Birational geometry of complete intersections

Hans-Georg Rackwitz Martin-Luther-Universität Halle-Wittenberg FB Mathematik und Informatik 06099 Halle (Saale), Germany

Abstract For any smooth projective variety Y ⊆ ℙ_{k}^{n} the finite-dimensional vector space H^{0}(Y, S^{r}Ω^{1}) of global symmetric differential forms on Y is a birational invariant of Y . In the case of smooth complete intersections Y some vanishing theorems are proved. Moreover, explicit formulas for the dimensiones dim_{k} H^{q}(Y, S^{r}Ω^{1}(p)) ( q ∈ {0, 1, dim Y − codim Y, dim Y − r}) and for the Euler-Poincaré characteristic χ(Y, S_{Y}^{r} (p)) are given.

Introduction

One of the difficult problems of the algebraic geometry is the classification problem - to classify varieties up to birational equivalence. We have some classical birational invariants of algebraic varieties Y ⊆ ℙ_{k}^{n}, for example the geometric genus, the plurigenera or the Hodge-numbers h^{p0}. We also know that for any smooth projective variety Y the finite-dimensional vector space H^{0}(Y, S^{r}Ω^{1}) of global symmetric differential forms on Y is a birational invariant of Y [6]. Many of these invariants are scarcely explicit computable in practice. We will study symmetric differential forms on smooth complete intersections Y ⊆ ℙ_{k}^{n} in this paper. At first we show vanishing theorems . Part a) of Theorem 3 already appears in [3] but only under the premise that H_{1} ∩ H_{2} ∩ … ∩ H_{i} is a smooth complete intersection for any i ∈ {1, 2, …,codim Y }. Let T be a Young tableau with r cells. In the special case depth T = 1 one gets the vanishing theorems 3a) and 4a) also from Theorem 4 in [5]. For the sake of completeness we include also a new short proof of this results . In the second section of this paper we prove formulas for the explicit computation of the dimensions of some cohomology groups H^{q}(Y, S^{r}Ω^{1}(p)). The following example shows among others that the birational invariants dim_{k} H^{0}(X, S^{r}Ω^{1} ⊗ ω_{X}^{s}) (s ∈ ℕ) are independent of the classical invariants : Let X ⊆ ℙ_{k}^{n} be a smooth d-dimensional complete intersection of multi-degree 𝔪 = (m_{1}, m_{2}, …, m_{c}) with d ≥ 2 and m_{i} ≥ 2 ∀i (c = n − d). From the Riemann-Roch- theorem follows that the classical birational invariants depend only on the elementary symmetrical polynomials σ_{1}, σ_{2}, …, σ_{d} and σ_{c} of m_{1}, m_{2}, …, m_{c} if d < c. Using this fact and the computer we got two 6-tuples of integers (13, 20, 22, 45, 45, 57) and (15, 15, 27, 38, 52, 55) with the same elementary symmetrical polynomials σ_{1}, σ_{2}, σ_{3}, σ_{4} and σ_{6}. By Bertini’s theorem nonsingular complete intersections X, Y ⊆ ℙ_{k}^{10} of multi-degree 𝔪_{X} = (13, 20, 22, 45, 45, 57) and 𝔪_{Y} = (15, 15, 27, 38, 52, 55) respectively, exist. Then we have :

P_{f}(X) = P_{f}(Y) = 5, 501, 925 ·
(6, 654, 316, 805 · f^{4} − 13, 308, 633, 610 · f^{3}
+10, 742, 195, 260 · f^{2} − 4, 087, 878, 455 · f
+612, 522, 757) (f > 1);
p_{g}(X) = p_{g}(Y) = 3, 370, 054, 269, 807, 224;
h^{r0}(X) = h^{r0}(Y) = 0 (0 < r < 4);
χ(X, S^{2}Ω^{1} ⊗ ω_{X}) = χ(Y, S^{2}Ω^{1} ⊗ ω_{Y}) = 114, 167, 699, 169, 059, 250,

but

dim_{k} H^{0}(X, S^{2}Ω^{1} ⊗ ω_{X}) = 114, 087, 667, 319, 096, 650

and

dim_{k} H^{0}(Y, S^{2}Ω^{1} ⊗ ω_{Y}) = 114, 086, 912, 974, 477, 150.

