# On conic-line arrangements with nodes, tacnodes, and ordinary triple points

Alexandru Dimca  Université Côte d’Azur, CNRS, LJAD, France and Simion Stoilow
Institute of Mathematics, P.O. Box 1-764, RO-014700 Bucharest, Romania
<EMAIL_ADDRESS>and  Piotr Pokora  Department of Mathematics, Pedagogical
University of Krakow, Podchora̧żych 2, PL-30-084 Kraków, Poland
<EMAIL_ADDRESS>
###### Abstract.

In the present paper, we study conic-line arrangements having nodes, tacnodes,
and ordinary triple points as singularities. We provide combinatorial
constraints on such arrangements and we give the complete classification of
free arrangements in this class.

###### Key words and phrases:

conic-line arrangements, nodes, tacnodes, freeness, nearly freeness

###### 2010 Mathematics Subject Classification:

Primary 14N20; Secondary 14C20, 32S22

##  1\. Introduction

In the present paper we study a class of conic-line arrangements in the
complex projective plane $\mathbb{P}^{2}_{\mathbb{C}}$, with special attention
to the free arrangements in this class. The theory of free line arrangements
is rather rich and we have many results which provide (at least) a partial
characterization of the freeness. In that subject, the ultimate goal is to
understand whether Terao’s Conjecture is true in its whole generality. On the
other hand, line arrangements show up naturally in algebraic geometry. For
example, Hirzebruch’s inequality appreciated very much in combinatorics is
motivated by many extreme problems in algebraic geometry and it is obtained
with its methods. Based on that, it seems to be quite natural to extend this
set-up to higher degree curves. From our perspective it seems very natural to
start working on arrangement consisting of rational curves in the plane. Here
we study arrangements of smooth conics and lines in the plane. The first main
motivation is that conic-line arrangements admit non-ordinary singularities,
so we can study arrangements having, for instance, tacnodes as singularities.

In general, singularities of such arrangements are in general not quasi-
homogeneous which makes their study quite complicated. By [14, Example 4.1],
we know that Terao’s conjecture does not hold in general for such
arrangements. Let us recall the following counterexample from the
aforementioned paper.

######  Example 1.1.

Consider the following conic-line arrangement

$\mathcal{CL}_{1}\,:\,xy\cdot(y^{2}+xz)\cdot(y^{2}+x^{2}+2xz)=0.$

The intersection point $P=(0:0:1)$ has multiplicity $4$ and it is quasi-
homogeneous (although it is not ordinary). One can show that
$\mathcal{CL}_{1}$ is free with the exponents $(2,3)$. If we perturb a bit
line $y=0$, taking for instance $x-13y=0$, we obtain a new conic-line
arrangement

$\mathcal{CL}_{2}\,:\,x\cdot(x-13y)\cdot(y^{2}+xz)\cdot(y^{2}+x^{2}+2xz)=0.$

In this new arrangement, the intersection point $P=(0:0:1)$ has multiplicity
$4$, but it is not longer quasi-homogeneous, and $\mathcal{CL}_{2}$ is not
free. In fact, the arrangement $\mathcal{CL}_{2}$ is nearly free, as defined
in [6]. Note that in many papers on arrangements of plane curves the
hypothesis that all the singularities are quasi-homogeneous plays a key role,
see for instance [4] and [15].

In the present paper we focus on conic-line arrangements in the plane such
that their singularities are nodes, tacnodes, and ordinary triple points. This
assumption is mostly related with our main scope, to verify Terao’s Conjecture
for as large as possible class of conic-line arrangements. One of our results,
Proposition 4.7, tells us that if $C$ is a free reduced plane curve of degree
$m$ having only nodes, tacnodes, and ordinary triple points, then $m\leq 9$.
Based on that combinatorial restriction, we can perform a detailed search in
order to find conic-line arrangements with nodes, tacnodes, and ordinary
triple points that are free. Our main result, Corollary 5.10, tells us that
the so-called Numerical Terao’s Conjecture holds for our class of conic-line
arrangements. As it was mentioned at the beginning of this section, curve
arrangements attract researchers working both in algebraic geometry and
combinatorics. Due to these reasons, we provide combinatorial constraints on
the weak combinatorics of conic-line arrangements with nodes, tacnodes, and
ordinary triple points. We deliver a Hirzebruch-type inequality for such
arrangements, see Theorem 2.1, and this theorem is in the spirit of results
presented in [13]. Then, using the properties of spectra of singularities as
in the seminal paper by Varchenko [17], we provide bounds on the number of
tacnodes and ordinary triple points, see Theorem 3.1.

##  2\. Hirzebruch-type inequality for conic-line arrangements

We start with presenting our set-up. Let
$\mathcal{CL}=\\{\ell_{1},...,\ell_{d},C_{1},...,C_{k}\\}\subset\mathbb{P}^{2}_{\mathbb{C}}$
be an arrangement consisting of $d$ lines and $k$ smooth conics. We assume
that our conic-line arrangements have $n_{2}$ nodes, $t$ tacnodes, and $n_{3}$
ordinary triple points. We have the following combinatorial count

(2.1) $4\binom{k}{2}+2kd+\binom{d}{2}=n_{2}+2t+3n_{3}.$

###### Proof.

Observe that the left-hand side is the number of pairwise intersections of
curves contained in $\mathcal{CL}$. The right-hand side, according to Bézout’s
theorem, is based on the intersection indices. If $p$ is a node, then the
intersection index of curves meeting at that point is equal to $1$. If $p$ is
a tacnode, then the intersection index of curves meeting at that point is
equal to $2$. Finally, if $p$ is an ordinary triple points, that the
intersection index of curves meeting at that point is equal to $3$. This
completes our justification. ∎

The first result of the present paper is the following Hirzebruch-type
inequality.

######  Theorem 2.1.

Let
$\mathcal{CL}=\\{\ell_{1},...,\ell_{d},C_{1},...,C_{k}\\}\subset\mathbb{P}^{2}_{\mathbb{C}}$
be an arrangement of $d$ lines and $k$ smooth conics and such that $2k+d\geq
12$. Assume that $\mathcal{CL}$ has only $n_{2}$ nodes, $t$ tacnodes, and
$n_{3}$ ordinary triple points. Then

(2.2) $20k+n_{2}+\frac{3}{4}n_{3}\geq d+4t.$

We will prove the above theorem using Langer’s variation on the Miyaoka-Yau
inequality [10] which involves the local orbifold Euler numbers $e_{orb}$ of
singular points. We recall basics on them in a concise way. Let
$(\mathbb{P}^{2}_{\mathbb{C}},\alpha C)$ be an effective and log-canonical
pair, where $C$ is a boundary divisor having only nodes, tacnodes, and
ordinary triple points as singularities. Then

  * •

if $q$ is a node, then the local orbifold Euler number is equal to
$e_{orb}(p,\mathbb{P}^{2}_{\mathbb{C}},\alpha C)=(1-\alpha)^{2}$ provided that
$0\leq\alpha\leq 1$,

  * •

if $q$ is a tacnode, then $e_{orb}(q,\mathbb{P}^{2}_{\mathbb{C}},\alpha
C)=(1-2\alpha)$ provided that $0\leq\alpha\leq\frac{1}{4}$

  * •

if $q$ is an ordinary triple point, then
$e_{orb}(q,\mathbb{P}^{2}_{\mathbb{C}},\alpha
C)\leq\bigg{(}1-\frac{3\alpha}{2}\bigg{)}^{2}$ provided that
$0\leq\alpha\leq\frac{2}{3}$

Here we are read to present our proof of Theorem 2.1.

