# Lecture 15: The $L\dash$Polynomial

## Big List of Important Facts 

Recall that we had $Z(t) + F(t) + G(t)$:

\[  
(q-1) F(t) 
&= \sum_{0 \leq \deg C \leq 2g-2 } q^{\ell(C)} t^{\deg(C)} \\
(q-1)G(t)
&= h \qty{ {q^g t^{2g-1} \over 1-qt} - {1 \over 1-t} }
.\]

Note that $F(t)$ is a polynomial of degree at most $2g-2$, and clearing denominators in $G(t)$ yields a polynomial of degree at most $2g$


:::{.definition title="The $L\dash$polynomial"}
The $L\dash$polynomial is defined as
\[  
L(t) \da (1-t)(1-qt) Z(t) = (1-t)(1-qt) \sum_{n=0}^\infty A_n t^n \in \ZZ[t]
.\]
:::

It turns out that the degree bound of $2g$ is sharp, and the coefficients closer to the middle are most interesting:

:::{.theorem title="?"}
Let $K/\FF_q$ be a function field of genus $g\geq 1$, then

a. $\deg L = 2g$.
b. $L(1) = h$
c. $L(t) = q^g t^{2g} L\qty{1\over qt}$.
d. Writing $L(t) = \sum_{j=1}^{2g} a_j t^{j}$,

  - $a_0 = 1$ and $a_{2g} = q^g$.
  - For all $0\leq j \leq g$, we have $a_{2g-j} = q^{g-j}a_j$.
  - $a_1 = \abs{\Sigma_1(K/\fq)} - (q+1)$, which notably does not depend on $g$.

- Write $L(t) = \prod_{j=1}^{2g} (1 - \alpha_j t) \in \CC[t]$ [^why_this_expansion]

e. The $\alpha_j \in \bar{\ZZ}$ [^algebraic_ints_note]
(which were *a priori* in $\CC$) and can be ordered such that for all $1\leq j \leq g$, we have $a_j a_{g+j} = q$. [^hint_at_rh]

f. If $L_r(t) = (1-t)(1-q^rt) Z_r(t)$ then $L_r(t) = \prod_{j=1}^{2g}(1-\alpha_j^r t)$, where $K_r$ is the constant extension $K \fqr /\fqr$

[^hint_at_rh]: This is the first hint at the Riemann hypothesis: if for example they all had the same complex modulus, this would force $\abs{a_j} = \sqrt q$.
Thus proving that they all have the same absolute value is 99% of the content!

[^algebraic_ints_note]: $\bar \ZZ$ denotes the algebraic integers.

[^why_this_expansion]: The polynomial isn't monic, but rather has a constant coefficient, so this expansion is somewhat more natural than (say) $\prod (t-\alpha)$.

:::

Note that the $\alpha_j$ are reciprocal roots.

## Proofs

### The degree of $L$ and $L(1)$

:::{.proof title="of a"}
We saw from $Z(t) = F(t) + G(t)$ that $\deg L \leq 2g$.
Equality will follow from the proof of (d) part 1, since this would imply that $a_{2g} = q^g \neq 0$.
:::

:::{.proof title="of b"}
Our formula $Z(t) = F(t) + G(t)$ and Schmidt's theorem (showing $\delta = 1$) gives
\[  
L(t) = (1-t) (1-qt) F(t) + {h \over q-1} \qty{ q^g t^{2g-2} (1-t) - (1-qt)}
,\]
where we've expanded $G$ but not $F$ because it involves various $\ell(D)$ which are difficult to compute. 
It is some polynomial though, and we can evaluate $L$ at 1 to get $L(1) = h$.
Thus the class number is the sum of the coefficients!
:::

### Functional Equation

:::{.proof title="of c"}
This follows easily from the functional equation for $Z(t)$, which we already established using the Riemann-Roch theorem:
\[  
Z(t) = q^{g-1} t^{2g-2} Z\qty{1\over qt}
.\]
We can compute
\[  
q^g t^{2g} L\qty{1\over qt} 
&= q^g t^{2g} \qty{1 - {1\over qt}} \qty{1 - {1\over t}} Z\qty{1\over qt} \\
&= q^{g-1} t^{2g-2} (1-t) (1-qt) Z\qty{1\over qt} \\
&= (1-t) (1-qt) Z(t) \\
&\da L(t)
,\]
where we've distributed one $q$ and two $t$s in the first steps.
:::

