On the monodromy of curve singularities

A. Némethi^{1}, Department of Mathematics, The Ohio State University, Columbus, OH 43210, U.S.A.

J.H.M. Steenbrink, Department of Mathematics, University of Nijmegen, Toernooiveld, 6525 ED Nijmegen, The Netherlands.

Abstract. We consider an analytic germ f : (X, x) → (C, 0) defined on a normal surface singularity (X, x). Let T_{λ} be the algebraic monodromy of f restricted to its generalized λ–eigenspace. With some restrictions for f and λ, we establish criteria for the finiteness of T_{λ}. Examples show that these restrictions are natural.

Introduction

Let (X, x) be a normal surface singularity and f : (X, x) → (C, 0) the germ of an analytic function. Choose an embedding (X, x) ↪ (C^{N}, 0). Take ε, η ∈ R with 0 < η ≪ ε ≪ 1 and put

E^{∗} = {z ∈ X : |z| < ε and |f(z)| = η}.

Then f|_{E^{∗}} : E^{∗} → {t ∈ C : |t| = η} is a C^{∞} fibre bundle. Let F denote its typical fiber and h : F → F its characteristic homeomorphism. Then h induces the monodromy transformation T : H^{1}(F, C) → H^{1}(F, C). The following result is a particular case of a well known general theorem (see [3,8,10]).

Monodromy Theorem. The eigenvalues of T are roots of unity. Moreover, the Jordan blocks of T are of size at most 2. If X is smooth at x then the Jordan blocks for eigenvalue 1 are of size at most 1.

For irreducible curves there is a stronger result.

Theorem.[9] Assume that X is smooth and f is locally irreducible at x. Then the monodromy T is of finite order.

1Supported by the Netherlands Organisation for the Advancement of Scientific Research N.W.O.

In this theorem there are two crucial assumptions. If one drops the irreducibility condition, then the result is not true anymore; the first example was constructed by A’Campo [1], who proved that in the monodromy transformation of f = (x^{2} + y^{3})(x^{3} + y^{2}) there is a 2 × 2 Jordan block with eigenvalue = −1.

On the other hand, the smoothness of X is essential too. In order to explain this, we introduce some notations.

Choose an embedded resolution π : X̃ → X of f, i.e. X̃ is smooth, π is bimeromorphic and (f ∘ π)^{−1}(0) is a divisor with normal crossings on X̃. Denote by Γ its resolution graph. Let h_{X} = rankH_{1}(Γ) be the number of independent cycles of Γ (alternatively, h_{X} = rankH_{1}(E) − rankH_{1}(Ẽ), where Ẽ is a smooth model for E = π^{−1}(x)). Our first result sharpens the result of Durfee [4], which says that the monodromy is finite if and only if the resolution of any cyclic covering of X ramified along {f = 0} contains no cycles.

Theorem A. Let X be a normal surface singularity and let T be the monodromy transformation of a non-zero holomorphic germ f : (X, x) → (C, 0). Then the number of 2 × 2 Jordan blocks with eigenvalue 1 is exactly h_{X}.

In particular, if we want to obtain, by a general theorem, the finiteness of T, we have to assume that Γ is a tree (i.e. h_{X} = 0) and f^{−1}(0) ⊂ X is irreducible (as a Weil divisor).

Let K_{X} = {z ∈ X : r(z) = ε} (0 < ε ≪ 1) be the link of X. Let d_{X} be the order of the torsion part of H_{1}(K_{X}, Z). If A denotes the intersection matrix of the irreducible exceptional divisors provided by a resolution of X, then d_{X} = | det A| (see, for example, [7]).

Our second result is the following generalisation of Lê’s Theorem:

Theorem B. Let (X, x) be a normal surface singularity with h_{X} = 0 and f : (X, x) → (C, 0) an analytic germ such that (f^{−1}(0), x) ⊂ (X, x) is an irreducible Weil divisor.

a) If there exists a 2 × 2 Jordan block of T with eigenvalue of order k, then there exist a positive integer α and a prime p such that p^{α} divides k and T has a 2 × 2 Jordan block for some eigenvalue of order p^{α}.

b) Assume that there is a 2×2 Jordan block of T with eigenvalue of order k. Then there exists a prime divisor p of k such that p^{2} divides d_{X}.

In particular, if d_{X} is squarefree, then T has finite order.

The proof of Theorem A and B is contained in the second subsection. The next examples show that the monodromy T can be of infinite order

provided that d_{X} is divisible by p^{2} for some prime p.

