How to calculate the local factor at the infinite place of a function field?

Question

First of all, my apologies for the long-winded nature of this question!

Yesterday, Mr. Barquero asked an excellent question regarding function fields and number theory: Why is it "easier" to work with function fields than with algebraic number fields?

I had a follow-up question of my own: Starting with the simplest kind of function field, i.e. $\mathbb F_q(T)$ where $q$ is a prime, we can calculate the zeta function over the ring $\mathbb F_q[T]$ to be:$$\Gamma_{\mathbb F_q[T]}(s)=\sum_{\mathfrak a \neq 0}N\mathfrak a^{-s}=\sum_{d=0}^\infty q^d(q^d)^{-s}=1/(1-q^{1-s})$$ In order to write down a functional equation, I need to include a local factor corresponding to the infinite place.

My question is: How to do that correctly, in the machinery used in Tate's thesis, and without direct recourse to the Riemann-Roch theorem?

I know the answer I am looking to get to is this:$$\xi(s)=\frac{q^{-s}}{(1-q^{-s})(1-q^{1-s})}$$

I can get part of this. Similarly to the calculation in the global field $\mathbb Q$, $$\int_{\mathbb R^\times} e^{-\pi x^2}|x|^{s-1}dx=\pi^{-s/2}\Gamma(s/2),$$

I can write down the local zeta factor:$$\sum_{d=0}^\infty q^{-sd}=(1-q^{-s})^{-1},$$ but I'm not sure how to arrive at the factor $q^{-s}$ in the numerator of the expression for $\xi(s)$. Is this a local epsilon factor? Is there an elementary way to calculate it without invoking the machinery of Riemann-Roch?

Answer 1

I wanted to follow up Professor Garrett's excellent answer with the explicit calculation. I hope it will be useful since most of the texts covering Tate's thesis concentrate on the number field case, without really giving example calculations for function fields. Also, I'll point out a little error in one of the frequently used texts for studying Tate's thesis.

Our field here is $K=\mathbb F_p(T)$, with $p$ a prime. This is the most basic example of a function field. The zeta function given above only includes the finite places. We need another local factor for the infinite place in order to write down a completed zeta function $\xi(s)$ that will have a functional equation $\xi(s)=\xi(1-s)$.

To get a feel for what happens in the usual number field case, see Terry Tao's blog post: http://terrytao.wordpress.com/2008/07/27/tates-proof-of-the-functional-equation/

The absolute value associated to the infinite place of $K=\mathbb F_p(T)$ is $|f/g|_\infty=p^{deg(f)-deg(g)}$, which is non-Archimedean. The completion of $K$ at the infinite place is the field of Laurent series: $$K_\infty=\mathbb F_p((T^{-1}))=\{\sum_{n=-\infty}^ra_nT^n|r\in \mathbb Z, a_n\in \mathbb F_p, a_r \neq 0\}$$ The local ring of integers at the infinite place is the ring of power series $\mathcal O_\infty=\mathbb F_p[[T^{-1}]]$.

The next step is to write down the local zeta function: $$Z(f,\chi)=\int_{K_\infty^\times}f(x)\chi(x)d^\times x$$ with $f(x)$ an appropriately chosen function that will be its own Fourier transform (like $e^{-\pi x^2}$ in the usual number field case). We'll put $\chi(x)=|x|_\infty^s$. The measure $d^\times x$ is a multplicative Haar measure normalized to give volume $1$ on $\mathcal O_\infty^\times$.

Now, in order to choose the right $f(x)$, we need to be able to take it's Fourier transform, and in order to do that, we need an additive character $\psi_\infty(x)$ on $K_\infty$. The motivation for the choice of this character is given in Professor Garrett's answer. We define $\psi_\infty(x)$ to be: $$\psi_\infty(\sum_{-\infty}^r a_nT^n)=e^{2\pi ia_{-1}/p}$$

(Here I should note the mistake on Ch 7, Exercise 3(b), pg 298 of Ramakrishnan's and Valenza's Fourier Analysis on Number Fields, where they give $a_1$ instead of $a_{-1}$. I'd recommend Weil's Basic Number Theory which gives a good account of this on pg 67.)

