok now ah the next thing that one has to do is that one has to show is the j is injective ok so for this ah i stated the fallowing thing last time so let me expand on it so we have you see we have ah so that is this region in the upper half plane so this is the tau plane ah the complex numbers being thought of as ah the z plane being thought of as a tau plane and well you take ah so you take ah this point which is ah minus half this is zero this is half and this is one this is the point minus one and that draw circle centre at zero radius one ok so i am going to get i will only worry about the upper semicircle ok and then i draw this line ah vertical line passing through a minus half and another vertical line passing through half so so i get something like this so this is the line here and well and this another one here and the fact is that we called ah we call this ah we call this this open region i mean this open side which is bounded by t my this portion of the imaginary axis this portion of the arc of the unit circle and this vertical line we call that is delta ok and we call its mirror image by the imagine ah by the imaginary axis as delta prime and then we define the the domain d to be ah delta it is just the ah closure of delta prime union a delta and delta prime to union of the closures but then you just take out you just take out this ah this line away and then take out this arc away ok so this is this it is this set minus take out ah all the ah ah so so this line is the set of all tau is the real par[t] real part of the tau is ah is half ok so what you do is you take out ah set of all tau such that tau in u such that real part of tau is ah half and of course the imaginary part of tau is greater than equal to root three by two ok so ah this this point will correspond to a complex cube root of unity ok so ah minus one plus i root three by two so the y coordinate is root three by two ok so real part of tau is half imaginary part of imaginary part of tau is ah greater than or equal to ah root three by two ok so that is that is to delete this whole line ok so in particular i also deleted this point ok and then i also want to delete this arc of the unit circle ok ah but i would like to retain this point i ok so so i also from so i also take away ah from from this union ok i am taking away this and

i am also taking away ah this union the other one is this arc which is ah the set of all ah e power i theta with theta varying from ah theta greater than pi by two and lesser than or equal to two pi by three ok so but of course ah one point is common but anyway let me write that ah pi by two is less than theta is less than or equal to two pi by three ok so that i wants to taking away this arc so finally what i get is well this should be real part of tau is minus half correct ah ah i want to take off this line ah thanks so is real part of tau is equal to minus half right it’s this line it is this it is this ray that i want to take off ok uh thanks so you see this is what i want to so i take this out so see the net net effect is you know after i take it out the if i if i take the region along with the boundary see what i will get is so let me let me kind of you know ah make this a dotted arc ok so that and let me also make this dotted so that you know it just to tell you that that i thrown out this ah this ray and also this arc ok so so this whole thing is d so if you if i if i draw it i mean let me draw it once more ah the region that i am i am worried about is ah is a fallowing ah may be i draw it may be i draw it here no matter so it’s it’s the fallowing region so this is well this is half this is one this is minus half and this is minus one and here is line and region i am worried about is then this is dotted then i start from here then it go here this is not here ok and this is my d this is what this is what d is this is your this is the region d ok this a region d and ah so what i ah what i just trying to tell ah last time is that the to prove the injectivity of j is is more difficult ok so ah and i said that ah the the first thing is to show injectivity of j telda in fact bijectivity of j telda when restricted to d ok so in fact i said ah i i said the fallowing thing i said ah we want to prove so let me write it somewhere here we want to show ah to show that ah j telda restricted to d is bijective ok among other things ok but before if you see further let me ah make a few remarks about what are called as fundamental regions ok for mappings and for groups ok so ah so let me make a definition so you see ah fundamental regions so ah let me explain what are what fundamental regions are so ah at least ah at least in the in the context that is relevant to us ok ah so so you see suppose suppose x suppose you have x to y you have map f is a is a holomorphic map be a holomorphic map of riemann surfaces suppose is a holomorphic map of riemann surfaces and and ah g a subgroup ah subgroup of the holomorphic automorphisms of the riemann surface x ok so x is a riemann surface y is a riemann surface f is a holomorphic mapping and ah suppose

