#  Brill-Noether loci.

montserrat Teixidor i Bigas  Mathematics Department, Tufts University, 177
College Avenue, Medford MA 02155, USA<EMAIL_ADDRESS>
###### Abstract.

Brill-Noether loci ${\mathcal{M}}^{r}_{g,d}$ are those subsets of the moduli
space ${\mathcal{M}}_{g}$ determined by the existence of a linear series of
degree $d$ and dimension $r$. We provide a new proof of the non-emptiness of
the loci when the expected codimension of the Brill-Noether loci satisfies
$0<-\rho(g,r,d)\leq g-2$. We show that for a generic point of a component of
this locus, the Petri map is onto and the maximal rank conjecture for quadrics
is satisfied. We also show that Brill-Noether loci of the same codimension are
distinct when the codimension is not too large

Let $C$ be a curve of genus $g$. The Brill-Noether locus $G^{r}_{d}(C)$
parameterizes linear series of degree $d$ and dimension $r$ on $C$. When the
Brill-Noether number $\rho(g,r,d)\geq 0$, the Brill-Noether locus is non-empty
on every curve and irreducible if the number is strictly positive. On the
other hand, when the Brill-Noether number is negative, the locus is empty for
$C$ generic but may be non-empty on special curves. One can then define
${\mathcal{M}}^{r}_{g,d}$ as the locus of curves that posses a $g^{r}_{d}$.

The loci ${\mathcal{M}}^{r}_{g,d}$ have played an important role in the study
of the geometry of ${\mathcal{M}}_{g}$ having allowed for example to show that
${\mathcal{M}}_{g}$ is of general type if $g$ is sufficiently large. Other
interesting loci are defined by the rank of natural maps among spaces of
sections of line bundles. One can look for instance at the maximal rank loci
determined when the hypersurfaces of the ambient space do not cut a linear
series of the expected dimension or the loci of curves where the Petri map
fails to be injective. When $\rho<0$, by dimensionality reasons, it is
impossible for the Petri map to be injective but could still be of maximal
rank if it were onto.

The first question to ask about ${\mathcal{M}}^{r}_{g,d}$ is whether these
loci are non-empty. In [P], Pflueger showed that ${\mathcal{M}}^{r}_{g,d}$ is
non-empty and has a component of the expected dimension $-\rho$ when
$-\rho\leq g-3$ . Here, we give a simpler proof for this result for $-\rho\leq
g-2$. We also show that on the generic point of that component of
${\mathcal{M}}^{r}_{g,d}$ the Petri map is onto, fitting with the philosophy
of expected maximal rank (see Theorem 2.1). In addition, with some more
restricted numerical conditions, we prove that in this locus, the first step
of the maximal rank conjecture holds, that is the symmetric map
$S^{2}H^{0}(C,L)\to H^{0}(C,L^{2})$ is injective (see Theorem 2.2).

Probably the most natural question after non-emptiness of the loci
${\mathcal{M}}^{r}_{g,d}$ is whether the different values of $r,d$ produce
distinct loci. By Serre duality,
${\mathcal{M}}^{r}_{g,d}={\mathcal{M}}^{g-d+r-1}_{g,2g-2-d}$, therefore, it
suffices to study the case $r+1\leq g-d+r$ or equivalently $d\leq g-1$. It was
proved in [CKK], [CK], that when the codimension is either 1 or 2 and two of
the loci ${\mathcal{M}}^{r_{1}}_{g,d_{1}},{\mathcal{M}}^{r_{2}}_{g,d_{2}}$ are
of the same dimension and $d_{i}\not=2g-2-d_{j}$, then the loci are distinct.
Here we extend this result to higher codimension (see Theorem 3.1).

In a recent paper [AH] Asher and Haburcak study the question of inclusions of
the different ${\mathcal{M}}^{r}_{g,d}$. From a linear series, by adding a
fixed point one can always create another linear series of higher degree and
the same dimension. Similarly, by removing one point, one can create a linear
series of both degree and dimension one less. Therefore, there are natural
inclusions ${\mathcal{M}}^{r}_{g,d}\subseteq{\mathcal{M}}^{r}_{g,d+1}$,
${\mathcal{M}}^{r}_{g,d}\subseteq{\mathcal{M}}^{r-1}_{g,d-1}$. In particular,
if one is interested on maximal Brill-Noether loci ${\mathcal{M}}^{r}_{g,d}$,
it suffices to consider those for which $\rho(g,r,d)<0$ and
$\rho(g,r-1,d-1)\geq 0,\rho(g,r,d+1)\geq 0$. Asher and Haburcak conjecture
that when these conditions are satisfied and $g\neq 7,8,9$, the loci
${\mathcal{M}}^{r}_{g,d}$ are maximal, that is, not contained on any other
${\mathcal{M}}^{r^{\prime}}_{g,d^{\prime}}$. They prove the conjecture for
genus up to 19 and 22, 23. Our methods can be used to deal with particular
cases of these inclusions (see Example 3.2) although we will not pursue this
question any farther here.

##  1\. Limit Linear Series on a chain of elliptic curves.

We will use chains of elliptic curves: given elliptic curves $E_{1},\dots
E_{g}$ with marked points $P_{i},Q_{i}\in E_{i}$, glue $Q_{i}$ to
$P_{i+1},i=1,\dots,g-1$ to form a nodal curve $X$ of arithmetic genus $g$.
When the curves and pairs of points on each curve are generic, we will say
that we have a generic chain. In this paper, we will often require that
$P_{i}-Q_{i}$ be a torsion point of a certain order.

Recall that a limit linear series of degree $d$ and dimension $r$ on $X$
consists of the data of a limit linear series of degree $d$ and dimension $r$
on each component so that the orders of vanishing of the sections
$u^{i}_{0}<\dots<u^{i}_{r},\ v^{i}_{0}>\dots>v^{i}_{r}$ at $P_{i},Q_{i}$
respectively satisfy $u^{i+1}_{k}+v^{i}_{k}\geq d,k=0,\dots,r,\
i=1,\dots,g-1$. The series is refined if we have equality in these
inequalities. Given a family of curves with central fiber $X$ and a space of
linear series of degree $d$ and dimension $r$, one obtains as the limit on $X$
(after perhaps adding to $X$ a few chains of rational curves) a refined limit
linear series.

In this section, we present a way to parameterize limit linear series on a
chain of elliptic curves that is not necessarily generic. Representation of
Brill-Noether loci on generic chains of elliptic curves by using Young
Tableaux was introduced by Edidin ([E] ) in the case where the Brill-Noether
number is 0. It was much later used, still in the case of $\rho=0$ in the
tropical context in in [CDPR]. This was generalized to any $\rho$ for
$W^{r}_{d}$ in [LT] and then Young Tableaux represent whole components of the
Brill-Noether locus instead of individual limit linear series. In the case of
special chains, the geometry has been exploited in [JR] in the tropical
context. In [T1], we describe completely the space of limit linear series
$G^{r}_{d}$ on a chain of generic elliptic curves with ramification at two
fixed points. A similar result is true when the chains are not generic with
only a few changes that we describe below. We do not need the case of added
ramification at two points, so we will not include it, although ramification
at two points does not make the proof any harder. Recall that in the case of
curves of genus 1, Riemann-Roch’s Theorem gives a complete answer to the
existence of linear series:

######  Lemma 1.1.

Let $C$ be an elliptic curve, $P,Q\in C$, $L$ a line bundle of degree $d$ on
$C$, $V\subseteq H^{0}(C,L)$ an $(r+1)$-dimensional space of sections and
$u_{0}<\dots<u_{r},\ v_{0}>\dots>v_{r}$ the distinct orders of vanishing of
the sections of $V$ at at $P,Q$ respectively. Then $u_{k}+v_{k}\leq
d,k=0,\dots,r$. If equality holds for a particular $k$, then
$L={\mathcal{O}}(kP+(d-k)Q)$. Therefore, if $L$ is generic, $u_{k}+v_{k}\leq
d-1,k=0,\dots,r$ while if $u_{k_{1}}+v_{k_{1}}=d,u_{k_{2}}+v_{k_{2}}=d$, then
$P-Q$ is a torsion element of order $u_{k_{2}}-u_{k_{1}}$ in the group
structure of $C$.

