#  $a$-numbers of cyclic degree $p^{2}$ covers of the projective line

Huy Dang  National Center for Theoretical Sciences, Mathematics Division, No.
1, Sec. 4, Roosevelt Rd., Taipei City 106, Taiwan Room 503, Cosmology
Building, National Taiwan University<EMAIL_ADDRESS>and  Steven
R. Groen  Department of Mathematics, Lehigh University, 17 Memorial Drive E,
Bethlehem, Pennsylvania 18018<EMAIL_ADDRESS>
(Date: September 2023)

###### Abstract.

We investigate the $a$-numbers of $\mathbb{Z}/p^{2}\mathbb{Z}$-covers in
characteristic $p>2$ and extend a technique originally introduced by Farnell
and Pries for $\mathbb{Z}/p\mathbb{Z}$-covers. As an application of our
approach, we demonstrate that the $a$-numbers of “minimal”
$\mathbb{Z}/9\mathbb{Z}$-covers can be deduced from the associated branching
datum.

##  1\. Introduction

Consider an algebraically closed field $k$ of characteristic $p>2$, and let
$\pi:Y\rightarrow X$ be a Galois branched cover of smooth, projective,
connected curves. A prevalent research direction is to investigate to what
extent invariants of $Y$ are determined by the branching datum of the cover
$\pi$. See Section 2.1 for a precise definition of the branching datum. For
instance, if $Y$ is an elliptic curve, then the branching datum of $\pi$ does
not determine whether $Y$ is ordinary or supersingular.

Of particular interest is the case when $p$ divides $\lvert G\rvert$, i.e.
when the cover is wildly ramified. Then the ramification of $\pi$ is more
complicated, due to the additional structure of the filtration of higher
ramification groups. In this paper we restrict to the case when $G$ is a
cyclic $p$-group, and we write $G=\mathbb{Z}/p^{n}\mathbb{Z}$.

Some well-known invariants for curves in characteristic $p$ include the genus,
$p$-rank, and $a$-number. Of particular interest to us is the $a$-number,
which is defined by

$a_{Y}:=\text{ker}\left(\mathcal{C}:\operatorname{H}^{0}(Y,\Omega_{Y}^{1})\to\operatorname{H}^{0}(Y,\Omega_{Y}^{1})\right),$

where $\mathcal{C}$ is the $p^{-1}$-linear _Cartier operator_. The size of
$a_{Y}$ measures the failure of a curve to be _ordinary_. The genus of $Y$ is
determined by the branching datum of $\pi$ via the Riemann-Hurwitz formula
[Har77, IV. Corollary 2.4]. Similarly, since $G$ is a $p$-group, the Deuring-
Shafarevich formula [Sub75, Theorem 4.1] dictates what the $p$-rank of $Y$ is
in terms of the branching datum of $\pi$. In contrast, the $a$-number of $Y$
is not determined by the branching datum of $\pi$. Understanding the
possibilities for the $a$-number of a curve has implications in the Schottky
problems, as it helps illuminate which Abelian varieties are isomorphic to the
Jacobian of a curve.

In recent years, there has been a growing interest in studying the $a$-number
of cyclic covers of $p$-power degree in characteristic $p$. In particular, the
authors of [FP13] focused on the special case where $X=\mathbb{P}^{1}$ and
$G=\mathbb{Z}/p\mathbb{Z}$. They derived a formula for the $a$-number of $Y$
when the ramification jumps divide $p-1$. In particular, in this case the
$a$-number is determined by the branching datum of $\pi$. Booher and Cais
expanded the study in [BC20] to include general $X$ and established bounds for
$a_{Y}$ based on $a_{X}$ and the branching datum of $\pi$. In [BC23], the
authors consider the (higher) $a$-numbers in towers of
$\mathbb{Z}/p^{n}\mathbb{Z}$-covers of a base curve $X$, which in key cases is
the projective line. In particular, they formulate far-reaching conjectures
for the behavior of the $a$-numbers as a function of $n$ and provide
theoretical and experimental evidence for these conjectures.

In this paper, we investigate covers of $\mathbb{P}^{1}$ with Galois group
$G=\mathbb{Z}/p^{2}\mathbb{Z}$ whose conductors are “minimal” (see Definition
2.14). This is analogous to the condition that the conductors divide $p-1$, as
assumed in [FP13]. Extending the approach in [FP13], we assign to each basis
differential of $\operatorname{H}^{0}(Y,\Omega_{Y}^{1})$ a _key term_ , which
is non-zero in $\mathcal{C}(\omega)$. Using this, Theorem 3.18 proves a
conditional lower bound for the rank of the Cartier operator, which is
equivalent to giving an upper bound on the $a$-number. If we furthermore
assume $p=3$, we prove this upper bound is sharp, resulting in the following
theorem.

######  Theorem 1.1.

The $a$-number of a minimal $\mathbb{Z}/9\mathbb{Z}$-cover in characteristic
$3$ is determined by its branching datum.

See Theorem 4.1 for a precise formula of $a_{Y_{2}}$ in terms of the branching
datum.

As far as we are aware, this yields the only known example of a family of
curves in odd characteristic that has constant genus, $p$-rank and $a$-number,
apart from the Artin-Schreier curves treated in [FP13].

######  Remark 1.2.

In forthcoming work, Booher, Cais, Kramer-Miller, and Upton conduct parallel
research on the $a$-numbers of $\mathbb{Z}/p^{n}\mathbb{Z}$-covers using a
$p$-adic analytic approach, which differs from the method employed in this
paper.

###  1.1. Outline

The rest of the paper is organized as follows. In Section 2, we provide
background information on Artin-Schreier-Witt theory and the ramification of
cyclic covers of the projective line in characteristic $p$. Additionally, we
review the Cartier-Manin matrix and its relationship with the $a$-number of a
curve. We also discuss the motivation behind our question and make comparisons
with existing results in this section.

Section 3 introduces the concept of key terms for
$\mathbb{Z}/p^{2}\mathbb{Z}$-covers. The key terms represent leading entries
of the Cartier-Manin matrix, generalizing the concept introduced in [FP13] for
$\mathbb{Z}/p\mathbb{Z}$-covers. It is shown in Theorem 3.18 that, under
certain conditions, the number of key terms is a lower bound or the rank of
the Cartier-Manin matrix, as differentials with a key term contribute to the
rank. This yields a lower bound on the $a$-number.

Finally, in Section 4, we demonstrate that the provided upper bound is sharp
when $p=3$, by showing the remaining differentials do not contribute to the
rank of the Cartier operator. This results in Theorem 1.1.

###  1.2. Acknowledgements

The authors thank Jeremy Booher, Bryden Cais, Joe Kramer-Miller, Rachel Pries,
and Damiano Testa for helpful discussions.

##  2\. Background

###  2.1. Artin-Schreier-Witt theory

According to the Artin-Schreier-Witt theory (as discussed in references such
as [Lan02]), for any field $K$ with characteristic $p$, the following group
isomorphism holds:

$\operatorname{H}^{1}(G_{K},\mathbb{Z}/p^{n}\mathbb{Z})\cong
W_{n}(K)/\wp(W_{n}(K)).$

In the above equation, $G_{K}$ represents the absolute Galois group of $K$,
$W_{n}(K)$ denotes the ring of length-$n$ Witt vectors over $K$, and
$\wp(\underline{x})=F(\underline{x})-\underline{x}$ corresponds to the Artin-
Schreier-Witt isogeny. We say two Witt vectors $\underline{f}$ and
$\underline{g}$ in $W_{n}(K)$ belong to the same Artin-Schreier-Witt class if
there exists a Witt vector $\underline{h}\in W_{n}(K)$ such that
$\underline{f}=\underline{g}+\wp(\underline{h})$.

Therefore, any $\mathbb{Z}/p^{n}\mathbb{Z}$-cover
$\phi_{n}:Y_{n}\xrightarrow{}\mathbb{P}^{1}$ can be represented by an Artin-
Schreier-Witt equation of the form

$\wp(y_{1},\ldots,y_{n})=(f_{1},\ldots,f_{n}),$

where $(f_{1},\ldots,f_{n})\in W_{n}(k(x))$ and $f_{1}\not\in\wp(k(x))$.

######  Remark 2.1.

For $1\leq i<n$, the unique $\mathbb{Z}/p^{i}$-sub-cover
$\phi_{i}:Y_{i}\rightarrow\mathbb{P}^{1}$ of $\phi_{n}$ is given by the Artin-
Schreier-Witt equation:

$\wp(y_{1},\ldots,y_{i})=(f_{1},\ldots,f_{i}).$

Alternatively, the cover $\phi_{n}$ can be expressed as a series of Artin-
Schreier extensions, which is advantageous for computational purposes, as
demonstrated later in this paper.

######  Proposition 2.2 ([OP10]).

The $\mathbb{Z}/p^{n}\mathbb{Z}$-cover associated with
$(f_{1},\ldots,f_{n})\in W_{n}(K)$ can be represented by the following system
of equations:

(2.1) $\begin{split}y_{1}^{p}-y_{1}&=f_{1}\\\
y_{2}^{p}-y_{2}&=g_{1}(y_{1})+f_{2}\\\ &\vdots\\\
y_{n}^{p}-y_{n}&=g_{n-1}(y_{1},\ldots,y_{n-1})+f_{n},\end{split}$

where $g_{i}\in\mathbb{F}_{p}[f_{1},\ldots,f_{i},y_{1},\ldots,y_{i}]$ are
explicitly defined. Specifically, $g_{i}$ can be given recursively in
$\mathbb{Z}[f_{1},\ldots,f_{i},y_{1},\ldots,y_{i}]$ as follows:

(2.2)
$g_{i}(f_{1},\ldots,f_{i},y_{1},\ldots,y_{i})=f_{i}+y_{i}+\sum_{d=1}^{i-1}\frac{1}{p^{i-d}}(f_{d}^{p^{i-d}}+y_{d}^{p^{i-d}}-g_{d-1}^{p^{i-d}}).$

In the case where $n=1$, we can eliminate all terms of $f_{1}$ with degrees
divisible by $p$ by adding rational functions of the form $h^{p}-h$. This
process can be generalized to the case $n>1$ using an induction technique
similar to the proof presented in [Lor08, §26, Theorem 5]. For a more
comprehensive proof, please refer to [Ach17, Lemma A.2.3]. To summarize this
result, we state the following proposition.

######  Proposition 2.3.

Every $\underline{f}\in W_{n}(k(x))$ belongs to the same Artin-Schreier-Witt
class as a vector $\underline{h}=(h_{1},\ldots,h_{n})$, where none of the
terms in the partial fraction decomposition of $h_{i}$ has degree divisible by
$p$.

We refer to such a vector $\underline{h}$ as reduced. For the remainder of our
discussion, we assume that $\underline{f}$ is reduced.

Let $B_{n}={P_{1},\ldots,P_{r}}$ denote the set of poles of the $f_{i}$’s,
which is also the branch locus of $\phi_{n}$. At each ramified point $Q_{j}$
over $P_{j}$, $\phi_{n}$ induces an exponent-$p^{n}$ cyclic extension of the
complete local ring
$\hat{\mathcal{O}}_{Y_{n},Q_{j}}/\hat{\mathcal{O}}_{\mathbb{P}^{1},P_{j}}$.
Consequently, we can derive the upper ramification filtration of $\phi_{n}$ at
the branch point $P_{j}$ in a canonical manner.

Assuming that the inertia group of $Q_{j}$ is $\mathbb{Z}/p^{m}\mathbb{Z}$
(where $n\leq m$), for $i\leq n-m$, we define the $i$-th ramification break
(jump) of $\phi_{n}$ at $P_{j}$ as $-1$. For $i>n-m$, the $i$-th ramification
break of $\phi_{n}$ at $P_{j}$ corresponds to the $(i-n+m)$-th break in
$\hat{\mathcal{O}}_{Y_{n},Q_{j}}/\hat{\mathcal{O}}_{\mathbb{P}^{1},P_{j}}$. We
denote the $i$-th upper ramification break of $\phi_{n}$ at $P_{j}$ as
$u_{j,i}$. The $i$-th conductor of $\phi_{n}$ at $P_{j}$ is defined as
$e_{j,i}:=u_{j,i}+1$. The following formula provides an explicit calculation
of the ramification filtration of $\phi_{n}$ in terms of $\underline{f}$.

######  Theorem 2.4 ([Gar02, Theorem 1]).

With the assumptions and the notations as above, we have

(2.3) $u_{j,i}=\max\\{p^{i-l}\deg_{(x-P_{j})^{-1}}(f_{l})\mid
l=1,\ldots,i\\},$

for $i>n-m$.

######  Proposition 2.5.

With the above settings, the genus of the curve $Y_{i}$ is

(2.4)
$g_{Y_{i}}=1-p^{i}+\frac{\sum_{l=1}^{i}(\sum_{j=1}^{r}e_{j,l})(p^{j}-p^{j-1})}{2}$

###### Proof.

Applying the Grothendieck-Ogg-Shafarevich formula [Ill77, Exp. X formula 7.2]
to $\phi_{i}:Y_{i}\xrightarrow{}X$, we obtain the relation

$2g_{Y_{i}}-2=\deg(\phi_{i})(2g_{X}-2)+\sum_{j=1}^{r}\deg\big{(}\mathscr{D}_{P{j,i}}\big{)},$

where is $\mathscr{D}_{P{j,i}}$ the different of $\phi_{i}$ at $P_{j}$ [Ser79,
IV]. Additionally, [Pop14, Fact 2.3] asserts that

$\deg\big{(}\mathscr{D}_{P{j,i}}\big{)}=\sum_{l=1}^{i}e_{j,l}(p^{l}-p^{l-1}).$

This and the fact that $g_{X}=0$ immediately implies the claim about the genus
of $Y_{i}$. ∎

To record the branching datum of the cover, we use an $r\times n$ matrix of
the form

(2.5) $\begin{bmatrix}e_{1,1}&e_{1,2}&\ldots&e_{1,n}\\\
e_{2,1}&e_{2,2}&\ldots&e_{n,n}\\\ \vdots&\vdots&\ddots&\vdots\\\
e_{r,1}&e_{r,2}&\ldots&e_{r,n}\\\ \end{bmatrix},$

which is referred to as the branching datum of $\phi_{n}$. This notion is used
to stratify the moduli space of cyclic covers of curves in [DH23]. The
following is an immediate consequence of Theorem 2.4.

######  Proposition 2.6.

A matrix of the form (2.5) is the branching datum of a
$\mathbb{Z}/p^{n}\mathbb{Z}$-cover if and only if the following conditions
hold:

  1. 1.

$e_{i,1}\not\equiv 1\pmod{p}$,

  2. 2.

$e_{i,j}\geq pe_{i,j-1}-p+1$, and

  3. 3.

if $e_{i,j}>pe_{i,j-1}-p+1$, then $e_{i,j}=pe_{i,j-1}-p+1+a_{j}$, where
$p\nmid a_{j}$.

######  Remark 2.7.

The branching datum of $\phi_{i}$ is the $r\times i$ matrix that contains the
first $i$ columns of (2.5).

######  Example 2.8.

Suppose $k$ is an algebraically closed field of characteristic $3$, and
$\phi_{2}:Y_{2}\xrightarrow{}\operatorname{Proj}k[x,z]$ is a
$\mathbb{Z}/9\mathbb{Z}$-cover given by the following affine equation:

$\wp(y_{1},y_{2})=\bigg{(}\frac{1}{x}+x,\frac{1}{x^{5}}-\frac{1}{x-1}\bigg{)}=:(f_{1},f_{2})=:\underline{f}.$

Since none of the terms in $f_{i}$’s are a power of $3$, $\underline{f}$ is
considered reduced. According to Theorem 2.4, $\phi$ branches at $x=0$, $x=1$,
and $x=\infty$, with the following branching datum:

$\begin{bmatrix}2&6\\\ 0&2\\\ 2&4\end{bmatrix}.$

The theorem also reveals that the ramification index of each ramified point
above $x=1$ is $3$, while the unique point above $x=1$ or $x=\infty$ has a
ramification index of $9$.

Finally, applying Proposition 2.2 allows us to write $\phi_{2}$ as a system of
Artin-Schreier equations as follows:

$\begin{split}y_{1}^{3}-y_{1}&=\frac{1}{x}+x=f_{1},\\\
y_{2}^{3}-y_{2}&=g_{1}(y_{1})+\frac{1}{x^{5}}-\frac{1}{x-1},\end{split}$

where
$g_{1}(y_{1})=\frac{1}{3}(f_{1}^{3}+y_{1}^{3}-(f_{1}+y_{1})^{3}))=-y_{1}^{7}+y_{1}^{5}$.

###  2.2. A basis for the space of regular differentials

We present a restatement of [Mad78, Lemma 5] using our established
conventions. Consider a $\mathbb{Z}/p^{n}\mathbb{Z}$-Galois cover
$Y_{n}\xrightarrow{\phi_{n}}\mathbb{P}^{1}$ whose branching datum is given by
Equation 2.5. Let $B_{n}=\\{P_{1},\ldots,P_{r}\\}\subset\mathbb{P}^{1}$ denote
the branch locus of $\phi_{n}$, where $P_{i}$ corresponds to the $i$-th row of
the branching datum. Without loss of generality, assume that $P_{1}$ is the
point at infinity and $e_{1,1}>0$, so that $P_{1}$ has index $p^{n}$. We
define $x_{i}:=\frac{1}{x-P_{i}}$ for $1<i\leq r$.