i have a this group g which is set in subgroup of holomorphic automorphisms of the source ah riemann surface ok ah and let’s assume that f is surjective so let me put that so i let me put is a surjective ok suppose suppose that f is g invariant suppose you have a g invariant map ok so so that is that is you see f of g of x is equal to f of x for all x in x and for all ah for every g in g ok so this is ah saying that f is g invariant is a same as saying that f is constant on g orbits ok it is a same as saying f is constant on g orbits f restricted to every orbit of g is constant ok so that is f is f restricted to each orbit of a g [noise[ in x is a constant ok suppose you have a g invariant map alright now ah of course you know ah if you have a g invariant map like this it means that this map factors through the set of orbits ok so ah so we have so we have a factorization of maps so you have x ah well x to f is from x to y and you have got x to x mod g x mod g is the set of orbits of g in x ok and this is a natural map ah i put a double arrow because it is just suggestion on to set of orbits and ah well ah you have also ah f is surjective ok so the fact that f is constant on orbits means that f goes down to a map like this ok so ah maybe i can call this as f hat if you want ok and of course of course in good situations ah x mod g could be a riemann surface we we know already that you know for example if g is acting properly discontinuously on x then x to x mod g also riemann surface i mean then x mod g is a riemann surface so that x to x mod g is natural projection is holomorphic and there in that case f hat will become also a holomorphic map but in general ah f if x mod g doesn’t have if actually of a riemann surface that is if g doesn’t act nicely then this is you just think of it as a set i am just saying that saying that f is g invariant is the same as saying that ah f it goes down to a map from set theoretic map at least from x mod g ok fine so [noise[ given such a map what is it we are interested in ah we will call by a fundamental region for that map f what we mean is an open it’s a region in a x namely it consist of an open subset of x so open connected subset a domain ok an open connected subset of a x together with the part of the boundary ok so it’s not open the interior is open but you have added part of the boundary such that f restricted to that set is actually ah surjective ok and you see the ah the the translates of that set by g cover all of x ok so you see so so in other words you see ah so let me write this down ah so a a subset r of x is called a fundamental region for f if number one well so let me write the fallowing ah r con[sist] r consist of a r consist of an open connected set open connected set together

with a part of its boundary that is the first condition the second condition is ah the the the the translates of a r by g cover ah x ok so this is the condition that ah this is the condition that you know ah every g orbit hits r ok if if the translates of r by g cover x what you are saying is that ah if you take the orbit of every point in r and take then it should you should get all the points of x that means every orbit essentially has a representative in r every g orbit in a x has a representative in r and ah i will put the extra condition that ah that representative is unique ok so the translates of r by g cover x and ah each g orbit of ah x has a unique representative in r ok of course when i say if i if i just say each g orbit of x has a unique representative in r that that already means that the translates of r by g cover x ok but this is more i want only one representative right and the third ah the third thing ah so so this is the first thing the third thing is f if i take f and restricted to r ok mind you that r is not a ah it’s not a open set or a close set ok because it’s an open set ah the interior of r is of course an open set but there’s a part of the boundary that is added so f restricted to r ah for example you cannot talk about f restricted r being holomorphic because r is not a an open subset ok but the point i want to make is f restricted to r should be injective ah it should be injective as well as surjective ok so f restricted to r is both ah injective and surjective ok so this so these are the conditions that define what is meant by a the fundamental region for a map alright and ah so you see ah let me try to give examples of this so ah so if you take so examples are well number one you take ah omega one omega two are ah two complex numbers non zero complex numbers such that ah omega two by omega one is not a real number ok then you take the lattice ah you take the lattice l of omega one comma omega two which is lattice ah consisting of integral ah integral combinations of omega one and omega two ok lattice generated by omega one and omega two so it is in omega one plus m omega two where n and m very integers then you know that i can define a torus using this lattice and namely i take ah i take t omega one comma omega two to be complex the complex plane modular this lattice ok ah so this gives me the complex torus that is defined by omega one and omega two and ah the so if you if you now look at if you look at this map you see to ah ah t omega one comma omega two if you look at this let me call this map as pi the natural projection map the quotient map ok ah of course you know that this is a in fact universal holomorphic universal covering with fundamental group of the torus being identified with the subgroup of mobius transformations that i given by translations by elements of the lattice ok you of course know that and the bind i want to make is that the if you take a portion of the fundamental parallelogram that is ah generated by omega one and omega two then that is a fundamental that is a fundamental ah region for this map ok this a holomorphic

