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

Question

I am reading Tate's Thesis. Tate derives a theorem which he calls "the number-theoretic analogue of the Riemann-Roch theorem" from an abstract Poisson summation formula. I am accustomed to thinking of the Riemann-Roch theorem as a statement about the dimension of global sections of invertible sheaves over a nonsingular projective curve.

Can you help me understand the connection between Tate's theorem and the theorem for curves?

The theorem for curves: Let $k$ be an algebraically closed field. Let $C$ be a nonsingular projective curve over $k$. Let $\mathcal{L}$ be an invertible sheaf on $C$. Let $\Omega^1$ be the invertible sheaf of 1-forms on $C$. Then

$$h^0(C,\mathcal{L}) - h^0(C,\Omega^1\otimes \check{\mathcal{L}}) = d-g+1$$

where $d$ is the degree of $\mathcal{L}$, and $g$ is the genus of $C$, defined as $h^0(C,\Omega^1)$.

Tate's theorem: Let $k$ be a number field and let $V$ be its adele ring. Let $U$ be the idele group (i.e. the units of $V$). Let $D$ be a fundamental domain of $V$ for the discrete action of $k$. Let $f:V\rightarrow\mathbb{C}$ be a continuous, $L^1$ function. Let $\hat f$ be its fourier transform. Let $|\cdot|$ be the canonical absolute value on $U$, which is the product of the local absolute values, each appropriately normalized so the product is trivial on $k$. If $f$ satisfies

• $\sum_{\xi\in k}f(a(x+\xi))$ is convergent for all $a\in U, x\in V$, and convergence is uniform for $x\in D$
• $\sum_{\xi\in k} |\hat f(a\xi)|$ converges for all $a\in U$

then

$$\frac{1}{|a|}\sum_{\xi\in k} \hat f(\xi / a) = \sum_{\xi\in k} f(a\xi)$$

How are these two theorems related?

Thoughts: Is there a way to think of an invertible sheaf as a continuous $L^1$ function on the adele ring of the curve? If so, I guess the fourier transform is related to the dual sheaf?

Answer 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.)