######  Definition 2.9.

We define $N\in\mathbb{M}_{r\times n}(\mathbb{Z})$ as follows:

$N(i,j)=\begin{cases}e_{i,j}-1&\text{if }e_{i,j}>0\\\ 0&\text{if
}e_{i,j}=0\end{cases}$

For each $i=1,2,\ldots,r$ and $j=1,\ldots,n$ such that $N(i,j)>0$, let

$E^{j}_{i}:=j-\min\\{l\mid e_{i,l}\neq 0\\}+1.$

######  Remark 2.10.

The positive integer $E^{j}_{i}$ represents the ramification index of each
ramified point of $Y_{j}\xrightarrow{\phi_{j}}\mathbb{P}^{1}$ above $P_{i}$.

######  Definition 2.11.

We define $\lambda(i,j)$ recursively as follows:

$\lambda(i,j)=p^{E^{j}_{i}}N(i,j)-(p-1)\sum_{l=1}^{j-1}p^{l+E^{j}_{i}-j-1}N(i,l).$

Now we can rephrase [Mad78, Lemma 5] using our notations.

######  Proposition 2.12.

With the above settings, the $k$-vector space
$\operatorname{H}^{0}(Y_{n},\Omega^{1}_{Y_{n}/k})$ has a basis given by
$\bigcup_{i=1}^{r}W_{i}$, where

(2.6) $\displaystyle W_{1}:=\begin{cases}y_{n}^{a_{n}}\ldots
y_{1}^{a_{1}}x^{v}dx\;|\;0\leq a_{i}<p,\\\ 0\leq
p^{n}v\leq(p-1)d_{1}-A_{1}-p^{n}-1\end{cases}\Bigg{\\}},$

and, for $i>1$,

(2.7) $\displaystyle W_{i}:=\begin{cases}y_{n}^{a_{n}}\ldots
y_{1}^{a_{1}}x_{i}^{v}dx\;|\;0\leq a_{i}<p,\\\
0<p^{e_{i}}v\leq(p-1)d_{i}-A_{i}+p^{e_{i}}-1\end{cases}\Bigg{\\}},$

where

$d_{i}:=\sum_{\nu=1}^{n}\lambda(i,\nu),\text{ and
}A_{i}:=\sum_{\nu=1}^{n}a_{\nu}\lambda(i,\nu).$

###  2.3. $a$-number of a curve

Suppose $C$ is a curve of genus $g$ over field $k$. We let $\sigma$ denote the
$p$-power Frobenius automorphism of $k$, which induces a pull-back through the
absolute Frobenius
$F^{\ast}_{X}:\operatorname{H}^{1}(X,\mathcal{O}_{X})\xrightarrow{}\operatorname{H}^{1}(X,\mathcal{O}_{X})$.
The Cartier operator is a $\sigma^{-1}$-linear map

$\mathcal{C}:\operatorname{H}^{0}(C,\Omega^{1}_{C/k})\rightarrow\operatorname{H}^{0}(C,\Omega^{1}_{C/k}).$

It is the dual of $F^{\ast}_{X}$ through Grothendieck–Serre duality. The
$a$-number of a curve $C$, denoted as $a_{C}$, is defined as the dimension of
the kernel of $\mathcal{C}$.

Assuming we have a basis $\beta=\\{\omega_{1},\ldots,\omega_{g}\\}$ of
$\operatorname{H}^{0}(C,\Omega^{1}_{C/k})$, for each $\omega_{j}$ we can find
coefficients $m_{j,k}\in k$ such that

$\mathcal{C}(\omega_{j})=\sum_{i=1}^{g}m_{i,j}\omega_{i}.$

The resulting $(g\times g)$-matrix $M=(m_{i,j})$ is known as the Cartier-Manin
matrix. Consequently, the $a$-number can be computed as
$a_{C}=g_{C}-\text{rank}(M)$.

######  Example 2.13.

  1. (1)

An ordinary elliptic curve has $p$-rank $1$ and $a$-number $0$.

  2. (2)

A supersingular elliptic curve has $p$-rank $0$ and $a$-number $1$.

Generalizing the case of elliptic curves, a curve is said to be ordinary if
its $a$-number is zero. Therefore, the $a$-number of a curve can be seen as a
measure of how far the curve is from being ordinary.

Unlike the genus and the $p$-rank, the $a$-number of a curve cover is not
solely determined by the base curve and the branching datum. An example of a
pair of $\mathbb{Z}/p\mathbb{Z}$-covers with identical branching datum, but
different $a$-number, is given in [BC20, Example 4.6].

###  2.4. $\mathbb{Z}/p^{n}\mathbb{Z}$-covers with minimial jumps

######  Definition 2.14.

A $\mathbb{Z}/p^{n}\mathbb{Z}$-cover with branching datum (2.5) is said to be
_minimal_ if it satisfies the following conditions:

  1. (1)

If $j=\min\\{k\mid e_{i,k}\neq 0\\}$, then $(e_{i,j}-1)\mid(p-1)$, and

  2. (2)

$e_{i,l}-1=p^{l-j}(e_{i,j}-1)$ for $j<l\leq n$.

The $a$-number of a minimal $\mathbb{Z}/p\mathbb{Z}$-cover is known [FP13,
Theorem 1.1]. Our computed examples lead to the following question:

######  Question 2.15.

Let $\phi_{n}$ be a minimal branched cover of $\mathbb{P}^{1}$ with Galois
group $\mathbb{Z}/p^{n}\mathbb{Z}$. Is the $a$-number of $\phi_{n}$ uniquely
determined by its branching datum?

In this paper, we focus on the case $n=2$. That allows us to only consider
covers with branching data of the form

$\begin{bmatrix}2&\ldots&a_{i}&\ldots&0&\ldots&0&\ldots\\\
p+1&\ldots&pa_{i}-p+1&\ldots&2&\ldots&a_{i}&\ldots\\\
\end{bmatrix}^{\intercal},$

where $(a_{i}-1)\mid(p-1)$. In Section 3, we generalize the methods in [FP13],
based on key terms, for $\mathbb{Z}/p^{2}\mathbb{Z}$-covers. As an application
of this new method, we prove that the answer is affirmative for the case when
$p=3$ and $n=2$.

######  Theorem 2.16.

The answer to Question 2.15 is affirmative when $p=3$ and $n=2$ (see Theorem
4.1).

######  Remark 2.17.

Note that the answer to Question 2.15 is certainly negative for covers that
are not minimal, as is demonstrated in [BC20, Example 7.2].

###  2.5. Comparison with the known results

In this section, we compare our results and conjectures with those that are
already known for Artin-Schreier covers. Consider a
$\mathbb{Z}/p^{n}\mathbb{Z}$-Galois cover
$\phi_{n}:Y_{n}\xrightarrow{}\ldots\xrightarrow{}Y_{1}\xrightarrow{}Y_{0}$.
Since $\phi_{n/(n-1)}:Y_{n}\xrightarrow{}Y_{n-1}$ is an Artin-Schreier cover,
we can establish bounds on the $a$-number of $Y_{n}$ inductively, building
upon what is known for $\mathbb{Z}/p\mathbb{Z}$-covers.

Suppose $P\in Y_{0}$ is a branch point with ramification index $p^{n}$. Let
$(m_{1},\ldots,m_{n})$ denote the upper ramification jumps of $\phi_{n}$ at
$P$. By employing a Hasse-Arf relation, it is possible to deduce the only jump
of $\phi_{n/(n-1)}$ at the branch point $P_{n-1}\in Y_{n-1}$ that lies above
$P$.

######  Proposition 2.18.

With the above notations, the jump of the intermediate Artin-Schreier-
extension $Y_{n}\xrightarrow{}Y_{n-1}$ at $P_{n-1}$ is

$\tilde{m}_{n}=p^{n-1}m_{n}-\sum_{i=1}^{n-1}m_{i}(p-1)p^{i-1}.$

###### Proof.

It follows from [OW16, Lemma 6.17]. ∎

Applying the above result to the case $n=2$ and $m_{2}=pm_{1}$ yields:

######  Corollary 2.19.

In the situation of Theorem 2.16, the jump of $Y_{2}\xrightarrow{}Y_{1}$ at
$P_{1}$ is given by

$\widetilde{m}=pm_{2}-m_{1}(p-1)=m_{1}(p^{2}-p+1).$

######  Remark 2.20.

The jump $\tilde{m}$ does not divide $p-1$, hence is not covered by [FP13].

Let’s now recall the following results for the $a$-number of a
$\mathbb{Z}/p\mathbb{Z}$-cover using the language of branching data.

######  Theorem 2.21 ([BC20, Theorem 1.1]).

Suppose $\phi:Y\rightarrow X$ is a $\mathbb{Z}/p\mathbb{Z}$-Galois cover with
branching datum

$\begin{bmatrix}d_{1,1}+1&d_{2,1}+1&\ldots&d_{r,1}+1\\\ \end{bmatrix}^{\top}.$

Then the $a$-number of $Y$ is bounded as follows

$\sum_{l=1}^{r}\sum_{i=j}^{p-1}\bigg{(}\bigg{\lfloor}\frac{id_{l,1}}{p}\bigg{\rfloor}-\bigg{\lfloor}\frac{id_{l,1}}{p}-\bigg{(}1-\frac{1}{p}\bigg{)}\frac{jd_{l,1}}{p}\bigg{\rfloor}\bigg{)}\leq
a_{Y}\leq
pa_{X}+\sum_{l=1}^{r}\sum_{i=1}^{p-1}\bigg{(}\bigg{\lfloor}\frac{id_{l,1}}{p}\bigg{\rfloor}-(p-i)\bigg{\lfloor}\frac{id_{l,1}}{p^{2}}\bigg{\rfloor}\bigg{)}.$

######  Proposition 2.22 ([FP13]).

Suppose a $\mathbb{Z}/p\mathbb{Z}$-cover $\phi:Y\rightarrow\mathbb{P}^{1}$ has
branching datum

$\begin{bmatrix}d_{1,1}+1&d_{2,1}+1&\ldots&d_{r,1}+1\\\ \end{bmatrix}^{\top}.$

Suppose moreover that $d_{i,1}\mid(p-1)$ for all $i=1,2,\ldots,r$. Then the
$a$-number of $Y$ is

$a_{Y}=\sum_{i=1}^{r}a_{i}\text{, where
}a_{i}=\begin{cases}\frac{(p-1)d_{i,1}}{4}&\text{if $d_{i,1}$ is even}\\\
\frac{(p-1)(d^{2}_{i,1}-1)}{4}&\text{if $d_{i,1}$ is odd.}\\\ \end{cases}$

We now can compute the lower bound of the $a$-number for the case
$Y_{2}\xrightarrow{}Y_{0}$ is minimal-one-point cover whose first jump equals
$1$.

######  Proposition 2.23.

Suppose $p=2q+1$. Then

$\sum_{i=q+1}^{p-1}\bigg{(}\bigg{\lfloor}\frac{i(p^{2}-p+1)}{p}\bigg{\rfloor}-\bigg{\lfloor}\frac{i(p^{2}-p+1)}{p}-\bigg{(}1-\frac{1}{p}\bigg{)}\frac{(q+1)(p^{2}-p+1)}{p}\bigg{\rfloor}\bigg{)}$

is equal to $\frac{p(p-1)^{2}}{4}$.

###### Proof.

By a straightforward computation, we can indeed see that each summand on the
left-hand side is equal to $q(2q+1)$. Since there are $q$ such summands, the
result follows immediately. ∎

Combining Theorem 2.21 and Proposition 2.23, we obtain the following.

######  Corollary 2.24.

Suppose $p=2q+1$ and $X$ is ordinary. Suppose $\phi_{2}:Y_{2}\to X$ is a
minimal $\mathbb{Z}/p^{2}\mathbb{Z}$-cover branched at one point, with first
jump $d_{1,1}=1$. Then

$\sum_{i=q+1}^{p-1}\bigg{(}\bigg{\lfloor}\frac{i\tilde{d}_{1,1}}{p}\bigg{\rfloor}-\bigg{\lfloor}\frac{i\tilde{d}_{1,1}}{p}-\bigg{(}1-\frac{1}{p}\bigg{)}\frac{j\tilde{d}_{1,1}}{p}\bigg{\rfloor}\bigg{)}\leq
a_{Y_{2}}\leq
pa_{Y_{1}}+\sum_{i=1}^{p-1}\bigg{(}\bigg{\lfloor}\frac{i\tilde{d}_{1,1}}{p}\bigg{\rfloor}-(p-i)\bigg{\lfloor}\frac{i\tilde{d}_{1,1}}{p^{2}}\bigg{\rfloor}\bigg{)}.$

In particular, when $d_{1,1}=1$, hence $\tilde{d}_{1,1}=p^{2}-p+1$, and
$X\cong\mathbb{P}^{1}$, hence $a_{Y_{1}}=0$, we have

$\frac{p(p-1)^{2}}{4}\leq a_{Y_{2}}.$

Since we expect the $a$-number of a minimal $\mathbb{Z}/p^{n}\mathbb{Z}$-cover
of $\mathbb{P}^{1}$ to be determined by the branching datum, and this lower
bound is attained in examples, we conjecture that this lower bound is always
sharp for this class of curves. On the other hand, in the case $d_{1,1}>1$ the
lower bound for the $a$-numbers of minimal $\mathbb{Z}/p^{2}\mathbb{Z}$-covers
obtained by applying 2.21 twice does not seem to be sharp.

##  3\. Key terms

Let $\phi_{2}:Y_{2}\to\mathbb{P}^{1}$ be a minimal
$\mathbb{Z}/p^{2}\mathbb{Z}$ cover, with subcover $\phi_{1}:Y_{1}\
to\mathbb{P}^{1}$. Using the identification $k(\mathbb{P}^{1})\cong k(x)$ and
Proposition 2.2, we obtain the models

(3.1) $\displaystyle Y_{1}:y_{1}^{p}-y_{1}$ $\displaystyle=f(x)$ (3.2)
$\displaystyle Y_{2}:y_{2}^{p}-y_{2}$ $\displaystyle=g_{1}(y_{1})+h(x).$

For brevity we write $g(y_{1}):=g_{1}(y_{1})$. Explicitly this polynomial is
given by

(3.3)
$g(y_{1}):=\sum_{i=1}^{p-1}(-1)^{i}\frac{(p-1)!}{i!(p-i)!}y_{1}^{p(p-i)+i}.$

Equivalently, the cover $Y_{2}\to\mathbb{P}^{1}$ is represented by the Witt
vector $(f,h)\in W_{2}(k(x))$, with $f$ and $h$ reduced in the sense of
Proposition 2.3.

In this section, we define for certain regular differentials
$\omega\in\operatorname{H}^{0}(Y_{2},\Omega_{Y_{2}}^{1})$ a _key term_
$\kappa(\mathcal{C}(\omega))$, analogous to [FP13, Definition 3.2]. An
analysis of the key terms leads to a lower bound for the rank of the Cartier-
Manin matrix, or equivalently an upper bound for the $a$-number $a_{Y_{2}}$.
This upper bound is given in Theorem 3.18.

Denote by $B_{2}$ the branch locus of the cover
$\phi_{2}:Y_{2}\to\mathbb{P}^{1}$. For $P\in B_{2}$, let $x_{P}\in k(x)$ be
such that $\text{ord}_{P}(x_{P})=-1$. Then the space
$\operatorname{H}^{0}(Y_{2},\Omega_{Y_{2}}^{1})$ is spanned by differentials
of the form $y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx$, as described in
Proposition 2.12. We consider two cases, depending on whether the cover
$\phi_{1}:Y_{1}\to\mathbb{P}^{1}$ is already branched at $P$.

If $P$ is a branch point of $\phi_{1}:Y_{1}\to\mathbb{P}^{1}$, with
ramification jump $d_{P}$, then $f$ has a pole at $P$ of order $d_{P}$. To
simplify proofs, we pick $f$ to have no further poles. Since the cover
$\phi_{2}:Y_{2}\to\mathbb{P}^{1}$ is minimal, it follows that $d_{P}|p-1$.
This case is discussed in Section 3.1.

If $\phi_{1}:Y_{1}\to\mathbb{P}^{1}$ is étale at $P$, then $f$ is regular at
$P$. By the assumption that $\phi_{2}:Y_{2}\to\mathbb{P}^{1}$ is branched at
$P$, the function $h\in k(x)$ has a pole at $P$. The order of this pole, which
we denote by $e_{P}$, equals the ramification jump at $P$ of the cover
$Y_{2}\to Y_{1}$. This case is treated in Section 3.2.

Finally, combining both types of key terms yields the conditional upper bound
on the $a$-number proved in Theorem 3.18. This conditional upper bound is
strong enough to answer Question 2.15 in the case $p=3$ and $n=2$.

