On the Hilbert vector of the Jacobian module of a plane curve

TL;DR

This paper characterizes the Hilbert vector of the Jacobian module for specific classes of plane curves, providing explicit formulas and bounds using vector bundle cohomology techniques.

Contribution

It identifies classes of curves with fully determined Jacobian module Hilbert vectors and establishes a sharp lower bound for the initial degree under semistability.

Findings

Complete determination of Hilbert vectors for 3-syzygy, maximal Tjurina, and nodal rational curves.

A sharp lower bound for the initial degree of the Jacobian module under semistability.

Application of Hartshorne's cohomology results to curve Jacobian modules.

Abstract

We identify several classes of curves $C:f=0$, for which the Hilbert vector of the Jacobian module $N(f)$ can be completely determined, namely the 3-syzygy curves, the maximal Tjurina curves and the nodal curves, having only rational irreducible components. A result due to Hartshorne, on the cohomology of some rank 2 vector bundles on $\mathbb{P}^2$, is used to get a sharp lower bound for the initial degree of the Jacobian module $N(f)$, under a semistability condition.

Equations

$$\mathrm{Syz}(J_f)=\{(a,b,c)\in S^3:a{f}_x+b{f}_y+c{f}_z=0\}.$$

$$1\le d_1\le d_2\le \dots \le d_m.$$

$$n(f{)}_0\le n(f{)}_1\le \dots \le n(f{)}_{\lfloor \frac{T}{2}\rfloor -1}\le n(f{)}_{\lfloor \frac{T}{2}\rfloor }\ge n(f{)}_{\lfloor \frac{T}{2}\rfloor +1}\ge \dots \ge n(f{)}_T.$$

$$\sigma (C):=\min \left\{j:n(f{)}_j\ne 0\right\}=\mathrm{indeg}(N(f)),$$

$$n(f{)}_j=n(f{)}_{T-j},$$

$$0\to {\oplus }_{i=1}^{m-2}S(-e_i)\to {\oplus }_{i=1}^{m}S(1-d-d_i)\to S^3(1-d)\to S,$$

$$e_j=d+d_{j+2}-1+{ϵ}_j,$$

$$0\to {\oplus }_{i=1}^{m-2}S(-e_i)\to {\oplus }_{i=1}^{m}S(-{\ell }_i)\to {\oplus }_{i=1}^{m}S(d_i-2(d-1))\to {\oplus }_{i=1}^{m-2}S(e_i-3(d-1)),$$

$$\sigma (C)=3(d-1)-e_{m-2}=2(d-1)-d_m-{ϵ}_{m-2}.$$

$$n(f{)}_j=\{\begin{array}{l}3(d^{\prime }{)}^2-(j-3d^{\prime }+2)(j-3d^{\prime }+1)-\tau (C)\text{mbox}ford=2d^{\prime }+1\\ 3(d^{\prime }{)}^2-3d^{\prime }+1-(j-3d^{\prime }+3{)}^2-\tau (C)\text{mbox}ford=2d^{\prime }\end{array}$$

$$n(f{)}_j=\{\begin{array}{l}\nu (C)-(j-\lfloor \frac{T}{2}\rfloor )(j-\lceil \frac{T}{2}\rceil )\text{mbox}ford=2d^{\prime }+1\\ \nu (C)-(j-\frac{T}{2}{)}^2\text{mbox}ford=2d^{\prime }\end{array}$$

$$ct(f)=\max \{q:\dim M(f{)}_k=\dim M(f_s{)}_k\text{ for all }k\le q\},$$

$$0\to S(-e)\to {\oplus }_{i=1}^{3}S(1-d-d_i)\to {\oplus }_{i=1}^{3}S(d_i+2-2d)\to S(e+3-3d),$$

