On cohomology of complete intersections with coefficients in the twisted sheaf of differential forms in the case of prime characteristic

By P. Brückmann and H.-G. Rackwitz

Abstract The cohomology groups of nonsingular projective complete intersections with coefficients in Ω^{r}(t) depend on the fact which of the defining degrees are divisible by char k . The k-dimensions of these groups will be calculatet, and for some higher cohomology groups k-basises are explicitely given. The regarded phenomenons occur only for complete intersections of dimension ≥ 3.

Introduction. In 1954 F. Hirzebruch [10] as an application of the Riemann-Roch theorem computed in the complex case the Euler characteristic χ(Y, Ω_{Y}^{r} (t)) and the dimensions of the cohomology groups H^{q}(Y, Ω_{Y}^{r}) of a nonsingular projective complete intersection Y ⊆ ℙ_{ℂ}^{n}. In 1977 I. Naruki [13] by means of the theory of isolated singularities calculated all dimensions dim_{ℂ} H^{q}(Y, Ω_{Y}^{r} (t)). In particular one knows dim_{ℂ} H^{q}(Y, Ω_{Y}^{r} (t)) = δ_{q,r} · δ_{t,0} with the Kronecker-δ if 0 < q < dim Y and q+r ≠ dim Y. We will see that even some of these dimensions may be arbitrarily great if the characteristic is positive. The object of this paper is to calculate via a purely algebraic approach all dimensions dim_{k} H^{q}(Y, Ω_{Y}^{r} (t)). Moreover k-basises and explicit formulas for the k-dimensions of these cohomology groups will be given. It will be shown that dim_{k} H^{q}(Y, Ω_{Y}^{r} (t)) depends only on q, r, t and on the dimension d, the codimension c and the multidegree (m_{1}, m_{2}, …, m_{c}) of Y and finally on char k. More precisely the formulas depend only on the fact, wich of the integers m_{1}, m_{2}, …, m_{c} are divisible by char k. A very simple example is a 3-dimensional quadric Y ⊂ ℙ_{k}^{4} with H^{0}(Y, Ω_{Y}^{2} (2)) = 0 for char k ≠ 2 , but H^{0}(Y, Ω_{Y}^{2} (2)) ≅ k for char k = 2. We will see that the Hilbert-polynomials χ(Y, Ω_{Y}^{r} (t)), the dimensions h^{q}(Y, Ω_{Y}^{r}) (without twist), h^{q}(Y, O_{Y} (t)), h^{q}(Y, Ω_{Y}^{d} (t)), h^{0}(Y, Ω_{Y}^{1} (t)) and h^{d}(Y, Ω_{Y}^{d−1}(t)) are independent of char k. The same is true for all dimensions h^{q}(Y, Ω_{Y}^{r} (t)) if Y is a curve or surface. If not one integer m_{j} is divisible by char k or if t is not divisible by char k then dim_{k} H^{q}(Y, Ω_{Y}^{r} (t)) is given by the same formulas like in the complex case (cp. Corollary 10). Note that for instance the birational invariants dim_{k} H^{0}(Y, Ω_{Y}^{r} ⊗ ω_{Y}^{s}) = dim_{k} H^{0}(Y, Ω_{Y}^{r} (s · n_{Y})) in general depend on char k and that the plurigenera are independent of char k. For the vanishing of the higher cohomology groups H^{q}(Y, Ω_{Y}^{r} (t)) with 0 < q < d and q + r ≠ d we receive a purely arithmetical necessary and sufficient condition. Some of these results already appear in [3] but only under the additional premise that (F^{(1)}, F^{(2)}, …, F^{(j)}) is a nonsingular irreducible ideal theoretic complete intersection for each j ∈ {1, 2, …, c} where (F^{(1)}, F^{(2)}, …, F^{(c)}) denotes the ideal of Y.

0. Prelimimaries

Let r > 0 be a fixed integer, k be an algebraically closed ground field with arbitrary characteristic p and let Y ⊆ ℙ_{k}^{n} be a smooth d-dimensional ideal theoretic complete intersection of multi-degree 𝔪 = (m_{1}, m_{2}, …, m_{c}) with m_{i} ≥ 2 ∀i, Id(Y) = (F^{(1)}, F^{(2)}, …, F^{(c)}) and degF^{(i)} = m_{i} (c = n − d). For any combination K_{s} = (j_{1}, j_{2}, …, j_{s}) of {1, 2, …, c}, s at a time, we set m_{K_{s}} := m_{j_{1}} + … + m_{j_{s}}. For simplification we set furthermore:

Ω_{ℙ^{n}}^{0} := O_{ℙ^{n}} , Ω_{Y}^{0} := O_{Y} E^{∗} := ⨁_{i=1}^{c}O_{ℙ^{n}}(−m_{i}) , N^{∗} := ⨁_{i=1}^{c}O_{Y} (−m_{i}) = O_{Y} ⊗E^{∗} ,

A^{j} := O_{Y} ⊗ Ω_{ℙ^{n}}^{j} (j = 0, 1, …, r), F_{0}^{r} := Ω_{ℙ^{n}}^{r} and

F_{s}^{r} := ^{⋀s}E^{∗} ⊗ Ω_{ℙ^{n}}^{r} = ⨁_{K_{s}}Ω_{ℙ^{n}}^{r}(−m_{K_{s}}),

where for any s ∈ {1, 2, …, c} K_{s} runs through all combinations of {1, 2, …, c}, s at a time, without repetition. The multiplicity of the canonical class of Y is n_{Y} = ∑_{i=1}^{c} m_{i} − n − 1 and one has ω_{Y} = O_{Y} (n_{Y}) - the canonical bundle on Y. 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_{2}}} F_{1}^{r} ^{\overline{}^{φ_{1}}} Ω_{ℙ^{n}}^{r} ^{\overline{}^{φ_{0}}} A^{r} \overline{}0  (1)

From the short exact sequence

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

with the conormal bundle N^{∗} of Y we get by considering the r-fold exterior product of the complex 0 \overline{}N^{∗} \overline{}A^{1} the exact sequence

0 \overline{}⨁_{V_{r}} O_{Y} (−m_{V_{r}}) ^{\overline{}^{γ_{r}}} … ^{\overline{}^{γ_{2}}} ⨁_{V_{1}} A^{r−1}(−m_{V_{1}}) ^{\overline{}^{γ_{1}}} A^{r} ^{\overline{}^{γ_{0}}} Ω_{Y}^{r} \overline{}0  (2)

with  ⨁_{V_{s}} A^{r−s}(−m_{V_{s}}) = ⨁_{1≤j_{1}≤...≤j_{s}≤c}O_{Y} (−m_{j_{1}} − … − m_{j_{s}}) ⊗ Ω_{ℙ^{n}}^{r−s},

i.e., V_{s} = (j_{1}, j_{2}, …, j_{s}) runs through all combinations of {1, 2, …, c} taken s at a time with repetitions. Throughout let t ∈ ℤ. Using (1) and (2) we will calculate the k-dimensions of H^{q}(Y, Ω_{Y}^{r} (t)). By Serre duality we know also

(H^{q}(Y, Ω_{Y}^{r} (t)))^{∗} ≅ H^{d−q}(Y, Ω_{Y}^{d−r}(−t))

and

(H^{q}(Y, A^{r}(t)))^{∗} ≅ H^{d−q}(Y, A^{n−r}(m_{1} + m_{2} + … + m_{c} − t)).

The cohomology groups H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)) are well known. For 0 < q < n one has independently of the ground field :

dim_{k} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)) = δ_{q,r} · δ_{t,0}  (3)

with the Kronecker-δ. The cohomology class of the cocycle ω^{(q)} with

ω_{i_{0}i_{1}...i_{q}}^{(q)} = \frac{x_{i_{0}}}{x_{i_{q}}} · d\frac{x_{i_{1}}}{x_{i_{0}}} ∧ d\frac{x_{i_{2}}}{x_{i_{1}}} ∧ … ∧ d\frac{x_{i_{q}}}{x_{i_{q−1}}}  (4)

is a k-basis of H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{q}). In the case q = 0 and 0 < r ≤ n one has [2], [3]

h^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)) = \binom{t − 1}{r} · \binom{t + n − r}{n − r} if t > r  (5)

and h^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)) = 0 otherwise. Using the Serre duality one gets for 0 ≤ r < n

h^{n}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)) = \binom{−t − 1}{n − r} · \binom{−t + r}{r} if t < r − n  (6)

and h^{n}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)) = 0 otherwise. Finally

h^{0}(ℙ^{n}, O_{ℙ^{n}}(t)) = \binom{t + n}{n} for t ≥ 0, h^{n}(ℙ^{n}, Ω_{ℙ^{n}}^{n}(t)) = \binom{−t + n}{n} for t ≤ 0.

χ(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)) = \frac{1}{n!} · \binom{n}{r} · ∏_{{i=r−n \atop i≠0}}^{r}(t − i) for 0 ≤ r ≤ n.

For t > r > 0 the global differential forms

x_{s}^{r+1} · ∑_{0≤i_{1}<...<i_{r}≤s−1} Q_{i_{1}...i_{r}}(x_{0}, x_{1}, …, x_{s}) · d\frac{x_{i_{1}}}{x_{s}} ∧ d\frac{x_{i_{2}}}{x_{s}} ∧ … ∧ d\frac{x_{i_{r}}}{x_{s}}

constitude a k-basis of H^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)), where s runs through the set {r, r+1, …, n}, and the Q_{i_{1}...i_{r}}(x_{0}, x_{1}, …, x_{s}) are homogeneous polynomials of the (t − r − 1)-th degree in the indicated variables [3].

1. Some vanishing theorems

We will need the following elementary lemma :

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

Proof. We cut the exact sequence above :

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

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

Now we define some useful sets G_{a}^{s} of integers (a, s ∈ ℤ) : For s ≥ 0 and a ≥ 0 we set G_{a}^{s} := {λ_{1}m_{1} + … + λ_{c}m_{c} | λ_{i} ∈ ℕ_{0}, ∑_{i=1}^{c} λ_{i} = s − a and ∑_{i=1}^{c} sgnλ_{i} ≥ s − 2a }. Then G_{a}^{s} = ∅ for 2a > s or c < s − 2a and G_{a}^{s} = {m_{K_{s−2a}} + m_{V_{a}}} where K_{s−2a} runs through all combinations of {1, 2, …, c} taken s − 2a at a time, without repetition and V_{a} runs through all combinations of {1, 2, …, c} taken s at a time, with repetitions. In particular one has G_{0}^{s} = {m_{K_{s}}} where K_{s} runs through all combinations of {1, 2, …, c} taken s at a time, without repetition. Finally let G_{a}^{s} := ∅ if s < 0 or a < 0.

