on the tjurina number of plane curve singularities · on the tjurina number of plane curve...

On the Tjurina Number of PlaneCurve Singularities

Masahiro Watari

Saitama University

Graduate School of Science and Engineering

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Today’s Contents

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Today’s Contents

§ 1 Introduction

§ 2 Preliminaries

§ 3 The determination of Tjurina number

§ 4 Problems

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§ 1 Introduction

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§ 1 Introduction

For an irreducible element f ∈ (x, y) ⊂ C[[x, y]], set

C := {u · f | u is a unit of C[[x, y]]}.

We call C a plane curve singularity.

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§ 1 Introduction




The Milnor number μ and Tjurina number τ of C aredefined by

μ := dimC (C[[x, y]]/ (fx, fy)) ,

τ := dimC (C[[x, y]]/ (f, fx, fy)) .

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§ 1 Introduction





μ := dimC (C[[x, y]]/ (fx, fy)) ,

τ := dimC (C[[x, y]]/ (f, fx, fy)) .

We easily see that μ ≥ τ.

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§ 1 Introduction





μ := dimC (C[[x, y]]/ (fx, fy)) ,

τ := dimC (C[[x, y]]/ (f, fx, fy)) .

We easily see that μ ≥ τ. Set r := μ − τ .

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Fact (Ebey, Zariski)

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If r > 0, then, up to analytic equivalence, C can be given bythe parametrization

⎧⎨⎩

x = tn,

y = tm + t– +∑i2G

aiti,

where n is the multiplicity of C, m > n and λ, λ + n /∈ S.

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⎧⎨⎩

x = tn,

y = tm + t– +∑i2G

aiti,


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⎧⎨⎩

x = tn,

y = tm + t– +∑i2G

aiti,


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⎧⎨⎩

x = tn,

y = tm + t– +∑i2G

aiti,


The integer λ is an analytic invariant. It is called the Zariski

invariant of C.

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Picture

x

y

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Picture

x

y

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Characteristic

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Characteristic

We define the sequence (βj) associated to aparametrization of C as follows:

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Characteristic


e0 = β0 = n,

βj = min{i | i �≡ 0 mod ej`1 and ai �= 0},

ej = gcd(ej`1, βj).

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Characteristic


e0 = β0 = n,

βj = min{i | i �≡ 0 mod ej`1 and ai �= 0},

ej = gcd(ej`1, βj).

Definition (Zariski)

We call the set Ch(C) := {β0, . . . , βg} and the positive

integer g the characteristic of C and the genus of C.

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Example 1

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Example 1

C : y2 − x3

x

y

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Example 1

C : y2 − x3

x

y

The parametrization of C is

x = t2, y = t3.

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Example 1

C : y2 − x3

x

y


x = t2, y = t3.

Ch(C) = {2, 3}

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Example 1

C : y2 − x3

x

y


x = t2, y = t3.

Ch(C) = {2, 3} =⇒ g = 1.

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Example 2

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Example 2

C : −y4 + 2x3y2 − x6 + 4x5y + x7

x

y

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Example 2

C : −y4 + 2x3y2 − x6 + 4x5y + x7

x

y


x = t4, y = t6 + t7.

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Example 2

C : −y4 + 2x3y2 − x6 + 4x5y + x7

x

y


x = t4, y = t6 + t7.

Ch(C) = {4, 6, 7}

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Example 2

C : −y4 + 2x3y2 − x6 + 4x5y + x7

x

y


x = t4, y = t6 + t7.

Ch(C) = {4, 6, 7} =⇒ g = 2.

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History

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History

• 1966 · Zariski determined C with r = 0.

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History

• 1966 · Zariski determined C with r = 0.• 1969 · Tjurina showed that the dimension of the base

space of a semi-universal deformation of ahypersurface singularity is equal to τ .

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History



• 1990 · Luengo and Pfister determined τ of C withCh(C) = {2p, 2q, β}.

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History




• 2001 · Bayer and Hefez classified C with r = 1, 2.

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History




• 2001 · Bayer and Hefez classified C with r = 1, 2.• 2005 · I classified C with r = 3.

· Nishiyama and I determined τ of C withn = 3, 4.

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History




• 2001 · Bayer and Hefez classified C with r = 1, 2.

•2005 I classified C with r = 3.

Nishiyama and I determined τ of C withn = 3, 4.

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Theorem (Zariski, 1966)

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♠ r = 0 =⇒ C is given by yn − xm where gcd(n, m) = 1.

