Dear All,

This question may appear elementary to all the experts in number theory , but forgive me. I really wanted to know how did the $L$-functions came into existence, especially the Hasse-Weil L-functions . Do they have some specific meaning in their formulation or they are just framed heuristically to build some thing else , as scaffolding .

I do know that Zeta functions and L-functions of the curve act as spies in collecting the secret information about the local part of curves and embed that information inside them, but its really a great trouble in understanding the formulation. I referred to many books and they have started saying " Let $L(s,E)$ be the ...." in an assuming manner .

I just wanted to know , why should one consider $$\zeta_{C/\mathbb{F_q}}(u)=\exp \bigg(\ \sum_{n=1}^{\infty}\frac{ | C(\mathbb{F_{q^n}})|}{n} u^{n} \bigg)$$ where $C$ is a projective curve with non-negative genus over finite field $\ \mathbb{F_q}$. Here are my pointers :

  • I didn't understand about the reason behind introducing exponential function on the right side .

  • I understood that there is some measure of points taking a ratio of the cardinality ( on R.H.S ) of the solutions, but why is the ratio needed ? I got this doubt when I looked at some other heuristic consideration $\prod\frac{N_p}{p}$ ( Where $N_p$ is the cardinality of solution set at some prime $p$ ) , why is the need to take the ratio ? Isn't it not sufficient to look at just $N_p$ ? We get the cardinality directly, why should we find the ratio again by dividing it with $p$ ?

Similarly , why is the formulation of local part of $L$-series ( Hasse Weil L-function ) appear as $L_p(T)=1-a_pT+pT^2$ when the curve has good reduction at $p$ ( here $a_p=p+1-N_p$ and has some other formulation like $L_p(T) = 1-T$ and $1+T $ when the curve has split multiplicative and non-split multiplicative reductions at $p$ respectively , and $L_p(T)=1$ when the curve has additive reduction at $p$.

How was the quadratic equation on R.H.S ( i.e $1-a_pT+pT^2$ ) formulated ? Was it a scaffolding to get some heuristic output later , or it has a specific meaning derived from something, or what ? Same with $1-T$ and $1+T$ .

Please do explain me , I am sorry my learned friends, if I have wasted your time, but every book I referred starts with Let, and I thought that its just a setting . If you want me to suggest some book that does the same task of explaining what I asked, you are welcome to suggest me .


Shanmukha Srinivasan.

  • $\begingroup$ If you are trying to understand the Hasse-Weil functions before the local zeta function of an elliptic curve, that is pretty much trying to run before you can walk. $\endgroup$ – Charles Matthews Nov 6 '12 at 18:46
  • $\begingroup$ @Charles Matthews : Mr.Charles, The thing you said makes more sense for me. I just wanted to know about the local-zeta, but it was my colleagues who wanted the explanation of Hasse-Weil L-function too ! I thought why to open another question, and hence sandwiched both into the same question. I will only take the local-zeta part, in what you are going to answer ( if you are going to ) and leave the rest for other knowledgeable persons, who are just having the same confusion . Is it a mistake to ask something, for others ? $\endgroup$ – Shanmukha_Srinivasan Nov 6 '12 at 18:53
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    $\begingroup$ Try Serre's article in Arithmetical Algebraic Geometry. Proceedings of a Conference held at Purdue University, December 5-7, 1963. Edited by O. F. G. Schilling, Harper & Row, Publishers, New York 1965. And en.wikipedia.org/wiki/Local_zeta-function which you are supposed to read before posting here, in fact. Sandwiches which are too think can dislocate your jaw. Split multiplicative has the local zeta of the projective line minus two points: think about that for intuition. $\endgroup$ – Charles Matthews Nov 6 '12 at 19:29
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    $\begingroup$ Take a look at Ireland and Rosen's nice book "A classical introduction to modern number theory", and in particular chapters 10 and 11 dealing with equations over finite fields and the definition of zeta functions. $\endgroup$ – Abhinav Kumar Nov 6 '12 at 19:42
  • $\begingroup$ @Charles Matthews, @Abhinav Kumar : Thanks a lot for your suggestions, I will try to look at them surely . $\endgroup$ – Shanmukha_Srinivasan Nov 6 '12 at 20:40

There is an excellent reason why the exponential term and the division by $n$ are there, although they look a bit mysterious at first.

Firstly, a correction to your formula: it should be $|C(\mathbb{F}_{q^n})|$, the number of solutions over the field with q elements, not $|C(\mathbb{F}_{q})|$. (Notice that this means that the $n$th term really depends on $n$ and $C$ in a subtle way, because it "knows" how many points C has over every extension of the original field.)

