A counterexample to Durfee’s conjecture

Dmitry Kerner and András Némethi

Abstract. An old conjecture of Durfee [Durfee1978] bounds the ratio of two basic invariants of complex isolated complete intersection surface singularities: the Milnor number and the singularity (or geometric) genus. We give a counterexample for the case of non-hypersurface complete intersections, and we formulate a weaker conjecture valid in arbitrary dimension and codimension. This weaker bound is asymptotically sharp. In this note we support the validity of the new proposed inequality by its verification in certain (homogeneous) cases. In our subsequent paper we will prove it for several other cases and we will provide a more comprehensive discussion.

Une conjecture vieux de Durfee [Durfee1978] limite le ratio de deux invariants fondamentales des singularités des surfaces complexes qui sont intersections complètes: le nombre de Milnor et la genre (géométrique) de la singularité. Nous presentons un contre-exemple pour le cas des intersections complètes (non-hypersurface), et nous formulons une conjecture plus faible en dimensions et codimensions arbitraires. Cette bond est forte asymptotiquement. Dans cette note nous appuyons la validité de l’inégalité nouvelle par sa vérification dans certains cases homogènes. Dans le papier prochaine nous allons le prouver pour plusieurs cases et nous fournirons une analyse plus complète.

1. Introduction

Let (X, 0) ⊂ (ℂ^{N}, 0) be the analytic germ of a complex isolated complete intersection singularity (ICIS). Among various singularity invariants the two most basic ones are:

• the Milnor number μ, which measures the change of the local topological/homological Euler characteristic in smoothing deformation;

• the singularity genus (called also geometric genus) p_{g}, which measures the change of the local analytic Euler characteristic (Todd index) when we replace (X, 0) by its resolution.

Their definitions and some of their properties are given in §2.

The relation of these two invariants was investigated intensively giving rise to several open problems as well. In particular, in [Durfee1978] two conjectures were formulated:

• strong inequality: if (X, 0) is an isolated complete intersection surface singularity, then μ ≥ 6p_{g};

• weak inequality: if (X, 0) is a normal surface singularity (not necessarily ICIS) which admits a smoothing with Milnor number (second Betti number of the fiber) μ, then μ + μ_{0} ≥ 4p_{g} (where μ_{0} is the rank of the kernel of the intersection form).

Quite soon a counterexample to the weak conjecture was given in [Wahl1981, page 240] providing a normal surface singularity (not ICIS) with μ = 3, μ_{0} = 0 and p_{g} = 1.

On the other hand, the ‘strong inequality’ valid for an ICIS was believed to be true (though hard to prove) and was verified in many particular hypersurfaces. For example,

[Tomari1993] proved 8p_{g} < μ for (X, 0) of multiplicity 2,

[Ashikaga1992] proved 6p_{g} ≤ μ − 2 for (X, 0) of multiplicity 3,

[Xu-Yau1993] proved 6p_{g} ≤ μ − mult(X, 0) + 1 for quasi-homogeneous singularities,

[Némethi98, Némethi99] proved 6p_{g} ≤ μ for suspension type singularities {g(x, y) + z^{k} = 0} ⊂ (ℂ^{3}, 0), [Melle-Hernández2000] proved 6p_{g} ≤ μ for absolutely isolated singularities.

Moreover, for arbitrary dimension, [Yau-Zhang2006] proved the inequality μ ≥ (n + 1)!p_{g} for isolated weightedhomogeneous hypersurface singularities in (ℂ^{n+1}, 0). The natural expectation was that the same bound (n + 1)! holds for any ICIS of any dimension n and any codimension.

Date: January 16, 2012.

A.N. is partially supported by OTKA grant of the Hungarian Academy of Sciences.

Part of the work was done in Mathematische Forschungsinstitute Oberwolfach, during D.K.’s stay as an OWL-fellow.

In §3 we give a counterexample to the strong conjecture in any codimension r ≥ 2. In fact, for r ≥ 2, the bound (n + 1)! is wrong even asymptotically. Therefore, in §4 we propose a weaker bound. It is based on the Stirling number

of the second kind: \left\{{n + r \atop r}\right\} := \frac{1}{r!} ∑_{j=0}^{r}(−1)^{j}\binom{r}{j}(r − j)^{n+r}, cf. [Abramowitz-Stegun, §24.1.4].

Conjecture. Let (X, 0) ⊂ (ℂ^{N}, 0) be an ICIS of dimension n and codimension r = N − n. Then

• for n = 2 and r = 1 one has μ ≥ 6p_{g},

• for n = 2 and arbitrary r one has μ > 4p_{g},

• for n ≥ 3 and fixed r one has μ ≥ C_{n,r} · p_{g}, where the coefficient C_{n,r} is defined by

C_{n,r} := \frac{\binom{n+r−1}{n}(n + r)!}{\left\{{n + r \atop r}\right\}r!}.