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Furthermore, we have

μ = τ = (n − 1)m − n + 1.

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μ = τ = (n − 1)m − n + 1,

and

Ch(C) = {n, m}.

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μ = τ = (n − 1)m − n + 1,

and

Ch(C) = {n, m}.

∴ g = 1.

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Theorem (Luengo and Pfister, 1990)

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A plane curve singularitiy C has

Ch(C) = {2a, 2b, β},

where gcd(a, b) = 1 and β is odd.

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A plane curve singularitiy C has

Ch(C) = {2a, 2b, β},

where gcd(a, b) = 1 and β is odd.

=⇒ We have τ = β + 3ab − a − 3b.

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Theorem (Bayer and Hefez, 2001)

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♠ r = 1

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♠ r = 1 =⇒ C is given by

x = tn, y = tm + t–,

where λ = (n − 1)m − 2n and g = 1.

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♠ r = 2

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♠ r = 2 =⇒ C is given by the following two cases.

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♠ r = 2 =⇒ C is given by the following two cases.

Case (1) g = 2

(a) x = t4, y = t6 + t–, where λ (> 6) is odd.

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Case (2) g = 1.

(b) x = tn, y = tm + t–,

where λ = (n − 1)m − 3n.

(c) x = t4 y = tm + t– +3m − 8

2mt3m`16 + at3m`12,

where λ = (n − 2)m − 2n, m > 8.

(d) x = tn y = tm + t– + at(n`1)m`3n,

where λ = (n − 2)m − 2n, n ≥ 5, m > 2n/(n − 3).

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Main Theorem 1

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Main Theorem 1

Write m = pn + q where q < n.

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Main Theorem 1


♠ r = 3 =⇒g = 1

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Main Theorem 1


♠ r = 3 =⇒g = 1 and the following three cases occur:

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Type (i): λ = (n − 1)m − 4n

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Type (i): λ = (n − 1)m − 4n

(A) x = tn, y = tm + t–, where n ≥ 3, p ≥ 2.

(B) x = tn, y = tm + t– + at(n`2)m`2n,

where n ≥ 5 and p = 1.

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Type (ii): λ = (n − 2)m − 2n

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Type (ii): λ = (n − 2)m − 2n

(C) x = tn, y = tm + t– + at(n`1)m`4n + bt(n`1)m`3n,

where n ≥ 5, p ≥ 2 and a (�= 0).

(D) x = t4, y = tm + t– + at3m`16 + bt3m`12,

where p ≥ 2 and a (�= (3m − 8)/2m).

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Type (iii): λ = (n − 3)m − 2n

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Type (iii): λ = (n − 3)m − 2n

(E)

x = tn, y = tm + t– +p∑

i=1

(aitmi + bit

ni) +2p∑

i=p+1

bitni ,

where ⎧⎪⎨⎪⎩

n > 2q, n ≥ 5, m > 2n/(n − 4),mi = (n − 2)m − (p + 3 − i)n,

ni = (n − 1)m − (2p + 3 − i)n.

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Type (iii): λ = (n − 3)m − 2n

(F)

x = tn, y = tm + t– +p∑

i=1

(aitmi + bit

ni) +2p+1∑

i=p+1

aitmi ,

where ⎧⎪⎨⎪⎩

n < 2q, n ≥ 5, m > 2n/(n − 4),mi = (n − 1)m − (2p + 4 − i)n,

ni = (n − 2)m − (p + 4 − i)n.

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Main Theorem 2 (with K.Nishiyama)

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Let C be a plane curve singularity with the multiplicity

n = 3 or 4. Then τ of C is given by the following table:

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Let C be a plane curve singularity with the multiplicity

n = 3 or 4. Then τ of C is given by the following table:

n Ch(C) μ λ τ No.3 {3, m} 2(m − 1) − 2(m − 1) (3.1)

2m − 3λ0 2m − λ0 − 1 (3.2)− 3(m − 1) (4.1)

4 {4, m} 3(m − 1) 3m − 4λ0 3m − λ0 − 2 (4.2)2m − 4λ0 see Next (4.3)

{4, 2α, β} 4α + β − 3 β 3α + β − 2 (4.4)

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Case (4.3)

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Case (4.3)

Up to analytic equivalence, C is given by

x = t4, y = tm+t–+p∑

i=1

bit3m`4(p+–0+1`i), where p =

[m

4

].

