 So, that is one thing I want to do. Another thing I wanted to do was to very quickly give you a overview of the Fermat's law theorem 2. That goes via this modular forms. Modular forms essentially come out of the zeta function. When you keep on generalizing this properties of zeta function you actually end up with modular forms. So, their connection with firstly elliptical and then secondly with the Fermat's law theorem. So, let us see how much I can cover. So, today we will start with our first step towards generalizing whatever we have done that would be for Dirichlet's theorem. This state equals what would you expect. So, a and q are two numbers which are relatively prime to each other and define psi a q psi x a q is the number of not quite number but sum over all primes p less than equal to x such that p equals a modulo q. We are not counting all the primes. We are only counting primes which are a modulo q and are less than equal to x and the summing over of the weight we sum with again same. So, what would you expect this quantity to be. So, now all the primes less than equal to x fall into different conjugacy classes with respect to q. How many such conjugacy classes are there? a must be relatively prime to q. So, if q is prime then q minus 1 but if q is not prime then it is 5 here it is a Euler function. So, that many conjugacy classes. So, if everything all the primes fall roughly equally in these one of those then you would expect this to be the leading term of psi to be divided by 5 of q because now you falling in different these many different classes. And the Dirichlet theorem says that that is basically what you get up to n r. So, this exactly same as prime number theorem is actually generalization of prime number theorem because it is you know now dividing the primes into different conjugacy classes. And saying that the count in each conjugacy class is what you would expect if things were uniformly distributed. Now, this can be proven essentially you know fall not essentially completely following or almost completely following the tools we developed. The entire you know strategy of writing this psi in front in the form of what did we do? We had started with psi then connected with the zeta function count to psi we wrote in terms of an integral which involved zeta prime over zeta. And then we did a contour integral and such it is that contour integral is potentially this sum of the poles of the zeta function and some other ones and that gives this estimate. So, we do exactly the same thing for this the difference is that now this becomes a little messier sum because we are counting only primes which are a modulo cube not all primes. We need some way of this pinpointing primes a modulo cube. So, for that what we will do is we will add some new ideas into this and then follow essentially the whole methodology. So, the new ideas are well one of them actually only one new idea is the characteristic modulo cube. So, chi we say is a map which is a homomorphism from z q star to complex numbers. Now, this psi this is not pi for pi you just divide by log x. So, chi is a map from z q star which is set of all numbers less than relatively prime to q two complex numbers. And z q star happens to be a group under multiplication modulo cube and chi is character if it is multiplicative there is chi of a b is chi of a times chi of. Now, some properties of characters are pretty evident first is that chi of a by q th root of infinity that is because for any a and z q star a to the phi q is one because the group size of this group is phi of q. So, a to the phi q is one therefore, since chi is multiplicative chi one is one. So, therefore, chi of a to the phi q is one which means that chi of a is phi q th root of infinity which obviously, implies that the characters always map to unit circle. And if you let g be the set of all characters and g is a group. So, that is multiplication of two characters is another character and it follows simply by the fact that firstly multiplication of two characters obviously, multiplicative and it is again maps z q star to complex more interestingly size of g is also phi of q. So, it is a group of same size as a group on which it operates I am not going to prove this stuff you guys you can work out this none of this is difficult to prove and then if you if you some take a character chi of a and sum over all a's in z q star then one of the two things happen one is that and this is very straight forward to see because these are phi q th roots of unity different a's will be map to different phi q th roots of unity or if they are map to the same then they will be same amount of mapping to the same one. So, this is either phi q if chi 0 where chi 0 this is a trivial character which maps every element to one. So, that is the case and obviously, it is something is to phi q that is not the case then just basically you are summing over all roots of unit phi q th root of unity and they will all cancel each other out this is standard. And finally, sum over all characters in the group g of chi of a again to one of the two things happen if it is psi of q phi of q if a is 1 0 otherwise. So, these are fairly basic facts about characters which sum of which we will use. So, how does this relate to the problem we have well if you recall we wanted to identify the those prime which are exactly equal to a model of q. Now, one way of you know thinking about it is a is the residue class model of q and in which is z a is an element of z q star. So, the characters operating on z q star which send it to complex numbers which is that essentially the domain that we are going to work on eventually. So, they can take this transform this problem of looking at residue class of a model of q to that of looking at certain roots of unity in the complex. So, that is a broad picture and that is how it does work out. In fact, what I am going to do now is to define what we call l functions which are generalizations of zeta functions before that I should have defined chi of n s. So, this is a generalization of zeta function almost generalize not it can say it is a generalization of zeta function where now an additional parameter which it takes is chi or as defined with respect to chi. The sum is again over n greater than equal to 1 the numerator is now chi of n, but now here I am using chi of n on all numbers chi I have just defined only on z q star. So, I have to extend that definition and that is whatever the obvious definition we use it chi of n is equal to chi of a. So, if q and n are not relatively prime then chi of n is if they are relatively prime then just take n go model of q take the residue class in which it belongs in the chi of a the value of chi. So, it is basically cycles the value of chi cycle in a cycle length of q chi of 0 you can also define chi of 0 is 0 chi of 1 is chi of 2 is whatever chi of 2 is you go up to any point where you hit a number which is relatively not relatively prime to q chi of that number is 0 otherwise chi of that number is defined by the definition reach up to q minus 1 then you go to q chi of q is 0 and then you cycle back into this exactly the same pattern and that pattern keeps on repeating. And now this l functions play the same role as the 0 function. So, when we translate that chi of x a q as a integral over complex plane we find this l function I will show you that maybe not today tomorrow, but let us spend some time and trying to understand what these l functions look like. So, what about how does l z chi 0 look like which is the