L-function of a curve and L-function of its jacobian

Question

When are the L-function of a curve and the L-function of its jacobian equal?

Answer 1

I assume you're working with $C/K$ some number field, a geometrically integral projective smooth curve. Note that it suffices to think about individual $L$-factors, and so we might as well examine the $\zeta$-function of $C$ and $\text{Jac}(C)$ over a finite field $\mathbb{F}_q$.

The answer then is, as you may have expected, only when $C$ is a genus $1$ curve. Indeed, recall from Grothendieck's trace formula (which is easier for curves and abelian varieties!) that:

$$\zeta(C,t)=\frac{P_1(t)}{P_0(t)P_2(t)}$$

where

$$P_{0,C}(t)=1-t, \qquad P_{2,C}(t)=1-qt$$

and

$$P_{1,C}(t)=\det(1-\text{Fr}\,t\mid H^1_{\acute{e}\text{t}}(C_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell))$$

where $\text{Fr}$ is the geometric Frobenius acting on the $G_{\mathbb{F}_q}$-module $H^1_{\acute{e}\text{t}}(C_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell)$.

Similarly, if $g$ is the genus of $C$, so the dimension of $\text{Jac}(C)$, we can write

$$\zeta(\text{Jac}(C),t)=\frac{P_{1,\text{Jac}(C)}(t)\cdots P_{2g-1,\text{Jac}(C)}(t)}{P_{0,\text{Jac}(C)}(t)\cdots P_{2g,\text{Jac}(C)}(t)}$$

where

$$P_{i,\text{Jac}(C)}(t)=\det(1-\text{Fr}\, t\mid H^i(\text{Jac}(C_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell))$$

with notation as before.

Now, note that one can compare some of these terms. Namely, it's well known that

$$H^1(C_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell)=(V_\ell\text{Jac}(C_{\overline{\mathbb{F}_q}}))^\vee=H^1(\text{Jac}(C_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell)$$

so that

$$P_{1,C}(t)=P_{1,\text{Jac}(C)}(t)$$

Much easier to see, is that

$$P_{0,C}(t)=1-t=P_{1,\text{Jac}(C)}(t)$$

Now, if $g>1$, then there's no way that these two series could be equal. Indeed, one has more terms than the other. Of course, one might be concerned that some amount of 'cancellation' might happen on $\zeta(\text{Jac}(C),t)$ to equal the two out. But, the Weil Conjectures (which are also much easier for curves an abelian varieties!) says that this is not the case.

So, if $g>1$, can't happen. If $g=1$ then $C\cong \text{Jac}(C)$ non-canonically. Indeed, this is clear as long as $C$ has a point. But, by the Weil estimates we know that

$$|C(\mathbb{F}_q)-q-1|\leqslant 2\sqrt{q}$$

in particular, we can't have $C(\mathbb{F}_q)$ else we'd have that

$$q+1\leqslant 2\sqrt{q}$$

which, I'm pretty sure you can check, never happens. :)