map it is the surjective holomorphic map and so you see ah this is a surjective holomorphic map is to riemann surfaces ok ah the the l omega one and omega two can be thought of as a subgroup of holomorphic automorphisms of see namely translations ok you think of every element of here as a translation by that element and then ah if you take ah of course this map ah this this map is a constant on orbits because it’s actually the quotient map ok so it is con[stant] so it is invariant and real omega one and omega two ok it is constant on orbits by this action ok and ah for this map the fundamental region is part of the fundamental parallelogram you throw away to adjust into edges of the fundamental parallelogram and you keep the rest ah then that will be a fundamental region ok then ah the ah the parallelogram ah ah defined by edges one omega one omega one plus omega two omega two so if i draw a diagrams well so here is when if this is omega one and that this is omega two then you get this parallelogram here this is omega one plus omega two ok [noise[ and ah you take ah the parallelogram with edges zero ah omega one omega one plus omega two and omega two ok and what you do ah is you take the whole interior ok and take for example take ah throw away this edge and this edge ok then that’s the fundamental region ok ah in fact let me say closed parallelogram the closed parallelogram defined by these edges ah defined by the ah not edges i should say vertices defined with a defined with vertices the closed parallelogram defined with vertices which means i the i take all all the interior parallelogram and take also the edges the boundary ok and then throw out to ah two quoted minus edges namely for example throw out ah the edge containing this edge and this edge ok minus ah with ah minus the edge ah omega two two omega one plus omega two ah and omega one two omega one plus omega two ok so you take this so you throw away so you throw away this so you throw away this and mind you when throw away this this point is also thrown out ok and you throw away this edge so this point is also thrown out and then you take you take this ah region ok then this is ah ah this is a fundamental region for the map pi is a fundamental region the map pi because you see first of all you will see that a every every orbit has a unique representative here ah that is clear the second thing is that ah this map restricted to this is both i[njective] the the map is restricted that is both injective and surjective and if i take translates of this by the lattice i will cover the whole complex plane so all the three conditions are satisfied all the three conditions are satisfied so this is the fundamental that’s a reason why it’s called ah the fundame[ntal] the fundamental parallelogram ok but of course you want to really make it a fundamental region you have to throw out so obviously you know ah one is throwing out this because you know if i include this then for every point here there is another point there which is another representative so i throw out this edge so that i get a unique representative ah for points here and i throw out this edge so that i get a unique representative points here and of course the lattice point ah will have all the four all the four points will be equivalent and relatives so i want only one of them so i so i but i already thrown out these three so i get only one ok fine

so ah now ah so this is one thing and ah you see ah if you look at if you go back and ah ah if you go back to this ah so so so let’s go back to this diagram lets go back to this diagram these also rei[mann] this is the riemann surfaces this is the riemann surface in this a holomorphic map ok and well ah you see ah the of course what is acting on top is psl two z ok ah which is a subgroup of holomorphic automorphisms of u which is the the all the holomorphic automorphism of u or psl two r and psl two z which is a subgroup is a unimodular subgroup ok and you can and the fact is for this map ok this is surjective holomorphic map for this map the fundamental ah region is actually this d ok that is a theorem that we are going to prove ok so i am just trying to refresh everything in terms of ah that language so so what i am trying to say is that ah sorry so the second and third statements i am going to write down actually claims we are going to prove them so a second statement is that this script d is a fundamental region for this quotient map ok two ah we will show that script d is a fundamental region for the quotient map ah u to u mod psl two z ok we will prove this this has to be prove ok and putting it as an example but we have to prove it ok and then so that is ah so that concern this map on the other hand look at this map this is also a map between ah riemann surfaces this is also holomorphic and surjective and again here i have this subgroup psl two z which is subgroup of holomorphic automorphisms of u and j telda is psl two z invariant ok that claim is for j telda the very same region d is again a fundamental region ok so three we will show that ah d is a fundamental region for j telda ok and once you prove these two it will fallow immediately that ah it fallow immediately that j is surjective j is an isomorphism j is see because you see ah so ah you can think you can think about it for a minute you see if there if there a two points here so first of all j is already surjective i have to only show injectivity if i have two points here ok which are map