User victor rotger - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:15:16Z http://mathoverflow.net/feeds/user/18643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae Axiomatizing Gross-Zagier formulae Victor Rotger 2012-10-08T14:31:07Z 2012-10-25T14:29:11Z <p>Let $\pi$ be a global automorphic representation of some reductive group over a number field, and let $L(\pi,s)$ denote its L-function. Assume $L(\pi,s)$ extends meromorphically to the complex plane and satisfies a functional equation of the form $$L(\pi,s)= \varepsilon(\pi,s) L(\pi^\star,1-s),$$ where $\pi^\star$ denotes the contragredient dual of $\pi$.</p> <p>Assume $L(\pi,1/2)=0$ and $L'(\pi,1/2)\ne 0$.</p> <p><strong>Question 1:</strong><br> Under what circumstances do we expect the existence of an algebraic null-homologous cycle $D$ on some variety $V$ for which its height $h(D)$ is meaningful and well-defined, and the equality (up to a non-zero, well-understood, fudge factor) $$L'(\pi,1/2) \overset{\cdot}{=} h(D)$$ holds?</p> <p><strong>Question 2:</strong> Assume the answer to the first question is expected to be "yes" for a given $\pi$, and <em>assume</em> we are also given a good, conjectural, candidate for $D$. Is it possible to axiomatize what one needs to show about the L-function $L(\pi,s)$, the height pairing $h$ and the cycle $D$ in order to prove the above <em>Gross-Zagier formula</em>?</p> <p>My feeling is that the answer to Q1 should be <em>yes</em> at least when $\pi$ is self-dual, that is to say, $\pi \simeq \pi^\star$. But I do not know whether such a formula is to be expected also for <em>non self-dual</em> $\pi$.</p> <p>Let me also clarify that I am <strong>not</strong> asking about the difficulties of constructing a suitable candidate for $D$. Not because it is an uninteresting question, but rather to focus the discussion. In the first question I just ask for whether there exists a cycle satisfying the Gross-Zagier formula, but I am not asking who $D$ is. In the second question I am assuming $D$ is given, and I am asking what properties it should satisfy, but I again don't care who $D$ is.</p> <p>Example 1: Let me explain the basic scenario I have in mind. Let $E/\mathbb{Q}$ be an elliptic curve and $K$ an imaginary quadratic field. If the pair $(E,K)$ satisfies the Heegner hypothesis, then the order of vanishing of the (self-dual) L-function $L(E/K,s)$ at its central critical value $s=1$ is odd, and Gross-Zagier proved that there exists a certain (Heegner) point $P_K\in E(K)$ such that $$L'(E/K,1) \overset{\cdot}{=} h(P_K).$$</p> <p>Here $P_K$ plays the role of $D$ in the general question. And we are evaluating the L-function at $s=1$ instead of $s=1/2$ just because we re-normalized it so that the functonal equation relates the values at $s$ and $2-s$. </p> <p>And let me explain now some examples in which I do not know the answers.</p> <p>Example 2: Let $E/\mathbb{Q}$ be an elliptic curve of prime conductor $p$ and $K$ a <em>real</em> quadratic field in which $p$ remains inert. Then the order of vanishing of the (self-dual) L-function $L(E/K,s)$ at its central critical value $s=1$ is odd. Henri Darmon has constructed a point $P_K\in E(K_p)$, rational over the completion of $K$ at $p$, which he conjectures to be actually rational over $K$. I am not asking how to prove this statement here, but rather: assume as a black box that $P_K$ indeed lies in $E(K)$. What one would need to know about $L(E/K,s)$ and $P_K$ in order to prove that $L'(E/K,1) \overset{\cdot}{=} h(P_K)$?