The dimensions h^{0}(X, S^{2}Ω^{1} ⊗ω_{X}) and h^{0}(Y, S^{2}Ω^{1} ⊗ω_{Y}) have been computed by the formula of Theorem 5. Finally we make a short remark about the application of Serre duality to our results.

0. Preliminaries

Let r ≥ 0 be an integer, k be an algebraically closed ground field of char k = 0 or char k > r. Y = H_{1} ∩ H_{2} ∩ … ∩ H_{c} ⊆ ℙ_{k}^{n} denotes a d-dimensional (d > 1) smooth complete intersection by the hypersurfaces H_{i} ⊆ ℙ_{k}^{n} =: ℙ^{n}(c := n − d). We set m_{i} := deg H_{i} (i = 1, 2, …, c) and without loss of generality we assume 2 ≤ m_{1} ≤ … ≤ m_{c}. The multiplicity of the canonical class is n_{Y} = ∑_{i=1}^{c} m_{i} − n − 1 and one has ω_{Y} = O_{Y} (n_{Y})- the canonical bundle on Y. The restriction of the bundle ⨁_{i=1}^{c} O_{ℙ^{n}}(m_{i}) to the complete intersection Y is its normal bundle. Throughout let p ∈ ℤ. For simplification we set :

S_{ℙ^{n}}^{r} := S^{r}Ω_{ℙ^{n}}^{1} , S_{Y}^{r} := S^{r}Ω_{Y}^{1} , S_{ℙ^{n}}^{0} := O_{ℙ^{n}} , S_{Y}^{0} := O_{Y} ,

E^{∗} := ⨁_{i=1}^{c}O_{ℙ^{n}}(−m_{i}) , N^{∗} := ⨁_{i=1}^{c}O_{Y} (−m_{i}) ,

F_{i}^{r} := ⋀^{i}E^{∗} ⊗ S_{ℙ^{n}}^{r} (i = 1, 2, …, c) , F_{0}^{r} := S_{ℙ^{n}}^{r} ,

A^{r} := O_{Y} ⊗ S_{ℙ^{n}}^{r}.

Knowing the Koszul-resolution of O_{Y} we receive the following exact sequence :

0 \overline{}F_{c}^{r} ^{\overline{}^{ϕ_{c}}} F_{c−1}^{r} ^{\overline{}^{ϕ_{c−1}}} … ^{\overline{}^{ϕ_{1}}} S_{ℙ^{n}}^{r} \overline{}A^{r} \overline{}0 (1)

We have the short exact sequence

0 \overline{}N^{∗} \overline{}A \overline{}Ω_{Y}^{1} \overline{}0

where N is the normal bundle of Y . Therefore by considering the r-fold symmetric product of the complex 0 \overline{}N^{∗} \overline{}A we get in the case c ≤ r the exact sequence

0 \overline{}⋀^{c}N^{∗} ⊗ S_{ℙ^{n}}^{r−c} ^{\overline{}^{γ_{0}}} ⋀^{c−1}N^{∗} ⊗ S_{ℙ^{n}}^{r−c+1} ^{\overline{}^{γ_{1}}} … ^{\overline{}^{γ_{c−1}}} A^{r} \overline{}S_{Y}^{r} \overline{}0. (2)

The case c ≥ r leads to the exact sequence

0 \overline{}⋀^{r}N^{∗} ^{\overline{}^{γ_{c−r}}} ⋀^{r−1}N^{∗} ⊗ Ω_{ℙ^{n}}^{1} ^{\overline{}^{γ_{c−r+1}}} … ^{\overline{}^{γ_{c−1}}} A^{r} \overline{}S_{Y}^{r} \overline{}0. (3)

Additionaly we will make use of the following known facts (n > 1) (comp.[2] in the case char k = 0 and [3] in the case char k > r) :

(4)  H^{q}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = 0 for 1 < q < n,
(5)  H^{0}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = 0 iff p < 2r,
(6)  H^{1}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = 0 iff p ≥ 2r − 1 or p ≤ r − 2,
(7)  H^{n}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = 0 iff p ≥ r − n,
(8)  h^{0}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = \frac{1}{n}\binom{r + n − 1}{n − 1}\binom{p − 2r + 1}{1}\binom{p + n − r}{n − 1},
(9)  h^{1}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = \frac{1}{n}\binom{r + n − 1}{n − 1}\binom{2r − p − 1}{1}\binom{p + n − r}{n − 1},
(10)  h^{n}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = \frac{1}{n}\binom{r + n − 1}{n − 1}\binom{2r − p − 1}{1}\binom{r − p − 2}{n − 1},
(11)  χ(Y, S_{Y}^{r} (p)) = \frac{1}{n!}\binom{r + n − 1}{n − 1} · (p − 2r + 1) · ∏_{i=2}^{n}(p − r + i).

