on the tjurina number of plane curve singularities · on the tjurina number of plane curve...
TRANSCRIPT
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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
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.
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
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.
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)) .
We easily see that μ ≥ τ.
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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.
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)) .
We easily see that μ ≥ τ. Set r := μ − τ .
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Fact (Ebey, Zariski)
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Fact (Ebey, Zariski)
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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Fact (Ebey, Zariski)
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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Fact (Ebey, Zariski)
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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Fact (Ebey, Zariski)
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.
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
We define the sequence (βj) associated to aparametrization of C as follows:
e0 = β0 = n,
βj = min{i | i �≡ 0 mod ej`1 and ai �= 0},
ej = gcd(ej`1, βj).
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Characteristic
We define the sequence (βj) associated to aparametrization of C as follows:
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
The parametrization of C is
x = t2, y = t3.
Ch(C) = {2, 3}
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Example 1
C : y2 − x3
x
y
The parametrization of C is
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
The parametrization of C is
x = t4, y = t6 + t7.
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Example 2
C : −y4 + 2x3y2 − x6 + 4x5y + x7
x
y
The parametrization of C is
x = t4, y = t6 + t7.
Ch(C) = {4, 6, 7}
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Example 2
C : −y4 + 2x3y2 − x6 + 4x5y + x7
x
y
The parametrization of C is
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
• 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 .
• 1990 · Luengo and Pfister determined τ of C withCh(C) = {2p, 2q, β}.
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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 .
• 1990 · Luengo and Pfister determined τ of C withCh(C) = {2p, 2q, β}.
• 2001 · Bayer and Hefez classified C with r = 1, 2.
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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 .
• 1990 · Luengo and Pfister determined τ of C withCh(C) = {2p, 2q, β}.
• 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
• 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 .
• 1990 · Luengo and Pfister determined τ of C withCh(C) = {2p, 2q, β}.
• 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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Theorem (Zariski, 1966)
♠ r = 0 =⇒ C is given by yn − xm where gcd(n, m) = 1.
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Theorem (Zariski, 1966)
♠ r = 0 =⇒ C is given by yn − xm where gcd(n, m) = 1.
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Theorem (Zariski, 1966)
♠ r = 0 =⇒ C is given by yn − xm where gcd(n, m) = 1.
Furthermore, we have
μ = τ = (n − 1)m − n + 1.
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Theorem (Zariski, 1966)
♠ r = 0 =⇒ C is given by yn − xm where gcd(n, m) = 1.
Furthermore, we have
μ = τ = (n − 1)m − n + 1,
and
Ch(C) = {n, m}.
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Theorem (Zariski, 1966)
♠ r = 0 =⇒ C is given by yn − xm where gcd(n, m) = 1.
Furthermore, we have
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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Theorem (Luengo and Pfister, 1990)
A plane curve singularitiy C has
Ch(C) = {2a, 2b, β},
where gcd(a, b) = 1 and β is odd.
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Theorem (Luengo and Pfister, 1990)
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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Theorem (Bayer and Hefez, 2001)
♠ r = 1
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Theorem (Bayer and Hefez, 2001)
♠ r = 1 =⇒ C is given by
x = tn, y = tm + t–,
where λ = (n − 1)m − 2n and g = 1.
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Theorem (Bayer and Hefez, 2001)
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Theorem (Bayer and Hefez, 2001)
♠ r = 2
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Theorem (Bayer and Hefez, 2001)
♠ r = 2 =⇒ C is given by the following two cases.
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Theorem (Bayer and Hefez, 2001)
♠ 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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Theorem (Bayer and Hefez, 2001)
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Theorem (Bayer and Hefez, 2001)
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
Write m = pn + q where q < n.
♠ r = 3 =⇒g = 1
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Main Theorem 1
Write m = pn + q where q < n.
♠ 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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Main Theorem 2 (with K.Nishiyama)
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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Main Theorem 2 (with K.Nishiyama)
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
Fix the notations as follows:
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
Fix the notations as follows:
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
Now we have OC∼= C[[tn, ϕ(t)]].
Since OC∼= C[[t]], we have OC ⊂ OC .
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The Semigroup of C
Now we have OC∼= C[[tn, ϕ(t)]].
Since OC∼= C[[t]], we have OC ⊂ OC .
The semigroup S of C is defined by
S := ν(OC) = {ν(A)| A ∈ OC}.
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The Semigroup of C
Now we have OC∼= C[[tn, ϕ(t)]].
Since OC∼= C[[t]], we have OC ⊂ OC .
The semigroup S of C is defined by
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
Now we have OC∼= C[[tn, ϕ(t)]].
Since OC∼= C[[t]], we have OC ⊂ OC .
The semigroup S of C is defined by
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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The relation between Ch(C) and S
Define the integers ni by
n0 = 1 and ei`1 = niei, (i = 1, . . . , g).
