The ‘corrected Durfee’s inequality’ for homogeneous complete intersections

Dmitry Kerner and András Némethi

Abstract. We address the conjecture of [Durfee1978], bounding the singularity genus p_{g} by a multiple of the Milnor number μ for an n–dimensional isolated complete intersection singularity. We show that the original conjecture of Durfee, namely (n + 1)! · p_{g} ≤ μ, fails whenever the codimension r is greater than one. Moreover, we proposed a new inequality C_{n,r} · p_{g} ≤ μ, and we verify it for homogeneous complete intersections. In the homogeneous case the inequality is guided by a ‘combinatorial inequality’, that might have an independent interest.

1. Introduction

1.1. Let (X, 0) ⊂ (ℂ^{N}, 0) be the analytic germ of an n–dimensional complex isolated complete intersection singularity (ICIS). One of the most important goals of the local singularity theory is the clarification of the subtle connections between the two basic numerical invariants, the Milnor number μ and singularity (geometric) genus p_{g}.

The surface case, n = 2, is already rather exotic and hard. In this case, if X_{F} is the Milnor fiber of (X, 0), and (μ_{+}, μ_{0}, μ_{−}) are the Sylvester invariants of the symmetric intersection form in the middle integral homology H_{2}(X_{F} , ℤ), then 2p_{g} = μ_{0} + μ_{+} [Durfee1978], while, obviously, μ = μ_{+} + μ_{0} + μ_{−}. Hence, numerical relations between μ and p_{g} can be rewritten in terms of the Sylvester invariants too. In topology one also uses the signature σ := μ_{+} − μ_{−} as well. In fact, for compact complex surfaces, the Euler number, Todd class and the signatura are the most important index–theoretical numerical invariants; their local analogs are the above integers μ, p_{g} and σ. For more about these invariants see the monographs [Milnor-book, AGV-book, AGLV-book, Looijenga-book] or articles [Laufer1977, Looijenga1986, Saito1981]. For various formulae regarding Milnor number of weighted homogeneous complete intersections see [Greuel1975, Greuel-Hamm1978, Hamm1986, Hamm2011] and for the geometric genus see [Khovanskii1978, Morales1985].

Examples show that for a local surface singularity μ_{−} should be ‘large’ compared with the other Sylvester invariants, or equivalently, p_{g} small with respect to μ, or, σ rather negative.

This was formulated more precisely in Durfee’s Conjectures [Durfee1978] as follows:

(A) Strong inequality: if (X, 0) is an isolated complete intersection surface singularity, then 6p_{g} ≤ μ. (B) 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 4p_{g} ≤ μ + μ_{0}, or equivalently, σ ≤ 0. (C) Semicontinuity of σ: if {(X_{t}, 0)}_{t∈(ℂ,0)} is a family of isolated surface singularities than σ(X_{t=0}) ≤ σ(X_{t≠0}).

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

A counterexample to semicontinuity of the signature was found much later, in [Kerner-Némethi2009].

On the other hand, the strong inequality, valid for an ICIS, was believed to be true and was verified in many particular hypersurface singularities (X, 0) in (ℂ^{3}, 0):

[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 n ≥ 2, [Yau-Zhang2006] proved the inequality (n + 1)!p_{g} ≤ μ for isolated weightedhomogeneous hypersurface singularities in (ℂ^{n+1}, 0). The natural expectation was that the same inequality holds for any ICIS of any dimension n and any codimension r := N − n.

1.2. The main results of the present article are the following, cf. [Kerner-Némethi.a, Kerner-Némethi.b]:

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.

I. For homogeneous ICIS of multidegree (p_{1}, …, p_{r}) we provide new formulae for μ and p_{g}; their special form allows us to compare them. II. Using these formulae one sees rather easily that (n + 1)! · p_{g} ≤ μ is not true whenever r ≥ 2 and n ≥ 2, already for p_{1} = ⋯ = p_{r} sufficiently large (that is, the strong inequality fails even asymptotically). III. We propose a new set of conjectured inequalities with new bounds for μ/p_{g}. For any n ≥ 1 and r ≥ 1 consider the Stirling number of the second kind, cf. [Abramowitz-Stegun, §24.1.4], and the coefficient C_{n,r} defined by

\left\{{n + r \atop r}\right\} := \frac{1}{r!} ∑_{j=0}^{r}(−1)^{j}\binom{r}{j}(r − j)^{n+r},

(1)  C_{n,r} := \frac{\binom{n+r−1}{n}(n + r)!}{\left\{{n + r \atop r}\right\}r!} = \frac{|K_{n,r}|}{∑_{\underline{k}∈K_{n,r}} ∏{r \atop i=1} \frac{1}{(k_{i}+1)!}} ,

where |K_{n,r}| = \binom{n+r−1}{n} is the cardinality of the set

(2)  K_{n,r} := \{\underline{k} = (k_{1}, …, k_{r}) : k_{i} ≥ 0 for all i, and ∑_{i} k_{i} = n\}.

The second equality of (1) follows from [Jordan1965, pages 176-178]. One also shows, cf. Corollary 4.2, that

(3)  C_{n,1} > C_{n,2} > ⋯ > C_{n,r} > ⋯ > lim_{r→∞} C_{n,r}.

E.g., for small r and for r → ∞ one gets

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

The limit can be computed using the asymptotical grows 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}} with limit 2^{n} as r → ∞.

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

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

• for n = 2 and arbitrary r one has 4p_{g} < μ (see last section too),

• for n ≥ 3 and arbitrary r one has C_{n,r} · p_{g} ≤ μ.

IV. We show that the proposed inequality 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}. V. We support the above conjecture by its proof for any homogeneous ICIS with any multidegree (p_{1}, …, p_{r}).