Lemma 2. From 0 ≤ s ≤ c, 0 < q < d + s and t ∉ G_{0}^{s+r−q} follows H^{q}(ℙ^{n}, Im φ_{s}(t)) = 0 (cp. (1)).

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

H^{q+j}(ℙ^{n}, F_{j+s}^{r}(t)) = 0 (j = 0, 1, …, c − s).

From 0 < q < n − c + s follows 0 < q + j < n ∀j ∈ {0, 1, …, c − s}. In the case q +j ≠ r we are ready (cp. (3)) . Now let be q +j = r. With our premise t ∉ G_{0}^{s+r−q} we have t − m_{K_{s+r−q}} ≠ 0 and we can also use (3). □ An inverse lemma will be proved later (Corollary 2).

Corollary 1.

a) H^{q}(Y, A^{r}(t)) = 0 for 0 < q < d and t ∉ G_{0}^{r−q}.

b) H^{q}(Y, A^{r}(t)) = 0 for 0 < q < d and (r < q or q < r − c).

c) H^{q}(ℙ^{n}, Im φ_{s}(t)) = 0 for 0 ≤ s ≤ c, 0 < q < d + s and (r < q or q < r − c + s).

Proof. a) follows immedeately from Lemma 2 (A^{r} = Im φ_{0}).

b) follows from c). In order to prove c) we assume ∃j ∈ {0, 1, …, c − s} with q + j = r. Then one has (r ≥ q and r = q + j ≤ q + c − s), i.e. (r ≥ q and q ≥ r − c + s) in opposition to (r < q or q < r − c + s). □

Now we will prove the vanishing theorems (cp. Theorem 6 for a precision).

Theorem 1. Let be 0 ≤ r ≤ d.

a) H^{q}(Y, Ω_{Y}^{r} (t)) = 0 for 0 < q < d − r and t ∉ G_{a}^{r−q} ∀a ∈ \{0, 1, …, [\frac{r − q}{2} ]\}.
b) H^{q}(Y, Ω_{Y}^{r} (t)) = 0 for d − r < q < d and − t ∉ G_{a}^{q−r} ∀a ∈ \{0, 1, …, [\frac{q − r}{2}]\}.

Proof. From (2) we obtain the exact sequence

H^{q}(Y, A^{r}(t)) \overline{}H^{q}(Y, Ω_{Y}^{r} (t)) \overline{}H^{q+1}(Y, Im γ_{1}(t)).

Under the premise of a) we get H^{q}(Y, A^{r}(t)) = 0 from Corollary 1. Therefore it is sufficient to prove that H^{q+1}(Y, Im γ_{1}(t)) = 0 . Using (2) we obtain the exact sequence

0 \overline{}⨁_{V_{r}} O_{Y} (−m_{V_{r}}) ^{\overline{}^{γ_{r}}} … ^{\overline{}^{γ_{2}}} ⨁_{V_{1}} A^{r−1}(−m_{V_{1}}) ^{\overline{}^{γ_{1}}} Im γ_{1} \overline{}0.

Because of Lemma 1 we have to show H^{q+a}(Y, _{⨁V_{a}} A^{r−a}(t − m_{V_{a}})) = 0 ∀a ∈ {1, 2, …, r}. Corollary 1a) is applicable. From q > 0, q + r < d and a ∈ {1, 2, …, r} follows 0 < q + a < d. If we suppose that t − m_{V_{a}} ∈ G_{0}^{r−q−2a} we get the following : t − m_{V_{a}} = μ_{1}m_{1} + μ_{2}m_{2} + … + μ_{c}m_{c} with μ_{i} ≥ 0 ∀i, ∑_{i=1}^{c} μ_{i} = r − q − 2a ≥ 0 and ∑_{i=1}^{c} sgn(μ_{i}) ≥ r − q − 2a, i.e. t = λ_{1}m_{1} + λ_{2}m_{2} + … + λ_{c}m_{c} with ∑_{i=1}^{c} λ_{i} = r − q − a and ∑_{i=1}^{c} sgn(λ_{i}) ≥ r − q − 2a. Furthermore holds 2a ≤ r − q, so 0 ≤ a ≤ [\frac{r − q}{2} ] . Thus t ∈ G_{a}^{r−q} in contradiction to our premise t ∉ G_{a}^{r−q}. With t − m_{V_{a}} ∉ G_{0}^{r−q−2a} and Corollary 1a) follows part a) of the theorem. From the Serre duality one has the formula h^{q}(Y, Ω^{r}(t)) = h^{d−q}(Y, Ω^{d−r}(−t)). Therefore b) follows from a). □

Theorem 2. Without loss of generality we assume 2 ≤ m_{1} ≤ … ≤ m_{c}. Then one has H^{q}(Y, Ω^{r}(t)) = 0 if one of the following sufficient conditions is satisfied.

1. q + r < d, q > r,

2. q + r < d, 0 < q ≤ r, t < [\frac{r − q + 1}{2}] · m_{1},

3. q + r < d, q ≤ r, t < r − q, in particular H^{0}(Y, Ω^{r}) = 0 for 0 < r < d,

4. q + r < d, 0 < q ≤ r, t > min\{r − q, [\frac{r − q + c}{2}]\} · m_{c},

5. q + r = d, 0 < q, t > max{(2r − d) · m_{c}, n_{Y} + r · m_{c}},

6. q + r > d, q < r,

7. q + r > d, r ≤ q < d, t > − [\frac{q − r + 1}{2}] · m_{1},

8. q + r > d, q ≥ r, t > r − q, in particular H^{d}(Y, Ω^{r}) = 0 for 0 < r < d,

9. q + r > d, r ≤ q < d, t < − min\{q − r, [\frac{q − r + c}{2}]\} · m_{c},

10. q + r = d, q < d, t < − max{(2q − d) · m_{c}, n_{Y} + q · m_{c}} .

Proof. Let be q + r < d. From q > r follows G_{a}^{r−q} = ∅, therefore we get 1. from

Theorem 1a). If t < [\frac{r − q + 1}{2}] · m_{1} = (r − q − [\frac{r − q}{2} ]) · m_{1} we have t ∉ G_{a}^{r−q}

∀a ∈ \{0, 1, …, [\frac{r − q}{2} ]\} and assertion 2. follows from Theorem 1a). In the special

case q ≠ 0 we get 3. from 2. (m_{1} ≥ 2). If q = 0 and r = 0 one has H^{0}(Y, O(t)) = 0 because of t < 0. Let be q = 0 and r > 0 . From (1) and (2) we get the exact sequences

H^{0}(ℙ^{n}, Ω^{r}(t)) \overline{}H^{0}(Y, A^{r}(t)) \overline{}H^{1}(ℙ^{n}, Im φ_{1}(t)),
H^{0}(Y, A^{r}(t)) \overline{}H^{0}(Y, Ω_{Y}^{r} (t)) \overline{}H^{1}(Y, Im γ_{1}(t)).

Because of t < r is valid evidently H^{0}(ℙ^{n}, Ω^{r}(t)) = 0 (cp. (5)) and t ∉ G_{0}^{r}. Using Lemma 2 one has H^{1}(ℙ^{n}, Im φ_{1}(t)) = 0. The short exact sequence above shows H^{0}(Y, A^{r}(t)) = 0. It remains to show H^{1}(Y, Im γ_{1}(t)) = 0 . Using sequence (2) and Lemma 1 one has to show that H^{a}(Y, _{⨁V_{a}} A^{r−a}(t − m_{V_{a}})) = 0 ∀a ∈ {1, 2, …, r}. So we can apply Corollary 1a). First we see 0 < a < d because of r < d . Assume ∃a ∈ {1, 2, …, r} with t − m_{V_{a}} ∈ G_{0}^{r−2a} . Then one has t − m_{V_{a}} = μ_{1}m_{1} + μ_{2}m_{2} + … + μ_{c}m_{c}, μ_{i} ≥ 0 ∀i, ∑_{i=1}^{c} μ_{i} = r − 2a ≥ 0 and ∑_{i=1}^{c} sgn(μ_{i}) ≥ r − 2a ≥ 0. So also t = λ_{1}m_{1} + λ_{2}m_{2} + … + λ_{c}m_{c}, ∑_{i=1}^{c} λ_{i} = r − a and ∑_{i=1}^{c} sgn(λ_{i}) ≥ r − 2a. Furthermore is valid 2a ≤ r, so 0 ≤ a ≤ [\frac{r}{2}] . So we have t ≥ 2 · (r − a) (m_{i} ≥ 2), t ≥ 2r − 2 · [\frac{r}{2}] ≥ r in opposition to t < r . That means t − m_{V_{a}} ∉ G_{0}^{r−2a} and with it H^{1}(Y, Im γ_{1}(t)) = 0, so we are done. In like manner one proves statement 5. We make a remark to the case r = 0 : From t > max{−d · m_{c}, n_{Y} }, follows t > n_{Y} and therefore H^{d}(Y, O(t)) ≈ H^{0}(Y, Ω^{d}(−t)) = H^{0}(Y, O(n_{Y} − t)) = 0. Now we prove 4.: t ≤ (∑_{i=1}^{c} λ_{i})·m_{c} = (r−q−a)·m_{c} ≤ (r−q)·m_{c} is evidently for any t ∈ G_{a}^{r−q}. From t > (r −q)·m_{c} follows therefore t ∉ G_{a}^{r−q} and our assertion. For any t ∈ G_{a}^{r−q} we have by definition ∑_{i=1}^{c} sgn(λ_{i}) ≥ r − q − 2a and ∑_{i=1}^{c} λ_{i} = r − q − a,

i.e. r − q − 2a ≤ c, 2(r − q − a) ≤ r − q + c, ∑_{i=1}^{c} λ_{i} ≤ [\frac{r − q + c}{2}] ,

t = ∑_{i=1}^{c} λ_{i} · m_{i} ≤ [\frac{r − q + c}{2}] · m_{c}. Thus from t > [\frac{r − q + c}{2}] · m_{c}

we have t ∉ G_{a}^{r−q} ∀a ∈ \{0, 1, …, [\frac{q − r}{2}]\} and the statement 4 follows from Theorem 1, a). By Serre duality is valid h^{q}(Y, Ω^{r}(t)) = h^{d−q}(Y, Ω^{d−r}(−t)). Therefore the statements 6. - 10. follow from the statements 1. - 5. of the theorem. □

Theorem 3. Assume m_{j} ≥ 3 for all j and 0 < r < d. Then

a) H^{0}(Y, Ω^{r}(t)) = 0 for t ≤ r ,

b) H^{d}(Y, Ω^{r}(t)) = 0 for t ≥ r − d.

Proof. One proves a) like statement 3. of Theorem 2. Then b) follows from a) by Serre duality. Note that from H^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t)) ≠ 0 ∀t > r follows H^{0}(Y, Ω_{Y}^{r} (t)) ≠ 0 ∀t > r (cp.(5)). □

Theorem 4. dim_{k} H^{q}(Y, Ω_{Y}^{r}) is independent of the ground field k.