Example 1. Consider the decorated graph

[figure: 1; 1; 2; 2; 1; 1; 1; (-2); (-2); (-2); (-2); (-2); (-3)]

By [6, p. 367] there is a normal surface singularity (X, x) and a germ f : (X, x) → (C, 0) such that (one of) the resolution graphs of f has the above form with all irreducible exceptional divisors rational. The numbers in brackets denote the selfintersection numbers, the other numbers are the multiplicities of f along the corresponding divisor. Since all multiplicities are positive and the intersection matrix A is negative definite, Grauert’s algebraicity condition works. By a computation one has d_{X} = | det A| = 4.

In fact, (X, x) can be taken as the germ at 0 of the surface with equation x^{10} + x^{3}y^{2} + y^{3} + z^{2} = 0 in C^{3} and f(x, y, z) = x.

Now it can be proved, that the monodromy transformation contains (exactly) one Jordan block of size 2 with eigenvalue = −1. To see this, use the facts in the proof of Theorem B. Or, a different argument could be the following. Consider the double covering (X_{2}, x_{2}) →^{c} (X, x) ramified along {f = 0}, and put f_{2} = f ∘ c. The resolution graph of f_{2} is:

[figure: 1; 1; 1; 1; (-2); (-3); (-2)]

Then X_{2} is the germ at 0 of the zero set of x^{20} + x^{6}y^{2} + y^{3} + z^{2}. Notice that the monodromy of f_{2} is the square of the monodromy of f, therefore the result follows from Theorem A.

Example 2. Let p be a prime number. Consider the following resolution graph:

[figure: 1; 1; 1; p; p; 1; 1; 1; (-p); (-p); (-2); (-2); (-p); (-p); ⋮; ⋮]

The notations are the same as in the first example. In the left hand side there are p nodes and in the right hand side p − 1 nodes (all together 2p + 1 nodes and one arrow).

Then d_{X} = p^{2p−2}, and the monodromy transformation of f has (exactly) one Jordan block of size 2 with eigenvalue exp(2πik/p) for any k = 1, …, p − 1.

Example 3. Assume that (X, x) is a singularity of type A_{k}. Then d_{X} = k + 1. The first case not covered by Theorem B is A_{3}. If we represent it by the equation x^{2} + xy^{2} + y^{5} + z^{2} = 0 and take f(x, y, z) = x we find an irreducible curve singularity on it with T of infinite order.

The proof of Theorem A and B

Let π : X̃ → X be aa embedded resolution of f and X̃_{0} = (f ∘ π)^{−1}(0). Let D be the set of irreducible exceptional divisors {E_{d}}_{d∈D}, let r denote the cardinality of D and for each d ∈ D, let m_{d} denote the multiplicity of f ∘ π along the divisor E_{d}. Let g_{d} be the genus of E_{d} and take g = ∑_{d∈D} g_{d}. Let s be the number of connected components of f^{−1}(0) ∩ K_{X}, i.e. the number of branches of f^{−1}(0) at x.

Lemma. Let H_{1} denote the generalized eigenspace of T for the eigenvalue 1. Then

dim(H_{1}) = 2g + 2h_{X} + s − 1.

Proof. From A’Campo’s formula for the characteristic polynomial of T (cf.[2] Thm. 4) we deduce that dim(H_{1}) = 1 − e(U) where U is the regular locus of (X̃_{0})_{red}, and e stands for Euler number. Note that h_{X} = δ − r + 1 where δ is the number of double points of (X̃_{0})_{red}. We have e(U) = ∑_{d∈D}(2 − 2g_{d}) − s − 2δ and the lemma follows.

Lemma. dim ker(T − I)) = 2g + h_{X} + s − 1.

Proof. By the Wang exact sequence

0 → C → H^{1}(K_{X} \ {f = 0}) → H^{1}(F) →^{T −1} H^{1}(F)

one has dim ker(T − 1) = dim H^{1}(K_{X} \ {f = 0}) − 1. First notice that K_{X} can be identified with the boundary of X̃, therefore H^{q}(X̃, K_{X}) = H_{4−q}(X̃) by Lefschetz duality theorem ([11], p. 198). Since H_{k}(X̃) = H_{k}(X̃_{0}), the above duality gives H^{1}(X̃, K_{X}) = 0. Consider the exact sequence of the pair (X̃, K_{X}):

0 → H^{1}(X̃) → H^{1}(K_{X}) → H^{2}(X̃, K_{X}) →^{j} H^{2}(X̃) →^{i} H^{2}(K_{X}).

Since the intersection form on X̃ is (negative) definite j is an isomorphism. Thus dim H^{1}(K_{X}) = dim H^{1}(X̃) = 2g + h_{X}. Moreover, i is trivial.