###### Proof.

From the definition of $v_{k}$, the subspace of $V$ of sections with order of
vanishing at $Q$ at least $v_{k}$ has dimension at least $k+1$. Similarly, the
space of sections vanishing with order at least $u_{k}$ at $P$ has dimension
at least $r-k+1$. Two subspaces of $V$ of dimensions at least $k+1$ and
$r-k+1$ necessarily intersect. Therefore, there is a section in $V$ that
vanishes with order at least $u_{k}$ at $P$ and at least $v_{k}$ at $Q$. As
the degree of $L$ is $d$, $u_{k}+v_{k}\leq d$ and equality implies that
$L={\mathcal{O}}(u_{k}P+v_{k}Q)$. ∎

We present first the simpler description of the space of limit linear series
on a non-generic chain of elliptic curves. This is the case we will mostly
use.

######  Definition 1.2.

Given a collection of boxes arranged forming a rectangle, an admissible
filling  assigns to each box precisely one number in $1,\dots,g$ so that
numbers in both rows and columns appear in strictly increasing order. Some of
the numbers $1,\dots,g$ may be used multiple times.

######  Proposition 1.3.

Choose $g,r,d$ positive integers. Let $X$ be a chain of $g$ elliptic curves
generic except for components $C_{i_{1}},\dots,C_{i_{e}}$ where

$l_{1}(P_{i_{1}}-Q_{i_{1}})=0,\dots,l_{e}(P_{i_{e}}-Q_{i_{e}})=0.$

There is a one-to-one correspondence between the two following set of data

  * •

Components of the Brill-Noether locus of refined limit linear series of degree
$d$ and dimension $r$ on $X$ such that vanishing at $P_{1},Q_{g}$ is
$(0,\dots,r),(r,\dots,0)$ and vanishing at $P_{i},Q_{i}$ satisfy
$u_{j}^{i}+v_{j}^{i}\geq d-1$.

  * •

Admissible fillings of an $(r+1)\times(g-d+r)$ rectangle so that only the
numbers $i_{1},\dots,i_{e}$ may appear more than once and when $i_{k}$ appears
both on row $r_{1}$ and column $c_{1}$ and on row $r_{2}$ and column $c_{2}$,
then $l_{k}$ divides $r_{2}-r_{1}+c_{2}-c_{1}$.

######  Remark 1.4.

Linear series in which $u_{j}^{1}+v_{j}^{i}\leq d-2$ are in a sense wasteful
and less likely to be possible when the Brill-Noether number is negative. They
give rise to positive dimensional subsets of the Brill-Noether locus in which
the line bundle is fixed and only the space of sections vary (see [CLPT]) We
will consider this possibility in 1.7 and then see in 1.9 that when the
general case is possible for a chain, so is the special case of
$u_{j}^{1}+v_{j}^{i}\geq d-1$. An example for both types of filling and
correspondences is shown in Example1.8. Some readers may prefer to look at the
example before the proofs that follow.

###### Proof.

Fix an admissible filling of the $(r+1)\times(g-d+r)$ rectangle as listed. We
construct a limit linear series associated to it:

First define vanishing orders at the nodes inductively as follows

$\text{For }j=0,\dots,r,\ u_{j}^{1}=j,\
u_{j}^{i}=\begin{cases}u_{j}^{i-1}&i-1\text{ is on column }j\\\
u_{j}^{i-1}+1&\text{ otherwise}\end{cases}\ i=2,\dots,g.$ $\text{For
}j=0,\dots,r,\ v_{j}^{i}=d-u_{j}^{i+1},\ \ i=1,\dots g-1,\ \ v_{j}^{g}=d-j.$

In particular, $u_{j}^{i}=j+i-1-|\\{a<i|a\text{ is in column }j\\}|$.

The condition on the fillings of the rectangle that rows are in strictly
increasing order guarantees that $u_{j}^{i}<u_{j+1}^{i}$. Moreover,

$(*)\ \ u_{j}^{i}+1=u_{j+1}^{i}\Leftrightarrow|\\{a<i|a\text{ is in column
}j\\}|=|\\{a<i|a\text{ is in column }j+1\\}|.$

When (*) holds, as rows are strictly increasing, $i$ cannot be placed in
column $j+1$. This ensures that in all cases, $v_{j}^{i}>v_{j+1}^{i}$.

If $i\in\\{1,2,\dots,g\\}$ does not appear in the filling of the Tableau, take
$L_{i}$ a generic line bundle of degree $d$. In that case, for every $j$,
$u_{j}^{i+1}=u_{j}^{i}+1$ and therefore,
$v_{j}^{i}=d-u_{j}^{i+1}=d-u_{j}^{i}-1$. There is then a well determined
$(r+1)$-dimensional space of sections of $L_{i}$ with the giving vanishing at
$P_{i},Q_{i}$.

If $i\in\\{1,2,\dots,g\\}$ appears in at least one spot in the filling of the
Tableau in column $j$, $u_{j}^{i+1}=u_{j}^{i}$. Therefore,
$v_{j}^{i}=d-u_{j}^{i+1}=d-u_{j}^{i}$. Define
$L_{i}={\mathcal{O}}(u_{j}^{i}P_{i}+v_{j}^{i}Q_{i})$ which is a line bundle of
degree $d$. If $i$ appears in the Tableau at least twice, on row $r_{1}$ and
column $c_{1}$ and on row $r_{2}$ and column $c_{2}$

$u_{c_{2}}^{i}-u_{c_{1}}^{i}=c_{2}-c_{1}+|\\{a<i|a\text{ is in column
}c_{2}\\}|-|\\{a<i|a\text{ is in column }c_{1}\\}|=c_{2}-c_{1}+r_{2}-r_{1}$

The conditions $l_{i}(P_{i}-Q_{i})=0$, $l_{i}$ divides
$r_{2}-r_{1}+c_{2}-c_{1}$ guarantee that
${\mathcal{O}}(u_{c_{2}}^{i}P_{i}+v_{c_{2}}^{i}Q_{i})={\mathcal{O}}(u_{c_{1}}^{i}P_{i}+v_{c_{1}}^{i}Q_{i})$.
Therefore, $L_{i}$ is well defined in all instances. As before, take the space
of sections of $L_{i}$ with the given vanishing at $P_{i},Q_{i}$.

Conversely, assume given the data of a limit linear series on $X$ satisfying
$u_{j}^{i}+v_{j}^{i}\geq d-1$ and whose vanishing at $P_{1},Q_{g}$ is
$(0,\dots,r),(r,\dots,0)$. We fill the rectangle by adding successively the
indices $1,\dots,g$ in the boxes on the first empty spot of column $j$ if
$u_{j}^{i}+v_{j}^{i}=d$. In particular, if this happens, as the line bundles
have degree $d$, $L_{i}={\mathcal{O}}(u_{j}^{i}P_{i}+v_{j}^{i}Q_{i})$. Then,
the condition that $P_{i},Q_{i}$ are generic on $C_{i}$ except on
$C_{i_{1}},\dots,C_{i_{e}}$ where $l_{i_{j}}(P_{i_{j}}-Q_{i_{j}})=0$ means
that only the indices $i_{1},\dots,i_{e}$ may need to be placed in more than
one spot. Moreover, if $i_{k}$ needs to be placed on row $r_{1}$ and column
$c_{1}$ and on row $r_{2}$ and column $c_{2}$, then
$u_{c_{1}}^{i_{k}}+v_{c_{1}}^{i_{k}}=d,u_{c_{2}}^{i_{k}}+v_{c_{2}}^{i_{k}}=d$.
This requires
$L_{i_{k}}={\mathcal{O}}(u_{c_{1}}^{i_{k}}P_{i_{k}}+v_{c_{1}}^{i_{k}}Q_{i_{k}})={\mathcal{O}}(u_{c_{2}}^{i_{k}}P_{i_{k}}+v_{c_{2}}^{i_{k}}Q_{i_{k}})$.
From the genericity of the curve except on the components $C_{i_{k}}$, $l_{k}$
divides $u_{c_{1}}^{i_{k}}-u_{c_{2}}^{i_{k}}$. As the series is refined,
$u_{j}^{t+1}=d-v_{j}^{t}$. From the vanishing $(0,\dots,r)$ at $P_{1}$ and the
condition $v_{j}^{t}=d-u_{j}^{t}-1$ unless the index $t$ is placed in column
$j$, in which case $v_{j}^{t}=d-u_{j}^{t}$, we get
$u_{c_{1}}^{i_{k}}=c_{1}+i_{k}-r_{1},v_{c_{1}}^{i_{k}}=d-c_{1}-i+r_{1}$.
Therefore, $u^{i_{k}}_{c_{2}}-u^{i_{k}}_{c_{1}}=c_{1}+r_{1}-c_{2}-r_{2}$ and
the repeated indices are placed according to the required distances.