by j to the same point then i can find two representatives there which are map by j telda to the same point but then those two representatives i can move them by psl two z into d because d is a fundamental region and its translates cover the whole upper half plane ok so i will get two points here i will get two points here and ah j telda taking the same value ok but j telda but you see ah script d is also a fundamental region for j telda so j telda restricted to be d d has to be injective so that will tell you the those two points are are actually equal and that will give me the infectivity of j ok so ah so in that lang[uage] so this is the language we are going to use to prove that actually j is a holomorphic isomorphism ok so ah fine so so let me ah having given you this language ah let me first go and and show ah the first the second statement ok namely that ah ah u that script d is a fundamental region for psl two z for the unimod[ular] unimodular group so in this in this definition i want to say something ah i just want to say that you know ah if you apply this definition

to the case ah of a quotient that is for this quotient map when this quotient is a riemann surface ok ah if you get a fundamental region for this map then that is also called the fundamental region for g ok so the fundamental region for ah subgroup of holomorphic automorphisms of a ah riemann surface is defined if it is exist it is defined to be d ah fundamental region for the quotient map with then you will have to assume that x mod g is also a riemann surface in this is a holomorphic map ok so ah in other words ah statement two is actually ah in statement two ah we are actually saying that for psl two z ah script d is a fundamental region ok right so ah ok so ah let me try to go to that we now prove [ ah script d ah is a fundamental region for r u to u mod psl two z let me call this map as pi ah don’t confuse this map with the pi in the first example ok let me call this map is pi ok we will proves ok so ah so the first thing that ah so the first thing that i will need to show is that of course ah the first condition is satisfied by d it is an open connected set together with the part of its boundary ok so d is ah d d is an open connected set and what i have included ah apart from the interior is part of the boundary namely this arc of the unit circle fallowed by this ray ok so it is so condition one is ah satisfied its condition ah two and three that i will have to verify so first thing ah let me verify that ah ah every every orbit of psl two z meets a script d i will first prove that then i will prove that it makes in script d exactly at one point ok p we only need to show a one ah let me put roman letter one every psl two z orbit in u meets ah meets script d number two ah every psl two z orbit in u meets d exactly at one point ok so ah of course two is ah in princ[iple] i mean the way i severed it to stronger than one i mean if i say every psl two z orbit in u meets the exactly at one point it also meets it meets it it does meet ah d at one point but essentially what i will to prove to what i will essentially prove is that if the orbit has two points in d then i will show the two points are equal so this is existence is uniqueness ok is existence is uniqueness so how does one go about ah proving one so for this we will have to again go back and take a small detour into ah some properties as a lattice ah so you see ah ah so first of all ah lets write out what are the conditions that characterize the point of t ok so ah d is characterized by number one ah mod tau is greater than or equal to one ok it is ah its so this region is ah on the exceed of the unit circle then the second thing is ah real part of tau ah lies ah between you know minus half and plus half

so its its here ok the of course ah tau is in u so imaginary part of tau ah is greater than zero right this is all in the upper half plane and ah then you see i want include ah i want include this arc ok and i want to throw i want to throw out that arc and i want to throw out this line ok so ah so the so the condition is that slightly more restrictive condition in part two namely that the real part of tau is ah always greater than minus half ok so real part of tau is greater than minus half which means you know ah i could have written two and four together and set the real part of tau actually lies between is greater than minus half but is less than or equal to half ok ok ah that throws out ah ah this this line segment alright and ah then i also want to throw out this arc ok so for that i will say that if mod tau is one which is the condition for tau would lie on the unit circle if mod tau is one then i will say that real part of tau is greater than or equal to zero ok so if mod tau is equal to one then real part of tau is greater than equal to zero ok so ah so these are the ah conditions that characterize when a point tau the complex plane is in d alright and what is what is that i have to prove i will have to prove that give me any point in the you start with the point tau ah now start with a point ah let us say tau one ok in the upper half plane i will have to show that ah the in the orbit of that point there is a point in ah script d ok so i have to find a tau to which is the translate of which which is the image of tau one under psl two z element and sets the tau to satisfies all these conditions and i should this for any tau one in u ok so starting starting with with tau one in