Note that C_{n,r} · p_{g} ≤ μ automatically implies

μ ≥ min_{\underline{k}∈K_{n,r}}\{∏_{i=1}^{r}(k_{i} + 1)!\} · p_{g} ≥ 2^{n} · p_{g}.

Some more comments are in order.

• If n = 1 then p_{g} corresponds to the delta invariant δ, and C_{1,r} = \frac{1}{2}. On the other hand, 2δ = μ + b(X, 0) − 1, where b(X, 0) is the number of locally irreducible branches. Hence μ ≤ 2δ; and the analogue of the strong inequality fails. Moreover, for n = 2 too, the inequality μ(X, 0) ≥ C_{n,r} · p_{g}(X, 0), in general, is not satisfied, and the asymptotically sharp inequality of this form is impossible. For more comments regarding n = 2 see section 5.

These facts also show that, in order to prove the conjectured inequalities, a ‘naive induction’ over n is impossible.

• Homogeneous singularities (considered in V) are rather particular ones. Nevertheless, it turns out that they are the building blocks for many other singularity types. In the forthcoming paper we use the statement of V to prove the above conjectured inequalities from III for more general families of complete intersections (e.g., for absolutely isolated singularities), cf. [Kerner-Némethi.c].

• Even in this particular case of homogeneous germs the proof involves a non-trivial combinatorial inequality, which ‘guides’ the inequality C_{n,r} · p_{g} ≤ μ, see the ‘Combinatorial Inequality’ below. Although its proof in its current form is relatively short, we believe it is far from being straightforward. It is also possible that it covers some probabilistic interpretation of the inequality C_{n,r} · p_{g} ≤ μ in terms of Dirichlet distributions.

1.3. Fix the integers n ≥ 1, r ≥ 1 and ℓ ≥ 0. We define \underline{y}_{\underline{k},ℓ} := ∏_{i} \frac{1}{(k_{i}+ℓ)!} and for free positive variables x_{1}, …, x_{r} we

set \underline{x}\underline{k} = x_{1}^{k_{1}} ⋯ x_{r}^{k_{r}}. By convention 0! = 1.

Combinatorial Inequality. For any n, r and ℓ as above, and for any positive x_{1}, ⋯, x_{r} one has:

(5)  (I_{n,r,ℓ}) :  \frac{1}{|K_{n,r}|} · ∑_{\underline{k}∈K_{n,r}} \underline{y}_{\underline{k},ℓ} · ∑_{\underline{k}∈K_{n,r}} \underline{x}\underline{k} ≥ ∑_{\underline{k}∈K_{n,r}} \underline{y}_{\underline{k},ℓ} · \underline{x}\underline{k}.

For our application we only need the ℓ = 1 case, nevertheless its proof uses all the ℓ values (the induction over n involves larger ℓ’s for smaller n’s).

1.4. We wish to thank G.-M. Greuel, H. Hamm, A. Khovanskii, M. Leyenson, L. Lovász, P. Milman, E. Shustin for advices and important discussions.

2. The μ and p_{g} formulae

Let (X, 0) = {f_{1} = ⋯ = f_{r} = 0} be a homogeneous ICIS of multidegree (p_{1}, …, p_{r}), that is deg(f_{i}) = p_{i}.

Proposition 2.1. (1) μ = ( ∏_{i=1}^{r} p_{i}) ∑_{j=0}^{n}(−1)^{j}( ∑_{\underline{k}∈K_{n−j,r}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}}) − (−1)^{n},

(2) p_{g} = ∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r} \binom{p_{i}}{k_{i}+1}.

The reader is invited to consult [Greuel1975, Greuel-Hamm1978, Hamm1986, Hamm2011] for μ and [Khovanskii1978, Morales1985, Hamm2011] for p_{g} of a weighted homogeneous ICIS. These formulae usually are rather different than ours considered above; nevertheless, both formulae (1) and (2) can be derived from expressions already present in the literature (though we have found them in a different way).

Proof. (1) We determine the Euler characteristic χ = (−1)^{n}μ + 1 of the Milnor fiber. For a power series Z := ∑_{i≥0} a_{i}x^{i} we write [Z]_{n} for the coefficient a_{n} of x^{n}. By 3.7(c) of [Greuel-Hamm1978]

χ = ∏_{i=1}^{r} p_{i} · [\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 [Morales1985, Theorem 2.4] (and computation of the lattice points under the ‘homogeneous Newton diagram’)

(6)  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 (6) 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. Hence, p_{g} = −rez_{z=1}F (z) too. By the change of variables z ↦ 1/z, this last expression transforms into

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}} = ∑_{\underline{k}∈K_{n,r}} \binom{p_{1}}{k_{1} + 1} ⋯ \binom{p_{r}}{k_{r} + 1}.

Consider the particular case p_{1} = ⋯ = p_{r} = p. Then the above formulae read as

(7)

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} = ∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r} \binom{p}{k_{i}+1}.

Therefore, for p large, μ and p_{g} asymptotically behave as follows:

μ = p^{N} \binom{N − 1}{n} + O(p^{N−1}) and p_{g} = p^{N} · ∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r} \frac{1}{(k_{i} + 1)!} + O(p^{N−1}).

Thus, asymptotically, \frac{μ}{p_{g}} = C_{n,r} + O(\frac{1}{p}). Note that C_{n,r} < C_{n,1} = (n + 1)! for any r ≥ 2, hence the strong Durfee’s inequality is violated for any p sufficiently large whenever r ≥ 2. The inequality C_{n,r} < (n + 1)! follows from the fact that \frac{1}{(n+1)!} is the smallest of all elements of type ∏_{i} \frac{1}{(k_{i}+1)!}, for \underline{k} ∈ K_{n,r}.