It remains to check that the whole rectangle is filled. By assumption,
$v_{j}^{g}=r-j$. By construction, $v_{j}^{g}=d-u_{j}^{1}-g+|\\{i|i\text{ is
placed in column }j\\}|=d-j-g+|\\{i|i\text{ is placed in column }j\\}|$.
Therefore, $|\\{i|i\text{ is placed in column }j\\}|=g-d+r$, meaning that all
the rows on column $j$ get filled. ∎

######  Definition 1.5.

Given integers $g\geq 2$,
$i_{1},\dots,i_{a}\in\\{1,\dots,g\\},l_{1},\dots,l_{a},0\leq a\leq g,l_{i}\geq
2$ and a rectangle consisting of $\alpha\times\beta$ squares, a filling of the
rectangle assigns to each box on the vertical strip containing the rectangle
(including boxes above and below it) numbers among $1,\dots,g$, each with a
positive or negative weight of 1. The $i$-weight $\mathbf{w^{i}(a,b)}$ of a
box is the sum of the weights of the numbers $i^{\prime}\leq i$ that appear on
it By definition, the 0-weight of a box is 1 for any squares above the given
rectangle and 0 for any other box. We call the filling an admissible filling
if it satisfies

  1. (a)

Only the numbers $i_{1},\dots,i_{a}$ appear more than once and when $i_{k}$
appears both on row $r_{1}$ and column $c_{1}$ and on row $r_{2}$ and column
$c_{2}$, then $l_{k}$ divides $r_{2}-r_{1}+c_{2}-c_{1}$. Any number may appear
multiple times with weight $-1$.

  2. (b)

Each box has at most one occurrence of a given number (including both positive
and negative weight).

  3. (c)

For each choice of an $i,0\leq i\leq g$, the $i$-weight of a box is either 0
or 1.

  4. (d)

The $i$-weight of any given box is greater than or equal to the weight of a
box to its right.

  5. (e)

The $i$-weight of any given box is greater than or equal to the weight of a
box below.

  6. (f)

The $g$-weight of a box is 1 for the boxes lying inside or above the rectangle
and 0 for those lying below that lower side of the rectangle.

######  Remark 1.6.

We are assuming that the filling can be placed in rows above and below the
rectangle. The weights of any number placed outside the rectangle cancel each
other.

######  Proposition 1.7.

Choose $g,r,d$. Let $X$ be a chain of $g$ elliptic curves generic except for
components $C_{i_{1}},\dots,C_{i_{e}}$ where

$l_{1}(P_{i_{1}}-Q_{i_{1}})=0,\dots,l_{e}(P_{i_{e}}-Q_{i_{e}})=0.$

There is a one-to-one correspondence between the components of the Brill-
Noether locus of limit linear series of degree $d$ and dimension $r$ on $X$
and admissible fillings of an $(r+1)\times(g-d+r)$ rectangle as described in
1.5.

###### Proof.

The proof is analogous to the proof of Theorem 3.4 in [T1] except that now we
do not assume the pairs of points $P_{i},Q_{i}$ on each elliptic component to
be generic. This allows us to place the index of that component in multiple
spots as was the case in Proposition 1.3 above. ∎

######  Example 1.8.

  1. (a)

We describe the limit linear series corresponding to the positive filling on
the left panel of Figure 1. Take a chain of $10$ elliptic curves, generic
except for the choice of the gluing points on $C_{5}$ where
$3(P_{5}-Q_{5})=0$. Here $r=1,d=7$. On the curves $C_{1},C_{2},C_{7}$ take
generic line bundles. Take

$L_{3}={\mathcal{O}}(2P_{3}+5Q_{3}),L_{4}={\mathcal{O}}(2P_{4}+5Q_{4}),\
L_{5}={\mathcal{O}}(2P_{5}+5Q_{5})={\mathcal{O}}(5P_{5}+2Q_{5}),L_{6}={\mathcal{O}}(2P_{6}+5Q_{6}),$
$L_{8}={\mathcal{O}}(7P_{8}),L_{9}={\mathcal{O}}(7P_{9}),L_{10}={\mathcal{O}}(7P_{10})$

Writing $s^{i}_{a,b},t^{i}_{a,b}$ for a section of $L_{i}$ vanishing at
$P_{i},Q_{i}$ with orders $a,b,a+b=d$ or $a+b=d-1$, we take the space of
sections spanned by the following sections

$(s^{1}_{0,6},t^{1}_{1,5}),\ (s^{2}_{1,5},t^{2}_{2,4}),\
(s^{3}_{2,5},t^{2}_{3,3}),\ (s^{4}_{2,5},t^{4}_{4,2}),\
(s^{5}_{2,5},t^{5}_{5,2}),\ (s^{6}_{2,5},t^{6}_{5,1}),\
(s^{7}_{2,4},t^{7}_{6,0}),\ (s^{8}_{3,3},t^{7}_{7,0}),\
(s^{9}_{4,2},t^{9}_{7,0}),\ (s^{10}_{5,1},t^{10}_{7,0})\ $

1 $2^{-}$ 3 4556$7^{-}$ 869103 45586910 Figure 1.  An arbitrary filling(right)
and corresponding positive filling(left).

  2. (b)

We describe the limit linear series corresponding to the filling on the right
panel of Figure 1. Take a chain of $10$ elliptic curves, generic except for
the choice of the gluing points on $C_{5}$ where $3(P_{5}-Q_{5})=0$ and on
$C_{6}$ where $3(P_{6}-Q_{6})=0$. Again, $r=1,d=7$. On the curves
$C_{2},C_{7}$ take generic line bundles. Take

$L_{1}={\mathcal{O}}(7Q_{1}),L_{3}={\mathcal{O}}(2P_{3}+5Q_{3}),L_{4}={\mathcal{O}}(2P_{4}+5Q_{4}),\
L_{5}={\mathcal{O}}(2P_{5}+5Q_{5})={\mathcal{O}}(5P_{5}+2Q_{5}),$
$L_{6}={\mathcal{O}}(2P_{6}+5Q_{6})={\mathcal{O}}(5P_{6}+2Q_{6}),L_{8}={\mathcal{O}}(7P_{8}),L_{9}={\mathcal{O}}(7P_{9}),L_{10}={\mathcal{O}}(7P_{10})$

Writing $s^{i}_{a,b},t^{i}_{a,b}$ for a section of $L_{i}$ vanishing at
$P_{i},Q_{i}$ with orders $a,b$, we take spaces of sections spanned by the
following sections

$(s^{1}_{0,7},t^{1}_{1,5}),\ (s^{2}_{0,5},t^{2}_{2,4}),\
(s^{3}_{2,5},t^{2}_{3,3}),\ (s^{4}_{2,5},t^{4}_{4,2}),\
(s^{5}_{2,5},t^{5}_{5,2}),\ (s^{6}_{2,5},t^{6}_{5,2}),\
(s^{7}_{2,4},t^{7}_{5,0}),\ (s^{8}_{3,3},t^{7}_{7,0}),\
(s^{9}_{4,2},t^{9}_{7,0}),\ (s^{10}_{5,1},t^{10}_{7,0})\ $

######  Proposition 1.9.

Choose $g,r,d$. Let $X$ be a chain of $g$ elliptic curves generic except for
components $C_{i_{1}},\dots,C_{i_{a}}$ where

$l_{1}(P_{i_{1}}-Q_{i_{1}})=0,\dots,l_{a}(P_{i_{a}}-Q_{i_{a}})=0.$

If there is a limit linear series on $X$, then there is a limit linear series
whose vanishing at $P_{1},Q_{g}$ is $(0,\dots,r),(r,\dots,0)$ and
$P_{i},Q_{i}$ satisfy $u_{j}^{i}+v_{j}^{i}\geq d-1$.