u we seek to to produce a tau two in d such that tau two is equal to ah ah ah tau to is the image of tau one under a a mobius transformation in psl two z namely a uni[modular] unimodular transformation ok this what you have to do you start with ah the once you do this what you are saying is that so your just saying that the psl two z orbit of tau one which contains tau two intersects d in tau two ok this what you have to do so ah so the point is that you have to you have to cook up this ah a suitable mobius transformation and so that the image of tau one under that ah unimodular mobius transformation is actually equal to tau two and that tau two ah lies in script d ok now so to beginning with to do that ah so i will ah so we begin as fallows so the first thing that we do is that ah ah we so you know this is so let me say something ah before i begin so you know this is ah on the same lines of the proof that i gave ah several lectures ago when i was trying to show that you know ah a discrete is at sub module of ah c is either zero or it is generated by a single complex number non zero complex number or it is generated by integer multiples of two complex numbers with non real ratio ok i proved this statement long back ok ah and it’s the proof is actually again an analysis of that proof ok and ah and using that proof a little the ideas of that proof little cleverly ok so ah so ah we we said such that mod omega one is least positive ok so you see ah so

what you do is that you look at that lattice generated by ah one and tau and you take the take a member of the lattice which is closes to the origin ok then there may be more than one but you can take one of them ok so call omega one to be that alright so call this to be omega one then what you do is now you look at all those elements in the lattices in the lattice which are not an ah real multiple of omega one namely ah not an integer multiple of omega one and among those choose omega two to be the closest one of to one that is one that is closest to the origin ok so omega two is equal to n two plus m to tau such that ah omega two is not a ah real multiple real of course it has to be integer multiple of omega one ok ah and ah mod omega two is least positive ok so ah this is how i choose omega one and omega two and now the fact is that you should take the lattice generated by omega one and omega two then that lattice is exactly the same as l tau ok so what i have done is i have instead of taking the basis one comma tau for the lattice over z i have found a new basis omega one comma omega two of the same lattice over z ok so ah how do how does one establish that so claim the first claim is that the lattice generated by omega one and omega two is actually the lattice generated by tau of course l tau is lattice generated by one and tau ok so the first claim is the lattice generated by so if you that is i am saying if you take all integer linear combinations of omega one and omega two you will get all integer linear combinations of one and tau of course l tau is l one tau ok l one tau ok so so how does one prove this so what one does is ah if if if z is a complex number ok if z is the complex number ah you see ah d ah ah the first thing that you need to notice is that you know omega one and omega two they are not ah omega two is not a multiple of omega one by the varied definition because you are choosing omega two among the members in the lattice l tau which are not real multiples of omega one so you see omega omega two ah by omega one is a non real ratio ok so that’s that that is already ah that’s already there in the definition here ok so so these are two complex numbers ah one which is not a real multiple of the other therefore these two complex numbers ah see it is they will form a basis for the complex vector space c over r over the real numbers so which means that any complex number can be written as a real linear combination of omega one and omega two ok so since ah omega one and omega two have non real ratio that exist lambda mu in r such that is z is lambda omega one plus mu times omega two ok you can write any complex number in terms of these two with real quotients ok because they form a basis of complex number of of of the field c over as a vector space over r ok now ah in particular if you take for z and an a member of l tau ok so if further ah z belong to l of tau suppose z is in l of tau ok alright choose ah choose lambda telda and

mu telda integers ok so that ah the the distance between lambda telda and lambda ah is less than or equal to half distance between mu telda and mu less than or equal to half after all lambda and mu are real numbers and i am saying choose the closest integer to ah lambda and call it lambda telda ok ah and choose the closest integer to mu and call it as mu telda ok that could be two of them if ah if if the if lambda or mu happens to be midpoint of an integer ah an interval with n points integers ok fine so choose this now ah consider z prime to be well z minus lambda telda omega one minus mu telda omega two look at this element look at this element now you see ah you see this element ah see zeta is in l tau ok and ah if you if you look at all these things ah notice that omega one is in l tau omega two is in l tau therefore the lattice generated by omega one and omega two is certainly in l tau the hard part that is the part that needs to proved is that l tau is contained inside this ok so ah this whole quantity namely z prime is in