With this correction made, a miracle occurs: the quantity $\zeta_{C / \mathbb{F}_q}$ -- a priori just some formal power series -- is a rational function.

To get some idea of the magic that's going on here, let's consider some simple examples. Firstly, you can take $C$ to be $\mathbb{P}^1$. That's not a very interesting curve, I know, but it's a curve. Then $C$ has exactly $q^n + 1$ points over $\mathbb{F}_{q^n}$ -- one for each element of the field, together with the point at $\infty$ -- and we get

$$ \zeta_{C / \mathbb{F}_q} = \exp( \sum_{n \ge 1} \frac{q^n + 1}{n} u^n) = \exp(-\log (1-u) - \log (1 - qu)) = \frac{1}{(1 - u)(1 - qu)}.$$

As promised, this is a rational function! If the funny exponential term and the $1/n$ factor hadn't been there in the definition, we wouldn't have got anything so nice.

It turns out that this happens for any curve $C$ (this was proved by Andre Weil) and in fact for higher-dimensional varieties too (this was proved by Dwork).

PS. If $C$ is an elliptic curve, then one can show (e.g. this is in Silverman's book "The Arithmetic of Elliptic Curves", in section V.5) that $\zeta_{C/\mathbb{F}_q}$ is given by $$ \zeta_{C/\mathbb{F}_q}(u) = \frac{1 - a u + q u^2}{(1 - u)(1 - qu)}$$ where $a = q + 1 - |C(\mathbb{F}_q)|$. So this quadratic term appearing here really appears for a reason; it wasn't just plucked out of midair. I hope that answers another part of your question, which is where this quadratic comes from.

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  • $\begingroup$ Your answer sounds wonderful ! Thank you David, I edited it . $\endgroup$ – Shanmukha_Srinivasan Nov 6 '12 at 20:48

One justification for this is the Euler product expression. To find the Euler product expression for the Hasse-Weil function, you have to ask yourself what the appropriate analogue of a prime is. It turns out to be a closed point on the variety. The analogue of the size of the prime is the size of the residue field of the closed point.

Without doing any calculation you can already see why the exp makes sense, because the Euler product will be a product over points, and counting points will be a sum over points, so to turn the one into the other you need to take an exponential.

To figure out why you need to divide, you need to compute more carefully.

You want to take the product of, for each closed point $x$, of residue degree $d_x$, $1/\left(1-\left(q^{d_x}\right)^{-s}\right)$.

Setting $u=q^{-s}$, you want to take the product of $1/(1-u^{d_x})$.

Since the product is exp of the sum of the log, you want to take the exp of the sum of minus the log of $1-u^{d_x}$. If you expand the log out as a power series, you get $\sum_x \sum_k u^{kd_x}/k$.

Each $\mathbb F_{q^n}$-point comes from a unique closed point $x$. That point must have a degree $d_x$ dividing $n$, and each closed point of degree $d$ dividing $n$ corresponds to $d$ $\mathbb F_{q^n}$-points.

So you can write $u^{kd_x}/k$ as coming from the $d_x$ $\mathbb F_{q^{kd_x}}$-points, divided by $kd_x$.

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    $\begingroup$ @ Will - thank you. This is the first time that I've seen this worked out before and I find it very enlightening. The only thing sticking point for me at the moment is that the analog of the size of a prime is the size of the residue field of the closed point. I'll think about this. $\endgroup$ – Jonah Sinick Nov 7 '12 at 7:06
  • $\begingroup$ Hmm...I'm confused, what's the ring of definition of the variety? At first I thought it was $\mathbb{Z}_{p}$ or $\mathbb{Q}_{p}$, but then I remembered that you wrote "Euler product expression for the Hasse-Weil function. Do you have in mind mathbb{Q}$ as the field of definition? If so, I'm not seeing how to assign a well defined residue degree to a point on the variety... $\endgroup$ – Jonah Sinick Nov 7 '12 at 7:22
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    $\begingroup$ Oh I just meant the Weil zeta function. For the Hasse-Weil zeta function, you have to do this over $\mathbb Z$. Then the $\mathbb Q$-rational points are not closed points over $\mathbb Z$ - they have specializations! Every closed point lies over some finite field, which is why the Hasse-Weil zeta function is a product of local factors. $\endgroup$ – Will Sawin Nov 7 '12 at 7:41
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    $\begingroup$ Well a prime is a prime ideal, and a closed point is also a prime ideal. The size of $p$ is certainly also the number of elements in the residue field of $p$, that is, in $\mathbb Z/p$. That's the connection. $\endgroup$ – Will Sawin Nov 7 '12 at 7:41
  • $\begingroup$ Very New perspective , Thanks @Will Sawin $\endgroup$ – Shanmukha_Srinivasan Nov 7 '12 at 10:37

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