Why are functional equations important? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:08:09Z http://mathoverflow.net/feeds/question/2040 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2040/why-are-functional-equations-important Why are functional equations important? Qiaochu Yuan 2009-10-23T05:51:15Z 2010-01-19T15:58:55Z <p>People who talk about things like modular forms and zeta functions put a lot of emphasis on the existence and form of functional equations, but I've never seen them used as anything other than a technical tool. Is there a conceptual reason we want these functional equations around? Have I just not seen enough of the theory? </p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/2044#2044 Answer by Ricardo for Why are functional equations important? Ricardo 2009-10-23T06:19:35Z 2009-10-23T06:19:35Z <p>Functional equations are useful to understand the "symmetries" of a given function, pretty much as periodicity. For example, they may be used in the continuation of functions to the complex plane. The functional equation for the Gamma function, Gamma(z+1) = zGamma(z), for instance, let us understand the continuation of this function to the left of the line Re(z)=0.</p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/2051#2051 Answer by David Lehavi for Why are functional equations important? David Lehavi 2009-10-23T07:47:00Z 2009-10-23T07:47:00Z <p>Sometimes you get very concrete algebro-geometric facts from functional equations:</p> <p>Example 1: The functional equation relating Weierstrass P (for a lattice L in C) and it's derivative is the equation representing the elliptic curve C/L in Weierstrass form.</p> <p>Example 2: if F is a modular form on the Siegel upper half plane (the universal cover of the moduli of Abelian varieties Ag) with weight W, then the class of (F)_0 in Pic(Ag) is W times the Hodge class.</p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/2076#2076 Answer by Eric Chopin for Why are functional equations important? Eric Chopin 2009-10-23T12:06:44Z 2009-10-23T12:06:44Z <p>Hi Qiaochu,</p> <p>I would say that functional equations of zeta-like function are popular because it is one of the simplest non trivial example of functions having a symmetry only when analytically continued. But in itself the impact of such a functional equation is quite limited, even though some people are convinced that this equation is a key for solving the Riemann Hypothesis, which is misleading (this feeling is supported also by the Hamburger theorem that states that any Dirichlet function having an analytic continuation like zeta and obeying the same functional equation is proportional to zeta, but when considering L-functions it becomes false and the Davenport-Heilbronn L-function is an example of Dirichlet function with a L-function-like functional equation but also with infinitely many zeroes in any vertical strip included in the critical strip, therefore failing to obey a Reimann hypothesis).</p> <p>However, as for me these functional equations hide a much deeper phenomenon. The functional equation is an artefact of the action of the Fourier transform on zeta functions. What's more, the same symmetry which can be proved in a simple manner for local zeta functions extends to global ones, which is a highly non-trivial result (proved for general Hecke zeta functions by John Tate in his thesis). In addition, you can see the Fourier transform as the image of a particular linear operator by the Weil representation. And looking at the action of the Weil representation in a more general context leads to some generalized Poisson Formulas, and most probably there is still some unexplored stuff to discover in this field (but that's a personal point of view...). </p> <p>Best regards,</p> <p>Eric </p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/2081#2081 Answer by David Speyer for Why are functional equations important? David Speyer 2009-10-23T12:31:19Z 2009-10-23T12:31:19Z <p>It was my understanding that, in some cases, the functional equation is the only known way to prove that the analytic continuation of one of these number theoretic functions is entire. Even in the case of zeta, I don't know another argument that zeta continues past Re(s)=0. (I can get into the critical strip by writing </p> <p>zeta(s) = 1/(s-1) + \sum [ n^{-s} - (n^{-s+1} - (n+1)^{-s+1})/(s-1) ],</p> <p>the sum now converges uniformly for Re(s)>epsilon>0.)