###  3.1. Key terms at poles of $f$

Let $B_{1}\subset\mathbb{P}^{1}$ be the branch locus of
$\phi_{1}:Y_{1}\to\mathbb{P}^{1}$ and let $P\in B_{1}$. After using an
automorphism of $\mathbb{P}^{1}$ if necessary, we may assume $\infty\in
B_{1}$. Since the cover is minimal (see Definition 2.14), the pole orders at
the points $P\in B_{1}$, defined by $d_{P}=-\text{ord}_{P}(f)$, all divide
$p-1$. Furthermore, since the upper jumps are minimal by Definition 2.14, one
may assume the pole order of $h$ at $P$ is bounded:
$\text{ord}_{P}(h)>-pd_{P}$. We write $x_{\infty}=x$ and $x_{P}=(x-P)^{-1}$.
Then partial fraction decomposition allows us to write

(3.4) $f(x)=\sum_{P\in B}f_{P}(x_{P}),$

where $f_{P}\in k[x_{P}]$ is a polynomial of degree $d_{P}$. Let $u_{P}\in
k^{\times}$ denote the leading coefficient of $f_{P}$. Without loss of
generality, we assume $u_{\infty}=1$.

Using Proposition 2.12, we obtain the following formulas for $W_{P}$.

(3.5) $\displaystyle
W_{\infty}:=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x^{v}dx\;|\;0\leq
a_{1},a_{2}<p,\\\ 0\leq p^{2}v\leq
pd_{P}(p-1-a_{1})+d_{P}(p^{2}-p+1)(p-1-a_{2})-p^{2}-1\end{cases}\Bigg{\\}}.$

(3.6) $\displaystyle
W_{P}:=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\;|\;0\leq
a_{1},a_{2}<p,\\\ 0<p^{2}v\leq
pd_{P}(p-1-a_{1})+d_{P}(p^{2}-p+1)(p-1-a_{2})+p^{2}-1\end{cases}\Bigg{\\}}.$

Following [FP13], define $\epsilon_{P}=-1$ if $P=\infty$ and $\epsilon_{P}=1$
otherwise. Then, as the Cartier operator on $k(x)dx$ is well understood, we
obtain

$\mathcal{C}(x_{P}^{ap+\epsilon_{P}}dx)=x_{P}^{a+\epsilon_{P}}dx.$

The approach we take in this paper relies essentially on reducing the action
of the Cartier operator on $\operatorname{H}^{0}(Y_{2},\Omega_{Y_{2}}^{1})$ to
the Cartier operator on $k(x)dx$, through the models of $Y_{2}$ and $Y_{1}$.

Using $a_{1},a_{2}\geq 0$, we get

(3.7)
$v\leq\frac{1}{p^{2}}(pd_{P}(p-1)+d_{P}(p^{2}-p+1)(p-1)+\epsilon_{P}p^{2}-1)<pd_{P}+\epsilon_{P}.$

The following lemma allows us to reduce computations involving the Cartier
operator on $Y_{2}$ to computations involving the Cartier operator on $Y_{1}$.

######  Lemma 3.1.

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx$ be an element of $W$. Then
we have

(3.8) $\mathcal{C}(\omega)=\sum_{0\leq j+l\leq
a_{2}}(-1)^{a_{2}-j}\binom{a_{2}}{j\,l}y_{2}^{j}\mathcal{C}(g(y_{1})^{l}h(x)^{a_{2}-j-l}y_{1}^{a_{1}}x_{P}^{v}dx).$

###### Proof.

Substituting $y_{2}=y_{2}^{p}-g(y_{1})-h(x)$, from Equation (3.2) and
expanding gives

$\displaystyle\omega$
$\displaystyle=(y_{2}^{p}-g(y_{1})-h(x))^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx$
$\displaystyle=\sum_{0\leq j+l\leq
a_{2}}(-1)^{a_{2}-j}\binom{a_{2}}{j\,l}y_{2}^{pj}g(y_{1})^{l}h(x)^{a_{2}-j-l}y_{1}^{a_{1}}x_{P}^{v}dx.$

Then, using the $p^{-1}$-linearity of the Cartier operator, we can pull out
the $y_{2}$ to obtain

$\displaystyle\mathcal{C}(\omega)$ $\displaystyle=\mathcal{C}\left(\sum_{0\leq
j+l\leq
a_{2}}(-1)^{a_{2}-j}\binom{a_{2}}{j\;l}y_{2}^{pj}g(y_{1})^{l}h(x)^{a_{2}-j-l}y_{1}^{a_{1}}x_{P}^{v}dx\right)$
$\displaystyle=\sum_{0\leq j+l\leq
a_{2}}(-1)^{a_{2}-j}\binom{a_{2}}{j\,l}y_{2}^{j}\mathcal{C}(g(y_{1})^{l}h(x)^{a_{2}-j-l}y_{1}^{a_{1}}x_{P}^{v}dx),$

as desired. ∎

In order to compute the $a$-number of $Y_{2}$, we bound the rank of
$\mathcal{C}$ from below. We do so using a differential we’ll call the _key
term_. Recall that we have assumed that $d_{P}|p-1$, allowing us to define
$\gamma_{P}:=\frac{p-1}{d_{P}}\in\mathbb{Z}$.

######  Definition 3.2.

Suppose $P\in B_{1}$. Let

$\displaystyle\alpha_{P}(v)$
$\displaystyle:=\left\lfloor\frac{\gamma_{P}(v-\epsilon_{P})}{p}\right\rfloor$
$\displaystyle\beta_{P}(v)$
$\displaystyle:=\gamma_{P}(v-\epsilon_{P})-p\alpha_{P}(v),$

such that $0\leq\beta_{P}(v)<p$. We define

$H_{P}:=\\{y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in W_{P}\;|\;\beta_{P}(v)\leq
a_{1}+pa_{2}\\}$

For $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H_{P}$, we define the
associated _key term_ $\kappa(\mathcal{C}(\omega))$ as follows:

(3.9)
$\kappa(\mathcal{C}(\omega)):=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}-\beta_{P}(v)}x_{P}^{v-d_{P}\alpha_{P}(v)}dx&\hbox{{if}
$a_{1}\geq\beta_{P}(v)$}\\\
y_{2}^{a_{2}-1}y_{1}^{a_{1}-\beta_{P}(v)+p}x_{P}^{v-d_{P}\alpha_{P}(v)}dx&\hbox{{if}
$a_{1}<\beta_{P}(v)$}.\end{cases}$

Note that $v-d_{P}\alpha_{P}(v)$ cannot be negative. In fact, it is
straightforward to verify $\kappa(\mathcal{C}(\omega))\in W_{P}$ for each
$\omega\in H_{P}$.

Our goal is to show that in certain cases the coefficient of
$\kappa(\mathcal{C}(\omega))$ is non-zero in $\mathcal{C}(\omega)$. The
following lemma gives a formula for this coefficient.

######  Lemma 3.3.

Let $\omega=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H_{P}$. Then the
coefficient of $\kappa(\mathcal{C}(\omega))$ in $\mathcal{C}(\omega))$ is

(3.10)
$c_{\omega}:=\begin{cases}(-1)^{\beta_{P}(v)}u_{P}^{\beta_{P}(v)/p}\binom{a_{1}}{\beta_{P}(v)}&\hbox{{if}
$a_{1}\geq\beta_{P}(v)$}\\\
u_{P}^{\beta_{P}(v)/p}\sum_{i=\beta_{P}(v)-a_{1}}^{p-1-a_{1}}(-1)^{\beta_{P}(v)+i+1}\frac{(p-1)!}{i!(p-i)!}\binom{a_{1}+i}{\beta_{P}(v)}&\hbox{{if}
$a_{1}<\beta_{P}(v)$.}\end{cases}$

###### Proof.

By Equation (3.8), we obtain

$\mathcal{C}(\omega)=\sum_{0\leq j+l\leq
a_{2}}(-1)^{a_{2}-j}\binom{a_{2}}{j\,l}y_{2}^{j}\mathcal{C}(g(y_{1})^{l}h(x)^{a_{2}-j-l}y_{1}^{a_{1}}x_{P}^{v}dx).$

We split the proof into the two cases of Definition 3.2: $a_{1}\geq\beta(v)$
and $a_{1}<\beta(v)$.

First, suppose $a_{1}\geq\beta(v)$. To obtain the exponent of $y_{2}$ in
$\kappa(\mathcal{C}(\omega))$, we must specialize to $j=a_{2}$ in Equation
(3.8), yielding the term

$y_{2}^{a_{2}}\mathcal{C}(y_{1}^{a_{1}}x_{P}^{v}dx)=\sum_{m=0}^{a_{1}}(-1)^{m}\binom{a_{1}}{m}y_{2}^{a_{2}}y_{1}^{a_{1}-m}\mathcal{C}(f(x)^{m}x_{P}^{v}dx).$

For the right exponent of $y_{1}$, we specialize to $m=\beta_{P}(v)$. Then the
leading term of $f_{P}(x_{P})$ (recall Equation (3.4)) gives the term

$\displaystyle\mathcal{C}(u_{P}^{\beta_{P}(v)}x_{P}^{d_{P}\beta_{P}(v)+v}dx)$
$\displaystyle=u_{P}^{\beta_{P}(v)/p}\mathcal{C}(x_{P}^{d_{P}(\gamma_{P}(v-\epsilon_{P})-p\alpha_{P}(v))+v}dx)$
$\displaystyle=u_{P}^{\beta_{P}(v)/p}\mathcal{C}(x_{P}^{(p-1)(v-\epsilon_{P})+v-pd_{P}\alpha_{P}(v)}dx)$
$\displaystyle=u_{P}^{\beta_{P}(v)/p}\mathcal{C}(x_{P}^{p(v-d_{P}\alpha_{P}(v)-\epsilon_{P})+\epsilon_{P}}dx)$
$\displaystyle=u_{P}^{\beta_{P}(v)/p}x_{P}^{v-d_{P}\alpha_{P}(v)}dx.$

This term is exactly $\kappa(\mathcal{C}(\omega))$. It is clear that this is
the only contribution in $\mathcal{C}(\omega)$ to
$\kappa(\mathcal{C}(\omega))$. Therefore the coefficient of
$\kappa(\mathcal{C}(\omega))$ in this case is

$(-1)^{\beta(v)}u_{P}^{\beta_{P}(v)/p}\binom{a_{1}}{\beta(v)}.$

Now let us suppose $a_{1}<\beta(v)$. In that case we must set $j=a_{2}-1$ in
Equation (3.8) to obtain the right exponent of $y_{2}$. Note that $a_{2}\geq
1$ since $\omega$ is an element of $H$. Now that $j=a_{2}-1$ has been
specified, $l$ can still be $0$ or $1$, giving two terms:

$y_{2}^{a_{2}-1}\left(-\mathcal{C}(y_{1}^{a_{1}}h(x)x_{P}^{v}dx)-\mathcal{C}(g(y_{1})y_{1}^{a_{1}}x_{P}^{v}dx)\right).$

Since $a_{1}<a_{1}-\beta(v)+p$ and $\mathcal{C}$ cannot increase the exponent
of $y_{1}$, the first term cannot contribute to $\kappa(\mathcal{C}(\omega))$.
For the second term, recall Equation (3.3) for $g(y_{1})$. We substitute to
get

$\displaystyle\sum_{i=1}^{p-1}(-1)^{i+1}\frac{(p-1)!}{i!(p-i)!}\mathcal{C}(y_{1}^{p(p-i)+i+a_{1}}x_{P}^{v}dx)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{p-1}(-1)^{i+1}\frac{(p-1)!}{i!(p-i)!}y_{1}^{p-i}\mathcal{C}(y_{1}^{i+a_{1}}x_{P}^{v}dx)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{p-1}(-1)^{i+1}\frac{(p-1)!}{i!(p-i)!}\sum_{m=0}^{i+a_{1}}(-1)^{m}\binom{a_{1}+i}{m}y_{1}^{p-m+a_{1}}\mathcal{C}(f(x)^{m}x_{P}^{v}dx).$

For the exponent of $y_{1}$, we must set $m=\beta_{P}(v)$. For the binomial
coefficients to be non-zero, we must have $\beta_{P}(v)-a_{1}\leq i\leq
p-1-a_{1}$. This choice of $m$ and the leading term of $f_{P}(x_{P})$ yields
the term

$\sum_{i=\beta_{P}(v)-a_{1}}^{p-1-a_{1}}\frac{(p-1)!}{i!(p-i)!}(-1)^{\beta_{P}(v)+i+1}\binom{a_{1}+i}{\beta_{P}(v)}y_{2}^{a_{2}-1}y_{1}^{a_{1}-\beta(v)+p}\mathcal{C}(u_{P}^{\beta_{P}(v)}x_{P}^{d_{P}\beta(v)+v}dx).$

As before, $\mathcal{C}(x_{P}^{d_{P}\beta(v)+v}dx)=x^{v-d_{P}\alpha(v)},$
which finishes the proof that this is a contribution to
$\kappa(\mathcal{C}(\omega))$. Again, one goes through the same steps to show
that there is no other contribution to $\kappa(\mathcal{C}(\omega))$. ∎

To make future calculations easier, we give a sufficient criterion for
$c_{\omega}$ to be non-zero.

######  Lemma 3.4.

Let $P\in B_{1}$ and let $\omega=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx$ be a
differential in $H_{P}$. If $\beta_{P}(v)\leq a_{1}$ or $\beta_{P}(v)\geq
p-2$, then $c_{\omega}$ is non-zero.

###### Proof.

If $\beta_{P}(v)\leq a_{1}$, it is clear from Lemma 3.3 that $c_{\omega}\neq
0$. If $\beta_{P}(v)=p-1$, then $c_{\omega}$ is a sum consisting of only one
non-zero term, so again $c_{\omega}\neq 0$.

In the case $\beta_{P}(v)=p-2$, $c_{\omega}$ is a sum with two terms. To be
precise, $c_{\omega}$ vanishes if and only if

$\sum_{i=p-2-a_{1}}^{p-1-a_{1}}\frac{(-1)^{i}(a_{1}+i)!}{i!(p-i)!\beta_{P}(v)!(a_{1}+i-\beta_{P}(v))!}=0.$

This sum vanishes precisely when its two terms are each other’s negative:

$\displaystyle\frac{(p-2)!}{(p-2-a_{1})!(a_{1}+2)!\beta_{P}(v)!(p-2-\beta_{P}(v))!}$
$\displaystyle=$
$\displaystyle\frac{(p-1)!}{(p-1-a_{1})!(a_{1}+1)!\beta_{P}(v)!(p-1-\beta_{P}(v))!},$
whence
$\displaystyle(p-2)!(p-1-a_{1})!(a_{1}+1)!\beta_{P}(v)!(p-1-\beta_{P}(v))!$
$\displaystyle=$
$\displaystyle(p-1)!(p-2-a_{1})!(a_{1}+2)!\beta_{P}(v)!(p-2-\beta_{P}(v))!.$

Using the assumption $\beta_{P}(v)=p-2$ yields

$(p-2-\beta_{P}(v))!=(p-1-\beta_{P}(v))!=1.$

Then dividing both sides by $(p-2)!(p-2-a_{1})!(a_{1}+1)\beta_{P}(v)!$ gives

$\displaystyle(p-1-a_{1})$ $\displaystyle=(p-1)(a_{1}+2)$ $\displaystyle-
a_{1}-1$ $\displaystyle=-(a_{1}+2)$ $\displaystyle-1$ $\displaystyle=-2.$

From this contradiction we infer that $c_{\omega}\neq 0$ in the case
$\beta_{P}(v)=p-2$. ∎

###  3.2. Key terms at poles of $h$

Let $B_{2}\subset\mathbb{P}^{1}$ denote the branch locus of the cover
$\phi_{2}:Y_{2}\to\mathbb{P}^{1}$. We consider a point $P\in B_{2}\setminus
B_{1}$. This implies that $f$ is regular at $P$ but $h$ has a pole at $P$. We
denote the order of this pole by $e_{P}$. By minimality, we have $e_{P}|p-1$
and define $\delta_{P}=\frac{p-1}{e_{P}}$.

Proposition 2.12 yields the set

(3.11) $\displaystyle
W_{P}:=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\;|\;0\leq
a_{1},a_{2}<p,\\\ 0<pv\leq e_{P}(p-1-a_{2})+p-1\end{cases}\Bigg{\\}}.$

For a differential $\omega=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in W_{P}$,
Lemma 3.1 holds, as the proof is independent of whether or not $f$ has a pole
at $P$. This allows us to define the key term of $\omega$.

######  Definition 3.5.

Let $\alpha_{p}(v):=\left\lfloor\frac{\delta_{P}(v-1)}{p}\right\rfloor$ and
$\beta_{P}(v):=\delta_{P}(v-1)-p\alpha_{P}(v)$. Define

$H_{P}:=\\{y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in W_{P}\;|\;\beta_{P}(v)\leq
a_{2}\\}$

For $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H_{P}$ we define the
associated key term $\kappa(\mathcal{C}(\omega))$ as

$\kappa(\mathcal{C}(\omega)):=y_{2}^{a_{2}-\beta_{P}(v)}y_{1}^{a_{1}}x_{P}^{v-e_{P}\alpha_{P}(v)}dx.$

######  Lemma 3.6.

Let $\omega=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}\in H_{P}$. Then the
coefficient $c_{\omega}$ of $\kappa(\mathcal{C}(\omega))$ in
$\mathcal{C}(\omega))$ is non-zero.