Proof. This is well known for r = 0 (Ω_{Y}^{r} = O_{Y}) and for r = d (Ω_{Y}^{d} = O_{Y} (n_{Y})). For 0 < r < d we have H^{0}(Y, Ω_{Y}^{r}) = 0, H^{d}(Y, Ω_{Y}^{r}) = 0 and for 0 < q < d, q ≠ r and q+r ≠ d also H^{q}(Y, Ω_{Y}^{r}) = 0. In the case d = 2r one has only H^{r}(Y, Ω_{Y}^{r}) ≠ 0 and thus h^{r}(Y, Ω_{Y}^{r}) = (−1)^{r} · χ(Y, Ω_{Y}^{r}). The Euler-Poincaré characteristic χ(Y, Ω_{Y}^{r} (t)) is independent of the ground field k ∀t ∈ ℤ (cp. (1), (2)). Now let be 0 < 2r < d. We have the exact sequences

H^{r}(Y, Im γ_{1}) \overline{}H^{r}(Y, A^{r}) \overline{}H^{r}(Y, Ω_{Y}^{r}) \overline{}H^{r+1}(Y, Im γ_{1}),
H^{r}(ℙ^{n}, Im φ_{1}) \overline{}H^{r}(ℙ^{n}, Ω_{Y}^{r}) \overline{}H^{r}(Y, A^{r}) \overline{}H^{r+1}(ℙ^{n}, Im φ_{1}).

Because of Lemma 2 resp. Corollary 1 one gets H^{r}(ℙ^{n}, Im φ_{1}) = 0 resp. H^{r+1}(ℙ^{n}, Im φ_{1}) = 0, that means h^{r}(Y, A^{r}) = 1. Like in the proof of Theorem 1 one shows H^{r}(Y, Im γ_{1}) = 0, H^{r+1}(Y, Im γ_{1}) = 0 and therefore h^{r}(Y, Ω_{Y}^{r}) = 1 independently of k. In the case 0 < 2r < d = q + r we obtain h^{d−r}(Y, Ω_{Y}^{r}) = (−1)^{d−r} · χ(Y, Ω_{Y}^{r}) − (−1)^{d} · h^{r}(Y, Ω_{Y}^{r}). If d < 2r < 2d the theorem follows from the Serre duality. □

2. Cohomology of the sheaves A^{r}(t)

Let Y ⊆ ℙ_{k}^{n} be as above. The cohomology groups of Y with coefficients in A^{r}(t) = O_{Y} (t) ⊗ Ω_{ℙ^{n}}^{r} will be computed using the exact sequence (1). We will show that the dimensions h^{q}(Y, A^{r}(t)) are independent of char k. At first we calculate the cohomology groups of the sheaves Im φ_{s}(t) using the short exact sequences

0 \overline{}Im φ_{s+1} \overline{}⨁_{K_{s}} Ω_{ℙ^{n}}^{r}(−m_{K_{s}}) \overline{}Im φ_{s} \overline{}0 (0 ≤ s < c).  (7)

The index K_{s} runs through all combinations of {1, 2, …, c}, s at a time, without repetition. One has

Im φ_{c} ≅ Ω_{ℙ^{n}}^{r}(−m_{1} − m_{2} − … − m_{c}) and  (8)
Im φ_{0} = A^{r}.  (9)

Let 𝔘 be the familiar open covering of ℙ_{k}^{n} by affine spaces U_{i} : x_{i} ≠ 0. We work

with the following cochains : ϱ^{(j)} with ϱ_{i}^{(j)} = x_{i}^{m_{j}} · d\frac{F^{(j)}}{x_{i}^{m_{j}}} = x_{i} · ∑_{ν=0}^{n} \frac{∂F^{(j)}}{∂x_{ν}} · d\frac{x_{ν}}{x_{i}} is a

global section of A^{1}(m_{j}). Understanding ϱ^{(j)} as cochain in C^{0}(𝔘, Ω_{ℙ^{n}}^{1}(m_{j}) one has (dϱ^{(j)})_{ik} = ϱ_{k}^{(j)} −ϱ_{i}^{(j)} = −m_{j} ·F^{(j)} ·ω_{ik}^{(1)} (cp.(4)), i.e., in the case p | m_{j} ϱ^{(j)} is even a global section of Ω_{ℙ^{n}}^{1}(m_{j}). Let P_{0}^{(j)}, P_{1}^{(j)}, …, P_{n}^{(j)} be any homogeneous polynomials with F^{(j)} = ∑_{ν=0}^{n} x_{ν} · P_{ν}^{(j)} for j ∈ {1, 2, …, c} and let σ^{(j)} ∈ C^{0}(𝔘, Ω_{ℙ^{n}}^{1}(m_{j})) be the

cochain with σ_{i}^{(j)} = x_{i} · ∑_{ν=0}^{n} P_{ν}^{(j)}d\frac{x_{ν}}{x_{i}}. Then σ_{k}^{(j)} − σ_{i}^{(j)} = −F^{(j)} · ω_{ik}^{(1)}, that means

σ^{(j)} is also a global section of A^{1}(m_{j}). Lemma 3. For 0 ≤ s ≤ c, 0 < q < d + s, r − c + s ≤ q ≤ r we get

1. H^{q}(ℙ^{n}, Im φ_{s}(t)) ≅ ⨁_{K_{s+r−q}} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{q}(t−m_{K_{s+r−q}})), were K_{s+r−q} runs through

all combinations of {1, 2, …, c}, s + r − q at a time, without repetition.

2. h^{q}(ℙ^{n}, Im φ_{s}(t)) = ∑_{K_{s+r−q}} δ_{t,m_{K_{s+r−q}}} (δ − Kroneckerdelta),

3. dim_{k} ⨁_{t∈ℤ} H^{q}(ℙ^{n}, Im φ_{s}(t)) = \binom{c}{s + r − q} .

Corollary 2. Let be 0 ≤ s ≤ c, 0 < q < d + s and q ≤ r. Then we have H^{q}(ℙ^{n}, Im φ_{s}(t)) = 0 iff t ∉ G_{0}^{s+r−q}. (cp. Lemma 2) Proof. Let be t ∈ G_{0}^{s+r−q} = {m_{K_{s+r−q}}}. Then there exists a combination K_{s+r−q} with t = m_{K_{s+r−q}}. Because of Lemma 3 follows H^{q}(ℙ^{n}, Im φ_{s}(t)) ≠ 0. □ Proof of Lemma 3. 2. and 3. follows from 1. and (3). If s = c then we have 0 < q = r < n, K_{s+r−q} = K_{c} = (1, 2, …, c) and we get 1. from (8). For s < c one has q + 1 < d + c = n and the exact cohomology sequence

H^{q}(ℙ^{n}, Im φ_{s+1}(t)) ^{\overline{}^{α}} ⨁_{K_{s}} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{s}})) \overline{}H^{q}(ℙ^{n}, Im φ_{s}(t)) \overline{}

\overline{}H^{q+1}(ℙ^{n}, Im φ_{s+1}(t)) ^{\overline{}^{β}} ⨁_{K_{s}} H^{q+1}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{s}})).  (10)

In the case s < c , q < r holds H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t−m_{K_{s}})) = 0 , if s < c and q = r we get H^{q}(ℙ^{n}, Im φ_{s+1}(t)) ≅ ⨁_{K_{s+1}} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{q}(t − m_{K_{s+1}})) by induction assumption. The homomorphism

α : ⨁_{K_{s+1}} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{q}(t − m_{K_{s+1}})) \overline{}⨁_{K_{s}} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{q}(t − m_{K_{s}}))

is defined as multiplication by the polynomials F^{(1)}, F^{(2)}, …, F^{(c)} . Because of (3) every product F^{(j)} · ω^{(q)} is cohomologous to zero. Thus Im α = 0 for q ≤ r and s < c. If q < r and s < c then Im β = 0 follows analogously. Therefore in the case q < r and s < c we obtain H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{s}})) = 0 and

H^{q}(ℙ^{n}, Im φ_{s}(t)) ≅ H^{q+1}(ℙ^{n}, Im φ_{s+1}(t))  (11)
≅ ⨁_{K_{s+r−q}} H^{q+1}(ℙ^{n}, Ω_{ℙ^{n}}^{q+1}(t − m_{K_{s+r−q}}))
≅ ⨁_{K_{s+r−q}} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{q}(t − m_{K_{s+r−q}}))

by induction assumption. For q = r and s < c we get H^{q+1}(ℙ^{n}, Im φ_{s+1}(t)) = 0 from Corollary 1c) and because of Im α = 0 also

H^{q}(ℙ^{n}, Im φ_{s}(t)) ≅ ⨁_{K_{s}} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{q}(t − m_{K_{s}})). □

With assumtions as in Lemma 3 one can explicite specify a k-basis of the vector space ⨁_{t∈ℤ}H^{q}(ℙ^{n}, Im φ_{s}(t)). One has a one to one mapping between the basis vectors and the combinations of {1, 2, …, c}, s + r − q at a time, without repetitions (cp. Lemma 3., 3.). In the following let K be such a combination and let be t = m_{K}. Because of the surjection

⨁_{K_{s}} C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})) ↠ C^{q}(𝔘, Im φ_{s}(m_{K}))

it is enough to define a suitable cochain τ_{K}^{(q)} ∈ ⨁_{K_{s}} C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})). In the case K_{s} ⊈ K , i.e., not all integers in K_{s} belong to K, we set the component τ_{K,K_{s}}^{(q)} ∈ C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})) equal to zero. If K_{s} ⊆ K we need for the present the complementary combination K \ K_{s} = (j_{1}, j_{2}, …, j_{r−q}). Here we set (τ_{K,K_{s}}^{(q)})_{i_{0}i_{1}...i_{q}} = ω_{i_{0}i_{1}...i_{q}}^{(q)} ∧ σ_{i_{0}}^{(j_{1})} ∧ σ_{i_{0}}^{(j_{2})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})}. Since m_{j_{1}} + m_{j_{2}} + … + m_{jr−q} = m_{K} − m_{K_{s}} we have τ_{K,K_{s}}^{(q)} ∈ C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})).

Lemma 4. With assumtions as in Lemma 3 the image of the cochain τ_{K}^{(q)} ∈ ⨁_{K_{s}} C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})) in C^{q}(𝔘, Im φ_{s}(m_{K})) is a cocycle wich is not cohomologous to zero. The cohomology classes of these cocycles form a k-basis of ⨁_{t∈ℤ}H^{q}(ℙ^{n}, Im φ_{s}(t)) and they are independent of the choise of the polynomials P_{ν}^{(j)}.