###### Proof.

In view of Propositions 1.3 , 1.7, it suffices to see that the existence of an
admissible filling of $(r+1)\times(g-d+r)$ rectangle as described in 1.5
implies the existence of that same rectangle with only positive weights. This
can be built by removing from the filling of each square of the rectangle all
but the last index that appears with positive weight 1.

The right pane of Figure 1 shows an arbitrary filling of the rectangle whose
associated positive filling is on the left pane. ∎

######  Lemma 1.10.

Assume that $\alpha\leq\beta$,

$e\leq\frac{(\alpha+2)(\alpha-1)}{2}\text{ if }\alpha<\beta;\ \
e\leq\frac{\alpha^{2}-2}{2}\text{ if }\alpha=\beta.$

Define $k,j$ by

$\frac{k(k+1)}{2}\leq e<\frac{(k+1)(k+2)}{2},\ \ j=e-\frac{k(k+1)}{2}.$

Consider an admissible filling of an $\alpha\times\beta$ rectangle as in 1.2
such that the only indices that appear twice are $i_{1},\dots i_{e}$. If
$i_{t}$ appears in row $r^{i_{t}}_{1}$ and column $c^{i_{t}}_{1}$ and on row
$r^{i_{t}}_{2}$ and column $c^{i_{t}}_{2}$. Then,

$\sum_{t=1}^{e}(|r^{i_{t}}_{2}-r^{i_{t}}_{1}|+|c^{i_{t}}_{2}-c^{i_{t}}_{1}|)\leq
e(\alpha+\beta-2)-2(\frac{k^{3}-k}{3}+jk).$

There are admissible fillings of the rectangle with the numbers
$1,2,\dots,\alpha\beta-e$ for which this inequality is an equality.

###### Proof.

We are trying to maximize the grid distance between the pairs of spots where
we place repeated indices.

As there are $\beta$ rows and $\alpha$ columns, $1\leq
r^{i}_{1}<r^{i}_{2}\leq\beta,1\leq c^{i}_{2}<c^{i}_{1}\leq\alpha$. Therefore

$|r^{i}_{2}-r^{i}_{1}|\leq\beta-1,\ |c^{i}_{2}-c^{i}_{1}|\leq\alpha-1.$

So an upper bound for the grid distance between two squares of the rectangle
is $\alpha+\beta-2$. The highest bound for both pairs of inequalities occurs
when the two squares are placed in diagonally opposite spots of the rectangle.
There are two spots close to each of the corners where one of the indices for
the row and column $r,c$ stays at the max(resp min) while the other decreases
(resp increases) by one. There are three spots where the difference with the
max or min is 3, …, there are $t$ squares close to each corner giving a
distance of $t-1$ to the optimal distancing up to $t=\alpha-1$. For
$t\geq\alpha$, there are only $\alpha-1$ spots with optimal distance as we
cannot place a double index close to the bottom left corner on the last column
or a double index close to the top right corner on the first column. Note that
$1+2+3+\dots+k=\frac{k(k+1)}{2}$,
$[1+2+3+\dots+(\alpha-1)]+(\alpha-1)=\frac{(\alpha+2)(\alpha-1)}{2}$. By
assumption $e\leq\frac{(\alpha+2)(\alpha-1)}{2}$ if $\alpha<\beta$ while
$e\leq\frac{\alpha(\alpha-1)}{2}$ if $\alpha=\beta$. By definition of $k$,
$\frac{k(k+1)}{2}\leq e<\frac{(k+1)(k+2)}{2}$. Therefore $k\leq\alpha-1$ and
also by definition of $j$, $j\leq k-1$. If $k=\alpha-1,\alpha=\beta$ , we
imposed the stronger condition $j\leq\frac{\alpha-2}{2}$. This is needed
because in this case the diagonal at distance $\alpha-1$ from the two corners
is the same for both corners. So the spots in that diagonal need to be shared
between the two corners, only $\frac{\alpha-2}{2}$ are available to each of
them.

To maximize the sum of grid distances, we should use all the optimal spots up
to distance $k$ and $j$ of those at distance $k+1$. This gives

$e(\alpha+\beta-2)-2[1\times 0+2\times
1+\dots+k(k-1)+jk]=e(\alpha+\beta-2)-2[\frac{k^{3}-k}{3}+jk].$

Because we are using grid distance, it does not matter in which way we pair
the squares at the two corners. Adding the maximum distances, we obtain the
upper bound.

To show that the bound is attained, we need to manufacture an admissible
filling that fits the spots with the optimal separation. The case
$\beta=\alpha,k=\alpha-1,j>0$ will require special atention.

If $j\not=0$, and $k+1<\beta$ we choose the last $j$ spots among those on the
$k+1$ diagonal with the largest possible column value. Write

$\epsilon_{i}=\begin{cases}0&\text{ if }i\leq\min(k-j+1,\alpha-1-j)\\\
1&\text{ if }\min(k-j+1,\alpha-1-j)<i\leq\min(k+1,\alpha-1)\end{cases}$

If $j\not=0$, and $k+1=\alpha=\beta$ we choose $j$ alternating spots among
those on the $k+1$ diagonal with the largest possible column value. Write

$\epsilon_{i}=\begin{cases}1&\text{ if }i=\alpha-2t,1\leq t\leq j\\\ 0&\text{
otherwise}\end{cases}$

On column $i$, we use $a_{i}=k-i+1+\epsilon_{i}$ rows from the bottom and
$b_{i}=k-\alpha+i+\epsilon_{i-1}$ rows from the top for the repeated indices.

Together, they use at most $2k-\alpha+i+\epsilon_{i}+\epsilon_{i+1}$ which
under our conditions is at most $\beta$. Therefore, the chosen spots for the
double indices do not overlap.

12712183481319569142071011152112131617221819202123136813247101435911156781216101112131714151617181246923571145681278910131011121314
Figure 2.  Examples with $\alpha=5,\beta=6,e_{1}=7,e_{2}=12$ (so
$k_{1}=3,j_{1}=1,k_{2}=4,j_{2}=2$) and with $\alpha_{3}=\beta_{3}=5,e_{3}=11$
(so $k_{3}=4,j_{3}=1$) .

Fill first the rectangle formed by the first $\alpha-k$ columns and $\beta-k$
rows in any admissible way with the different numbers
$1,2,\dots,(\alpha-k)(\beta-k)$. Fill then the first spot in row $\beta-k+1$
and the first spot in column $\alpha-k+1$ with the same next number. Finish
filling row $\beta-k+1$ and column $\alpha-k+1$ with different numbers in an
admissible way

Next fill the first spot in row $\beta-k+2$ and the first spot in column
$\alpha-k+2$ with the same next number, Fill the second spot in row
$\beta-k+2$ and the second spot in column $\alpha-k+2$ both with the same
following number. Finish filling row $\beta-k+2$ and column $\alpha-k+2$ with
different numbers in an admissible way.

Keep filling row $\beta-k+s$ and column $\alpha-k+s,s\leq k-j,k<\alpha-1$
where the first $s$ spots of the row match with first $s$ spots of the column
while the rest are different. For $s=k-j+1,\dots,k,,k<\alpha-1$ take matching
numbers in $s+1$ of the spots instead of just $s$. These choices, attain the
maximum distance that we computed above.

If $k=\alpha-1$ and $\alpha\not=\beta$ proceed as above up to the filing of
row $\beta-j-1$ and column $\alpha-j-1$. When filling row $\beta-j$ in
addition to the double spots from column $\alpha-j$ as before, use the bottom
spot of column $\alpha-j+1$ and then complete the filling of the rectangle
$(\alpha-j)\times(\beta-j)$.