3. The inequality C_{n,r} · p_{g} ≤ μ in the homogeneous case

Note that if r = 1, then C_{n,1} · p_{g} = (n + 1)! · p_{g} ≤ μ for any isolated homogeneous germs and for any n ≥ 2. This follows from (n + 1)!p_{g} = (n + 1)!\binom{p}{n+1} ≤ (p − 1)^{n+1} = μ, where p is the degree of the germ. If r > 1 then the n = 2 case is rather special, and it will be discussed in the last section. Hence, we start the case n ≥ 3. We prove:

Theorem 3.1. Assume that (X, 0) ⊂ (ℂ^{n+r}, 0) is an ICIS of dimension n > 2 and codimension r, equisingular to a homogeneous ICIS of multidegree (p_{1}, …, p_{r}). Then C_{n,r} · p_{g} ≤ μ.

In the next discussions we assume p_{i} ≥ 2 for all i; if p_{i} = 1 for some i then one can reduce the setup to the smaller (r − 1)–codimensional case. In addition, as all the formulas are symmetric in {p_{i}}_{i}, we will sometimes assume that p_{r} is largest among all the p_{i}’s. Hence we wish to prove:

(8) LHS := ∏ p_{i} ∑_{j=0}^{n}(−1)^{j}( ∑_{\underline{k}∈K_{n−j,r}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}}) − (−1)^{n} ≥ C_{n,r} · ∏ p_{i} · ∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1} =: RHS.

The proof consists of several steps.

Theorem 3.2. (1) The inequality (8) is the consequence of the next inequality:

∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}} ≥ C_{n,r}( ∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1} + ∑_{\begin{matrix}\underline{k}∈K_{n,r} \\ k_{r}>0 \end{matrix}} \binom{p_{r} − 2}{k_{r} − 1}\frac{1}{k_{r}(k_{r} + 1)} ∏_{i=1}^{r−1} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1}).

This inequality is the consequence of two further inequalities, listed in part (2) and (3):

(2) For n ≥ 3 the following inequality holds:

∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r} \frac{(p_{i} − 1)^{k_{i}}}{(k_{i} + 1)!} ≥ ∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1} + ∑_{\begin{matrix}\underline{k}∈K_{n,r} \\ k_{r}>0 \end{matrix}} \binom{p_{r} − 2}{k_{r} − 1}\frac{1}{k_{r}(k_{r} + 1)} ∏_{i=1}^{r−1} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1}.

(3) For any n ≥ 2 and r ≥ 1 the following inequality holds:

∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r}(p_{i} − 1)^{k_{i}} ≥ C_{n,r} ∑_{\underline{k}∈K_{n,r}} ∏_{i=1}^{r} \frac{(p_{i} − 1)^{k_{i}}}{(k_{i} + 1)!} .

Here the third statement (3) is the most complicated one, it follows from the general Combinatorial Inequality from the Introduction, and it is proved separately in the next section.

Proof. (1) Expand the LHS of (8) in terms with decreasing r:

(9)  LHS = (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_{s}) := ∑_{\underline{k}∈K_{n,s}} ∏_{i=1}^{s}(p_{i} − 1)^{k_{i}}, for 1 ≤ s ≤ r. It is natural to expand the right hand side of (8) similarly.

For this, define D_{n}(p_{1}, …, p_{s}) := ∑_{\underline{k}∈K_{n,s}} ∏_{i=1}^{s} \binom{p_{i}−1}{k_{i}} \frac{1}{k_{i}+1}. E.g., D_{1}(p_{1}) = \binom{p_{1}−1}{n} \frac{1}{n+1}. Then, we write RHS/C_{n,r} as

((∏_{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, namely:

(p_{s} − 1)(∏_{i=1}^{s−1} p_{i})C_{n}(p_{1}, …, p_{s}) ≥ C_{n,r}((∏_{i=1}^{s} p_{i})D_{n}(p_{1}, …, p_{s}) − (∏_{i=1}^{s−1} p_{i})D_{n}(p_{1}, …, p_{s−1})), for all 1 ≤ s ≤ r.

Since C_{n,r} ≤ C_{n,s}, cf (3), it is enough to prove the last inequality with coefficient C_{n,s} instead of C_{n,r}, or equivalently,

(p_{r} − 1)(∏_{i=1}^{r−1} p_{i})C_{n}(p_{1}, …, p_{r}) ≥ C_{n,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 all r ≥ 1, n ≥ 1.

Further, split the right hand side of this last inequality into two parts:

(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})),

and rewrite D_{n}(p_{1}, …, p_{r}) − D_{n}(p_{1}, …, p_{r−1}) into

∑_{\begin{matrix}\underline{k}∈K_{n,r} \\ k_{r}>0 \end{matrix}} ∏_{i=1}^{r} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1} = (p_{r} − 1) ∑_{\begin{matrix}\underline{k}∈K_{n,r} \\ k_{r}>0 \end{matrix}} \binom{p_{r} − 2}{k_{r} − 1}\frac{1}{k_{r}(k_{r} + 1)} ∏_{i=1}^{r−1} \binom{p_{i} − 1}{k_{i}} \frac{1}{k_{i} + 1}.

This provides precisely the expression of part (1).