$$\sigma (C)=3(d-1)-(d_1+d_2+d_3).$$

$$0\to S(-d-d_3)\to S(-d-d_3+1)\oplus S(-k_1-2)\oplus S(-k_2-2)\to$$

$$\to S(-k_1-1)\oplus S(-k_2-1)\oplus S(-k_3-1)\to S(-k_3).$$

$$n(f{)}_k=\{\begin{array}{l}0\text{mbox}fork<\sigma \\ \left(\genfrac{}{}{0ex}{}{k-\sigma +2}{2}\right)\text{mbox}for\sigma \le k<k_3\\ \left(\genfrac{}{}{0ex}{}{k-\sigma +2}{2}\right)-\left(\genfrac{}{}{0ex}{}{k-k_3+2}{2}\right)\text{mbox}for{k}_3\le k<k_2\\ \left(\genfrac{}{}{0ex}{}{k-\sigma +2}{2}\right)-\left(\genfrac{}{}{0ex}{}{k-k_3+2}{2}\right)-\left(\genfrac{}{}{0ex}{}{k-k_2+2}{2}\right)\text{mbox}for{k}_2\le k<T_0,\text{par}\end{array}$$

$$T_0=\{\begin{array}{l}{k}_1-1\text{mbox}if{d}_1\ge \frac{d}{2}\\ d+d_1-2\text{mbox}if{d}_1<\frac{d}{2}.\end{array}$$

$$\sigma =3(d-1)-(d_1+d_2+d_3)<3(d-1)-(d-1)-d_3=2(d-1)-d_3=k_3.$$

$$0\to S(-e)\to {\oplus }_{j=1}^{3}S(-{\ell }_j)\to {\oplus }_{i=1}^{3}S(-k_i)\to S(e-3(d-1)),$$

$$n(f{)}_k=\dim S_{k-\sigma }=(\genfrac{}{}{0ex}{}{k-\sigma +2}{2}).$$

$$n(f{)}_k=\dim S_{k-\sigma }-\dim S_{k-k_3}=(\genfrac{}{}{0ex}{}{k-\sigma +2}{2})-(\genfrac{}{}{0ex}{}{k-k_3+2}{2}).$$

$$n(f{)}_k=(\genfrac{}{}{0ex}{}{k-\sigma +2}{2})-(\genfrac{}{}{0ex}{}{k-k_3+2}{2})-(\genfrac{}{}{0ex}{}{k-k_2+2}{2}).$$

$$e=\sum _{i=1}^3{d}_i=127,\sigma =59,k_3=62,k_2=68.$$

$$e=\sum _{i=1}^3{d}_i=47,\sigma =10,k_3=k_2=19.$$

$$\tau (C)=(d-1)(d-r-1)+r^2-\left(\genfrac{}{}{0ex}{}{2r-d+2}{2}\right),$$

$$\sigma (C)=2d-r-3.$$

$$n(f{)}_j=\{\begin{array}{l}3(d^{\prime }{)}^2-(j-3d^{\prime }+2)(j-3d^{\prime }+1)-\tau (C)\text{mbox}ford=2d^{\prime }+1\\ 3(d^{\prime }{)}^2-3d^{\prime }+1-(j-3d^{\prime }+3{)}^2-\tau (C)\text{mbox}ford=2d^{\prime }\end{array}$$

$$n(f{)}_k=\{\begin{array}{l}m(f_s{)}_k-|\mathcal{N}|\text{mbox}ford-3<k\le T/2\\ m(f_s{)}_k-|\mathcal{N}|+n(C)-1\text{mbox}fork=d-3.\end{array}$$

$$n(f{)}_k=m(f{)}_k-d(f{)}_k,$$

$$n(f{)}_k=m(f_s{)}_k-d(f{)}_k,$$

Full text

1 This work has been partially supported by the French government, through the $\rm UCA^{\rm JEDI}$ Investments in the Future project managed by the National Research Agency (ANR) with the reference number ANR-15-IDEX-01

