# The Natural Riemann Surface Structure on an Algebraic Affine Nonsingular Plane Curve all right so ah so let me recall we are trying to ah show that every complex torus is actually in algebraic elliptical curve so let me continue with ah what i was doing there previous lecture ah so if you recall ah we started with we started with ah tau in the upper half plane and the then of course we had the lattice ah l of tau ah and we also have the torus t ah sub tau which is just t complex plane modulo this lattice and associated with ah tau is the weierstrass phi function phi tau of z and the this is ah this phi function ah is defined ah in the i mean in the complement of the lattice on the complex plane and its a holomorphic function and at the point of the lattice it has ah poles these are ah double pole at each of these points of order its ah pole of order two and the with some of its reduce zero and phi tau ah satisfies differential equation ah first order differential equation ah phi prime tau of z ah whole cube ah whole squared is equal to four times t tau of z cube minus three minus ah g two tau or phi tau of z minus g three tau and in fact what we did was ah in the last lecture ah we notice that ah this can be treated as ah you can look of look at this as a polynomial in two variables so you put ah put so you look at the polynomial y squared is equal to four x cube minus g two tau x minus g three tau ok and then what you do is you look at the locus of zeros of the polynomial in a c two so we look at this locus z two sub tau ah this is a set of all lambda come out mu in ah six c two cross c such that ah lambda mu ah satisfy this equation this polynomial equation when i plug in for x lambda and y mu ok so ah well so what i do is ah i will i just cause i will i will define f two tau xy to be ah this polynomial namely four x cube minus g two tau x minus g three tau minus y squared and then ah if lambda mu satis lambda mu satisfies this polynomial with ah lambda substitute for x and mu substitute for y if and only if lambda mu is zero of the polynomial so so i will just write this ah f two tau of lambda comma mu is zero so the so this ah this z this ah z two tau is ah is called ah the ah it is called the affine elliptic curve affine elliptic algebraic curve associated to tau ok so it lives inside this this lives insides the c two ah c two it lives inside c two which c cross c and ah what we did yesterday was we proved that you can define a map phi sub two ah so ah so so let me ah make that statement more precisely so ah on ah the complex plane minus the ah lattice points from there to ah c two ah c two ah there is a you can you can define following map z going to phi tau of z come of a prime sub tau sub tau of z and ah i call this map is p sub two and ah in and this map is of course ah invariant ah under l tau because phi tau and phi prime tau are invariant under l tau they are w periodic function elliptic functions and ah therefore this map goes down to ah map of the tours minus the point x naught the point x naught being the image of the ah the lattice all the point of lattice go to ah single point because we form a single orbit we have single equivalent class and i call that point is x naught and ah and this map is just restriction of the natural projection from the complex plane to the ah to the tours ok and of course this is a this is an open set ah and this also an open set here i have just deleted a point and ah so what happen is that ah inside the c two ah we have z two ah tau seating inside this this affine algebraic elliptic curve is seating inside c two and the by definition the map ah the map goes into ah c two because the these two phi and phi primes satisfies this polynomial due to this differential equation and therefore you get ah map ah which also by abusive notation i call it as phi sub two ah and what we proved yesterday was that phi sub two ah is a bijective continuous map ok so ah so so let me write here ah is a bijective ah continuous so bijective continuous map now what i am planning to do today is i am i am i want to tell you more about this map so i want to explain how there is a natural structure of riemann surface on z on this ah e two tau ok and the ah and i will also explain how this map becomes when e two tau is ah given that natural riemann surface structure i will explain why this map becomes a bi holomorphic so therefore the aim of the lecture will be to tell you there is natural riemann surface structure on this there is natural riemann surface structure on this such that this map phi two is actually ah phi two is actually a bi holomorphic map holomorphic isomorphism so ah so indeed the ah so that the so the short of the whole story is that this differential equation ah that ah first order ah degree three deferential equation that phi satisfies ah gives rise to an algebraic equation a cubic equation and that cubic equations defines its zero locus in c two defines c defines an elliptic algebraic curve and this elliptic algebraic curve is naturally a riemann surface and this identification ah of the torus minus ah this point this open subset of the torus with that fine elliptic curve is actually a holomorphic isomerism ok ah so thats what i am going to do in this lecture and let me also tell you what i want to do latter on latter on i will explain how this ah this affine elliptic curve can be compactified by adding a single point at infinity so called point that infinity as a discalled algebraic geometry and the then and