###### Proof.

Let $C:=\ell_{1}+...+\ell_{d}+C_{1}+...+C_{k}$ be a divisor associated with
$\mathcal{CL}$ and such that $m={\rm deg}\,C=d+2k\geq 12$ – we will see in a
moment why the last assumption is crucial. First of all, we need to choose
$\alpha$ in such a way that $K_{\mathbb{P}^{2}_{\mathbb{C}}}+\alpha C$ is
effective and log canonical. In order to obtain the effectivity of the pair,
one needs to satisfy the condition that $-3+\alpha(2k+d)\geq 0$ which implies
$\alpha\geq 3/(2k+d)$. On the other hand, our pair is log-canonical if
$\alpha\leq{\rm min}\\{1,2/3,1/4\\}$, so $\alpha\leq 1/4$. Due to these two
reasons, we get $\alpha\in\bigg{[}3/(2k+d),1/4\bigg{]}$, and this condition is
non-empty provided that $2k+d\geq 12$. From now on we take
$\alpha=\frac{1}{4}$, and then apply the following inequality

(2.3) $\sum_{p\in{\rm
Sing}(C)}3\bigg{(}\alpha(\mu_{p}-1)+1-e_{orb}(p,\mathbb{P}^{2}_{\mathbb{C}},\alpha
C)\bigg{)}\leq(3\alpha-\alpha^{2})m^{2}-3\alpha m,$

where $\mu_{p}$ is the Milnor number of a singular point $p\in{\rm Sing}(C)$.
This gives us

$3n_{2}\bigg{(}\frac{1}{4}(1-1)+1-(1-1/4)^{2}\bigg{)}+3t\bigg{(}\frac{1}{4}(3-1)+1-(1-1/2)\bigg{)}+3n_{3}\bigg{(}\frac{1}{4}(4-1)+1-(1-3/8)^{2}\bigg{)}$
$\leq\sum_{p\in{\rm
Sing}(C)}3\bigg{(}\alpha(\mu_{p}-1)+1-e_{orb}(p,\mathbb{P}^{2}_{\mathbb{C}},\alpha
C)\bigg{)}.$

After some simple manipulations, we obtain

$\frac{21}{16}n_{2}+3t+\frac{261}{4}n_{3}\leq\frac{11}{16}m^{2}-\frac{3}{4}m.$

Using (2.1), we have $m^{2}=2n_{2}+4t+6n_{3}+4k+d$, and this implies

$11m^{2}-12m=22n_{2}+44t+66n_{3}+44k+11d-24k-12d=22n_{2}+44t+66n_{3}+20k-d.$

Combining the above computations together, we get

$21n_{2}+48t+\frac{261}{4}n_{3}\leq 22n_{2}+44t+66n_{3}+20k-d,$

which completes the proof. ∎

##  3\. Spectra of singularities and constraints on conic-line arrangements

Let $F_{\bullet}:\,H=F^{0}\supset...\supset F^{p}\supset F^{p+1}\supset...$ be
the Hodge filtration on the (reduced) vanishing cohomology
$H=H^{n}(X_{\infty})$ of an isolated singularity $f$, where $X_{\infty}$
denotes the canonical Milnor fibre. The filtration $F_{\bullet}$ is invariant
with respect to the action of the semisimple part of the monodromy $T_{s}$.
Hence $T_{s}$ acts on ${\rm Gr}_{F}^{p}H=F^{p}/F^{p+1}$ and ${\rm
Gr}_{F}^{p}H=\bigoplus_{\lambda}({\rm Gr}_{F}^{p})_{\lambda}$, where $({\rm
Gr}_{F}^{p})_{\lambda}={\rm Gr}_{F}^{p}H_{\lambda}$ is the eigensubspace
corresponding to $\lambda$. Denote by

$\mu_{p}={\rm dim}\,{\rm Gr}_{F}^{p},\,\quad\mu_{\lambda}^{p}={\rm dim}({\rm
Gr}_{F}^{p})_{\lambda}.$

Then $\sum_{p}\mu^{p}=\mu$ is the Milnor number,
$\sum_{\lambda}\mu_{\lambda}^{p}=\mu^{p}$, and
$\sum_{p}\mu_{\lambda}^{p}=\mu_{\lambda}$ is the multiplicity of an eigenvalue
$\lambda$, where $\mu_{\lambda}={\rm dim}H_{\lambda}$. Now to each eigenvalue
$\lambda$ one defines

$\alpha=-(1/2\pi\iota){\rm log}(\lambda),$

where $\iota^{2}=-1$. Since $\lambda$ is a root of the unity, $\alpha$ is a
rational number defined modulo an integer. We normalize $\alpha$ according to
the level $p$ of $\lambda$ with respect to $F_{\bullet}$ by the condition

$\alpha=-\frac{1}{2\pi\iota}{\rm log}\,\lambda,\quad n-p-1<\alpha\leq n-p,$

where this $\lambda$ comes from the action of $T_{s}$ on ${\rm Gr}_{F}^{p}H$.
In this way, one obtains an element of the group $\mathbb{Z}^{\mathbb{Q}}$ of
the form

${\rm
Sp}(f)=(\alpha_{1})+...+(\alpha_{\mu})=\sum_{\alpha}n_{\alpha}\cdot(\alpha),$

with $n_{\alpha}=\mu_{\lambda}^{p}$, which is called the spectrum of the
singularity. The numbers $\alpha$ are called spectral numbers, and the
coefficients $n_{\alpha}$ are the spectral multiplicities. Let us recall some
basic properties of spectral numbers for isolated singularities of
hypersurfaces.

  1. (1)

$\alpha_{j}\in(0,n)$ if $n=\dim X$.

  2. (2)

The spectrum is an invariant of a singularity.

  3. (3)

Symmetry: $\alpha_{i}=\alpha_{\mu-i}$.

  4. (4)

_Thom-Sebastiani Principle_ : If $f\in\mathbb{C}\\{x_{0},\ldots x_{m}\\}$ and
$g\in\mathbb{C}\\{y_{0},\ldots,y_{n}\\}$ are two series in separate sets of
variables, the expression

$f\oplus g=f(x_{0},\ldots
x_{m})+g(y_{0},\ldots,y_{n})\in\mathbb{C}\\{x_{0},\ldots
x_{m},y_{0},\ldots,y_{n}\\}$

is called the Thom-Sebastiani sum of $f$ and $g$. Then

${\rm Sp}(f\oplus g)=\\{\alpha+\beta\,:\,\alpha\in{\rm Sp}(f),\beta\in{\rm
Sp}(g)\\}.$

  5. (5)

${\rm Sp}(x^{m})=\\{\frac{1}{m},\frac{2}{m},...,\frac{m-1}{m}\\}.$

Using the formulae above, we can compute spectral numbers for nodes $(A_{1})$,
tacnodes $(A_{3})$, and ordinary triple points $(D_{4})$.

  1. $(A_{1})$:

This singular point can be locally described by $x^{2}+y^{2}=0$, so ${\rm
Sp}(x^{2})=\\{\frac{1}{2}\\}$, ${\rm Sp}(y^{2})=\\{\frac{1}{2}\\}$, and then
${\rm Sp}(A_{1})=1\cdot 1$.

  2. $(A_{3})$:

This singular point can be locally described by $y^{2}+x^{4}=0$, so ${\rm
Sp}(y^{2})=\\{\frac{1}{2}\\}$, ${\rm
Sp}(x^{4})=\\{\frac{1}{4},\frac{1}{2},\frac{3}{4}\\}$, and we obtain ${\rm
Sp}(A_{3})=1\cdot\frac{3}{4}+1\cdot 1+1\cdot\frac{5}{4}$.

  3. $(D_{4})$:

This singular point can be locally described by $x^{3}+y^{3}=0$, so ${\rm
Sp}(x^{3})={\rm Sp}(y^{3})=\\{\frac{1}{3},\frac{2}{3}\\}$, and ${\rm
Sp}(D_{4})=1\cdot\frac{2}{3}+2\cdot 1+1\cdot\frac{4}{3}$.

Now we present the main result of this section.