### Coefficients $a_j$ for $j=0, 1, 2g$ and Duality

:::{.proof title="of d"}
Using the functional equation from (c), we can write
\[  
L(t) = q^g t^{2g}  L\qty{1\over qt} = \qty{a_{2g} \over q^g} + \qty{a_{2g-1} \over q^{g-1}}t + \cdots +  \qty{a_0 q^g} t^{2g}
,\]
where we're correcting by enough in $t$ but not enough in $q$ and seeing what we get.
Equating coefficients, for $0\leq j \leq g$ we have
\[  
a_{2g-j} = q^{g-j} a_j
\label{eq:sym_formula_proofc}
.\] 

Using the fact that $A_0$ is the number of effective degree zero divisors, which is only zero, we have $A_0 = 1$ and we can multiply formal power series to obtain 
\[  
L(t) = a_0 + a_1 t + \cdots + a_{2g} t^{2g} 
&= (1-t)(1-qt) \sum_{n=0}^\infty A_n t^n \\
&= \qty{ 1 - (q+1)t + qt^2 }(1 + A_1 t + A_2 t^2 + \cdots)\\
&= 1 + \qty{A_1 - (q+1) }t + \cdots
.\]
From this, we can read off

- $L(0) = a_0 = 1$
- $a_1 = A_1 - (q+1) = \Sigma_1(K/k) - (q+1)$
- $a_{2g} = a_{2g-0} = q^{g-0}a_0 = a^g$ by taking $j=0$ in \cref{eq:sym_formula_proofc}, and thus $\deg L = 2g$.

:::

### Absolute Values of Roots / RH

:::{.proof title="of e (the most interesting!)"}
Consider the **reciprocal polynomial** 
\[  
L^{\perp}(t) \da t^{2g} L\qty{1\over t}
= t^{2g} + a_1 t^{2g-1} + \cdots + q^g
.\]
The original polynomial had $\ZZ$ coefficients and constant term 1, so this polynomial is monic and has a nonzero constant term.
Thus its roots are patently nonzero algebraic integers in $\bar{\ZZ}\nonzero$.
If $L^{\perp}(t) = \prod_{j=1}^{2g} (t-\alpha_j)$, then 
\[  
L(t) = t^{2g} L^{\perp}\qty{1\over t} = \prod_{j=1}^{2g} (1 - \alpha_j t)
\]
and if the roots of $L(t)$ are $r_j$, then the roots of $L^{\perp}(t)$ are the reciprocal roots $1/r_j$ and vice-versa.
This shows the first assertion that $r_j \in \bar{\ZZ}$ as well.

The most interesting part is what follows.
Making the substitution $t=qu$ and using (c) we get
\[  
L^{\perp}(t)
&= \prod_{j=1}^{2g} (t- \alpha_j) \\
&\da t^{2g} L\qty{1\over t} \\
&= q^{2g} u^{2g} L\qty{1\over qu} && \text{by (c)}
.\]

Using $u = t/q$, we can write 
\[  
q^g L(u) 
&= q^g \prod_{j=1}^{2g} (1 - \alpha_j u) \\
&= q^g \prod_{j=1}^{2g} \qty{ 1 - {\alpha_j \over q}t} \\
&= q^g \prod_{j=1}^{2g} {\alpha_j \over q} \prod_{j=1}^{2g}\qty{ t - {1\over \alpha_j} } \\
&= \prod_{j=1}^{2g} \qty{t - {q\over \alpha_j}}
,\]
where we've pulled out a factor of $-\alpha_j/q$ and
in the last step we've used that $\prod_{j=1}^{2g} \alpha_j = q^g$. 
This follows because the $\alpha_j$ are the roots of $L^{\perp}$, which has even degree, so the product of all of the roots is equal to the constant term of $L^{\perp}$, which is the leading term of $L$, which we showed was $q^g$.