l tau ok this is in l tau ok and compute the modulus of z prime if you compute the modulus of z prime well and plug in for zeta lambda omega one plus mu omega two so i will get modulus of lambda omega one plus mu omega two minus lambda telda omega one minus mu telda omega two this is just modulus of lambda minus lambda telda into omega one plus mu minus mu telda into omega two ok and this is strictly less than modulus of lambda minus lambda telda into mod omega one plus modulus of mu minus mu telda into mod omega two this is actually less than or equal to by triangle inequality and you know in triangle inequality you will get an equality only when these two are ah one is a one is a real multiple of the other ok which it is not which it is not omega one is not a real multiple of omega two ok they have non real ratio therefore it’s a strict inequality in the triangle inequality and well if you and you know now mod lambda minus lambda telda minus lambda is less than or equal to half so what it i will get is this is this is less than or equal to half mod omega one plus half mod omega two ok but you see the way i chosen ah mod omega one is a least and mod omega two is great greater than or equal to mod omega one so this is less than or equal to mod ah ah ah mod omega two because mod omega one is less than or equal to mod omega two ok as mod omega one ok by every choice so you see so now if you look at it i am i am having an element of l tau ok whose ah this element ah it is having ah length i mean it is having modulus lesser than ah mod omega two ok therefore this element has to be here integer multiple of it it has to be an integer multiple of omega one z prime is ah is an integer multiple multiple of omega one that’s what it means ok and ah ah see if and now go back to this equation if z prime is an int[eger] integer multiple of omega one and you push z to one side that will you that z is in the lattice ah generated by omega one and omega two so i start with a z in l tau and i have established that z is in the lattice generated by ah l ah by omega one and omega two ok therefore these equality ok so ok so so this so this essentially shows that ah l omega one comma omega two contains l

tau which was the hard point this is already contained in l tau that it contains l tau based on here so they are equal so this claim is settle so what does this tell you this tells you that omega one omega two is another basis for the over z for the lattice generated by tau ok now ah now keep that in mind and ah well i will do one i will do something ah so what i can use it for reference later let me move ah all these conditions here it may write then here d is characterized by ah number one imaginate part of tau is positive mod tau is greater than or equal to one number two i will write minus half less than or equal to strictly less than real part of tau less than or equal to half that combines this and this ah these two have combined as one and there’s only one more condition mod tau equal to one implies real part of tau greater than or equal to zero ok so ah let me let me put that here so now let’s go back and look at what is written here so so you see ah so you see omega one comma omega two is ah another z base for l tau just like ah well one comma tau ok one comma tau is of course a z basis and what we have just shown is that omega one comma omega two is also z basis now you just write this ah write this expressions for omega one and omega two see you will see omega two is ah m two tau plus n two and omega one is m one tau plus n one you can write it compactly as omega two omega one in metric form is equal to m two n two m one n one tau one ok you can write it in this form ok and ah so this would tell you that this matrix which will this this is this will give you a z linear map from l tau to l tau ok this matrix will give you a z linear map from l tau to l tau ok which is same as l tau is same as l omega one omega two and this matrix is mapping the basis one comma tau to the basis omega two comma omega one ok so ah so the so the moral of the story is that this matrix is an is a z isomorphism ok therefore its its its its it so it is invertible in z therefore is determinant plus or minus one ok so so so let me write that here ah hence ah m two n two m one n one has ah determinant ah plus or minus one ok because you see it’s a mat[rix] see it is ah it is a map of l tau to l tau which is a z linear map ok and it is taking a basis to ok therefore ah is an isomorphism ok so you see ah you can think about this ok ah we have use the similar argument earlier a specially we use such an argument to ah show how ah we should in fact we used to show that if you take tau one and tau two two different you know elements half the upper half plane then the complex tori i defined by tau one and tau two are holomorhpically isomorphic if and only if tau can be moved to tau two by means of a psl two z element ok so it’s the same kind of reasoning that will have to essentially use here ok so ah this has determine plus or minus one ok and ah well so so what this tells you is that you know if i so now what we do is now define tau two to be so you do the fallowing thing so what you do is you define it has ah omega two by omega one ok if ah if if the if the