######  Theorem 3.1.

Let
$\mathcal{CL}=\\{\ell_{1},...,\ell_{d},C_{1},...,C_{k}\\}\subset\mathbb{P}^{2}_{\mathbb{C}}$
be an arrangement of $d\geq 0$ lines and $k\geq 0$ smooth conics. Assume that
$\mathcal{CL}$ has only $n_{2}$ nodes, $t$ tacnodes, and $n_{3}$ ordinary
triple points. Let $C=\ell_{1}+...+\ell_{d}+C_{1}+...+C_{k}$ and write
$m:={\rm deg}\,C=d+2k$ as $m=3m^{\prime}+\epsilon$ with
$\epsilon\in\\{1,2,3\\}$. Then one has

$t+n_{3}\leq\binom{m-1}{2}+k-\frac{m^{\prime}(5m^{\prime}-3)}{2}$

and

$n_{3}\leq(m^{\prime}+1)(2m^{\prime}+1).$

###### Proof.

We are going to use the theory of spectra of singularities. Recall that if
$(X,0)$ is the union of $m$ lines passing through the origin of
$\mathbb{C}^{2}$, then the corresponding spectrum is

${\rm
Sp}(X,0)=\sum_{j=1}^{m-1}j\cdot\frac{j+1}{m}+\sum_{j=2}^{m-1}(m-j)\cdot\frac{m+j-1}{m}.$

We apply the semicontinuity property of the spectrum in the form presented by
Steenbrink in [16] (see also [9, Theorem 8.9.8]) for the semicontinuity domain
$B=(\frac{1}{3},\frac{4}{3}]$. If $L$ is a generic line in
$\mathbb{P}^{2}_{\mathbb{C}}$, then the trace of the arrangement
$\mathcal{CL}$ on the complement
$\mathbb{C}^{2}=\mathbb{P}^{2}_{\mathbb{C}}\setminus L$ can be identified with
a deformation $X_{s}$ of a singularity of type $(X,0)$ introduced above.
Therefore we get the following equation

(3.1) ${\rm deg}_{B}\sum_{y}{\rm Sp}(X_{s},y)=n_{2}+3t+4n_{3},$

where the above sum is over singular points $y\in X_{s}$, and ${\rm
deg}_{B}\,\sum_{y}{\rm Sp}(X_{s},y)$ denotes the sum of all spectral
multiplicities for spectral numbers that are contained in domain $B$.

On the other hand, the total degree of the spectrum ${\rm Sp}(X,0)$ is equal
to the Milnor number $\mu(X,0)=(m-1)^{2}$. To get the degree for the
restriction of the spectrum ${\rm Sp}(X,0)$ to the interval
$B=(\frac{1}{3},\frac{4}{3}]$, we have to subtract the sum $S_{1}$ of the
multiplicities of the spectral numbers $\alpha$ such that
$\alpha\leq\frac{1}{3}$, and the sum $S_{2}$ of the multiplicities of the
spectral numbers $\alpha>\frac{4}{3}$. By the symmetry property of the
spectrum, the last case can be replaced by $\alpha<\frac{2}{3}$.
The first sum $S_{1}$ is at least equal to

$S_{1}=1+2+\ldots+(m^{\prime}-1)=\frac{m^{\prime}(m^{\prime}-1)}{2}.$

The second sum $S_{2}$ is at least equal to

$S_{2}=1+2+\ldots+(2m^{\prime}-1)=m^{\prime}(2m^{\prime}-1).$

It follows that

(3.2) ${\rm deg}_{B}\,{\rm
Sp}(X,0)\leq(m-1)^{2}-S_{1}-S_{2}=(m-1)^{2}-\frac{m^{\prime}(5m^{\prime}-3)}{2}.$

Therefore the semicontinuity theorem implies that

(3.3) $n_{2}+3t+4n_{3}\leq(m-1)^{2}-\frac{m^{\prime}(5m^{\prime}-3)}{2}.$

Observe that the combinatorial count (2.1) can be rewritten as

(3.4) $n_{2}+2t+3n_{3}=\binom{m}{2}-k.$

By the above, we can conclude that

$t+n_{3}\leq(m-1)^{2}-\binom{m}{2}+k-\frac{m^{\prime}(5m^{\prime}-3)}{2}=$
$=\binom{m-1}{2}+k-\frac{m^{\prime}(5m^{\prime}-3)}{2}.$

For the second inequality, we choose the semicontinuity domain
$B=(-\frac{1}{3},\frac{2}{3}]$, and for this choice of $B$ we have

(3.5) ${\rm deg}_{B}\,\sum_{y}{\rm Sp}(X_{s},y)=n_{3},$

where the sum is taken over all the singular points $y\in X_{s}$. On the other
hand, we have

(3.6) ${\rm deg}_{B}\,{\rm Sp}(X,0)\leq
1+2+\ldots+(2m^{\prime}+1)=(m^{\prime}+1)(2m^{\prime}+1).$

This completes the proof. ∎

######  Example 3.2.

These bounds are rather good, at least in some cases. In order to see this for
the bound involving $t+n_{3}$, consider Figure 2 where we present a conic-line
arrangement with $d=3$ and $k=2$ having $t=5$ and $n_{3}=3$. In this case
$m=7$, hence $m^{\prime}=2$ and the first inequality in Theorem 3.1 is

$8=t+n_{3}\leq 10.$

Next, consider the dual Hesse arrangement given by

$(x^{3}-y^{3})(y^{3}-z^{3})(x^{3}-z^{3})=0,$

which has $n_{3}=12$ triple points. In this case $m=9$, $m^{\prime}=2$, and
the second inequality in Theorem 3.1 gives us

$12=n_{3}\leq 15.$

######  Remark 3.3.

Since $m^{\prime}=(m-\epsilon)/3$ and $k\leq m/2$, it follows that we have

$t+n_{3}\leq\frac{1}{18}\bigg{(}4m^{2}+m(10\epsilon-9)-5\epsilon^{2}-9\epsilon+18\bigg{)}\approx\frac{2}{9}m^{2}+O(m).$

##  4\. Combinatorial constraints on the freeness of reduced curves

We begin with a general introduction to the subject. Let $C$ be a reduced
curve $\mathbb{P}^{2}_{\mathbb{C}}$ of degree $m$ given by $f\in
S:=\mathbb{C}[x,y,z]$. We denote by $J_{f}$ the Jacobian ideal generated by
the partials derivatives $\partial_{x}f,\,\partial_{y}f,\,\partial_{z}f$.
Moreover, we denote by $r:={\rm mdr}(f)$ the minimal degree of a relation
among the partial derivatives, i.e., the minimal degree $r$ of a triple
$(a,b,c)\in S_{r}^{3}$ such that

$a\cdot\partial_{x}f+b\cdot\partial_{y}f+c\cdot\partial_{z}f=0.$

We denote by $\mathfrak{m}=\langle x,y,z\rangle$ the irrelevant ideal.
Consider the graded $S$-module $N(f)=I_{f}/J_{f}$, where $I_{f}$ is the
saturation of $J_{f}$ with respect to $\mathfrak{m}=\langle x,y,z\rangle$.

######  Definition 4.1.

We say that a reduced plane curve $C$ is _free_ if $N(f)=0$.

Let us recall that for a reduced curve $C:f=0$ we define the Arnold exponent
$\alpha_{C}$ which is the minimum of the Arnold exponents of the singular
points $p$ in $C$. Using the modern language, the Arnold exponents of singular
points are nothing else than the log canonical thresholds of singularities.

######  Definition 4.2.

Let $C\,:\,f=0$ be a reduced curve in $\mathbb{C}^{2}$ which is singular at
$0\in\mathbb{C}^{2}$. Denote by $\phi:Y\rightarrow\mathbb{C}^{2}$ the standard
minimal resolution of singularities, i.e., the smallest resolution that has
simple normal crossings (which exists and it is unique). We write then
$K_{Y}=\phi^{*}K_{\mathbb{C}^{2}}+\sum_{i}a_{i}E_{i}$ and
$\phi^{*}C=\phi_{*}^{-1}C+\sum_{i}b_{i}E_{i}$, where $=$ means the linear
equivalence. Then the log canonical threshold of $C$ in $\mathbb{C}^{2}$ is
defined as

$c_{0}(f)={\rm min}_{i}\bigg{\\{}\frac{a_{i}+1}{b_{i}}\bigg{\\}}.$

Using this local (analytical) description, the Arnold exponent $\alpha_{C}$ of
$C$ is then the minimum over all log canonical thresholds of singular points.
In order to compute the actual values of the log canonical thresholds, we can
us the following result – see for instance [2, Theorem 4.1].

######  Theorem 4.3.

Let $C$ be a reduced curve in $\mathbb{C}^{2}$ which has degree $m$. Then
$c_{0}(f)\geq\frac{2}{m}$, and the equality holds if and only if $C$ is a
union of $m$ lines passing through $0$.

######  Remark 4.4.

If $p=(0,0)\in\mathbb{C}^{2}$ is an ordinary singularity of multiplicity $r$
determined by $C\,:\,f=0$, then $c_{0}(f)=\frac{2}{r}$.

Now we need to compute the log canonical threshold for tacnodes. Since
tacnodes are quasi-homogeneous singularities, then we can use the following
pattern (cf. [4, Formula 2.1]).