Proof. If s = c then from the assumtions of Lemma 3 follows 0 < q = r < n just as K = K_{s} = (1, 2, …, c). In this case the k-vector space

⨁_{t∈ℤ}H^{r}(ℙ^{n}, Im φ_{c}(t)) ≅ ⨁_{t∈ℤ} H^{r}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K})) = H^{r}(ℙ^{n}, Ω_{ℙ^{n}}^{r})

is one-dimensional and from the construction above we get the known basis - the cohomology class of ω^{(r)} ∈ C^{r}(𝔘, Ω_{ℙ^{n}}^{r}). If s < c and q = r then from K_{s} ⊆ K follows K_{s} = K and the k-vector space

⨁_{t∈ℤ}H^{r}(ℙ^{n}, Im φ_{s}(t)) ≅ ⨁_{t∈ℤ} ⨁_{K_{s}} H^{r}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{s}})) ≅ ⨁_{K_{s}} H^{r}(ℙ^{n}, Ω_{ℙ^{n}}^{r})

is \binom{c}{s}-dimensional. The construction of the basis elements leads again to the cochains ω^{(r)}. If finally s < c and q < r (11) is worth and by induction assumption one knows the

basis of ⨁_{t∈ℤ}H^{q+1}(ℙ^{n}, Im φ_{s+1}(t)) :

Let K be any combination of (s + 1) + r − (q + 1) = s + r − q elements from {1, 2, …, c}. The corresponding basis element is given by a cochain τ_{K}^{(q+1)} ∈ ⨁_{K_{s+1}} C^{q+1}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s+1}})) . In order to prove the lemma we explore the following commutative diagram :

⨁_{K_{s}} C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})) \overline{}C^{q}(𝔘, Im φ_{s}(m_{K}))
d  ⟲  _{}d
C^{q+1}(𝔘, Im φ_{s+1}(m_{K})) ↪^{ι} ⨁_{K_{s}} C^{q+1}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})) \overline{}C^{q+1}(𝔘, Im φ_{s}(m_{K}))
⟲  [figure]φ_{s+1}

⨁_{K_{s+1}} C^{q+1}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s+1}}))

We fix a combination K_{s} with K_{s} ⊆ Kand K \ K_{s} = (j_{1}, j_{2}, … j_{r−q}) and calculate the coboundary of the cochain τ_{K,K_{s}}^{(q)} ∈ C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})) :

(dτ_{K,K_{s}}^{(q)})_{i_{0}...i_{q+1}} = (τ_{K,K_{s}}^{(q)})_{i_{1}...i_{q+1}} − (τ_{K,K_{s}}^{(q)})_{i_{0}i_{2}...i_{q+1}} ± …
= ω_{i_{1}...i_{q+1}}^{(q)} ∧ σ_{i_{1}}^{(j_{1})} ∧ σ_{i_{1}}^{(j_{2})} ∧ … ∧ σ_{i_{1}}^{(j_{r−q})}
−ω_{i_{0}i_{2}...i_{q+1}}^{(q)} ∧ σ_{i_{0}}^{(j_{1})} ∧ σ_{i_{0}}^{(j_{2})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})} ± …
+(−1)^{q+1} · ω_{i_{0}i_{1}...i_{q}}^{(q)} ∧ σ_{i_{0}}^{(j_{1})} ∧ σ_{i_{0}}^{(j_{2})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})}
= ω_{i_{1}...i_{q+1}}^{(q)} ∧ (σ_{i_{0}}^{(j_{1})} − F^{(j_{1})} · ω_{i_{0}i_{1}}^{(1)}) ∧ … ∧ (σ_{i_{0}}^{(j_{r−q})} − F^{(j_{r−q})} · ω_{i_{0}i_{1}}^{(1)})
+( − ω_{i_{0}i_{2}...i_{q+1}}^{(q)} ± … + (−1)^{q+1} · ω_{i_{0}i_{1}...i_{q}}^{(q)}) ∧ σ_{i_{0}}^{(j_{1})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})}
= −ω_{i_{1}...i_{q+1}}^{(q)} ∧ F^{(j_{1})} · ω_{i_{0}i_{1}}^{(1)} ∧ σ_{i_{0}}^{(j_{2})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})}
−ω_{i_{1}...i_{q+1}}^{(q)} ∧ σ_{i_{0}}^{(j_{1})} ∧ F^{(j_{2})} · ω_{i_{0}i_{1}}^{(1)} ∧ σ_{i_{0}}^{(j_{3})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})} − …
−ω_{i_{1}...i_{q+1}}^{(q)} ∧ σ_{i_{0}}^{(j_{1})} ∧ σ_{i_{0}}^{(j_{2})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q−1})} ∧ F^{(j_{r−q})} · ω_{i_{0}i_{1}}^{(1)}
= (−1)^{q+1} · ω_{i_{0}...i_{q+1}}^{(q+1)} ∧ (F^{(j_{1})} · σ_{i_{0}}^{(j_{2})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})}
−F^{(j_{2})} · σ_{i_{0}}^{(j_{1})} ∧ σ_{i_{0}}^{(j_{3})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})} ± …
+(−1)^{r−q−1} · F^{(j_{r−q})} · σ_{i_{0}}^{(j_{1})} ∧ … ∧ σ_{i_{0}}^{(j_{r−q−1})})  (12)

We note that φ_{s+1} is defined by multiplication with the polynomials F^{(j)} at which alternating signs as in (12) are used. Hence follows that the cochain dτ_{K}^{(q)} is equal to the image of the cochain τ_{K}^{(q+1)} ∈ ⨁_{K_{s+1}} C^{q+1}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s+1}})) by the homomorphism φ_{s+1} (except the sign). dτ_{K}^{(q)} = −φ_{s+1}(τ_{K}^{(q+1)}) has an image at the embedding ι. Thus the image of the cochain τ_{K}^{(q)} in C^{q}(𝔘, Im φ_{s}(m_{K})) is a cocycle. Furthermore the cohomology classes of these cocycles form a k-basis of ⨁_{t∈ℤ}H^{q}(ℙ^{n}, Im φ_{s}(t)) since by using the isomorphism (11) they will even send to the basis elements of ⨁_{t∈ℤ}H^{q+1}(ℙ^{n}, Im φ_{s+1}(t)) . We also see by induction that these cohomology classes do not depend on the choise of the polynomials P_{ν}^{(j)}. Lemma 4 is proved. □ Applying Lemma 3, Lemma 4 and Corollary 2 to A^{r} = Im φ_{0} one gets

Corollary 3. Let be 0 < q < d and r ≥ 0.

1. If r < q or q < r − c then H^{q}(Y, A^{r}(t)) = 0 ∀t ∈ ℤ

2. If r − c ≤ q ≤ r then H^{q}(Y, A^{r}(t)) ≅ ⨁_{K_{r−q}} H^{q}(ℙ^{n}, Ω_{ℙ^{n}}^{q}(t − m_{K_{r−q}})) and

h^{q}(Y, A^{r}(t)) = ∑_{K_{r−q}} δ_{t,m_{K_{r−q}}} , where K_{r−q} runs through all combinations of {1, 2, …, c}, r − q at a time, without repetition.

3. H^{q}(Y, A^{r}(t)) = 0 iff t ∉ G_{0}^{r−q}.

4. dim_{k} ⨁_{t∈ℤ}H^{q}(Y, A^{r}(t)) = \binom{c}{r − q} for r − c ≤ q ≤ r.

The cohomology classes of the cocycles

ω^{(q)} ∧ σ^{(j_{1})} ∧ σ^{(j_{2})} ∧ … ∧ σ^{(j_{r−q})} ∈ C^{q}(𝔘, A^{r}(m_{j_{1}} + m_{j_{2}} + … + m_{j_{r−q}}))

(1 ≤ j_{1} < j_{2} < … < j_{r−q} ≤ c) form a k-basis of ⨁_{t∈ℤ}H^{q}(Y, A^{r}(t)).

We know that these cohomology classes are independent of the choise of the polynomials P_{ν}^{(j)} with whose help σ^{(j)} is defined. Therefore in the case p ∤ m_{j} one can

choose P_{ν}^{(j)} = \frac{1}{m_{j}} · \frac{∂F^{(j)}}{∂x_{ν}}, i.e. σ^{(j)} = \frac{1}{m_{j}} · ϱ^{(j)}.

The essential distinction between σ^{(j)} and ϱ^{(j)} in the case p | m_{j} gives rise to define further cochains. If char k = p is positive, we assume without loss of generality that p divides the integers m_{1}, m_{2}, …, m_{u} and only these (0 ≤ u ≤ c). Like in Lemma 3 let be 0 ≤ s ≤ c, 0 < q < n − c + s and r − c + s ≤ q ≤ r. Furthermore we fix one j ∈ {1, 2, …, u} (with p | m_{j} ) and a combination K of {1, 2, …, c}, s + r − q at a time, with j ∈ K. Then let (j)λ_{K}^{(q)} ∈ ⨁_{K_{s}} C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(m_{K} − m_{K_{s}})) be the following cochain: First we set (j)λ_{K,K_{s}}^{(q)} = 0 if K_{s} ⊈ K or j ∈ K_{s}. In the case K_{s} ⊆ K and j ∉ K_{s} let stand for the present K \ K_{s} = (j_{1}, …, j, …, j_{r−q}) with 1 ≤ j_{1} < … < j < … < j_{r−q} ≤ c and (^{(j)}λ_{K,K_{s}}^{(q)})_{i_{0}i_{1}...i_{q}} = ω_{i_{0}i_{1}...i_{q}}^{(q)} ∧ σ_{i_{0}}^{(j_{1})} ∧ … ∧ ϱ_{i_{0}}^{(j)} ∧ … ∧ σ_{i_{0}}^{(j_{r−q})} . If ^{(j)}λ_{K,K_{s}}^{(q)} ≠ 0 then the distinction between ^{(j)}λ_{K,K_{s}}^{(q)} and τ_{K,K_{s}}^{(q)} is only the fact that the global section ϱ^{(j)} ∈ Γ(ℙ^{n}, Ω_{ℙ^{n}}^{1}(m_{j})) is used instead of σ^{(j)}.

Lemma 5. The image φ_{s}(^{(j)}λ_{K}^{(q)}) of ^{(j)}λ_{K}^{(q)} in C^{q}(𝔘, Im φ_{s}(m_{K})) is cohomologous to zero.