###### Proof.

Recall Equation (3.8):

$\mathcal{C}(\omega)=\sum_{0\leq j+l\leq
a_{2}}(-1)^{a_{2}-j}\binom{a_{2}}{j\,l}y_{2}^{j}\mathcal{C}(g(y_{1})^{l}h(x)^{a_{2}-j-l}y_{1}^{a_{1}}x_{P}^{v}dx).$

In order to obtain the exponent $a_{2}-\beta_{p}(v)$ of $y_{2}$, we must pick
$j=a_{2}-\beta_{P}(v)$. Picking $l=0$ results in a term

$y_{2}^{a_{2}-\beta_{P}(v)}\mathcal{C}(y_{1}^{a_{1}}h(x)^{\beta_{P}(v)}x_{P}^{v}dx)=\sum_{m=0}^{a_{1}}(-1)^{m}\binom{a_{1}}{m}y_{2}^{a_{2}-\beta_{P}(v)}y_{1}^{a_{1}-m}\mathcal{C}(h(x)^{\beta_{P}(v)}f(x)^{m}x_{P}^{v}dx).$

We pick $m=0$. The differential $\mathcal{C}(h(x)^{\beta_{P}(v)}x_{P}^{v}dx)$
has a non-zero term

$\displaystyle\mathcal{C}(x_{P}^{e_{P}\beta_{P}(v)+v}dx)$
$\displaystyle=\mathcal{C}(x_{P}^{e_{P}(\delta_{P}(v-1)-p\alpha_{P}(v))+v}$
$\displaystyle=x_{P}^{v-e_{P}\alpha_{P}(v)}dx.$

We do not get such a term if we pick $l>0$. This shows that there is a single
non-zero contribution to the coefficient of $\kappa(\mathcal{C}(\omega))$ in
$\mathcal{C}(\omega)$, implying that the coefficient is non-zero. ∎

###  3.3. An upper bound on the $a$-number

We now use the machinery of key terms to provide a conditional upper bound on
the $a$-number $a_{Y_{2}}$. We do this by showing the Cartier operator is
injective when restricted to a suitable subspace of
$\operatorname{H}^{0}(Y_{2},\Omega_{Y_{2}}^{1})$, whose dimension bounds the
rank of the Cartier operator from below.

First we define $W:=\bigcup_{P\in B_{2}}W_{P}$, such that $W$ is a basis of
$\operatorname{H}^{0}(Y_{2},\Omega_{Y_{2}}^{1})$. Let $H:=\bigcup_{P\in
B_{2}}H_{P}$ be the subset of $W$ consisting of differentials that have a key
term. Finally, let $K$ be the set of key terms. As opposed to the case
described in [FP13], it is now possible for two differentials to have the same
key term. The following lemma describes exactly when this happens.

######  Lemma 3.7.

If two differentials $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H_{P}$
and $\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{Q}^{w}dx\in H_{Q}$ have the
same key term, then $P=Q$. In the case $P\in B_{1}$, then $\omega$ and
$\omega^{\prime}$ have the same key term if and only if the following
equations are satisfied:

(3.12) $\displaystyle v+d_{P}(a_{1}+pa_{2})$
$\displaystyle=w+d_{P}(b_{1}+pb_{2})$ (3.13) $\displaystyle
v-d_{P}\alpha_{P}(v)$ $\displaystyle=w-d_{P}\alpha_{P}(w).$

In the case $P\in B_{2}\setminus B_{1}$, $\omega$ and $\omega^{\prime}$ have
the same key term if and only if the following equations are satisfied:

(3.14) $\displaystyle v+e_{P}a_{2}$ $\displaystyle=w+e_{P}b_{2}$ (3.15)
$\displaystyle a_{1}$ $\displaystyle=b_{1}$ (3.16) $\displaystyle
v-e_{P}\alpha_{P}(v)$ $\displaystyle=w-e_{P}\alpha_{P}(w).$

###### Proof.

Assume $\kappa(\mathcal{C}(\omega))=\kappa(\mathcal{C}(\omega^{\prime}))$.
First we show that $P=Q$. Note that, if $P\neq Q$ then the key terms of
$\omega$ and $\omega^{\prime}$ must not involve $x_{P}$ and $x_{Q}$, such that

$v-d_{P}\alpha_{P}(v)=w-d_{Q}\alpha_{Q}(v)=0.$

However, for $P\neq\infty$ we have

$v-d_{P}\alpha_{P}(v)=v-d_{P}\left\lfloor\frac{\gamma_{P}(v-1)}{p}\right\rfloor\geq
v-\frac{d_{P}\gamma_{P}(v-1)}{p}=v-\frac{(p-1)(v-1)}{p}>0.$

This implies that $P=Q=\infty$ and hence $P$ and $Q$ are not distinct.

Assume $P\in B_{1}$. We show that Equations (3.12) and (3.13) hold. Equating
the exponents of $x_{P}$ in $\kappa(\mathcal{C}(\omega))$ and
$\kappa(\mathcal{C}(\omega^{\prime}))$ immediately gives Equation (3.13), so
the rest of the proof revolves around showing Equation (3.12). By equating the
exponent of $y_{2}$ in the key terms, we obtain
$b_{2}\in\\{a_{2}-1,a_{2},a_{2}+1\\}$. Without loss of generality, we assume
$b_{2}\geq a_{2}$, such that only the cases $b_{2}=a_{2}$ and $b_{2}=a_{2}+1$
remain.

In the case $b_{2}=a_{2}$, equating the exponent of $x_{P}$ in the key terms
gives

(3.17) $v=w+d_{P}(\alpha_{P}(v)-\alpha_{P}(w)).$

Substituting this into the equation for the exponents of $y_{1}$ yields

$\displaystyle a_{1}-\beta_{P}(v)$ $\displaystyle=b_{1}-\beta_{P}(w)$
$\displaystyle a_{1}-\gamma_{P}(v-\epsilon_{P})+p\alpha_{P}(v)$
$\displaystyle=b_{1}-\gamma_{P}(w-\epsilon_{P})+p\alpha_{P}(w)$ $\displaystyle
a_{1}-\gamma_{P}(w+d_{P}(\alpha_{P}(v)-\alpha_{P}(w)-\epsilon_{P})+p\alpha_{P}(v)$
$\displaystyle=b_{1}-\gamma_{P}(w-\epsilon_{P})+p\alpha_{P}(w)$ $\displaystyle
a_{1}-(p-1)(\alpha_{P}(v)-\alpha_{P}(w))+p\alpha_{P}(v)$
$\displaystyle=b_{1}+p\alpha_{P}(w)$
$\displaystyle\alpha_{P}(v)-\alpha_{P}(w)$ $\displaystyle=b_{1}-a_{1}.$

Substituting this back into Equation (3.17) then gives
$v=w+d_{P}(b_{1}-a_{1})$ and hence $v+d_{P}a_{1}=w+d_{P}b_{1}$, as desired.

On the other hand, suppose $b_{2}=a_{2}+1$. In that case, Equation (3.17)
still follows from equating the exponent of $x_{P}$. Equating the exponent of
$y_{1}$ yields $a_{1}-\beta_{P}(v)=b_{1}-\beta_{P}(w).$ Following the same
steps as in the case $b_{2}=a_{2}$ gives
$\alpha_{P}(v)-\alpha_{P}(w)=b_{1}-a_{1}+p$ and hence
$v=w+d_{P}(b_{1}-a_{1}+p)$. Rearranging this and using $b_{2}-a_{2}=1$ gives
the desired Equation (3.12).

For the opposite direction, assume Equations (3.12) and (3.13) hold. By
Equation (3.13), $\kappa(\mathcal{C}(\omega))$ and
$\kappa(\mathcal{C}(\omega^{\prime}))$ have the same exponent of $x_{P}$.
Substituting $v=w+d_{P}(\alpha_{P}(v)-\alpha_{P}(w))$ in Equation (3.12)
yields

$\alpha_{P}(v)-\alpha_{P}(w)=b_{1}-a_{1}+p(b_{2}-a_{2}).$

This implies

$\displaystyle b_{1}-\beta_{P}(w)+p(b_{2}-a_{2})$
$\displaystyle=a_{1}+\alpha_{P}(v)-\alpha_{P}(w)-\beta_{P}(w)$
$\displaystyle=a_{1}+\alpha_{P}(v)-\alpha_{P}(w)-\gamma_{P}(w-\epsilon_{P})+p\alpha_{P}(w)$
$\displaystyle=a_{1}+\alpha_{P}(v)-\alpha_{P}(w)$ $\displaystyle\hskip
14.22636pt-\gamma_{P}(v+d_{P}(\alpha_{P}(w)-\alpha_{P}(v))-\epsilon_{P})+p\alpha_{P}(w)$
$\displaystyle=a_{1}+\alpha_{P}(v)-\alpha_{P}(w)-\gamma_{P}(v-\epsilon_{P})$
$\displaystyle\hskip
14.22636pt-(p-1)(\alpha_{P}(w)-\alpha_{P}(v))+p\alpha_{P}(w)$
$\displaystyle=a_{1}-\gamma_{P}(v-\epsilon_{P})+p\alpha_{P}(v)$
$\displaystyle=a_{1}-\beta_{P}(v).$

Using this, one verifies that in all cases for $b_{2}$, the exponents of
$y_{1}$ and $y_{2}$ in $\kappa(\mathcal{C}(\omega))$ and
$\kappa(\mathcal{C}(\omega^{\prime}))$ agree. Thus $\omega$ and
$\omega^{\prime}$ have the same key term.

We now consider the case $P\in B_{2}\setminus B_{1}$. Equation (3.16) follows
immediately from equating the exponent of $x_{P}$ in the key term. Similarly,
Equation (3.15) follows from equating the exponents of $y_{1}$. For Equation
(3.14), we write $v=w+e_{P}r$ and $\alpha_{P}(v)=\alpha_{P}(w)-r$ for some
integer $r$. By equating the exponents of $y_{2}$ in the key term, we obtain

$\displaystyle a_{2}-b_{2}$ $\displaystyle=\beta_{P}(v)-\beta_{P}(w)$
$\displaystyle=\delta_{P}(v-1)-p\alpha_{P}(v)-(\delta(w-1)-p\alpha_{P}(w))$
$\displaystyle=\delta_{P}(v-w)+p(\alpha_{P}(v)-\alpha_{P}(w))$
$\displaystyle=\delta_{P}e_{P}r+pr=-r.$

Thus Equation (3.14) also holds.

For the opposite direction, assume Equations (3.14), (3.15) and (3.16) hold.
Then we only need to show that $y_{2}$ has the same exponent in both key
terms. This is achieved by writing $a_{2}=b_{2}-r$ and $v=w+e_{P}r$. Then a
similar calculation as above shows

$a_{2}-\beta_{P}(v)=b_{2}-\beta_{P}(w),$

as desired. ∎

To display a minor of the Cartier-Manin matrix in a convenient form, we define
a partial order on $W$. We first order the branch points in such a way that
$\infty$ is the smallest.

######  Definition 3.8.

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{Q}^{w}dx$. We say
$\omega^{\prime}\prec\omega$ in any of the following cases:

  1. (i)

$Q\neq P$ and $b_{1}+pb_{2}<a_{1}+pa_{2}$;

  2. (ii)

$Q<P$ and $b_{1}+pb_{2}=a_{1}+pa_{2}$;

  3. (iii)

$Q=P\in B_{1}$ and $w+d_{P}(b_{1}+pb_{2})<v+d_{P}(a_{1}+pa_{2})$;

  4. (iv)

$Q=P\in B_{1}$ and $w+d_{P}(b_{1}+pb_{2})=v+d_{P}(a_{1}+pa_{2})$ and
$w-d_{P}\alpha_{P}(w)>v-d_{P}\alpha_{P}(v)$.

  5. (v)

$Q=P\in B_{2}\setminus B_{1}$ and $w+e_{P}a_{2}<v+e_{P}a_{2}$.

  6. (vi)

$Q=P\in B_{2}\setminus B_{1}$ and $w+e_{P}b_{2}=v+e_{P}a_{2}$ and
$b_{1}<a_{1}$.

  7. (vii)

$Q=P\in B_{2}\setminus B_{1}$ and $w+e_{P}b_{2}=c+e_{P}a_{2}$ and
$b_{1}=a_{1}$ and $w-e_{P}\alpha_{P}(w)>v-d_{P}\alpha_{P}(v)$.

######  Remark 3.9.

By comparing Definition 3.8 to Lemma 3.7, note that two differentials $\omega$
and $\omega^{\prime}$ are incomparable by the order $\prec$ if and only if
they have the same key term.

The goal of introducing the partial order $\prec$ is to prove the following
lemma.

######  Lemma 3.10.

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{Q}^{w}dx\in W$ with
$\omega^{\prime}\prec\omega$. Then the coefficient of the key term
$\kappa(\mathcal{C}(\omega))$ is zero in $\mathcal{C}(\omega^{\prime})$.

In order to prove Lemma 3.10, we prove seven lemmas, corresponding to the
cases of Definition 3.8. In each case, we have

(3.18) $\mathcal{C}(\omega^{\prime})=\sum_{0\leq j+l\leq
b_{2}}(-1)^{b_{2}-j}\binom{b_{2}}{j\,l}y_{2}^{j}\mathcal{C}(g(y_{1})^{l}h(x)^{b_{2}-j-l}y_{1}^{b_{1}}x_{Q}^{w}dx).$

Recall also the formula for the key term $\kappa(\mathcal{C}(\omega)):$

(3.19)
$\kappa(\mathcal{C}(\omega)):=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}-\beta_{P}(v)}x_{P}^{v-d_{P}\alpha_{P}(v)}dx&\hbox{{if
} $P\in B_{1}$ { and } $a_{1}\geq\beta_{P}(v)$}\\\
y_{2}^{a_{2}-1}y_{1}^{a_{1}-\beta_{P}(v)+p}x_{P}^{v-d_{P}\alpha_{P}(v)}dx&\hbox{{if
} $P\in B_{1}$ { and } $a_{1}<\beta_{P}(v)$}\\\
y_{2}^{a_{2}-\beta_{P}(v)}y_{1}^{a_{1}}x_{P}^{v-e_{P}\alpha_{P}(v)}dx&\hbox{{if
} $P\in B_{2}\setminus B_{1}$.}\end{cases}$

######  Lemma 3.11 (Case (i) of Definition 3.8).

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{Q}^{w}dx\in W$ with $Q\neq P$
and $b_{1}+pb_{2}<a_{1}+pa_{2}$. Then the coefficient of
$\kappa(\mathcal{C}(\omega))$ is zero in $\mathcal{C}(\omega^{\prime})$.

###### Proof.

Since $0\leq a_{1},b_{1}<p$, the assumption implies $b_{2}\leq a_{2}$. First
assume that $P\in B_{1}$. Then it is clear from Equation (3.18) and Equation
(3.19) that we are done in the case $b_{2}<a_{2}-1$. In the case
$b_{2}=a_{2}-1$, we must have $a_{1}<\beta_{P}(v)$. In Equation (3.18), we
need to specialize to $j=a_{2}-1$ and $l=0$ to get the right exponent of
$y_{2}$. This leaves us with a term

$y_{2}^{a_{2}-1}\mathcal{C}(y_{1}^{b_{1}}x_{Q}^{w}dx)=\sum_{m=1}^{b_{1}}y_{2}^{a_{2}-1}y_{1}^{b_{1}-m}\mathcal{C}(f(x)^{m}x_{Q}^{w}dx).$

To get the right exponent of $y_{1}$, we must pick
$m=b_{1}-a_{1}+\beta_{P}(v)-p$. Then the differential
$f(x)^{b_{1}-a_{1}+\beta_{P}(v)-p}x_{Q}^{w}dx$ has a pole of order at most
$d_{P}(b_{1}-a_{1}+\beta_{P}(v)-p)$ at $P$. However, a pole of order
$d_{P}\beta_{P}(v)+v$ is needed to produce the exponent $v-d_{P}\alpha_{P}(v)$
of $x_{P}$. Since $b_{1}-a_{1}-p<0$, it follows that this exponent is not
attained and hence $\kappa(\mathcal{C}(\omega))$ is zero in
$\mathcal{C}(\omega^{\prime})$.