Recall that the germ $(C,p)$ is weighted homogeneous of type $(w_{1},w_{2};1)$
with $0<w_{j}\leq 1/2$ if there are local analytic coordinates $y_{1},y_{2}$
centered at $p=(0,0)$ and a polynomial
$g(y_{1},y_{2})=\sum_{u,v}c_{u,v}y_{1}^{u}y_{2}^{v}$ with
$c_{u,v}\in\mathbb{C}$, where the sum is taken over all pairs
$(u,v)\in\mathbb{N}^{2}$ with $uw_{1}+vw_{2}=1$. In this case, we have

$c_{0}(g)=w_{1}+w_{2}.$

######  Remark 4.5.

Let $g=y^{2}+x^{4}$, so $g$ defines a tacnode at $p=(0,0)$. Then
$w_{1}=\frac{1}{2},w_{2}=\frac{1}{4}$, and hence we have
$c_{0}(g)=\frac{3}{4}$

In order to show our main result for this section, recall the following [4,
Theorem 2.1].

######  Theorem 4.6 (Dimca-Sernesi).

Let $C\,:\,f=0$ be a reduced curve of degree $m$ in
$\mathbb{P}^{2}_{\mathbb{C}}$ having only quasi-homogeneous singularities.
Then

${\rm mdr}(f)\geq\alpha_{C}\cdot m-2.$

Since nodes, tacnodes, and ordinary triple points are quasi-homogeneous
singularities, then we can prove the following result.

######  Proposition 4.7.

Let $C\,:\,f=0$ be a reduced curve of degree $m$ in
$\mathbb{P}^{2}_{\mathbb{C}}$ having only nodes, tacnodes, and ordinary triple
points as singularities. Then

${\rm mdr}(f)\geq\frac{2}{3}m-2.$

In particular, if $C$ is free, then $m\leq 9$.

###### Proof.

Since $C$ has nodes, tacnodes, and ordinary triple points as singularities,
then

$\alpha_{C}={\rm
min}\bigg{\\{}1,\frac{3}{4},\frac{2}{3}\bigg{\\}}=\frac{2}{3},$

so by the above result we have

${\rm mdr}(f)\geq\frac{2}{3}m-2.$

If $C$ is a free curve, then

$\frac{2}{3}m-2\leq{\rm mdr}(f)\leq\frac{m-1}{2},$

which gives $m\leq 9$. ∎

######  Remark 4.8.

If we restrict our attention to reduced free curves with nodes and tacnodes,
then analogous computations as above give $m\leq 5$, and this bound is sharp
according to what we shall see in Example 4.14 in the forthcoming section. In
fact, we can show that for every $m\in\\{3,4,5\\}$ there exists a conic-line
arrangement having nodes and tacnodes with $k\geq 1$ that is free.

######  Remark 4.9.

Observe that Proposition 4.7 is sharp in the class of reduced free curves –
the dual Hesse arrangement of $9$ lines and $12$ triple points considered in
Example 3.2 is free.

Let
$\mathcal{CL}=\\{\ell_{1},...,\ell_{d},C_{1},...,C_{k}\\}\subset\mathbb{P}^{2}_{\mathbb{C}}$
be a free arrangement of $d\geq 1$ lines and $k\geq 1$ conics. We are going to
use the following homological characterization of the freeness, see for
instance [6], which can be checked on specific examples using `Singular` [3],
or other computer algebra software.

######  Theorem 4.10.

Let $C\subset\mathbb{P}^{2}_{\mathbb{C}}$ be a reduced curve of degree $m$ and
let $f=0$ be its defining equation. Then $C$ is free if and only if the
minimal free resolution of the Milnor algebra $M(f)=S/J_{f}$ has the following
form:

$0\rightarrow S(-d_{1}-(m-1))\oplus S(-d_{2}-(m-1))\rightarrow
S^{3}(-m+1)\rightarrow S\rightarrow M(f)\rightarrow 0$

with $d_{1}+d_{2}=m-1$. In particular, if $d_{1}\leq d_{2}$, then ${\rm
mdr}(f)=d_{1}\leq\frac{m-1}{2}$.

We will need additionally the following lemma, see for instance [4, Lemma
4.4].

######  Lemma 4.11.

If $C$ is a free plane curve of degree $m$, then the exponents $(d_{1},d_{2})$
are positive integers satisfying the following system of equations:

$d_{1}+d_{2}=m-1,\quad\quad d_{1}d_{2}=(m-1)^{2}-\tau(C),$

where $\tau(C)$ denotes the total Tjurina number.

If now $\mathcal{CL}$ is a conic-line arrangement having degree $m=2k+d$ with
$n_{2}$ nodes, $t$ tacnodes, and $n_{3}$ ordinary triple points, then the
above lemma can be rewritten as

(4.1) $d_{1}+d_{1}=m-1,\quad\quad
d_{1}^{2}+d_{2}^{2}+d_{1}d_{2}=n_{2}+3t+4n_{3},\quad\quad d_{1}\leq d_{2}.$

So our problem reduces to possible geometrical realizations of some positive
integer solutions to (4.1).

######  Example 4.12.

Let $m=3$, so we have one conic and one line. An easy inspection tells us that
$d_{1}=d_{2}=1$, $n_{2}=n_{3}=0$, and $t=1$ satisfies (4.1). Let us consider

$\mathcal{CL}_{3}:\quad(x-z)\cdot(x^{2}+y^{2}-z^{2})=0.$

Observe that $\mathcal{CL}_{3}$ is exactly a line tangent to a conic. We can
compute the minimal free resolution of the Milnor algebra $M(f)=S/J_{f}$ which
has the following form:

$0\rightarrow S^{2}(-3)\rightarrow S^{3}(-2)\rightarrow S\rightarrow
M(f)\rightarrow 0,$

so $\mathcal{CL}_{3}$ is free with the exponents $(1,1)$.

######  Example 4.13.

Consider now the case with $m=4$, and note that there are many solutions to
(4.1). Take $d_{1}=1$ and $d_{2}=2$, then we can find the following
Diophantine solution, namely $t=2$ and $n_{2}=1$. Consider

$\mathcal{CL}_{4}:\quad(x^{2}-z^{2})\cdot(x^{2}+y^{2}-z^{2})=0,$

hence two lines tangent to the same conic. Then the minimal free resolution of
the Milnor algebra $M(f)=S/J_{f}$ has the following form:

$0\rightarrow S(-5)\oplus S(-4)\rightarrow S^{3}(-3)\rightarrow S\rightarrow
M(f)\rightarrow 0,$

so $\mathcal{CL}_{3}$ is indeed free with the exponents $(1,2)$.

######  Example 4.14.

Consider now the case with $m=5$. One positive integer solution to (4.1) is
$d_{1}=d_{2}=2$ and $n_{2}=t=3$, $n_{3}=0$. Consider the following conic-line
arrangement

$\mathcal{CL}_{5}:(y-z)\cdot(x^{2}-z^{2})\cdot(x^{2}+y^{2}-z^{2})=0.$

The minimal free resolution of the Milnor algebra $M(f)=S/J_{f}$ has the
following form:

$0\rightarrow S^{2}(-6)\rightarrow S^{3}(-4)\rightarrow S\rightarrow
M(f)\rightarrow 0,$

so $\mathcal{CL}_{5}$ is free.
Another positive integer solution to (4.1) is $d_{1}=d_{2}=2$ and $n_{2}=t=0$,
$n_{3}=3$. Consider now the second case and the following arrangement

$\mathcal{CL}_{5}^{{}^{\prime}}:y\cdot(x+y-4z)\cdot(x-y+4z)\cdot(x^{2}+y^{2}-16z^{2})=0.$

Observe that $\mathcal{CL}_{5}^{\prime}$ has $n_{2}=t=0$ and $n_{3}=3$. The
minimal free resolution of the Milnor algebra $M(f)=S/J_{f}$ has the same form
as above, namely

$0\rightarrow S^{2}(-6)\rightarrow S^{3}(-4)\rightarrow S\rightarrow
M(f)\rightarrow 0,$

so $\mathcal{CL}_{5}^{{}^{\prime}}$ is free.

Figure 1. Free conic-line arrangements with $m=5$ and $k=1$.

##  5\. Classification of free conic-line arrangements with nodes, tacnodes
and triple points

After the above warm-up, we are ready to present the first classification
result. It gives a complete characterization of free arrangements having
$d\in\\{1,2,3\\}$ lines and $k\geq 1$ smooth conics. We start with a general
discussion about free conic-line arrangements with nodes, tacnodes, and
ordinary triple points.