(2) We start to compare individually the particular summands indexed by \underline{k} ∈ K_{n,r} of both sides. First, we consider some \underline{k} ∈ K_{n,r} with k_{r} > 1. Then the corresponding individual inequality is true. Indeed,

\frac{∏_{i=1}^{r} \frac{(p_{i}−1)^{k_{i}}}{k_{i}!} − ∏_{i=1}^{r} \binom{p_{i}−1}{k_{i}}}{∏_{i=1}^{r}(k_{i}+1)} − \binom{p_{r}−2}{k_{r}−1} \frac{1}{k_{r}(k_{r}+1)} ∏_{i=1}^{r−1} \binom{p_{i}−1}{k_{i}} \frac{1}{k_{i}+1} ≥ ^{\frac{\frac{(pr −1)^{kr −1}}{(kr −1)!} −\binom{pr −2}{kr −1}}{k_{r}(k_{r}+1)}} ∏_{i=1}^{r−1} \binom{p_{i}−1}{k_{i}} \frac{1}{k_{i}+1} ≥ 0.

Here and in the sequel we constantly use (p_{i} − 1)^{k_{i}} − (p_{i} − 1) · (p_{i} − 2) ⋯ (p_{i} − k_{i}) ≥ (p_{i} − 1)^{k_{i}−1}, valid for k_{i} > 1. Next, we assume that \underline{k} has the following properties: k_{r} = 1, but k_{j} > 1 for some j. Then again

\frac{p_{r}−1}{2} · ^{\frac{∏_{i=1}^{r−1} \frac{(p_{i}−1)^{k_{i}}}{k_{i}!} − ∏_{i=1}^{r−1} \binom{p_{i}−1}{k_{i}}}{∏_{i=1}^{r}(k_{i}+1)}} − \frac{1}{2} ∏_{i=1}^{r−1} \binom{p_{i}−1}{k_{i}} \frac{1}{k_{i}+1} ≥ \frac{p_{r}−1}{2} · \frac{(p_{j}−1)^{k_{j} −1}}{(k_{j}+1)!} ∏_{\begin{matrix}1≤i<r \\ i≠j \end{matrix}} \frac{(p_{i}−1)^{k_{i}}}{(k_{i}+1)!} − \frac{1}{2} ∏_{i=1}^{r−1} \binom{p_{i}−1}{k_{i}} \frac{1}{k_{i}+1} ≥ 0.

Here we used the initial assumption that p_{r} ≥ p_{i} for any i.

Now, we assume that k_{r} = 1 and k_{i} ≤ 1 for all i. This can happen only for n ≤ r. Let i_{1}, …, i_{n−1} be those indices different than r for which k_{i} = 1, that is, k_{r} = k_{i_{1}} = ⋯ = k_{i_{n−1}} = 1 and all the other k_{i}’s are zero.

In this case ∏_{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}, hence the individual inequality corresponding to \underline{k} fails. Therefore, we will group this term by some other terms with k_{r} = 0. More precisely, we will group this \underline{k} together with terms which correspond to those \underline{k}’s which satisfy k_{r} = 0, k_{i_{j}} = 2 for exactly one j ∈ \{1, …, n−1\}, k_{i_{l}} = 1 if l ∈ \{1, …, n−1\}\\{j\}, and all the other k_{i}’s are zero.

Note that if k_{r} = 0 then there is no contribution from the sum ∑_{\underline{k}∈K_{n,r}, k_{r}>0}. Therefore, the n individual inequality

corresponding to the above \underline{k}’s altogether provide

∑_{l=1}^{n−1} \frac{(p_{i_{l}} − 1) − (p_{i_{l}} − 2)}{3} · ∏_{l=1}^{n−1} \frac{p_{i_{l}} − 1}{2} − \frac{1}{2} ∏_{l=1}^{n−1} \frac{p_{i_{l}} − 1}{2} = \frac{2n − 5}{6} · ∏_{l=1}^{n−1} \frac{p_{i_{l}} − 1}{2}.

For n ≥ 3 this is positive, hence the statement.

Any other remaining \underline{k} ∈ K_{n,r} can again be treated individually: for all of them k_{r} = 0 and \frac{(p−1)^{k}}{(k+1)!} ≥ \binom{p−1}{k} \frac{1}{k+1}.

4. Proof of the Combinatorial Inequality

In the proof we will use some further notations. We will consider the following partition of K_{n,r}: for any s ∈ {0, …, r − 1} we define

(10)  K_{n,r}^{s} := \{\underline{k} ∈ K_{n,r} : |\{i : k_{i} = 0\}| = s\}.

Note that for s < r−n one has K_{n,r}^{s} = ∅. Corresponding to these sets, we consider the arithmetic mean X_{n,r} and X_{n,r}^{s} of the elements \underline{x}\underline{k} indexed by the sets K_{n,r} and K_{n,r}^{s} respectively. In parallel, Y_{n,r,ℓ} and Y_{n,r,ℓ}^{s} denote the arithmetic mean of elements \underline{y}_{\underline{k},ℓ} indexed by the same sets K_{n,r} and K_{n,r}^{s} respectively.

Theorem 4.1. With the above notation one has (a) X_{n,r}^{0} ≤ X_{n,r}^{1} ≤ ⋯ ≤ X_{n,r}^{r−1}

(b) Y_{n,r,ℓ}^{0} > Y_{n,r,ℓ}^{1} > ⋯ > Y_{n,r,ℓ}^{r−1}

(c) (a) and (b) imply the Combinatorial Inequality from the Introduction.

Proof. (a) By definition, X_{n,r}^{s} is the arithmetic mean of \binom{r}{s}\binom{n−1}{r−s−1} monomials. Let X̃_{n,r}^{s}(x_{1}, …, x_{r}) be the sum of these monomials, that is, X̃_{n,r}^{s}(x_{1}, …, x_{r}) = X_{n,r}^{s} · \binom{r}{s}\binom{n−1}{r−s−1}.