Introduction

Let $S=\mathbb{C}[x,y,z]$ be the graded polynomial ring in three variables $x,y,z$ with complex coefficients. Let $C:f=0$ be a reduced curve of degree $d$ in the complex projective plane $\mathbb{P}^{2}$. We denote by $J_{f}=(f_{x},f_{y},f_{z})$ the Jacobian ideal, i.e. the homogeneous ideal in $S$ spanned by the partial derivatives $f_{x},f_{y},f_{z}$ of $f.$ Since $C$ is reduced, the singular subscheme of $C$, which is defined by the Jacobian ideal $J_{f}$, is 0-dimensional, and its degree is denoted by $\tau(C)$, and is called the global Tjurina number of $C$. Consider the graded $S-$module of Jacobian syzygies of $f$, namely

[TABLE]

Let $mdr(f):=\min\{k:\operatorname{Syz}(J_{f})_{k}\neq(0)\}$ be the minimal degree of a Jacobian syzygy for $f$; in this paper we will assume $mdr(f)\geq 1$ unless otherwise specified. In fact, if $mdr(f)=0$, the curve $C$ is a pencil of lines.

We say that $C:f=0$ is a $m-$syzygy curve if the $S-$module $\operatorname{Syz}(J_{f})$ is minimally generated by $m$ homogeneous syzygies, $r_{1},r_{2},\ldots,r_{m}$, of degree $d_{i}=\deg r_{i}$, ordered such that

[TABLE]

The multiset $(d_{1},d_{2},\ldots,d_{m})$ is called the exponents of the plane curve $C$ and $\{r_{1},r_{2},\ldots,r_{m}\}$ is said to be a minimal set of generators for the $S-$module $\operatorname{Syz}(J_{f}).$ Some of the $m-$syzygy curves have been carefully studied. We recall that:

• •

a $2-$syzygy curve $C$ is said to be free, since then the $S-$module $\operatorname{Syz}(J_{f})$ is a free module of the rank $2$, see [3, 5, 10, 27, 28, 29];

• •

a $3-$syzygy curve is said to be nearly free when $d_{3}=d_{2}$ and $d_{1}+d_{2}=d$, see [11, 2, 3, 5, 6, 23];

• •

a $3-$syzygy line arrangement is said to be a plus-one generated line arrangement of level $d_{3}$ when $d_{1}+d_{2}=d$ and $d_{3}\geq d_{2}$, see [1]. By extension, a $3-$syzygy curve $C$ is said to be a plus-one generated curve of level $d_{3}$ when $d_{1}+d_{2}=d$ and $d_{3}\geq d_{2}$, see [14].

The Jacobian module of $f$, or of the plane curve $C:f=0$, is the quotient module $N(f)=\hat{J_{f}}/J_{f}$, with $\hat{J_{f}}$ the saturation of the ideal $J_{f}$ with respect to the maximal ideal $\mathbf{m}=(x,y,z)$ in $S$. The Jacobian module $N(f)$ coincides with $H^{0}_{\mathbf{m}}(S/J_{f})$, see [26]. Let $n(f)_{j}=\dim N(f)_{j}$, $T=3(d-2)$ and recall that the Jacobian module $N(f)$ enjoys a weak Lefschetz type property, see [7] for this result, and [19, 20, 22] for Lefschetz properties of Artinian algebras in general. More precisely, we have

[TABLE]

We consider the following two invariants for a curve $C:f=0$

[TABLE]

The self duality of the graded $S-$module $N(f)$, see [21, 26, 30], implies that

[TABLE]

for any integer $j$, in particular $n(f)_{k}\neq 0$ exactly for $k=\sigma(C),\ldots,T-\sigma(C).$