If $\mathcal{CL}$ is free with the exponents $(d_{1},d_{2})$, then we must
have

$d_{1}+d_{2}=2k+d-1$

and

$d_{1}d_{2}=(2k+d-1)^{2}-n_{2}-3t-4n_{3}=(2k+d-1)^{2}-(n_{2}+2t+3n_{3})-t-n_{3}.$

Using the combinatorial count (2.1), we obtain

(5.1) $d_{1}d_{2}=2k^{2}-2k+1+2kd+\frac{d^{2}-3d}{2}-t-n_{3}.$

If we multiply equation (5.1) by $2$, add it to (2.1), then we obtain

$2d_{1}d_{2}=2k^{2}+2k(d-1)+\frac{(d-1)(d-4)}{2}+n_{2}+n_{3}.$

Since $d_{1}d_{2}\leq(2k+d-1)^{2}/4$, we get

(5.2) $n_{2}+n_{3}\leq\frac{3}{2}(d-1).$

Using once again $(\ref{comb:naiv})$, we obtain

(5.3) $t\geq k^{2}-k+kd+\frac{(d-1)(d-3)}{4}-n_{3}.$

If $k\geq 2$, two conics $C_{1}$ and $C_{2}$ from the arrangement
$\mathcal{CL}$ can be in one of the following $3$ situations:

  1. (1)

$|C_{1}\cap C_{2}|=4$, and then all the intersection points are nodes.

  2. (2)

$|C_{1}\cap C_{2}|=3$, and then one intersection point produces a tacnode in
$\mathcal{CL}$, and the other two intersection points will give nodes.

  3. (3)

$|C_{1}\cap C_{2}|=2$, and then two intersection points produce two tacnodes
in $\mathcal{CL}$.

Let $m_{j}$ with $j\in\\{2,3,4\\}$ be the number of pairs of conics in
$\mathcal{CL}$ such that $|C_{1}\cap C_{2}|=j$. Then the number of tacnodes
coming from the contact of $2$ conics is

(5.4) $t^{\prime}=2m_{2}+m_{3}.$

We also have, by counting pairs of conics in two different ways,

(5.5) $m_{2}+m_{3}+m_{4}=\binom{k}{2}$

and

(5.6) $2m_{3}+4m_{4}\leq n_{2}+n_{3}.$

The last inequality follows from the fact that the nodes coming from
$C_{1}\cap C_{2}$ will give either nodes or triple points in $\mathcal{CL}$.
To evaluate the number of tacnodes created by a line in $\mathcal{CL}$, we
need the following.

######  Lemma 5.1.

Let $C_{1}$ and $C_{2}$ be two smooth conics in $\mathcal{CL}$ such that
$|C_{1}\cap C_{2}|=2$. Then any line $L$ in $\mathcal{CL}$ is tangent to at
most one of the conics $C_{1}$ and $C_{2}$.

###### Proof.

Since the pair of conics $C_{1}$ and $C_{2}$ gives rise to two tacnodes, then
up-to a linear change of coordinates one can take

$C_{1}:x^{2}+y^{2}-z^{2}=0\text{ and }C_{2}:x^{2}+y^{2}-r^{2}z^{2}=0$

where $r\in\mathbb{C}$, $r\neq 0,\pm 1$ – please consult [12, Proposition 3].
It follows that the dual $C_{1}^{\vee}$ is given by $x^{2}+y^{2}-z^{2}=0$,
while the dual $C_{2}^{\vee}$ is given by
$x^{2}+y^{2}-\frac{1}{r^{2}}z^{2}=0$. A common tangent for $C_{1}$ and $C_{2}$
corresponds to a point in the intersection $C_{1}^{\vee}\cap C_{2}^{\vee}$,
and hence it is given by equation $T_{\pm}:x\pm y=0$. However, both these
lines pass through one of the two tacnodes situated at $(1:\pm 1:0)$. Hence
$L$ cannot be tangent to two conics, since the curve ${\mathcal{C}}$ has only
nodes, ordinary triple points and tacnodes as singularities. ∎

######  Theorem 5.2.

Let $\mathcal{CL}$ be an arrangement of $0\leq d\leq 3$ lines and $k\geq 1$
smooth conics with $n_{2}$ nodes, $t$ tacnodes, and $n_{3}$ ordinary triple
points. Assume that $\mathcal{CL}$ is free, then the following pairs are
admissible:

$(d,k)\in\\{(1,1),(2,1),(3,1),(3,2)\\}.$

###### Proof.

We need to consider some cases.

Case $d=0$. Then (5.2) implies $n_{2}+n_{3}<0$, which is clearly impossible.

Case $d=1$. Then $n_{2}+n_{3}\leq 0$, and hence $n_{2}=n_{3}=0$. Using the
combinatorial count (2.1) we get $t=k^{2}$. The case $k=1$ is clearly
possible, i.e., a conic plus a tangent line form a free curve, see Example
4.12. We show now that the case $k>1$ is impossible. Using (5.5), we see that
$m_{3}=m_{4}=0$, and hence any two conics in $\mathcal{CL}$ meet in two
points, as in Lemma 5.1 above.

There are $\binom{k}{2}$ pairs of conics, hence the number of tacnodes
obtained as intersection of two conics is $t^{\prime}=k^{2}-k$. Recall also
that by [7] one has $k\in\\{2,3,4\\}$. The only possibility to have $t=k^{2}$
tacnodes is that the unique line, call it $L$, is tangent to all the conics
simultaneously. In the light of Lemma 5.1, $L$ cannot be tangent to each conic
in $\mathcal{CL}$. This completes the proof in this case.

Case $d=2$. The case $k=1$ is possible, see Example 4.13. We show that the
cases $k\geq 2$ are impossible. One has $n_{2}+n_{3}\leq 1$ and using (5.3)
above, we get

$t\geq k^{2}-k+(2k-\frac{1}{4}-n_{3}).$

When $k\geq 2$, one has $2k-\frac{1}{4}-n_{3}>2$, and hence

$t\geq k^{2}-k+3.$

Combining the condition $n_{2}+n_{3}\leq 1$ and inequality (5.6), we get
$m_{3}=m_{4}=0$. It means, in particular, that $m_{2}=\binom{k}{2}$,
$t^{\prime}=k^{2}-k$, and $k\in\\{2,3\\}$. An easy inspection, performed along
the lines of Lemma 5.1, shows that a line can add at most one tacnode, and in
this case one must have

$t\leq k^{2}-k+2,$

hence we have a contradiction.

Case $d=3$. Observe that by Proposition 4.7 we have $k\in\\{1,2,3\\}$. The
case $k=1$ is possible, see Example 4.14. We show now that $k=3$ is
impossible. In this case one has $n_{2}+n_{3}\leq 3$ and using formula (5.3)
above, we get

(5.7) $t\geq k^{2}-k+(3k-n_{3}).$

Then (5.6) and $n_{2}+n_{3}\leq 3$ lead us to $m_{3}\leq 1$ and $m_{4}=0$. If
$m_{3}=0$, then we use Lemma 5.1 to get a contradiction since $3k-n_{3}\geq
6>d=3$. Assume that $m_{3}=1$, so let’s say that $|C_{1}\cap C_{2}|=3$. The
maximal number of tacnodes coming from the contact of $3$ lines and conics in
such an arrangement is $5$. Indeed, two lines can be tangents to both $C_{1}$
and $C_{2}$, giving $4$ tacnodes, and the third line can be tangent to at most
one conic in $\mathcal{CL}$. It follows that

$t\leq 2m_{2}+m_{3}+5=k(k-1)+4.$

Since $k=3$ one has $9-n_{3}\geq 6>4$, and this contradiction proves the
claim.