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Case (4.3)

Up to analytic equivalence, C is given by

x = t4, y = tm+t–+p∑

i=1

bit3m`4(p+–0+1`i), where p =

[m

4

].

Then τ of C is determined as in the following table:

τ Conditionsp = λ0, b1 = (3m − 4λ0)/2m

3m − 2λ0 − 1 p > λ0, bi = 0 (1 ≤ i ≤ p − λ0),bp`–0+1 = (3m − 4λ0)/2m

p = λ0, b1 �= (3m − 4λ0)/2m

3m − 2λ0 − 2 p > λ0, bi = 0 (1 ≤ i ≤ p − λ0),bp`–0+1 �= (3m − 4λ0)/2m

3m − λ0 − (p − d + 3) p > λ0, ∃bi �= 0 (1 ≤ i ≤ p − λ0)d := min{i| bi �= 0}

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§ 2 Preliminaries

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§ 2 Preliminaries

Fix the notations as follows:

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§ 2 Preliminaries


C : a plane curve singularity,

C : the nonsingular model of C,

OC : the local ring of C,

OeC : the local ring of C,

ν : the order function of OeC ,

Ω1C : the differential module of OC ,

Ω1eC

: the differential module of OeC ,

(x, y) = (tn, ϕ(t)) : the parametrization of C.

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§ 2 Preliminaries


C : a plane curve singularity,

C : the nonsingular model of C,

OC : the local ring of C,

OeC

∼= C[[t]] : the local ring of C,

ν : the order function of OeC ,

Ω1C : the differential module of OC ,

Ω1eC

: the differential module of OeC ,

(x, y) = (tn, ϕ(t)) : the parametrization of C.

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The Semigroup of C

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The Semigroup of C

Now we have OC∼= C[[tn, ϕ(t)]].

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The Semigroup of C


Since OC∼= C[[t]], we have OC ⊂ OC .

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The Semigroup of C



The semigroup S of C is defined by

S := ν(OC) = {ν(A)| A ∈ OC}.

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The Semigroup of C




S := ν(OC) = {ν(A)| A ∈ OC}.

We call G := N \ S the set of gaps of S.

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The Semigroup of C




S := ν(OC) = {ν(A)| A ∈ OC}.

We call G := N \ S the set of gaps of S.

It is known that if the genus of C is g, then ∃v0, v1, · · · , vg

such thatS = 〈v0, v1, · · · , vg〉.

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Example 3

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Example 3

C : y2 − x3.

The parametrization of C is x = t2, y = t3.

Ch(C) = {2, 3}.

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Example 3

C : y2 − x3.

The parametrization of C is x = t2, y = t3.

Ch(C) = {2, 3},

OC = C[[t2, t3]],

S = {0, 2, 3, 4, 5, 6, 7, 8, 9, . . .} = 〈2, 3〉,

G = {1}.

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Example 4

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Example 4

C : −y4 + 2x3y2 − x6 + 4x5y + x7

The parametrization of C is x = t4, y = t6 + t7.

Ch(C) = {4, 6, 7},

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Example 4

C : −y4 + 2x3y2 − x6 + 4x5y + x7

The parametrization of C is x = t4, y = t6 + t7.

Ch(C) = {4, 6, 7},

OC = C[[t4, t6 + t7]],

S = {0, 4, 6, 8, 10, 12, 13, 14, 16, . . .} = 〈4, 6, 13〉,

G = {1, 2, 3, 5, 7, 9, 11, 15}.

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The relation between Ch(C) and S

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Define the integers ni by

n0 = 1 and ei`1 = niei, (i = 1, . . . , g).

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n0 = 1 and ei`1 = niei, (i = 1, . . . , g).

It is known that the minimally generators of S are given by

v0 = n and vi = ni`1vi`1 + βi − βi`1 (i = 1, . . . , g).

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n0 = 1 and ei`1 = niei, (i = 1, . . . , g).

It is known that the minimally generators of S are given by

v0 = n and vi = ni`1vi`1 + βi − βi`1 (i = 1, . . . , g).

We easily see that v1 = m and v0 < v1 < · · · < vg.

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The properties of the differential module

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For ζ ∈ Ω1eC, we have

ζ = H(t)dt for some H(t) ∈ OeC.

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The order of ζ is naturally defined by ν(ζ) := ν(H(t)).

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We have π˜ (Ω1

C

) ⊂ Ω1eC

where π˜ is the pull back.