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The relation between Ch(C) and S
Define the integers ni by
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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The relation between Ch(C) and S
Define the integers ni by
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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The properties of the differential module
For ζ ∈ Ω1eC, we have
ζ = H(t)dt for some H(t) ∈ OeC.
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The properties of the differential module
For ζ ∈ Ω1eC, we have
ζ = H(t)dt for some H(t) ∈ OeC.
The order of ζ is naturally defined by ν(ζ) := ν(H(t)).
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The properties of the differential module
For ζ ∈ Ω1eC, we have
ζ = H(t)dt for some H(t) ∈ OeC.
The order of ζ is naturally defined by ν(ζ) := ν(H(t)).
We have π˜ (Ω1
C
) ⊂ Ω1eC
where π˜ is the pull back.
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The properties of the differential module
For ζ ∈ Ω1eC, we have
ζ = H(t)dt for some H(t) ∈ OeC.
The order of ζ is naturally defined by ν(ζ) := ν(H(t)).
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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The properties of the differential module
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The properties of the differential module
A differential ξ = dA for some A ∈ OC is called exact .
SetdOC :=
{ξ ∈ Ω1
C
∣∣ ξ is exact}
.
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The properties of the differential module
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 dual description of the difference r
The following important Theorem was proved by Zariski.
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The dual description of the difference r
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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§ 3 The determination of τ
In this section, we always consider the plane curvesingularities of genus 1.
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§ 3 The determination of τ
In this section, we always consider the plane curvesingularities of genus 1.
We have the following relations:
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§ 3 The determination of τ
In this section, we always consider the plane curvesingularities of genus 1.
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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Angermuller’s Lemma
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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Angermuller’s Lemma
Lemma (Angermüller)
If g = 1, then any integer t is expressed
uniquely ast = t1m − t0n
where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.
It follows form this lemma that t ∈ S ⇐⇒ t0 ≤ 0.
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Angermuller’s Lemma
Lemma (Angermüller)
If g = 1, then any integer t is expressed
uniquely ast = t1m − t0n
where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.
It follows form this lemma that t ∈ S ⇐⇒ t0 ≤ 0.
Furthermore we see that
G = {t = t1m − t0n| 1 ≤ t1 ≤ n − 1, t0 > 0}.
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Angermuller’s Lemma
Lemma (Angermüller)
If g = 1, then any integer t is expressed
uniquely ast = t1m − t0n
where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.
It follows form this lemma that t ∈ S ⇐⇒ t0 ≤ 0.
Furthermore we see that
G = {t1m − t0n| 1 ≤ t1 ≤ n − 1, t0 > 0}.
In particular, we find that max{G} = (n − 1)m − n.
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Angermuller’s Lemma
Lemma (Angermüller)
If g = 1, then any integer t is expressed
uniquely ast = t1m − t0n
where t0, t1 ∈ Z and 0 ≤ t1 ≤ n − 1.
It follows form this lemma that t ∈ S ⇐⇒ t0 ≤ 0.
Furthermore we see that
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
If a plane curve singularity C is given,
The parametrization of C is determined.
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Flow chart
If a plane curve singularity C is given,
The parametrization of C is determined.
⇐
The characteristic Ch(C) is determined.
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Flow chart
If a plane curve singularity C is given,
The parametrization of C is determined.
⇐
The characteristic Ch(C) is determined.
⇐
The semigroup S is determined.
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Flow chart
If a plane curve singularity C is given,
The parametrization of C is determined.
⇐
The characteristic Ch(C) is determined.
⇐
The semigroup S is determined.
⇐The Milnor number C is determined.
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Flow chart
If a plane curve singularity C is given,
The parametrization of C is determined.
⇐
The characteristic Ch(C) is C determined.
⇐
The semigroup S is 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
Recall that λ /∈ S. So we express λ as
λ = λ1m − λ0n for some λ0, λ1.
It follows from λ > m and λ /∈ S that
2 ≤ λ1 ≤ n − 1 and λ0 ≥ 2.
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Zariski invariant
Recall that λ /∈ S. So we express λ as
λ = λ1m − λ0n for some λ0, λ1.
It follows from λ > m and λ /∈ S that
2 ≤ λ1 ≤ n − 1 and λ0 ≥ 2.
Zariski showed the following fact:
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Zariski invariant
Recall that λ /∈ S. So we express λ as
λ = λ1m − λ0n for some λ0, λ1.
It follows from λ > m and λ /∈ S that
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,
where 1 ≤ R ≤ p, (p = [m/n]) and gcd(n, m) = 1, then wehave r = R.
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,
where 1 ≤ R ≤ p, (p = [m/n]) and gcd(n, m) = 1, then wehave r = R.
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
Now we have OC = C[[tn, tm + t–]].