For $b_{2}=a_{2}$ the proof is similar. In the case $a_{1}\geq\beta_{P}(v)$,
we must pick $j=a_{2}$ and $l=0$ in Equation (3.18), yielding the term

$y_{2}^{a_{2}}\mathcal{C}(y_{1}^{b_{1}}x_{Q}^{w}dx)=\sum_{m=0}^{b_{1}}(-1)^{m}\binom{b_{1}}{m}y_{2}^{a_{2}}y_{1}^{b_{1}-m}\mathcal{C}(f(x)^{m}x_{Q}^{w}dx).$

In order to achieve the right exponent of $y_{1}$, we must specialize
$m=b_{1}-a_{1}+\beta_{P}(v)$. Since $\omega^{\prime}\prec\omega$, we have
$b_{1}<a_{1}$ and it follows that the right exponent of $x_{P}$ is not
achieved. In the case $a_{1}<\beta_{P}(v)$, we must specialize to $j=a_{2}-1$
in Equation (3.18). The possibilities $l=0$ and $l=1$ result in the terms
$y_{2}^{a_{2}-1}\mathcal{C}(g(y_{1})y_{1}^{b_{1}}x_{Q}^{w}dx)$ and
$y_{2}^{a_{2}-1}\mathcal{C}(y_{1}^{b_{1}}h(x)x_{Q}^{w}dx).$ The second term
could never attain the right exponent of $y_{1}$, since $b_{1}<a_{1}$. On the
other hand, for the first term we substitute the definition of $g(y_{1})$ to
obtain

$\displaystyle\mathcal{C}(g(y_{1})y_{1}^{b_{1}}x_{Q}^{w}dx)$
$\displaystyle=\sum_{i=1}^{p-1}(-1)^{i}\frac{(p-1)!}{i!(p-i)!}y_{1}^{p-i}\mathcal{C}(y_{1}^{b_{1}+i}x_{Q}^{w}dx)$
$\displaystyle=\sum_{i=1}^{p-1}\frac{(p-1)!}{i!(p-i)!}\sum_{m=0}^{b_{1}+i}(-1)^{m+i+1}\binom{b_{1}+i}{m}y_{1}^{p-m+b_{1}}\mathcal{C}(f(x)^{m}x_{Q}^{w}dx).$

For any $i$, we must pick $m=b_{1}-a_{1}+\beta_{P}(v)$ to obtain the right
exponent of $y_{1}$. Then the differential
$f(x)^{b_{1}-a_{1}+\beta_{P}(v)}x_{Q}^{w}dx$ has a pole of order at most
$d_{P}(b_{1}-a_{1}+\beta_{P}(v))$ at $P$. Since $b_{1}<a_{1}$, this is smaller
than the required pole order $d_{P}\beta_{P}(v)+v$ to obtain the right
exponent of $x_{P}$. Hence the coefficient of $\kappa(\mathcal{C}(\omega))$ is
again zero in $\mathcal{C}(\omega^{\prime})$.

Now, assume that $P\in B_{2}\setminus B_{1}$. In Equation (3.18), we must pick
$j=a_{2}-\beta_{P}(v)$. In the case $b_{1}<a_{1}$, the required exponent of
$y_{1}$ can never be obtained. On the other hand, in the case $b_{2}<a_{2}$,
the required exponent $\beta_{P}(v)$ of $h(x)$ cannot be obtained. In both
cases, $\mathcal{C}(\omega^{\prime})$ cannot have a non-zero term
$y_{2}^{a_{2}-\beta_{P}(v)}y_{1}^{a_{1}}x_{P}^{v-e_{P}\alpha_{P}(v)}dx$, which
finishes the proof. ∎

######  Lemma 3.12 (Case (ii) of Definition 3.8).

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{Q}^{w}dx\in W$ with $Q<P$ and
$b_{1}+pb_{2}=a_{1}+pa_{2}$. Then the coefficient of
$\kappa(\mathcal{C}(\omega))$ is zero in $\mathcal{C}(\omega^{\prime})$.

###### Proof.

The proof is similar to the proof of Lemma 3.11. Note that we have
$b_{2}=a_{2}$ and $b_{1}=a_{1}$. First suppose $P\in B_{1}$. After attaining
the right exponents of $y_{2}$ and $y_{1}$, we are left with a differential on
$\mathbb{P}^{1}$ that has a pole of order at most $d_{P}\beta_{P}(v)$ at $P$,
whereas a pole of order $d_{P}\beta_{P}(v)+v$ is needed. Then, it follows from
the assumption $Q<P$ that $P\neq\infty$ (since $\infty$ is the smallest branch
point in our ordering) and therefore $v>0$. Thus the required pole order is
not attained and therefore the coefficient of $\kappa(\mathcal{C}(\omega))$ is
zero in $\mathcal{C}(\omega^{\prime})$.

Similarly, suppose $P\in B_{2}\setminus B_{1}$. In Equation (3.18), we must
choose $j=a_{2}-\beta_{P}(v)$, such that the resulting differential of
$\mathbb{P}^{1}$ has a pole of order at most $e_{P}\beta_{P}(v)$, whereas a
pole of order $e_{P}\beta_{P}(v)+v$ is needed. Again, noting $v>0$ finishes
the proof. ∎

######  Lemma 3.13 (Case (iii) of Definition 3.8).

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{P}^{w}dx\in W$ with $P\in
B_{1}$ and

(3.20) $w+d_{P}(b_{1}+pb_{2})<v+d_{P}(a_{1}+pa_{2}).$

Then the coefficient of $\kappa(\mathcal{C}(\omega))$ is zero in
$\mathcal{C}(\omega^{\prime})$.

###### Proof.

The same principle as before is used: we assume there is a term in
$\mathcal{C}(\omega^{\prime})$ that contributes to
$\kappa(\mathcal{C}(\omega))$. We first specialize to the terms that have the
required exponent of $y_{2}$ and $y_{1}$. Then we are left with a differential
on $\mathbb{P}^{1}$ whose pole order at $P$ is too small to give the required
exponent of $x_{P}$.

Equation (3.20) and Equation (3.7) together imply that $b_{2}\leq a_{2}+1$.
Moreover, by Equations (3.18) and (3.19) it follows that we are done if
$b_{2}<a_{2}-1$. We split the rest of the proof up into the two possibilities
for $\kappa(\mathcal{C}(\omega))$: either $a_{1}\geq\beta_{P}(v)$ or
$a_{1}<\beta_{P}(v)$.

In the case $a_{1}\geq\beta_{P}(v)$, we have

$\kappa(\mathcal{C}(\omega))=y_{2}^{a_{2}}y_{1}^{a_{1}-\beta_{P}(v)}x^{v-d_{P}\alpha_{P}(v)}dx.$

This forces $b_{2}\geq a_{2}$, so $b_{2}$ equals either $a_{2}$ or $a_{2}+1$.
In the case $b_{2}=a_{2}$, possible contributions to
$\kappa(\mathcal{C}(\omega))$ of $\mathcal{C}(\omega^{\prime})$ come from
setting $j=a_{2}$ and $l=0$ in Equation (3.18). This leaves the term

$y_{2}^{a_{2}}\mathcal{C}(y_{1}^{b_{1}}x_{Q}^{w}dx)=\sum_{m=0}^{b_{1}}(-1)^{m+1}\binom{b_{1}}{m}y_{2}^{a_{2}}y_{1}^{b_{1}-m}\mathcal{C}(f(x)^{m}x_{P}^{w}dx).$

For the right exponent of $y_{1}$, we need $m=b_{1}-a_{1}+\beta_{P}(v)$. This
results in the term $\mathcal{C}(f(x)^{b_{1}-a_{1}+\beta_{P}(v)}x_{P}^{w})$,
which has a pole of order $d_{P}(b_{1}-a_{1}+\beta_{P}(v))+w$, which is
smaller than the required $d_{P}\beta_{P}(v)+v$ by Equation (3.20). The case
$b_{2}=a_{2}+1$ is handled similarly. In Equation (3.18), possible
contributions come from $j=a_{2}$ and $l\in\\{0,1\\}$. In both cases for $l$,
the resulting differential $f(x)^{m}x_{P}^{w}dx$ does not have sufficiently
large pole order at $P$ to contribute to $\kappa(\mathcal{C}(\omega))$.

In the case $a_{1}<\beta_{P}(v)$, we have

$\kappa(\mathcal{C}(\omega))=y_{2}^{a_{2}-1}y_{1}^{a_{1}-\beta_{P}(v)+p}x_{P}^{v-d_{P}\alpha_{P}(v)}dx.$

This leaves three possibilities for $b_{2}$: $a_{2}-1$, $a_{2}$ and $a_{2}+1$.
We treat only the case $b_{2}=a_{2}$, as the other cases are similar. In
Equation (3.18) we must specialize to $j=a_{2}-1$. The possibilities $l=0$ and
$l=1$ give the two terms
$y_{2}^{a_{2}-1}\mathcal{C}(g(y_{1})y_{1}^{b_{1}}x_{P}^{w}dx)$ and
$y_{2}^{a_{2}-1}\mathcal{C}(y_{1}^{b_{1}}h(x)x_{P}^{w}dx).$ For the first
term, we expand

$\displaystyle\mathcal{C}(g(y_{1})y_{1}^{b_{1}}x_{P}^{w}dx)$
$\displaystyle=\sum_{i=1}^{p-1}(-1)^{i}\frac{(p-1)!}{i!(p-i)!}y_{1}^{p-i}\mathcal{C}(y_{1}^{b_{1}+i}x_{P}^{w}dx)$
$\displaystyle=\sum_{i=1}^{p-1}\frac{(p-1)!}{i!(p-i)!}\sum_{m=0}^{b_{1}+i}(-1)^{m+i+1}\binom{b_{1}+i}{m}y_{1}^{p-m+b_{1}}\mathcal{C}(f(x)^{m}x_{P}^{w}dx).$

To obtain the right exponent of $y_{1}$, setting $m=b_{1}-a_{1}+\beta_{P}(v)$
is needed. Then Equation (3.20) implies that
$\mathcal{C}(f(x)^{m}x_{P}^{w}dx)$ does not have a sufficiently large pole
order at $P$. For the second term, we expand

$\mathcal{C}(y_{1}^{b_{1}}h(x)x_{P}^{w}dx)=\sum_{m=1}^{b_{1}}(-1)^{i}\binom{b_{1}}{m}y_{1}^{b_{1}-m}\mathcal{C}(h(x)f(x)^{m}x_{P}^{w}dx).$

The right exponent of $y_{1}$ results only from
$m=b_{1}-a_{1}+\beta_{P}(v)-p$. Then since the pole order of $h(x)$ at $P$ is
smaller than $pd_{P}$ (this is implied by the assumption that the upper jumps
are minimal), the differential $h(x)f(x)^{m}x_{P}^{w}dx$ has pole order
smaller than $d_{P}(b_{1}-a_{1}+\beta_{P}(v))+w$. By Equation (3.20), this is
smaller than the required $v+d_{P}\beta_{P}(v)$. This implies that the
coefficient of $\kappa(\mathcal{C}(\omega))$ is zero in
$\mathcal{C}(\omega^{\prime})$. ∎

######  Lemma 3.14 (Case (iv) of Definition 3.8).

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{P}^{w}dx\in W$ with $P\in
B_{1}$ and $w+d_{P}(b_{1}+pb_{2})=v+d_{P}(a_{1}+pa_{2})$ and
$w-d_{P}\alpha_{P}(w)>v-d_{P}\alpha_{P}(v)$. Then the coefficient of
$\kappa(\mathcal{C}(\omega))$ is zero in $\mathcal{C}(\omega^{\prime})$.

###### Proof.

From the first condition, we infer $w\equiv v\bmod{d_{P}}$, so we write
$w=v+ed_{P}$ and $b_{1}+pb_{2}=a_{1}+pa_{2}-e$. By the second condition, we
must have

$\displaystyle\alpha_{P}(w)$
$\displaystyle=\alpha_{P}(v+ed_{P})=\alpha_{P}(v)+e-1.$ This implies
$\displaystyle\beta_{P}(w)$
$\displaystyle=\beta_{P}(v+ed_{P})=\beta_{P}(v)-e+p.$

Note that this forces $e>\beta_{P}(v)$. We have

(3.21)
$\omega^{\prime}=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}-e}x_{P}^{v+ed_{P}}dx&\hbox{if
$a_{1}\geq e$}\\\ y_{2}^{a_{2}-1}y_{1}^{a_{1}-e+p}x_{P}^{v+ed_{P}}dx&\hbox{if
$a_{1}<e$}.\end{cases}$

We again consider the two possibilities determining the key term of $\omega$:
$a_{1}\geq\beta_{P}(v)$ and $a_{1}<\beta_{P}(v)$.

We treat the case $a_{1}\geq\beta_{P}(v)$ first, in which the key term is

$\kappa(\mathcal{C}(\omega))=y_{2}^{a_{2}}y_{1}^{a_{1}-\beta_{P}(v)}x_{P}^{v-d_{P}\alpha_{P}(v)}dx.$

To achieve this as a term in $\mathcal{C}(\omega^{\prime})$, we need
$b_{2}\geq a_{2}$. Appealing to Equation (3.21), only the case $a_{1}\geq e$
remains. In Equation (3.18) we need $j=a_{2}$ and $l=0$ to obtain the required
exponent of $y_{2}$. This gives the term

$y_{2}^{a_{2}}\mathcal{C}(y_{1}^{a_{1}-e}x_{P}^{v+ed_{P}}dx)=\sum_{m=0}^{a_{1}-e}(-1)^{m}\binom{a_{1}-e}{m}y_{2}^{a_{2}}y_{1}^{a_{1}-e-m}\mathcal{C}(f(x)^{m}x_{P}^{w}dx).$

In order to obtain the exponent of $y_{1}$, setting $m=\beta_{P}(v)-e$ is
required, but this is negative as we have ascertained $e>\beta_{P}(v)$. Thus
the right exponent of $y_{1}$ cannot be achieved.

On the other hand, assume $a_{1}<\beta_{P}(v)$. Note that this implies
$a_{1}<e$, since $\beta_{P}(v)<e$. By Equation (3.21), this determines
$\omega^{\prime}$. In Equation (3.18), we have to specialize to $j=a_{2}-1$
and $l=0$, yielding the term

$y_{2}^{a_{2}-1}\mathcal{C}(y_{1}^{a_{1}-e+p}x_{P}^{w}dx)=\sum_{m=0}^{a_{1}-e+p}(-1)^{m}\binom{a_{1}-e+p}{m}y_{2}^{a_{2}-1}y_{1}^{a_{1}-e+p-m}\mathcal{C}(f(x)^{m}x_{P}^{w}dx).$

Again the required exponent of $y_{1}$ can only be attained if
$m=\beta_{P}(v)-e$, which is negative. ∎

######  Lemma 3.15 (Case (v) of Definition 3.8).

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{P}^{w}$ with $P\in
B_{2}\setminus B_{1}$ and $w+e_{P}b_{2}<v+e_{P}a_{2}$. Then the coefficient of
$\kappa(\mathcal{C}(\omega))$ is zero in $\mathcal{C}(\omega^{\prime})$.

###### Proof.

In order to obtain the key term $\kappa(\mathcal{C}(\omega))$, we have to take
$j=a_{2}-\beta_{P}(v)$. Then the exponent of $h(x)$ can be at most
$b_{2}-a_{2}+\beta_{P}(v)$, making the exponent of $x_{P}$ at most
$e_{P}(b_{2}-a_{2}+\beta_{P}(v))+w$, which by our assumption is smaller than
the required $e_{P}\beta_{P}(v)+v$. ∎

######  Lemma 3.16 (Case (vi) of Definition 3.8).

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{P}^{w}$ with $P\in
B_{2}\setminus B_{1}$, $w+e_{P}b_{2}=v+e_{P}a_{2}$ and $b_{1}<a_{1}$. Then the
coefficient of $\kappa(\mathcal{C}(\omega))$ is zero in
$\mathcal{C}(\omega^{\prime})$.

###### Proof.

In Equation (3.18), we must take $j=a_{2}-\beta_{P}(v)$ and $l=0$ to get the
right exponents of $y_{2}$ and $x_{P}$. Then the differential

$\mathcal{C}(y_{1}^{b_{1}}h(x)^{b_{2}-a_{2}+\beta_{P}(v)}x_{P}^{w}dx)$

cannot have a term in which exponent of $y_{1}$ is $a_{1}$. This finishes the
proof. ∎

######  Lemma 3.17 (Case (vi) of Definition 3.8).

Let $\omega:=y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\in H$ and
$\omega^{\prime}:=y_{2}^{b_{2}}y_{1}^{b_{1}}x_{P}^{w}$ with $P\in
B_{2}\setminus B_{1}$, $w+e_{P}b_{2}=v+e_{P}a_{2}$, $b_{1}=a_{1}$ and
$w-e_{P}\alpha_{P}(w)>v-e_{P}\alpha_{P}(v)$. Then the coefficient of
$\kappa(\mathcal{C}(\omega))$ is zero in $\mathcal{C}(\omega^{\prime})$.

###### Proof.

The proof is similar to the proof of Lemma 3.14. Write $b_{2}=a_{2}-r$ and
$w=v+re_{P}$ for some integer $r$. From the inequality, we infer

$\displaystyle\alpha_{P}(w)$
$\displaystyle=\alpha_{P}(v+re_{P})=\alpha_{P}(v)+r-1$
$\displaystyle\beta_{P}(w)$
$\displaystyle=\beta_{P}(v+re_{P})=\beta_{P}(v)-r+p.$

This forces $r>\beta_{P}(v)$. Then in Equation (3.18), we need to take
$j=a_{2}-\beta_{P}(v)$ in order to obtain the right exponent of $y_{2}$. But
this is impossible as it exceeds $b_{2}=a_{2}-r$. Therefore there can be no
contribution to $\kappa(\mathcal{C}(\omega))$ in
$\mathcal{C}(\omega^{\prime})$. ∎

Lemma 3.10 now follows.