Note that for any curve singularity (i.e. n = 1), p_{g} is the delta invariant δ and μ ≤ 2δ, (with equality exactly for irreducible ones); see example 2.1. For n = 2 the inequality μ(X, 0) ≥ C_{n,r} · p_{g}(X, 0), in general, is not satisfied (see Proposition 4.2 and the comments following it).

We show that the bound of the conjecture is asymptotically sharp, i.e. for any fixed n and r there exists a sequence of isolated complete intersections for which the ratio \frac{μ}{p_{g}} tends to C_{n,r}. In our subsequent longer article [Kerner-Némethi] we verify it for several cases; here we exemplify only some ‘homogeneous’ situations. Moreover, we list some elementary properties of the sequence {C_{n,r}}_{n,r}, cf. §4. For example, we show:

(1)  (n + 1)! = C_{n,1} ≥ C_{n,2} = \frac{(n + 2)!(n + 1)}{2^{n+2} − 2} ≥ ⋯ ≥ lim_{r→∞} C_{n,r} = 2^{n}.

We wish to thank H. Hamm, A. Khovanskii, M. Leyenson, P. Milman, E. Shustin for advises and important discussions. Moreover, we thank the referee for his careful proof–reading.

2. Background

2.1. The Milnor number. Let (X, 0) ⊂ (ℂ^{N}, 0) be an isolated complete intersection of dimension n ≥ 1 defined as the germ of the zero set of an analytic germ f : (ℂ^{N}, 0) → (ℂ^{r}, 0). Let B ⊂ ℂ^{N} be a small ball centered at the origin. By the classical theory (see e.g. [Looijenga-book]) the ‘Milnor fiber’ f^{−1}(ϵ) ∩ B (0 < ϵ ≪ the radius of B) is smooth and has the homotopy type of a bouquet of n–spheres. Their number μ is called the Milnor number of (X, 0).

For general formulae of μ see e.g. [Milnor-book] and [Looijenga-book, §9.A].

2.2. The singularity (geometric) genus. Let (X, 0) ⊂ (ℂ^{N}, 0) be an ICIS as above with n > 1. Let (X̃, E) → (X, 0) be one of its resolutions, i.e. a birational morphism with X̃ smooth, X̃ \E ⟶^{∼} X \\{0\} and \overline{X̃ \ E} = X̃. The singularity genus reflects the cohomological non-triviality of the structure sheaf on the germ (X̃, E), [Artin1966]:

(2)  p_{g}(X, 0) := h^{n−1}OX̃.

This number does not depend on the choice of the resolution.

Example 2.1. • If (X, 0) is a (reduced) curve singularity then the analogue of singularity genus is classical delta invariant: δ = dim_{ℂ} OX̃/O_{(X,0)}. It is related to the Milnor number via 2δ = μ + ♯ − 1, where ♯ is the number of local branches; cf. [Milnor-book], [Buchweitz-Greuel1980], [Looijenga1986].

• Assume that (X, 0) is a normal surface singularity which admits a smoothing. If F is the Milnor fiber of the smoothing, let (μ_{+}, μ_{0}, μ_{−}) be the Sylvester invariants of the symmetric intersection form in the middle integral homology H_{2}(F, ℤ). Then 2p_{g} = μ_{0} + μ_{+} [Durfee1978, Proposition 3.1].

• Let (X, 0) be a Gorenstein normal surface singularity which admits a smoothing. Then, in fact, the Milnor number of the smoothing is independent of the smoothing. Indeed, if (X̃, E) → (X, 0) is any resolution with relative canonical class KX̃, then μ + 1 = χ_{top}(E) + K2 ; X̃ + 12p_{g}. This formula was proved in [Laufer1977] for hypersurface surface singularities and in [Looijenga1986] for normal Gorenstein smoothable singularities.

• Let (X, 0) ⊂ (ℂ^{n+1}, 0) be an isolated hypersurface singularity. Let Sp(X, 0) ⊂ (−1, n) be its singularity spectrum. Then p_{g} is the cardinality of Sp ∩ (−1, 0], cf. [Saito1981, Steenbrink1981].

For the general introduction to singularities we refer to [AGLV-book], [Dimca-book], [Looijenga-book], [Seade-book].

3. The counterexample

Let Y = {f_{1} = ⋯ = f_{r} = 0} ⊂ ℙ^{N−1} be a smooth projective complete intersection defined by homogeneous polynomials of degrees deg(f_{i}) = p_{i}. Define the corresponding ICIS as the cone over Y: (X, 0) = Cone(Y) ⊂ (ℂ^{N}, 0).