The main aim of this paper is to identify classes of curves $C:f=0$ for which the Hilbert vector $(n(f)_{j})$ of the Jacobian module $N(f)$ can be completely determined. In [13, Theorem 3.1, Theorem 3.2], recalled below in Theorem 1.1, there is a description of the dimensions $n(f)_{j}$ for a certain range of $j$. Moreover, in [14, Theorem 3.9, Corollary 3.10], recalled below in Theorem 1.3 and Corollary 1.4, there are descriptions of the minimal resolution of $N(f)$, when $C:f=0$ is a $3-$syzygy curve, and respectively a plus-one generated curve of degree $d\geq 3$. Using these results, we first give a general formula for the Hilbert vector $(n(f)_{j})$ of the Jacobian module of a $3-$syzygy curve in Theorem 2.1, as well as a graphic representation of its behavior. Then we determine the Hilbert vector $(n(f)_{j})$ of the Jacobian module $N(f)$ when $C:f=0$ is a maximal Tjurina curve, see Proposition 3.1. Next we get some information on the Hilbert vector $(n(f)_{j})$ when $C:f=0$ is a nodal curve, which is complete if in addition all the irreducible components of $C$ are rational, see Theorem 3.2.

In the final section we use a result due to Hartshorne, see [18, Theorem 7.4], to relate the cohomology of some rank 2 vector bundles on $\mathbb{P}^{2}$ to the Hilbert vector $(n(f)_{j})$ of the Jacobian module $N(f)$. More precisely, we get in this way a sharp lower bound for the initial degree $\sigma(C)$ of the Jacobian module $N(f)$, under the condition $mdr(f)\geq(d-1)/2$, see Theorem 4.2.

We would like to thank the referee for his careful reading of our manuscript and for his many very useful suggestions to improve the presentation.

1. Preliminaries

We recall some notations and results. Let $C:f=0$ be a reduced complex plane curve of $\mathbb{P}^{2}$, assumed not free, and consider the Milnor algebra $M(f)=S/J_{f}$, where $J_{f}=(f_{x},f_{y},f_{z})$ is the Jacobian ideal. The general form of the minimal resolution for the Milnor algebra $M(f)$ of such a curve $C:f=0$ is

[TABLE]

with $e_{1}\leq e_{2}\leq\ldots\leq e_{m-2}$ and $1\leq d_{1}\leq d_{2}\leq\cdots\leq d_{m}$. It follows from [21, Lemma 1.1] that one has

[TABLE]

for $j=1,\ldots,m-2$ and some integers $\epsilon_{j}\geq 1.$ The minimal resolution of $N(f)$ obtained from (3), by [21, Proposition 1.3], is

[TABLE]

where $\ell_{i}=d+d_{i}-1$. It follows that

[TABLE]

The following result describes the central part of the Hilbert vector of $N(f)$.

Theorem 1.1**.**

Let $C:f=0$ be a reduced, non free curve of degree $d$ and set $r=mdr(f)$. Then one has the following.

1. (i)

if $r\geq\frac{d}{2}$ and $2d-4-r\leq j\leq d-2+r$, then

[TABLE] 2. (ii)

if $r<\frac{d}{2}$ and $d+r-3\leq j\leq 2d-r-3$, then $n(f)_{j}=\nu(C)$. Moreover $n(f)_{d+r-4}=n(f)_{2d-r-2}=\nu(C)-1$.

Proof.

See [13, Theorem 3.1 and Theorem 3.2]. ∎

By Theorem 1.1, in case (i), the points $(j,n(f)_{j})$ lie on an upward pointing parabola. Moreover, using the formulas (9) and (10) and Remark 4.1, the claim (5) can be written:

[TABLE]

with $T=3(d-2)$ as above. On the other hand, in the case (ii), the points $(j,n(f)_{j})$ lie on a horizontal line segment, with a one-unit drop at the extremities, as represented in Figure 1 below.

Recall the following definition, see [5, 10].

Definition 1.2**.**

For a plane curve $C:f=0$, the coincidence threshold of $f$ is the integer

[TABLE]

with $f_{s}$ a homogeneous polynomial in $S$ of the same degree $d$ as $f$ and such that $C_{s}:f_{s}=0$ is a smooth curve in $\mathbb{P}^{2}$.