</p> <p>So maybe the question should be "do we care about the values of zeta (and other L-functions) to the left of the critical strip?"</p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/2085#2085 Answer by Andreas Holmstrom for Why are functional equations important? Andreas Holmstrom 2009-10-23T13:00:14Z 2009-10-23T13:07:11Z <p>There are many reasons that functional equations are important. Some background: To most varieties/schemes occurring in arithmetic geometry, you can associate a zeta function/L-function. There are two main ways of constructing these, either from l-adic cohomology, or from counting solutions in various finite fields. Usually the word L-function is used for functions associated to varieties over number fields, constructed from l-adic cohomology, and the word zeta function is used for functions associated to schemes over Z or some other ring of integers, or over a finite field. However, there is some confusion about the terminology, and there is also some overlap between the two, since there is a close relation between the L-function of a variety and the zeta function of an integral model for the variety.</p> <p>Over finite fields, the functional equation is part of the famous Weil conjectures, proved by Deligne. One reason that the functional equation is cool is that it reflects Poincare duality in the etale cohomology of the variety, so it is in some sense a deep geometric statement. For background on the Weil conjectures, see for example Freitag and Kiehl: Etale cohomology and the Weil conjectures, and the survey of Mazur: Eigenvalues of Frobenius acting on algebraic varieties over finite fields, in some conference proceedings.</p> <p>For the definition of L-functions and zeta functions and lots of other background, see Manin and Panchishkin: Introduction to modern number theory, chapter 6. There are at least two deep reasons for being interested in the functional equations for these functions. Firstly, the existence of a functional equation seems to always be directly related to the L-function coming from an automorphic representation, and the idea that "every L-function from algebraic geometry (aka motivic L-function) also comes from an automorphic representation" is in some sense the number-theoretic incarnation of the global Langlands program. See for example Bump et al: An introduction to the Langlands program. The most famous cases where this has been proved is (1) Tate's thesis, which treats Hecke L-functions, and where the corresponding automorphic representation is one-dimensional, i.e. a character on the ideles, and (2) the work by Wiles and others related to Fermat's last theorem, where they show that the L-function associated to an elliptic curve over Q also comes from a modular form, and hence satisfies the expected functional equation. See the book of Diamond and Shurman on modular forms.</p> <p>The other deep reason for thinking about the functional equation is that some optimistic people dream of an "arithmetic cohomology theory", which would allow us to mimick the proof of the Weil conjectures, but for zeta functions over Z or L-functions over Q. Then the functional equation should be related to Poincare duality for this cohomology. All this is related to the Riemann hypothesis, noncommutative geometry, and the field with one element. See for example Deninger: Motivic L-functions and regularized determinants, the more recent survey available <a href="http://front.math.ucdavis.edu/0505.5354" rel="nofollow">here</a>, some <a href="http://people.math.jussieu.fr/~fpaugam/documents/transparents-expose-equa-fonc-baltimore.pdf" rel="nofollow">slides of Paugam</a> on the functional equation, and also the later chapters of Manin-Panchishkin. Some of the key names if you want to find more references: Deninger, Connes, Consani, Marcolli; most of them have lots of stuff on their webpages and on the arXiv.</p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/2090#2090 Answer by Rob Harron for Why are functional equations important? Rob Harron 2009-10-23T14:08:05Z 2009-10-23T14:08:05Z <p>As was mentioned in other answers, the functional equations are related to duality. In general, however, the "dual" object need not be the original object. It just so happens that the objects that give the Riemann zeta function and L-functions of modular forms are self-dual (up to a twist). For a general motive, M, for example, the functional equation (conjecturally) relates values of the L-function of M to values of the L-function of M*, the dual of M (see for example an article of Panchishkin's <a href="http://www.ams.org/mathscinet-getitem?mr=1306547" rel="nofollow">link text</a>). For automorphic representations, its the contragredient (see section 14 of Borel's article on L-functions in Corvallis <a href="http://www.ams.org/online%5Fbks/pspum332/" rel="nofollow">link text</a>). So, conceptually, the functional equation is just an expression of this duality.