Step 1. We show that all the inequalities can be deduced from the first one: X_{n,r}^{1} ≥ X_{n,r}^{0}. Indeed, by definition, X̃_{n,r}^{s}(x_{1}, …, x_{r}) can be written as the summation over all the subsets of {1, …, r} with (r − s) elements:

(11) \begin{matrix}X̃_{n,r}^{s}(x_{1}, …, x_{r}) = ∑_{\{i_{1},...,i_{r−s}\}⊂\{1,...,r\}} x_{i_{1}} ⋯ x_{i_{r−s}} X̃_{n−r+s,r−s}(x_{i_{1}}, …, x_{i_{r−s}}) \\ = ∑_{\{i_{1},...,i_{r+1−s}\}⊂\{1,...,r\}} \frac{1}{s} ∑_{j∈\{i_{1},...,i_{r+1−s}\}} \frac{x_{i_{1}} ···x_{i_{r+1−s}}}{x_{j}} · X̃_{n−r+s,r−s}(x_{i_{1}}, …, x̂_{j}, …, x_{i_{r+1−s}}). \end{matrix}

Here in the second line the notation x̂_{j} means that the variable x_{j} is omitted. Note that in the second line the summation is as the summation in X̃_{n,r}^{s−1}(x_{1}, …, x_{r}), hence these terms can be combined. From (11) one gets

X_{n,r}^{s}(x_{1}, …, x_{r}) = ∑_{\{i_{1},...,i_{r+1−s}\}⊂\{1,...,r\}} ∑_{j∈\{i_{1},...,i_{r+1−s}\}} \frac{x_{i_{1}} ⋯ x_{i_{r+1−s}}}{x_{j}} · \frac{X̃_{n−r+s,r−s}(x_{i_{1}}, …, x̂_{j}, …, x_{i_{r+1−s}})}{s\binom{r}{s}\binom{n−1}{r−s−1}},

and by similar argument X_{n,r+1−s}^{1}(x_{i_{1}}, …, x_{i_{r+1−s}}) equals

∑_{\{i_{1},...,i_{r+1−s}\}⊂\{i_{1},...,i_{r+1−s}\}} ∑_{j∈\{i_{1},...,i_{r+1−s}\}} \frac{x_{i_{1}} ⋯ x_{i_{r+1−s}}}{x_{j}} · \frac{X̃_{n−r+s,r−s}(x_{i_{1}}, …, x̂_{j}, …, x_{i_{r+1−s}})}{\binom{r+1−s}{1}\binom{n−1}{r−s−1}}.

These two identities and s\binom{r}{s}\binom{n−1}{r−s−1} = \binom{r+1−s}{1}\binom{n−1}{r−s−1}\binom{r}{s−1} provide

X_{n,r}^{s}(x_{1}, …, x_{r}) = ∑_{\{i_{1},...,i_{r+1−s}\}⊂\{1,...,r\}} \frac{1}{\binom{r}{s−1}} · X_{n,r+1−s}^{1}(x_{i_{1}}, …, x_{i_{r+1−s}}).

Using the first line of (11), by similar comparison we get

X_{n,r}^{s−1}(x_{1}, …, x_{r}) = ∑_{\{i_{1},...,i_{r+1−s}\}⊂\{1,...,r\}} \frac{1}{\binom{r}{s−1}} · X_{n,r+1−s}^{0}(x_{i_{1}}, …, x_{i_{r+1−s}}).

Therefore,

X_{n,r}^{s} − X_{n,r}^{s−1} = ∑_{\{i_{1},...,i_{r+1−s}\}⊂\{1,...,r\}} \frac{1}{\binom{r}{s−1}}(X_{n,r+1−s}^{1}(x_{i_{1}}, …, x_{i_{r+1−s}}) − X_{n,r+1−s}^{0}(x_{i_{1}}, …, x_{i_{r+1−s}})).

Thus, if the inequality X_{n′,r′}^{1} ≥ X_{n′,r′}^{0} is satisfied for any n′ ≤ n and r′ ≤ r, then X_{n′,r′}^{s} ≥ X_{n′,r′}^{s−1} is also satisfied for any n′ ≤ n, r′ ≤ r and 0 ≤ s ≤ r′ − 1.

Step 2. We prove X_{n,r}^{1} ≥ X_{n,r}^{0}, or, equivalently, \frac{X̃_{n,r}^{1}}{r(r−1)} ≥ \frac{X̃_{n,r}^{0}}{n−r+1}.

Note that X̃_{n,r}^{0} = (∏_{i=1}^{r} x_{i})X̃_{n−r,r} and similarly X̃_{n,r}^{1} = (∏_{i=1}^{r} x_{i}) ∑_{j=1}^{r} \frac{X̃_{n−r+1,r−1}(x_{1},...,x̂_{j},...,x_{r})}{x_{j}}. Both X̃_{n−r,r}

and X̃_{n−r+1,r−1} can be decomposed further according to the s–types: X̃_{n−r,r} = ∑_{s=0}^{r−1} X̃_{n−r,r}^{s} and X̃_{n−r+1,r−1} = ∑_{s=0}^{r−2} X̃_{n−r+1,r−1}^{s}. We set

X̃_{n,r}^{0,s} := (∏_{i=1}^{r} x_{i})X̃_{n−r,r}^{s} and X̃_{n,r}^{1,s} := (∏_{i=1}^{r} x_{i}) ∑_{j=1}^{r} \frac{X̃_{n−r+1,r−1}^{s}(x_{1}, …, x̂_{j}, …, x_{r})}{x_{j}}.

We claim that for any 0 ≤ s ≤ r − 2 one has:

(12)  (r − s − 1) · X̃_{n,r}^{1,s} ≥ (r − s − 1)(s + 1) · X̃_{n,r}^{0,s+1} + (s + 1)(s + 2) · X̃_{n,r}^{0,s+2}.