It is known that $ct(f)\geq d-2+mdr(f)$.

Note that $C:f=0$ is a free curve if and only if $\nu(C)=0$, hence $N(f)=0$. In other words, the Jacobian ideal is satured in this case, i.e. $\hat{J}_{f}=J_{f}$, see [8, 29]. The first nontrivial case is that of nearly free curves; indeed, by [11, Corollary 2.17], for a nearly free curve $C:f=0$, one has $N(f)\neq 0$ and $\nu(C)=1$. Moreover $\sigma(C)=d+d_{1}-3$ and this describes completely the Hilbert vector of the Jacobian module of a nearly free curve. In particular it has the shape described on the left side of Figure 2.

Recall that a nearly free curve is exactly a plus-one generated curve with exponents satisfying $d_{2}=d_{3}$. For the more general case of the $3-$syzygy curves, we recall the following result.

Theorem 1.3**.**

[14, Theorem 3.9]** Let $C:f=0$ be a $3-$syzygy curve with exponents $(d_{1},d_{2},d_{3})$ and set $e=d_{1}+d_{2}+d_{3}$. Then the minimal free resolution of $N(f)$ as a graded $S-$module has the form

[TABLE]

where the leftmost map is the same as in the resolution (3), when $m=3$. In particular,

[TABLE]

This implies the following.

Corollary 1.4**.**

[14, Corollary 3.10]** Let $C:f=0$ be a plus-one generated curve of degree $d\geq 3$ with $(d_{1},d_{2},d_{3})$, which is not nearly free, i.e. $d_{2}<d_{3}.$ Set $k_{j}=2d-d_{j}-3$ for $j=1,2,3$. Then one has the following minimal free resolution of $N(f)$ as a graded $S-$module:

[TABLE]

In particular $\sigma(C)=k_{3}<k_{2}\leq\frac{T}{2}$ and the Hilbert vector of $N(f)$ is given by following formulas:

1. (1)

$n(f)_{j}=0$* for $j<k_{3}$;* 2. (2)

$n(f)_{j}=j-k_{3}+1$* for $k_{3}\leq j\leq k_{2}$;* 3. (3)

$n(f)_{j}=d_{3}-d_{2}+1=\nu(C)$* for $k_{2}\leq j\leq\frac{T}{2}.$*

By above corollary, the Hilbert vector of the Jacobian module of a plus-one generated curve of degree $d$ and level $d_{3}$ has the shape given on the right hand side of Figure 2, where we have drawn only the part corresponding to $j\leq\frac{T}{2}$, due to the symmetry (2).

2. Results on the Hilbert vector of $N(f)$ for 3-syzygy curves

As a simple example of a 3-syzygy curve which is not a plus-one generated curve, let $C:f=0$ be a smooth curve of degree $d\geq 3$, where $d_{1}=d_{2}=d_{3}=d-1$. It is known that the Hilbert function of the Milnor algebra $M(f)$ is in this case $\left(\frac{1-t^{d-1}}{1-t}\right)^{3}$. For a smooth curve we have $N(f)=M(f)$, hence $n(f)_{j}=\dim M(f)_{j}$ and the Hilbert vector of the Jacobian module $N(f)$ has the shape described in Figure 3. It is interesting to notice the change in convexity when we pass through the value $j=d-1$.

For a general $3-$syzygy curve, we have the following result.

Theorem 2.1**.**

Let $C:f=0$ be a $3-$syzygy curve of degree $d$, not plus-one generated, with exponents $d_{1}\leq d_{2}\leq d_{3}$. Set $e=d_{1}+d_{2}+d_{3}$ and $k_{i}=2(d-1)-d_{i}$ for $i=1,2,3$. Then the following hold.

[TABLE]

where $\sigma=\sigma(C)=3(d-1)-e$ and

[TABLE]

Note that $n(f)_{k}$ is known for $T_{0}\leq k\leq\frac{T}{2}$ in view of Theorem 1.1, hence the information on the Hilbert vector of $N(f)$ is complete in this situation.