determinant above is one and you define it as minus omega two by omega one if determinant is minus one ok you you make you define tau to in this way now the claim is once you define it like this then my claim is ah so already the way we have defined tau two tau two is already the image of tau one under psl two z element ok so tau two ah so you see tau two is actually ah so tau two is in this case if the determinant is one ok it is ah well its its m two tau plus n two by m one tau plus n one ok with ah with this ah with the mobius transformation ah is z going to ah you know ah m two z plus n two by m one z plus n one thought of as an element of psl two z ok so this is ah this is a first case when the determinant is one and or it may be you have to put a minus sign ok so that if you put a minus sign then ah the first row ah its only the first row that gets so you have to observe the minus into the omega two or to omega one so if you do that only one of the rows gets multiplied by minus therefore the determinant gets multiplied by minus so if the determinant is minus one you will get plus one so so it will be well minus m two tau ah minus n two by m one tau plus n one for example ah which is where you know ah and and z going to minus into z plus n two ah when this should be minus by m one z plus n one this is a again in psl two z ok so these are the two cases so so what this tells you is that tau two is in the psl two z orbit of tau one ok so so tau two is in the psl two z orbit of tau one ok now the beautiful thing is that ah tau two is almost in d ok ah except that you will have to little bit work too ah very little work has to be done too ah get all these condition satisfy ok ah so i will explain that so we will have to show tau two is satisfy most of the conditions for being in script d so for that i will do a ah translation of a ah so i will have to again go back so you see it’s a little involved i will have to again go back to the way in which omega one and omega two were chosen ok so ah ah re recall that omega one and omega two were chosen so that ah well ah mod omega one is less than or equal to mod omega two this is ah this is particular condition because we chose mod omega one to be the closest point in the lattice defined by tau to the origin and then ah we took away the ah the although a lattice points on the line that joint omega one to the origin and from the remaining we try to ah we took omega two to be the closest point to the origin ok so this is a this condition is already satisfied then so this is a and then there are two other conditions the the condition is ah the other two conditions are mod omega two is less than or equal to modulus of omega one plus or minus omega two ok so this is a second condition ok so ah well ah i think this is this is fairly ah this fairly straight forward because you see if you take mod ome[ga] if you take omega if you take omega one plus or minus omega two that is not a ah that is not a real multiple of omega one ok so ah among those which are not a real multi[ple] which are not re[al] tho[se] among those members of the lattice l tau which are not real multiples of omega one omega two is of the smallest modulus therefore this ok now ah this is ah this in terms of omega one and omega two translate

everything in terms of tau two with this definition ok so ah in terms of a tau two if you translate it what will get is a fallowing you translate everything in terms of tau two ok so this gives ah this gives a ah i will call this as well a again ah you will get mod tau is mod tau two is greater than or equal to one because tau two is either plus its plus or minus omega two be omega one ok and ah mod omega mod tau two will be there therefore mod omega two by omega one just greater than or equal to one so you see already this this condition is satisfied ok then ah b is well ah these two conditions are equivalent to real modulus of real part of tau less than or equal to half ok ah ok which is essentially ah this condition ok except that ah i need to ah do something to throw away ah the case when real part of tau is is equal to minus half ok so ah so how does one i i think it’s pretty easy to see ah essentially ah probably only ah it is just can you see in the fallowing so this is tau two well how do you see this ah to see b right tau two ah tau two as ah x two plus i y two ok right tau two is a x two plus i y two and ah then you see ah b b gives b gives well if you divide throughout by ah mod if you divide throughout by mod omega one ok then you will get ah mod tau mod tau two is a less than or equal to one plus or minus ah mod tau two and ah now you plug in this if you plug in this you will get ah in fact b is equiva[lent] b is of course equivalent to this ok b is equivalent to this and that is equivalent to well x well i square it x two squared plus y two squared is less than or equal to ah well i will get two things i will get let me take the first then i will get x two plus one the whole square plus ah ah plus y two squared and i will also get x two squared plus y squared is less than or equal to ah one minus x two the whole squared ah ah plus y two square ok this is what i will get and and of course you know if i now expand everything out see i will get ah the x two square and y two square both sides will cancel the first one will give me two x two plus it will give me x two greater than or equal to minus half the second one will give me x two less than or equal to half so this will be equivalent to x two ah mod x two less than or equal to half and that is and of course mod x two is mod modulus of real part of tau two so it’s fairly simple that these two translate to