and and you can send x naught ah to that point and this can be done in such a way so that even after adding that extra point that that that one point compactification that also becomes a ah naturally riemann surface and this map then the extended map then become isomerism and then we call the corresponding ah curve as the projective ah algebraic elliptic curve associated to the to the to tau to that torus and the the fact is that curve will actually live in ah projective two space so that will require me to explain to you ah how ah what projective two spaces ok so anyway so let get get long with what we need to do in this in this lecture namely to show that g two sub tau e two sub tau is naturally riemann surface ok so ah we shall show ah so ah we shall show ah e two sub tau is naturally riemann surface so that ah phi sub two is bi holomorphic so ah so the point is that somehow ah ah more generally the question is that if you if you give me algebraic equation like this in two variables then i can look at set of zeros of that equation ok ah that will be a the set of zeros will be a sub set of c two and the question is ah under what circumstancesah can i naturally make it into ah riemann surface so this will this will this will be done using the implicit function theorem ah for complex variables ah along with the fact that ah the graph of a holomorphic function is automatically a riemann surface so that is the starting point ok so we will do is we so what i will do is we first show we first show that the the graph of a holomorphic function of a holomorphic function is naturally riemann surface the point is that once you prove that the graph holomorphic function is naturally riemann naturally riemann surface then ah what you can do is that you know given a given a polynomial equation in two variables like this given a two variable polynomial you know ah i can using the implicit function theorem i can locally get in a explicit ah function of of the second variable in terms of the first variable and then ah the fact is that this ah and infact the implicit function theorem will tell you that locally this will look like a graph and since ah locally ah since every graph is already a riemann surface so you will get on the zero locus of polynomial ah like this ah locally a riemann surface and then you will have to show that the there is compatibility between the charts and once you do that you get a globally riemann surface structure on this ok ah but this will not happen for any any polynomial the polynomial has to satisfy certain hypothysis of an so call hypothysis of non singularity ah which are which are the hypothysis that i needed for the implicit function theorem ok which i will make clear and ah so so so so start with ah g from v to c where holomorphic holomorphic that is analytic complex analytic and ah v is an open subset of the complex plane so you will look at a you look at the function g define on a open subset v of the complex plane ok and taking complex values right what is the graph of g the graph of g is well it is the set of all points ah e z comma g z where e z belongs to v this is the graph of this a function the graph is a subset of ah v cross c ok so ah so ah pictorially ah though you know its ha its its hard to visualize the graph of a complex value function complex variable because essentially it will be four real dimensions ok but never the less we do it in a very suggestive kind of way we we we map ah we we put v here we put v here ah ah or rather we put c here and we put c here and think of this is c two ok and then we we think of v the open set as ah patch here and the then you know we ah we draw the graph so this is just like you draw the graph of a real valued function ah in first quadrent if it has positive values so you know so you so you end up drawing something like this and and every point so given a point z here then i get this point which is ah this point is e z comma g z and the this so so this project down to z and to z ah and that project down to g z ok so this is a subset of you can see it is a subset of c cross c all right and ah the point is how do you make this graph into naturally into holomorphic i mean how do you make it naturally into riemann surface see the fact is that you need ah homomorphism of this graph with the with an open substitute of the complex plane and that obviously given by the first projection so what you do is you this is what you do you define you consider p two ah mean rather p one from ah c cross c to c that you just send any lambda comma mu to lambda this is the first projection ok and then what you do is you restrict this first projection to the graph you restrict the first projection to the graph of g that will go from the graph of g of g to v ok so ah now now what you will have to notice is that you will have to notice that this ah this is actually homomorphism in in fact this is ah you can see that ah you can see obviously that ah you know ah see you are just sending z comma g z back to z ok and you know its obviously surjective you because given any z i have the corresponding point of the graph z comma g z and whose image will go back to z so its surjective and it obviously injective because if z one comma g z one and z two comma g z two go to the same z one equal to z two then if z equal to z two then g z one equal to g z two trivialy therefore this is a bijective map ok and the also this map as an this map has an inverse namely you just send ah has has inverse ah ah from v to graph