 So, let us finish this off. Yesterday we saw that the Mobius transforms keep up a half of complex plane to itself. And I also mentioned that this is the characterization of the transforms that preserve angles between two curves. Now, how are these guys related to what everything that we are doing? Well that is defined. So, let me define a particular tau is gamma 1 not to be confused with the gamma function. This is essentially the set of all matrices. So, these are precisely the matrices we are just investigated the transforms given by tau. I am now going to define subgroups certain subgroups of this group. So, by the way this is the group under multiplication because for obvious reasons. So, I am going to define certain subgroups here. The condition the additional condition is take all the matrices in gamma 1 and c should be congruent to 0 mod 1. So, c should be divisible by the number n which is supplied separately. And the number n defines an appropriate subgroup of this. Why is this subgroup? Simply the only additional condition is the bottom left entry should be divisible by n. If you take two such elements two such matrices and take the bottom left multiplication that would be c 1 times something plus d 1 times c 2 c 2. Now, c 1 c 2 both are divisible by n. So, therefore, we will continue with that. And now we define. So, we say the function f is a modular form of height 1 and level n if the following two conditions hold. And one is the primary condition really two is just a slight twist on the one. The condition is that you apply this any of this transforms in gamma n on z and apply f on it. What you get? The value is f z times c z plus d whole square denominator of the transform comes up as a multiplier in degree 2. And earlier what I wrote about height higher we can define for higher height when this degree instead of 2 is more. But we will not be interested in that. And second condition is that f of minus 1 over n z this is not quite this does not quite fall into this. This determinant of this transform is n z goes to minus 1 over n z. So, here a is 0 b is minus 1 c is n d 0. So, it is 0 minus 1 n 0. So, that determinant is n. So, it is not does not belong to this. But it is very closely related to this group. In fact, just a simple observation if you let matrix w be 0 minus 1 n 0. So, essentially the matrix w commutes with the group that commutes in the sense not with individual elements, but with the whole group. So, this is very closely related to this group. And what we are saying is that in fact, what one can show is if this condition is true then f of minus 1 over n z must be a linear combination of plus n z square f z and minus n z square f z. And what this condition is saying is that it should be just one of the two that plus or minus. So, that is makes the function a function f which is remember that is the function of the kind we the Fourier series kind that we are looking at. That is the function we started with. This is the function associated with the elliptic curve rational number which is Fourier series periodic with the period of 1. And then we are looking at at least I said that we look at this transformations on the upper half plane and see how the function behaves there. And that is the behavior of the function not all of them, but that is the behavior of the function we are interested in and this function is called model of forms. Now, this may seem like a very strange condition to why could such a transform be equal to f z first of all. It is very hard to believe that it should be the case. And even if something like this whole what use that is it have well there is a very interesting theorem which says that f e this is the Fourier series associated with the elliptic curve e over the rational number is a you can only f that see if I can get this expression right. So, this relates this Fourier series representation with the Dirichlet series representation that the Fourier series being modular is equivalent to the corresponding Dirichlet series satisfying a function. And the of course, I fix the height to be 1, but this is more general theorem holds actually if the height is whatever is the height a height is k in this number 2 which is occurring here and here becomes 2 k. So, the symmetry becomes along a real that equals k. And level of course, occurs at this level really represents the multiplicative factor. And this is the central connection with the things that we have studied. So, far and the things that I have new things that I have introduced that if you look recall you go back. So, I gave the started this proof sketch right. So, you looked at this curve then there is a point of order and it does not have a point if f is modular to the f is the that specific elliptic curve. And the phrase theorem said if f is modular then it does not have a point of order equal to 6 right. And then we jumped into modular curve. So, what are modular curves really are we say curve is modular let me just define it also. So, elliptic curve e we call a modular elliptic curve if the corresponding Fourier series is a modular form of height 1 and some level. So, now everything connects up that what phrase showed was that if the elliptic curve specific elliptic curve f was modular then it does not have a point of order greater than equal to 6. Which means that the discriminant of it that cannot be greater than equal to 6 which in turn means that the solution to that Fermat's equation does not hold. And finally, the last piece in the puzzle was put in by Henry wise then others later on that which showed that all in fact, Henry wise showed it not for all elliptic curve, but only a subclass of elliptic curve which was good enough because that particular elliptic curve f did fall in that subclass called semi stable because that is not important. And then later on other mathematicians worked on that proof and generalized it to include all elliptic curves. So, now we know that all elliptic curves are modular which means the corresponding Fourier series have our modular forms which is equivalent to saying the corresponding zeta function associated with those elliptic curve satisfy the function. But this required huge amount of effort that Henry wise proof is originally of course, was more than 100 pages long it was later on brought down then simplified. But still it is at least 50 pages whereas, if you remember we derived functional equation for our zeta function very easily required some still required some bit of work actually if you remember we had to do this going to the Fourier analysis. We did this big sum of e to the minus pi n square z and then did something with it and it came down, but still it was not too difficult. And in fact, this if you now go back and look at that proof and keep in mind that what you are see this keep in mind this equivalence between the zeta function functional equation and the corresponding Fourier series is modularity. So, with