Remark : Let \binom{a}{b} = 0 iff a < b (a, b ∈ ℤ, b ≥ 0). I would like to thank P. Brückmann for many inspiring and helpful discussions.

1. Vanishing theorems for symmetric tensor forms on complete intersections

We start by recording two elementary lemmas.

Lemma 1. Let 0 \overline{}C_{m} ^{\overline{}^{ϕ_{m}}} C_{m−1} ^{\overline{}^{ϕ_{m−1}}} … ^{\overline{}^{ϕ_{1}}} C_{0} \overline{}E \overline{}0 be an exakt sequence of sheaves on some variety X. Then H^{q}(X, E) = 0 as soon as for every i ∈ {0, 1, …, m} H^{q+i}(X, C_{i}) = 0.

Proof. We cut the exact sequence above :

0 \overline{}Im ϕ_{1} \overline{}C_{0} \overline{}E \overline{}0,
0 \overline{}Im ϕ_{i} \overline{}C_{i−1} \overline{}Im ϕ_{i−1} \overline{}0, (i = 2, 3, …, m − 1),
0 \overline{}C_{m} \overline{}C_{m−1} \overline{}Im ϕ_{m−1} \overline{}0.

The Lemma follows immediately from the corresponding exact cohomology sequences. □

Lemma 2.

a) H^{q}(Y, A^{r}(p)) = 0 for 1 < q < d;
b) H^{1}(Y, A^{r}(p)) = 0 for p ≥ 2r − 1 or p ≤ r − 2;
c) H^{d}(Y, A^{r}(p)) = 0 for p ≥ r + n_{Y} + 1.

Proof. We prove c). Because of sequence (1) and Lemma 1 we have to show that

H^{d+i}(ℙ^{n}, F_{i}^{r}(p)) = 0 (i = 0, 1, …, c).

We have presently d + i > 1 (d > 1) . Using (4) we see that only the case d + i = n, e.g. i = n − d = c is interesting. With p ≥ r + n_{Y} + 1 and (7) we obtain H^{n}(ℙ^{n}, F_{c}^{r}(p)) = 0. a) and b) are proved in the same way. □ Now we can prove our vanishing theorems.

Theorem 3.

a) H^{q}(Y, S_{Y}^{r} (p)) = 0 for 1 < q < d − min{c, r};
b) H^{q}(Y, S_{Y}^{r} (p)) = 0 for 1 < q, q ≥ d − min{c, r} and

p > n_{Y} + ∑_{i=n−2d+q+1}^{c} m_{i} + r + q − d;

c) H^{1}(Y, S_{Y}^{r} (p)) = 0 for p ≥ max{2r − 1, n_{Y} + ∑_{i=n−2d+2}^{c} m_{i} + r − d + 2}.

Proof. a) We cut the exact sequence (2) (or (3)) and get the following exact cohomology sequence :

(∗) H^{q}(Y, A^{r}(p)) \overline{}H^{q}(Y, S_{Y}^{r} (p)) \overline{}H^{q+1}(Y, Im γ_{c−1}(p)).

From 1 < q < d − min{c, r} follows 1 < q < d, e.g. H^{q}(Y, A^{r}(p)) = 0 ( Lemma 2a)). By (∗) it is enough to prove that H^{q+1}(Y, Im γ_{c−1}(p)) = 0 . Thus we use (2) (or (3)) and obtain the exact sequence

0 \overline{}⋀^{c}N^{∗} ⊗ S_{ℙ^{n}}^{r−c} ^{\overline{}^{γ_{0}}} ⋀^{c−1}N^{∗} ⊗ S_{ℙ^{n}}^{r−c+1} ^{\overline{}^{γ_{1}}} … ^{\overline{}^{γ_{c−1}}} Im γ_{c−1} \overline{}0

( 0 \overline{}⋀^{r}N^{∗} ^{\overline{}^{γ_{c−r}}} ⋀^{r−1}N^{∗} ⊗ Ω_{ℙ^{n}}^{1} ^{\overline{}^{γ_{c−r+1}}} … ^{\overline{}^{γ_{c−1}}} Im γ_{c−1} \overline{}0 ).