If $k=\alpha-1,\alpha=\beta,j>0$, we proceed with the filling as before until
we reach row and column $\alpha-2j$. Then the spot in row and column
$\alpha-2j$ is matched with the element in the first row in column
$\alpha-2j+1$. Then proceed to filling row $\alpha-2j+1$ and column
$\alpha-2j+1$ using the first spot in row $\alpha-2j+2$ and the first spot in
column $\alpha-2j+2$(in this order). Continue similarly with the remaining
rows and columns using alternatively the bottom element in the next column or
both the bottom element in the next column and the left most element on the
next row. ∎

######  Lemma 1.11.

Let $\alpha,\beta,g$ be integers such that $\alpha\leq\beta$, and
$\frac{\alpha\beta}{2}+1\leq g\leq\alpha\beta$. There exists an admissible
filling of the $\alpha\times\beta$ rectangle with the indices $1,\dots,g$
where $\alpha\beta-g$ of the indices appear exactly twice and the remainder
appear exactly once.

###### Proof.

We build a filling such that that the repeated indices occupy the spots closer
to the top right and bottom left corners of the rectangle forming stair case
shapes. The empty space (if any) between the two stair cases will be filled
with non-repeated indices, generalizing the construction in Lemma 1.10.

We first determine the spots for the repeated indices, depending on the values
of $g,\alpha,\beta$. Write $e=\alpha\beta-g$. Define $t_{0}$ to approximate
half the difference between $\beta$ and $\alpha$, that is
$t_{0}=\lfloor{\frac{\beta-\alpha+1}{2}}\rfloor$.

For $\alpha=\beta$, the result was proved in the full range of $g$ allowed by
our condtions in Lemma 1.10. For $\alpha<\beta$, the result was proved for
$e\leq\frac{(\alpha+2)(\alpha-1)}{2}$. We can therefore assume that
$\alpha<\beta,e>\frac{(\alpha+2)(\alpha-1)}{2}$.

Define $t,l,$ and $\epsilon_{i},i=1,\dots,\alpha-1$ as follows:

  1. (a)

If $e\leq\frac{\alpha(\alpha-1)}{2}+t_{0}(\alpha-1)$, define $t$ by

$\frac{(\alpha+2t)(\alpha-1)}{2}=\frac{\alpha(\alpha-1)}{2}+t(\alpha-1)\leq
e<\frac{\alpha(\alpha-1)}{2}+(t+1)(\alpha-1)=\frac{(\alpha+2t+2)(\alpha-1)}{2}$

in particular, $1\leq t\leq t_{0}$ and $0\leq
e-\frac{(\alpha+2t)(\alpha-1)}{2}=j<\alpha-1$. Take $l=0$ and define
$\epsilon_{i},i=1,\dots,\alpha-1$ by

$\epsilon_{i}=\begin{cases}0&\text{ if }i\leq\alpha-1-j\\\ 1&\text{ if
}\alpha-j\leq i\leq\alpha-1\end{cases}$

Note that

$e=\frac{\alpha(\alpha-1)}{2}+(\alpha-1)t+\sum\epsilon_{i},l=0.$

  2. (b)

If $\frac{\alpha(\alpha-1)}{2}+t_{0}(\alpha-1)\leq e$, then $t=t_{0}$.

If $\beta-\alpha$ is odd, then $\epsilon_{i}=0,i=1,\dots,\alpha-1$.

If $\beta-\alpha$ is even and
$j=\min(e-\frac{(\alpha+2t)(\alpha-1)}{2},\frac{\alpha-1}{2})$ we choose $j$
alternating spots among those on the $\alpha+t_{0}$ diagonal with the largest
possible column value. Write

$\epsilon_{i}=\begin{cases}1&\text{ if }i=\alpha-2t,1\leq t\leq j\\\ 0&\text{
otherwise}\end{cases}$

Take

$l=e-\frac{(\alpha+2t_{0})(\alpha-1)}{2}-\sum\epsilon_{i}.$

The spots for the repeated indices in the lower left corner will be as
follows,

  * •

On column one, take the bottom $a_{1}=\alpha-1+t+\epsilon_{1}+l$ rows.

  * •

On column $i,i=2,\dots,\alpha-1$, take the bottom
$a_{i}=\alpha-i+t+\epsilon_{i}$ rows.

On the top right corner, the chosen spots for the repeated indices are the
first $b_{i}$ rows from the top with $b_{i}$ given by

  * •

On column $i,i=2,\dots,\alpha-1$, take the top $b_{i}=i-1+t+\epsilon_{i-1}$
rows.

  * •

On column $\alpha$ take the first
$b_{\alpha}=\alpha-1+t+\epsilon_{\alpha-1}+l$ rows.

By definition of the $\epsilon_{i}$ and $l$,

(1) $e=a_{1}+\dots+a_{\alpha-1}=b_{2}+\dots+b_{\alpha}$

Therefore, the number of spots chosen to carry a double index is $e$ on each
of the two corners.

By construction,

(2) $a_{1}\leq a_{2}\leq\dots\leq a_{\alpha-2}\leq a_{\alpha-1},\ b_{2}\geq
b_{3}\geq\dots\geq b_{\alpha-1}\geq b_{\alpha}$

Note that

$a_{i}+b_{i}=\begin{cases}\alpha+2t-1+\epsilon_{i}+\epsilon_{i+1}&\text{if
}i\not=1,\alpha\\\ \alpha+t-1+\epsilon_{1}+l&\text{if }i=1\\\
\alpha+t-1+\epsilon_{\alpha-1}+l&\text{if }i=\alpha\end{cases}$

As $t\leq t_{0}$, using the definition of $t_{0}$ and $\epsilon_{i}$, we
obtain $a_{i}+b_{i}\leq\beta,i=2,\dots,\alpha-1$. In particular, this means
that the chosen double spots fit in their respective columns without
overlapping. Moreover, if $l\not=0$, then
$a_{i}+b_{i}=\beta,i=2,\dots,\alpha-1$. Then, from the definition of $l$

$2l=2e-2\sum_{i=2}^{\alpha-1}a_{i}-2(\alpha-1+t_{0}+\epsilon_{1})=2e-\sum_{i=2}^{\alpha-1}a_{i}-\sum_{i=2}^{\alpha-1}b_{i}-2(\alpha-1+t_{0})-\epsilon_{1}-\epsilon_{\alpha-1}.$

By assumption, $g\geq\frac{\alpha\beta}{2}+1$. Therefore,
$e=\alpha\beta-g\leq\frac{\alpha\beta}{2}-1$. Then, for $l\not=0$,

$2l\leq\alpha\beta-2-(\alpha-2)\beta-2(\alpha-1+t_{0})-\epsilon_{1}-\epsilon_{\alpha-1}-=2(\beta-\alpha-
t_{0})-\epsilon_{1}-\epsilon_{\alpha-1}\leq 2t_{0}-\epsilon_{1}$

In particular

$a_{1}=\alpha-1+t_{0}+\epsilon_{1}+l\leq\alpha+2t_{0}\leq\beta-1$

This again guarantees that the spots chosen on the first column fit in the
rectangle with at least one space left over. Moreover, if there is exactly one
space left over, then $\beta-\alpha=2t_{0}+1$ is odd so $\epsilon_{i}=0$ for
all $i$. As

$b_{\alpha}=\alpha-1+t_{0}+\epsilon_{\alpha-1}=a_{1}-\epsilon_{1}+\epsilon_{\alpha-1}$

the last column also fits with at least one space left over.

1591126101537111648121759131861014191112152016171821125102361134712458136791489101511121316141516171258132369143479154581116679121710111213181415161719
Figure 3.  Examples with $\alpha=4,\beta=8,e_{1}=11,e_{2}=15=g-2$ and with
$\alpha_{3}=5,\beta_{3}=7,e_{3}=16=g-3$ .

Fill now the rectangle as follows.

  * •

Fill the first $\beta-a_{1}$ spots in column one with different indices
$1,\dots,\beta-a_{1}$. This means filling the spots in column one that have
not been reserved for repeated indices.

  * •

Fill the first $b_{2}$ spots in column two and the next $b_{2}$ spots in
column one using the next $b_{2}$ indices available, (the same index at the
bottom of column two and the first portion of the spots chosen for repeated
index in column one). This is possible, as the conditions in equation (2)
imply that $b_{2}\leq a_{1}$. This fills column one up to height $\beta-
a_{1}+b_{2}$.