Now, consider the exact sequence of the pair (K_{X}, K_{X} \ {f = 0}). Then H^{q} := H^{q}(K_{X}, K_{X} \ {f = 0}) = ⊕_{i=1}^{s}H^{q}(D^{2} × S^{1}, S^{1} × S^{1}) = 0 if q = 1, and = Z^{s} if q = 2. Moreover, the inclusion K_{X} ↪ X̃ induces an isomorphism H^{2} = L^{2}, where L^{2} := H^{2}(X̃, X̃ \ {f = 0}) = ⊕_{i=1}^{s}H^{q}(D^{2} × D^{2}, S^{1} × D^{2}). Therefore, the natural map H^{2} → H^{2}(K_{X}), via this isomorphism, is the composed map L^{2} → H^{2}(X̃) →^{i} H^{2}(K_{X}), i.e. it is trivial. Hence dim H^{1}(K_{X} \ {f = 0}) = dim H^{1}(K_{X}) + s.

¿From these two lemmas it follows that there are exactly h_{X} Jordan blocks of T_{1} of size 2.

Now, we continue with the proof of Theorem B, in particular, we assume that f^{−1}(0) is irreducible (i.e. s = 1). Let X_{k} for k ∈ N be the k-fold cyclic cover of X ramified over f = 0, i.e. the fibre product of f : X → C and λ_{k} : C → C where λ_{k}(z) = z^{k}. Let X̃_{k} be the normalization of the fibre product of fπ : X̃ → C and λ_{k}; this is a k-fold cyclic cover of X̃ ramified over X̃_{0}. Its fibre Y_{k} over 0 has only ordinary double points and X̃_{k} has cyclic quotient singularities at these double points. We let r_{k} denote the number of compact irreducible components of Y_{k} and δ_{k} its number of double points. Because X̃_{k} has only cyclic quotient singularities, h_{X_{k}} is equal to the number of (independent) cycles of the graph given by X̃_{k}, i.e. h_{X_{k}} = δ_{k} − r_{k} + 1.

The idea of the proof of Theorem B is the following: the Milnor fibres of f : X → C and of z : X_{k} → C are the same, but the monodromy of z is the k-th power of T. Hence the presence of 2 × 2 Jordan blocks of T associated to k-th roots of unity becomes visible as cycles in the dual graph of X̃_{k}. We have to control the number h_{X_{k}} of these cycles. As all resolution graphs involved are connected,

h_{X_{k}} = |E_{k}| − |D_{k}| + 1.

For simplicity, we will assume that the irreducible exceptional divisors have no selfintersection points. Let E be the set of intersection points of the exceptional divisors of X̃ → X and E_{k} the same for X̃_{k} → X_{k}. Also, let D_{k} be the set of compact irreducible components of Y_{k}.

For e = (d, d′) ∈ E we put m_{e} := gcd(m_{d}, m_{d′}). Let d^{∗} ∈ D be the exceptional divisor which has non–empty intersection with the strict transform of f^{−1}(0). Let D^{∗} = D \ {d^{∗}}. For any d ∈ D^{∗}, let D(d) denote the set of exceptional divisors which have non–empty intersection with E_{d}. For e =∈ E let d(e) be the vertex of e which has biggest distance in Γ from d^{∗}; i.e. d(e) and d^{∗} are in two different components of Γ \ {e}. Then e ↦ d(e) defines a bijection of E and D^{∗}. The map X̃_{k} → X̃ defines surjections D_{k} → D and E_{k} → E. By a local computation one checks that the number of elements of E_{k} which map to e = (d, d′) ∈ E is equal to gcd(k, m_{e}). Let us put m′_{d} = gcd(m_{d′} : d′ ∈ D(d)). Then the same computation shows that the number of elements of D_{k} which map to d ∈ D^{∗} (respectively to d^{∗}) is equal to gcd(k, m′_{d}) (respectively to one). So we find

h_{X_{k}} = ∑_{e∈E}(gcd(k, m_{e}) − gcd(k, m′_{d(e)}))  (1)

and observe that m′_{d(e)} divides m_{e} for all e ∈ E.

Consider a 2 × 2 Jordan block with eigenvalue λ and let k be the order of λ. Then by theorem A one has h_{X_{k}} > 0. This means that there exist e = (d(e), d′(e)) ∈ E, a prime p, and an integer α ≥ 1 such that

p^{α}|gcd(k, m_{e}), but p^{α} |̸ m_{v} for some v ∈ D_{d(e)} \ {d′(e)}.  (2)

This last result, together with (1) applied for k = p^{α}, gives the first part of the theorem. For the second part, we will show that p^{2}|d_{X}.

Let A be the intersection matrix of Γ, i.e.

(A_{dd′})_{d,d′∈D} = \{ \begin{matrix}E_{d}^{2} & if d = d′, \\ 1 & if d ≠ d′, (d, d′) ∈ E \\ 0 & otherwise. \end{matrix}

We order D such that d^{∗} is its first element. Let m be the column vector of the multiplicities {m_{d}}_{d∈D} and e_{1} = (1 0 ⋯ 0)^{t}. Then the relation (f^{−1}(0), E_{d}) = 0, for any d ∈ D, reads as Am + e_{1} = 0. Take B = −A, then m = B^{−1}e_{1}.