This says that if we take these roots $\alpha_j$ as a multiset and replace each $\alpha_j$ with $q/\alpha_j$, we get the same multiset back.
I.e., this multiset is stable under the involution
\[  
\CC\units &\to \CC\units \\
z &\mapsto {q\over z}
.\]
This almost pairs up the elements of this finite set of roots, except it may have fixed points.
The complex numbers $\alpha$ such that $\alpha = q/\alpha$ are precisely $\pm \sqrt q$.
So group the $\alpha_i^{-1}$ into

- $k$ **pairs** of nonfixed points, where $\alpha_i \neq q/\alpha_i$,
- $m$ points such that $\alpha_i = \sqrt q$,
- $n$ points such that $\alpha_i = -\sqrt q$.

So we'd like to show that $m$ and $n$ are both even, so when we're pairing roots with reciprocals these get paired with themselves.
We know $2k + m + n = 2g$, so $m+n$ is even.
We also know that 
\[  
q^g 
&= \prod_{j=1}^{2g} \alpha_j \\
&= q^k \qty{\sqrt{q}}^m \qty{-\sqrt q}^n  \\
&= (-1)^n q^{k + {m \over 2} + {n\over 2}} \\
&= (-1)^n q^g
.\]
This forces $n$ to be even, and since $m = 2g-2k-n$, $m$ must be even as well.
:::

:::{.proof title="of f"}
We used Dirichlet's character-style decomposition of $Z(t)$ in Schmidt's theorem, and we'll use it again here.
Write 
\[  
L_r(t^r) 
&= (1-t^r) (1-q^r t^r) Z_r(t^r) \\
&= (1-t^r) (1-q^r t^r) \prod_{\xi \in \mu_r} Z(\xi t) \\
&= (1-t^r) (1-q^r t^r) \prod_{\xi \in \mu_r} {L(\xi t) \over (1-\xi t)(1-q\xi t) } \\
&= \prod_{\xi \in \mu_r} L(\xi t) 
,\]

where we've used that
\[  
\prod_{\xi \in \mu_r} {1\over 1 - \xi t} &= 1-t^r \\
\prod_{\xi \in \mu_r} {1\over 1 - q\xi t} &= 1-q^rt^r \\
\]
which leads to all of the denominators canceling.
We can then expand $L_r(t^r)$ as a product to compute
\[  
L_r(t^r) 
&= \prod_{\xi \in \mu_r} L(\xi t) \\
&= \prod_{\xi\in \mu_r} \prod_{j=1}^{2g} (1- \alpha_j qt) \\
&= \prod_{j=1}^{2g} \prod_{\xi\in \mu_r} (1- \alpha_j qt)  && \text{since these are finite products}\\
&= \prod_{j=1}^{2g} (1 - \alpha_j^r t^r)
.\]

From this we can conclude that $L_r(t) = \prod_{j=1}^{2g} (1- \alpha_j^r t)$, since $t^r$ is just an indeterminate and these are all identities of polynomials.
:::

## Applications and Corollaries


### Counting Rational Points

:::{.corollary title="?"}
Suppose $K/\fq$ is genus $g\geq 1$ and $L(t) = \prod_{j=1}^{2g}(1- \alpha_j t)$.
Then for all $r\in \ZZ^{\geq 0}$, we have a nice expression for $N_r$:
\[  
N_r \da \abs{\Sigma_1(K_r/\fqr)} = q^r + 1 - \sum_{j=1}^{2g} \alpha_j^r
.\]
:::

:::{.proof title="?"}
Let $L_r(t) = \sum_{j=1}^{2g} a_{j, r} = \prod_{j=1}^{2g} (1 - \alpha_j^r t)$,
so $a_{1, r} = -\sum_{j=1}^{2g} \alpha_j^r$.
Then using (d) part 3, we can write
\[  
\abs{\Sigma_1(K_r/\fqr)} = q^r + 1 + a_{1, r} = q^r + 1 - \sum_{j=1}^{2g} \alpha_j^r
.\]
This follows from consider $\prod (1-\alpha_j^r t)$, where extracting the $t^1$ coefficient involves choosing $-\alpha_j^r$ once and 1 from all of the remaining terms, and then you sum over the disjoint possibilities.
:::