Proposition 3.1. (1) μ = ( ∏_{i=1}^{r} p_{i}) ∑_{j=0}^{n}(−1)^{j}( ∑_{\begin{matrix}(k_{1},..,k_{r}), k_{i}≥0 \\ ∑_{i} k_{i}=n−j \end{matrix}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}}) − (−1)^{n},

(2)  p_{g} = ∑_{\begin{matrix}(k_{1},..,k_{r}), k_{i}≥0 \\ ∑_{i} k_{i}=n \end{matrix}} ∏_{i=1}^{r} \binom{p_{i}}{k_{i}+1}.

For a systematic study of the Milnor number of weighted homogeneous complete intersections the reader is invited to consult [Greuel1975, Greuel-Hamm1978], where several formulae were proved. Later, even some more formulae were found, see e.g. [Hamm1986, Hamm2011]. For formulae regarding the geometric genus we refer to [Khovanskii1978, Morales1985, Hamm2011]. These formulae usually are rather different than ours considered above. Nevertheless, the above expressions can be derived from them: here for the Milnor number we will use [Greuel-Hamm1978], and for the geometric genus [Morales1985].

Proof. (1) We will determine the Euler characteristic χ = (−1)^{n}μ + 1 of the Milnor fiber. For a power series Z := ∑_{i≥0} a_{i}x^{i} write [Z]_{n} for the coefficient a_{n} of x^{n}. Also, set P := ∏_{i=1}^{r} p_{i}. Then, by formula 3.7(c) of [Greuel-Hamm1978]

χ = P · [\frac{(1 + x)^{N}}{∏_{i}(1 + p_{i}x)}]_{n}.

Rewrite 1 + p_{i}x as (1 + x)(1 − \frac{(1−p_{i})x}{1+x}), hence

[\frac{(1 + x)^{N}}{∏_{i}(1 + p_{i}x)}]_{n} = [(1 + x)^{n} · ∏_{i} ∑_{k_{i}≥0} (\frac{(1 − p_{i})x}{1 + x})^{k_{i}}]_{n} =

[ ∑_{k_{1}≥0,..,k_{r}≥0} x^{∑ k_{i}}(1 + x)^{n−∑ k_{i}} ∏_{i}(1 − p_{i})^{k_{i}}]_{n} = ∑_{k_{1}≥0,..,kr ≥0 ; ∑ k_{i}≤n} ∏_{i}(1 − p_{i})^{k_{i}}.

(2) By Theorem 2.4 of [Morales1985] (and computation of the lattice points under the ‘homogeneous Newton diagram’) we get

(3)  p_{g} = \binom{∑_{k} p_{k}}{N} − ∑_{1≤i≤r} \binom{(∑_{k} p_{k}) − p_{i}}{N} + ∑_{1≤i<j≤r} \binom{(∑_{k} p_{k}) − p_{i} − p_{j}}{N} − …

Using the Taylor expansion \frac{1}{(1−z)^{N+1}} = ∑_{l≥N} \binom{l}{N}z^{l−N}, the right hand side of (3) is [^{\frac{∏_{i}(1−z^{p_{i}})}{(1−z)^{N+1}}}]_{∑_{k} p_{k}−N}. Thus

p_{g} = rez_{z=0}F (z), where F (z) := \frac{∏_{i}(1 − z^{p_{i}})}{z^{∑ p_{i}−N+1}(1 − z)^{N+1}} .

Note that rez_{z=∞}F (z) = 0 since F (1/z)/z^{2} is regular at zero. As F (z) is a rational function the sum of its residues vanishes: ∑_{pt} rez_{z=pt}F (z) = 0. Since F (z) has poles at z = 0 and z = 1 only, we have p_{g} = −rez_{z=1}F (z). By the change of variables z ↦ 1/z we get

p_{g} = rez_{z=1} \frac{∏_{i=1}^{r}(z^{p_{i}} − 1)}{(z − 1)^{N+1}}.

Since z^{p} − 1 = ∑_{k≥0} \binom{p}{k+1}(z − 1)^{k+1}, we obtain

p_{g} = rez_{z=1} \frac{1}{(z − 1)^{n+1}} · ∏_{i=1}^{r} ∑_{k_{i}≥0} \binom{p_{i}}{k_{i} + 1}(z − 1)^{k_{i}} = ∑_{{k_{1},...,kr ≥0 \atop k_{1}+···+kr =n}} \binom{p_{1}}{k_{1} + 1} ⋯ \binom{p_{r}}{k_{r} + 1}.

This ends the proof.

Example 3.2. Consider the particular case p_{1} = ⋯ = p_{r} = p. Then the formulae of Proposition 3.1 read as

(4) μ = (−1)^{n}(p^{r} ∑_{j=0}^{n}(1 − p)^{j}\binom{j+r−1}{j} − 1), (see also [Greuel-Hamm1978, 3.10(b)]),

p_{g} = ∑\begin{matrix}(k_{1},..,k_{r}), k_{i}≥0 \\ ∑_{i} k_{i}=n \end{matrix} ∏_{i=1}^{r} \binom{p}{k_{i}+1}.