Note that ν(dOC) = {s − 1| s ∈ S}.
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Proof of Theorem 3
Now we have OC = C[[tn, tm + t–]].
Note that ν(dOC) = {s − 1| s ∈ S}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
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Proof of Theorem 3
Now we have OC = C[[tn, tm + t–]].
Note that ν(dOC) = {s − 1| s ∈ S}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
Set V + := {α + 1| α ∈ V }
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Proof of Theorem 3
Now we have OC = C[[tn, tm + t–]].
Note that ν(dOC) = {s − 1| s ∈ S}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
Set V + := {α + 1| α ∈ V }=⇒ min{V +} = 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}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
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}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
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
Now we have OC = C[[tn, tm + t– + · · · ]].Note that ν(dOC) = {s − 1| s ∈ S}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
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
Now we have OC = C[[tn, tm + t– + · · · ]].Note that ν(dOC) = {s − 1| s ∈ S}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
Set V + := {α + 1| α ∈ V }=⇒ min{V +} = λ + n.Put G0 = {γ ∈ G| γ ≥ λ + n} =⇒ V + ⊆ G0 � G.�
�
�
���
��
G
G0
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Proof of Theorem 3
Now we have OC = C[[tn, tm + t– + · · · ]].Note that ν(dOC) = {s − 1| s ∈ S}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
Set V + := {α + 1| α ∈ V }=⇒ min{V +} = λ + n.Put G0 = {γ ∈ G| γ ≥ λ + n} =⇒ V + ⊆ G0 � G.�
�
�
���
��
G ×λ
G0
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Proof of Theorem 3
Now we have OC = C[[tn, tm + t– + · · · ]].Note that ν(dOC) = {s − 1| s ∈ S}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
Set V + := {α + 1| α ∈ V }=⇒ min{V +} = λ + n.Put G0 = {γ ∈ G| γ ≥ λ + n} =⇒ V + ⊆ G0 � G.�
�
�
���
����
��V +
G ×λ
G0
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Proof of Theorem 3
Now we have OC = C[[tn, tm + t– + · · · ]].Note that ν(dOC) = {s − 1| s ∈ S}.
So V = ν(Ω1C) \ ν(dOC) = {γ − 1| for some γ ∈ G}.
Set V + := {α + 1| α ∈ V }=⇒ min{V +} = λ + n.Put G0 = {γ ∈ G| γ ≥ λ + n} =⇒ V + ⊆ G0 � G.�
�
�
���
����
��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
All elements of G0 are
(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
All elements of G0 are
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
All elements of G0 are
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
All elements of G0 are
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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§ 4 ProblemsReducible case
What about τ for reducible plane curve singularities?
⇐
The polar degree
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§ 4 ProblemsReducible case
What about τ for reducible plane curve singularities?
⇐
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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§ 4 ProblemsReducible case
What about τ for reducible plane curve singularities?
⇐
The polar degree
F : a reduced homogeneus polynomial in C[x, y, z].
C: a projective plane curve defined by F.
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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§ 4 ProblemsReducible case
What about τ for reducible plane curve singularities?
⇐
The polar degree
F : a reduced homogeneus polynomial in C[x, y, z].
C: a projective plane curve defined by F.
For this C, consider the polar map
ϕC : P2(C) −→ P2(C) defined by p �−→ (Fx(p), Fy(p), Fz(p)
).
We call the degree of ϕC the polar degree of C.
The polar degree of is denoted by Pdeg C.
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§ 4 ProblemsReducible case
What about τ for reducible plane curve singularities?
⇐
The polar degree
F : a reduced homogeneus polynomial in C[x, y, z].
C: a projective plane curve defined by F.
For this C, consider the polar map
ϕC : P2(C) −→ P2(C) defined by p �−→ (Fx(p), Fy(p), Fz(p)
).
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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The characteristic of L.P. type
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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The characteristic of L.P. type
Set
Ch(C) = {2g`1a0, 2g`1a1, . . . , 2g`iai, . . . , 2ag`1, ag},
where gcd(ai, ai+1) = 1, g ≥ 2 and ag is odd.
We call this Ch(C) the Luengo and Pfister type.
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The characteristic of L.P. type
Set
Ch(C) = {2g`1a0, 2g`1a1, . . . , 2g`iai, . . . , 2ag`1, ag},
where gcd(ai, ai+1) = 1, g ≥ 2 and ag is odd.
We call this Ch(C) the Luengo and Pfister type.
When g = 2, the Tjurina number τ depends on Ch(C).
(Luengo and Pfister’s Theorem)
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The characteristic of L.P. type
Set
Ch(C) = {2g`1a0, 2g`1a1, . . . , 2g`iai, . . . , 2ag`1, ag},
where gcd(ai, ai+1) = 1, g ≥ 2 and ag is odd.
We call this Ch(C) the Luengo and Pfister type.
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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