###### Proof of Lemma 3.10.

We have treated all seven cases of $\omega^{\prime}\prec\omega$ given in
Definition 3.8 in Lemmas 3.11, 3.12, 3.13, 3.14, 3.15, 3.16 and 3.17. In every
case, the coefficient of $\kappa(\mathcal{C}(\omega))$ is zero in
$\mathcal{C}(\omega^{\prime})$. ∎

We now finally use key terms to prove a lower bound on the rank of the Cartier
operator. Recall that $K$ is the set of key terms and that $H\subset W$ is the
set of basis regular differentials for which key terms are defined.

######  Theorem 3.18.

Assume that for each $\omega^{\prime}\in K$, there is an $\omega\in H$ such
that $\kappa(\mathcal{C}(\omega))=\omega^{\prime}$ and $c_{\omega}\neq 0$.
Then we have $\text{rk}(\mathcal{C})\geq\\#K$.

###### Proof.

By the assumptions, we can find a subset $H^{\prime}\subseteq H$ such that
every $\omega^{\prime}\in K$ has a unique $\omega\in H^{\prime}$ with the
properties $\kappa(\mathcal{C}(\omega))=\omega^{\prime}$ and $c_{\omega}\neq
0$. Note that $H^{\prime}$ and $K$ have the same cardinality. By Lemma 3.7 and
Remark 3.9, the partial order $\prec$ is a linear order when restricted to
$H^{\prime}$. We restrict the Cartier operator to $\text{span}_{k}H^{\prime}$
and then project to $\text{span}_{k}K$:

$\mathcal{C}|_{H^{\prime}}:\text{span}_{k}H^{\prime}\to\text{span}_{k}K.$

By Lemma 3.10, the matrix respresenting $C|_{H^{\prime}}$ with respect to the
basis $H^{\prime}$ ordered by $\prec$ has only zeros below the diagonal.
Furthermore, by Lemma 3.3, the diagonal entries are $c_{\omega}$ in the case
$P\in B_{1}$. These are non-zero by assumption. In the case $P\in
B_{2}\setminus B_{1}$, $c_{\omega}$ is non-zero unconditionally by Lemma 3.6.
Hence $\mathcal{C}|_{H^{\prime}}$ is invertible, implying that
$\text{rk}(\mathcal{C})\geq\\#K.$ ∎

######  Remark 3.19.

Optimistically one might hope to prove the equality
$\text{im}(\mathcal{C})=\text{span}_{k}K$. However, both inclusions fail
already in the case $p=3$. See Remark 4.2.

##  4\. The case $p=3$

In this section we use the machinery from the preceding sections to answer
Question 2.15 for $\mathbb{Z}/9\mathbb{Z}$-covers of $\mathbb{P}^{1}$ in
characteristic $3$. In Theorem 4.1, we prove that the upper bound from Theorem
3.18 is sharp for $\mathbb{Z}/9\mathbb{Z}$ covers of $\mathbb{P}^{1}$.

Let $\phi_{2}:Y_{2}\to\mathbb{P}^{1}$ be such a cover, with branching datum

$\left[\begin{array}[]{rrrrrrrr}3&\ldots&2&\ldots&0&\ldots&0&\ldots\\\
\makebox[0.0pt][l]{$\smash{\underbrace{\phantom{\begin{matrix}7&\ldots\end{matrix}}}_{\text{$n_{1}$}}}$}7&\ldots&\makebox[0.0pt][l]{$\smash{\underbrace{\phantom{\begin{matrix}4&\ldots\end{matrix}}}_{\text{$n_{2}$}}}$}4&\ldots&\makebox[0.0pt][l]{$\smash{\underbrace{\phantom{\begin{matrix}3&\ldots\end{matrix}}}_{\text{$n_{3}$}}}$}3&\ldots&\makebox[0.0pt][l]{$\smash{\underbrace{\phantom{\begin{matrix}{2}&\ldots\end{matrix}}}_{\text{$n_{4}$}}}$}{2}&\ldots\\\
\end{array}\right]^{\intercal}.$

Recall the models

(4.1) $\displaystyle Y_{1}:y_{1}^{3}-y_{1}$ $\displaystyle=f(x)$ (4.2)
$\displaystyle Y_{2}:y_{2}^{3}-y_{2}$ $\displaystyle=g(y_{1})+h(x),$

where

$g(y_{1})=\sum_{i=1}^{2}(-1)^{i}\frac{(2)!}{i!(3-i)!}y_{1}^{3(3-i)+i}=-y_{1}^{7}+y_{1}^{5}.$

As before, let $B_{1}$ be the branch locus of
$\phi_{1}:Y_{1}\to\mathbb{P}^{1}$. For $P\in B_{1}$, let $d_{P}$ be the unique
break in the lower-numbering ramification filtration at $P$. Minimality of the
cover (see Definition 2.14) implies $d_{P}|p-1$. We may assume $f$ has a pole
of order $d_{P}$ at every $P\in B_{1}$ and no poles elsewhere. By Theorem 2.4,
the pole order of $h$ at $P$ can be assumed to be less than $pd_{P}$. Without
loss of generality, we assume $\infty\in B_{1}$.

Similarly, let $B_{2}$ be the branch locus of the cover
$\phi_{2}:Y_{2}\to\mathbb{P}^{1}$. For $P\in B_{2}\setminus B_{1}$, let
$e_{P}$ be the ramification jump of the cover $Y_{2}\to Y_{1}$ at $P$. By
minimality, we have $e_{P}|p-1$.

In Theorem 4.1 we prove that the $a$-number of $Y_{2}$ depends only on the
ramification jumps $d_{P}$ and $e_{P}$. We do this by showing that the lower
bound provided by Theorem 3.18 is sharp. We compute the basis of differentials
$W_{P}$ given in Equation (3.5) and Equation (3.6). This results in six cases.
We distinguish between $P=\infty$, $P\in B_{1}\setminus\\{\infty\\}$ and $P\in
B_{2}\setminus B_{1}$. Moreover, we distinguish different cases depending on
whether the ramification jump ($d_{P}$ or $e_{P}$) equals $1$ or $2$. These
cases are treated in the following subsections. All of this information is
combined to prove Theorem 4.1.

In the case $p=3$, Lemma 3.4 guarantees that the coefficients $c_{\omega}$ are
non-zero. Note that by construction we have $0\leq\beta_{P}(v)\leq 2$. In the
case $\beta_{P}(v)=0$, the condition $\beta_{P}(v)\leq a_{1}$ is satisfied.
Otherwise, the condition $\beta_{P}(v)\geq p-2=1$ is satisfied, so Lemma 3.4
always applies. This is used throughout this section.

###  4.1. A pole of $f$ of order $1$ at infinity

This section concerns the action of $\mathcal{C}$ on $W_{\infty}$, assuming
$f$ has a pole at $\infty$ of order $d_{\infty}=1$. The definition of
$W_{\infty}$, from Equation (3.5), becomes

$\displaystyle W_{\infty}$ $\displaystyle=\begin{aligned}
\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x^{v}dx\;|\;0\leq a_{1},a_{2}<3,\\\
0\leq 9v\leq 3(2-a_{1})+7(2-a_{2})-10\end{cases}\Bigg{\\}}\end{aligned}$
$\displaystyle=\\{dx,xdx,y_{1}dx,y_{1}^{2}dx,y_{2}dx,y_{2}y_{1}dx\\}.$

We study the restricted Cartier operator
$\mathcal{C}_{W_{\infty}}:\text{span}_{k}W_{\infty}\to\operatorname{H}^{0}(Y_{2},\Omega_{Y_{2}}^{1})$,
using machinery from Section 3. For each element of $W_{\infty}$, we check
whether it is in $H_{\infty}$. For the differentials $\omega$ that lie in
$H_{\infty}$, we compute the key term $\kappa(\mathcal{C}(\omega))$ and the
coefficient $c_{\omega}$. This results in Table 1.

$a_{2}$ | $a_{1}$ | $v$ | $\omega$ | $\alpha_{\infty}(v)$ | $\beta_{\infty}(v)$ | $\kappa(\mathcal{C}(\omega))$ | $c_{\omega}$
---|---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $dx$ |  |  |  |
$0$ | $0$ | $1$ | $xdx$ |  |  |  |
$0$ | $1$ | $0$ | $y_{1}dx$ |  |  |  |
$0$ | $2$ | $0$ | $y_{1}^{2}dx$ | $0$ | $2$ | $dx$ | $\binom{2}{2}=1$
$1$ | $0$ | $0$ | $y_{2}dx$ | $0$ | $2$ | $y_{1}dx$ | $-\frac{2!}{2!1!}\binom{2}{2}=-1$
$1$ | $1$ | $0$ | $y_{2}y_{1}dx$ | $0$ | $2$ | $y_{1}^{2}dx$ | $\frac{2!}{1!2!}\binom{2}{2}=1$
Table 1. The computation of key terms in the case $P=\infty$ and
$d_{\infty}=1$.

There are three different key terms, which contribute $3$ to the rank of
$\mathcal{C}$ via Theorem 3.18. In order to show later that the remainig
elements of $W_{\infty}$ do not contribute to the rank, we apply the Cartier
operator to the three elements of $W_{\infty}\setminus H_{\infty}$.

$\displaystyle\mathcal{C}(dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(xdx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(y_{1}dx)$
$\displaystyle=\mathcal{C}((y_{1}^{p}-f(x))dx)=y_{1}\mathcal{C}(dx)-\mathcal{C}(f(x)dx)$
$\displaystyle=-\sum_{Q\in
B}\mathcal{C}(f_{Q}(x_{Q})dx)\in\text{span}_{k}\\{x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}.$

###  4.2. A pole of $f$ of order $1$ away from infinity

We now consider the case $d_{P}=1$, where $P\neq\infty$. We compute the basis

$\displaystyle
W_{P}=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\;|\;0\leq
a_{1},a_{2}<3,\\\ 0<9v\leq 3(2-a_{1})+7(2-a_{2})+8\end{cases}\Bigg{\\}}.$

Table 2 shows the $14$ differentials and their key terms, if they exist. The
coefficients $c_{\omega}$ are omitted, as Lemma 3.4 provides that they are
non-zero.

$a_{2}$ | $a_{1}$ | $v$ | $\omega$ | $\alpha_{P}(v)$ | $\beta_{P}(v)$ | $\kappa(\mathcal{C}(\omega))$
---|---|---|---|---|---|---
$0$ | $0$ | $1$ | $x_{P}dx$ | $0$ | $0$ | $x_{P}dx$
$0$ | $0$ | $2$ | $x_{P}^{2}dx$ | $0$ | $2$ |
$0$ | $0$ | $3$ | $x_{P}^{3}dx$ | $1$ | $1$ |
$0$ | $1$ | $1$ | $y_{1}x_{P}dx$ | $0$ | $0$ | $y_{1}x_{P}dx$
$0$ | $1$ | $2$ | $y_{1}x_{P}^{2}dx$ | $0$ | $2$ |
$0$ | $2$ | $1$ | $y_{1}^{2}x_{P}dx$ | $0$ | $0$ | $y_{1}^{2}x_{P}dx$
$0$ | $2$ | $2$ | $y_{1}^{2}x_{P}^{2}dx$ | $0$ | $2$ | $x_{P}^{2}dx$
$1$ | $0$ | $1$ | $y_{2}x_{P}dx$ | $0$ | $0$ | $y_{2}x_{P}dx$
$1$ | $0$ | $2$ | $y_{2}x_{P}^{2}dx$ | $0$ | $2$ | $y_{1}x_{P}^{2}dx$
$1$ | $1$ | $1$ | $y_{2}y_{1}x_{P}dx$ | $0$ | $0$ | $y_{2}y_{1}x_{P}dx$
$1$ | $1$ | $2$ | $y_{2}y_{1}x_{P}^{2}dx$ | $0$ | $2$ | $y_{1}^{2}x_{P}^{2}dx$
$1$ | $2$ | $1$ | $y_{2}y_{1}^{2}x_{P}dx$ | $0$ | $0$ | $y_{2}y_{1}^{2}x_{P}dx$
$2$ | $0$ | $1$ | $y_{2}^{2}x_{P}dx$ | $0$ | $0$ | $y_{2}^{2}x_{P}dx$
$2$ | $1$ | $1$ | $y_{2}^{2}y_{1}x_{P}dx$ | $0$ | $0$ | $y_{2}^{2}y_{1}x_{P}dx$
Table 2. The computation of key terms in the case $P\in
B_{1}\setminus\\{\infty\\}$ and $d_{P}=1$.

The table shows that all the $11$ key terms are different. The basis elements
without a key term require more attention.

$\displaystyle\mathcal{C}(x_{P}^{2}dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(x_{P}^{3}dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(y_{1}x_{P}^{2}dx)$
$\displaystyle=y_{1}\mathcal{C}(x_{P}^{2}dx)-\mathcal{C}(x_{P}^{2}f(x)dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{x_{P}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}&\hbox{if $d_{\infty}=1$}\\\
\text{span}_{k}\\{dx,x_{P}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}&\hbox{if $d_{\infty}=2$}.\end{cases}$

###  4.3. A pole of $f$ of order $2$ at infinity

Assume $f$ has a pole of order $2$ at $P=\infty$. We then compute the basis

$\displaystyle
W_{\infty}=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x^{v}dx\;|\;0\leq
a_{1},a_{2}<3,\\\ 0\leq 9v\leq
6(2-a_{1})+14(2-a_{2})-10\end{cases}\Bigg{\\}}.$

Table 3 shows these $16$ basis differentials and their key terms, if they
exist.

$a_{2}$ | $a_{1}$ | $v$ | $\omega$ | $\alpha_{\infty}(v)$ | $\beta_{\infty}(v)$ | $\kappa(\mathcal{C}(\omega))$
---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $dx$ | $0$ | $1$ |
$0$ | $0$ | $1$ | $xdx$ | $0$ | $2$ |
$0$ | $0$ | $2$ | $x^{2}dx$ | $1$ | $0$ | $dx$
$0$ | $0$ | $3$ | $x^{3}dx$ | $1$ | $1$ |
$0$ | $1$ | $0$ | $y_{1}dx$ | $0$ | $1$ | $dx$
$0$ | $1$ | $1$ | $y_{1}xdx$ | $0$ | $2$ |
$0$ | $1$ | $2$ | $y_{1}x^{2}dx$ | $1$ | $0$ | $y_{1}dx$
$0$ | $2$ | $0$ | $y_{1}^{2}dx$ | $0$ | $1$ | $y_{1}dx$
$0$ | $2$ | $1$ | $y_{1}^{2}xdx$ | $0$ | $2$ | $xdx$
$0$ | $2$ | $2$ | $y_{1}^{2}x^{2}dx$ | $1$ | $0$ | $y_{1}^{2}dx$
$1$ | $0$ | $0$ | $y_{2}dx$ | $0$ | $1$ | $y_{1}^{2}dx$
$1$ | $0$ | $1$ | $y_{2}xdx$ | $0$ | $2$ | $y_{1}xdx$
$1$ | $1$ | $0$ | $y_{2}y_{1}dx$ | $0$ | $1$ | $y_{2}dx$
$1$ | $1$ | $1$ | $y_{2}y_{1}xdx$ | $0$ | $2$ | $y_{1}^{2}xdx$
$1$ | $2$ | $0$ | $y_{2}y_{1}^{2}dx$ | $0$ | $1$ | $y_{2}y_{1}dx$
$2$ | $0$ | $0$ | $y_{2}^{2}dx$ | $0$ | $1$ | $y_{2}y_{1}^{2}dx$
Table 3. The computation of key terms in the case $P=\infty$ and
$d_{\infty}=2$.

Note that $12$ differentials have a key term, but there are only $9$ distinct
key terms. The differentials outside $H_{\infty}$ and the pairs of
differentials with the same key term require extra attention.