Because of Lemma 1 we have to show

H^{q+j}(Y, ⋀^{j}N^{∗} ⊗ S_{ℙ^{n}}^{r−j}(p)) = 0, for j = 1, 2, …, min{c, r}.

This is evident through 1 < q + j < d and Lemma 2a).

b) From the proof of part a) we know that we must show

H^{q+i}(Y, ⋀^{i}N^{∗} ⊗ S_{ℙ^{n}}^{r−i}(p)) = 0, for i = 0, 1, …, min{c, r}, (⋀^{0}N^{∗} = O_{Y}).

Because of q > 1 the only interesting case for us is now q + j = d (j = 0, 1, …, min{c, r}) : With our hypothesis

p > n_{Y} + ∑_{i=n−2d+q+1}^{c} m_{i} + r + q − d
we have  p > n_{Y} + ∑_{s=1}^{d−q} m_{i_{s}} + r + q − d
and p − ∑_{s=1}^{d−q} m_{i_{s}} > n_{Y} + r (q ≤ d).

We would like to remind that

⋀^{j}N^{∗} ⊗ S_{ℙ^{n}}^{r−j}(p) = ⨁_{1≤i_{1}<...<i_{j}≤c}O_{Y} (p − m_{i_{1}} − m_{i_{2}} − … − m_{i_{j}}) ⊗ S_{ℙ^{n}}^{r−j}.

Application of Lemma 2c) proves b).

c) We use (∗) with q = 1. Let p ≥ max{2r − 1, n_{Y} + ∑_{i=n−2d+2}^{c} m_{i} + r − d + 2} From Lemma 2b) follows H^{1}(Y, A^{r}(p)) = 0. H^{2}(Y, Im γ_{c−1}(p)) = 0 we obtain from Lemma 2c) like above, which completes the proof. □

Remark . Theorem 3a) shows, that the vanishing of H^{q}(Y, S_{Y}^{r} (p)) in case 1 < q < d − min{r, c} is independent from the twist p ∈ ℤ of the sheaf S_{Y}^{r} . In like manner one proves the following theorem.

Theorem 4.

a) H^{0}(Y, S_{Y}^{r} (p)) = 0 for min{c, r} < d and p < r + min{r, m_{1} − 2};
b) H^{1}(Y, S_{Y}^{r} (p)) = 0 for min{c, r} < d − 1 and (p ≤ r − 2 or p ≥ 2r − 1).
2. Some formulas for dim_{k} H^{q}(Y, S_{Y}^{r}(p))

In the following part of this paper we will prove formulas for

h^{i}(Y, S_{Y}^{r} (p)) = dim_{k} H^{i}(Y, S_{Y}^{r} (p)) (i ∈ {0, 1, d − c, d − r})

and for the Euler-Poincaré characteristic χ(Y, S_{Y}^{r} (p)). We use the same notations as above. Furthermore we set L_{i} := ⋀^{i} E^{∗} (i ∈ ℕ, L_{0} = O_{ℙ^{n}}).

Theorem 5. If min{r, c} < d − 1 and r < m_{1} , then

a) h^{0}(Y, S_{Y}^{r} (p)) = ∑_{i=0}^{min{c,r}} ∑_{j=0}^{c}(−1)^{i+j} · h^{0}(ℙ^{n}, F_{j}^{r−i}(p) ⊗ L_{i})
− ∑_{i=0}^{min{c,r}} ∑_{j=1}^{c}(−1)^{i+j} · h^{1}(ℙ^{n}, F_{j}^{r−i}(p) ⊗ L_{i})
− ∑_{i=1}^{min{c,r}}(−1)^{i} · h^{1}(ℙ^{n}, F_{i}^{r−i}(p)).
b) h^{1}(Y, S_{Y}^{r} (p)) = h^{1}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = \frac{1}{n}\binom{r + n − 1}{n − 1}\binom{2r − p − 1}{1}\binom{p + n − r}{n − 1}.