Consider the remaining case $k=2$. First we assume that $m_{3}=0$. In order
the have $2d-n_{3}\leq d$, we must have $n_{3}=3$ and hence $n_{2}=0$.
However, any line, say $L_{1}$, is tangent to one of the two conics, say to
$C_{1}$, and it is secant to the other, so $L\cap C_{2}=\\{p,q\\}$. The point
$p$ cannot be a node, so one of the remaining lines, say $L_{2}$, passes
through $p$ and is tangent to $C_{1}$, and the other line, say $L_{3}$, is
passing through $q$, and is again tangent to $C_{1}$. The third triple point
should be

$r=C_{2}\cap L_{2}\cap L_{3}.$

This configuration is geometrically realizable, we can take for instance

$\mathcal{CL}_{7}:(x^{2}+y^{2}-z^{2})\cdot(x^{2}+y^{2}-4z^{2})\cdot(x-z)\bigg{(}y+\frac{\sqrt{3}}{3}x+\frac{2\sqrt{3}}{3}\bigg{)}\cdot\bigg{(}y-\frac{\sqrt{3}}{3}x-\frac{2\sqrt{3}}{3}\bigg{)}=0.$

Moreover, it is unique up to a projective transformation. Indeed, we use again
[12, Proposition 3] which implies that one may assume that the two conics
$C_{1}$ and $C_{2}$ are concentric circles, $C_{1}$ with radius $r_{1}=1$, and
$C_{2}$ with radius $r_{2}\in\mathbb{C}$, $r_{2}\neq 0,\pm 1$. Then we note
that a triangle in which the center of the inscribed circle coincides with the
center of the circumscribed circle is necessarily equilateral. This implies
that $r_{2}=2$, and the corresponding equation is given above, where the
vertices of the equilateral triangle in the affine plane $z=1$ are $(1,0)$ and
$(-\frac{1}{2},\pm\frac{\sqrt{3}}{2})$. Using `Singular` we can compute the
minimal free resolution of the Milnor algebra of $\mathcal{CL}_{7}$, it has
the following form

$0\rightarrow S^{2}(-9)\rightarrow S^{3}(-6)\rightarrow S\rightarrow
M(f)\rightarrow 0,$

so $\mathcal{CL}_{7}$ is free.

Finally, we consider the case $k=2$ and $m_{3}=1$, i.e., the two conics are
tangent in one point. In order to satisfy (5.7), we have to use the 3 lines to
create many tacnodes and many triple points, in fact we need $t+n_{3}\geq 8$
of such singular points. We can use the first two lines to get two tangents to
both conics, hence $4$ new tacnodes. The third line can create either $2$
triple points and $2$ double points, one triple point and $4$ nodes, or a
tacnode and $4$ nodes. Neither possibility satisfies $t+n_{3}\geq 8$, and this
completes the proof. ∎

Figure 2. The unique free arrangement of $2$ conics and $3$ lines with $t=5$
and $n_{3}=3$.

Now we are going to discuss non-freeness for conic-line arrangements with
$d\in\\{4,5,6,7\\}$ – please bear in mind that the upper bound on the number
of lines follows from Proposition 4.7.

######  Proposition 5.3.

Let $\mathcal{CL}$ be an arrangement of $d=4$ lines and $k\geq 1$ smooth
conics having only nodes, tacnodes, and ordinary triple points as
singularities. Then $\mathcal{CL}$ is never free.

###### Proof.

If we assume that ${\mathcal{C}}$ is free, then formula (5.2) implies that

(5.8) $n_{2}+n_{3}\leq 4$

and by (5.3) we obtain

(5.9) $t\geq k^{2}-k+4k+\frac{3}{4}-n_{3}.$

Using (5.6), it follows that

(5.10) $2m_{3}+4m_{4}\leq 4.$

In particular, we have $m_{3}+m_{4}\leq 2$. Consider the graph
$\Gamma({\mathcal{C}})$ associated to the conics in ${\mathcal{C}}$, whose
vertices are these conics, and two vertices $C_{i}$ and $C_{j}$ are connected
by an edge (resp. by a double edge) if and only if $|C_{i}\cap C_{j}|=3$
(resp. $|C_{i}\cap C_{j}|=4$) – please consult [12] for more details. It
follows that this graph has at most two edges. To get the maximum number of
tacnodes, Lemma 5.1 implies that the $4$ lines have to be tangent lines to the
conics connected by an edge to other conic. There are either $3$ conics in a
chain connected by edges, or two pairs of conics, each pair connected by an
edge. A simple analysis show that the maximum number of tacnodes created in
this way is $8$. Hence the number of tacnodes satisfies

(5.11) $t\leq t^{\prime}+8=k^{2}-k-m_{3}-2m_{4}+8\leq k^{2}-k+8.$

Combining this with inequality (5.9) we get

$4k-3\leq 4k+1-n_{3}\leq 8,$

which implies $k\leq 2$.

Case: $k=2$.

If $m_{2}=1$, then any line is tangent to at most one conic, by Lemma 5.1,
hence we have $t\leq 6$ and (5.9) yields a contradiction.

If $m_{3}=1$, the inequality (5.9) implies $t+n_{3}\geq 11$. We can use $2$
lines, say $L_{1}$ and $L_{2}$ to create $4$ new tacnodes ($5$ in total) and a
new node ($3$ in total). The third line $L_{3}$ can produce $2$ triple points
if it passes through the $2$ nodes in the intersection $C_{1}\cap C_{2}$, but
it will create $2$ new nodes as the intersections with $L_{1}$ and $L_{2}$.
This stands in contradiction with (5.8). If $L_{3}$ passes through the node
$L_{1}\cap L_{2}$, then it will be secant to both $C_{1}$ and $C_{2}$, thus
creating $4$ new nodes, again it stands in contradiction with (5.8). Finally,
if $L_{3}$ is tangent to one of the conics, it would be secant for the other,
creating $4$ new nodes, $2$ on that second conic and $2$ on $L_{1}\cap L_{2}$,
a contradiction with respect to (5.8).

If $m_{4}=1$, $C_{1}\cap C_{2}$ gives rise to $4$ nodes. As soon as we add $2$
lines, a new node will be created (and some old nodes transformed in triple
points), but in any case we get a contradiction with (5.8).

Hence the case $k=2$ is impossible.

Case: $k=1$.

Since $k=1$, we clearly have $t\leq 4$. The only possibility to have (5.9)
satisfied is that $n_{3}\geq 1$. If we look on the (sub)arrangement
$\mathcal{A}$ of the $4$ lines in $\mathcal{CL}$, we have $2$ cases to
discuss.

Case 1: $\mathcal{A}$ has $3$ concurrent lines.

Let $L_{1},L_{2}$ and $L_{3}$ be the lines meeting at a point $p$, and $L_{4}$
is a secant line, meeting $L_{j}$ at $q_{j}$ for $j=1,2,3$.

If $n_{3}=1$, we get from (5.9) that $t\geq 4$, which is impossible. Indeed,
only two lines among $L_{1},L_{2}$ and $L_{3}$ can be tangent to the conic, so
we get $t\leq 3$.

If $n_{3}=2$, the second triple point is at the intersection of a secant line
to the conic passing through $p$, say $L_{2}$, the conic and the line $L_{4}$.
In this case, we get from (5.9) that $t\geq 3$, which is impossible, since
$L_{2}$ and $L_{4}$ cannot be tangent to the conic.

If $n_{3}=3$, the new triple point is at the intersection of a new secant line
to the conic passing through $p$, say $L_{1}$, the conic and the line $L_{4}$.
In this case, we get from (5.9) that $t\geq 2$, which is impossible, since
$L_{1}$, $L_{2}$ and $L_{4}$ cannot be tangent to the conic.

Finally, if $n_{3}=4$, then the 3 triple points distinct from $p$ are all
situated on the conic and on the line $L_{4}$, which is impossible.

Case 2: $\mathcal{A}$ is a nodal arrangement.

It follows that now all the triple points are on the conic. More precisely,
the conic passes through $n_{3}$ nodes of the line arrangement $\mathcal{A}$.
Hence $\mathcal{CL}$ has $n_{3}$ triple points and at least $6-n_{3}$ double
points. It follows that $n_{2}+n_{3}\geq 6$, a contradiction with (5.8). ∎

######  Proposition 5.4.

Let $\mathcal{CL}$ be an arrangement of $d=5$ lines and $k\geq 1$ smooth
conics having only nodes, tacnodes, and ordinary triple points. Then
$\mathcal{CL}$ is never free.

###### Proof.

If we assume that $\mathcal{CL}$ is free. Using Proposition 4.7, we get
$m=2k+5\leq 9$, and hence $k\leq 2$. Then formula (5.2) implies that

(5.12) $n_{2}+n_{3}\leq 6$

and by (5.3) we get

(5.13) $t\geq k^{2}-k+5k+2-n_{3}.$

Case $k=2$.

If $m_{2}=1$, then any line is tangent to at most one conic, which follows
from Lemma 5.1, so we have $t\leq 7$ and (5.13) yields a contradiction.