Now, the adjoint matrix (C_{dd′}) of B is given by the following formula (see [5], (20.2)): let Γ_{1}, …, Γ_{s} be the components of the graph obtained from Γ by deleting the path \overline{dd′} in Γ from d to d′ and all incident edges. Then

C_{dd′} = ∏_{i=1}^{s} det(−A(Γ_{i})).

For simplicity, for any graph Γ, we use the notation d_{Γ} = det(−A(Γ)). Now, take e = (d(e), d′(e)) ∈ E as above. Let

Γ \ \overline{d^{∗}d′(e)} = Γ_{1} ∪ ⋯ ∪ Γ_{s} ∪ Γ̃, where d(e) ∈ Γ̃; and
Γ \ \overline{d^{∗}d(e)} = Γ_{1} ∪ ⋯ ∪ Γ_{s} ∪ Γ′_{1} ∪ ⋯ ∪ Γ′_{r}.

Then:

m_{d′(e)} = (det B)^{−1} · ∏_{i=1}^{s} d_{Γ_{i}} · dΓ̃,

m_{d(e)} = (det B)^{−1} · ∏_{i=1}^{s} d_{Γ_{i}} · ∏_{i=1}^{r} d_{Γ′_{i}}.  (3)

Now, let U (resp. V) be the matrix obtained from B by deleting all the lines and columns which corresponds to d ∈ {d′(e)} ∪ {d : d ∈ Γ̃} (resp. d ∈ Γ̃). Then:

det B = det V · dΓ̃ − det U · ∏_{i} d_{Γ′_{i}}.

The above identities give:

m_{d′(e)} · det V − m_{d(e)} · det U = ∏_{i=1}^{s} d_{Γ_{i}}.  (4)

Notice that each of the graphs Γ′_{j} has precisely one vertex v_{j} ∈ D_{d(e)} \{d′(e)}. By the same argument as above, for any j = 1, …, r, there is an integer n_{v_{j}} such that

m_{v_{j}} = (det B)^{−1} · ∏_{i=1}^{s} d_{Γ_{i}} · \frac{∏_{i=1}^{r} d_{Γ′_{i}}}{d_{Γ′_{j}}} · n_{v_{j}}.  (5)

Recall that there exists v ∈ {v_{1}, …, v_{r}} such that p^{α} |̸ m_{v}, therefore from (5) one has that p^{α} |̸ (det B)^{−1} ∏_{i=1}^{s} d_{Γ_{i}}. This together with (3) gives that

p| ∏_{i=1}^{r} d_{Γ′_{i}}.  (6)

The relation ∑_{i=1}^{r} m_{v_{i}} + m_{d′(e)} + E_{d(e)}^{2}m_{d(e)} = 0 ensures that there are at least two elements v ∈ {v_{1}, …, v_{r}} such that p^{α} |̸ m_{v}. Therefore, by (6), there is an element v_{j} ∈ {v_{1}, …, v_{r}} with p^{α} |̸ m_{v_{j}} and p|(∏_{i=1}^{r} d_{Γ′_{i}})/d_{Γ′_{j}}. Consider now the relation (5) for this v_{j}. Since by (4) p^{α}| ∏_{i=1}^{s} d_{Γ_{i}} one has p^{α+1}|(det B) · m_{v_{j}}, which implies p^{2}| det B. This ends the proof.

Remark. In the case g = 0 we can reprove the second part of theorem B as follows. One has an exact sequence

0 → H_{2}(K_{X}) → H_{1}(F, ∂F) →^{V} H_{1}(F) → H_{1}(K_{X}) → 0

where V is the variation map, cf. [12]. If j : H_{1}(F) → H_{1}(F, ∂F) is the natural map, then T − I = V j. If f^{−1}(0) is irreducible, j is an isomorphism, so H_{1}(K_{X}) ≃ coker(T − I). If h_{X} = g = 0 then d_{X} =| Δ(1) | where Δ(t) is the characteristic polynomial of T.

Write Δ(t) = ∏_{d} Φ_{d}(t)^{r_{d}} with Φ_{d}(t) the d-th cyclotomic polynomial; then Δ(1) = ∏_{d} Φ_{d}(1)^{r_{d}}. Now, if d is not a prime power, then Φ_{d}(1) = 1 whereas Φ_{d}(1) = p if d is a power of the prime p. We conclude that, if T has a 2 × 2 Jordan block then it has also a double eigenvalue of order a power of a prime p (cf. theorem B.a), hence d_{X} must be divisible by p^{2}.

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