Proof.

Note that $\sigma\geq 0$, since, by [14, Theorem 2.4] we have $d_{j}\leq d-1$ for $j=1,2,3$. Then, by [14, Theorem 2.3] we have $d_{1}+d_{2}>d>d-1$ and hence

[TABLE]

By Theorem 1.3, the minimal resolution of $N(f)$ is

[TABLE]

where $\ell_{j}=d-1+d_{j}$. We note that $k_{3}\leq k_{2}\leq k_{1}$ and also $k_{2}\leq T_{0}$. If we fix $k$ with $\sigma\leq k<k_{3}$, the minimal resolution of $N(f)$ above yields

[TABLE]

Now we consider the case $k_{3}\leq k<k_{2}$. We have $\ell_{1}>k_{2}$, since $d_{1}+d_{2}>d$ as we have seen above. It follows that

[TABLE]

This difference is a linear form in $k$, and the coefficient of $k$ is given by $(k_{3}-\sigma)$. Note that $k_{3}-\sigma=2(d-1)-d_{3}-3(d-1)+e=d_{1}+d_{2}-(d-1)\geq 2$. To continue, we need to discuss the position of $\ell_{1}$ with respect to $k_{1}$. Note that $\ell_{1}>k_{1}$ if and only if $d_{1}\geq d/2$. Hence we have to consider two cases.

Case 1: $d_{1}\geq d/2$. In this case, we can compute the value $n(f)_{k}$ for $k\leq k_{1}-1$ exactly as above, and we get

[TABLE]

Note that in this case $T_{0}=k_{1}-1=2(d-1)-d_{1}-1>2d-4+d_{1}$ and hence all the Hilbert vector $(n(f)_{j})$ is known by using Theorem 1.1 (i).

Case 2: $d_{1}<d/2$. In this case $\ell_{1}\leq k_{1}$, and we can compute the value $n(f)_{k}$ for $k\leq\ell_{1}-1$ exactly as above, obtaining the same formula. Note that in this case $T_{0}=\ell_{1}-1=d-2+d_{1}>d-3+d_{1}$, and hence again all the Hilbert vector $(n(f)_{j})$ is known by using Theorem 1.1 (ii).

∎

Example 2.2**.**

Let $C:f=(x^{9}+y^{4}z^{5})^{7}+xz^{62}=0$, a singular curve of degree $d=63$. It is a $3-$syzygy curve, not plus-one generated, with $d_{1}=9$, $d_{2}=56$ and $d_{3}=62$. We have

[TABLE]

Since $d_{1}<\frac{d}{2}$, $T_{0}=d+d_{1}-2=70.$ The first quadratic part is for $k\in[59,61]$, the middle linear part is for $k\in[62,67]$, and the second quadratic part is for $k\in[68,70]$. This second quadratic part is too short, containing only 3 points $(j,n(f)_{j})$, to be seen in a graphical representation of the corresponding Hilbert vector. Note also that one has $n(f)_{k}=\nu(C)=27$ for $k\in[69,91]$, where $91=\lfloor T/2\rfloor$. In particular, the last two points on the second quadratic part are in fact situated on this horizontal line segment.

Example 2.3**.**

Let $C:f=(x+y)^{2}(x-y)^{2}(x+2y)^{2}(x-2y)^{2}(x+3y)^{2}(x-3y)^{2}(x+4y)^{2}(x-4y)^{2}(x+5y)^{2}(x-5y)^{2}+z^{20}=0$, a singular curve of degree $d=20$. It is a $3-$syzygy curve not plus-one generated, with $d_{1}=9$ and $d_{2}=d_{3}=19$. We have

[TABLE]

Since $d_{1}<\frac{d}{2}$, $T_{0}=d+d_{1}-2=27=T/2.$ The first quadratic part is for $k\in[10,18]$, the middle linear part is missing since $k_{2}=k_{3}$, the second quadratic part is for $k\in[19,27]$ and one has $n(f)_{k}=\nu(C)=81$ for $k\in[26,27]$, where $27=T/2=T_{0}$.

