Smooth rational surfaces of degree 11
and sectional genus 8 in P5 .
Abdul Moeed Mohammad
University of Oslo
7th June, 2010
Motivation
Classification problem
Given a numerical polynomial
P(n) = a2 n2 + a1 n + a0 .
(a2 6= 0)
Does there exists a smooth projective surface with P(n) as its Hilbert
polynomial?
Hilbert polynomial depends upon a projective embedding of the surface.
Let H be a very ample divisor on S such that |H| : S ֒→ Pdim |H| is an
embedding.
Then the Hilbert polynomial of S ⊂ Pdim |H| is
1 2
H2 2
n +
H + 1 − πH n + χ(OS ).
χ(OS (nH)) =
2
2
Abdul Moeed Mohammad (University of Oslo)
June 2010
2/1
Choice of invariants
Let S ⊂ P5 be a smooth rational surface. We choose dS = 11 and πS = 8.
Reason: Arrondo and Gross have classified rational surfaces of degree ≤ 10
contained in smooth quadrics. Our choice does not interfere with their
classification.
Main problem
Does there exist a smooth linearly normal rational surface in P5 with
n
P(n) = 11
+ 4n + 1
2
as its Hilbert polynomial?
Answer: Yes.
Abdul Moeed Mohammad (University of Oslo)
June 2010
3/1
Adjunction theory I
Suppose H is very ample divisor on S such that H 2 = 11 og pa (H) = 8.
Can we find an explicit description of H?
Key idea
Consider H + KS .
Sommese and Van de Ven prove that |H + KS | is basepoint free except
when S ≃ P2 or S ≃ F0 or S ≃ F1 .
Our case: May assume |H + KS | er basepoint free. Obtain a morphism
ϕ : S → Pdim |H+KS | .
Abdul Moeed Mohammad (University of Oslo)
June 2010
4/1
Adjunction theory II
If |H + KS | is basepoint free, then Sommese has proved:
(1). dim ϕ(S) = 0 ⇒ H ≡ −KS .
(2). dim ϕ(S) = 1 ⇒ S is ruled in conics.
(3). dim ϕ(S) = 2 ⇒ degϕ ∈ {1, 2}.
Case dim ϕ(S) = 2 and deg ϕ = 2, Sommese and Van de ven, proved that
S ≃ P̃2 (x1 , .., x7 ) or S ≃ P̃2 (x1 , .., x8 ). They describe H as well.
i.
ii.
iii.
iv.
v.
vi.
Input: H 2 and pa (H)
Describe H by (1) and (2).
Describe H when dim ϕ(S) = 2 and deg ϕ = 2.
Case dim ϕ(S) = 2 and deg ϕ = 1:
If possible, use earlier classifications.
Else, run i. with input: (H + KS )2 and pa (H + KS ).
Algorithm terminates when H 2 = 11 and pa (H) = 8.
Abdul Moeed Mohammad (University of Oslo)
June 2010
5/1
Possibly very ample divisors I
Theorem
Suppose there exists a smooth linearly normal rational surface S of degree
11 and sectional genus 8, with an embedding i : S ֒→ P5 . Let
OS (H) ≃ i ∗ OP5 (1). At least one of the following holds:
(1). |H + nKS | maps S to a curve, for some n ≥ 1.
(2). |H + nKS | gives a 2 : 1 map of S to some surface, for some n > 1.