</p> <p>Example 3: Let $E/\mathbb{Q}$ be an elliptic curve. Let $\chi$ be a Dirichlet character and $\mathbb{Q}_\chi$ be the abelian extension of $\mathbb{Q}$ cut out by $\chi$. Assume $L(E,\chi,1)=0$ and $L'(E,\chi,1) \ne 0$. The conjecture of Birch and Swinneton-Dyer predicts the existence of a <em>non-zero</em> point $P_{\chi} \in E(\mathbb{Q}_{\chi})\otimes \mathbb{C}$ lying in the $\chi$-eigenpart of the Modell-Weil group under the Galois action. </p> <p>Do we expect the equality $L'(E,\chi,1) \overset{\cdot}{=} h(P_\chi)$ to hold up to some conceptually well-understood fudge factor? (Note that if we assume both sides to be non-zero, the formula obviously holds by setting the fudge factor to be $L'(E,\chi,1)/h(P_\chi)$, and this is not what one would call a well-understood fudge factor!) </p> <p>Example 4: Let $f\in S_2(N,\chi)$ be a (cuspidal, normalized) newform of weight $2$, level $N$ and nebentype character $\chi$. Then the field $\mathbb{Q}_f$ generated by the fourier coefficents of $f$ is a finite extension of $\mathbb{Q}$, say of degree $d$. The Eichler-Shimura construction yields an abelian variety $A_f/\mathbb{Q}$.</p> <p>On the geometric side, we again have a natural construction of Heegner points: $A$ is a quotient of the jacobian $J_1(N)$ of $X_1(N)$. Given an imaginary quadratic field $K$, the theory of complex multiplication allows us to construct Heegner points $P$ on $X_1(N)$ which are rational over a suitable abelian extension $H/K$. This has been extensively studied for $X_0(N)$, in which case $H$ is a ring class field. But is also well-known for $X_1(N)$, where $H$ is no longer anticyclotomic; it contains for instance the abelian extension of $\mathbb{Q}$ cut out by $\chi$. </p> <p>In any case, one can construct a Heegner point $P_K\in A(K)$ by tracing down $P$ from $H$ to $K$. And if $\psi$ is a character of $\mathrm{Gal}(H/K)$, one can also define $$P_\psi = \sum_{\tau\in \mathrm{Gal}(H/K)} \psi^{-1}(\tau)P^\tau \in E(H)\otimes \mathbb{C},$$ which lies in the $\chi$-eigenpart of $E(H)\otimes \mathbb{C}$.</p> <p>On the L-function side, $L(A/\mathbb{Q},s)$ factors as $$L(A/\mathbb{Q},s) = \prod L(f^\sigma,s)$$ where $\sigma$ ranges over the $d$ different embeddings of $\mathbb{Q}_g$ into $\mathbb{C}$.</p> <p>While $L(A/\mathbb{Q},s)$ is self-dual, each of the individual factors $L(f^\sigma,s)$ is self-dual if and only if $\chi$ is the trivial character. If $f^\star$ denotes the modular form obtained from $f$ by complex conjugating its fourier coefficients, then the functional equation of $L(f,s)$ relates it to $L(f^*,2-s)$.</p> <p>A similar discussion holds for the base change of $A$ to $H$. The L-function of $A\times H$ is self-dual, but it factors as the product of L-series of the type $L(f/K,\psi,s)$ where $\psi$ ranges over the characters of $\mathrm{Gal}(H/K)$. Each of the individual L-functions are not always self-dual (regarding $\chi$ and $\psi$ adelically over $\mathbb{Q}$ and $K$ respectively, $L(f/K,\psi,s)$ is self-dual if and only if the restriction of $\psi$ to the ideles of $\mathbb{Q}$ is the inverse of $\chi$.)</p> <p>Gross-Zagier formulas are proved in the self-dual setting by Zhang and his collaborators, and also by Howard. And Olivier reminded us that such a formula is <em>not</em> to be expected if we insist to use the point $P_\psi$. So the question is: for arbitrary pairs $(\chi,\psi)$, does there exist a point $P\in (E(H)\otimes \mathbb{C})^{\psi}$ for which $L'(f/K,\psi,1)\overset{\cdot}{=} h(P)$ up to a well-understood non-zero fudge factor?</p> http://mathoverflow.net/questions/78563/is-the-square-of-a-curve-minus-its-diagonal-affine Is the square of a curve minus its diagonal affine? Victor Rotger 2011-10-19T10:33:29Z 2012-09-28T17:09:45Z <p>Let $X$ be a smooth irreducible projective algebraic curve of genus $g\geq 1$ and $S=X^2$ the surface one obtains as the cartesian product of $X$ with itself. Let $\Delta$ be the diagonal in $S$, that is to say, a copy of $X$ embedded diagonally in $X^2$. </p> <p>Is the quasi-projective surface $U=S\setminus \Delta$ affine? </p> <p>One criterion for showing affinness is showing that $\Delta$ is an ample divisor in $S$. But the self-intersection of $\Delta$ is $\Delta^2=2-2g&lt;0$ and therefore $\Delta$ can not be ample. Does this already imply that $U$ is not affine? I guess not.