3. Maximal Tjurina curves and nodal curves

We assume in this section that $r=d_{1}\geq d/2$.

A reduced plane curve $C:f=0$ of degree $d$ is called a maximal Tjurina curve if the global Tjurina number $\tau(C)$ equals the du Plessis-Wall upper bound, namely if

[TABLE]

see [15, 16, 17]. We know that a reduced plane curve $C:f=0$ of degree $d$ is a maximal Tjurina curve if and only if one has $d_{1}=d_{2}=\dots=d_{m}=r$, $e_{1}=e_{2}=\dots=e_{m-2}=d+r$ and $m=2r-d+3$, see [15, Theorem 3.1]. Using now the equality (4), it follows that in this case

[TABLE]

Theorem 1.1 yields then the following result.

Proposition 3.1**.**

Let $C:f=0$ be a maximal Tjurina curve of degree $d$ with $r=d_{1}\geq d/2$. Then the Hilbert vector of the Jacobian module $N(f)$ is given by the following

[TABLE]

for $2d-3-r\leq j\leq d-3+r$ and $n(f)_{j}=0$ otherwise.

Consider now an arbitrary nodal curve $C:f=0$ of degree $d$ in $\mathbb{P}^{2}$. Let ${\mathcal{N}}$ denote the set of nodes of the curve $C$ and $n(C)$ the number of irreducible components of $C$. For such curves we have the following result.

Theorem 3.2**.**

Let $C:f=0$ is a nodal curve in $\mathbb{P}^{2}$ of degree $d\geq 4$. Then one has the following, with $f_{s}$ as in Definition 1.2.

[TABLE]

Moreover, when all the irreducible components of $C$ are rational, one has in addition $n(f)_{k}=0$ for $k\leq d-3$.

Proof.

For any reduced plane curve $C:f=0$, one clearly has

[TABLE]

where $m(f)_{k}=\dim M(f)_{k}$ and $d(f)_{k}=\dim S_{k}/(\hat{J_{f}})_{k}$. Since we have to determine $n(f)_{k}$ only for $k\leq T/2$ by symmetry, and since $ct(f)\geq d-2+r>T/2$ when $r=d_{1}\geq d/2$, it follows that

[TABLE]

with $f_{s}$ as in Definition 1.2 and $k\leq T/2$. In particular, for such curves, we have to determine only the values $d(f)_{k}$ for $k\leq T/2$. On the other hand, we know that

[TABLE]

for $k\geq T-ct(C)$, see [4, Proposition 2]. In particular, this equality holds for $k\geq 3(d-2)-(d-2+r)=2d-4-r$, see also the proof of [13, Theorem 3.1]. Assume now that $C:f=0$ is a nodal curve in $\mathbb{P}^{2}$. Then $r=d_{1}\geq d-2\geq d/2$ for $d\geq 4$, see [8, Example 2.2 (i)]. Let $\operatorname{def}S_{k}({\mathcal{N}})$ denote the defect of the set of nodes ${\mathcal{N}}$ with respect to the linear system $S_{k}$. Then it is known that

[TABLE]

see [4]. On the other hand, [9, Corollary 1.6] implies that $\operatorname{def}S_{k}({\mathcal{N}})=0$ for $k>d-3$ and $\operatorname{def}S_{k}({\mathcal{N}})=n(C)-1$ for $k=d-3$. If all the irreducible components of $C$ are rational, then [12, Theorem 2.7] shows that $n(f)_{k}=0$ for $k\leq d-3$. These facts imply our claims. ∎