In some low dimensional cases we have:

⋆ for n = 1: p_{g} = \frac{r(p−1)p^{r}}{1!2^{1}}

⋆ for n = 2: p_{g} = \frac{r(p−1)p^{r}}{2!2^{2}} (r(p − 1) + \frac{p−5}{3})

⋆ for n = 3: p_{g} = \frac{r(p−1)p^{r}}{3!2^{3}} (pr − 2 − r)(pr − 3 + p − r).

Analyzing (4) one sees that in the case of μ, the leading term in p (for p large) comes from the last summand (j = n) and μ = p^{N} \binom{N−1}{n} + O(p^{N−1}).

The leading term for p_{g} is more complicated. For p ≫ 0 one has:

p_{g} = p^{N} · ∑_{\begin{matrix}(k_{1},..,k_{r}), k_{i}≥0 \\ ∑_{i} k_{i}=n \end{matrix}} ∏_{i=1}^{r} \frac{1}{(k_{i} + 1)!} + O(p^{N−1}).

Thus, asymptotically, \frac{μ}{p_{g}} = \binom{N−1}{n}_{∑ ; \begin{matrix}(k_{1},..,k_{r}), k_{i}≥0 \\ ∑_{i} k_{i}=n \end{matrix} ∏_{i=1}^{r} \frac{1}{(k_{i}+1)!}} + O(\frac{1}{p}), which in low dimensions gives:

(5)  n = N − r = 2 : \frac{μ}{4p_{g}} = \frac{r + 1}{r + \frac{1}{3}} + O(\frac{1}{p});  n = N − r = 3 : \frac{μ}{8p_{g}} = \frac{r + 2}{r} + O(\frac{1}{p}).

Since 4(r + 1)/(r + \frac{1}{3}) < 6 for all r ≥ 2, in the case n = 2 the strong Durfee bound is violated even asymptotically — that is, for any p sufficiently large — whenever r ≥ 2. Similarly, for n = 3, 8(r + 2)/r < 4! whenever r ≥ 2.

4. The new conjecture with the new weaker bound

In the previous section we have seen that for the singularity which is the cone over a smooth projective complete intersection with p_{1} = ⋯ = p_{r}, the ratio μ/p_{g} tends to the numerical factor

(6)  C_{n,r} := _{\begin{matrix}( & {N−1 \atop n} & ) \\ ∑_{\begin{matrix}(k_{1},..,k_{r}), k_{i}≥0 \\ ∑_{i} k_{i}=n \end{matrix}} & ∏_{i=1}^{r} & \frac{1}{(k_{i}+1)!} \end{matrix}} .

Next we list some properties of these numbers.

Lemma 4.1. (I) C_{n,r} = _{\frac{\binom{n+r−1}{n}(n+r)!}{\left\{{n + r \atop r}\right\}r!}} (see also [Jordan1965, pg.176-178])

(II) (n + 1)! = C_{n,1} ≥ C_{n,2} ≥ ⋯ ≥ lim_{r→∞} C_{n,r} = 2^{n}.

Proof. (I) Let S_{n,r} := ∑\begin{matrix}\{k_{i}≥0\} \\ ∑_{i} k_{i}=n \end{matrix} ∏_{i=1}^{r} \frac{1}{(k_{i}+1)!}. First we prove

(7)  \frac{r^{n+r}}{(n + r)!} = S_{n,r} + rS_{n+1,r−1} + \binom{r}{2}S_{n+2,r−2} + · · · .

Consider the expansion r^{n+r} = (1 + ⋯ + 1)^{n+r} = ∑_{k_{1},..,k_{r}≥0}(∏_{i=1}^{r} 1^{k_{i}})\frac{(n+r)!}{∏_{i=1}^{r} k_{i}!}. This gives

(8)  \frac{r^{n+r}}{(n + r)!} = ∑_{\begin{matrix}k_{1},..,k_{r}≥0 \\ ∑_{i} k_{i}=n+r \end{matrix}} \frac{1}{∏_{i=1}^{r} k_{i}!}.

For each 0 ≤ j ≤ r and subset I(j) ∈ {1, ⋯, r} of cardinality j set

K_{I(j)} := \{\underline{k} = (k_{1}, …, k_{r}) with k_{i} = 0 for i ∈ I(j) and ∑_{i} k_{i} = n + r.\}

Then the sum ∑\underline{k} from the right hand side of (8) can be replaced by ∑_{j=0}^{r} ∑_{I(j)} ∑_{\underline{k}∈K_{I(j)}}. Since for each fixed I(j)

∑_{\underline{k}∈K_{I(j)}} \frac{1}{∏_{i=1}^{r} k_{i}!} = S_{n+j,r−j}

(by a change k_{j} ↦ k_{j} − 1 for those indices when k_{j} ≠ 0), and for each j there are \binom{r}{j} subsets I(j) of cardinality j, (7) follows.