This follows from the ‘elementary’ inequality (I_{i_{1}i_{2}}^{k}) : x_{i_{1}}^{k} + x_{i_{2}}^{k} ≥ x_{i_{1}}^{k−1}x_{i_{2}} + x_{i_{1}}x_{i_{2}}^{k−1}, where {i_{1}, i_{2}} ⊂ {1, …, r} and k ≥ 2. Indeed, for any fixed pair (i_{1}, i_{2}) consider all the monomials of type M = ∏_{i=1}^{r} x_{i}^{m_{i}}/(x_{i_{1}}^{m_{i_{1}}} x_{i_{2}}^{m_{i_{2}}} ) with m_{i} > 0 and of degree n − k. Then (x_{i_{1}}^{k} + x_{i_{2}}^{k}) · M ∈ X̃_{n,r}^{1,s}. Moreover, each monomial \underline{x}\underline{k} ∈ X̃_{n,r}^{1,s} can be realized by exactly (r − 1 − s) ways, where (r − 1 − s) stays for the number of k_{i}’s with k_{i} ≥ 2.

Consider next the same monomial M as before. Then \overline{M} := (x_{i_{1}}^{k−1}x_{i_{2}} + x_{i_{1}}x_{i_{2}}^{k−1}) · M ∈ X̃_{n,r}^{0}. If k > 2 then the number of exponents in \overline{M} which equal 1 is s + 1, hence \overline{M} ∈ X̃_{n,r}^{0,s+1}. If k = 2 then \overline{M} ∈ X̃_{n,r}^{0,s+2}.

Any monomial \underline{x}\underline{k} ∈ X̃_{n,r}^{0,s+1} can be realized by exactly (r − 1 − s)(s + 1) ways, which is |\{i : k_{i} ≥ 2\}| · |\{i : k_{i} = 1\}|, the number of possible candidates for the pair (i_{1}, i_{2}) for (I_{i_{1}i_{2}}^{k}). Furthermore, any monomial \underline{x}\underline{k} ∈ X̃_{n,r}^{0,s+2} can be realized by exactly (s + 1)(s + 2) ways, the possible ordered pairs of {i : k_{i} = 1}.

Hence all the possible inequalities (I_{i_{1}i_{2}}^{k}) multiplied by the possible monomials provide exactly (12). Note that the above monomial counting is compatible with the identity obtained from |K_{n,r}^{s}| = \binom{r}{s}\binom{n−1}{r−s−1} and

r · |K_{n−r+1,r−1}^{s}| = (s + 1) · |K_{n−r,r}^{s+1}| + \frac{(s + 1)(s + 2)}{r − 1 − s} · |K_{n−r,r}^{s+2}|.

Now, by taking the sum over s in (12), and by regrouping the right hand side we obtain

X̃_{n,r}^{1} ≥ (∏_{i=1}^{r} x_{i}) ∑_{s=0}^{r−2} ((s + 1)X̃_{n−r,r}^{s+1} + \frac{(s + 2)(s + 1)}{r − s − 1}X̃_{n−r,r}^{s+2}) = (∏_{i=1}^{r} x_{i}) ∑_{s=1}^{r−1} \frac{sr}{r − s + 1}X̃_{n−r,r}^{s}

Note that in the last sum the term with s = 0 can also be included, as its coefficient vanishes. Thus (for some c > 0):

(13) c(X_{n,r}^{1} − X_{n,r}^{0}) = \frac{n − r + 1}{r(r − 1)} · X̃_{n,r}^{1} − X̃_{n,r}^{0} ≥ (∏_{i=1}^{r} x_{i}) ∑_{s=0}^{r−1} (\frac{s(n − r + 1)}{(r − s + 1)(r − 1)} − 1)\binom{r}{s}\binom{n − r − 1}{r − s − 1} · X_{n−r,r}^{s}.

Next, the right hand side of (13) is non-negative by the following generalization of the Chebyshev’s sum inequality (which basically is the summation ∑_{s,t}(α_{s}α_{t}β_{s} − α_{s}α_{t}β_{t})(x_{s} − x_{t}) ≥ 0)

(14)  ( ∑_{s} α_{s}β_{s})( ∑_{s} α_{s}x_{s}) ≤ ( ∑_{s} α_{s})( ∑_{s} α_{s}β_{s}x_{s})

whenever x_{s} and β_{s} are both decreasing (or both increasing) sequences and α_{s} > 0. In the present case take α_{s} := \binom{r}{s}\binom{n−r−1}{r−s−1}, β_{s} := \frac{s(n−r+1)}{(r−s+1)(r−1)} and x_{s} := X_{n−r,r}^{s}. Clearly β_{s} is increasing, x_{s} is increasing by induction, and ∑_{s} α_{s} = ∑_{s} β_{s}α_{s} by a computation based on ∑_{s} \binom{p}{s}\binom{q}{m−s} = \binom{p+q}{m}. Hence, via (13), X_{n,r}^{1} ≥ X_{n,r}^{0}.

(b) As in part (a), we first reduce the general statement to Y_{n,r,ℓ}^{0} > Y_{n,r,ℓ}^{1}, and then we prove this particular case too. As in the previous case, set Ỹ_{n,r,ℓ}^{s} := ∑_{\underline{k}∈K_{n,r}^{s}} y_{\underline{k},ℓ} = \binom{r}{s}\binom{n−1}{r−s−1}Y_{n,r,ℓ}^{s}. We wish to prove:

(15)  \frac{Ỹ_{n,r,ℓ}^{s}}{(s + 1)(n − r + s + 1)} > \frac{Ỹ_{n,r,ℓ}^{s+1}}{(r − s)(r − s − 1)}.