4. Relation to a result by Hartshorne

Let $C:f=0$ be a curve of degree $d$ in $\mathbb{P}^{2}$, and let $r=mdr(f)$ be the minimal degree of a Jacobian syzygy for $f$. In this section we give some informations about the invariant $\sigma(C)$, using a result by Hartshorne, namely [18, Theorem 7.4]. We recall that the sheafification of $\operatorname{Syz}(J_{f})$, denoted by $E_{C}:=\widetilde{\operatorname{Syz}(J_{f})}$, is a rank two vector bundle on $\mathbb{P}^{2}$, see [2, 25, 26]. We set

[TABLE]

for any integer $m$. Associated to the vector bundle $E_{C}$ there is the normalized vector bundle $\mathcal{E}_{C}$, which is the twist of $E_{C}$ such that $c_{1}(\mathcal{E}_{C})\in\{-1,0\}$. More precisely,

**when $d=2d^{\prime}+1$ is odd: **

[TABLE]

and

**when $d=2d^{\prime}$ is even: **

[TABLE]

see [13, Section 2].

Remark 4.1**.**

The vector bundle $E_{C}$ is stable if and only if $\mathcal{E}_{C}$ has no sections, see [24, Lemma 1.2.5]. This is equivalent to $r=mdr(f)\geq\frac{d}{2},$ see [26, Proposition 2.4]. Moreover by [13, Theorem 2.2] and using the formulas (9) and (10), we have that for a stable vector $E_{C}$, $c_{2}(\mathcal{E}_{C})=\nu(C).$ Moreover, the vector bundle $E_{C}$ is semistable if and only if $r=mdr(f)\geq(d-1)/2$, see again [24, Lemma 1.2.5], a condition that occurs in our Theorem 4.2 below.

The important key point is the identification

[TABLE]

for any integer $k$, see [26, Proposition 2.1]. Hence the study of the Hilbert vector of the Jacobian module $N(f)$ is equivalent to the study of the dimension of $H^{1}(C,E_{C}(k))$.

Theorem 4.2**.**

Let $C:f=0$ be a curve of degree $d$, and let $r=mdr(f)$ be the minimal degree of a Jacobian syzygy for $f$. Assume that $r\geq(d-1)/2$, in other words that the rank 2 vector bundle $E_{C}$ is semistable. Then we have the following.

1. (1)

If $d=2d^{\prime}+1$ is odd, then

[TABLE] 2. (2)

If $d=2d^{\prime}$ is even, then

[TABLE]

The above inequalities are sharp, in particular they are equalities when $C$ is a maximal Tjurina curve with $r\geq d/2$.

Proof.

We discuss only the case $d=2d^{\prime}+1$, the other case being completely similar. One has

[TABLE]

Moreover $h^{0}(\mathbb{P}^{2},\mathcal{E}_{C}(t))=h^{0}(\mathbb{P}^{2},E_{C}(t+d^{\prime}))\neq 0$ if and only if $t+d^{\prime}\geq r$. Hence the minimal $t$ satisfying this condition is $t_{m}=r-d^{\prime}\geq 0$. Then [18, Theorem 7.4] implies that $n(f)_{k}=0$ when

[TABLE]

Using the formula for $t_{m}$ above, and the formula for $c_{2}(\mathcal{E})$ given in the equations (9), we get that $n(f)_{k}=0$ when

[TABLE]

which clearly implies our claim (1). The fact that the inequality in (1) is in fact an equality when $C$ is a maximal Tjurina curve with $r\geq d/2$ follows by a direct computation. Indeed, using the above definition of a maximal Tjurina curve of degree $d=2d^{\prime}+1$, namely the equality (7), we see that

[TABLE]

Hence

[TABLE]

where the last equality follows from (8). ∎

Example 4.3**.**

Let $C:f=0$ be a curve of degree $d=2d^{\prime}+1$, having a unique node as singularities. Then it is known that $r=d-1=2d^{\prime}$, and $\tau(C)=\sigma(C)=1$. The inequality in Theorem 4.2 (1) is in this case

[TABLE]

hence the two terms in this inequality can be far apart in some cases.