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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 special rational point on $E$.

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:

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: $$\zeta'(\Delta, 0) = \chi(M) \times C, \qquad C\neq 0.$$ But there is a more suggestive way of writing the Euler characteristic. The diagonal $D \subseteq M \times M$ determines a cohomology class 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). 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 an explicit non-trivial cohomology class which "accounts" for the non-vanishing (in the first case, viewing a rational point on $E$ as a degree $0$ Galois cohomology class).

As René points out, $D$ is always nontrivial, for trivial reasons. What is interesting to me is not so much the nontriviality of $D$, but rather that one can deduce this nontriviality from $\zeta'(\Delta, 0)\neq 0$.

Another important difference that I should point out betweeen the two situations is that $\zeta'(\Delta, 0)\neq 0$ does not imply $\zeta(\Delta, 0)= 0$, because there is no analogue of BSD. In fact, $\zeta(\Delta, 0)$ is essentially $\mathrm{vol}(M)$.

Nevertheless, I am still curious.

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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Are there examples where the class of $D \subset M \times M$ in the cohomology of $M \times M$ is trivial? (Since it has non-zero intersection number with $M \times \{ \operatorname{pt} \}$, I am tempted to think the answer can't be yes.) –  René Mar 2 at 16:59
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@René Good point. In any case, it is the resemblance between the two formulas which I am really curious about. –  Bruno Joyal Mar 2 at 19:14
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