(3). H linearly equivalent to at least one of:
Abdul Moeed Mohammad (University of Oslo)
June 2010
6/1
Possibly very ample divisors II
6L −
P2
i =1
5B + 5F −
2Ei −
P9
i =1
P11
P19
j=3
2Ei −
Ej
P16
j=10
P17
i =1 2Ei −
j=8 Ej
P
P
P
9L − 6i =1 3Ei − 8j=7 2Ej − 16
k=9 Ek
P5
P10
P15
9L − i =1 3Ei − j=6 2Ej − k=11 Ek
P
8L − 13
i =1 2Ei − E14
P
P
10L − 8i =1 3Ei − 12
j=9 2Ej − E13
P6
P9
P
12L − i =1 4Ei − j=7 3Ej − 11
k=10 2Ek
P
P
P12
12L − 5i =1 4Ei − 10
3E
−
j
j=6
k=11 2Ek
P9
13L − i =1 4E1 − 3E10 − 2E11 − E12
P
14L − 5E1 − 11
i =2 4Ei
P8
P
16L − 6E1 − i =2 5Ei − 4E9 − 11
j=10 3Ej
P9
19L − i =1 6Ei − 5E10 − E11
P
25L − 9E1 − 8i =2 8Ei − 7E9 − 6E10
7L −
Ej
P15
8L − 3E1 − i =2 2Ei − j=12 Ej
P
P
10L − 4E1 − 8i =2 3Ei − 2E9 − 15
j=10 Ej
P9
P14
10L − i =1 3Ei − 2E10 − k=11 Ek
P
P13
7B + 7F − 10
i =1 3Ei − 2E11 −
j=12 Ej
P
11L − 4E1 − 11
3E
−
2E
12
i
i =2
P
P
13L − 5E1 − 8i =2 4E1 − 3E9 − 12
j=10 2Ej
P
P
P
16L − 6E1 − 8j=2 5Ej − 10
4E
− 12
k
k=9
t=11 Et
P5
P10
15L − i =1 5Ei − j=6 4Ej − 3E11
P
16L − 9i =1 5Ei − 4E10 − 2E11
P
P
24L − 5i =1 8Ei − 10
j=6 7Ej
P7
−
Question: Does every divisor class above give an embedding S ֒→ P5 ?
Answer: No, for 20 classes. Yes, for 1 class. Maybe, for 3 classes.
Abdul Moeed Mohammad (University of Oslo)
June 2010
7/1
Reducing the possibilities
Curves on the latter slide satisfy h2 (OS (H)) = 0.
Riemann-Roch: χ(OS (H)) = 12 H(H − KS ) + 1 = 5.
h0 (OS (H)) = 6 if and only if the speciality h1 (OS (H)) = 1.
Key idea
Let S ≃ P̃2 (x1 , .., xr ). Speciality of rational surfaces "measures" how far
x1 , .., xr are from being in general position.
Example
H
≡
≡
P
11L − 4E1 − 11
i =2 3Ei − 2E12
10
10
X
X
Ei + 8L − 3E1 −
2Ei − 3E11 − 2E12
3L −
|
{zi =1 }
A
i =2
|
{z
B
}
A effective on S. Points x1 , .., x10 lie on a cubic. H not very ample.
Abdul Moeed Mohammad (University of Oslo)
June 2010
8/1
Possibly very ample divisors III
Theorem
Suppose there exists a smooth linearly normal rational surface S of degree
11 and sectional genus 8, with an embedding i : S ֒→ P5 . Let
OS (H) ≃ i ∗ OP5 (1). At least one of the following holds:
(1). |H + nKS | maps S to a curve, for some n ≥ 1.
(2). |H + nKS | gives a 2 : 1 map of S to some surface, for some n > 1.
(3). H linearly equivalent to at least one of:
P2
P19
6L − i=1 2Ei − j=3 Ej .
P7
P17
7L − i=1 2Ei − j=3 Ej .
P6
P8
P16
9L − i=1 3Ei − j=7 2Ej − k=9 Ek .
P15
P8
10L − 4E1 − i=2 3Ei − 2E9 − j=10 Ej .
Abdul Moeed Mohammad (University of Oslo)
June 2010
9/1
Example: Lifting arguments
H
≡
≡
P
P
10L − 4E1 − 8i =2 3Ei − 2E9 − 15
j=10 Ej
15
9
8
X
X
X
Ej + 3L −
2Ei −
Ei
7L − 3E1 −
i =2
|
{z
A
j=10
}
|
(1). Suppose h0 (OS (H)) = 6. Then A effective on S.
{zi =1 }
B
(2). OA (H) ≃ ωA .