of g this is a natural map you just send z to e z comma g z this is the inverse map so let me call this ah let me call this inverse as phi let me call this as phi sub g ok so so so if i take v comma phi sub g ah i should rather take ah yes so i should take so so if i take graph of g comma phi sub g is a chart is a global chart on ah is a global chart on global chart for ah graph g ok and in fact ah this is ah this will this is since we have there is only one chart ok ah since there is only one chart ah there is no compatibility to be checked therefore this this makes the graph of g in to ah riemann surface you just identify graph of g with v which is the which is the domain of the function g all right and the ah so this single ah chart will do converted into ah riemann surface and of course you know there are ah there are ah the rather little points ah for a riemann surface first of topological space needs to be if you if you insistent you want it to be connected ah then you had better assume that this ah phi is connected subset ah even if if if not then you will have to you will always you be able to break down b into connected components and each component to be a will be converted into riemann surface ah then the other thing is you also want to be housed of second controversial which is obvious because ah you see ah because of this because of this ah because of this homeomorphism ah graph g is homeomorphism v and subset of the complex plane is of course housed of in second countable ok so ah if you stick by the definition that riemann surface structure is a complex artless ah ok that is a system of coordinate chart which are compatible each other on manifold which is ah housed of second countable and connected then all this conditions are satisfied ah of course connectedness you will have you assume for v ok so in any case this is the way in which you can make the graph of function into a in to a riemann surface ah in a and then now so once we do this for the ah graph of a holomorphic function then we can go down and do it for ah function of two variable ok so so i will ah so i will state the implicit function theorem this is exactly this is exactly the complex analog of the implicit function theorem that you would have in ah real variables ah and the philosophy of the prof is perhaps nearly the same ah so ah so ah let ah let f of capital small f of x comma x be a polynomial ah with complex coefficients coefficients ok ah and of course when i write x comma y you must always beware x and y are complex variable is not real variable ok ah probably it would have been better if i use z and w but ah never mind ah please that capital x and capital y are complex variables ok we are we are already in the complex set up so the our scleral is always complex numbers ok so ah so ah let ah z of a f comma z of x f of x comma y ah be the zero locus of f of x comma y in c two ok look at look at the zero locus ok so so coming back to think of it good i didnt z and w because now i want z to cannot the set of zero so so that is what do it mean it means z of f of x comma y is the set up all lambda comma mu in c two such that f of lambda comma mu is zero is zero this is the zero locus and the point is ah the implicit function theorem says that you can solve ah you can solve ah the equation the equation f of x comma y equal to zero can be solve locally as y is equal to g of z g of x ah locally provided there is the condition on the derivative just as in the real implicit function theorem so ah so let me ah so let me see this ah let ah lambda comma mu be let lambda comma mu be a point in the zero locus such that ah if i differentiate partially ah f with respect to y second variable and evaluate it at lambda comma mu the the resulting value is non zero ok so the ah so basically if you so what is the implicit function theorem it says if you have a if we are try to solve f of x comma y equal to zero which is an implicit relationship between x and y then you can solve for it as y is equal to g x at a point where the partial derivative of f with respect to y does not vanish this is the implicit function theorem ok so that is a condition i have put here alright so what it say is so let me let me write that ah then there exists an open neighborhood there exists an open neighborhood ah u lambda or let me call it as v lambda of ah lambda in in c and a holomorphic map g from v lambda to c such that following things happen the first thing is g solves g solves f of x comma y ah for all ah for all x in v lambda ok so let me write that ah g y is equal to g x solves f of x comma y equal to zero for all x in v lambda ok that is that is for every lambda prime in v lambda f of lambda prime comma g lambda prime zero this is the so this is the the explicit the the implicit equation is solve by an explicit equation in an neighborhood of ah of the variable in a neighborhood of a point where the ah where the partial derivative with respect to the explicit ah dependent dependent variable is is non zero ok the second condition is of course ah what will happen is that ah the and g is holomorphic so what its derivative the derivative can be ah d g by d ah x is actually minus dou f by dou x divided by dou f by dou y this is this is derivative ah in ah in v lambda ok this is the so this this is the identity that you get ah the you you take the total derivative of this ok if you take the total derivative of this and think treat y as a function of x then you will or you will automatically get this