Proof. For s = c follows 0 < q = r < n and K_{s} = K = (1, 2, …, c) consequently j ∈ K_{s} and ^{(j)}λ_{K,K_{s}}^{(q)} = 0. Now let s < c and q = r. Then from K_{s} ⊆ K follows K_{s} = K and therefore j ∈ K_{s} and also in this case ^{(j)}λ_{K}^{(q)} = 0. Finally we have the isomorphism (11) in the case s < c and q < r. Like in the proof of Lemma 4 follows that the isomorphism (11) sends the cohomology class of the image of (j)λ_{K}^{(q)} in C^{q}(𝔘, Im φ_{s}(m_{K})) to the cohomology class of the image of ^{(j)}λ_{K}^{(q+1)} in C^{q+1}(𝔘, Im φ_{s+1}(m_{K})). This image is cohomologous to zero by induction assumption. □

Corollary 4. Let 0 < q < dim Y, q < r and let j_{1} ∈ {1, 2, …, u}, i.e. p | m_{j_{1}}. Then for any j_{2}, j_{3}, …, j_{r−q} ∈ {1, 2, …, c} the cocycle ω^{(q)} ∧ ϱ^{(j_{1})} ∧ σ^{(j_{2})} ∧ … ∧ σ^{(j_{r−q})} ∈ C^{q}(𝔘, A^{r}(m_{j_{1}} + m_{j_{2}} + … + m_{j_{r−q}})) is cohomologous to zero.

Finally we have to calculate H^{0}(Y, A^{r}(t)) and H^{d}(Y, A^{r}(t)).

Lemma 6. Let 0 ≤ s ≤ c. t ∈ ℤ, 0 < r. Then holds

h^{0}(ℙ^{n}, Im φ_{s}(t)) = ∑_{μ=s}^{c}(−1)^{μ−s}·∑_{K_{μ}} h^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t−m_{K_{μ}}))+ ∑_{μ=s+r}^{c}(−1)^{μ−s−r}·∑_{K_{μ}} δ_{t,m_{Kμ}}.

K_{μ} runs through all combinations of {1, 2, …, c} , μ at a time, without repetition. Proof. If s = c we get the formula from (8). The following cohomology sequence is exact for s < c (cp. (7)):

0 \overline{}H^{0}(ℙ^{n}, Im φ_{s+1}(t)) \overline{}⨁_{K_{s}} H^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{s}})) \overline{}H^{0}(ℙ^{n}, Im φ_{s}(t)) \overline{}

\overline{}H^{1}(ℙ^{n}, Im φ_{s+1}(t)) ^{\overline{}^{α}} ⨁_{K_{s}} H^{1}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{s}}))

with Im α = 0 (cp. proof of Lemma 3). Now one has

h^{0}(ℙ^{n}, Im φ_{s}(t)) = ∑_{μ=s}^{c}(−1)^{μ−s} ∑_{K_{μ}} h^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{μ}})) +

+ ∑_{μ=s+1}^{c}(−1)^{μ−s−1} · h^{1}(ℙ^{n}, Im φ_{μ}(t)).

Using Lemma 3, 2. completes the proof. □ Because of Im φ_{0} = A^{r} follows (cp. (5)) Corollary 5. Let r > 0. Then

h^{0}(Y, A^{r}(t)) = ∑_{μ=0}^{c}(−1)^{μ} · ∑_{K_{μ}} h^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{μ}})) + ∑_{μ=r}^{c}(−1)^{μ−r} · ∑_{K_{μ}} δ_{t,m_{Kμ}}.

K_{μ} runs through all combinations of {1, 2, …, c} , μ at a time, without repetition. \underline{Remark:} The next formula follows by Serre duality (r < n): h^{d}(Y, A^{r}(t)) = h^{0}(Y, A^{n−r}(m_{1} + m_{2} + … + m_{c} − t)) =

= ∑_{μ=0}^{c}(−1)^{c−μ} · ∑_{K_{μ}} h^{0}(ℙ^{n}, Ω_{ℙ^{n}}^{n−r}(m_{K_{μ}} − t)) + ∑_{μ=n−r}^{c}(−1)^{μ−n+r} · ∑_{K_{μ}} δ_{m_{1}+...+m_{c}−t,m_{Kμ}}

= ∑_{μ=0}^{c}(−1)^{c−μ} · ∑_{K_{μ}} h^{n}(ℙ^{n}, Ω_{ℙ^{n}}^{r}(t − m_{K_{μ}})) + ∑_{μ=0}^{r−d}(−1)^{r−d−μ} · ∑_{K_{μ}} δ_{t,m_{Kμ}}

Easy to see

h^{0}(Y, O_{Y} (t)) = ∑_{μ=0}^{c}(−1)^{μ} · ∑_{K_{μ}} h^{0}(ℙ^{n}, O_{ℙ^{n}}(t − m_{K_{μ}}))

3. Cohomology of Ω_{Y}^{r} (t)

For the explicite notation of the dimensions h^{q}(Y, Ω_{Y}^{r} (t)) we need the following arithmetical functions Δ and Z : Let be 𝔪 = (m_{1}, m_{2}, …, m_{c}) the multidegree of Y and v, l, s, t ∈ ℤ with 0 ≤ v ≤ c. In the case v > 0 we set

Δ(v, 𝔪, l, s, t) = card\{𝔱 ∈ ℤ^{v}|t_{ν} ≥ 0 ∀ν, l = ∑_{ν=1}^{v} t_{ν}, s = ∑_{ν=1}^{v} [\frac{t_{ν}}{2} ] , t = ∑_{ν=1}^{v} [\frac{t_{ν} + 1}{2}]·m_{ν}\}.

If v = 0 then let be Δ(v, 𝔪, l, s, t) = δ_{l,0}·δ_{s,0}·δ_{t,0}. Obviously one has Δ(v, 𝔪, l, s, t) = 0 if l < 0 or s < 0 or t < 0 or s > \frac{l}{2} or s < \frac{l−v}{2}. Furthermore Δ satisfies the following recursion formula with v > 0 :

Δ(v, 𝔪, l, s, t) = Δ(v − 1, 𝔪, l, s, t) + Δ(v − 1, 𝔪, l − 1, s, t − m_{v}) +
+Δ(v, 𝔪, l − 2, s − 1, t − m_{v})

Using the Kronecker-δ one gets the following explicite formula: If v > 0, 0 ≤ s ≤ \frac{l}{2}, s ≥ \frac{l−v}{2}, then

Δ(v, 𝔪, l, s, t) = ∑_{K_{l−2s}} ∑_{V_{s}} δ_{t,m_{K_{l−2s}}+m_{Vs}}

where K_{l−2s} runs through all combinations of {1, 2, …, v} , l−2s at a time, without repetition and V_{s} runs through all combinations of {1, 2, …, v} , s at a time, with repetitions. So one has the following equation for v > 0, 0 ≤ s ≤ \frac{l}{2}, s ≥ \frac{l−v}{2} :

∑_{t∈ℤ}Δ(v, 𝔪, l, s, t) = \binom{v}{l − 2s} · \binom{v + s − 1}{s}

In the case m_{1} = m_{2} = … = m_{v} (and v > 0, 0 ≤ s ≤ \frac{l}{2}, s ≥ \frac{l−v}{2}) is valid

Δ(v, 𝔪, l, s, t) = \binom{v}{l − 2s} · \binom{v + s − 1}{s} · δ_{t,(l−s)·m_{1}}.

The function Z is defined by

Z(v, 𝔪, l, s, t) = \begin{matrix}[ & \frac{l}{2} & ] \\ ∑ \\ σ=s \end{matrix} Δ(v, 𝔪, l, σ, t) if s ≥ 0 and Z(v, 𝔪, l, s, t) = 0 if s < 0.

Z satisfies the recursion formula

Z(v, 𝔪, l, s, t) = Z(v − 1, 𝔪, l, s, t) + Z(v − 1, 𝔪, l − 1, s, t − m_{v}) +

+Z(v, 𝔪, l − 2, max{0, s − 1}, t − m_{v}) with v > 0 and s ≥ 0.

Obviously Z(0, 𝔪, l, s, t) = Δ(0, 𝔪, l, s, t) = δ_{l,0} · δ_{s,0} · δ_{t,0}. If v > 0 and s ≥ 0 one obtains

Z(v, 𝔪, l, s, t) = ∑_{σ=s}^{[ \frac{l}{2} ]} ∑_{K_{l−2σ}} ∑_{V_{σ}} δ_{t,m_{K_{l−2σ}} +m_{Vσ}} .

Note that Z(v, 𝔪, l, s, t) = Δ(v, 𝔪, l, s, t) = 0 if t is not divisible by g.c.d.(m_{1}, m_{2}, … m_{v}). \underline{Remark:} With Lemma 3 and Corollary 3 one gets now h^{q}(ℙ^{n}, Im φ_{s}(t)) = Δ(c, 𝔪, s + r − q, 0, t) for 0 ≤ s ≤ c, 0 < q < d + s and r − c + s ≤ q ≤ r resp. h^{q}(Y, A^{r}(t)) = Δ(c, 𝔪, r − q, 0, t) for 0 < q < d, r − c ≤ q ≤ r. Now we use the exact sequence (2). The homomorphism γ_{s} in this sequence is defined as multiplication (at the left) with the global sections ϱ^{(j)} ∈ Γ(Y, A^{1}(m_{j})). Let be 0 < r < n, 0 ≤ s ≤ r and 0 < q < dim Y. For simplification we write

Θ^{q,s} := ⨁_{t∈ℤ}H^{q}(Y, A^{s}(t))  (13)

We know from Corollary 1 and Corollary 3 that Θ^{q,s} is a \binom{c}{s−q} - dimensional k - vector space. From sequence (2) we get the following complex of finite-dimensional vector spaces :

Θ : 0 \overline{}⨁_{V_{r−q}}Θ^{q,q} ^{\overline{}^{γ_{r−q}^{(q)}}} ⨁_{V_{r−q−1}}Θ^{q,q+1} ^{\overline{}^{γ_{r−q−1}^{(q)}}} … ^{\overline{}^{γ_{2}^{(q)}}} ⨁_{V_{1}} Θ^{q,r−1} ^{\overline{}^{γ_{1}^{(q)}}} Θ^{q,r} ^{\overline{}^{γ_{0}^{(q)}}} 0 (14)
dim_{k} ⨁_{V_{s}} Θ^{q,s} = \binom{c + s − 1}{s} · \binom{c}{s − q}.  (15)

We will calculate the homology of this complex. Again we assume p divides the integers m_{1}, m_{2}, …, m_{u} and only these (0 ≤ u ≤ c). At first we look at the complex

K : 0 \overline{}A ^{\overline{}^{α}} B \overline{}0

with A = _{⨁V_{1}}Θ^{q,q} = ⨁_{i=1}^{c} Θ^{q,q}, dim A = c and B = Θ^{q,q+1}, dim B = c. Because of Corollary 4 we know that the multiplication with ϱ^{(j)} gives a cocycle which is cohomologous to zero if 1 ≤ j ≤ u. Thus Ker α ≅ k^{u} and Coker α ≅ k^{u}. We get by considering of the (r−q)-fold exterior product of K the following complex of k-vector spaces :