Now let $f(x)$ be the characteristic function on $T^{-1}\mathcal O_\infty=T^{-1}\mathbb F_p[[T^{-1}]]$. The Fourier transform of $f(x)$ is: $$\hat{f}(\xi)=\int_{K_\infty}f(x)\overline{\psi_\infty}(\xi x)dx$$ It can be checked that this gives again the characteristic function on $T^{-1}\mathcal O_\infty$, i.e. $f(x)$ is self-dual.

Now we can calculate the local zeta factor:$$Z(f,\chi)=\int_{T^{-1}\mathcal O_\infty \setminus 0}|x|_\infty^sd^\times x$$ Since $$T^{-1}\mathcal O_\infty \setminus 0= \bigsqcup_{j=1}^\infty T^{-j} \mathcal O_\infty^\times$$ we have, $$Z(f,\chi)=\sum_{j=1}^\infty|T^{-j}|_\infty^s=\sum_{j=1}^\infty p^{-js}=\frac{p^{-s}}{1-p^{-s}}$$ So finally we have the completed zeta function, $$\xi(s)=Z(f,\chi)\Gamma_{\mathbb F_p[T]}(s)=\frac{p^{-s}}{(1-p^{-s})(1-p^{1-s})}$$ which satisfies the functional equation $\xi(s)=\xi(1-s)$.

Answer 2

Yes, it fits in the Tate (-Iwasawa) pattern: the net issue is getting the family of normalizations over all primes consistent. The local ring of integers "at infinity", $\mathbb F_q[T^{-1}]$, by itself is indistinguishable from $\mathbb F_q[T]$, but that's potentially misleading (e.g., won't get the numerator as the question notes). The family of Haar measures should give the adeles measure 1, but any flaw there won't give a $q^{-s}$. Normalization/choice of the additive measures won't give that, either, tho' might change the details on what it takes for a local Schwartz function to be mapped to itself under F.T. (for "optimal" choice).

Edit: oop, and the local additive characters, too!

Edit: in response to further comment. There is certainly much opportunity for cognitive dissonance with $\mathbb F_q[T^{-1}]$. Let's see whether this works out straightforward-ly.

A reasonable collection of local characters at finite places corresponding to irreducible monic $P_v$ should be the "residue" of $f$ at $v$. In the function-field case, unlike the number field case, we have canonical representatives for $\mathbb F_q[T]/P_v$, namely, polynomials of lesser degree. Then the "residue" of a Laurent expansion $\sum_{n\ge -N} q_n/P_v^n$ is the coefficient of $T^{\deg P_v - 1}$ in $q_{-1}$. Then take Galois trace to the prime field $\mathbb F_p$, and feed the result into $x\rightarrow e^{2\pi ix/p}$. These characters are trivial on the local integers $o_v$ at all finite places, and non-trivial on $P_v^{-1}o_v$, and match expectations.

(There is a geometric heuristic, too, that the sum of residues of a meromorphic function on projective 1-space is $0$, but this requires suitable interpretation of "residue... Still, in principle, this would explain what the character $\psi_\infty$ should be, too.)

Thus, apart from the degree-equal-characteristic case, $\psi_T(T^{-1})\not=1$. Since $T$ is a local unit at all other finite places, necessarily $\psi_\infty(T^{-1})\not=1$ for $\prod_{v\le\infty}\psi_v(x)=1$ for $x\in\mathbb F_q(T)$. That's pretty decisive.

But/and $\psi_T(T^{-2})=1$, which explains (modulo a small exercise) why $\psi_\infty$ is trivial on $(T^{-1})^2\cdot o_\infty$, with $o_\infty$ the local ring of integers at infinity (= formal power series in $T^{-1}$).

Thus, the FT of the char fcn of $o_\infty$ cannot be itself again (nor a constant multiple).

Hopefully this better indicates the origin of the numerator. The clincher (for me) is that the additive char at infinity cannot be quite what one might have thought. (And, again, I think that the char fcn of $T^{-1}o_\infty$ is, up to a measure constant, its own FT, but not so for the char fcn of $o_\infty$.)

Edit 2: Perhaps it is worth noting that the "residue of differential form" viewpoint does also suggest the correct normalization at $\infty$, namely, the differential form is $f(x)\,dz$, properly. Thus, at $\infty$, the $dz$ must be rewritten in terms of the local coordinate, $1/z$! Thus, $f(1/z)d(1/z)$ near $z=0$ is $f(1/z)\frac{-1}{z^2}dz$, which suggests both the sign flip and the "unexpected" shift in the additive character.