but then to bring it to this form you must have that the derivative dou f by dou y is non zero so ah so you must ah you must remember that f is a small f is polynomial so dou f by dou y is also polynomial its always continues and if ah and you and if continues function vanish it a point you can always chose a neighborhood ah where is ah never going to vanish ok so so this what happen so thats thats thats what allows me to divide by dou f by dou y ok and the third thing is ah most important ah ez ah it is zero z of f of x comma y ah locally add the point lambda comma mu looks like like the graph the graph of ah the graph of g ok so so in other words ah so the so the first point is ah for a every lambda prime v lambda we have f of lambda prime comma g lambda prime is zero that is one thing the other thing is ah because lambda if you take a point lambda prime comma mu prime in the in the zero set of f x y that it must happen that mu prime is actually g of lambda prime that is what it says ok so so in other words if i take so let me draw let me draw a diagram to explain this ah more graphically so you see you have this is c two c cross c and what is happening is that you have you have this zero setup f of x comma y ok and then you are choosing a point ah lambda comma mu with the property that the partial differs partial derivative of small f with respect to y at that point does not vanish ok then what happens is that you are able to so then your project down here you get this point lambda ah and then you are able to find a v lambda v lambda is a ah open neighborhood of lambda and on v lambda there is a ah there is a holomorphic map if you draw the graph of this map namely you take the set of all lambda prime comma g lambda prime here then that is exactly this graph so you know if i if i just extend it like this and if i extend it like this then this then this guy is exactly graph of a of g ok so in particular it means that you know if you take a any if if if you take any lambda prime comma mu prime such that lambda prime is v lambda the mu prime has to be g of lambda prime ok so there is a ah so it locally graph of the that function ok and ah if you grant so so ah i am i am not trying to ah i am not going to going to giving a proof of this because its quite standard you you can look it up ah and but the point is that i want to say that once you have the implicit function theorem immediate corollary is that if you give me if you look at the zero set of a polynomial then locally already it is a graph of a holomorphic function therefore locally already a riemann surface because we already i already explain to you how very easily the graph of a holomorphic function is a riemann surface so what is the the short of this result is the movement of polynomial satisfy this condition ok ah then at in a neighborhood of that point ah its a riemann surface you can you can make ah neighborhood of that point on the zero locus riemann surface the question is what happen if it doesnt satisfy this condition then it might satisfy the condition the same condition similar condition with respect to the first variable ok ah ok and then you can ah use projection out of second variable and the you know instead of writing ah y is a function of x you will able to write x function of y if dou f by dou x ah at a given point is non zero ok so all these put together the the point is that if you have a polynomial which is so called nonsingular namely which satisfy the condition that one of the partial derivatives of the polynomial never vanishes at each point ok then its very clear that the graph the the set of zero of that polynomial is naturally locally a riemann surface ok and the only thing that one has one one has to check there is globally riemann surface is to check that all these ah local riemann surface structure agree you will have to just check that all these charts that you that you got by the graph construction ah there all compatible once you check that then its very clear that ah the zero set will be riemann surface ok so ah and of course this zero set being a it a close subset of c two so it automatically ah housed of and second countable ok any sub any sub space of housed of space and house and ah any subspace of a second countable spaces second countable so those conditions are automatic ah of course you might want it to be connected and the fact is that ah you will so if you put this non singularity condition ah then you are all then automatically means its connected ah and in fact ah more generally you could have assume that ah the or or rather to to be on the safer side one sufficient condition you can put on f is the it is irreducible as a polynomial ok so ah thats a technicality that one need not vary about at this stage but at the worst if you take a polynomial who is first partial derivative simultaneously do not vanish ah at any any given point then it becomes a riemann surface ok so let me write let me write the rest of it ah so if so if ah if dou f by dou x ah at lambda comma mu is not equal to zero then there exists and open neighborhood neighborhood ah w mu of mu in c and a holomorphic function ah h from w mu to c such that ah z f of f x comma y is locally ah for that is for ah x in for y in w mu the graph of of h so this is condition on the other variable ok so ah so the so well ah so that is the so so you know you have to rewrite i am not writing everything down ok ah i am i am just stating the case when the first when the first partial with respect