0 \overline{}S^{r−q}A ^{\overline{}^{α_{r−q}}} S^{r−q−1}A ⊗ B ^{\overline{}^{α_{r−q−1}}} S^{r−q−2}A ⊗^{⋀2}B ^{\overline{}^{α_{r−q−2}}} … ^{\overline{}^{α_{3}}}
\overline{}^{α_{3}} S^{2}A ⊗^{⋀r−q−2}B ^{\overline{}^{α_{2}}} A ⊗^{⋀r−q−1}B ^{\overline{}^{α_{1}}} ^{⋀r−q}B ^{\overline{}^{α_{0}}} 0 \overline{}…

with Ker α_{i}/ Im α_{i+1} ≅ S^{i}(Ker α) ⊗ ⋀^{r−q−i}(Coker α) and dim (Ker α_{i}/ Im α_{i+1}) = \binom{u}{r − q − i} · \binom{u + i − 1}{i}. In the case u = 0 or char k = 0 this complex is acyclic. From Corollary 3 and Corollary 4 follows that the complexes ⋀^{r−q} K and Θ are isomorphic. Using this isomorphism one finds k-basises of the homology groups Ker γ_{s}^{(q)}/Im γ_{s+1}^{(q)} : Lemma 7. Let be char k > 0, u > 0 and let L^{q,r−s} ⊆ Θ^{q,r−s} be the \binom{u}{r−q−s}- dimensional subspace with the basis ω^{(q)} ∧ σ_{j_{1}} ∧ σ_{j_{2}} ∧ … ∧ σ_{j_{r−q−s}} (1 ≤ j_{1} < j_{2} < … < j_{r−q−s} ≤ u). Then are valid (cp. (14))

1. Ker γ_{s}^{(q)} = Im γ_{s+1}^{(q)} ⨁ ⨁_{V′_{s}} L^{q,r−s} where V′_{s} runs through all combinations of {1, 2, … u}, s at a time, with repetitions.

2. dim Ker γ_{s}^{(q)}/Im γ_{s+1}^{(q)} = \binom{u}{r − q − s} · \binom{u + s − 1}{s} .

For the description of the cohomology groups of the sheaves Im γ_{s} (cp.(2)) and especially of Im γ_{0} = Ω_{Y}^{r} we will need further cochains: Let be p = char k > 0 and p | m_{j} (1 ≤ j ≤ u), thus ∑_{ν=0}^{n} x_{ν} · \frac{∂F^{(j)}}{∂x_{ν}} = 0. Then there exist homogenious polynomials Q_{νμ}^{(j)} ∈ k[x_{0}, …, x_{n}]_{m_{j}−2} with Q_{νμ}^{(j)} = −Q_{μν}^{(j)}, Q_{νν}^{(j)} = 0 and

\frac{∂F^{(j)}}{∂x_{ν}} = ∑_{μ=0}^{n} Q_{νμ}^{(j)} · x_{μ}. For 0 < 2s ≤ n let ^{(j)}ω^{(0,2s)} ∈ C^{0}(𝔘, Ω_{ℙ^{n}}^{2s}(s · m_{j})) be the

cochain with

(j)ω_{i}^{(0,2s)} = x_{i}^{2s} · ∑_{0≤i_{1}<...<i_{2s}≤n} ∑_{π} Q_{j_{1}j_{2}}^{(j)} · Q_{j_{3}j_{4}}^{(j)} · . . . · Q_{j_{2s−1}j_{2s}}^{(j)} · d\frac{x_{j_{1}}}{x_{i}} ∧ d\frac{x_{j_{2}}}{x_{i}} ∧ … ∧ d\frac{x_{j_{2s}}}{x_{i}}

where π runs through all permutations \binom{i_{1}i_{2}...i_{2s}}{j_{1}j_{2}...j_{2s}} with j_{2ν−1} < j_{2ν} ∀ν ∈ \{1, 2, …, s\} and j_{2ν−1} < j_{2ν+1} ∀ν ∈ {1, 2, …, s − 1} . One has :

(j)ω_{k}^{(0,2)} − ^{(j)}ω_{i}^{(0,2)} = ω_{ik}^{(1)} ∧ ϱ_{i}^{(j)} and
(j)ω_{k}^{(0,2s)} − ^{(j)}ω_{i}^{(0,2s)} = ω_{ik}^{(1)} ∧ ϱ_{i}^{(j)} ∧^{(j)} ω_{i}^{(0,2s−2)} for s > 1.

Furthermore we set ^{(j)}ω^{(0,1)} = σ^{(j)}, ^{(j)}ω^{(0,2s+1)} = σ^{(j)} ∧ ^{(j)}ω^{(0,2s)} if s > 0, (j)ω^{(0,0)} = 1, ^{(j)}ω^{(0,t_{j})} = 0 if t_{j} < 0. Then holds

(j)ω_{k}^{(0,t_{j})} − ^{(j)}ω_{i}^{(0,t_{j})} = ω_{ik}^{(1)} ∧ ϱ_{i}^{(j)} ∧^{(j)} ω_{i}^{(0,t_{j}−2)} ∀t_{j} ∈ ℤ.  (16)

Finally we set for q ≥ 0 : ω_{i_{0}...i_{q}}^{(q,𝔱)} = ω_{i_{0}...i_{q}}^{(q)} ∧^{(1)} ω_{i_{0}}^{(0,t_{1})} ∧ … ∧^{(u)} ω_{i_{0}}^{(0,t_{u})} with 𝔱 = (t_{1}, …, t_{u}) ∈ ℤ^{u}. Then ω^{(q,𝔱)} = 0 if there exists a ν with t_{ν} < 0. On the other hand obviously is valid ω^{(q,𝔱)} ∈ C^{q}(𝔘, Ω_{ℙ^{n}}^{r}(t)) with

r = q + t_{1} + … + t_{u}, t = ∑_{ν=1}^{u} [\frac{t_{ν} + 1}{2}] · m_{ν}.  (17)

For the restriction of the cochain ω^{(q,𝔱)} to A^{r} (resp. Ω_{Y}^{r}) we will also write ω^{(q,𝔱)}. Now we define some cocycles in C^{q}(𝔘, Im γ_{s}(t)). Because of the surjection

⨁_{V_{s}} C^{q}(𝔘, A^{r−s}(t − m_{V_{s}})) ↠ C^{q}(𝔘, Im γ_{s}(t))  (18)

it is enough to define a suitable cochain τ^{(q,s,𝔱)} ∈ ⨁_{V_{s}} C^{q}(𝔘, A^{r−s}(t − m_{V_{s}})). Assume 0 ≤ s ≤ r, 0 < q < dim Y − r + s, 𝔱 = (t_{1}, …, t_{u}) ∈ ℤ^{u}, s + r − q = t_{1} + … + t_{u}, t = ∑_{ν=1}^{u}[\frac{t_{ν}+1}{2}] · m_{ν} : If 1 ≤ j_{1} ≤ … ≤ j_{s} ≤ u and V_{s} = (j_{1}, …, j_{s}) let the corresponding component of τ^{(q,s,𝔱)} be ω^{(q,𝔱_{Vs})} with 𝔱_{V_{s}} = 𝔱−2·(𝔢_{j_{1}} +…+𝔢_{j_{s}}) and 𝔢_{j} = (0, …, 0, 1, 0, …, 0) ∈ ℤ^{u}. If the combination V_{s} contains a j with p ∤ m_{j} (u < j ≤ c) then let the corresponding component of τ^{(q,s,𝔱)} be equal to zero. Lemma 8. Let be p = char k > 0, u > 0, 0 < r ≤ n, 0 ≤ s ≤ r, 0 < q < dim Y − r + s, t ∈ ℤ, 𝔱 = (t_{1}, …, t_{u}) ∈ ℤ^{u}, with ∑_{ν=1}^{u} t_{ν} = s + r − q and t = ∑_{ν=1}^{u}[\frac{t_{ν}+1}{2}] · m_{ν}. Then holds

1. The cochain τ^{(q,s,𝔱)} ∈ ⨁_{V_{s}} C^{q}(𝔘, A^{r−s}(t − m_{V_{s}})) is different from zero if and only if s ≤ ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ] and t_{ν} ≥ 0 ∀ν.

2. The image \overline{τ}^{(q,s,𝔱)} of the cochain τ^{(q,s,𝔱)} under the surjection (18) is a cocycle in C^{q}(𝔘, Im γ_{s}(t)).

3. If s < r and s < ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ] then the image of the cohomology class of the cocycle \overline{τ}^{(q,s,𝔱)} under the homomorphism d : H^{q}(Y, Im γ_{s}(t)) ⟶ H^{q+1}(Y, Im γ_{s+1}(t)) is (except the sign) the cohomology class of the cocycle \overline{τ}^{(q+1,s+1,𝔱)}.

4. If s = ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ] then the cochain τ^{(q,s,𝔱)} is a cocycle in ⨁_{V_{s}} C^{q}(𝔘, A^{r−s}(t − m_{V_{s}})). The corresponding cohomology class goes under the homomorphism

δ : ⨁_{V_{s}} H^{q}(Y, A^{r−s}(t − m_{V_{s}})) ⟶ H^{q}(Y, Im γ_{s}(t))

to the cohomology class of \overline{τ}^{(q,s,𝔱)}.

Proof. From s > ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ] follows that for each combination V_{s} there is a negative component of 𝔱_{V_{s}} ∈ ℤ^{u}. Thus ω^{(q,𝔱_{Vs})} = 0 and τ^{(q,s,𝔱)} = 0. In the case s = ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ] and t_{ν} ≥ 0 ∀ν there exists one and only one combination V_{s} such that the corresponding component ω^{(q,𝔱_{Vs})} of τ^{(q,s,𝔱)} is not equal to zero. If for instance t_{1}, …, t_{w} are odd and t_{w+1}, …, t_{u} are even (0 ≤ w ≤ u), then one gets for this combination V_{s} : 𝔱_{V_{s}} = (1, …, 1, 0, …, 0). Because of ^{(j)}ω^{(0,1)} = σ^{(j)} holds ω^{(q,𝔱_{Vs})} = ω^{(q)} ∧σ^{(1)} ∧…∧σ^{(w)} so τ^{(q,s,𝔱)} is a cocycle in ⨁_{V_{s}} C^{q}(𝔘, A^{r−s}(t−m_{V_{s}})) and statement 4. is proved. Note that these cocycles are independent of the polynomials

Q_{νμ}^{(j)} with ∑_{μ=0}^{n} x_{μ} · Q_{νμ}^{(j)} = \frac{∂F^{(j)}}{∂x_{ν}} . If s < ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ] and t_{ν} ≥ 0 ∀ν then one sees also

τ^{(q,s,𝔱)} ≠ 0, this shows statement 1. In order to prove 2. we look at the following commutative diagram with surjections β, β′ and β′′ :

⨁_{V_{s}} C^{q}(𝔘, A^{r−s}(t − m_{V_{s}})) ^{\overline{}^{β}}  C^{q}(𝔘, Im γ_{s}(t))
d  ⟲  _{}d
C^{q+1}(𝔘, Im γ_{s+1}(t)) ↪^{α} ⨁_{V_{s}} C^{q+1}(𝔘, A^{r−s}(t − m_{V_{s}})) ^{\overline{}^{β′′}} C^{q+1}(𝔘, Im γ_{s}(t))
β′  ⟲ [figure]γ_{s+1}

⨁_{V_{s+1}} C^{q+1}(𝔘, A^{r−s−1}(t − m_{V_{s+1}})).