Now, we prove that S_{n,r} = ^{\frac{∑_{i=0}^{r}(r−i)^{n+r}(−1)^{i}\binom{r}{i}}{(n+r)!}}. Note that the recursion (7) defines S_{n,r} uniquely. Hence it is enough

to verify that the proposed solution satisfies this recursive relation; that is, r^{n+r} = ∑_{j=0}^{r} \binom{r}{r−j} ∑_{i=0}^{j}(j −i)^{n+r}(−1)^{i}\binom{j}{i}.

Set j̃ = j − i. Then the sum can be rewritten as ∑_{j̃=0}^{r} j̃^{n+r}\binom{r}{j̃} ∑_{i=0}^{r−j̃}(−1)^{i}\binom{r−j̃}{i}. Note that ∑_{i=0}^{r−j̃}(−1)^{i}\binom{r−j̃}{i} =

(1 − 1)^{r−j̃} = 0 unless r = j̃. Hence the above formula for S_{n,r} follows.

This formula can be compared with that one satisfied by the Stirling numbers (see introduction), hence we get

S_{n,r} = \left\{{n + r \atop r}\right\}\frac{r!}{(n+r)!}, which proves the statement for C_{n,r}.

(II) The limit of C_{n,r} is computed using the asymptotics of Stirling numbers of the second kind, [Abramowitz-Stegun,

§24.1.4]: \left\{{n + r \atop r}\right\} ∼ \frac{r^{2n}}{2^{n}n!}. This gives: C_{n,r} ∼ 2^{n} \frac{(n+r−1)!(n+r)!}{(r−1)!r!r^{2n}} → 2^{n}.

The following proof showing that C_{n,r} is non-increasing sequence in r was communicated to us by D. Moews. Write the inequality C_{n,r} ≥ C_{n,r+1} in terms of Stirling numbers of the second kind:

(9)  \left\{{n + r + 1 \atop r + 1}\right\}r(r + 1) ≥ \left\{{n + r \atop r}\right\}(n + r)(n + r + 1).

Set N := n + r as usual, and write the inequality in terms of generating functions:

(10)  \frac{r(r + 1)}{x} ∑_{N≥r} \left\{{N + 1 \atop r + 1}\right\}\frac{x^{N+1}}{(N + 1)!} ⪰ x∂_{x} ∑_{N≥r} \left\{{N \atop r}\right\}\frac{x^{N}}{N!}

with the convention: ∑ a_{n}x^{n} ⪰ ∑ b_{n}x^{n} if and only if a_{n} ≥ b_{n} for each n ≥ 0.

Recall that Stirling numbers are the coefficients of Taylor series: \frac{(e^{x}−1)^{r}}{r!} = ∑_{N≥r} \left\{{N \atop r}\right\}\frac{x^{N}}{N!} . Therefore the in-

equality to be proved can be rewritten as: \frac{(e^{x}−1)^{r+1}}{(r−1)!x} ⪰ \frac{xe^{x}(e^{x}−1)^{r−1}}{(r−1)!} . We claim that this inequality follows from (e^{x} − 1)^{2} ⪰ x^{2}e^{x}. Indeed, if ∑ a_{n}x^{n} ⪰ ∑ b_{n}x^{n} and the Taylor expansion of g(x) has only positive coefficients, then g(x) ∑ a_{n}x^{n} ⪰ g(x) ∑ b_{n}x^{n}. Hence it is enough to prove (e^{x} − 1)^{2} ⪰ x^{2}e^{x}. For the same reason, this last statement will follow from (e^{x} − 1) ⪰ xe^{\frac{x}{2}} , which is immediate.

The coefficients C_{n,r} satisfies the following identities and inequalities (some more will be listed in [Kerner-Némethi], where the general inequality μ ≥ C_{n,r}p_{g} will also be discussed).

Proposition 4.2. Let (X, 0) = ∩_{i=1}^{r}{f_{i}(x_{1}, …, x_{N}) = 0} ⊂ (ℂ^{N}, 0) be an isolated complete intersection singularity, where each f_{i} is homogeneous of degrees p_{i}. Set P := ∏_{i=1}^{r} p_{i}. Then:

For n = 1: μ + P − 1 = C_{1,r} · p_{g} = 2p_{g}. (Here p_{g} := δ.) For n = 2: μ + P · (\frac{r−1}{3r+1} ∑_{i=1}^{r}(p_{i} − 1) − ∑_{i<j} \frac{(p_{i}−p_{j})^{2}}{3r+1} − 1) + 1 = C_{2,r} · p_{g} = 4\frac{r+1}{r+ \frac{1}{3}} · p_{g}.

