What does extra zero of an $L$-function mean?

Question

This is a very vague question. What does an extra zero of an $L$-function mean? There are lots of papers written on this topic, investigating the extra/exceptional zeros of various $p$-adic $L$-functions. What is all this about? Thank you!!

Answer 1

An equally vague response: there are a series of conjectures (BSD, Bloch-Beilinson, Bloch-Kato, etc.) that predict, among other things, that the order of vanishing of the L-function attached to a motive at its central point is equal to the rank of some higher Chow group or (Bloch-Kato) Selmer group.

In order to study this, you might attempt to interpolate the complex L-values to get a corresponding $p$-adic L-function. In this setting you might expect the same behavior, specifically that the $p$-adic L-function vanishes to the same order. Most of the time you'd be right, but occasionally you get "exceptional zeros" coming from some extra factor in the interpolation process, and so the $p$-adic L-function vanishes to higher order.

In a simple setting, suppose $L(E,s)$ is the L-function attached to an elliptic curve $E/\mathbf{Q}$. The BSD conjecture predicts that $\mathrm{rk} \, E(\mathbf{Q}) = \mathrm{ord}_{s=1} L(E,s)$. The $p$-adic L-function $L_p(E,s)$ is constructed in such a way that $$ L_p(E, 1) = \left(1 - \frac{1}{a_p}\right) \frac{L(E,1)}{\Omega_E}, $$ where $\Omega_E$ is a nonzero constant, and $a_p = p + 1 - \#E(\mathbf{F}_p)$ for a prime $p$ of good reduction, or $a_p = 1$ (resp. $-1$) for primes of split (resp. non-split) multiplicative reduction. In particular, if $E$ has split multiplicative reduction at $p$, then the extra interpolation factor is always zero, and so one expects $$\mathrm{ord}_{s=1}L_p(E,s) = \mathrm{ord}_{s=1} L(E,s) + 1.$$ This is the "exceptional zero conjecture" of Mazur-Tate-Teitelbaum, later proved by Greenberg-Stevens. Since then, several more exceptional zero conjectures have been formulated precisely (for other objects), and some have been (partially) solved.