If $m_{3}=1$, the inequality (5.13) implies $t+n_{3}\geq 14$. We can use $2$
lines, say $L_{1}$ and $L_{2}$, to create $4$ new tacnodes ($5$ in total) and
a new node ($3$ in total). The third line $L_{3}$ can produce $2$ triple
points if it passes through the $2$ nodes in the intersection $C_{1}\cap
C_{2}$, but will create $2$ new nodes as intersections with $L_{1}$ and
$L_{2}$. It means that the remaining $2$ lines should not create any new node
or triple point, which is impossible. If $L_{3}$ passes through the node
$L_{1}\cap L_{2}$, then it will be secant to both $C_{1}$ and $C_{2}$, thus
creating $4$ new nodes, a contradiction with (5.8). Finally, if $L_{3}$ is
tangent to one of the conics, it would be secant for the other, creating $4$
new nodes, $2$ on that second conic and $2$ on $L_{1}\cap L_{2}$. This
contradicts (5.8).

If $m_{4}=1$, $C_{1}\cap C_{2}$ gives rise to 4 nodes. As soon as we add $3$
lines, at least $3$ new nodes will be created (and some old nodes transformed
in triple points), but in any case we get a contradiction with (5.8).

Hence the case $k=2$ is impossible.

Case: $k=1$.

The above shows that the only possibility to have (5.12) and (5.13) satisfied
is that $k=1$ and $n_{3}\geq 2$. If we look on the (sub)arrangement
$\mathcal{A}$ of the $5$ lines in $\mathcal{CL}$, there are $3$ cases to
discuss.

Case 1: $\mathcal{A}$ has $2$ triple points.

If we denote by $p_{1}$ and $p_{2}$ the two triple points, then the line
determined by these points must be in $\mathcal{A}$. We denote this line by
$L$. The other $2$ lines passing through $p_{1}$ ($p_{2}$, respectively) are
denoted by $L_{1}$ and $L_{1}^{\prime}$ ($L_{2}$ and $L_{2}^{\prime}$,
respectively). There are $4$ nodes in the arrangement $\mathcal{A}$, the
intersections $L_{1}\cap L_{2}$, $L_{1}\cap L_{2}^{\prime}$,
$L_{1}^{\prime}\cap L_{2}$, and $L_{1}^{\prime}\cap L_{2}^{\prime}$.

If $n_{3}=2$, we get from (5.13) that $t\geq 5$, which is impossible since
$L$, $L_{1}$ and $L_{1}^{\prime}$ cannot be all tangent to the conic.

If $n_{3}=3$, the new triple point $q_{1}$ is at the intersection of two
secant lines with the conic. In this case, we get from (5.9) that $t\geq 4$,
which is impossible, since the $2$ lines meeting at $q_{1}$ cannot be tangent
to the conic.

If $n_{3}=4$, then there are $2$ triple points $q_{1}$ and $q_{2}$ situated on
the conic, coming from the intersection of at least $3$ lines from
$\mathcal{A}$ with the conic. In this case, we get from (5.9) that $t\geq 3$,
which is impossible since the lines passing through $q_{1}$ or $q_{2}$ cannot
be tangent to the conic.

If $n_{3}=5$, then there are $3$ triple points $q_{1}$, $q_{2}$ and $q_{3}$
situated on the conic, coming from the intersection of at least $4$ lines from
$\mathcal{A}$ with the conic. In this case, we get from (5.9) that $t\geq 2$,
which is impossible since the lines passing through $q_{1}$, $q_{2}$ or
$q_{3}$ cannot be tangent to the conic.

Finally, if $n_{3}=6$, then $n_{2}=0$, the conic, call it $Q$, passes through
all the $4$ nodes of $\mathcal{A}$ and it is tangent to the line $L$. The
conics passing through the $4$ nodes form a pencil of conics, determined by
the degenerate conics $Q_{1}$ and $Q_{2}$, where $Q_{1}$ (resp. $Q_{2}$) is
the union $L_{1}\cup L_{1}^{\prime}$ (resp. $L_{2}\cup L_{2}^{\prime}$). Note
that these two degenerate conics $Q_{1}$ and $Q_{2}$ meet the line $L$ in one
point, namely $Q_{1}\cap L=p_{1}$ and $Q_{2}\cap L=p_{2}$. The conic $Q$ is in
this pencil ${\alpha}Q_{1}+{\beta}Q_{2}$, and meets the line $L$ in one point
as well. This is a contradiction, since in a pencil only two members can meet
a given line in a single point. To check this claim, we assume that $L$ is
given by $x=0$. Then the condition that ${\alpha}Q_{1}+{\beta}Q_{2}$ meets
line $L$ in one point is a quadratic form in $y,z$, with coefficients being
linear forms in ${\alpha},{\beta}$, such that it has zero discriminant, which
yields the vanishing of the quadratic form in ${\alpha},{\beta}$.

Case 2: $\mathcal{A}$ has a triple point.

It follows that $\mathcal{A}$ has $7$ nodes. To create $n_{3}$ triple points
in $\mathcal{CL}$, the conic passes through $n_{3}-1$ nodes of the line
arrangement ${\mathcal{A}}$. Hence $\mathcal{CL}$ has $n_{3}$ triple points
and at least $7-(n_{3}-1)$ double points. It follows that $n_{2}+n_{3}\geq 8$,
a contradiction with respect to (5.12).

Case 3: $\mathcal{A}$ is a nodal arrangement.

It follows that now $\mathcal{A}$ has $10$ nodes. To create $n_{3}$ triple
points in $\mathcal{CL}$, the conic passes through $n_{3}$ nodes of the line
arrangement $\mathcal{A}$. Hence $\mathcal{CL}$ has $n_{3}$ triple points and
at least $10-n_{3}$ double points. It follows that $n_{2}+n_{3}\geq 10$, a
contradiction with (5.12).

∎

######  Proposition 5.5.

Let $\mathcal{CL}$ be an arrangement of $d=6$ lines and $k\geq 1$ smooth
conics having only nodes, tacnodes, and ordinary triple points. Then
$\mathcal{CL}$ is never free.

###### Proof.

If we assume that $\mathcal{CL}$ is free, then using Proposition 4.7 we get
$m=2k+6\leq 9$, and hence $k=1$. Then the formula (5.2) implies that

(5.14) $n_{2}+n_{3}\leq 7$

and by (5.3) we have

(5.15) $t\geq 10-n_{3}.$

Since $t\leq 6$, the only possibility to have (5.14) and (5.15) been satisfied
is $n_{3}\geq 4$. Consider the (sub)arrangement $\mathcal{A}$ of $6$ lines in
$\mathcal{CL}$. Then $\mathcal{A}$ has only double and triple points, and let
us denote by $n_{2}^{\prime}$ and $n_{3}^{\prime}$ their respective numbers.
It is known that

(5.16) $n_{2}^{\prime}+3n_{3}^{\prime}=\binom{6}{2}=15.$

The conic in $\mathcal{CL}$ has to pass through $n_{3}-n_{3}^{\prime}$ nodes
of $\mathcal{A}$, and hence $n_{2}\geq n_{2}^{\prime}-(n_{3}-n_{3}^{\prime})$.
It follows that

$n_{2}+n_{3}\geq n_{2}^{\prime}+n_{3}^{\prime}=15-2n_{3}^{\prime}.$

It is well-known that the maximal number of triple points $n_{3}^{\prime}$ in
this case is $4$, and the arrangement is projectively equivalent to

${\mathcal{A}}_{0}:(x^{2}-y^{2})(x^{2}-z^{2})(y^{2}-z^{2})=0.$

Combining this fact with inequality in (5.14), it follows that we are exactly
in this case, that is $n_{2}^{\prime}=3$ and $n_{3}^{\prime}=4$. Moreover, the
intersection between the conic and the lines in $\mathcal{A}$ should not add
any new point to the set of seven multiple points of $\mathcal{A}$. Note that
each line contains exactly one double point. This would imply that the conic
is tangent to the 6 lines (in new points given rise to tacnodes), but this is
clearly impossible as we have seen above ($3$ concurrent lines cannot all be
tangent to the same conic). ∎

######  Proposition 5.6.

Let $\mathcal{CL}$ be an arrangement of $d=7$ lines and $k\geq 1$ smooth
conics having only nodes, tacnodes, and ordinary triple points as
singularities. Then $\mathcal{CL}$ cannot free.

