Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\pi$.

Assume $L(\pi,1/2)=0$ and $L'(\pi,1/2)\ne 0$.

Question 1:
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?

Question 2: Assume the answer to the first question is expected to be "yes" for a given $\pi$, and assume 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 Gross-Zagier formula?

My feeling is that the answer to Q1 should be yes 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 non self-dual $\pi$.

Let me also clarify that I am not 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.

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 $$ L'(E/K,1) \overset{\cdot}{=} h(P_K). $$

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 $2-s$.

And let me explain now some examples in which I do not know the answers.

Example 2: Let $E/\mathbb{Q}$ be an elliptic curve of prime conductor $p$ and $K$ a real 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 $L'(E/K,1) \overset{\cdot}{=} h(P_K)$?

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 non-zero 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.

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!)

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 $A_f/\mathbb{Q}$.

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 $\chi$.

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 $E(H)\otimes \mathbb{C}$.

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 $\mathbb{C}$.

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 $L(f^*,2-s)$.

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 $\chi$.)

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 not 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?

share|improve this question
3  
Is it obvious that $L(f/K, \psi, 1)$ is zero in the non-self-dual case? In the self-dual cases the sign in the functional equation is forcing $L(f/K, \psi, 1) = 0$ and I guess that is somehow the reason to expect that the next derivative is arithmetically meaningful, but in the non-self-dual cases that wouldn't happen, would it? –  David Loeffler Oct 8 '12 at 17:43
    
My recollection is that if you take level 122, nontrivial real character, weight 2 newform over $Q(i)$ -- then $L(f,1)=0$ but the sign is merely some random number on the unit circle. It is an example to investigate. I don't know whether $L'(f,1)$ is meaningful. Usually I expect, as per David Loeffler that the 0th derivative is nonvanishing. What does this mean in the Heegner construction context? –  Junkie Oct 8 '12 at 19:37
    
I don't know what happens for $K$ imaginary quadratic, but for $K=\mathbf{Q}$, the product of root numbers $w(f \otimes \chi) w(f \otimes \overline{\chi})$ is one, so there is no "obvious" vanishing forced by the functional equation. On the other hand the twisted $L$-value $L(f \otimes \chi,1)$ may indeed vanish, since elliptic curves may acquire points over abelian extensions of odd degree. –  François Brunault Oct 8 '12 at 19:38
    
Hi again! In response to David and François, 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 "we do not expect a Gross-Zagier formula to hold", these will surely be the conceptual reasons why. –  Victor Rotger Oct 8 '12 at 20:28
    
But nevertheless it just happens 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 $L'(f/K,\psi,1)$. –  Victor Rotger Oct 8 '12 at 20:36

2 Answers 2

UPDATE: I have updated this answer slightly to take into account Victor's remark.

I think that the precise questions being asked admit a straightforward answer. At the moment, no such formula is known and the proofs of Gross-Zagier, Waldspurger, Zhang et al. and Howard all absolutely and crucially require the hypothesis of self-duality. The reason for this is that the representation-theoretic part of the proof requires an understanding of test-vectors, as in the works of Tunell and Saito or as in the conjecture of Gross-Prasad. This is explained for instance in the article of Gross entitled Heegner points and representation theory as well as in Non-triviality of CM points in ring class field towers. Aflalo, Esther and Nekovář, Jan. Israel J. Math. 175 (2010), 225--284 (in which the formal setting is explored).

As for whether similar formula could hold, I am not too optimistic. A Gross-Zagier formula should involve the $\psi$-eigenpart of the action of $\textrm{Gal}(H/\mathbb Q)$ on the projection of a CM point on the $\pi(f)$-component of the Jacobian of a Shimura curve. However, comparing the Galois action on CM points with the adelic action on the Jacobian, we see that this $\chi$-eigenpart can be non-trivial only when the restriction of $\psi$ to $\mathbb A_{\mathbb Q}$ is equal to $\chi$, or equivalently only in the self-dual case. This is proved for instance in Cornut, Christophe; Vatsal, Vinayak Nontriviality of Rankin-Selberg L-functions and CM points. Lemma 4.6.

Note also that this is what we should expect from the conjectures on special values of $L$-functions applied to $L(f/K,\psi,s)$ when $f$ is not self-dual. In that case, the conjecture implies that $L'(f/K,\psi,s)$ should be related to cohomology classes in the dual of the motive of $f$. Only in the self-dual case does these collapse in a formula involving the height of a point. Finally, in the situation you describe, even though $L(f,\psi,s)$ might vanish at 1, it is is expected that it doesn't generically (say in a relevant $\mathbb Z_{p}$-extension), so the conjectural relation between $L(f/K,\psi,s)$ (or its Selmer group) and putative point could hold only "locally at the specialization corresponding to $f$" in a $p$-adic family of automorphic representation containing $\pi(f)\otimes\psi$. All the arguments that I know relating these objects would then simply vanish.