</p> <p>Besides, Serre's criterion provides a necessary and sufficient condition for $U$ to be affine: this is the case if and only if $H^i(U,\mathcal F)=0$ for all $i>0$ and all coherent sheaves $\mathcal F$ on $U$. But I don't know how to check this in this example. </p> http://mathoverflow.net/questions/106498/examples-of-q-expansions-in-a-hida-family Examples of q-expansions in a Hida family Victor Rotger 2012-09-06T11:10:31Z 2012-09-09T01:53:06Z <p>Let $p$ be a prime number and $N$ a positive integer not divisible by $p$. </p> <p>For some easy choices of $p$ and $N$, can anybody provide me with explicit examples of collections $${f_k,\quad 2\leq k \leq k_0 }$$ of q-expansions such that, for each $k$, $f_k$ is the $q$-series of a classical eigenform of weight $k$ and level $Np$, all being members of an ordinary $p$-adic Hida family of tame level $N$? Here $k_0$ is some reasonable upper bound for the weight, and I leave to the reader the meaning of the term "reasonable". Even if $k_0=4$ I will already be happy, and it is also fine if $k$ only runs among some of the integers between $2$ and $k_0$. But of course, the cardinal of ${ f_k}$ should be greater than $1$!</p> <p>I am interested in Hida families which are neither Eisenstein not CM, as explicit examples of those are already well-known.</p> <p>The simplest example seems to arise when one takes $N=1$ and $p=11$. In this case the associated Hida Hecke algebra is $\mathbf{T}=\Lambda:=\mathbb{Z}_p[[T]]$. I guess that people like Kevin Buzzard, Robert Pollack, William Stein and many others have computed the $q$-expansions $f_k$ in this case, up to some reasonable $k_0$, and I would be happy to have access to these computations.</p> <p>Yet another interesting example: reading the paper "How can we construct abelian Galois extensions of basic number fields?" by Barry Mazur, he reports on an example computed by Citro and Stein, where $N=1$ and $p=691$. They compute explicitly ${f_2,f_{12}}$ where $f_{12}$ is the $p$-stabilization of the discriminant modular form $\Delta$ and $f_2$ turns out to be the only weight 2 newform of level $p$ and character $\omega^{-10}$, where $\omega$ is the Teichmuller character. I would like to know the $q$-expansions $f_k$ in this case for some other integers in between $2$ and $12$. (For integers $k$ larger than $12$ one can possibly compute $f_k$ by multiplying $f_{12}$ by a suitable Eisenstein series of weight $k-12$ and applying Hida's ordinary projector.) </p> <p>And of course, any other collection of examples is welcome!</p> http://mathoverflow.net/questions/94332/beilinsons-formula-for-the-product-of-two-modular-curves Beilinson's formula for the product of two modular curves Victor Rotger 2012-04-17T22:16:56Z 2012-04-18T09:56:01Z <p>In his cellebrated 1984 paper "Higher regulators and values of L-functions", Beilinson proved (among many other exciting things) that the value at the non-critical point $s=2$ of the Rankin L-function $L(f\otimes g,s)$ of the convolution of two eigenforms of weight $2$ (say of the same level $N$) is related to the image under the complex regulator map of a certain diagonal element $\Delta$ in the $K_1$ of the surface $S=X_1(N)\times X_1(N)$. </p> <p>This happens in the pretty short section 6 of the above mentioned paper. However, I have not been able to spot the precise relationship between $L(f\otimes g,2)$ and $\mathrm{reg}(\Delta)$. I have also read several of the surveys on the subject existing in the literature, like Tony Scholl's "Integral elements in $K$-theory and products of modular curves" and a few others, again without luck: those articles rather focus on other aspects of the story. (To the best of my knowledge, of course, I'll be happy to be corrected.)</p> <p>My question is: is there any place in the literature where such a formula is precisely stated? By this I mean a formula where all the involved quantities are explicitly written down.</p> <p>With my collaborators I am working on $p$-adic analogues of this formula, that's why the interest. We can redo Beilinson's computations in order to write down the formula (as it surely is a formal consequence of the ideas of his paper, once everything is written down carefully enough), but I wonder whether this is already done explicitly elsewhere.