to the first variable is non zero ok that was the case when the first partial with respect to the second variable is not zero right then you are then you can write the second variable as the function first variable in this case you will be able to write the first function the first variable x as a ah function second variable so x will be h of y ok right so ah yeah so what is the ah so what is the ah of this this is that ah the if you if you if you give me a polynomial ah which satisfy one of these two conditions at each point of it zero locus then ah it there are riemann surface structures locally ok ah which come out of graph construction ok so let me write that now those if a polynomial f x y ah is such that ah there is no point there is no no solution to ah the system f of x comma y equal to zero dou f by dou x at x comma y is equal to zero ah dou f by dou y at x comma y is equal to zero in which case in which case we say we say ah f is non singular ok ah then ah the zero set of f of x comma y is locally a riemann surface ok so the up short of the thing is that ah the zero set of this polynomial ah which is non singular polynomial ah non singularity means that you should not able to find the point zero locus at which both first partial derivative vanish ok then ah f is ah such a polynomial call non singular and then the zero locus is locally riemann surface now i only need to ah verify that to verify a riemann surface i i have to only tell you that all this ah riemann surface structure locally they all glue together i mean they all agree namely i just ah what do i mean by saying it locally riemann surface locally i am i am getting charts which are given by the projection ah of because locally this are graphs of holomorphic functions and then ah i take projections on to the independent variable then i get ah the charts so to show that all these together make this into a riemann surface i will have to only check that they charts are compatible ok so if i do that then ah for a non singular polynomial i will have that its automatically a riemann surface ok then ah you know what i am going to do i am going to check that ah this particular polynomial that i having this cubic satisfy ah and this polynomial here ah i am going to show that this satisfies a non singularly hypothysis and therefore that will tell you that ah this e two tau is in fact ah a riemann surface alright so and we will also the way in which ah we have done it we will also see this map is naturally holomorphic because what will actually happen is that locally the coordinate chart i just given by projection and the first projection is going to be phi which is holomorphic the second projection is going to be phi primes is holomorphic therefore this map becomes holomorphic is a holomorphic bijective map so its a holomorphic isomorphism and we are done ok so ah so i will so let me so let me go head and try to show you a why all this riemann surface structure locally they all give same riemann surface structure on globally ok so ah ah we we verify that all these riemann surface of structures structures glue ah that is agree that is the the charts given by the graph construction the local charts given by the graph the graph construction are mutually compatible so basically there are there are three cases ok three cases ah the first case is ah when the ah local riemann surface structures comes from ah both projections are on the first variable ok basically i will have to take two ah riemann local riemann surface structures ok and show that the intersection the charts ah the transition function are holomorphic thats what i have to show so ah the ah t might happen that the riemann surface structure n these two overlapping pieces open pieces of the zero locus ah they both come from first projection or they both come from the second projection or one may come from the first projection and the other may come this from the second projection so there are three cases so let me draw ah diagram for each of this so you know so ah so this is first case namely ah so here is my this is my zero locus and the well basically ah i have i have point ah here let me call this is p one oh god no let me call this as something ah let me call this is as t one let me call this is as t two ok and ah so t one t one is t one is ah lambda one comma mu one and dou f by dou y at ah t one not zero so ah what happen is that you find ah a neighborhood lambda of lambda of lambda one in c which is ah which is given by v sub lambda one and you find a function g one holomorphic function from g one to c such that ah the the graph ah so z of f x comma y is equal to graph of g one ah ah for x in ah v sub lambda one ok so ah so in this case so basically ah i draw a diagram it its going to be look it going to look like this so here is my so this is my v one v three sub lambda one ok then similarly let me assume that so so this portion that i have drawn here ah so from here to here the ah z of z comma y is the graph of g one ok and the then you also have the same same conditions for t two ok so dou f ah so t two is again lambda two comma mu two and dou f by dou y at ah t two is non zero then you get ah and open neighborhood v sub lambda two of lambda two and ah function g two holomorphic function g two define on that with values in c such that z of a f of x comma y is actually ah graph of g two ah for x in v lambda two ok so what happen is that so this v lambda two so ah so this is v lambda one and well v lambda two could be something