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We have π˜ (Ω1

C

) ⊂ Ω1eC

where π˜ is the pull back.

So we define the order of ξ (∈ Ω1C) by ν(ξ) := ν(π˜(ξ)).

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A differential ξ = dA for some A ∈ OC is called exact .

SetdOC :=

{ξ ∈ Ω1

C

∣∣ ξ is exact}

.

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A differential ξ = dA for some A ∈ OC is called exact .

SetdOC :=

{ξ ∈ Ω1

C

∣∣ ξ is exact}

.

Definition

Define the set of orders V by

V := ν(Ω1

C

) \ ν (dOC),

where ν(Ω1

C

)=

{ν(ξ)| ξ ∈ Ω1

C

}, ν(dOC) = {ν(ξ)| ξ ∈ dOC}.

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The dual description of the difference r

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The following important Theorem was proved by Zariski.

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The following important Theorem was proved by Zariski.

Theorem (Zariski)

We have r = � (V ).

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§ 3 The determination of τ

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In this section, we always consider the plane curvesingularities of genus 1.

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We have the following relations:

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We have the following relations:

Fact

g = 1 ⇐⇒ gcd(n, m) = 1

⇐⇒ Ch(C) = {n, m}⇐⇒ S = 〈n, m〉

= {am + bn| a, b ∈ Z–0}

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Angermuller’s Lemma

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Lemma (Angermüller)

If g = 1, then any integer t is expressed

uniquely ast = t1m − t0n

where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.

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where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.

It follows form this lemma that t ∈ S ⇐⇒ t0 ≤ 0.

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where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.


Furthermore we see that

G = {t = t1m − t0n| 1 ≤ t1 ≤ n − 1, t0 > 0}.

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where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.



G = {t1m − t0n| 1 ≤ t1 ≤ n − 1, t0 > 0}.

In particular, we find that max{G} = (n − 1)m − n.

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where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.



G = {t1m − t0n| 1 ≤ t1 ≤ n − 1, t0 > 0}.

In particular, we find that max{G} = (n − 1)m − n.

Fact μ = (n − 1)m − n + 1.

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Flow chart

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Flow chart

If a plane curve singularity C is given,

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Flow chart


The parametrization of C is determined.

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Flow chart



⇐

The characteristic Ch(C) is determined.

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Flow chart



⇐


⇐

The semigroup S is determined.

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Flow chart



⇐


⇐


⇐The Milnor number C is determined.

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Flow chart



⇐

The characteristic Ch(C) is C determined.

⇐


⇐The Milnor number C is determined.

⇐

If r is determined, then τ = μ − r.

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Zariski invariant

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Zariski invariant

Recall that λ /∈ S.

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Zariski invariant

Recall that λ /∈ S. So we express λ as

λ = λ1m − λ0n for some λ0, λ1.

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Zariski invariant



It follows from λ > m and λ /∈ S that

2 ≤ λ1 ≤ n − 1 and λ0 ≥ 2.

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Zariski invariant




2 ≤ λ1 ≤ n − 1 and λ0 ≥ 2.

Zariski showed the following fact:

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Zariski invariant




2 ≤ λ1 ≤ n − 1 and λ0 ≥ 2.

Zariski showed the following fact:

Fact

For the differential ω := mydx − nxdy, we have

ν(ω) = λ + n − 1 = min{V }.

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Theorem 3

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Theorem 3

If C is given by

x = tn, y = tm + t(n`1)m`(R+1)n,

where 1 ≤ R ≤ p, (p = [m/n]) and gcd(n, m) = 1, then wehave r = R.

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Theorem 3

If C is given by

x = tn, y = tm + t(n`1)m`(R+1)n,


Namely, we have τ = (n − 1)m − n + 1 − R.

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Theorem 3

If C is given by

x = tn, y = tm + t(n`1)m`(R+1)n,


Namely, we have τ = (n − 1)m − n + 1 − R.

Remark

We have λ = (n − 1)m − (R + 1)n.

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Proof of Theorem 3

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Proof of Theorem 3

Now we have OC = C[[tn, tm + t–]].

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Proof of Theorem 3


Note that ν(dOC) = {s − 1| s ∈ S}.

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Proof of Theorem 3



So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.

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Proof of Theorem 3




Set V + := {α + 1| α ∈ V }

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Proof of Theorem 3




Set V + := {α + 1| α ∈ V }=⇒ min{V +} = λ + n.

