# BSD leading-term coefficient in terms of places without distinction

After reading this blog post, I learned the BSD conjectural formula for the coefficient of the leading term $a_0$ of the L-function of an elliptic curve $E$, namely $$a_0 \stackrel{?}{=} \frac{\Omega_E\cdot Reg_E \cdot \prod_p c_p \cdot \#Sha(E/\mathbb{Q})}{(\# E_{tors}(\mathbb{Q}))^2}$$ All the terms are defined, if people are interested and don't already know, at the above link. Now the factors in the numerator here come in several flavours: the real period $\Omega_E$ arises after looking at the curve over $\mathbb{R}$; the numbers $c_p$ are 1 for all but finitely many primes $p$, and come from looking at the curve over $p$-adic numbers, hence completions of $\mathbb{Q}$; $Reg_E$, the regulator, is the volume of a certain torus (not the curve itself!) given by comparing the rational and real points of $E$; the Sha group arises from comparing Galois cohomology over $\mathbb{Q}$ with all its completions at finite primes. Clearly the Archimedean place and the non-Archimedean ones behave differently, but one can often unify them in certain formalisms. (If one is willing to split the denominator, and invert the regulator, then it is a product of three ratios, each of which is something like (something about a completion)/(some sort of volume), but this just extremely flaky and ignorant, and best ignored)

My question is this: can we write this product (or perhaps the whole quotient) more uniformly via reinterpreting various terms in more abstract ways via places?

Please note I'm not trying to do anything with such a formulation, I'm just curious if it is known.

EDIT: the factor $\Omega_E/\# E_{tors}(\mathbb{Q})$ is the volume of the stack given by the action groupoid $E(\mathbb{Q})\otimes\mathbb{R}//E(\mathbb{Q})$. Do the other terms measure other geometric objects, such that the whole thing is the measure of some adelic object?

• Tate's original forumlation archive.numdam.org/ARCHIVE/SB/SB_1964-1966__9_/… using adelic integration uniformises $\Omega_E\cdot \prod c_p$ into one expression. Since the regulator measures something on the Mordell-Weil group modulo torsion, one could also consider the quotient of $\mathrm{Reg}_E/(\#E(\mathbb{Q})_{\mathrm{tors}}))^2$ as one thing. – Chris Wuthrich Jul 30 '14 at 9:26
• Yet, in the end, whatever formulation you will see, there will always be the input of several sides. The Bloch Kato conjecture will also put together comparison isomorphisms etc to get to a formulation. So I don't think there is a single expression that covers the full quotient. – Chris Wuthrich Jul 30 '14 at 9:28
• A single expression is perhaps too ambitious, but treating the different places in a uniform way is really what would be nice. – David Roberts Jul 30 '14 at 11:22
• First, you say denominator, but mean numerator. Second, I don't at all agree that the regulator compares the rational and the real points of $E$. But the canonical height (when written as a sum of local heights) is a sort of average measure of a rational point over how it sits in $E(\mathbb{Q}_v)$ for all places $v$, including the real place. But it's certainly not just the real place. I'm not sure whether this helps or hurts what you're looking for. – Joe Silverman Jul 30 '14 at 11:26
• What would be your uniform interpretation for the class number formula ? – Chris Wuthrich Jul 30 '14 at 11:28

## 2 Answers

The global Tamagawa number formula is pretty simple and it unifies BSD with the class number formula and the Siegel mass formula.

The Tamagawa number $\tau$ of an algebraic $G$ group over $\mathbb Q$ is the volume of $G(\mathbb A)/G(\mathbb Q)$, or a proxy thereof*. The conjecture is that it is Pic over Sha. This is a theorem for linear algebraic groups and equivalent to BSD for abelian varieties. This was conjectured by Weil in the simply connected case, where the answer is particularly nice: 1. But tori and abelian varieties do not have a simply connected cover; and the mass formula is about a group that is not simply connected.

What is the measure on the adelic group? It is a Haar measure, but that is unique only up to a real number. Tamagawa showed that there is a canonical normalization, Tamagawa measure. First choose the Haar measure additive group of the adeles so $\mathop{vol}(\mathbb A/\mathbb Q)=1$, which is also the obvious measure. For more general groups, an invariant differential form gives a local identification of the group with a power of the additive group, allowing the measure to be transferred. (Well, that's the heuristic, but I think it does not directly work with adeles, but only for local fields, which is enough.) General adelic differential forms still produce all Haar measures, but the product formula says that multiplying a differential form by an element of $\mathbb Q^*$ does not change the measure. Thus differential forms defined over $\mathbb Q$ yield a single canonical normalization of Haar measure.

The class number formula for a number field $K$ is equivalent to 1 being the Tamagawa number for a particular torus, (the Weil restriction of) the multiplicative group over $K$. To avoid infinities, you could replace that with the kernel of the norm map to $\mathbb G_{m,\mathbb Q}$, analogous to dividing the zeta function of $K$ by the zeta function of $\mathbb Q$. The Siegel mass formula can be reduced to the Tamagawa number for the appropriate orthogonal group, which I believe was Tamagawa's original motivation.

I don't know why the formula is Pic over Sha. That is a multiplicative factor necessary for good behavior in group extensions, including isogenies. Actually, I did figure out the role of Sha, namely that a better object than $G(\mathbb A_{\mathbb Q})/G(\mathbb Q)$ is $(G(\mathbb A_{\mathbb K})/G(K))^{\mathop{Gal}(K/\mathbb Q)}\cong G(\mathbb A_{\mathbb Q})/G(\mathbb Q)\times Ш_{G,\mathbb Q}$. I suspect that the explanation of Pic is that we should replace the space with a stack with that isotropy. Also, what shows up is in the abelian variety case is not the whole Picard group, but the torsion part.

* Proxy for measure: In the semisimple case, $G(\mathbb Q)$ is a lattice in $G(\mathbb A)$ and everything is great. In the case of $\mathbb G_m$, there is infinite covolume, and $\tau(\mathbb G_m)$ is set to $1$ by fiat. For abelian varieties of positive rank, the rational points are not discrete, so volume is problematic, but Bloch showed that there is a canonical extension of a rank $n$ abelian variety by $\mathbb G_m^n$ so that the Tamagawa number conjecture for that group is equivalent to BSD for the original. One often defines the Tamagawa number of the original abelian variety as the honest Tamagawa number of the extension.

• Instead of setting the Tamagawa number of $\mathbb G_m$ to 1, it is probably better to leave it infinite, but declare it a pole with residue 1. Then this formulation covers not just the BSD leading term, but also the order of vanishing. – Ben Wieland Aug 5 '14 at 17:06