  * •

Fill the next $\beta-a_{2}-b_{2}$ spots in column two with different indices.
For this to be possible, we need the height $h$ of already filled spots in
column 1 to be at least as tall as the height we want to fill in column 2.

On column one, we filled the first $\beta-a_{1}+b_{2}$ spots. On column two we
want to fill $\beta-a_{2}$ spots. We need to check that $\beta-
a_{1}+b_{2}\geq\beta-a_{2}$ or equivalently, $a_{1}\leq a_{2}+b_{2}$. If
$l=0$, this follows from the definitions of $a_{i},b_{i}$ while for $l\not=0$,
we already proved that $a_{1}\leq\beta-1\leq a_{2}+b_{2}$.

  * •

We proceed by induction. Assume that we have filled columns $1,\dots,i-1$ all
of them to height at least $\beta-\alpha+i$ and the first $b_{i}$ spots of
column $i$ in such a way that the spots reserved for repeated indices on the
top right corner are matched with corresponding spots on the left lower corner
and the spots in between use different indices.

We check now that we can carry out one more step: We are assuming $t\geq 1$,
therefore

$\beta-a_{i}=\beta-(\alpha-i+t+\epsilon_{i})\leq\beta-\alpha+i-1.$

By induction assumption, columns $1,\dots,i-1$ have height at least
$\beta-\alpha+i-1$. Therefore, they are high enough to allow filling column
$i$ from rows $b_{i}+1$ to row $\beta-a_{i}+i-1$ with distinct numbers. After
this, place indices in the first $b_{i+1}$ spots of column $i+1$ matching them
with spots reserved for repeated indices in columns $1,\dots,i$. This is
possible as up to this point, we had used $b_{1}+\dots+b_{i}$ spots designed
for double filling in rows $1,\dots,i-1$ and from equations (1), (2),

$a_{1}+\dots+a_{i}-(b_{2}+\dots+b_{i}+b_{i+1})\geq 0$

Moreover, by assumption, before adding the filling with double indices,
columns $1,\dots,i-1$ were filled to height at least $\beta-\alpha+i-1$ while
column $i$ was filled to height $\beta-a_{i}=\beta-\alpha-i+-t-\epsilon_{i}$
As we are adding $b_{i+1}=i+t+\epsilon_{i}$ indices to these columns, this
allows us to raise the height of the columns to at least $\beta-\alpha+i$
completing the induction step.

  * •

After filling columns $1,\dots,\alpha-1$ covering all unassigned spots in
these columns and the bottom $b_{\alpha}$ spots of column $\alpha$, we have
filled $b_{2}+\dots+b_{\alpha}=a_{1}+\dots+a_{\alpha-1}$ spots earmarked for
repeated indices. Therefore, columns $1,2,\dots,\alpha-1$ are completely
filled. Then, we can fill the remaining indices in column $\alpha$ with
different indices. As $e=\alpha\beta-g$ indices have been used twice and the
remaining indices have been used once, we used all indices $1,\dots,g$ and no
others.

∎

We recall the following result that we will need later. The statement in Lemma
1.7 and Proposition 1.8 of [T1] assumes that the chain of elliptic curves is
generic but this condition is not used in the proof

######  Proposition 1.12.

Given a chain of elliptic curves and a limit section of the canonical linear
series, there exists at least one elliptic component $C^{i}$ of the chain such
that $ord_{P^{i}}s^{i}+ord_{Q^{i}}s^{i}=2g-2$.

Given $(s_{0}^{i}),\dots(s_{\bar{r}}^{i})$ refined limit sections of the
canonical linear series such that for

$ord_{P^{i_{k}}}s_{k}^{i_{k}}+ord_{Q^{i_{k}}}s_{k}^{i_{k}}=2g-2,\ \
i_{m}\not=i_{n}\text{ if }m\not=n$

Then the $(s_{k}^{i}),k=0,\dots,\bar{r}$ are linearly independent limit linear
series.

##  2\. Existence of components of codimension $-\rho$.

We now use the results of the previous section to show the existence of
components of the Brill-Noether loci of the expected dimension. We will then
use the techniques developed in [LOTZ1], [LOTZ2], [T1] to show that at a
generic point of this component, the Petri map is onto and the generic maximal
rank conjecture is satisfied.

######  Theorem 2.1.

Fix values of the genus $g,r,d$ such that $0<-\rho(g,r,d)\leq g-2$. Then there
exists at least one component of ${\mathcal{M}}^{r}_{g,d}$ of codimension
$-\rho$ in ${\mathcal{M}}_{g}$. The Petri map at the generic point of this
component is onto.

###### Proof.

From Lemma 1.10, with the above condition on $-\rho=e$ (in the notation of
Lemma 1.10 ), there exists a filling of the $(r+1)\times(g-d+r)$-rectangle
with the indices $1,2,\dots,(r+1)\times(g-d+r)+\rho$. As there are
$(r+1)\times(g-d+r)$ spots in the rectangle, $-\rho$ of the indices in the
filling are repeated. From Proposition 1.3, such a filling corresponds to
limit linear series on chains of elliptic curves such that on $-\rho$ of the
curves in the chain (corresponding to the repeated indices) the difference of
the nodes is a torsion point on the elliptic curve (of order determined by the
grid distance between the two appearances of the index). Therefore, the chains
of elliptic curves on which such a linear series exists moves on a space of
codimension precisely $-\rho$ in the space of all chains of elliptic curves.
This is the expected codimension of the loci of curves that posses a linear
series of the given degree and dimension. From the determinantal nature of the
space of linear series, these series extend to nearby non-singular curves.
Therefore, there exists at least one component of ${\mathcal{M}}^{r}_{g,d}$ of
codimension $-\rho$ in ${\mathcal{M}}_{g}$.

We now need to show that the Petri map is onto, or equivalently, that it has
maximal rank.

Each column of the filled rectangle corresponds to a limit linear series
$s_{i},i=1,\dots,r+1$ such that

$ord_{P_{1}}s_{i}=i-1,ord_{Q_{g}}s_{i}=r-i+1,\
ord_{Q_{t}}s_{i}+ord_{P_{t+1}}s_{i}=d,\ \
ord_{P_{n}}s_{i}+ord_{Q_{n}}s_{i}=\begin{cases}d&\text{ if }n\text{ is in
column }\ \ \ \ i\\\ d-1&\text{ if }n\text{ is not in column }i\end{cases}$

Interchanging rows and columns for an admissible filled rectangle, we obtain
another admissible filled rectangle. We will call this rectangle the Serre
dual rectangle. The linear series of segree $d$ and dimension $r$ then has a
Serre dual linear series of degree $2g-2-d$ and dimension $g-d+r-1$. As grid
distances are preserved when interchanging rows and columns, the Serre dual
series works on the same special chain as the original one. We denote by
$t_{1},\dots,t_{g-d+r}$ the sections corresponding to the columns of the Serre
dual linear series. To simplify notations, we write $g-d+r=\bar{r}+1$

In the construction of an admissible filling from Lemma 1.11, the repeated
indices appear in the last $a_{i}$ spots of column $i,\ i=1,\dots,r$. Consider
then the collection of sections

$s_{1}t_{1},s_{1}t_{2},\dots,s_{1}t_{\bar{r}+1-a_{1}},s_{2}t_{1},s_{2}t_{2},\dots,s_{2}t_{\bar{r}+1-a_{2}},\dots,s_{r}t_{1},s_{r}t_{2},\dots,s_{r}t_{\bar{r}+1-a_{r}},s_{r+1}t_{1},s_{r+1}t_{2},\dots,s_{r+1}t_{\bar{r}+1}$

Let $k$ be the index that appears in column $i$ and row $j$. Then, the section
$s_{i}t_{j}$ satisfies

$\ ord_{P_{k}}s_{i}+ord_{Q_{k}}s_{i}=d,\
ord_{P_{k}}t_{j}+ord_{Q_{k}}t_{j}=2g-2-d$

therefore,

$\ ord_{P_{k}}(s_{i}t_{j})+ord_{Q_{k}}(s_{i}t_{j})=2g-2$

For the collection of sections that we chose above, all these values of $k$
are different. Then from Proposition 1.12, the sections $s_{i}t_{j}$ in the
above list are linearly independent. ∎

######  Theorem 2.2.

Fix values of the genus $g,r,d$ such that $r+1\leq g-d+r,\
0<-\rho(g,d,r)\leq\frac{r(r+1)}{2}$. Then there exists at least one component
of ${\mathcal{M}}^{r}_{g,d}$ of codimension $-\rho$ in ${\mathcal{M}}_{g}$
whose generic element satisfies the maximal rank conjecture for $m=2$..