For r = 1: μ − C_{n,1} · p_{g} = (p − 1)^{N} − \frac{p!}{(p−N)!} ≥ 0. In particular, μ ≥ (n + 1)! · p_{g}. For n = 3: μ > C_{3,r} · p_{g}.

We wish to emphasize the equation μ + P · E + 1 = C_{2,r} · p_{g} in the case n = 2. Note that the expression E is −1 (hence negative) if r = 1, but for r ≥ 2 and some choices of p_{i}’s (e.g. whenever they are all equal) E might be positive.

Proof. By Proposition 3.1 we have:

(11)  μ = P · ∑_{j=0}^{n}(−1)^{j}( ∑_{\begin{matrix}\{k_{i}≥0\}, \\ ∑_{i} k_{i}=n−j \end{matrix}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}}) − (−1)^{n},  p_{g} = P · ∑_{\begin{matrix}\{k_{i}≥0\} \\ ∑_{i} k_{i}=n \end{matrix}} ∏_{i=1}^{r} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1}.

We can assume p_{i} > 1, otherwise some of defining hypersurfaces are hyperplanes, so this reduces to a singularity of the same dimension and smaller codimension. Case n=1. Use

(12)  μ = P · ( ∑_{i=1}^{r}(p_{i} − 1) − 1) + 1,  p_{g} = P · ∑_{i=1}^{r} \frac{(p_{i} − 1)}{2},  C_{1,r} = 2.

Case n=2. We have

(13)  \begin{matrix}μ = P · ( ∑_{i}(p_{i} − 1)(p_{i} − 2) + ∑_{i<j}(p_{i} − 1)(p_{j} − 1) + 1) − 1, \\ p_{g} = P · ( ∑_{i} \binom{p_{i}−1}{2} \frac{1}{3} + ∑_{i<j} \frac{(p_{i}−1)(p_{j}−1)}{4}),  C_{2,r} = 4\frac{r+1}{r+ \frac{1}{3}} . \end{matrix}

Hence

(14)  \begin{matrix}μ − C_{2,r}p_{g} = P · ( ∑_{i}(p_{i} − 1)(p_{i} − 2)\frac{(r−1)}{3r+1} − \frac{2}{3r+1} ∑_{i<j}(p_{i} − 1)(p_{j} − 1) + 1) − 1 = \\ = P · ( ∑_{i<j} \frac{(p_{i}−1)^{2}−2(p_{i}−1)(p_{j}−1)+(p_{j}−1)^{2}}{3r+1} − ∑_{i}(p_{i} − 1)\frac{(r−1)}{3r+1} + 1) − 1, \end{matrix}

giving the needed equality. Note that from here one also obtains for r = 1 and n = 2:

(15)  6p_{g} = μ − P + 1 ≤ μ.

Case r = 1 is obvious. Case n = 3. It is enough to prove:

(16)  ∑_{j=0}^{3}(−1)^{j}( ∑_{\begin{matrix}\{k_{i}≥0\}, \\ ∑_{i} k_{i}=3−j \end{matrix}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}}) > \frac{8(r + 2)}{r} ∑_{\begin{matrix}\{k_{i}≥0\} \\ ∑_{i} k_{i}=3 \end{matrix}} ∏_{i=1}^{r} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1}

Present the difference as follows

(17) \begin{matrix}\underbrace{∑_{\begin{matrix}\{k_{i}≥0\}, \\ ∑_{i} k_{i}=3 \end{matrix}} ( ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}})(1 − \frac{8(r + 2)}{r ∏_{i=1}^{r}(k_{i} + 1)!})}_{P art I} + \underbrace{\frac{8(r + 2)}{r} ∑_{\begin{matrix}\{k_{i}≥0\} \\ ∑_{i} k_{i}=3 \end{matrix}} ( ∏_{i=1}^{r} \frac{(p_{i} − 1)^{k_{i}}}{(k_{i} + 1)!} − ∏_{i=1}^{r} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1})}_{P art II} + \\ + \underbrace{∑_{j=1}^{3}(−1)^{j}( ∑_{\begin{matrix}\{k_{i}≥0\}, \\ ∑_{i} k_{i}=3−j \end{matrix}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}})}_{P art III} . \end{matrix}

First we prove that P art I ≥ 0.

(18)  P art I = \frac{2r − 2}{3r} ∑_{i}(p_{i} − 1)^{3} + \frac{r − 4}{3r} ∑_{i≠j}(p_{i} − 1)^{2}(p_{j} − 1) − \frac{2}{r} ∑_{i<j<k}(p_{i} − 1)(p_{j} − 1)(p_{k} − 1).