</p> http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae/110665#110665 Comment by Victor Rotger Victor Rotger 2012-10-26T09:27:18Z 2012-10-26T09:27:18Z So nice. This gives support to an affirmative answer to Question 1 in a non-self dual situation: remember I'm not asking how to construct the points. Assuming the existence of the point (which in examples 2,3,4 is granted by BSD), I am asking what would one need to prove ( about the L-function and about the point) in order to prove a Gross-Zagier formula. Since the original proof of Gross-Zagier is a computation which, to a large extent, works independently on both sides, these are really two questions: what does one need to prove about $L(\pi,s)$ and what about the point or cycle? http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae/109180#109180 Comment by Victor Rotger Victor Rotger 2012-10-24T12:02:45Z 2012-10-24T12:02:45Z Ok, I just managed. http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae/109180#109180 Comment by Victor Rotger Victor Rotger 2012-10-24T11:08:52Z 2012-10-24T11:08:52Z But it's not displaying well, I don't understand what I'm typing wrong. http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae/109180#109180 Comment by Victor Rotger Victor Rotger 2012-10-23T13:20:13Z 2012-10-23T13:20:13Z I just edited the question in order to focus it on the aspects I would like to learn more about. http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae/109180#109180 Comment by Victor Rotger Victor Rotger 2012-10-08T21:05:10Z 2012-10-08T21:05:10Z Thanks, Olivier, this is the kind of ideas I was looking forward to discuss. The reasons you pose make me agree with you that &quot;it is not straight-forward&quot; to extend the current circle of ideas to non self-dual settings. And I agree even more with you in that it would be fun to learn whether some other ideas can be exploited to push these GZ formulae to some non self-dual scenario. On the geometric side of the formula, notice that I don't require the point to be a Heegner point. And I am an optimist as for whether some sort of GZ formula should hold (not that it'd be easy to prove!). http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae Comment by Victor Rotger Victor Rotger 2012-10-08T20:36:03Z 2012-10-08T20:36:03Z But nevertheless it just <i>happens</i> some times that $L(f/\mathbb{Q},\chi,1)$ or $L(f/K,\psi,1)$ vanish and the first derivative does not, and BSD is still in force! In the latter case, say, it predicts that the $\psi$-eigenpart of $E(H)\otimes \mathbb{C}$ has rank $1$ over $\mathbb{C}$, and is therefore generated by some point. The point $P_\psi$ of my question is one natural candidate, and there could be some other alternative constructions. In any case, one could aim to prove a Gross-Zagier formula showing that its height is related to $L'(f/K,\psi,1)$. http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae Comment by Victor Rotger Victor Rotger 2012-10-08T20:28:22Z 2012-10-08T20:28:22Z Hi again! In response to David and Fran&#231;ois, sure, in the non self-dual setting there is not such a neat criterion for the L-function to vanish at $s=1$. The reason is of course that $s=1$ is not any more the central critical value of the L-function, and the root number is just a complex number of absolute value $1$. If the answer to my question is &quot;we do not expect a Gross-Zagier formula to hold&quot;, these will surely be the conceptual reasons why. http://mathoverflow.net/questions/106498/examples-of-q-expansions-in-a-hida-family/106549#106549 Comment by Victor Rotger Victor Rotger 2012-09-06T21:49:43Z 2012-09-06T21:49:43Z Thanks! I'll wait for your paper for more details then! http://mathoverflow.net/questions/94332/beilinsons-formula-for-the-product-of-two-modular-curves/94378#94378 Comment by Victor Rotger Victor Rotger 2012-04-18T12:59:05Z 2012-04-18T12:59:05Z Merci beaucoup, Fran&#231;ois!