like this ah so let me not ah let me not write here so here is v lambda two so this lambda two so this projection is lambda two ok so from here to here this is ah v sub lambda two alright and ah so ah so what are the what are the ah so how do i check that this two charts are compatible so you see basically i have this open sets which is the image of v lambda one ok and then i have this ah open set which is the image of the holomorphic of v lambda two under the holomorphic map and this is the intersection so this intersection i can call this ah intersecting set as w one two rather let me call it as u one two ok and ah so what is the situation situation is i have to look at transaction function so i have from u one two on the on hand i have so there something that i have to correct here see the ah the the ah when i write a chart ok i the first member is an open set on the on the riemann topological space on which i want to get a complex ah coordinate the second one should be defined on that so actually this should not be phi sub g this should actually be p one ok so please correct this this should be p one so thats a global chart so you see so you know ah so i have so i have p one that is one p one which goes from u one two ah u one two is thought of a setting inside ah well you know if i call this whole thing as u one ok and if i call this whole open set as u two ok so this is if i think of it a sitting inside u one then my then my ah and u one is actually graph of g one this u one is actually graph of g one u one is actually graph of g one and ah and this is ah from u one it will go to v sub lambda one ok this is ah this is just p one and this is ah this is a this a holomorphic isomerism homeomorphism ok and the so i will go into p one of u one two so this is the ah which is which sits inside this ok ah so it will be it will be this it will be this ah region here i mean it will be this open set ok and you also have ah ah the the coordinate coming from ah u two by a g two so so you so you have also u two which is graph of g two and you have well again its p one ah so this is p one which is again ah an isomorphism with ah v sub lambda two so this will go to t one of u one two again ok ah now you know if i if i follow this sub so the transition function is to go is the composition of these two ok if i do that i actually i get the identity map see because if i start with a lambda so transition function what is the what is the transition function the transition function will be just lambda going to ah from here to here you will go to lambda comma g one of lambda ok you will go to lambda comma g one of lambda but then you see the ah and then if you project it again the first variable i simply get lambda so you can see this is transmission function is just identity map it is just the identity map on this intersection ok and ah the fact these two open set v lambda on v lambda two intersect is because u one and u two intersection and i am actually trying to check ah the the compatibility on u one two which is intersection of u one and u two ok so the transition function is just identity it is just identity map hence holomorphic ok so ah this settles the case that ah the charts the local charts that you get ah compatibility when when both ah both are graph of function first variable ok ah yes you can ah i wont be writing it down but ah if i do it with ah the assumption that ah both are both partial derivative with respect to x are not vanishing ok ah ok then i will have to take projection with respect to second variable so effectively the argument is you just you know you just take a mirror image of this you take a mirror image about the diagonal line that is kind of diagram you will get and essentially in that case also you will see that the transition function will continued to be identity so its holomorphic ok so the only case that i will have to worry about is when one of the partial derivative with respect to is with respect is with respect to x and known vanishing and the other is with respect to y is non vanishing and in in that case you will see that it will be one of those ah functions whos graph is being considered so it that case also it will be ah it will also be holomorphic ok so let me just ah write that down so similarly we can ah argue for the case when dou f by dou x at lambda comma mu is non zero and dou f by dou x ah at yeah at t one so let me use t one t two and t one is not zero and t two is not zero ok so if similarly you can argue for other case ah in which case also the transition function turns out to be identity so its holomorphic so the only case that is left is that one one of this is with respect to x and the other is with respect to y which also pretty easy ill write it down ah v v r left with with a compatibility checking checking for ah dou f by dou x lets let me write dou f by dou y at t one is not zero and dou f by dou x at t two is not zero so this this is the only case you have to check ok that means the ah the the the charts are coming from different projection projection on different variables ok so that case if i if i draw the same kind of diagram what i will get you can see it pretty easy ah so you see this is my zero set of f of f of x comma y and the ah so i have so i have this so here is my t one and and here is ah so this is my so i get this ah this my v sub lambda one ok and then ah and this portion ah this portion ah is ah you u one and u one is just the graph of u one is just the graph of u one is the just