these two conditions ok so so i have got mod so i have got i have got two of these conditions is only one problem namely ah i will have to ensure ah what happens when real part of tau is half and i also have to ensure that imaginary part of tau is positive so ah well the first thing is you see the way i have written tau two tau two is a psl two z translate of an of of of of so this i guess was ah ah i thought x one point is started with did we start with tau one or did we start with tau tau one ok so at some point i missed ah so this tau is supposed to be tau one ok so i hope it you don’t get confused so you know i started with that tau in the upper half plane i mean a tau in the upper half plane and i was trying to produce a tau two in script d which ah is in the psl two z orbit of tau one but somehow ah this tau one has changed to tau so please don’t get confused ok so ah this is the tau that i started within the upper half plane and i am trying to produce a tau two ok which is in the psl two z orbit of this tau and which is in script d so i am just trying to so you see since since tau two is ah the image of tau under a psl two z element and fills tau is in the upper half plane the way i defined it tau two is al[ready] already in the upper half plane so imaginary part of tau two is automatically greater than

zero ok ah clearly imaginary part of tau two is greater than zero that’s i mean that is a reason why we switch the we introduce the sign here then the determinant was minus one ok ok so ah so the condition that imaginary part of tau is greater than zero a sat[isfied] is now satisfied ah for tau equal to tau two mod tau two is also greater than or equal to one now i will have to look at only these two ah these two conditions so you see you do the fallowing thing ok so ah if ah ah so let’s look at this condition that ah ah so let let’s look at the last and condition ah if if mod tau two is one ok if mod tau two is one ah consider ah tau three to be minus one by tau two ok that is you are just applying another mobius transformation another psl two z element ah on the upper half plane ok ok so we are just applying the transformation is z going to minus one by z ok which is a psl two z element ah we are just moving it in the moving some in the within the upper half plane and you consider this ok then then tau three is also in the ah psl two z orbit of a tau because ah tau two is in the psl ah two z orbit of tau and tau three is the image of tau two under another psl ah two z element ok therefore tau three also in the psl two z orbit of tau and of course all these conditions that satisfied by ah ah tau two that so far we will also be satisfied by tau three so we will have we will have ah ah well mod tau three ah is equal to one by mod tau two and of course i assume mod tau to is one so this is one which is greater than or equal to one ok so i have this condition ok ah mod tau three is greater than or equal to one of course imaginary part of tau three is greater than zero because it’s in tau three is ah in the upper half plane because tau two is in upper half plane and this is also in the upper half plane then ok ah imaginary part of tau three is positive and ah of course you will have also and and you see if you look at the if you look at the ah if you calculate the real part of a tau three ah you will see that ah yeah so you see ah so real part of tau three yeah you have to just write it out this is going to be ah real part of tau three is minus one by tau two and tau two is x two plus i y two so if i calculate this i will get real part of well multiplying divide by x two minus i y two i will get minus x two plus i y two by x two squared plus y two squared this what i will get this and ah well this is going to be minus of a x two by x two squared plus y two squared and that is minus x two ok because x two squared plus y two squared is one mod tau two is one and this is ah going to be minus real part of ah tau two ok and ah so you see ah well and this is going to be ah and this has to be greater than or equal to zero ah it is it is it has to be greater than zero ok if ah so you see ah i will have to see ok so let me explain something see for the condition three what i have to do is i will have to worry ah fo[r] for the case when mod tau is one and real part of tau is negative ok so what i do is if mod tau is one ah and let me also and real part of tau two is negative then you do this then ah since real part of tau two is negative minus real part of tau two will be positive ok so i will modify tau two by a tau three by a psl two z element which will satisfy the condition that ah when mod tau three is one ah then the real part

of tau three is positive so the third condition is satisfy ok you know if tau is a i mean the the tau two that you constructed or the tau three that you con[structed] the tau two that you constructed if it was on this line on this ray then all you have to do is just translate it by one and translation by one is also psl two z element ok so that’s all you have to do so that so you can ensure therefore that in the ah given any tau in the upper half plane ah there is a representative of its psl two z orbit in script d ok so so the so probably i should ah stop here ok and in the next lecture what i will try to do is i will try to show that ah there is only one such ah representative in script d ok and that is what will make ah script d into a fundamental region for psl two z ok so i will stop here