###### Proof.

Assume first $r+1=g-d+r,\ 0<-\rho(g,d,r)=\frac{r(r+1)}{2}$. It suffices to
find a family of curves intersecting ${\mathcal{M}}^{r}_{g,d}$ in codimension
$-\rho$ and such that the generic curve in this intersection has a finite
number of $g^{r}_{d}$’s for which the map $S^{2}(H^{0}(C,L))\to
H^{0}(C,L^{2})$ is injective. We will take as the special curve and linear
series a chain of elliptic curves corresponding to a a filling of the
$(r+1)\times(r+1)$ rectangle with the indices $1,2,\dots,(r+1)^{2}+\rho$ as in
Lemma 1.10. This means that we fill the first $k-1$ spots of row $k$ and
column $k$ with the same indices while the indices in the diagonal appear only
once.

12471123581245691378910141112131415 Figure 4.

For an index $k,1\leq k\leq g$, define $a,t$ by

$\frac{a(a+1)}{2}<k\leq\frac{(a+1)(a+2)}{2},\ \ k=\frac{a(a+1)}{2}+t,1\leq
t\leq a+1.$

Then, the index $k$ gets placed on column $a+1$ row $t$ and row $a+1$ column
$t$. Prior to the placement of $k$, columns $1,\dots,t-1$ have $a+1$ spots
filled , columns $t,\dots,a$ have $a$ spots filled, column $a+1$ has $t-1$
spots filled and columns $a+1,\dots,r+1$ are empty. From the description of
the linear series associated to a filling in Proposition 1.3 , the vanishing
of the section $s_{i}$ at $P_{k}$ is

$i-2+k-\begin{cases}a+1&\text{ if }i<t\\\ a&\text{ if }t\leq i\leq a\\\
t-1&\text{ if }i=a+1\\\ 0&\text{ if }i>a+1\end{cases}$

the vanishing of the section $s_{i}$ at $Q_{k}$ is

$d-i+1-k+\begin{cases}a+1&\text{ if }i\leq t\\\ a&\text{ if }t<i\leq a\\\
t&\text{ if }i=a+1\\\ 0&\text{ if }i>a+1\end{cases}$

To prove the injectivity of the Symmetric cup-product map on the special
curve, we use the method introduced in [LOTZ1] , [LOTZ2]. The general process
is as follows: from the sections that we used already for the linear series ,
we form the products $s_{1}s_{1},s_{1}s_{2},\dots
s_{1}s_{r+1},s_{2}s_{2},\dots,s_{2}s_{r+1},\dots,s_{r}s_{r},s_{r}s_{r+1},s_{r+1}s_{r+1}$.
We assume we have a linear combination that is zero. Instead of concentrating
all the degree on a single elliptic component, as it is normally done in the
study of limit linear series, one can choose a particular distribution of the
degree on the chain of elliptic curves, so that the total degree $d^{\prime}$
is still as prescribed (for $L^{2}$, $d^{\prime}=2d$). If we assign degrees
$d_{1},\dots,d_{g},\sum d_{i}=d^{\prime}$ to the different components, a
section of $L^{2}$ will be non-zero on the curve $C_{i}$ only if it vanishes
at $P_{i}$ with order at least $d_{1}+\dots+d_{i-1}$ and at $Q_{i}$ with order
$d_{i+1}+\dots+d_{g}$. Looking then successively at components
$C_{1},\dots,C_{g}$ allows us to deduce that the coefficients appearing in the
linear combination are zero.

In our case, we choose the distribution of degrees that assigns degree 1 to
$C_{1}$, degree 2 to $C_{2},\dots,C_{g-1}$ and degree $2d-1-2(g-2)$ to
$C_{g}$. From our assumptions $r+1=g-d+r,-\rho(g,d,r)=\frac{r(r+1)}{2}$, one
can check that $2d-1-2(g-2)=1$.

Given a linear combination $\sum_{i\leq j}\lambda_{ij}s_{i}s_{j}=0$, we will
show the following:

Claim. Assume that $\lambda_{ij}=0\text{ if }\ i\leq j\leq a,\text{ or if }\
i<t,j=a+1.$ Then also $\lambda_{t,a+1}=0$

We compute the order of vanishing at $P_{k},Q_{k}$ of $s_{i}s_{j},i\leq j$ for
those pairs $i,j$ for which we are not already assuming that $\lambda_{ij}=0$.

We note that from our choice of degree distribution, the only non-zero
sections on $C_{k}$ are those such that they vanish at $P_{k}$ with order at
least $2k-3$ and at $Q_{k}$ with order at least $2g-2k-1=2(g-1)-2k+1=2d-2k+1$.
From our computations above, the section $s_{t}s_{a+1}$ vanishes to orders
$2k-2$ and $2d-2k+2$ at $P_{k}$ and $Q_{k}$ respectively. On the other hand,
the order of vanishing at $Q_{k}$ of $s_{i}s_{j}$ for those sections for which
we do not already know that the coeffficent $\lambda_{i,j}=0$ is given as

$2d-2k+2-i-j+\begin{cases}a+1&\text{ if }i\leq t,j>a+1\\\ a+t&\text{ if
}t<i\leq a,j=a+1\\\ a&\text{ if }t<i\leq a,j>a+1\\\ 2t&\text{ if
}j=i=a+1,\text{ (only if }t<a+1)\\\ t&\text{ if }i=a+1,j>a+1\\\ 0&\text{ if
}i,j>a+1\end{cases}$

In all cases, we check that this number is less than $2d-2k+2$. Therefore, the
only not trivially zero section on $C_{k}$ with the chosen distribution of
degrees is $s_{t}s_{a+1}$. This forces $\lambda_{t,a+1}=0$. Therefore, all of
the $\lambda_{i,j}=0$ and the sections are linearly independent.

If $r+1=g-d+r,\ 0<-\rho(g,d,r)\leq\frac{r(r+1)}{2}$, we can consider a chain
of elliptic curves corresponding to a filling of the square with
$g_{1}=(r+1)^{2}-\frac{r(r+1)}{2}$ indices and take the $g-g_{1}$ last curves
in the chain to be generic. Such a curve is more special than the one
corresponding to a filling using all the indices $1,\dots,g$. The proof in the
case $\rho(g,d,r)=\frac{r(r+1)}{2}$ shows that the map $S^{2}(H^{0}(C,L))\to
H^{0}(C,L^{2})$ is injective on the special curve. Therefore, it is also
injective on the more general curve.

When $r+1<g-d+r$, using the condition $\rho(g,r,d)\leq\frac{r(r+1)}{2}$, we
can concentrate the repeated indices on the top $(r+1)^{2}$ square. Then the
same proof as before shows that the map $S^{2}(H^{0}(C,L))\to H^{0}(C,L^{2})$
is injective. ∎

##  3\. Components of the same codimension are different

We want to look at possible equalities of loci ${\mathcal{M}}^{r}_{g,d}$ when
these loci are of the same expected dimension. It was proved in [CKK], [CK],
that when the codimension is 1 or 2, then the loci are distinct. We generalize
this result here to higher codimension. Note that by Serre duality,
${\mathcal{M}}^{r}_{g,d}={\mathcal{M}}^{r+g-d-1}_{g,2g-2-d}$, therefore,
assuming that $r+1\leq g-d+r$ is not restrictive.

######  Theorem 3.1.

Fix values of the genus $g,r_{i},d_{i}$ such that $\rho(g,r_{i},d_{i})<0$.
Write $e_{i}=-\rho(g,r_{i},d_{i})$.