###### Proof.

If we assume that $\mathcal{CL}$ is free, then using Proposition 4.7 we get
$m=2k+7\leq 9$, and hence $k=1$. Then the formula (5.2) implies that

(5.17) $n_{2}+n_{3}\leq 9$

and the formula (5.3) gives

(5.18) $t\geq 13-n_{3}.$

Since $t\leq 7$, the only possibility to have (5.17) and (5.18) been satisfied
is $n_{3}\geq 6$. Consider the (sub)arrangement $\mathcal{A}$ of $7$ lines in
$\mathcal{CL}$. Then $\mathcal{A}$ has only double and triple points, and let
us denote by $n_{2}^{\prime}$ and $n_{3}^{\prime}$ their respective numbers.
By the combinatorial count, we have

(5.19) $n_{2}^{\prime}+3n_{3}^{\prime}=\binom{7}{2}=21.$

The conic in $\mathcal{CL}$ has to pass through $n_{3}-n_{3}^{\prime}$ nodes
of $\mathcal{A}$, and hence $n_{2}\geq n_{2}^{\prime}-(n_{3}-n_{3}^{\prime})$.
It follows that

$n_{2}+n_{3}\geq n_{2}^{\prime}+n_{3}^{\prime}=21-2n_{3}^{\prime}.$

Hence $n_{3}^{\prime}\geq 6$. Using the classification of line arrangements
with ${\rm mdr}(f)\leq 2$ presented in [1], it follows that our arrangement
$\mathcal{A}$ satisfies ${\rm mdr}(f)\geq 3$. A calculation of the total
Tjurina number, under the assumption that $n_{3}^{\prime}\geq 6$, shows us
that $n_{3}^{\prime}=6$, $n_{2}^{\prime}=3$, ${\rm mdr}(f)=3$, and the
arrangement $\mathcal{A}$ has to be free. It follows that the arrangement
$\mathcal{A}$ is projectively equivalent to

${\mathcal{A}}_{0}:xyz(x+y)(x+z)(y-z)(x+y+z)=0,$

see [1, Theorem 2.6], where this arrangement occurs as IIIc. Moreover, exactly
as in the previous proof, the intersections between the conic and the lines in
$\mathcal{A}$ should not add any new point to the set of seven multiple points
of ${\mathcal{A}}$. Note that $4$ of the $7$ lines, denoted here by
$L_{2},L_{3},L_{4}$, and $L_{5}$, where $L_{j}$ means the line given by the
$j$-th factor in the equation of $\mathcal{A}$, contain each $3$ triple
points, while the remaining $3$ lines, $L_{1},L_{6}$ and $L_{7}$, contain each
$2$ triple points and $2$ double points. This implies that the conic is
tangent to $4$ lines $L_{2},L_{3},L_{4}$, and $L_{5}$ (in new points given
rise to tacnodes), and passes through the remaining $3$ double points, located
at

$(0:1:1),\ (0:-1:1)\text{ and }(-2:1:1).$

A direct computation shows that such a conic does not exist. ∎

In conclusion, we can state the following complete classification result.

######  Theorem 5.7.

Let $\mathcal{CL}$ be an arrangement of $d\geq 1$ lines and $k\geq 1$ smooth
conics having only nodes, tacnodes, and ordinary triple points as
singularities. Then $\mathcal{CL}$ is free if and only if one of the following
cases occur: In each case we list the numbers $n_{2}$, $t$, and $n_{3}$ of
nodes, tacnodes, and ordinary triple points, respectively.

  1. (1)

$d=k=1$ and $\mathcal{CL}$ consists of a smooth conic and a tangent line. In
this case, $n_{2}=n_{3}=0$, $t=1$.

  2. (2)

$d=2$, $k=1$ and $\mathcal{CL}$ consists of a smooth conic and two tangent
lines. In this case $n_{2}=1$, $n_{3}=0$, $t=2$.

  3. (3)

$d=3$, $k=1$ and either $\mathcal{CL}$ is a smooth conic inscribed in a
triangle, or $\mathcal{CL}$ is a smooth conic circumscribed in a triangle. In
the first case we have $n_{2}=3$, $n_{3}=0$, $t=3$, and in the second case we
have $n_{2}=t=0$, $n_{3}=3$.

  4. (4)

$d=3$, $k=2$ and $\mathcal{CL}$ consists of a triangle $\Delta$, a smooth
conic inscribed in $\Delta$, and another smooth conic circumscribed in
$\Delta$. In this case, $n_{2}=0$, $n_{3}=3$, $t=5$.

In particular, a free conic-line arrangement having only nodes, tacnodes, and
ordinary triple points is determined up to a projective equivalence by the
numerical data $n_{2}$, $n_{3}$ and $t$.

Before we formulate the final corollary for this section, we need the
following notations inspired by [11].

######  Definition 5.8.

We say that two conic-line arrangements in $\mathbb{P}^{2}_{\mathbb{C}}$ with
nodes, tacnodes, and ordinary triple points have the same weak combinatorics
if these arrangements have the same list of invariants $(m;n_{2},t,n_{3})$
with $m$ being degree of the arrangements.

######  Conjecture 5.9 (Numerical Terao’s Conjecture).

Let $\mathcal{CL}_{1},\mathcal{CL}_{2}\subset\mathbb{P}^{2}_{\mathbb{C}}$ be
two conic-line arrangements with nodes, tacnodes, and ordinary triple points.
Assume that $\mathcal{CL}_{1}$ is free and $\mathcal{CL}_{1}$,
$\mathcal{CL}_{2}$ have the same weak combinatorics, then $\mathcal{CL}_{2}$
is also free.

######  Corollary 5.10.

Numerical Terao’s Conjecture holds for conic-line arrangements with nodes,
tacnodes, and ordinary triple points.

###### Proof.

Note that the equation (3.4) implies that the list of invariants
$(m;n_{2},t,n_{3})$ determines the number $k$ of conics and the number
$d=m-2k$ of lines in the arrangements $\mathcal{CL}_{1}$ and
$\mathcal{CL}_{2}$. Then Theorem 5.7 implies that $k=1$ or $k=2$. In each
case, using the fact that $k$ and $d$ are very small, it is easy to see that
up to a projective transformation, the possibilities for $\mathcal{CL}_{2}$
are exactly those listed in Theorem 5.7, and hence the arrangement
$\mathcal{CL}_{2}$ is free as well. ∎

######  Remark 5.11.

Numerical Terao’s Conjecture can be formulated, in principle, for all reduced
singular plane curves. As it was showed in [11], Numerical Terao’s Conjecture
fails for some (triangular) line arrangements. On the other hand, it holds for
line arrangements having only points of multiplicity $\leq 3$. Indeed,
Proposition 4.7 shows that such a free line arrangement $\mathcal{A}:f=0$ has
to satisfy $m=\deg f\leq 9$. Then Theorem 4.10 implies that either $d_{1}={\rm
mdr}(f)\leq 3$ or $d_{1}={\rm mdr}(f)=4$ and $d=9$. Note that if
$\mathcal{A}^{\prime}:f^{\prime}=0$ has the same weak combinatorics as
${\mathcal{A}}:f=0$, then $\tau({\mathcal{A}})=\tau({\mathcal{A}}^{\prime})$,
which implies that $r^{\prime}={\rm mdr}(f^{\prime})\leq{\rm mdr}(f)$ using
the maximality of the Tjurina number for free reduced curves, see [8]. In the
first case, one concludes using the complete classification of line
arrangements with ${\rm mdr}(f)\leq 3$, see [1]. In the second case, we use
again the maximality of the Tjurina number of free curves according to [8] and
we conclude that $\tau({\mathcal{A}})=\tau({\mathcal{A}}^{\prime})=48$,
$n_{2}=0$ and $n_{3}=12$. The only line arrangement with these invariants is
the line arrangement in Example 3.2 above, which is indeed free with the
exponents $(4,4)$.

## Funding

The first author was partially supported by the Romanian Ministry of Research
and Innovation, CNCS - UEFISCDI, Grant PN-III-P4-ID-PCE-2020-0029, within
PNCDI III. The second author was partially supported by the National Science
Center (Poland) Sonata Grant Nr 2018/31/D/ST1/00177. We want to thank an
anonymous referee for comments on that paper.

## Data availability

Not applicable as the results presented in this manuscript rely on no external
sources of data or code.

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