Step 1. Use the decomposition K_{n,r}^{s} = ∐ K_{n,r−s}^{0}, the disjoint union of \binom{r}{s} copies, to get:

(16)  Ỹ_{n,r,ℓ}^{s} = ∑_{∑ k_{i}=n ; |\{i:k_{i}=0\}|=s} ∏_{i=1}^{r} \frac{1}{(k_{i} + ℓ)!} = \binom{r}{s}\frac{1}{(ℓ!)^{s}} Ỹ_{n,r−s,ℓ}^{0}.

Similarly, Ỹ_{n,r,ℓ}^{s+1} = \binom{r}{s+1} \frac{1}{(ℓ!)^{s+1}} Ỹ_{n,r−s−1,ℓ}^{0} = \binom{r}{s} \frac{1}{(ℓ!)^{s}} \frac{1}{s+1} Ỹ_{n,r−s,ℓ}^{1}. Here we used Ỹ_{n,r−s,ℓ}^{1} = \frac{r−s}{ℓ!} Ỹ_{n,r−s−1,ℓ}^{0}, cf. (16). Therefore, the inequality (15) for any s is equivalent to (15) for s = 0.

Step 2. Here we prove \frac{Ỹ_{n,r,ℓ}^{0}}{n−r+1} > \frac{Ỹ_{n,r,ℓ}^{1}}{r(r−1)}. We run induction on n: we assume the stated inequalities, indexed by (n, r, ℓ), is true for any (n′, r, ℓ′) with n′ < n (but ℓ′ can be larger than ℓ). In fact, we use (n − r, r, ℓ + 1) ⇒ (n, r, ℓ).

We consider exactly the same combinatorial set–decomposition as in Step 2 of (a), the only difference is that we replace the inequality (I_{i_{1}i_{2}}^{k}), written for k ≥ 2, by

(I_{i_{1}i_{2},ℓ}^{k}) : \frac{2}{(k + ℓ)!ℓ!} < \frac{2}{(k − 1 + ℓ)!(ℓ + 1)!}.

After a computation, we obtain the analogue of (13) (for the same positive constant c), namely

(17) c(Y_{n,r,ℓ}^{1} − Y_{n,r,ℓ}^{0}) = \frac{n − r + 1}{r(r − 1)} · Ỹ_{n,r,ℓ}^{1} − Ỹ_{n,r,ℓ}^{0} < ∑_{s=0}^{r−1} (\frac{s(n − r + 1)}{(r − s + 1)(r − 1)} − 1)\binom{r}{s}\binom{n − r − 1}{r − s − 1} · Y_{n−r,r,ℓ+1}^{s}.

The right hand side of (17) is non-positive by (14) with reversed inequality, valid for α_{s} > 0, β_{s} and x_{s} oppositely ordered. Indeed, α_{s} and β_{s} are the same as before with β_{s} is increasing, while x_{s} := Y_{n−r,r,ℓ+1}^{s} is a decreasing by induction. Hence, via (17), Y_{n,r,ℓ}^{1} < Y_{n,r,ℓ}^{0}.

(c) The proof is double induction over r and n. We assume that for any fixed r and n the inequality (I_{n,r′,ℓ}) is true for any n and ℓ and r′ < r, and (I_{n′,r,ℓ}) is true for any n′ < n and any ℓ. We wish to prove (I_{n,r,ℓ}).

First we write the left hand side of the inequality as a sum

∑_{\underline{k}∈K_{n,r}} \underline{y}_{\underline{k},ℓ} · \underline{x}\underline{k} = ∑_{s=0}^{r−1} ∑_{\underline{k}∈K_{n,r}^{s}} \underline{y}_{\underline{k},ℓ} · \underline{x}\underline{k} .

Note that corresponding to s = 0, after we factor out x_{1} ⋯ x_{r}, the sum over K_{n,r}^{0} can be identified with the left hand side of the inequality (I_{n−r,r,ℓ+1}) (multiplied by x_{1} ⋯ x_{r}). Hence, by the inductive assumption,

∑_{\underline{k}∈K_{n,r}^{0}} \underline{y}_{\underline{k},ℓ} · \underline{x}\underline{k} ≤ Y_{n,r,ℓ}^{0} · ∑_{\underline{k}∈K_{n,r}^{0}} \underline{x}\underline{k}.

For s = 1, the sum over K_{n,r}^{1} is a sum of r sums corresponding to the ‘missing’ coordinate x_{i}, and each of them can be identified (after factorization of a monomial) with the inequality (I_{n−(r−1),r−1,ℓ+1}). For an arbitrary s ≤ r − 2 one can apply in the similar way the inequality (I_{n−(r−s),r−s,ℓ+1}). In the case of s = r − 1 all coefficients \underline{y}_{\underline{k},ℓ} equal [(l′)^{r−1}(n + l)!]^{−1}. Therefore, by induction, we get

∑_{s=0}^{r−1} ∑_{\underline{k}∈K_{n,r}^{s}} \underline{y}_{\underline{k},ℓ} · \underline{x}\underline{k} ≤ ∑_{s=0}^{r−1} |K_{n,r}^{s}| · Y_{n,r,ℓ}^{s} · X_{n,r}^{s}.

But, using parts (a) and (b), by Chebyshev’s sum inequality ((14) with α_{s} = 1):

∑_{s=0}^{r−1} |K_{n,r}^{s}| · Y_{n,r,ℓ}^{s} · X_{n,r}^{s} ≤ |K_{n,r}| · Y_{n,r,ℓ} · X_{n,r},

whose right hand side is the left hand side of Combinatorial Inequality. This ends the proof of (c).

The above discussion and the statement of Theorem 4.1(b) imply the inequality (3) from the introduction as well. This is a proof of (3) in the spirit of the Combinatorial Inequality (based on Chebyshev’s type inequalities), for a different proof see [Kerner-Némethi.a].