First of all we compute d(τ^{(q,s,𝔱)}) . For the present we set 𝔱_{V_{s}} = (t′_{1}, …, t′_{u}). Then is valid :

(dω^{(q,𝔱_{Vs})})_{i_{0}...i_{q+1}} = ω_{i_{1}...i_{q+1}}^{(q,𝔱_{Vs})} − ω_{i_{0}i_{2}...i_{q+1}}^{(q,𝔱_{Vs})} ± … + (−1)^{q+1} · ω_{i_{0}...i_{q}}^{(q,𝔱_{Vs})} =

= ω_{i_{1}...i_{q+1}}^{(q)} ∧ ^{(1)}ω_{i_{1}}^{(0,t′_{1})} ∧ … ∧ ^{(u)}ω_{i_{1}}^{(0,t′_{u})} − (ω_{i_{0}i_{2}...i_{q+1}}^{(q)} − ω_{i_{0}i_{1}i_{3}...i_{q+1}}^{(q)} ± … + (−1)^{q} · ω_{i_{0}...i_{q}}^{(q)}) ∧ ^{(1)}ω_{i_{0}}^{(0,t′_{1})} ∧ … ∧ ^{(u)}ω_{i_{0}}^{(0,t′_{u})}.

Using (16) one sees

ω_{i_{1}}^{(0,t′_{j})} = ω_{i_{0}}^{(0,t′_{j})} + ω_{i_{0}i_{1}}^{(1)} ∧ ϱ_{i_{0}}^{(j)} ∧ ω_{i_{0}}^{(0,t′_{j}−2)},

with which equation one gets now

(dω^{(q,𝔱_{Vs})})_{i_{0}...i_{q+1}} =
= ω_{i_{1}...i_{q+1}}^{(q)} ∧ ω_{i_{0}i_{1}}^{(1)} ∧ (ϱ_{i_{0}}^{(1)} ∧ ^{(1)}ω_{i_{0}}^{(0,t′_{1}−2)} ∧ ^{(2)}ω_{i_{0}}^{(0,t′_{2})} ∧ … ∧ ^{(u)}ω_{i_{0}}^{(0,t′_{u})} + … +
+ ϱ_{i_{0}}^{(u)} ∧ ^{(1)}ω_{i_{0}}^{(0,t′_{1})} ∧ … ∧ ^{(u−1)}ω_{i_{0}}^{(0,t′_{u−1})} ∧ ^{(u)}ω_{i_{0}}^{(0,t′_{u}−2)})
= −ϱ_{i_{0}}^{(1)} ∧ ω_{i_{0}...i_{q+1}}^{(q+1,𝔱′−2𝔢_{1})} − ϱ_{i_{0}}^{(2)} ∧ ω_{i_{0}...i_{q+1}}^{(q+1,𝔱′−2𝔢_{2})} − … − ϱ_{i_{0}}^{(u)} ∧ ω_{i_{0}...i_{q+1}}^{(q+1,𝔱′−2𝔢_{u})}.

The homomorphism γ_{s+1} is defined as multiplication with the global sections ϱ^{(j)} ∈ Γ(ℙ^{n}, Ω_{ℙ^{n}}^{1}(m_{j})) so that the image of the cochain τ^{(q+1,s+1,𝔱)} under γ_{s+1} is equal to dτ^{(q,s,𝔱)} (except the sign). Therefore dτ^{(q,s,𝔱)} ∈ Im α = Ker β” and β(τ^{(q,s,𝔱)}) is a cocycle. Statement 2 is proved. Furthermore this calculation shows that the cohomology class of \overline{τ}^{(q,s,𝔱)} = β(τ^{(q,s,𝔱)}) goes under the homomorphism d to the cohomology class of \overline{τ}^{(q+1,s+1,𝔱)} = β′(τ^{(q+1,s+1,𝔱)}). It stands to reason that this is also right in the case s ≥ ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ], but then one has τ^{(q+1,s+1,𝔱)} = 0 by statement 1. □

From sequence (2) one gets the following long exact cohomology sequence :

… \overline{}Ψ^{q,r−s−1} ^{\overline{}^{α_{s+1}^{(q)}}}  ⨁_{V_{s}} Θ^{q,r−s}  ^{\overline{}^{β_{s}^{(q)}}} Ψ^{q,r−s}  ^{\overline{}^{d_{s}^{(q)}}}
\overline{}^{d_{s}^{(q)}} Ψ^{q+1,r−s−1}  ^{\overline{}^{α_{s+1}^{(q+1)}}} ⨁_{V_{s}} Θ^{q+1,r−s}  ^{\overline{}^{β_{s}^{(q+1)}}} …  (19)
with Ψ^{q,r−l} := ⨁_{t∈ℤ}H^{q}(Y, Im γ_{l}(t)).  (20)

Because of Imγ_{0} = Ω_{Y}^{r} one has Ψ^{q,r} = ⨁_{t∈ℤ} H^{q}(Y, Ω_{Y}^{r} (t)) . With the help of the cocycle τ^{(q,s,𝔱)} resp. \overline{τ}^{(q,s,𝔱)} we will define the following subspaces (q ≥ 0, 0 ≤ s ≤ r) : Let A(q, s) ⊆ ⨁_{V_{s}} Θ^{q,r−s} resp. B(q, s) ⊆ Ψ^{q,r−s} be the subspaces, generated by the cohomology classes of the cocycles τ^{(q,s,𝔱)} resp. \overline{τ}^{(q,s,𝔱)} with t_{ν} ≥ 0 ∀ν and ∑_{ν=1}^{u} t_{ν} = s + r − q and s = ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ] (cp. Lemma 8, 4.). Let C(q, s) ⊆ Ψ^{q,r−s} be the subspace, generated by the cohomology classes of the cocycles \overline{τ}^{(q,s,𝔱)} with t_{ν} ≥ 0 ∀ν, ∑_{ν=1}^{u} t_{ν} = s + r − q and s < ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ]. Note that [τ^{(q,s,𝔱)}] ∈ H^{q}(Y, Im γ_{s}(t)) with t = ∑_{ν=1}^{u}[\frac{t_{ν+1}}{2} ] · m_{ν}. Obviously the number of the cocycles \overline{τ}^{(q,s,𝔱)} with [\overline{τ}^{(q,s,𝔱)}] ∈ (B(q, s) + C(q, s)) ∩ H^{q}(Y, Im γ_{s}(t)) is equal to Z(u, 𝔪, s + r − q, s, t) with 𝔪 = (m_{1}, m_{2}, …, m_{u}). In the case s = 0 the cochain τ^{(q,0,𝔱)} has only one component ω^{(q,𝔱)} and [ω^{(q,𝔱)}] ∈ ⨁_{t∈ℤ} H^{q}(Y, Ω_{Y}^{r} (t)) . One has [ω^{(q,𝔱)}] ∈ B(q, 0) if t_{ν} ≤ 1 ∀ν and [ω^{(q,𝔱)}] ∈ C(q, 0) if there exists a ν with t_{ν} > 1. If q > r − s then A(q, s) = 0 and B(q, s) = 0. If q ≥ r − s then C(q, s) = 0. We will show later that the given generators of A(q, s), B(q, s) and C(q, s) are linearly independent and independent of the choise of the polynomials Q_{νμ}^{(j)} with ∑_{μ=0}^{n} x_{μ} · Q_{νμ}^{(j)} = \frac{∂F^{(j)}}{∂x_{ν}} .

The following commutative diagram of linear maps of k-vector spaces is equipped with one exact row and one exact column.

⨁_{V_{s+1}} Θ^{q,r−s−1} Ψ^{q−1,r−s+1} ⨁_{V_{s+1}} Θ^{q+1,r−s−1}

β_{s+1}^{(q)} [figure] γ_{s+1}^{(q)} _{}d_{s−1}^{(q−1)} _{}β_{s+1}^{(q+1)} [figure] γ_{s+1}^{(q+1)}

Ψ^{q−1,r−s} ^{\overline{}^{d_{s}^{(q−1)}}} Ψ^{q,r−s−1} ^{\overline{}^{α_{s+1}^{(q)}}} ⨁_{V_{s}} Θ^{q,r−s} ^{\overline{}^{β_{s}^{(q)}}} Ψ^{q,r−s} ^{\overline{}^{d_{s}^{(q)}}} Ψ^{q+1,r−s−1} ^{\overline{}^{α_{s+1}^{(q+1)}}} ⨁_{V_{s}} Θ^{q+1,r−s}

[figure] γ_{s}^{(q)}  _{}α_{s}^{(q)}

⨁_{V_{s−1}} Θ^{q,r−s+1}

β_{s−1}^{(q)}

Ψ^{q,r−s+1}

For 0 < q < dim Y holds: Θ^{q,r−s} ≠ 0 iff r − s − c ≤ q ≤ r − s (cp. Corollary 2). Lemma 9. Let be p =char k > 0, u > 0, 0 < r ≤ d, 0 ≤ s ≤ r, 0 < q < d − r + s. Then holds

1. dim_{k} A(q, s) = dim_{k} B(q, s) = \binom{u}{r − q − s} · \binom{u + s − 1}{s},

2. dim_{k} C(q, s) = ∑_{ν=s+1}^{[ \frac{s+r−q}{2} ]} \binom{u}{s + r − q − 2ν} · \binom{u + ν − 1}{ν},

3. Im α_{s}^{(q)} = Im γ_{s}^{(q)} ,

4. Ker γ_{s}^{(q)} = Im γ_{s+1}^{(q)} ⊕ A(q, s),

5. β_{s}^{(q)} : A(q, s) ⟶^{∼} B(q, s) is an isomorphism ,

6. d_{s}^{(q)} : C(q, s) ⟶^{∼} B(q + 1, s + 1) ⊕ C(q + 1, s + 1) is an isomorphism.

If one sets E(q, s) := Im γ_{s}^{(q)} and E(q, r + 1) = 0 then there exist subspaces

D(q, s) ⊆ ⨁_{V_{s}} Θ^{q,r−s} and D̃(q, s) ⊆ Ψ^{q,r−s} with

7. ⨁_{V_{s}} Θ^{q,r−s} = A(q, s) ⊕ D(q, s) ⊕ E(q, s + 1),

8. Ψ^{q,r−s} = B(q, s) ⊕ D̃(q, s) ⊕ C(q, s), such that the following homomorphisms are isomorphisms :

9. β_{s}^{(q)} : D(q, s) ⟶^{∼} D̃(q, s),

10. α_{s}^{(q)} : D̃(q, s) ⟶^{∼} E(q, s).