Now use the elementary inequalities (the algebraic mean compared to the geometric mean)

∑ a_{i}^{n} = ∑_{i≠j} \frac{la_{i}^{n} + (n − l)a_{j}^{n}}{n(r − 1)} ≥ \frac{1}{r − 1} ∑_{i≠j} a_{i}^{l}a_{j}^{n−l}

and (for 2k < n)

∑_{i≠j} a_{i}^{k}a_{j}^{n−k} = ∑_{i≠j≠m} \frac{(n − k − 1)a_{i}^{k}a_{j}^{n−k} + a_{i}^{k}a_{m}^{n−k}}{(n − k)(r − 2)} ≥ \frac{∑_{i≠j≠m} a_{i}^{k}a_{j}^{n−k−1}a_{m}}{r − 2}

to get:

(19)  P art I ≥ \frac{2 + r − 4}{3r} ∑_{i≠j}(p_{i} − 1)^{2}(p_{j} − 1) − \frac{2}{r} ∑_{i<j<k}(p_{i} − 1)(p_{j} − 1)(p_{k} − 1) ≥ 0.

Now simplify P art II:

(20)  P art II = \frac{8(r + 2)}{r}( ∑_{i} \frac{(p_{i} − 1)(3p_{i} − 5)}{4!} + ∑_{i≠j} \frac{(p_{i} − 1)}{3!} \frac{(p_{j} − 1)}{2}).

Thus, by direct check one gets: P art II + P art III > 0. Hence the statement.

Finally, we add a slightly weaker inequality, but which is valid for any n (including n = 2) and any r. Since \binom{N−1}{n}

is the number of terms in the sum from the denominator of C_{n,r}, we obtain that C_{n,r} ≥ min_{\begin{matrix}∑ k_{i}=n \\ k_{i}≥0 \end{matrix}}\{ ∏_{i=1}^{r}(k_{i} + 1)!\}.

Proposition 4.3. Let (X, 0) as in Lemma 4.2 and n ≥ 2. Then

μ ≥ min_{\begin{matrix}∑ k_{i}=n \\ k_{i}≥0 \end{matrix}}\{ ∏_{i=1}^{r}(k_{i} + 1)!\} · p_{g} ≥ 2^{n} · p_{g}.

Moreover, if n > r, let n = n_{1}r + r_{1} for n_{1}, r_{1} ≥ 0, r_{1} < r. Then μ > ((n_{1} + 1)!)^{r−r_{1}}((n_{1} + 2)!)^{r_{1}}p_{g}. In particular, for r = 1: μ > (n + 1)!p_{g}. For r = 2 and n-odd: μ ≥ (⌊\frac{n}{2}⌋ + 1)!(⌊\frac{n}{2}⌋ + 2)!p_{g}. For r = 2 and n-even:

μ ≥ ((⌊\frac{n}{2}⌋ + 1)!)^{2}p_{g}.

Proof. Step 1. We claim that (by Proposition 3.1) it is enough to prove the inequality

(21) L.H.S. := P · ( ∑_{j=0}^{n}(−1)^{j} ∑_{\begin{matrix}\{k_{i}≥0\} \\ ∑_{i} k_{i}=n−j \end{matrix}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}}) − (−1)^{n} > ∑_{\begin{matrix}k_{1},...,k_{r}≥0 \\ k_{1}+···+k_{r}=n \end{matrix}} ( ∏_{i=1}^{r} \binom{p_{i}}{k_{i} + 1}(k_{i} + 1)!) =: R.H.S.

Clearly this imply the first inequality of the proposition, while the second follows from (k + 1)! ≥ 2^{k}. Step 2. The left hand side can be written as

(22)  L.H.S. = (p_{r} − 1)(∏_{i=1}^{r−1} p_{i})C_{n}(p_{1}, .., p_{r}) + (p_{r−1} − 1)(∏_{i=1}^{r−2} p_{i})C_{n}(p_{1}, .., p_{r−1}) + .. + (p_{1} − 1)(p_{1} − 1)^{n},

where C_{n}(p_{1}, .., p_{r}) := ∑_{\begin{matrix}\{k_{i}≥0\} \\ ∑_{i} k_{i}=n \end{matrix}} ( ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}}). This is expansion in terms with decreasing r, hence it is natural to

expand the right hand side similarly.