Proof. We use the exact sequence (comp.(1)) 0 \overline{}H^{0}(ℙ^{n}, Im ϕ_{1}(p)) \overline{}H^{0}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) \overline{}H^{0}(Y, A^{r}(p)) \overline{} \overline{}H^{1}(ℙ^{n}, Im ϕ_{1}(p)) \overline{}H^{1}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) . From r < m_{1} follows that one of the two cohomology groups H^{1}(ℙ^{n}, Im ϕ_{1}(p)) and H^{1}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) is trivial. This is an immediate consequence of the exact sequence H^{1}(ℙ^{n}, F_{1}^{r}(p) \overline{}H^{1}(ℙ^{n}, Im ϕ_{1}(p)) \overline{}H^{2}(ℙ^{n}, Im ϕ_{2}(p)) . Let us first suppose that H^{1}(ℙ^{n}, F_{1}^{r}(p) = 0 . From the exact sequence

0 \overline{}F_{c}^{r} ^{\overline{}^{ϕ_{c}}} F_{c−1}^{r} ^{\overline{}^{ϕ_{c−1}}} … ^{\overline{}^{ϕ_{3}}} F_{2}^{r} ^{\overline{}^{ϕ_{2}}} Im ϕ_{2} \overline{}0

and from Lemma 1 one has H^{2}(ℙ^{n}, Im ϕ_{2}(p)) = 0, since H^{i+2}(ℙ^{n}, F_{i+2}^{r}(p)) = 0 ∀i = 0, 1, …, c − 2 (comp. (4)). Let us secondly suppose that H^{1}(ℙ^{n}, F_{1}^{r}(p) ≠ 0 . From (6) we get now p−m_{i} > r−2 for one i ∈ {1, 2, …, c}, i.e. p − m_{1} > r − 2. That means p ≥ 2r − 1 because of m_{1} > r and we have now H^{1}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)) = 0 by (6). As above one shows that H^{1}(ℙ^{n}, Im ϕ_{j}(p)) = 0 or H^{1}(ℙ^{n}, F_{j−1}^{r}(p)) = 0 for all j = 2, 3, …, c − 1 and one of the two cohomology groups H^{1}(ℙ^{n}, F_{c−1}^{r}(p)) and H^{1}(ℙ^{n}, F_{c}^{r}(p)) is trivial. For this reason and because of the corresponding cohomology sequences we obtain from (1), we can now write down the formula :

h^{0}(Y, A^{r}(p)) = ∑_{i=0}^{c}(−1)^{i} · h^{0}(ℙ^{n}, F_{i}^{r}(p) + ∑_{i=0}^{c−2}(−1)^{i} · h^{1}(ℙ^{n}, Im ϕ_{i+1}(p))

(12)

+ (−1)^{c−1} · h^{1}(ℙ^{n}, F_{c}^{r}(p)).

In the next step one shows as above

h^{1}(ℙ^{n}, Im ϕ_{j}(p)) = h^{1}(ℙ^{n}, F_{j}^{r}(p)) (j = 1, 2, …, c − 1)

and we obtain from (12)

(13) h^{0}(Y, A^{r}(p)) = ∑_{i=0}^{c}(−1)^{i} · h^{0}(ℙ^{n}, F_{i}^{r}(p)) − ∑_{i=1}^{c}(−1)^{i} · h^{1}(ℙ^{n}, F_{i}^{r}(p)).

From (2) (or (3)) we get the exact cohomology sequence

0 \overline{}H^{0}(Y, Im γ_{c−1}(p)) \overline{}H^{0}(Y, A^{r}(p)) \overline{}H^{0}(Y, S_{Y}^{r} (p)) \overline{}
\overline{}H^{1}(Y, Im γ_{c−1}(p)) \overline{}H^{1}(Y, A^{r}(p)) \overline{}H^{1}(Y, S_{Y}^{r} (p)).

Again from (2) (or (3)) follows by using Lemma 1 and Lemma 2 that one of the two cohomology groups H^{1}(Y, Im γ_{c−1}(p)) and H^{1}(Y, A^{r}(p)) is trivial. Therefore we have

h^{0}(Y, S_{Y}^{r} (p)) = h^{0}(Y, A^{r}(p)) − h^{0}(Y, Im γ_{c−1}(p)) + h^{1}(Y, Im γ_{c−1}(p)).

In order to prove our theorem we have to compute h^{0}(Y, Im γ_{c−1}(p)) and h^{1}(Y, Im γ_{c−1}(p)). Using the first computations and the exact sequences one obtains resumptive

h^{0}(Y, S_{Y}^{r} (p)) = ∑_{i=0}^{min{c,r}}(−1)^{i} · h^{0}(Y, ⋀^{i}N^{∗} ⊗ S_{ℙ^{n}}^{r−i}(p))
− ∑_{i=1}^{min{c,r}}(−1)^{i} · h^{1}(Y, ⋀^{i}N^{∗} ⊗ S_{ℙ^{n}}^{r−i}(p)).