$\displaystyle\mathcal{C}(dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(xdx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(x^{2}dx)$ $\displaystyle=dx$
$\displaystyle\mathcal{C}(x^{3}dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(y_{1}dx)$
$\displaystyle=-\mathcal{C}(f(x)dx)\in\text{span}_{k}\\{dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\mathcal{C}(y_{1}xdx)$
$\displaystyle=-\mathcal{C}(xf(x)dx)\in\text{span}_{k}\\{dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\mathcal{C}(y_{1}x^{2}dx)$
$\displaystyle=y_{1}dx-\mathcal{C}(x^{2}f(x)dx)\in\text{span}_{k}\\{dx,y_{1}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\mathcal{C}(y_{1}^{2}dx)$
$\displaystyle=y_{1}\mathcal{C}(f(x)dx)+\mathcal{C}(f(x)^{2}dx)\in\text{span}_{k}\\{dx,y_{1}dx,x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{Q}^{2}dx\>|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}$
$\displaystyle\mathcal{C}(y_{1}^{2}x^{2}dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(x^{2}dx)+y_{1}\mathcal{C}(x^{2}f(x)dx)+\mathcal{C}(x^{2}f(x)^{2}dx)$
$\displaystyle\in\text{span}_{k}\\{dx,xdx,y_{1}dx,y_{1}^{2}dx,x_{Q}dx,y_{1}x_{Q}dx,\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}$ $\displaystyle\mathcal{C}(y_{2}dx)$
$\displaystyle=\mathcal{C}(y_{1}^{7}dx)-\mathcal{C}(y_{1}^{5}dx)-\mathcal{C}(h(x)dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(y_{1}dx)-y_{1}\mathcal{C}(y_{1}^{2}dx)-\mathcal{C}(h(x)dx)$
$\displaystyle\in\text{span}_{k}\\{dx,xdx,y_{1}dx,y_{1}^{2}dx,x_{Q}dx,y_{1}x_{Q}dx,y_{1}^{2}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{Q}^{2}dx,y_{1}x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{R}dx\;|\;R\in B_{2}\setminus B_{1}\\}.$

###  4.4. A pole of $f$ of order $2$ away from infinity

We now consider the case when $f$ has a pole at $P\neq\infty$ of order
$d_{P}=2$. The basis is

$\displaystyle
W_{P}=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\;|\;0\leq
a_{1},a_{2}<3,\\\ 0<9v\leq 6(2-a_{1})+14(2-a_{2})+8\end{cases}\Bigg{\\}}.$

The $24$ differentials and their key terms, if they exist, are compiled in
Table 4.

$a_{2}$ | $a_{1}$ | $v$ | $\omega$ | $\alpha_{P}(v)$ | $\beta_{P}(v)$ | $\kappa(\mathcal{C}(\omega))$
---|---|---|---|---|---|---
$0$ | $0$ | $1$ | $x_{P}dx$ | $0$ | $0$ | $x_{P}dx$
$0$ | $0$ | $2$ | $x_{P}^{2}dx$ | $0$ | $1$ |
$0$ | $0$ | $3$ | $x_{P}^{3}dx$ | $0$ | $2$ |
$0$ | $0$ | $4$ | $x_{P}^{4}dx$ | $1$ | $0$ | $x_{P}^{2}dx$
$0$ | $0$ | $5$ | $x_{P}^{5}dx$ | $1$ | $1$ |
$0$ | $1$ | $1$ | $y_{1}x_{P}dx$ | $0$ | $0$ | $y_{1}x_{P}dx$
$0$ | $1$ | $2$ | $y_{1}x_{P}^{2}dx$ | $0$ | $1$ | $x_{P}^{2}dx$
$0$ | $1$ | $3$ | $y_{1}x_{P}^{3}dx$ | $0$ | $2$ |
$0$ | $1$ | $4$ | $y_{1}x_{P}^{4}dx$ | $1$ | $0$ | $y_{1}x_{P}^{2}dx$
$0$ | $2$ | $1$ | $y_{1}^{2}x_{P}dx$ | $0$ | $0$ | $y_{1}^{2}x_{P}dx$
$0$ | $2$ | $2$ | $y_{1}^{2}x_{P}^{2}dx$ | $0$ | $1$ | $y_{1}x_{P}^{2}dx$
$0$ | $2$ | $3$ | $y_{1}^{2}x_{P}^{3}dx$ | $0$ | $2$ | $x_{P}^{3}dx$
$0$ | $2$ | $4$ | $y_{1}^{2}x_{P}^{4}dx$ | $1$ | $0$ | $y_{1}^{2}x_{P}^{2}dx$
$1$ | $0$ | $1$ | $y_{2}x_{P}dx$ | $0$ | $0$ | $y_{2}x_{P}dx$
$1$ | $0$ | $2$ | $y_{2}x_{P}^{2}dx$ | $0$ | $1$ | $y_{1}^{2}x_{P}^{2}dx$
$1$ | $0$ | $3$ | $y_{2}x_{P}^{3}dx$ | $0$ | $2$ | $y_{1}x_{P}^{3}dx$
$1$ | $1$ | $1$ | $y_{2}y_{1}x_{P}dx$ | $0$ | $0$ | $y_{2}y_{1}x_{P}dx$
$1$ | $1$ | $2$ | $y_{2}y_{1}x_{P}^{2}dx$ | $0$ | $1$ | $y_{2}x_{P}^{2}dx$
$1$ | $1$ | $3$ | $y_{2}y_{1}x_{P}^{3}dx$ | $0$ | $2$ | $y_{1}^{2}x_{P}^{3}dx$
$1$ | $2$ | $1$ | $y_{2}y_{1}^{2}x_{P}dx$ | $0$ | $0$ | $y_{2}y_{1}^{2}x_{P}dx$
$1$ | $2$ | $2$ | $y_{2}y_{1}^{2}x_{P}^{2}dx$ | $0$ | $1$ | $y_{2}y_{1}x_{P}^{2}dx$
$2$ | $0$ | $1$ | $y_{2}^{2}x_{P}dx$ | $0$ | $0$ | $y_{2}^{2}x_{P}dx$
$2$ | $0$ | $2$ | $y_{2}^{2}x_{P}^{2}dx$ | $0$ | $1$ | $y_{2}y_{1}^{2}x_{P}^{2}dx$
$2$ | $1$ | $1$ | $y_{2}^{2}y_{1}x_{P}dx$ | $0$ | $0$ | $y_{2}^{2}y_{1}x_{P}dx$
Table 4. The computation of key terms in the case $P\in
B_{1}\setminus\\{\infty\\}$ and $d_{P}=2$.

The table displays that $20$ differentials have a key term, but there are only
$17$ distinct key terms. The differentials without a key term and the pairs of
differentials with the same key term require extra attention.

$\displaystyle\mathcal{C}(x_{P}^{2}dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(x_{P}^{3}dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(x_{P}^{4}dx)$ $\displaystyle=x_{P}^{2}dx$
$\displaystyle\mathcal{C}(x_{P}^{5}dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(y_{1}x_{P}^{2}dx)$
$\displaystyle=y_{1}\mathcal{C}(x_{P}^{2}dx)-\mathcal{C}(x_{P}^{2}f(x)dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{x_{P}dx,x_{P}^{2}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}&\hbox{if $d_{\infty}=1$}\\\
\text{span}_{k}\\{dx,x_{P}dx,x_{P}^{2}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}&\hbox{if $d_{\infty}=2$}\end{cases}$
$\displaystyle\mathcal{C}(y_{1}x_{P}^{3}dx)$
$\displaystyle=y_{1}\mathcal{C}(x_{P}^{3}dx)-\mathcal{C}(x_{P}^{3}f(x)dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{x_{P}dx,x_{P}^{2}dx,x_{Q}dx\;|\;\in
B_{1}\setminus\\{P,\infty\\}\\}&\hbox{if $d_{\infty}=1$}\\\
\text{span}_{k}\\{dx,x_{P}dx,x_{P}^{2}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}&\hbox{if $d_{\infty}=2$}\end{cases}$
$\displaystyle\mathcal{C}(y_{1}x_{P}^{4}dx)$
$\displaystyle=y_{1}\mathcal{C}(x_{P}^{4}dx)-\mathcal{C}(x_{P}^{4}f(x)dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{y_{1}x_{P}^{2}dx,x_{P}dx,x_{P}^{2}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}&\hbox{if $d_{\infty}=1$}\\\
\text{span}_{k}\\{y_{1}x_{P}^{2}dx,dx,x_{P}dx,x_{P}^{2}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}&\hbox{if $d_{\infty}=2$}\end{cases}$
$\displaystyle\mathcal{C}(y_{1}^{2}x_{P}^{2}dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(x_{P}^{2}dx)+y_{1}\mathcal{C}(x_{P}^{2}f(x)dx)+\mathcal{C}(x_{P}^{2}f(x)^{2}dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{y_{1}x_{P}^{2}dx,dx,x_{P}dx,x_{P}^{2}dx,y_{1}x_{P}dx,x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}\\\ \oplus\text{span}_{k}\\{x_{Q}^{2}dx\;Q\in
B_{1}\setminus\\{P,\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=1$}\\\
\text{span}_{k}\\{y_{1}x_{P}^{2}dx,dx,y_{1}dx,x_{P}dx,x_{P}^{2}dx,y_{1}x_{P}dx,x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}\\\ \oplus\text{span}_{k}\\{x_{Q}^{2}dx\;Q\in
B_{1}\setminus\\{P,\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=2$}\end{cases}$ $\displaystyle\mathcal{C}(y_{1}^{2}x_{P}^{4}dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(x_{P}^{4}dx)+y_{1}\mathcal{C}(x_{P}^{4}f(x)dx)+\mathcal{C}(x_{P}^{4}f(x)^{2})dx$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{y_{1}^{2}x_{P}^{2}dx,dx,x_{P}dx,x_{P}^{2}dx,x_{P}^{3}dx,y_{1}x_{P}^{2}dx,x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}\\\
\oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=1$}\\\
\text{span}_{k}\\{y_{1}^{2}x_{P}^{2}dx,dx,y_{1}dx,x_{P}dx,x_{P}^{2}dx,x_{P}^{3}dx,y_{1}x_{P}^{2},x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}\\\
\oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=2$}\end{cases}$ $\displaystyle\mathcal{C}(y_{2}x_{P}^{2}dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(y_{1}x_{P}^{2}dx)-y_{1}\mathcal{C}(y_{1}^{2}x_{P}^{2}dx)-\mathcal{C}(x_{P}^{2}h(x)dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{y_{1}^{2}x_{P}^{2}dx,dx,y_{1}dx,x_{P}dx,x_{p}^{2}dx,x_{P}^{3}dx,y_{1}x_{P}dx,y_{1}x_{P}^{2}dx,y_{1}^{2}x_{P}dx\\}\\\
\oplus\text{span}_{k}\\{x_{Q}dx,y_{1}x_{Q}dx,y_{1}^{2}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}\oplus\text{span}_{k}\\{x_{R}dx\>|\;R\in
B_{2}\setminus B_{1}\\}\\\
\oplus\text{span}_{k}\\{x_{Q}^{2}dx,y_{1}x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=1$}\\\
\text{span}_{k}\\{y_{1}^{2}x_{P}^{2}dx,dx,xdx,y_{1}dx,y_{1}^{2}dx,x_{P}dx,x_{p}^{2}dx,x_{P}^{3}dx,y_{1}x_{P}dx,y_{1}x_{P}^{2}dx,y_{1}^{2}x_{P}dx\\}\\\
\oplus\text{span}_{k}\\{x_{Q}dx,y_{1}x_{Q}dx,y_{1}^{2}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}\oplus\text{span}_{k}\\{x_{R}dx\>|\;R\in
B_{2}\setminus B_{1}\\}\\\
\oplus\text{span}_{k}\\{x_{Q}^{2}dx,y_{1}x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=2$.}\end{cases}$

###  4.5. A pole of $h$ of order $1$

Assume $P\in B_{2}\setminus B_{1}$ and $e_{P}=1$, meaning $h(x)$ has a pole of
order $1$ at $P$. The basis $W_{P}$ is given by

$\displaystyle
W_{P}=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\;|\;0\leq
a_{1},a_{2}<3,\\\ 0<3v\leq 4-a_{2}\end{cases}\Bigg{\\}}.$

The value of $a_{1}$ plays no role in whether $\omega$ is regular and in the
key term $\kappa(\mathcal{C}(\omega))$ as defined in Definition 3.5. In Table
5, the three possibilities $0\leq a_{1}\leq 2$ are combined into one line.

$a_{2}$ | $v$ | $\omega$ | $\alpha_{P}(v)$ | $\beta_{P}(v)$ | $\kappa(\mathcal{C}(\omega))$
---|---|---|---|---|---
$0$ | $1$ | $y_{1}^{a_{1}}x_{P}dx$ | $0$ | $0$ | $y_{1}^{a_{1}}x_{P}dx$
$1$ | $1$ | $y_{2}y_{1}^{a_{1}}x_{P}dx$ | $0$ | $0$ | $y_{2}y_{1}^{a_{1}}x_{P}dx$
Table 5. The computation of key terms in the case $P\in B_{2}\setminus B_{1}$,
and $e_{P}=1$.

Observe that these six differentials all have a unique key term.

###  4.6. A pole of $h$ of order $2$

Assume $P\in B_{2}\setminus B_{1}$ and $e_{P}=2$, meaning $h(x)$ has a pole of
order $2$ at $P$. The basis is

$\displaystyle
W_{P}=\begin{cases}y_{2}^{a_{2}}y_{1}^{a_{1}}x_{P}^{v}dx\;|\;0\leq
a_{1},a_{2}<3,\\\ 0<3v\leq 6-2a_{2}\end{cases}\Bigg{\\}}.$

The differentials in $W_{P}$ with their key terms, if they exist, are
presented in Table 6.

$a_{2}$ | $v$ | $\omega$ | $\alpha_{P}(v)$ | $\beta_{P}(v)$ | $\kappa(\mathcal{C}(\omega))$
---|---|---|---|---|---
$0$ | $1$ | $y_{1}^{a_{1}}x_{P}dx$ | $0$ | $0$ | $y_{1}^{a_{1}}x_{P}dx$
$0$ | $2$ | $y_{1}^{a_{1}}x_{P}^{2}dx$ | $0$ | $1$ |
$1$ | $1$ | $y_{2}y_{1}^{a_{1}}x_{P}dx$ | $0$ | $0$ | $y_{2}y_{1}^{a_{1}}x_{P}dx$
Table 6. The computation of key terms in the case $P\in B_{2}\setminus B_{1}$
and $e_{P}=2$.

The differentials without a key term require extra attention.

$\displaystyle\mathcal{C}(x_{P}^{2}dx)$ $\displaystyle=0$
$\displaystyle\mathcal{C}(y_{1}x_{P}^{2}dx)$
$\displaystyle=y_{1}\mathcal{C}(x_{P}^{2}dx)-\mathcal{C}(f(x)x_{P}^{2}dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{x_{P}dx\\}\oplus\text{span}_{k}\\{x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}&\hbox{if $d_{\infty}=1$}\\\
\text{span}_{k}\\{dx,x_{P}dx\\}\oplus\text{span}_{k}\\{x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}&\hbox{if $d_{\infty}=2$}\end{cases}$
$\displaystyle\mathcal{C}(y_{1}^{2}x_{P}^{2}dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(x_{P}^{2}dx)+y_{1}\mathcal{C}(f(x)x_{P}^{2}dx)+\mathcal{C}(f(x)^{2}x_{P}^{2}dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{dx,x_{P}dx,y_{1}x_{P}dx\\}\oplus\text{span}_{k}\\{x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}\\\ \oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=1$}\\\
\text{span}_{k}\\{dx,y_{1}dx,x_{P}dx,y_{1}x_{P}dx\\}\oplus\text{span}_{k}\\{x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}\\\ \oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=2$.}\end{cases}$

###  4.7. The main theorem

Let $Y_{2}\to\mathbb{P}^{1}$ be a $\mathbb{Z}/9\mathbb{Z}$-cover which is
minimal in the sense of Definition 2.14. We now prove a formula for the
$a$-number of $Y_{2}$, implying that the answer to Question 2.15 is “yes”
under the assumptions $p=3$ and $n=2$.

######  Theorem 4.1.

Suppose $Y_{2}\xrightarrow{}\mathbb{P}^{1}$ has branching datum

$\left[\begin{array}[]{rrrrrrrr}2&\ldots&3&\ldots&0&\ldots&0&\ldots\\\
\makebox[0.0pt][l]{$\smash{\underbrace{\phantom{\begin{matrix}4&\ldots\end{matrix}}}_{\text{$n_{1}$}}}$}4&\ldots&\makebox[0.0pt][l]{$\smash{\underbrace{\phantom{\begin{matrix}7&\ldots\end{matrix}}}_{\text{$n_{2}$}}}$}7&\ldots&\makebox[0.0pt][l]{$\smash{\underbrace{\phantom{\begin{matrix}2&\ldots\end{matrix}}}_{\text{$n_{3}$}}}$}2&\ldots&\makebox[0.0pt][l]{$\smash{\underbrace{\phantom{\begin{matrix}{3}&\ldots\end{matrix}}}_{\text{$n_{4}$}}}$}{3}&\ldots\\\
\end{array}\right]^{\intercal}$

Then the $a$-number of $Y_{2}$ satisfies the formula

$a_{Y_{2}}=3n_{1}+7n_{2}+0n_{3}+3n_{4}.$

###### Proof.

We apply Theorem 3.18 to obtain a lower bound for $\text{rk}(\mathcal{C})$ and
hence an upper bound for $a_{Y_{2}}$. Combining Sections 4.1, 4.2, 4.3 and
4.4, we find the number of distinct key terms is

$\\#K=11n_{1}+17n_{2}+6n_{3}+6n_{4}-8.$

The correcting factor $-8$ comes from the pole at infinity. Applying Theorem
3.18, using $g_{Y_{2}}=\\#W$, yields

$\displaystyle a_{Y_{2}}$ $\displaystyle\leq g_{Y_{2}}-\\#K$
$\displaystyle=3n_{1}+7n_{2}+0n_{3}+3n_{4}.$

The remainder of the proof revolves around showing that this upper bound is
sharp. We do so by showing two things. First, we show that differentials which
have no key term do not contribute to the rank of $\mathcal{C}$. Second, we
show that differentials with the same key term together contribute $1$ to the
rank of $\mathcal{C}$, such that the rank of $\mathcal{C}$ is exactly the
number of key terms. We do this locally at the points in $B_{2}$, by treating
the six cases corresponding to Sections 4.1, 4.2, 4.3, 4.4, 4.5 and 4.6.