(3). Suppose A is smooth on S. Then OA (B − KS ) ≃ OA .
(4). Vanishing of H 1 (OS (B − KS − A)) and cohomology of
0 −→ OS (B − KS − A) −→ OS (B − KS ) −→ OA −→ 0
implies that B − KS is effective on S.
But, H.(B − KS ) = 2 and pa (B − KS ) = 2 contradicts very ampleness of H.
Abdul Moeed Mohammad (University of Oslo)
June 2010
10 / 1
Drawback with last argument: Assumed A is smooth on S.
Not found a way to work around the assumption.
Indication of non-existence:
P
P
6L − 2i =1 2Ei − 19
E.
P8 j=3 j
P
10L − 4E1 − i =2 3Ei − 2E9 − 15
j=10 Ej .
What about
P
P
7L − 7i =1 2Ei − 17
E.
P6
P8j=3 j
P
9L − i =1 3Ei − j=7 2Ej − 16
k=9 Ek .
We believe 9L −
P6
i =1 3Ei
−
Abdul Moeed Mohammad (University of Oslo)
P8
j=7 2Ej
−
P16
k=9 Ek
gives an embedding.
June 2010
11 / 1
Constructing rational surfaces
Main strategy: Obtain S by blowing up P2 in x1 , .., xr . Realizing x1 , .., xr on
a complete intersection.
Main problem
Does there exist a smooth linearly normal rational surface in P5 with
n
P(n) = 11
+ 4n + 1
2
as its Hilbert polynomial?
Abdul Moeed Mohammad (University of Oslo)
June 2010
12 / 1
Answer to Main problem I
Choose points x1 , .., x5 ∈ P2 . No 3 collinear.
π1 : S1 −→ P2
(π1−1 (xi ) = Ei )
P
General curve in ∆1 := |6L − P5i =1 2Ei | is smooth and irreducible.
General curve in ∆2 := |4L − 5i =1 Ei | is smooth and irreducible.
Fixate A1 ∈ ∆1 smooth and irreducible.
Possible to choose B1 ∈ ∆2 smooth and irreducible, such that A1 ∩ B1
consists of y1 , y2 , y1′ , y2′ , z1 , .., z10 such that:
(1). A1 and B1 share common tangent direction yi′ in yi .
(2). A1 and B1 intersect transversally in zi .
π2 : S −→ S1
Abdul Moeed Mohammad (University of Oslo)
(π2−1 (yi ) = Fi , π2−1 (zi ) = Gi )
June 2010
13 / 1
Answer to Main problem II
Obtain effective divisors on S
P
P
P
A ≡ 6L − 5i =1 2Ei − 2j=1 Fj − 10
G
P5
P2
P10k=1 k
B ≡ 4L − i =1 Ei − j=1 Fj − k=1 Gk
H ≡ A + L − F1 − F2
|
{z
}
C
Tangent directions y1′ , y2′ ∈ S1 correspond to points on F1 , F2 ⊂ S.
Obtain OA (B) ≃ OA (F1 + F2 ).
This yields h0 (OS (A)) = 2 and h0 (OS (H)) = 6. For very ampleness we use
Alexander-Bauer lemma
Let H ≡ A + C be effective på S. If h0 (OS (A)) ≥ 2, the maps
H 0 (OS (H)) → H 0 (OA (H)) and H 0 (OS (H)) → H 0 (OC (H)) are surjective,
H|A very ample, H|C very ample. Then H very ample on S.
Abdul Moeed Mohammad (University of Oslo)
June 2010
14 / 1
Answer to Main problem III
Since h0 (OS (H)) = 6 and H very ample
Theorem
There exists a smooth linearly normal rational surface in P5 with
n
P(n) = 11
+ 4n + 1
2
as its Hilbert polynomial.
Abdul Moeed Mohammad (University of Oslo)
June 2010
15 / 1
End
Thank you for listening.
Abdul Moeed Mohammad (University of Oslo)
June 2010
16 / 1