Corollary 4.2. C_{n,r} > C_{n,r+1} for any n ≥ 1 and r ≥ 1.

Proof. By (1) the inequality C_{n,r} > C_{n,r+1} is equivalent to Y_{n,r} < Y_{n,r+1}. We drop the index ℓ = 1 from the notations (hence, we write e.g. Y_{n,r} := Y_{n,r,1}), and we set Y_{n,r}^{≥1} := ∪_{s≥1}Y_{n,r}^{s}, and similarly Ỹ_{n,r}^{≥1} and K_{n,r}^{≥1}. By 4.1, part (b), the mean Y_{n,r+1}^{0} is the largest among {Y_{n,r+1}^{s}}_{s}, hence Y_{n,r+1} > Y_{n,r}^{≥1}. Hence, we need to prove Y_{n,r}^{≥1} ≥ Y_{n,r}. They can be decomposed as sums over the same index set. Indeed, by similar arguments as in the previous proof part (b)

Ỹ_{n,r+1}^{≥1} = ∑_{s=0}^{r−1} \binom{r + 1}{s + 1} · Ỹ_{n,r−s}^{0} = ∑_{s=0}^{r−1} \begin{matrix}( & {r+1 \atop s+1} & ) \\ ( & {r \atop s} & ) \end{matrix} · Ỹ_{n,r}^{s} = ∑_{s=0}^{r−1} \begin{matrix}( & {r+1 \atop s+1} & ) \\ ( & {r \atop s} & ) \end{matrix} · \binom{r}{s}\binom{n − 1}{r − s − 1} · Y_{n,r}^{s}.

This can be rewritten as

(18)  Y_{n,r+1}^{≥1} = \frac{1}{|K_{n,r+1}^{≥1}|} · ∑_{s=0}^{r−1} \frac{r + 1}{s + 1}\binom{r}{s}\binom{n − 1}{r − s − 1} · Y_{n,r}^{s}.

Similarly,

(19)  Y_{n,r} = \frac{1}{|K_{n,r}|} · ∑_{s=0}^{r−1} \binom{r}{s}\binom{n − 1}{r − s − 1} · Y_{n,r}^{s}.

Hence, via (18) and (19), the inequality Y_{n,r} ≤ Y_{n,r}^{≥1} follows from (14) applied for α_{s} = \binom{r}{s}\binom{n−1}{r−s−1}, β_{s} = \frac{r+1}{s+1} and x_{s} = Y_{n,r}^{s}. Indeed, α_{s} > 0 while x_{s} and β_{s} are both decreasing sequences. For x_{s} this follows from Theorem 4.1(b).

5. The homogeneous case for n = 2

Assume that (X, 0) is a 2–dimensional ICIS with p_{i} ≥ 2 for all i. Then C_{2,r} = 4\frac{r+1}{r+ \frac{1}{3}} . Set P := ∏_{i} p_{i}, the multiplicity of (X, 0). Then using 2.1 one obtains

\frac{p_{g}}{P} = ∑_{i} \frac{(p_{i} − 1)(p_{i} − 2)}{6} + ∑_{i<j} \frac{(p_{i} − 1)(p_{j} − 1)}{4} and \frac{μ + 1 − P}{P} = ∑_{i} ((1 − p_{i}) + (1 − p_{i})^{2}) + ∑_{i<j}(1 − p_{i})(1 − p_{j}).

By a computation μ + P · E + 1 = C_{2,r} · p_{g}, where E := \frac{r−1}{3r+1} ∑_{i=1}^{r}(p_{i} − 1) − ∑_{i<j} \frac{(p_{i}−p_{j})^{2}}{3r+1} − 1.

If r = 1 then E = −1, but for r ≥ 2 and for some choices of p_{i}’s (e.g. whenever they are all equal) E might be positive, providing C_{2,r} · p_{g} ≥ μ + 1. We collect here the precise statements:

Theorem 5.1. (a) If r = 1 then 6p_{g} = μ + 1 − multiplicity.

(b) If r ≥ 2 then

C_{2,r} · p_{g} = 4 · \frac{r + 1}{r + 1/3} · p_{g} ≤ μ + 1

in general is \underline{not} true, although asymptotically \frac{μ}{p_{g}} tends to \frac{4(r+1)}{r+1/3}.

(c) For any r the inequality

4p_{g} ≤ μ + 1 − multiplicity

is valid, and if p_{i} = 2 is allowed then 4 is the sharpest bound whenever r > 1.

(d) If p_{i} ≥ d + 1 for all i then

4 · \frac{d(r − 1) + 2(d − 1)}{d(r − 1) + \frac{4}{3}(d − 1)} · p_{g} ≤ μ + 1 − multiplicity.

Proof. Using the above explicit formulae, all the statements are elementary. Let us give some hint for (d). Note that (4 + c)p_{g} ≤ μ + 1 − P reads as

(20)  (4 + c) ∑_{i} \frac{(p_{i} − 1)(p_{i} − 2)}{6} + c ∑_{i<j} \frac{(p_{i} − 1)(p_{j} − 1)}{4} ≤ ∑_{i}(1 − p_{i}) + ∑_{i}(1 − p_{i})^{2}.

Using ∑_{i=1}^{r} a_{i}^{2} ≥ \frac{2}{r−1} ∑_{i<j} a_{i}a_{j}, the inequality (20) follows from

(4 + c) ∑_{i} \frac{(p_{i} − 1)(p_{i} − 2)}{6} + c · \frac{r − 1}{8} ∑_{i}(p_{i} − 1)^{2} ≤ ∑_{i}(1 − p_{i}) + ∑_{i}(1 − p_{i})^{2}.

This is a sum over i of elementary quadratic inequalities whose discussion is left to the reader.

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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