If $d_{1},r_{1},d_{2},r_{2}$ are such that $e_{1}=e_{2}$ and for each $i$,
either $r_{i}+1<g-d_{i}+r_{i}$ and $e_{i}\leq\frac{(r_{i}+3)r_{i}}{2}$ or
$r_{i}+1=g-d_{i}+r_{i}$ and $e_{i}\leq\frac{(r_{i}^{2}-2r_{i}-1)}{2}$ then
either $(r_{1},d_{1})=(r_{2},d_{2})$ or
${\mathcal{M}}^{r_{1}}_{g,d_{1}}\not={\mathcal{M}}^{r_{2}}_{g,d_{2}}$.

###### Proof.

The conditions $\rho(g,r_{1},d_{1})=\rho(g,r_{2},d_{2})$, is equivalent to
$(r_{1}+1)(g-d_{1}+r_{1})=(r_{2}+1)(g-d_{2}+r_{2})$. If also
$(r_{1}+1)+(g-d_{1}+r_{1})=(r_{2}+1)+(g-d_{2}+r_{2})$, then, as the sum and
product of two numbers determine the two numbers up to the order, either
$r_{1}+1=r_{2}+1,g-d_{1}+r_{1}=g-d_{2}+r_{2}$ or
$r_{1}+1=g-d_{2}+r_{2},g-d_{1}+r_{1}=r_{2}+1$. The first option means that
$d_{1}=d_{2},r_{1}=r_{2}$. The second option translates into
$d_{1}=2g-2-d_{2},r_{1}+1=d_{1}+1-g+r_{2}+1$. So, the two Brill-Noether loci
correspond to the same values of degree and dimension or to the Serre dual
values.

Assume then that $(r_{1}+1)+(g-d_{1}+r_{1})\not=(r_{2}+1)+(g-d_{2}+r_{2})$. To
fix ideas, we can assume that
$(r_{1}+1)+(g-d_{1}+r_{1})>(r_{2}+1)+(g-d_{2}+r_{2})$. With the notation of
Lemma 1.10, as $e_{1}=e_{2}=e$, also $k_{1}=k_{2}=k,j_{1}=j_{2}=j$. From Lemma
1.10, there is an admissible fillings of the
$(r_{1}+1)\times(g-d_{1}+r_{1})$-rectangle with the numbers
$1,2,\dots,(r_{1}+1)(g-d_{1}+r_{1})-e$ for which the sums of grid distances
$|r^{i_{t}}_{1,2}-r^{i_{t}}_{1,1}1|+|c^{i_{t}}_{1,2}-c^{i_{t}}_{1,1}|$ between
the spots where the indices $i_{t},t=1,\dots,e$ appear satisfies

$\sum_{t=1}^{e}(|r^{i_{t}}_{1,2}-r^{i_{t}}_{1,1}|+|c^{i_{t}}_{1,2}-c^{i_{t}}_{1,1}|)=e(g+2r_{1}-d_{1}-2)-2[2\times
1+3\times 2+\dots+k(k-1)+jk]=A_{1}.$

On the other hand, for every filling of the
$(r_{2}+1)\times(g-d_{2}+r_{2})$-rectangle with the numbers
$1,2,\dots,(r_{2}+1)(g-d_{2}+r_{2})-e$ the sums of grid distances of repeated
digits satisfy

(3)
$\sum_{t=1}^{e_{2}}(|r^{i_{t}}_{2,2}-r^{i_{t}}_{2,1}|+|c^{i_{t}}_{2,2}-c^{i_{t}}_{2,1}|)\leq
e(g+2r_{2}-d_{2}-2)-2[2\times 1+3\times 2+\dots+k(k-1)+jk]<A_{1}.$

Consider a chain of elliptic curves for which there exists a limit linear
series of degree $d_{1}$ and rank $r_{1}$ with maximal sum of grid distances
$A_{1}$ among repeated indices. From inequality 3, such a curve does not
support a limit linear series of rank $r_{2}$ and degree $d_{2}$. Therefore,
${\mathcal{M}}^{r_{1}}_{g,d_{1}}\not={\mathcal{M}}^{r_{2}}_{g,d_{2}}$.∎

The methods we use in the previous proof can also be used to exclude certain
inclusions of Brill-Noether loci. While this tool is unlikely to prove Auel-
Haburcak conjecture, it provides some evidence that inclusion of potentially
maximal Brill-Noether loci are rare. We give only one example below:

######  Example 3.2.

Assume that
${\mathcal{M}}^{r_{1}}_{g,d_{1}}\subseteq{\mathcal{M}}^{r_{1}+1}_{g,d_{2}}$
with the two loci being of codimension 2 and 1 respectively in
${\mathcal{M}}_{g}$.

The conditions on codimension mean that $g-(r_{1}+1)(g-d_{1}+r_{1})=-2,\
g-(r_{2}+1)(g-d_{2}+r_{2})=-1$. We can construct a smoothable
$g^{r_{1}}_{d_{1}}$ on a chain of elliptic curves in which two of the curves
have the differences of their nodes to be torsion of order
$r_{1}+1+g-d_{1}+r_{1}-3$. If such a chain is in the boundary of
${\mathcal{M}}^{r_{2}}_{g,d_{2}}$, we need $r_{1}+1+g-d_{1}+r_{1}-3\leq
r_{2}+1+g-d_{2}+r_{2}-2$. Writing
$\alpha_{i}=r_{i}+1,\beta_{i}=g-d_{i}+r_{i}$, we have the equations

$\alpha_{1}\beta_{1}=\alpha_{2}\beta_{2}+1,\ \
\alpha_{1}+\beta_{1}\leq\alpha_{2}+\beta_{2}+1,\ \ \alpha_{2}=\alpha_{1}+1$

It follows that $\beta_{2}\leq 2\alpha_{1}-1$ and therefore $\beta_{1}\leq
2\alpha_{1}+1$. If we have equality in both inequalities, then

$\alpha_{1}\beta_{1}-(\alpha_{1}+1)\beta_{2}=1.$

The general solution $\beta_{1},\beta_{2}$ of the above diophantine equation
is

$\beta_{1}=2\alpha_{1}+1-t(\alpha_{1}+1),\beta_{2}=2\alpha_{1}-1-t\alpha_{1}$

for (positive) integral values of $t$. The only possible solutions are
$t=0,t=1$.

If $t=0$, we have a potential inclusion

${\mathcal{M}}^{\alpha-1}_{2\alpha^{2}+\alpha-2,2\alpha^{2}-4}\subseteq{\mathcal{M}}^{\alpha}_{2\alpha^{2}+\alpha-2,2\alpha^{2}-1}$

For $\alpha=2$, this give the known inclusion
${\mathcal{M}}^{1}_{8,4}\subseteq{\mathcal{M}}^{2}_{8,7}$ while for
$\alpha=3,4\dots$, it would lead to potential inclusions
${\mathcal{M}}^{2}_{19,14}\subseteq{\mathcal{M}}^{3}_{19,17},\
{\mathcal{M}}^{3}_{34,28}\subseteq{\mathcal{M}}^{4}_{34,31},\dots$

If $t=1$, we have a potential inclusion

${\mathcal{M}}^{\alpha-1}_{\alpha^{2}-2,\alpha^{2}-3}\subseteq{\mathcal{M}}^{\alpha}_{\alpha^{2}-2,\alpha^{2}-1}={\mathcal{M}}^{\alpha-2}_{\alpha^{2}-2,\alpha^{2}-5}$

where the last equality is obtained from Serre duality. We need the degree
$\alpha^{2}-3$ to be larger than 1, so we take $\alpha$ to be at least three.
For $\alpha=3$, we obtain the known inclusion
${\mathcal{M}}^{2}_{7,6}\subseteq{\mathcal{M}}^{3}_{7,8}={\mathcal{M}}^{1}_{7,4}$.
For $\alpha=4,5\dots$, it would lead to potential inclusions
${\mathcal{M}}^{3}_{14,13}\subseteq{\mathcal{M}}^{2}_{14,9},\
{\mathcal{M}}^{4}_{23,22}\subseteq{\mathcal{M}}^{2}_{23,20},\dots$. From the
work of [AH], the potential inclusions do not exist for values of $\alpha\leq
3$. Other inclusions will need to be excluded with other techniques. Still,
this method show that potential inclusions of Brill-Noether loci are few and
far between.

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