the graph of ah of g one while g one has been defined its a holomorphic map from v lambda one to c ok and then you also have ah t two here but with respect to t two its only they first variable partial derivative which is non zero so you have ah basically ah you know something like this so i get ah i get open neighborhood of ah of this projection of t two which is mu two and i get it w sub mu two ah which you open neighborhood mu two ok this of course open neighborhood of lambda two ok and the this ah this portion is ah u two and u two is actually u two is actually the graph of h two where h two is from w mu two to c ok ah this is lambda one sorry ok this is lambda one thank you this is lambda one so ah so then you see ah if i look at this pieces which is ah which is u want to ok and try to look at the transition function its pretty easy so what you will get is that ah if i write the same kind of diagram mash here so u one two is sitting inside u one ah and u one is ah from and then then you have the you have projection first projection of u one into v sub lambda one this is holomorphic isomorphism and under this projection you are ah looking at t one of u one two ah which is going to be a open sub set of ah is going to be a open set of v lambda one and this also of course homeomorphic ah and then from u one two ah you also have you one two is also sitting inside ah u two and from u two its p two its a second projection because the coordinate chart is in this direction ok so it t two and this goes to w sub mu two and ah well ah so this second projection will give me p p this was p two so its p two of u one two which is an open sub set of w mu two and now if i look at the transition function the transition function is ah well if i start with a lambda ah it will go to ah lambda comma g one of lambda ok and ah then if i take the second position it will go on to g one of lambda therefore lambda going to j one lambda which is holomorphic by definition therefore transition function is holomorphic so you are done with this case also so so which is holomorphic so so this this implies that the movement ah you take a polynomial f x by which is non singular then the zero set of that polynomial automatically a riemann surface ok so so let me write that down thus if f of x comma y is non singular its zero set is z f x y the sub set of c two is naturally a riemann surface so is naturally a riemann surface now ah i will verify i i just want to say that this ah e two tau i want to say is actually a riemann surface the only thing i have to verify that this polynomial capital f two sub tau is non singular so ah we so to ah ah ensure that ah f two tau ah sorry ah e two tau which is just the zero set of f two tau is a riemann surface surface we need we need to only check that capital f two sub tau is non singular is non singular so that is that is there are no no common there are no solutions to the system f two tau of x comma y is equal to zero ah dou f two tau by dou x ah is equal to zero dou f two tau by dou y is equal to zero ok so lets write this out its pretty easy and you will see that we will be using again ah the fact that ah e one e two e three are distinct ok e one tau e two tau and e three tau which are the which are the zeros of the derivative to the phi function you know they are the three distinct zeros of phi function and they distinctness is what is going to give you the non singularity in this case ok you will see we will see that so let so let do the computation so what will get is basically ah ah so i will get ah let ah let lambda comma mu be be a solution ok and we and we lets assume there is a solution get a contradiction ok let lambda comma mu be a solution and we arrive a contradiction at contradiction so what what it will means lambda comma mu satisfy that ah that equations ah so i will get ah four lambda q minus g lambda q minus g two tau lambda minus g three tau ah minus mu square zero this is this is this is the first equation the second equation is i differentiate dou f two tau by dou x and a this is by differentiate it with respect to x i am going to get twelve x squared twelve x squared minus g two tau ah thats all ah then if i so so this equal to zero at lambda comma mu will tell me that twelve lambda squared minus g two tau is zero ok and then ah dou f two tau by dou y is zero for that condition i differentiate partial with respect to y and i end up with my minus two y so the condition i will get is minus two ah mu is zero alright and this already tell you that mu zero ok so ah you will get ah so ah so the first thing i want to tell you is that ah you see if ah so let let let me go back to the first equation put mu equal to zero so i will get four lambda q ah minus g two tau lambda minus g three tau is zero ok ah so i will get this this and then i will add this equation also twelve lambda square is equal to g two tau ok so i will get this three equation ok and the ah i am just saying that this three equation will will give us the contradiction see first thing i want to tell is that ah you know ah the the ah so let me ah so as i told you the point come from on the fact that the the when you factories the right side the right side the equation in to three linear factors the zeros are e one tau e two e two e three tau ah there are the values of phi tau at half tau by two and one plus tau by two the three fundamental zeros of ah derivative of the phi function distinct already seen that so if you use that factorization you will actually