Let D_{n}(p_{1}, .., p_{r}) := ∑_{\begin{matrix}k_{1},...,k_{r}≥0 \\ k_{1}+···+k_{r}=n \end{matrix}} ∏_{i=1}^{r} \binom{p_{i}−1}{k_{i}}(k_{i})!. So D_{1}(p_{1}) = \binom{p_{1}−1}{n}n!. Write the R.H.S. as follows:

(23)

((∏_{i=1}^{r} p_{i})D_{n}(p_{1}, .., p_{r}) − (∏_{i=1}^{r−1} p_{i})D_{n}(p_{1}, .., p_{r−1})) + ((∏_{i=1}^{r−1} p_{i})D_{n}(p_{1}, .., p_{r−1}) − (∏_{i=1}^{r−2} p_{i})D_{n}(p_{1}, .., p_{r−2})) + ... + p_{1}D_{1}(p_{1}).

Thus it is enough to prove the inequality for each pair of terms in these expansions:

(24)  ∀r ≥ 1, n ≥ 2 : (p_{r} − 1)(∏_{i=1}^{r−1} p_{i})C_{n}(p_{1}, .., p_{r}) ≥ (∏_{i=1}^{r} p_{i})D_{n}(p_{1}, .., p_{r}) − (∏_{i=1}^{r−1} p_{i})D_{n}(p_{1}, .., p_{r−1}).

For example, for r = 1 we have: (p_{1} − 1)^{n+1} > p_{1}\binom{p_{1}−1}{n}n!. Present the right hand side of (24) in the form

(p_{r} − 1)(∏_{i=1}^{r−1} p_{i})D_{n}(p_{1}, .., p_{r}) + (∏_{i=1}^{r−1} p_{i})(D_{n}(p_{1}, .., p_{r}) − D_{n}(p_{1}, .., p_{r−1})).

Note that

(25) D_{n}(p_{1}, .., p_{r}) − D_{n}(p_{1}, .., p_{r−1}) = ∑_{\begin{matrix}\{k_{i}≥0\}, k_{r}>0 \\ k_{1}+···+k_{r}=n \end{matrix}} ∏_{i=1}^{r} \binom{p_{i} − 1}{k_{i}}(k_{i})! = (p_{r} − 1)D_{n−1}(p_{1}, .., p_{r−1}, (p_{r} − 1)).

Hence we have to prove the inequality: C_{n}(p_{1}, .., p_{r}) ≥ D_{n}(p_{1}, .., p_{r}) + D_{n−1}(p_{1}, .., p_{r−1}, (p_{r} − 1)). In fact, we will prove by induction on r the stronger inequality:

(26)  C_{n}(p_{1}, .., p_{r}) ≥ D_{n}(p_{1}, .., p_{r}) + D_{n−1}(p_{1}, .., p_{r}).

Step 3. Both parts are summations over (k_{1}, .., k_{r}), expand them in k_{r}. Then we have to prove: (27)

∑_{k_{r}=0}^{n} ((p_{r} − 1)^{k_{r}}C_{n−k_{r}}(p_{1}, .., p_{r−1}) − \binom{p_{r} − 1}{k_{r}}k_{r}!D_{n−k_{r}}(p_{1}, .., p_{r−1})) ≥ ∑_{k_{r}=1}^{n} \binom{p_{r} − 1}{k_{r} − 1}(k_{r} − 1)!D_{n−k_{r}}(p_{1}, .., p_{r−1}).

For k_{r} ≥ 2 one has: (p_{r} − 1)^{k_{r}} ≥ \binom{p_{r}−1}{k_{r}}k_{r}! + \binom{p_{r}−1}{k_{r}−1}(k_{r} − 1)!, by direct check. And C_{j}(p_{1}, .., p_{r−1}) ≥ D_{j}(p_{1}, .., p_{r−1}), for all values of j, r. Therefore we only need to check the terms for k_{r} = 0, 1, i.e. it is enough to prove:

(28) (p_{r} − 1)(C_{n−1}(p_{1}, .., p_{r−1}) − D_{n−1}(p_{1}, .., p_{r−1})) + C_{n}(p_{1}, .., p_{r−1}) − D_{n}(p_{1}, .., p_{r−1}) ≥ D_{n−1}(p_{1}, .., p_{r−1}).

Again, C_{n−1}(p_{1}, .., p_{r−1}) ≥ D_{n−1}(p_{1}, .., p_{r−1}), so the initial inequality reduces to C_{n}(p_{1}, .., p_{r−1}) ≥ D_{n}(p_{1}, .., p_{r−1}) + D_{n−1}(p_{1}, .., p_{r−1}). This completes the induction step from (r − 1) to r.

Finally, for r = 1 the initial inequality is: (p_{1} −1)^{n} ≥ \binom{p_{1}−1}{n}n!+\binom{p_{1}−1}{n−1}(n−1)!. By direct check it is true for n ≥ 2.

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Department of Mathematics, University of Toronto, 40 St. George Street, Toronto, Canada

E-mail address: dmitry.kerner@gmail.com

Rényi Institute of Mathematics, Budapest, Reáltanoda u. 13–15, 1053, Hungary

E-mail address: nemethi@renyi.hu