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(Edit: I have asked this question on MO.)

The Gross-Zagier formula and various variations of it form the starting point in most of the existing results towards the Birch and Swinnerton-Dyer conjecture. It relates the value at $1$ of the derivative of the $L$-function of an elliptic curve $E$ to the canonical height of a very special rational point on $E$ called a Heegner point.

Let me describe the Gross-Zagier formula in the simplest case.

Setup. Let $E/\mathbf Q$ be an elliptic curve. According to the modularity theorem, there exists a finite $\mathbf Q$-morphism $p :X_0(N) \to E$ mapping the cusp $\infty$ to the origin of $E$, where $N$ is the conductor of $E$. Let $K\subseteq \mathbf C$ be an imaginary quadratic field, other than $\mathbf Q[i]$ or $\mathbf Q[\sqrt{-3}]$, in which the prime factors of $N$ split. Then we can choose an ideal $\mathcal J$ in $\mathcal O_K$, such that $\mathcal O_K/J \simeq \mathbf Z/N\mathbf Z$. Viewing $\mathcal J$ as a lattice in $\mathbf C$, we can form the elliptic curves $C_\mathcal J = \mathbf C/\mathcal J$ and $C_K = \mathbf C/\mathcal O_K$. While they are a priori elliptic curves over $\mathbf C$, we know from the theory of complex multiplication that they are actually defined over the Hilbert class field $H$ of $K$. Anyways, the inclusion $\mathcal J \subseteq \mathcal O_K$ induces by passage to the quotient an isogeny $C_\mathcal J \to C_K$ of degree $N$, which is precisely the kind of gadget that $X_0(N)$ parametrizes as a moduli space. Thus we get a point in $X_0(N)(H)$. We can take the image of this point via the modular parametrization $p$ to get a point $y_J \in E(H)$. Taking the sum of its Galois conjugates down to $K$ we get a point $y_K \in E(K)$, which, it turns out, doesn't depend on the choice of $\mathcal J$, up to sign and up to torsion. This means that its Néron-Tate height $\hat{h}(y_K)$ is a well-defined non-negative real number, which, by the non-degeneracy of the height pairing, is zero if and only if $y_K$ is a torsion point on $E$.

With this setup, Gross and Zagier proved:

Theorem: $$L'(E/K, 1) = \hat{h}(y_K) \frac{\iint_{E(\mathbf C)} \omega \wedge i\omega}{\sqrt D},$$ where $\omega$ is the invariant differential on $E$ (suitably normalized), and $D$ is the discriminant of $K$.

Since the factor $\frac{\iint_{E(\mathbf C)} \omega \wedge i\omega}{\sqrt D}$ is never zero, we have:

Corollary: Suppose that $L(E/K, s)$ has a simple zero at $s=1$. Then $y_K$ is a point of infinite order in $E(K)$, and in particular, $\text{Rank}_\mathbf{Z}E(K)>0$.

Thus, we have an example of a situation where the vanishing of $L(E/K, s)$ at $s=1$ implies the existence of an (explicit!) point of infinite order in $E(K)$ -- a behavior which is, of course, predicted by the BSD conjecture (minus the "explicit" part, which comes as a surprise).

If we believe BSD, it is natural to make the following:

Conjecture (Gross-Zagier): Suppose that $L'(E/K, 1)$ is nonzero, or equivalently, that $\hat{h}(y_K)$ is nonzero, or still equivalently, that $y_K$ is non-torsion. Then $E(K)$ has rank exactly one, and $\left<y_K\right>$ has finite index in it.

This conjecture was proven by Kolyvagin.

In any case, since $\hat{h}(y_K) = \left<y_K, y_K\right>$ where $\left<,\right>$ is the Néron-Tate height pairing on $E$, we can rewrite the Gross-Zagier formula as

$$L'(E/K, 1) = \left<y_K, y_K\right> \times C$$ where $C$ is the nonzero constant of the formula.


Let me now tell you about something completely different, namely the Minakshisundaram-Pleijel zeta function. (I don't know enough about this, so please forgive any inaccuracies.) Given a compact Riemannian manifold $M$, one has the Laplace-Beltrami operator $\Delta$ on $M$ which generalizes the familiar Laplacian on $\mathbf R^n$. The spectrum of this operator is an important invariant of $M$, and it is encoded in the M-P zeta function $$\zeta(\Delta, s) = \sum_{n=1}^\infty \lambda_n^{-s}.$$ It admits a meromorphic continuation to the whole plane, and is holomorphic at $s=0$. Its Taylor coefficients at $0$ around contain geometric data about $M$. Let us specialize to the case where $M$ is a surface; then $$\zeta'(\Delta, 0) = \frac{1}{12}\int_M K dA$$ where $K$ is the Gaussian curvature. According to the Gauss-Bonnet theorem, this is essentially the Euler characteristic, so: $$\zeta'(\Delta, 0) = \chi(M) \times C.$$ But there is a more suggestive way of writing the Euler characteristic. The diagonal $D \subseteq M \times M$ determines a cohomology class (its Poincaré dual) on $M \times M$; moreover it lies in the piece of the cohomology of $M\times M$ which is self-dual with respect to Poincaré duality, so the expression $\left<D, D\right>$ makes sense (it is the "self-intersection" number of the diagonal). In fact, as is well-known, $\left<D, D\right>$ is precisely the Euler characteristic of $M$. So, we can write $$\zeta'(\Delta , 0) = \left<D, D\right> \times C.$$

Corollary: If $\zeta'(\Delta, 0) \neq 0$, then $D$ is a nontrivial cohomology class on $M\times M$.

Thus, we have a (completely different!) example of a situation where the non-vanishing of the derivative of a zeta function implies the existence of a non-trivial cohomology class which "accounts" the non-vanishing (viewing a rational point on $E(K)$ as a degree $0$ Galois cohomology class).

Questions:

  1. Is the similarity between the two formulas, and between their Corollaries, coincidental? (This question probably doesn't have a precise answer, but I feel that it's worth asking.) And, assuming that the answer to this question is not completely disappointing...

  2. Have things like this been pointed out before? Are there more examples of formulas outside of number theory which bear resemblance to the Gross-Zagier formula?

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    $\begingroup$ Nice question! Perhaps you should post this to MO? $\endgroup$ – user27126 Jan 2 '14 at 8:46
  • $\begingroup$ @Sanchez Thank you! I will post it there if it doesn't do well here. $\endgroup$ – Bruno Joyal Jan 2 '14 at 8:47
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    $\begingroup$ Great question and thanks for posting to M.SE. Although I certainly can't provide an answer I really enjoy seeing questions like these which I probably wouldn't see if they were only posted on MO. Looking forward to an answer. $\endgroup$ – Dylan Yott Jan 6 '14 at 0:39
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This question has been cross posted at https://mathoverflow.net/q/159136/5687 and has a couple of answers there.

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