Similarly one proves that H^{1}(Y, S_{Y}^{r} (p)) ≅ H^{1}(Y, A^{r}(p)) ≅ H^{1}(ℙ^{n}, S_{ℙ^{n}}^{r}(p)). Applying

(13) one completes the proof. □

It’s also interesting to know the dimensiones of H^{q}(Y, S_{Y}^{r} (p)) for q > 1 . In some special cases such results can be deduced from the exact sequences (1), (2) and (3).

Theorem 6. Let 1 < d − c, c ≤ r and p ≥ 2 · n_{Y} − m_{1} + d + r + 3. Then

a) h^{d−c}(Y, S_{Y}^{r} (p)) = \frac{1}{n}\binom{r + d − 1}{n − 1}\binom{2 · n_{Y} + 2(r + d) + 1 − p}{1}\binom{2 · n_{Y} + r + d + n − p}{n − 1}.
b) H^{d−c}(Y, S_{Y}^{r} (p)) = 0 iff r < c or p > 2 · n_{Y} + r + d + 1.

If r ≤ c, 1 < d − r and if p ≥ n_{Y} + ∑_{i=c−r+2}^{c} m_{i} + 2 then holds

c) h^{d−r}(Y, S_{Y}^{r} (p)) = h^{d}(Y, (⋀^{r}N^{∗})(p)) with h^{d}(Y, O_{Y} (p)) = f(n_{Y} − p) and
f(p) = \binom{p + n}{n} + ∑_{j=1}^{c}(−1)^{j} ∑_{1≤i_{1}<...<i_{j}≤c} \binom{p + n − m_{i_{1}} − … − m_{i_{j}}}{n}.

Proof. a) Using (2) one shows firstly

H^{d−c}(Y, S_{Y}^{r} (p)) ≅ H^{d}(Y, A^{r−c}(p − m_{1} − m_{2} − … − m_{c})).

In the second step one shows by inspection of (1) the isomorphism

H^{d}(Y, A^{r−c}(p − m_{1} … − m_{c})) ≅ H^{n}(ℙ^{n}, S_{ℙ^{n}}^{r−c}(p − 2 · n_{Y} + 2 (n − 1))).

Now we get a) from (10). b) is an immediate consequence of a). Part c) is done in the same way like a) by using sequences (3) and (1). □

Corollary. We have H^{d−c}(Y, S_{Y}^{r} (p)) ≠ 0 if c ≤ r and 3 − m_{1} ≤ p − 2 · n_{Y} − d − r ≤ 1. For this reason a non-empty interval I ⊂ ℤ with H^{d−c}(Y, S_{Y}^{r} (p)) ≠ 0 for p ∈ I exists.

The Euler-Poincaré characteristic χ(Y, S_{Y}^{r} (p)) is a polynomial in p with rational coefficients. We compute this polynomial :

Theorem 7. With our notations we have

χ(Y, S_{Y}^{r} (p)) = ∑_{i=0}^{min{c,r}} ∑_{j=0}^{c}(−1)^{i+j} · χ(ℙ^{n}, F_{j}^{r−i}(p) ⊗ L_{i}).

Proof. This is an immediate consequence of (1) and (2). □

Remark (Serre duality): From [4] we know, that H^{q}(Y, S_{Y}^{l} (p)) ≈ H^{d−q}(Y, Ω_{Y}^{T} ((1 − l) · n_{Y} − p)). Ω_{Y}^{T} is called the sheaf of germs of T -symmetrical tensor forms over Y and T is the Young tableau with d − 1 rows, l columns and r = (d − 1) · l cells ( T is a rectangle ( comp. [5] )). We remark, that for each integer a ≥ 0

dim_{k} H^{d}(Y, S_{Y}^{l} ((1 − l − a) · n_{Y})) = dim_{k} H^{0}(Y, Ω_{Y}^{T} (a · n_{Y})) and
dim_{k} H^{d}(Y, Ω_{Y}^{T} ((1 − l − a) · n_{Y})) = dim_{k} H^{0}(Y, S_{Y}^{l} (a · n_{Y}))

are also birational invariants of Y. In the case d = 2 one has Ω_{Y}^{T} = S_{Y}^{r}.

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