Blueprint of L-functions and need for introducing them ( Hasse-Weil L-functions ) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:52:11Z http://mathoverflow.net/feeds/question/111670 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111670/blueprint-of-l-functions-and-need-for-introducing-them-hasse-weil-l-functions Blueprint of L-functions and need for introducing them ( Hasse-Weil L-functions ) Shanmukha_Srinivasan 2012-11-06T18:17:38Z 2012-11-07T06:06:16Z <p>Dear All, </p> <p>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 . </p> <p>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 . </p> <p>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 :</p> <ul> <li><p>I didn't understand about the reason behind introducing exponential function on the right side .</p></li> <li><p>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 $p$ ? </p></li> </ul> <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 $p$. </p> <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 $1+T$ . </p> <p>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 .</p> <p>Cordially, </p> <p>Shanmukha Srinivasan.</p> http://mathoverflow.net/questions/111670/blueprint-of-l-functions-and-need-for-introducing-them-hasse-weil-l-functions/111673#111673 Answer by David Loeffler for Blueprint of L-functions and need for introducing them ( Hasse-Weil L-functions ) David Loeffler 2012-11-06T19:28:41Z 2012-11-06T19:35:06Z <p>There is an excellent reason why the exponential term and the division by $n$ are there, although they look a bit mysterious at first. </p> <p>Firstly, a correction to your formula: it should be <code>$|C(\mathbb{F}_{q^n})|$</code>, the number of solutions over the field with <em>q</em> elements, not <code>$|C(\mathbb{F}_{q})|$</code>. (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.) </p> <p>With this correction made, a miracle occurs: the quantity $\zeta_{C / \mathbb{F}_q}$ -- a priori just some formal power series -- is a <strong>rational function</strong>. </p> <p>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</p> <p><code>$$\zeta_{C / \mathbb{F}_q} = \exp( \sum_{n \ge 1} \frac{q^n + 1}{n} u^n) = \exp(-\log (1-u) - \log (1 - qu)) = \frac{1}{(1 - u)(1 - qu)}.$$</code></p> <p>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. </p> <p>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). </p> <p>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 <code>$\zeta_{C/\mathbb{F}_q}$</code> 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.</p> http://mathoverflow.net/questions/111670/blueprint-of-l-functions-and-need-for-introducing-them-hasse-weil-l-functions/111707#111707 Answer by Will Sawin for Blueprint of L-functions and need for introducing them ( Hasse-Weil L-functions ) Will Sawin 2012-11-07T06:06:16Z 2012-11-07T06:06:16Z <p>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.</p> <p>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.</p> <p>To figure out why you need to divide, you need to compute more carefully.</p> <p>You want to take the product of, for each closed point $x$, of residue degree $d_x$, $1/\left(1-\left(q^{d_x}\right)^{-s}\right)$. </p> <p>Setting $u=q^{-s}$, you want to take the product of $1/(1-u^{d_x})$.</p> <p>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 $\sum_x \sum_k u^{kd_x}/k$. </p> <p>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. </p> <p>So you can write $u^{kd_x}/k$ as coming from the $d_x$ $\mathbb F_{q^{kd_x}}$-points, divided by $kd_x$.</p>