</p> <p>As Andreas Holmstrom alluded to above, the functional equation is (conjecturally) related to which L-functions are automorphic. The type of result is called a "Converse theorem". A famous "classical" one is Weil's converse theorem which basically says that if you have two Dirichlet series such that for infinitely many Dirichlet characters, the twisted Dirichlet series have analytic continuation, are bounded in vertical strips AND are related by a functional equation, then they are Dirichlet series coming from a modular form (see Bump's book, theorem 1.5.1). For GL(n) see Cogdell's "L-functions and Converse theorems for GL(n)" <a href="http://www.math.ohio-state.edu/~cogdell/" rel="nofollow">link text</a>).</p> <p>To add to the above paragraph, the existence of a functional equation is thought to be one of the characterizing properties of L-functions as expressed by the notion of the "Selberg class" (<a href="http://en.wikipedia.org/wiki/Selberg%5Fclass" rel="nofollow">link text</a>)</p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/2102#2102 Answer by engelbrekt for Why are functional equations important? engelbrekt 2009-10-23T15:25:48Z 2009-10-23T15:25:48Z <p>Apart from their conceptual significance, functional equations are useful when doing concrete estimates in analytic number theory. The usual textbook proof by the method of contour integrals for the bound \vartheta \leq 1/3 in the Dirichlet divisor problem relies on the functional equation of the Riemann zeta function.</p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/3590#3590 Answer by Kevin Buzzard for Why are functional equations important? Kevin Buzzard 2009-10-31T20:32:34Z 2009-10-31T20:32:34Z <p>I am surprised no-one seems to have mentioned one key use for functional equations: they are a key input in converse theorems. If you have a power series in q that you're trying to show is a modular form of level 1, then one strategy is to show that the associated L-function has the right kind of functional equation (and a couple of other nice properties). The functional equation can translate into a relation between f(z) and f(-1/z), and the right kind of functional equation translates into the right kind of translation property for f, which is what you needed to show it was modular (the fact that it was a power series in q gives you the boundedness at infinity and f(z)=f(z+1)). </p> <p>This idea was generalsed by Weil, who proved that sufficiently many functional equations for your L-function and its twists will imply that it's the L-function of a modular form. Hence elliptic curves with complex multiplication are modular! (because their L-functions are L-functions of grossencharacters). That's an algebraic statement whose proof I outlined above makes essential use of the functional equation of the L-function.</p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/11874#11874 Answer by KConrad for Why are functional equations important? KConrad 2010-01-15T15:47:58Z 2010-01-17T20:16:28Z <p>The simplest reason functional equations have importance, for someone learning this stuff for the first time (not knowing about modular forms, automorphic representations, etc.) is that they can be used to verify the Riemann hypothesis numerically up to some height!</p> <p>To explain this, let's start off with a limitation of methods of complex analysis in detecting zeros of functions. There are theorems in complex analysis which tell you how to count zeros of an analytic function $f(s)$ inside a region by integrating $f'(s)/f(s)$ around the boundary (this is the argument principle). So we could integrate around the boundary of a box surrounding the critical strip up to some height and see there are, say, 10 zeros of the Riemann zeta-function up to that height (yeah, there's a pole on the boundary at $s = 1$ which messes up the argument principle integration, but don't worry about that right now). How can we prove the 10 zeros in the critical strip up to that height are on the critical line? Complex analysis provides no theorems that assure you an analytic function has zeros on a line!