Now an argument from ignorance is not a very good one, and I would very much like to be proven false, if only to learn something. Hidden behind all this is the question of the link between Kato's Euler system and rational points on modular varieties. The link is mysterious to me, but David Loeffler and Sarah Zerbes have some ideas.

share|improve this answer
    
Thanks, Olivier, this is the kind of ideas I was looking forward to discuss. The reasons you pose make me agree with you that "it is not straight-forward" 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!). –  Victor Rotger Oct 8 '12 at 21:05
    
Just to expand on one of Olivier's remarks: the cases where one can prove a formula for the derivative of an L-function are more or less restricted to ones where the L-function vanishes for "obvious reasons" (e.g. sign-induced vanishing at central critical values for self-dual motives with sign -1; or the trivial zeros of Rankin-Selberg L-functions at s=1 and of modular form L-functions at s=0, corresponding respectively to the Beilinson-Flach and Beilinson-Kato elements). This "obvious" vanishing will, as Olivier remarks, be robust enough to happen generically in a p-adic family. –  David Loeffler Oct 9 '12 at 9:13
1  
... When the L-value vanishes "coincidentally", it is probably going to be much more difficult to prove a formula for the derivative in terms of heights etc; it would probably more closely resemble the case of elliptic curves of rank $\ge 2$, where we certainly expect BSD to hold but nobody knows an explicit construction for the necessary points. –  David Loeffler Oct 9 '12 at 9:15
    
I just edited the question in order to focus it on the aspects I would like to learn more about. –  Victor Rotger Oct 23 '12 at 13:20
    
But it's not displaying well, I don't understand what I'm typing wrong. –  Victor Rotger Oct 24 '12 at 11:08

Here is a more detailed version of my comment above.

I will consider only the setting of your Example 3, namely $E/\mathbf{Q}$ is an elliptic curve and $\chi$ is a Dirichlet character such that $L(E \otimes \chi,1)=0$ and $L'(E \otimes \chi,1) \neq 0$.

Let $m \geq 1$ be an integer. The base change $L$-function $L(E \otimes \mathbf{Q}(\zeta_m),s)$ is the product of twisted $L$-functions $L(E \otimes \chi,s)$ where $\chi$ runs through the characters of conductor dividing $m$. The BSD conjecture for the base change $E \otimes \mathbf{Q}(\zeta_m)$ can be refined for each factor $L(E \otimes \chi,s)$. Roughly, the idea is that each arithmetical invariant appearing in the conjecture for $E \otimes \mathbf{Q}(\zeta_m)$ should factor in a way that reflects the decomposition of the motive $h^1(E \otimes \mathbf{Q}(\zeta_m)) = \bigoplus_{\chi} h^1(E \otimes \chi)$. The motive $M_\chi = h^1(E \otimes \chi)$ has dual $M_{\overline{\chi}}$, so it is self-dual only when $\chi$ is quadratic.

Assume $L(E \otimes \chi,s)$ vanishes at order one at $s=1$. The main invariant to consider is the discriminant of the Néron-Tate height pairing $\langle,\rangle$ on $E(\mathbf{Q}(\zeta_m)) \otimes \mathbf{R}$. We can extend this pairing to a positive definite hermitian form on $V=E(\mathbf{Q}(\zeta_m)) \otimes \mathbf{C}$. There is a decomposition of $V$ into isotypical components $V_\chi$ which are pairwise orthogonal with respect to the pairing. In this particular case we indeed expect $L'(E \otimes \chi,1) \sim \langle P_{\chi},P_{\chi} \rangle$ where $P_\chi$ is a generator of $V_\chi$. Usually $P_\chi$ takes the form $\sum_{\sigma} P^{\sigma} \otimes \chi(\sigma)$ for some $P \in E(\mathbf{Q}(\zeta_m))$.

In order to convince you this is reasonable, here is an example I computed using Magma. The rank $0$ elliptic curve $E=X_0(20):y^2 = x^3 + x^2 + 4x + 4$ acquires two independent points of infinite order over the cubic extension $\mathbf{Q}(a)$ with $a=2\operatorname{cos}(2\pi/9))$. Letting $P=(a+1,2a+3)$, we check numerically that $L'(E \otimes \chi,1) \sim \Omega_E \cdot \langle P_\chi,P_\chi \rangle$. The fudge factor appears to be an algebraic number of degree 6.

E:=EllipticCurve("20a1");
G:=FullDirichletGroup(9);
chi:=(G.1)^2; // Character of conductor 9 and order 6
LEchi:=TensorProduct(LSeries(E),LSeries(chi)); // L-Series L(E \otimes \chi,s)
Evaluate(LEchi,1);
//2.22329881577004394873515961159E-30
LEchi1:=Evaluate(LEchi,1:Derivative:=1);
LEchi1;
//2.78851510267155729197040153856 + 0.491690448714030907428058875920*$.1

Q<x>:=PolynomialRing(Rationals());
K<a>:=NumberField(x^3-3*x+1); // Cubic sufield of Q(zeta_9)
EK:=BaseChange(E,K);

G,_,map:=AutomorphismGroup(K); // Galois group of K
P:=[EK![(map(g))(a)+1,2*(map(g))(a)+3] : g in G]; // Set of conjugates of P=[a+1,2a+3] with a=2cos(2pi/9)
M:=HeightPairingMatrix(P);

C:=ComplexField();
PchiPchi:=C!M[1,1] + C!chi(2)*C!M[1,2] + C!chi(4)*C!M[1,3];

ratio:=LEchi1/(Periods(E)[1]*PchiPchi);
PowerRelation(ratio,6);
//27*x^6 - 9*x^3 + 1

I'm not sure about the situation for general abelian varieties $A_f/\mathbf{Q}$, but I guess some of the above might extend.

Of course, the real question seems to be whether or not these points $P_\chi$ are related in some way to Heegner points!

share|improve this answer
    
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? –  Victor Rotger Oct 26 '12 at 9:27
    
Yes, these are the questions. I'm not trained enough in GZ computations so I would still need to flesh out these questions in some more precise way... –  François Brunault Oct 26 '12 at 11:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.