Tate's Thesis: in what sense is Tate's Theorem 4.2.1 the Riemann-Roch theorem for curves?

Solution 1:

If you work in the function field case, i.e. replace the number field $K$ and its places $v$ by a finite extension of $k[x]$ for some finite field $k$, and work with its places, then this statement becomes the Riemann--Roch theorem.

The appearance of $1-g$ in the RR thm., and the role of the canonical bundle in forming the correct kind of dual, will here be absorbed into the the definition of the self-dual measure on the adeles.

To be a little more precise, you should imagine that your line bundle is of the form $\mathcal L(D)$ for some divisor $D$ (it always is, after all!); then $a$ will play the role of $D$ (or maybe $-D$). And you should take $f$ to the characteristic function of the integral adeles. Then one side of the equality will count rational functions $\xi$ for which $\xi a^{-1}$ has no demoninators (so global sections of $\mathcal L(D)$), and the other side will count rational functions $\xi$ such, roughly, $\xi a^{-1}$ is integral, which is the global sections of $\mathcal L(-D)$; except that $f$ is not quite self-dual. In the number field case the different comes in, and in the function field case we are considering here, the canonical bundle will come in (as well as a factor related to $1-g$).

Finally, to get the familiar statement about dimensions, take log of both sides and divide by $\log q$ (where $q = |k|$).

(You can check then that the $|a|^{-1}$ on the LHS, after taking logs, gives the $\deg D$ term.)