</p> <p>The functional equation comes to the rescue here. I'll illustrate for the Riemann zeta-function $\zeta(s)$. Its functional equation is most cleanly expressed in terms of </p> <p>$$Z(s) = \pi^{-s/2}\Gamma(s/2)\zeta(s)$$</p> <p>and is the following:</p> <p>$$Z(1-s) = Z(s).$$</p> <p>We also need another "symmetry": <code>$Z(s^*)^* = Z(s)$</code>, where <code>${}^*$</code> means complex conjugation. Where does this come from? For an entire function $f(s)$, the function <code>$f(s^*)^*$</code> is also entire: in fact its local power series expansion at any point a is the one whose coefficients are complex conjugate to the coefficients of $f(s)$ at $a^*$. Or you could directly prove <code>$f(s^*)^*$</code> is complex-differentiable when $f(s)$ is. The significance of this is that if $f(s)$ is <em>real-valued</em> for some interval of real numbers then <code>$f(s^*)^* = f(s)$</code> on that interval, so by the rigidity of analytic functions we must have <code>$f(s^*)^* = f(s)$</code> everywhere when it is true on a real interval (not one point intervals, obviously). Lesson: an entire function $f(s)$ that is real-valued on some interval of the real line satisfies the formula <code>$f(s^*)^* = f(s)$</code> for <em>all</em> complex numbers $s$. By the way, this also applies to meromorphic functions on $\mathbf C$ too (rigidity of meromorphic functions).</p> <p>Let's now return to the zeta-function. Because $\zeta(s), \pi^{-s/2}$, and $\Gamma(s/2)$ are real-valued for real $s > 1$, their product $Z(s)$ is real for $s > 1$, so <code>$Z(s^*)^* = Z(s)$</code> for all complex $s$. In particular, for a number $s = \frac12 + it$ on the critical line (here $t$ is real), we have the <em>key calculation</em></p> <p><code>$$Z(s)^* = Z(1/2 + it)^* = Z(1 - (1/2 + it))^* = Z(1/2 - it)^* = Z((1/2 + it)^* )^* = Z(1/2 + it) = Z(s),$$</code></p> <p>where we used $Z(s) = Z(1-s)$ in the second equation and <code>$Z(s^*)^* = Z(s)$</code> in the second to last equation.</p> <p>This tells us the function $Z(s)$ is <em>real-valued</em> on the critical line. The Riemann zeta-function is <em>not</em> real-valued on the critical line, but this modified (completed) zeta-function $Z(s)$ is. Moreover, because $Z(s)$ differs from $\zeta(s)$ by factors that are finite and nonzero inside the critical strip ($\pi^{-s/2}$ is a nowhere-vanishing entire function and $\Gamma(s/2)$ is meromorphic with no zeros and only has poles at $s = 0, -2, -4, \dots$), the zeros of $\zeta(s)$ and $Z(s)$ inside the critical strip are the same thing. (In fact, <em>nontrivial</em> zeros of $\zeta(s)$ are exactly the same thing as <em>all</em> zeros of $Z(s)$, which is one reason $Z(s)$ is a nicer object that $\zeta(s)$: the Riemann hypothesis is a statement about <em>all</em> zeros of $Z(s)$!) So what? Well, we just showed in the key calculation above that the function $Z(1/2 + it)$ is real when $t$ is real, so by computing we can provably detect zeros of $Z(s)$ on the critical line Re($s$) = $\frac12$ by looking for <em>sign changes</em> of $Z(1/2 + it)$ as t runs through the real numbers. </p> <p>So here is a two-step procedure for proving the RH numerically up to height $T$ (i.e., in the box in the critical strip from the real axis up to height $T$):</p> <ol> <li><p>Use complex analysis (the argument principle) to count how many zeros $Z(s)$ has in the critical strip up to height $T$ by integrating $Z'(s)/Z(s)$ around a box surrounding that region. (If the poles of $Z(s)$ at $s = 0$ and $s = 1$ bother you, recall the argument principle can account for poles or you might prefer to use $s(1-s)Z(s)$ in place of $Z(s)$ to be working with an <em>entire</em> function which satisfies the same functional equation as $Z(s)$ and is also real-valued on the critical line.) </p></li> <li><p>Count sign changes for $Z(1/2 + it)$ when $0 \le t \le T$. There is a zero between any two sign changes, so $Z(s)$ has at least as many zeros <em>on the critical line</em> as the number of sign changes that were found. (Finding a sign change is a computable thing: if a function value at a point is approximately positive or negative then it is provably so by checking the error in your computation well enough, whereas proving a function value at a point is exactly zero with a computer is basically impossible.) </p></li> </ol> <p>If the counts in steps 1 and 2 match, then voila: all zeros of $Z(s)$ up to height $T$ in the critical strip are on the critical line, which confirms the Riemann hypothesis up to height $T$. </p> <p>This method will <em>not</em> work if there are any multiple zeros on the critical line: the argument principle counts each zero with its multiplicity, so if for instance there is a double zero then the argument principle may tell us $Z(s)$ has 10 zeros (with multiplicity!) up to some height while we find only 9 sign changes because one zero is a double zero so it doesn't give us a sign change. (Or if there were a triple zero we get 10 zeros with multiplicity from the argument principle but we find only 8 sign changes.) A graph may <em>suggest</em> that the mismatch in the numbers in the two steps is coming from a multiple zero, but it doesn't rigorously prove anything. Fortunately, this has never happened in practice with the Riemann zeta-function: the two counts always match. In fact the conjecture is that all nontrivial zeros of $\zeta(s)$ are simple zeros. </p> <p>What about more general L-functions $L(s)$? By multiplying $L(s)$ by suitable exponential and Gamma functions, you get a function $\Lambda(s)$ whose functional equation is </p> <p><code>$$\Lambda(1-s^*)^* = w\Lambda(s),$$</code></p> <p>where $w$ is a constant with absolute value 1. (For the Riemann zeta-function, $\Lambda(s) = Z(s)$ and $w = 1$. For Dirichlet L-functions, $w$ is usually not equal to 1.) Let $u$ be one of the square roots of $w$, so $w = u^2$. Then using the above functional equation, the function </p> <p>$$\frac1u \Lambda(s)$$</p> <p>is real-valued on the critical line ($s = 1/2 + it$ for real $t$), so we can detect its zeros there by looking for sign changes. The same method described above for detecting zeros of $\zeta(s)$ in the critical strip by using $Z(s)$ and its functional equation can be applied also to $L(s)$. This is basically the way all variants on the Riemann hypothesis are checked numerically (modulo important details of practical calculation that I don't get into), and the functional equation is an essential ingredient in justifying the method.</p> <p>What is crucial here is not just the idea of sign changes, but also the expectation that the zeros are all simple (so we can find all the zeros by sign changes and the argument principle). As with the Riemann zeta-function, it is expected that the nontrivial zeros of Dirichlet $L$-functions are all simple. But there are examples of $L$-functions with a multiple zero on its critical line, thanks to ideas from the Birch and Swinnerton-Dyer conjecture. This does not wreck this approach to verifying the Riemann hypothesis for such L-functions, because such multiple zeros are supposed to happen only at one (known) point which we think we understand.</p> http://mathoverflow.net/questions/2040/why-are-functional-equations-important/12316#12316 Answer by Emerton for Why are functional equations important? Emerton 2010-01-19T15:58:55Z 2010-01-19T15:58:55Z <p>It seems worth mentioning another application of the functional equations. To fix ideas, I will suppose that $E$ is an elliptic curve over ${\mathbb Q}$, and let $L(E,s)$ be the $L$-function of $E$. There is then a functional equation relating $L(E,s)$ and $L(E,2-s)$; one feature of this functional equation is that it has a sign, i.e. has the form $L(E,s-2) = \pm (\text{ a positive constant })L(E,s),$ where $\pm$ is some fixed sign, depending on $E$, and easily computed (say using the modular form corresponding to $E$).</p> <p>Thus one find that the order of vanishing of $L(E,s)$ at $s = 1$ is odd (even) precisely if the sign is $-1$ ($+1$). </p> <p>Since the order of vanishing is conjectured to coincide with the rank of $E({\mathbb Q})$ (the BSD conjecture), this is pretty important arithmetic information which is obtained pretty easily from the functional equation. In particular, if the sign is $-1$, we always expect there to be a rational point of infinite order, and arranging things so that the sign <I>is</I> $-1$ (by twisting by a well-chosen character, say) is a common way of forcing the existence of rational points in various situations. (There is a large body of research focused around this expected relationship between signs of functional equations and the existence of rational points on elliptic curves; for example, it lies at the heart of the study of Heegner points by Gross and Zagier and Kolyvagin. Some important recent contributions are by Nekovar, Mazur and Rubin, and T. and V. Dokchitser.)</p>