respect to our zeta function the original Riemann zeta function also we can associate the corresponding Fourier series. And if you run through the proof of the functional part functional equation of the Riemann zeta function the functional structure of the Riemann zeta function that proof is actually showing also that the corresponding modular function not modular the corresponding Fourier form is modular not exactly in this, but more or less. So, let me see it shows that it is the corresponding Fourier series is modular of height half and level 2 half because you see this is 2 here. So, whatever is the height you multiply by 2 and then that is the line with along with the symmetry hold. So, it is half here and level is 2 because or level is 1 1 or no not 1 here root n by 2 what was it it was pi to the minus z if I remember right pi to the minus z gamma z by 2 zeta z. So, that gamma z by 2 actually comes play a role actually that is where this whole thing flips. So, here we are sticking with gamma z. So, there is some transformation that happens which brings in that level 2 I mean this very easy to work out if you just sit down and write it down. And this proof is actually fairly straight forward it is not at all fancy the equal as between modular forms and functional equation just write down the thing condition and then verify it in the brute force it is what for example, if you say which direction you want to prove let us start with this let us say suppose f is a model form. And then we want to show that this functional equation holds. So, let us start with the left hand side what is this equal to let us just write down everything what is gamma z that integral which is t to the z minus 1 t to the minus t dt. And zeta z is n greater than equal to 1 a n divide by n to the and I will be using that uniform convergence. So, swapping the infinite integral infinite sum freely. So, then this becomes sum n greater than equal to 1 a n and let us do a variable substitution u is then what you get t is e to the minus and then dt over t dt is u by u now take the integral summation inside again what is this sum what was f f of e what is f of e because sum over n a n e to the 2 pi I think I missed it 2 somewhere anyway 2 pi i n. So, this is going to be equal to f of thing I should stick it to somewhere u is t over 2 pi n. So, if you use u over 2 pi n then t over pi n is 2 what I will take care of yes I is not a problem. And d u by u that is there 2 cancels out. So, this is going to be equal to now split this integral into 2 parts going from 0 to 1 over root n and then 1 over root n to infinity and we look at the first part this is actually I am just maybe thinking this the functional form for zeta function there also we did this we did this exponential sum we replace it with there was a function w that we defined. And then we split the integral 0 to infinity to 0 to 1 and 1 to infinity and then 1 to infinity 0 to 1 we work with and use that property of w function to write it in terms of is 0 to 1 to infinity integral that is exactly what we are going to do. So, let us just look at the 0 to 1 over root and this integral what is f of i because f is modular we should f you let us use this one the second condition f of minus 1 over n z is this for i u we should use directly the other condition f of i u is going to be 2 things f is only make sense when its argument is on the upper half of complex plane and that is certainly the case f of i u u is positive. So, it is on upper half of complex plane so that it does make sense to work with this properties of f what is f of i u use that swapping of this let us use this and see how we can write it. So, that is i u we can actually easier to use the second form. So, let us pretend that let me first do a variable substitution here let u be 1 over n v then d u is minus 1 over n v square then i equals what happens to i when u is 0 v is infinite and when u is 1 over square root n. Then v is 1 over square root n so i is negative of the integral going from 1 over square root n to infinity and then what is square root n u 2 by square root n v f of i by n v d u is minus 1 by minus 1 goes away 1 by n v square and u is 1 over n v. So, that is d v by v and now we are going to flip this f of i over n v. So, treat z to be so f of i over n v is f of minus 1 over n i v i v is on the upper half of complex plane v is real it is in upper half complex plane. So, I can use the other form to write it as plus minus n v square which is i v square times f of i v. So, let us just stick that in i equals 1 over square root n to infinity 2 by square root n v to the z and what happens to this this is minus plus n v square f of i v. So, what comes of this that is perfect except for this power of 2 that is sticking out power of 2 should not have been there I might have goofed somewhere. See, but for this power of 2 if you see this integral 2 power that if you forget about square root n q to the z f i u d u by u the other integral is square root n v to the 2 minus z f i v d v by v. So, the same integral only thing that changes is the exponent here from z goes to 2 minus z and that gives the functional equation that is. So, basically what we saying is the total entire integral is square root n u to the z plus square root n u to the 2 minus z times something which is independent of z. So, when you flip z to 2 minus z it becomes a stage invariant except for the sign because this plus minus sign it can change the sign flipping z to 2 minus z can change the sign and that is what is occurring here where is the theorem. So, I should have said that there is flip on the sign here and I think there is the 2 is to be absorbed by here if you stick it to here then we start with this, this, this and you stick this then this 2 goes away good. So, that is it for today the take away from this of course, apart from whatever I have described is that for the zeta functions over rationals for elliptic curves we have not as much knowledge as we have for other zeta functions even the functional form we just got recently and we are not even close to proving the corresponding Riemann hypothesis for this. Again you can say the same things because of this functional form one thing that immediately follows is that the zeta function is meromorphic it is defined over the entire complex thing and now the middle line the symmetric around line real z equals 1. So, you want to the conjecture would be that all its zeros lie all non trivial zeros lie on the line real z equals 1 we have no clue how to prove this. In fact there is that is may be I can talk about it next time there is another very famous conjecture which is called Birch-Dyer conjecture which talks about specifically about this zeta function over elliptic curves over rationals and about its properties. So, it is a very famous conjecture. Birch-Dyer conjecture and it is very we have no idea how to prove this, but it remains one of the major open questions.