The orbifold Langer-Miyaoka-Yau inequality and Hirzebruch-type inequalities

Using Langer's variation on the Bogomolov-Miyaoka-Yau inequality \cite[Theorem 0.1]{Langer} we provide some Hirzebruch-type inequalities for curve arrangements in the complex projective plane.


Introduction
In 2003 A. Langer [8] has shown the following beautiful variation on the classical Bogomolov-Miyaoka-Yau inequality for a normal surface X with a boundary divisor D.
Theorem 1.1. Let (X, D) be a normal projective surface with a Q-divisor D = i a i D i with 0 a i 1. Assume that the pair (X, D) is log canonical and K X + D is Q-effective. Then (K X + D) 2 3e orb (X, D). (1) Moreover, if equality holds, then K X + D is nef.
In the above formulae, e orb (X, D) denotes the global orbifold number for (X, i a i D i ), i.e., and by e orb (x, X, D) we denote the local orbifold Euler number at x [8, Definition 3.1]. For us the most important property is that local orbifold Euler numbers are analytic in their nature. In the present note we would like to obtain some Hirzebruch-type inequalities for curve arrangements in the complex projective plane such that all irreducible components are smooth and have pairwise transversal intersection points, i.e., all singularities are ordinary and locally look like {x k = y k }. For those singularities, Langer [8,Theorem 8.7] computed their local orbifold Euler numbers using lines in C 2 .
Proposition 1.2. Let L 1 , ..., L n be n distinct lines in C 2 passing through 0. Set D = n i=1 a i L i , where 0 a 1 ... a n 1, and a = n i=1 a i . Then e orb (0; C 2 , D) = 0 if a > 2 (1 − a + a n )(1 − a n ) if 2a n a and e orb (0; Now we would like to look at the case of curves in the complex projective plane. If C is a reduced curve of degree d, then by [3, p. 820] we have where µ p denotes the Milnor number of a singular point p ∈ Sing(C).
It is easy to observe that if (P 2 C , αC) is a log canonical pair for a suitable α, then we can write (1) as follows [8,Section 11]: The main idea of this note is to provide some Hirzebruch-type inequalities for arrangements of curves in the complex projective plane. We will show that using Langer's variation on the Bogomolov-Miyaoka-Yau inequality [9] one can obtain rather elementary proofs of them, i.e., we do not need to pass to Hirzebruch's construction which involves abelian covers branched along arrangements of curves. It is worth pointing out that applied methods allow to deal with configurations of curves having different degrees of irreducible components in a much easier way than in Hirzebruch's construction, especially one can avoid very complicated conditions under which the resulting surfaces has non-negative Kodaira dimension, a crucial condition to apply the classical Bogomolov-Miyaoka-Yau inequality. Another motivation is to show that using Langer's approach one can obtain, somehow surprisingly, the so-called 'quadratic right-hand side in Hirzebruch's inequality' for a large class of curve arrangements.

Hirzebruch-type inequalities
In this section, we assume that all curve arrangements C ⊂ P 2 C have ordinary singularities and every irreducible component of C is smooth. For a given arrangement of curves C we denote by t r the number of r-fold points, i.e., points where r-curves from the arrangement meet. Moreover, we define for i ∈ {0, 1, 2} the following numbers and finally we will use the following elementary observations: where by m p we denote the multiplicity of p ∈ Sing(C) -in our situation this is equal to the number of curves passing through p. Now we present the main results. The first one is devoted to line-conic arrangements in the complex projective plane and this result, according to the author's knowledge, is the first result providing some constraints on the combinatorics of such arrangements. Moreover, these arrangements seem to be interesting in the context of a generalized Terao's conjecture. As we can see in [11,Example 4.2], it is possible to find two configurations of lines and conics with ordinary singularities, which are combinatorially identical, but only one of them is free.
Theorem 2.1. Let LC = {L 1 , ..., L l , C 1 , ..., C k } be an arrangement of l lines and k conics such that t r = 0 for r > 2(l+2k) 3 . Then one has Proof. For an arrangement LC let us denote by C = L 1 + ... + L l + C 1 + ... + C k the associated divisor. First of all, we need to choose α in such a way that K P 2 + αC is effective and log-canonical. Thus where r max denotes the maximal possible multiplicity of singular points in LC. This implies in particular that r max 2ℓ+4k 3 . Let us now choose α ∈ [ 3 ℓ+2k , 2 rmax ]. Our aim is to apply (2) in the above setting. We start with the left-hand side. Using the fact that for a singular point p we have µ p = (m p − 1) 2 (see for instance [4,Remark 2.8]), one has L : e orb (p, P 2 , αC).
We need to use Proposition 1.2. Since all a 1 = ... = a l+k = α, if p is a double point, then and for points p with multiplicities 3 m p = r r max one has e orb (p, This leads to e orb (p, P 2 , αC), and finally we obtain e orb (p, P 2 , αC).
Let us come back to the right-hand side. First of all, observe that assuming also that 0 2 = 1 2 = 0. With d = 2k + l we can rewrite the above combinatorial equality as Using this identity, we get Since L R, we obtain and this provides In particular, taking α −1 = (2k + l)/3 one has which completes the proof.
Now we consider the case when all components have the same degree.
Theorem 2.2. Let C = {C 1 , ..., C k } ⊂ P 2 C be an arrangement of k curves such that all irreducible components have degree d 1. Moreover, we assume that t r = 0 for r > 2dk 3 . Then one has Proof. For a given arrangement C let us denotes by C = C 1 + ... + C k the associated divisor. First of all, we need to choose α in such a way that K P 2 + αC is effective and log-canonical. Thus where r max denotes the maximal multiplicity of singular points in C. This implies in particular that r max The proof is analogical as in the previous case. Since the local orbifold Euler number is analytic in nature, thus the left-hand side has the following form: Now we focus on the right-hand side. Recall that we have the following combinatorial equality This leads to Using L R we get In particular, taking α −1 = dk 3 one has t 2 + 3 4 t 3 + d 2 k(dk − k − 1) r 5 r 2 4 − r t r .