simplest of these l functions chi 0 is 1, but it is not 1 everywhere. So, it is not quite the 0 function chi 0 is it is 1 precisely when n is not co prime to q. So, this is n greater than equal to 1 and q is 1 1 over into this. Now, is there a way of writing that it is certainly messier or more complicated than the zeta function, but can we write it as in the same form that multiplicative form that we could do for zeta function each of the prime factors p comma q is 1. So, it has although it is little more complicated than zeta function it still has that same multiplicative expression and that is really the one of the real key points of studying was one of the key point of studying zeta function. In fact, almost all the results we used about zeta function came out of this product form because then we took log of zeta which translated this into sum and then the derivative or just use the log itself. So, all of that comes only because we can have this multiplicative form which does exist in this case which is good to know that we can play around with this more easily. Similarly, let us take a general character does this have a multiplicative form does and that is exactly this as you would expect why this is a bit of justification. If you write this down rather write this as a infinite power sum you get 1 plus chi p over p to the z plus chi is p whole square by p to the 2 z and so on right chi of p whole square the same as chi of p square. So, that is infinite sum can be written as chi of p to the i by p to the i z some more and then when you multiply with this odd primes p which are literally prime to q you just get all numbers chi of n and divide by n to the z for all n which are literally prime to q chi n is 0 whenever n is not related. So, here we are crucially using the fact that character chi is multiplicative if it is not multiplicative this factorization does not work because it is multiplicative it is and that is second that is really the defining property in a sense or a key property in order to get such a factorization over primes for that infinite sum. So, later on we will see we will put in more complex things here than just the characters, but that is a key property we will always want to have that whatever we put in here factorizes. So, both of general in general l function it is such a factorization. Now, very quickly because again for zeta function we know that it has a pole at z equals 1 what about these they have poles at z equals 1. So, again let us go back to the first the trivial character as it is called chi 0 does not have pole at z equals 1. So, you just put z equals 1 in this sum. So, this is like a logarithm to the harmonic series, but some pieces are missing. So, what are the pieces which are missing? The pieces which are missing are for all n's for which n q is greater than 1. Now, does that make any significant difference in the that is a question. So, now all those n's in which I do not have those prime factors they can be only a they can be mapped to another set in which the total number prime all the primes infinite primes constant number of primes. Are you saying that the sum this sum is the full sum minus a finite number. No that is not true. Yes primes will be finite that is true, but the sum will be finite why the sum will be finite. No this is for example, let us say q is 2 or 3. So, what are we excluding? We are excluding all n's which are multiples of either 2 or 3. So, what we are excluding is 1 by summation 1 by 2 n. We are also excluding summation 1 over 3 n and we are we are counting their product twice. So, we have to add. So, this is like inclusion exclusion principle we have to add summation 1 by 6 n. So, let us let us do that. So, let say let p 1 p 2 to p l divide q and no other prime. So, all multiples n all n's which are multiples of any of these are excluded, but now we have to. So, we basically we can write as then we have to subtract what do we subtract summation i going from 1 to l 1 over p i times zeta 1, but now we are over over subtracting. So, we have to add back again zeta 1 minus. So, this is exclusion inclusion exclusion that goes on here. So, basically zeta 1 is always there that is ever present times 1 minus summation i 1 over p i plus summation i j is this a recognizable quantity as long as actually at least as long as we show this is not 0 we are done because zeta 1 we know is what it is and that is it problem solved. Now, what about non principal character or non trivial character then what about does this diverge zeta 1 times. Number of terms are same we are running over all n. So, instead of 1 we have a number greater than 1 it will be the original series plus some other positive series. So, it will diverge. So, why is this how does the original series come from no it is a complex number it is always on the unit circle on the complex plane it could be minus 1 actually it converges very simple proof look at the absolute value do I want to look at absolute value I need to be I do not know that is why I do not take look at absolute value of k i, but it is n no problem there is no zeta, but let us forget for the time being the denominator and suppose we are just summing of the numerator over all n does this some diverge diverge that is basically being undefined that is right, but since we are. So, what I am not saying it converges to a value I am asking you does it diverge. So, if you sum it up in the usual sense of using the prefixes this is always less than equal to 3 actually 5, because the first q when you sum of the first q the you get 0 these are roots of unit is non trivial roots of unit is. So, there is a circle around and when you add them up you get 0 the next q 0 again the next q 0 again. So, whatever is a segment which is left out actually I should try this is strictly speaking what I should write is put an upper bound here and then it is for any n this is bounded by this one. Now, let us come to this sum what can we say about this sum I want to show that this is also convergent there is only few of q non zero term in that segment of length q and I am just adding them all up of course, it is not case it will be less than q. So, now, when you stick a n in the denominator what happens to this should still be bounded what is the way to show. Let us take a segment of length q after several initial large number of initial ones. So, what happens to that let us say n q less than n less than equal to this would be if you see the denominator here it will vary from n q to n plus 1 q. So, denominators would roughly be the same when n is large and the difference is just between the denominator value is n q to n q plus q and the q is of course, the fixed quantity. So, it is going to be very minor difference in the denominator. So, if the denominators were equal then this sum is 0. So, the ratio of two successive terms in that case would still be less than 1 because that series diverges. Which series diverges? Without the n in the denominator that converges. So, now, if you look at the other series and compare the successive terms then I think the ratio can be written in terms of the original series times some quantity which is totally less than 1. That is less than 1 yes, but the original ones here were complex numbers. So, I do not know how to that was more for reals that you that holds complex number that does not work. So, I mean this is rough. This should be very close to 0 and as n grows bigger and bigger this should approach 0 quite rapidly I would say because let me write it in the following way.