We begin with the differentials in $W_{\infty}$ in the case $d_{\infty}=1$, as
described in Section 4.1. All the key terms occur uniquely, so we treat only
the differentials that have no key term. As the differentials $dx$ and $xdx$
are killed by $\mathcal{C}$, they clearly don’t contribute to the rank. Next,
we consider $y_{1}dx$, which also does not have a key term. We have

$\mathcal{C}(y_{1}dx)\in\text{span}_{k}\\{x_{Q}dx\;|\;Q\neq\infty\\}.$

However, for $Q\neq\infty$, we have $\mathcal{C}(x_{Q}dx)=x_{Q}dx$ and
$x_{Q}dx$ does have a key term. Hence $y_{1}dx$ also does not contribute to
the rank of $\mathcal{C}$. Thus we have exhibited $3$ linearly independent
differentials that don’t contribute to $\text{rk}(\mathcal{C})$, meaning they
contribute to $a_{Y_{2}}$.

Second, we consider the case $P\in B_{1}\setminus\\{\infty\\}$ and $d_{P}=1$.
Section 4.2 shows that all the key terms are unique. The differentials
$x_{P}^{2}dx$, $x_{P}^{3}dx$ and $y_{1}x_{P}^{2}dx$ have no key term. The
differentials $x_{P}^{2}dx$ and $x_{P}^{3}dx$ are killed by $\mathcal{C}$ and
therefore don’t contribute to $\text{rk}(\mathcal{C})$. Furthermore, we have
computed

$\mathcal{C}(y_{1}x_{P}^{2}dx)\in\begin{cases}\text{span}_{k}\\{x_{P}dx,x_{Q}dx\;|\;P\neq
Q\neq\infty\\}&\hbox{if $d_{\infty}=1$}\\\
\text{span}_{k}\\{dx,x_{P}dx,x_{Q}dx\;|\;P\neq Q\neq\infty\\}&\hbox{if
$d_{\infty}=2$}\end{cases}$

In the case $d_{\infty}=1$, we have already shown that this does not
contribute to $\text{rk}(\mathcal{C})$. In the case $d_{\infty}=2$, we have
$\mathcal{C}(x^{2}dx)=dx$. In both cases $y_{1}x_{P}^{2}dx$ does not
contribute to $\text{rk}(\mathcal{C})$. Again, the differentials in $W_{P}$
contribute exactly $3$ to $a_{Y_{2}}$.

Third, we consider the differentials in $W_{\infty}$ in the case
$d_{\infty}=2$. As shown in Section 4.3, the differentials $dx$, $xdx$ and
$x^{3}dx$ are killed by $\mathcal{C}$. $x^{2}dx$ and $y_{1}dx$ both have the
key term $dx$. We’ve seen that $\mathcal{C}(x^{2}dx)=dx$ and

$\mathcal{C}(y_{1}dx)\in\text{span}_{k}\\{dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}.$

We have already established that

$\text{span}_{k}\\{x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}\subset\text{im}(\mathcal{C}),$

so the differentials $x^{2}dx$ and $y_{1}dx$ together contribute $1$ to
$\text{rk}(\mathcal{C})$. It then immediately follows that $y_{1}xdx$, which
has no key term, does not contribute to $\text{rk}(\mathcal{C})$, since we
have computed

$\mathcal{C}(y_{1}xdx)\in\text{span}_{k}\\{dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}.$

Next, the two differentials $y_{1}x^{2}dx$ and $y_{1}^{2}dx$ both have the key
term $y_{1}dx$. We’ve computed

$\displaystyle\mathcal{C}(y_{1}x^{2}dx)$
$\displaystyle\in\text{span}_{k}\\{dx,y_{1}dx,x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\mathcal{C}(y_{1}^{2}xdx)$
$\displaystyle\in\text{span}_{k}\\{dx,y_{1}dx,x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}.$

We show that they contribute $1$ to $\text{rk}(\mathcal{C})$ together, by
showing that $y_{1}x_{Q}dx$ and $x_{Q}^{2}dx$ (for $d_{Q}=2$) are already in
the image of $\mathcal{C}$. We compute

$\displaystyle\mathcal{C}(y_{1}x_{Q}dx)$
$\displaystyle=y_{1}x_{Q}dx-\mathcal{C}(x_{Q}f(x)dx)$
$\displaystyle\in\text{span}_{k}\\{dx,y_{1}x_{Q}dx,x_{P}dx\;|\;P\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\mathcal{C}(x_{Q}^{4}dx)$
$\displaystyle=x_{Q}^{2}dx.$

The second computation assumes $d_{Q}=2$, so that $x_{Q}^{4}dx$ is indeed
regular. These computations together imply that the differentials
$y_{1}x^{2}dx$ and $y_{1}^{2}dx$, which have the same key term, together
contribute $1$ to $\text{rk}(\mathcal{C})$. Finally, we consider the
differentials $y_{1}^{2}x^{2}dx$ and $y_{2}dx$, which both have the key term
$y_{1}^{2}dx$. In Section 4.3, we have established

$\displaystyle\mathcal{C}(y_{1}^{2}x^{2}dx)$
$\displaystyle\in\text{span}_{k}\\{dx,xdx,y_{1}dx,y_{1}^{2}dx,x_{Q}dx,y_{1}x_{Q}dx,\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}.$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}$ $\displaystyle\mathcal{C}(y_{2}dx)$
$\displaystyle\in\text{span}_{k}\\{dx,xdx,y_{1}dx,y_{1}^{2}dx,x_{Q}dx,y_{1}x_{Q}dx,y_{1}^{2}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{Q}^{2}dx,y_{1}x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{R}dx\;|\;R\in B_{2}\setminus B_{1}\\}.$

It remains to be shown that $xdx$, $y_{1}^{2}x_{Q}dx$, $y_{1}x_{Q}^{2}dx$ (in
the case $d_{Q}=2$) and $x_{R}dx$ (with $R\in B_{2}\setminus B_{1}$) already
lie in the image of $\mathcal{C}$. We compute

$\displaystyle\mathcal{C}(y_{1}^{2}xdx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(xdx)+y_{1}\mathcal{C}(xf(x)dx)+\mathcal{C}(xf(x)^{2}dx)$
$\displaystyle\in\text{span}_{k}\\{dx,xdx,y_{1}dx,x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}$
$\displaystyle\mathcal{C}(y_{1}^{2}x_{Q}dx)$
$\displaystyle=y_{1}^{2}x_{Q}dx+y_{1}\mathcal{C}(x_{Q}f(x)dx)+\mathcal{C}(x_{Q}f(x)^{2}dx)$
$\displaystyle\in\text{span}_{k}\\{y_{1}^{2}x_{Q}^{2}dx,dx,y_{1}dx,x_{P}dx,y_{1}x_{P}dx\;|\;P\in
B_{1}\setminus\\{\infty\\}\\}$ $\displaystyle\hskip
14.22636pt\oplus\text{span}_{k}\\{x_{P}^{2}dx\;|\;P\in
B_{1}\setminus\\{\infty\\},d_{P}=2\\}$
$\displaystyle\mathcal{C}(y_{1}x_{Q}^{4}dx)$
$\displaystyle=y_{1}x_{Q}^{2}dx-\mathcal{C}(x_{Q}^{4}f(x)dx)$
$\displaystyle\in\text{span}_{k}\\{y_{1}x_{Q}dx,dx,x_{Q}^{2}dx,x_{P}dx\;|\;P\in
B_{1}\setminus\\{\infty\\}\\}.\mathcal{C}(x_{R}dx)$ $\displaystyle=x_{R}dx.$

These computations verify that indeed $xdx$, $y_{1}^{2}x_{Q}dx$,
$y_{1}x_{Q}^{2}dx$ and $x_{R}dx$ already lie in the image of $\mathcal{C}$,
without the help of $y_{1}^{2}x^{2}dx$ and $y_{2}dx$. Thus the differentials
$y_{1}^{2}x^{2}dx$ and $y_{2}dx$ together contribute $1$ to
$\text{rk}(\mathcal{C})$.

As the fourth step, we treat the case where $P\in B_{1}\setminus\\{\infty\\}$
and $d_{P}=2$. In Section 4.4, we have seen that $x_{P}^{2}dx$, $x_{P}^{3}dx$
and $x_{P}^{5}dx$ are killed by the Cartier operator. The differentials
$x_{P}^{4}dx$ and $y_{1}x_{P}^{2}$ have the key term $x_{P}^{2}dx$, but the
computations in Section 4.4, together with computations earlier in this proof,
imply that they together contribute $1$ to $\text{rk}(\mathcal{C})$. It then
follows immediately that $y_{1}x_{P}^{3}$, which has no key term, does not
contribute to $\text{rk}(\mathcal{C})$. Next we consider the differentials
$y_{1}x_{P}^{4}dx$ and $y_{1}^{2}x_{P}^{2}dx$, which both have the key term
$y_{1}x_{P}^{2}dx$. The computations in Section 4.4 show that they together
contribute $1$ to $\text{rk}(\mathcal{C})$; they both extend the imagee of
$\mathcal{C}$ only by $\text{span}_{k}\\{y_{1}x_{P}^{2}dx\\}$ and nothing
else. Finally, we consider the differentials $y_{1}^{2}x_{P}^{4}dx$ and
$y_{2}x_{P}^{2}dx$, which both have the key term $y_{1}^{2}x_{P}^{2}dx$. They
introduce not only $y_{1}^{2}x_{P}^{2}dx$, but also $x_{P}^{3}dx$ and
$y_{1}dx$. We show that these differentials already lie in the image of
$\mathcal{C}$ independently. For $x_{P}^{3}dx$, we compute

$\displaystyle\mathcal{C}(y_{1}^{2}x_{P}^{3})$
$\displaystyle=y_{1}^{2}\mathcal{C}(x_{P}^{3}dx)+y_{1}\mathcal{C}(x_{P}^{3}f(x)dx)+\mathcal{C}(x_{P}^{3}f(x)^{2}dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{x_{P}^{3}dx,dx,x_{P}dx,x_{P}^{2}dx,y_{1}x_{P}dx,y_{1}x_{P}^{2}dx\\}\\\
\oplus\text{span}_{k}\\{x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}\\\
\oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}\hskip 85.35826pt\hbox{if
$d_{\infty}=1$}\\\
\text{span}_{k}\\{x_{P}^{3}dx,dx,y_{1}dx,x_{P}dx,x_{P}^{2}dx,y_{1}x_{P}dx,y_{1}x_{P}^{2}dx\\}\\\
\oplus\text{span}_{k}\\{x_{Q}dx,y_{1}x_{Q}dx,y_{1}^{2}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{P,\infty\\}\\}\\\
\oplus\text{span}_{k}\\{x_{Q}^{2}dx,y_{1}x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}\hskip 48.36967pt\hbox{if
$d_{\infty}=2$}\end{cases}$

Hence $x_{P}^{3}dx$ was already in the image of $\mathcal{C}$. It remains to
be shown that $y_{1}dx$ is already in the image of $\mathcal{C}$ in the case
$d_{\infty}=1$. We compute

$\displaystyle\mathcal{C}(y_{1}^{2}dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(dx)+y_{1}\mathcal{C}(f(x)dx)+\mathcal{C}(f(x)^{2}dx)$
$\displaystyle\in$$\displaystyle\\{dx,x_{Q}dx,y_{1}x_{Q}dx\>|\;Q\in
B_{1}\setminus\\{\infty\\}\\}\oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}$ $\displaystyle\mathcal{C}(y_{2}dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(dx)-y_{1}\mathcal{C}(y_{1}^{2}dx)-\mathcal{C}(h(x)dx)$
$\displaystyle\in\text{span}_{k}\\{dx,y_{1}dx,x_{Q}dx,y_{1}x_{Q}dx,y_{1}^{2}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}$
$\displaystyle\oplus\text{span}_{k}\\{x_{Q}^{2}dx,y_{1}x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}.$

This shows that $y_{1}dx$ was already in the image of $\mathcal{C}$. It
follows that the differentials $y_{1}^{2}x_{P}^{4}dx$ and $y_{2}x_{P}^{2}dx$
together contribute $1$ to $\text{rk}(\mathcal{C})$, which concludes the
fourth step.

As the fifth step, we treat the case $P\in B_{2}\setminus B_{1}$ and
$e_{P}=1$, meaning $h$ has a pole of order $1$. In that case, Section 4.5
shows that the differentials in $W_{P}$ all have different key terms. Hence
these differentials do not contribute to the $a$-number.

As the sixth and last step, we treat the case $P\in B_{2}\setminus B_{1}$ and
$e_{P}=2$, meaning $h$ has a pole of order $2$. Section 4.6 shows that $6$ out
of the $9$ differentials in $W_{P}$ have a key term, and these are all unique.
This implies that the rank of the Cartier operator restricted to
$\text{span}_{k}W_{P}$ is at least $6$. We now show that it’s exactly $6$, by
analyzing the basis differentials that do not have a key term. Since
$\mathcal{C}(x_{P}^{2}dx)=0$, the differential $x_{P}^{2}dx$ does not
contribute to the rank of $\mathcal{C}$. Moving on to $y_{1}x_{P}^{2}dx$,
recall

$\displaystyle\mathcal{C}(yx_{P}^{2}dx)$
$\displaystyle=y\mathcal{C}(x_{P}^{2}dx)-\mathcal{C}(f(x)x_{P}^{2}dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{x_{P}dx\\}\oplus\text{span}_{k}\\{x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}&\hbox{if $d_{\infty}=1$}\\\
\text{span}_{k}\\{dx,x_{P}dx\\}\oplus\text{span}_{k}\\{x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}&\hbox{if $d_{\infty}=2$}\end{cases}$

Since $\mathcal{C}(x_{P}dx)=x_{P}dx$, these differentials are already in the
image of $\mathcal{C}$. Hence $y_{1}x_{P}^{2}dx$ does not contribute to the
rank of $\mathcal{C}$. Finally, we treat $y_{1}^{2}x_{P}^{2}dx$:

$\displaystyle\mathcal{C}(y_{1}^{2}x_{P}^{2}dx)$
$\displaystyle=y_{1}^{2}\mathcal{C}(x_{P}^{2}dx)+y_{1}\mathcal{C}(f(x)x_{P}^{2}dx)+\mathcal{C}(f(x)^{2}x_{P}^{2}dx)$
$\displaystyle\in\begin{cases}\text{span}_{k}\\{dx,x_{P}dx,y_{1}x_{P}dx\\}\oplus\text{span}_{k}\\{x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}\\\ \oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=1$}\\\
\text{span}_{k}\\{dx,y_{1}dx,x_{P}dx,y_{1}x_{P}dx\\}\oplus\text{span}_{k}\\{x_{Q}dx,y_{1}x_{Q}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\}\\}\\\ \oplus\text{span}_{k}\\{x_{Q}^{2}dx\;|\;Q\in
B_{1}\setminus\\{\infty\\},d_{Q}=2\\}\hskip 56.9055pt\hbox{if
$d_{\infty}=2$.}\end{cases}$

To show that this does not contribute to the rank of $\mathcal{C}$, we only
need to show that $y_{1}x_{P}dx$ was already in the image of $\mathcal{C}$.
This is clear since

$\mathcal{C}(y_{1}x_{P}dx)=y_{1}x_{P}dx.$

Thus $P$ contributes $3$ to the $a$-number of $Y_{2}$.

We have treated all six cases of $W_{P}$, depending on the ramification at $P$
and on whether $P=\infty$. We have shown that a pole of $f$ of order $2$
contributes $7$ to the $a$-number, while a pole of $f$ of order $1$ and a pole
of $h$ of order $2$ both contribute $3$ to the $a$-number. A pole of $h$ of
order $1$ contributes nothing to the $a$-number. In conclusion, we’ve proved
the formula

$a_{Y_{2}}=3n_{1}+7n_{2}+3n_{4}.$

∎

######  Remark 4.2.

Seeing this proof, it would be tempting to think there is an inclusion
$\text{im}(\mathcal{C})\subseteq\text{span}_{k}K$, where $K$ is the set of key
terms. However, this is not true, even in the case $p=3$. To show this, take
the differential $y_{2}^{2}dx$, assuming $d_{\infty}=2$. The differential
$\mathcal{C}(y_{2}^{2}dx)$ has a term $\mathcal{C}(h(x)^{2}dx)$. The rational
function $h(x)$ has a pole of order at most $5$ at infinity. This means for a
suitable choice of $h(x)$, we have a term $\mathcal{C}(x^{8}dx)=x^{2}dx$,
which does not occur as a key term.

Since $\text{im}(\mathcal{C})$ and $\text{span}_{k}K$ have the same dimension,
by virtue of Theorem 4.1, the opposite inclusion also fails to hold.

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