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Proof of Theorem 3




Set V + := {α + 1| α ∈ V }=⇒ min{V +} = λ + n.Put G0 = {γ ∈ G| γ ≥ λ + n}

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Proof of Theorem 3

Now we have OC = C[[tn, tm + t– + · · · ]].Note that ν(dOC) = {s − 1| s ∈ S}.


Set V + := {α + 1| α ∈ V }=⇒ min{V +} = λ + n.Put G0 = {γ ∈ G| γ ≥ λ + n} =⇒ V + ⊆ G0 � G.

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Proof of Theorem 3



Set V + := {α + 1| α ∈ V }=⇒ min{V +} = λ + n.Put G0 = {γ ∈ G| γ ≥ λ + n} =⇒ V + ⊆ G0 � G.�

�

�

�

G

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Proof of Theorem 3




�

�

��

��

G

G0

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Proof of Theorem 3




�

�

��

��

G ×λ

G0

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Proof of Theorem 3




�

�

��

��

��V +

G ×λ

G0

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Proof of Theorem 3




�

�

��

��

��V +

G ×λ

G0

Furthermore, we see that �(V +) = �(V ) = r.

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Proof of Theorem 3

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Proof of Theorem 3

All elements of G0 are

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Proof of Theorem 3


(n − 1)m − Rn = λ + n,

(n − 1)m − (R − 1)n,

(n − 1)m − (R − 2)n,

...

(n − 1)m − 2n,

(n − 1)m − n.

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Proof of Theorem 3


ν(ω) + 1 =(n − 1)m − Rn = λ + n,

ν(xω) + 1 =(n − 1)m − (R − 1)n,

ν(x2ω) + 1 =(n − 1)m − (R − 2)n,

...

ν(xR`2ω

)+ 1 =(n − 1)m − 2n,

ν(xR`1ω

)+ 1 =(n − 1)m − n.

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Proof of Theorem 3


ν(ω) + 1 =(n − 1)m − Rn = λ + n,

ν(xω) + 1 =(n − 1)m − (R − 1)n,

ν(x2ω) + 1 =(n − 1)m − (R − 2)n,

...

ν(xR`2ω

)+ 1 =(n − 1)m − 2n,

ν(xR`1ω

)+ 1 =(n − 1)m − n.

Hence V + = {ν(ω) + 1, ν(xω) + 1, . . . , ν(xR`1ω

)+ 1}.

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Proof of Theorem 3


ν(ω) + 1 =(n − 1)m − Rn = λ + n,

ν(xω) + 1 =(n − 1)m − (R − 1)n,

ν(x2ω) + 1 =(n − 1)m − (R − 2)n,

...

ν(xR`2ω

)+ 1 =(n − 1)m − 2n,

ν(xR`1ω

)+ 1 =(n − 1)m − n.

Hence V + = {ν(ω) + 1, ν(xω) + 1, . . . , ν(xR`1ω

)+ 1}.

By Zariski’s Theorem, we have r = �(V +) = R. �

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§ 4 Problems

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§ 4 ProblemsReducible case

What about τ for reducible plane curve singularities?

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⇐

The polar degree

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⇐

The polar degree

F : a reduced homogeneus polynomial in C[x, y, z].

C: a projective plane curve defined by F.

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⇐

The polar degree



For this C, consider the polar map

ϕC : P2(C) −→ P2(C) defined by p �−→ (Fx(p), Fy(p), Fz(p)

).

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⇐

The polar degree





).

We call the degree of ϕC the polar degree of C.

The polar degree of is denoted by Pdeg C.

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⇐

The polar degree





).

We call the degree of ϕC the polar degree of C.

The polar degree of is denoted by Pdeg C.

Problem: Classify C with Pdeg C = 2, 3, 4.

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The characteristic of L.P. type

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Set

Ch(C) = {2g`1a0, 2g`1a1, . . . , 2g`iai, . . . , 2ag`1, ag},

where gcd(ai, ai+1) = 1, g ≥ 2 and ag is odd.

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Set



We call this Ch(C) the Luengo and Pfister type.

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Set




When g = 2, the Tjurina number τ depends on Ch(C).

(Luengo and Pfister’s Theorem)

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Set




When g = 2, the Tjurina number τ depends on Ch(C).

(Luengo and Pfister’s Theorem)

What about the cases where g ≥ 3?

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on the tjurina number of plane curve singularities · on the tjurina number of plane curve...

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