actually what happens is that a the the four lambda cube minus g two tau lambda minus g three tau this factories as four times lambda minus e one tau into lambda minus e two tau into lambda minus e three tau ok and then if you now calculate so call discriminate of this cubic discriminant of a polynomial equation of one variable is just the ah square of the the difference of the distinct ah of of its roots ok ah square of the difference of its roots taken in some order ok and the and it it is it is an important tool in its an important tool in algebra because ah you can check that the polynomial as distinct roots namely there it is so call separable by checking that discriminant is ah non zero ok so you if you calculate the discriminant in this case this is the little bit of algebra which you should be able to do ah then you will find that its its a its not a difficult exercise you can do it then you will find that if i take discriminant discriminant is just product over i less than j e i tau minus e j tau these are the square of the differences ok and the you can check that this discriminant you compute it ah you can see that ah is discriminant when you compute it it turns out to be ah ah let me write it tell you what it is it is sixteen times g three tau sixteen times g three tau i think squared minus ah no cube minus twenty seven minus twenty seven g two tau tau the whole square so this is the discriminant and the discriminant is not equal to zero the discriminant not zero so ah the discriminant is not zero because all the e i is distinct ok and the ah so so you see you know already mu zero ah if lambda is zero ok if lambda is zero what will tell you it will tell you g two and g three are zero if g two and g three are zero then the discriminant is zero that a contradiction so the first thing you can tell you can conclude from this is lambda is not zero so you see lambda is not equal to zero ok so and and it also and once lambda is not zero its also clear that g two is not zero and g three is also not zero so ah so which which which implies that g two tau is not zero g three tau is not zero this two are not zero ok and ah then what i can do is litterly ah i can just eliminate lambda ah from ah from this equations ah and so you know i have ah twelve lambda square is g two tau ok and the ah and i have ah you know so i eliminate lambda from from these two ok and believe it or not you eliminate lambda from these two what you will get is g three ah cube tau minus twenty seven g two square tau equal to zero ok it will give a if you eliminate lambda you will exactly get determinant is zero which is not which is not true that that is contradiction ok so eliminating eli so basically i take a square root of lambda here ah and equated to cube root of lambda from here and then i race both sides to the power six and thats its so eliminating lambda gives ah twenty seven i mean the you will get g g three cube ah tau ah wait a minute i think its g two cube and g three square so i should be careful so this is g two cube and this is g three square let me check that yeah g two cubed and this is g three squared please ah correct it so eliminating will give you g three square minus twenty seven g two cube minus twenty seven g three square tau is zero a contradiction so you see the it is the distinctness of the g f of phi the derivative of the phi function that actually gives you the fact that ah the the corresponding polynomial that defines the affine algebraic elliptic curve is a non singular polynomial therefore it become a riemann surface and then so so the moral of the story is that ah ah this e two tau is elliptic curve is indeed riemann surface naturally ah by the construction that we have just seen using the graph construction the implicit function theorem and ah then ah now so so so e two tau is a riemann surface and the the the bijective continues map that i wrote down namely phi two it is from the torus minus special point which is the image of lattice two the ah two e to tau is is basically this map is e z going to well ah so its just e z so i put ah box here to show that this is equalance class because the torus is set of equalance class and this goes to phi z phi tau of z ah phi prime tau z ah this map ah is actually holomorphic because if you project and the first variable it gives a holomorphic map phi sub tau and if you project and second variable you get the holomorphic map phi prime sub tau is holomorphic hence ah an isomorphism hence by bi holomorphic ok so you ah ah because essentially ah how do how do i check that this is holomorphic the method is that i will have to take locally an open set which has a chart so locally it will be a graph and then for a graph we coordinate the the coordinate map is just projection on to either the first variable or the second variable so i will be essentially be locally getting t tau or p prime tau depending on whether i am projecting on the first variable on the second variable therefore they are holomorphic therefore this maps are holomorphic map ok and the that that shows you that the punctuate torus is ah naturally isomorphic riemann surface to the affine elleptic algebraic curve define by the by the polynomial equation that comes out by the differential equation satisfied by the weierstrass phifunction ok so so ill stop here the next lecture i am going to tell you can extended this map to from the torus to one pint compactification of this analytic curve which will live in projection two space ok so i stop here