:::{.remark}
We'd really like to compute the coefficients of the $L$ polynomials, since we can solve a polynomial equation to get the roots.
But the Galois groups of these polynomials may not be solvable, so the term $\sum \alpha_j^r$ will in general be some symmetric function in the complex roots.
Note that any symmetric polynomial in the roots is also a symmetric polynomial in the coefficients.
:::

### Relating Rational Points to Coefficients

:::{.corollary title="?"}
For $K/\fq$ a function field, define
\[  
S_r \da N_r - (q^r + 1) = - \sum_{j=1}^{2g} \alpha_j^r
.\]
Note that $N_r = \abs{\Sigma(K_r/\fqr)}$ is the number of $\fqr\dash$rational point.
Then

a. $L'(t)/L(t) = \sum_{r=1}^\infty S_r t^{r-1}$.
b. $a_0 = 1$, and for all $1\leq i \leq g$, 
\[  
ia_i = S_i a_0 + S_{i-1} a_1 + \cdots + S_1 a_{i-1}
.\]
:::

:::{.remark}
What's the usefulness here?
If you only have the coefficients of the $L$ polynomials, taking the logarithmic derivative gives access to these quantities $S_r$.
The second formula is a recursive expression for the $a_i$ in terms of the $S_i$.
So you can compute the coefficients of the $L$ polynomial by counting $\fqr\dash$rational points on your curve (or places on your function field) for $r=1,2,\cdots, g$.
Similarly, if you have all of the coefficients for a $Z$ polynomial, you can solve for the $S_i$.
:::

:::{.proof title="of a"}
Essentially just a computation.
Logarithmically differentiating both sides of $L(t) = \prod_{j=1}^{2g} (1-\alpha_j t)$ and expanding in a geometric series yields
\[  
{L'(t)  \over L(t) } 
&= \sum_{j=1}^{2g} {-\alpha_j \over a -\alpha_j t} \\
&= \sum_{j=1}^{2g} (-\alpha_j) \sum_{r=0}^\infty \qty{\alpha_j t}^r \\
&= \sum_{r=1}^{\infty} \qty{\sum_{j=1}^{2g} (-\alpha_j^r) }t^{r-1} \\
&= \sum_{r=1}^{\infty} S_r t^{r-1}
.\]
:::

:::{.proof title="of b"}
Clearing denominators and equating coefficients in $L'(t) = L(t) \sum_{r=1}^{\infty} S_r t^{r-1}$ yields the result immediately, since the $ia_i$ are what appear as coefficients in the derivative of a formal power series, whereas the RHS is a Cauchy product.
:::

:::{.remark}
The moral: to compute zeta functions, you don't have to enumerate divisors and compute dimensions of Riemann-Roch spaces.
Note that the Riemann-Roch theorem tells us something interesting about these dimensions, but doesn't compute the dimension outright!
Instead, it suffices to compute $\fqr\dash$rational points for $r\leq g$.

A few lectures ago we discussed the places on a hyperelliptic function field, including a place at infinity.
Computing the zeta function of a hyperelliptic curve involves plugging in $x\dash$values and determining if it is

- A nonzero non-square: no $y\dash$values,
- Zero: exactly one $y\dash$value,
- A nonzero square: two $y\dash$values.

This is what happens at the finite places.
To handle the place at $\infty$, there is a recipe for the degree of the polynomial in terms of the coefficients.
So for any hyperelliptic function field (and in particular, for any elliptic function field) we have a concrete algorithm for computing their zeta functions.
Note that this is not necessarily a *good* algorithm: it still involves plugging in many values and checking if things are squares in finite values.
It seems that most people who compute a lot of zeta functions mostly focus on hyperelliptic function fields.

How are you going to compute zeta functions or even places for more complicated function fields?
The Riemann-Hurwitz formula says that since any function field is a finite degree extension of a rational function field, the curve is given as a degree 2 branched cover of $\PP^1$, it suffices to compute the fibers of this cover in order to get point counts.
:::