Finally holds

11. dim_{k} D(q, s) = ∑_{ν=0}^{r−q−s}(−1)^{ν} · (\binom{c}{r−q−s−ν} · \binom{c+s−1+ν}{ν+s} − \binom{u}{r−q−s−ν} · \binom{u+s−1+ν}{ν+s}),

12. D(q, s) = 0 if s = 0 or u = c. \underline{Remarks:}

1. The statement remains true in the case u = 0 or char k = 0 but then we have A(q, s) = 0, B(q, s) = 0 and C(q, s) = 0.

2. Statements 1. , 2. , 5. and 6. remain true for q = 0 . Proof. We use induction on r−s : In the case s = r holds Im γ_{s} = ⨁_{V_{r}} O_{Y} (−m_{V_{r}}), A^{r−s} = O_{Y} (cp. (2)) and because of 0 < q < dim Y : Ψ^{q,r−s} = ⨁_{t∈ℤ} H^{q}(Y, Im γ_{s}(t)) = 0, Θ^{q,r−s} = ⨁_{t∈ℤ} H^{q}(Y, A^{r−s}(t)) = 0, Im α_{r}^{(q)} = Im γ_{r}^{(q)} = 0, A(q, r) = B(q, r) = C(q, r) = 0 and finaly E(q, r + 1) = 0. This shows 1. - 10. in the case of the basis of the induction. To prove Statement 4. we will show (cp. Lemma 7) ⨁_{V′_{s}} L^{q,r−s} = A(q, s) : For a combination V′_{s} = (k_{1}, …, k_{s}) and a basis element ω^{(q)} ∧ σ_{j_{1}} ∧ … ∧ σ_{j_{r−s−q}} of L^{q,r−s} let be 𝔱 = ∑_{ν=1}^{r−s−q} 𝔢_{j_{ν}} + 2 · ∑_{μ=1}^{s} 𝔢_{k_{μ}} and let τ^{(q,s,𝔱)} be the corresponding generator of A(q, s) . One has ∑_{ν=1}^{u} t_{ν} = s+r −q, ∑_{ν=1}^{u}[\frac{t_{ν}}{2} ] = s and also evidently an one-to-one mapping between these basis elements of ⨁_{V′_{s}} L^{q,r−s} resp. A(q, s) (cp. proof of Lemma 8). In particular the generators of A(q, s) are linearly independent and one has dim A(q, s) = \binom{u}{r−q−s} · \binom{u+s−1}{s}. This completes the proof of statement 4. Now let be 0 ≤ s < r < dim Y and 0 < q < dim Y − r + s. Then holds 0 < q + 1 < dim Y − r + s + 1, i.e.,1.-10. follows from the inductional assumption for both pairs (q, s + 1) and (q + 1, s + 1). So one has A(q, s) ∩ Im γ_{s+1}^{(q)} = 0 by statement 4. and

Im γ_{s+1}^{(q)} = Im α_{s+1}^{(q)} = Ker β_{s}^{(q)} = E(q, s + 1)  (21)

by inductional assumption , i.e. A(q, s) ∩ Ker β_{s}^{(q)} = 0 . Therefore the restriction of β_{s}^{(q)} onto A(q, s) is an isomorphism from A(q, s) onto B(q, s). This shows the statements 5. and 1. Under d_{s}^{(q)} the generators of C(q, s) go onto the generators of B(q + 1, s + 1) ⊕ C(q + 1, s + 1) (inductional assumption). So they are linearly independent and 6. and 2. are shown. Furthermore follows C(q, s) ∩ Ker d_{s}^{(q)} = C(q, s) ∩ Im β_{s}^{(q)} = 0 and with B(q, s) ⊆ Im β_{s}^{(q)} one gets B(q, s) ∩ C(q, s) = 0. So B(q, s) ⊕ C(q, s) is a subspace of Ψ^{q,r−s} and there exists a decomposition 8. with a suitable subspace D̃(q, s) ⊆ Ψ^{q,r−s}. Because of A(q, s)∩ Im γ_{s+1}^{(q)} = A(q, s) ∩ E(q, s + 1) = 0 the direct sum A(q, s) ⊕ E(q, s + 1) is a subspace of ⨁_{V_{s}} Θ^{q,r−s} and there exists a decomposition 7. with a suitable subspace D(q, s). Now we have to prove the statements 3. , 9. and 10. of the lemma : One has Ψ^{q,r−s} = B(q, s) ⊕ D̃(q, s) ⊕ C(q, s) ≅ Ker d_{s}^{(q)}⊕ Im d_{s}^{(q)} = Im β_{s}^{(q)}⊕ Im d_{s}^{(q)} and Ψ^{q+1,r−s−1} = B(q+1, s+1)⊕D̃(q+1, s+1)⊕C(q+1, s+1) (inductional assumption). C(q, s) goes under d_{s}^{(q)} isomorphic onto B(q + 1, s + 1) ⊕ C(q + 1, s + 1). Because of the inductional assumption the restriction of α_{s+1}^{(q+1)} onto D̃(q + 1, s + 1) is also an isomorphism. So we get Ker α_{s+1}^{(q+1)} = Im d_{s}^{(q)} = B(q + 1, s + 1) ⊕ C(q + 1, s + 1) ≅ C(q, s), Ker d_{s}^{(q)} = Im β_{s}^{(q)} ≅ B(q, s) ⊕ D̃(q, s). From (21) follows (A(q, s) ∩ D(q, s))∩ Ker β_{s}^{(q)} = 0, i.e. A(q, s) ⊕ D(q, s) ≅ Im β_{s}^{(q)} via β_{s}^{(q)} , with β_{s}^{(q)}(A(q, s)) = B(q, s). Without loss of generality one can choose D̃(q, s) so that D̃(q, s) is the isomorphic image of D(q, s) under β_{s}^{(q)} . From statement 4. follows Ker γ_{s}^{(q)} = A(q, s) ⊕ E(q, s + 1). So D(q, s) goes under γ_{s}^{(q)} isomorphic onto Im γ_{s}^{(q)} = E(q, s) , i.e. D̃(q, s) goes under α_{s}^{(q)} isomorphic onto E(q, s) . This shows 9. and 10. Statement 3. follows immediately because of Ψ^{q,r−s} = B(q, s) ⊕ D̃(q, s) ⊕ C(q, s) with B(q, s) ⊕ C(q, s) ⊆ Im d_{s−1}^{(q−1)} = Ker α_{s}^{(q)}, i.e. Im γ_{s}^{(q)} = Im α_{s}^{(q)} = E(q, s). Statement 11. follows easily from the statements 9., 10., 7., 1. and (15). Finally Statement 12. follows from 11., the lemma is proved. □

Corollary 6. For the function Z(u, 𝔪, l, s, t) one gets in the case u > 0

∑_{t∈ℤ}Z(u, 𝔪, s + r − q, s, t) = dim_{k} B(q, s) + dim_{k} C(q, s)
= ∑_{ν=s}^{[ \frac{s+r−q}{2} ]} \binom{u}{s + r − q − 2ν} · \binom{u + ν − 1}{ν}

and Z(u, 𝔪, s + r − q, s, t) = dim_{k}((B(q, s) ⊕ C(q, s)) ∩ H^{q}(Y, Imγ_{s}(t))).

Using Lemma 9. with s = 0 one gets the main result of the paper :

Theorem 5. Let be char k > 0, u > 0, 0 < r < d and 0 < q < d − r. Then one has

h^{q}(Y, Ω_{Y}^{r} (t)) = Z(u, 𝔪, r − q, 0, t),
⨁_{t∈ℤ}H^{q}(Y, Ω_{Y}^{r} (t)) = B(q, 0) ⊕ C(q, 0),
dim_{k} ⨁_{t∈ℤ}H^{q}(Y, Ω_{Y}^{r} (t)) = ∑_{ν=0}^{[ \frac{r−q}{2} ]} \binom{u}{r − q − 2ν} · \binom{u + ν − 1}{ν}.

The cohomology classes of the cocycles ω^{(q,𝔱)} with t_{ν} ≥ 0 ∀ν, ∑_{ν=1}^{u} t_{ν} = r − q form a k-basis of the vector space ⨁_{t∈ℤ} H^{q}(Y, Ω_{Y}^{r} (t)). Exactly : [ω^{(q,𝔱)}] ∈ B(q, 0) if t_{ν} ≤ 1 ∀ν . If there exists a ν with t_{ν} > 1 then [ω^{(q,𝔱)}] ∈ C(q, 0). One has [ω^{(q,𝔱)}] ∈ H^{q}(Y, Ω_{Y}^{r} (t)) with t = ∑_{ν=1}^{u}[\frac{t_{ν}+1}{2}] · m_{ν} (cp. (17)). The cohomology class of ω^{(q,𝔱)} is independent of the choise of the polynomials Q_{νμ}^{(j)}

with ∑_{μ=0}^{n} x_{μ} · Q_{νμ}^{(j)} = \frac{∂F^{(j)}}{∂x_{ν}} .

Proof. In the case 0 ≤ s = ∑_{ν=0}^{u}[\frac{t_{ν}}{2} ] the cochains τ^{(q,s,𝔱)} and \overline{τ}^{(q,s,𝔱)} are independent of the polynomials Q_{νμ}^{(j)} (cp. proof of Lemma 8). Therefore the given basis elements of A(q, s) and B(q, s) are independent of the Q_{νμ}^{(j)}. In the case 0 ≤ s < ∑_{ν=0}^{u}[\frac{t_{ν}}{2} ] Lemma 8 shows that the cohomology class of \overline{τ}^{(q,s,𝔱)} goes (except the sign) under the isomorphism d_{s}^{(q)} : C(q, s) ⟶^{∼} B(q+1, s+1)⊕C(q+1, s+1) to the cohomology class of \overline{τ}^{(q+1,s+1,𝔱)}. This cohomology class is independent of the choise of the polynomials Q_{νμ}^{(j)} by inductional assumption. Because of Im γ_{0} = Ω_{Y}^{r} follows the independence statement. For the proof of the first statement compare the definition of the function Z at the beginning of this chapter. □

Using the Serre duality one gets :

Corollary 7. If char k > 0 and u > 0, 0 < r < d and d − r < q < d, then

h^{q}(Y, Ω_{Y}^{r} (t)) = Z(u, 𝔪, q − r, 0, −t),
dim ⨁_{t∈ℤ}H^{q}(Y, Ω_{Y}^{r} (t)) = ∑_{ν=0}^{[ \frac{q−r}{2} ]} \binom{u}{q − r − 2ν} · \binom{u + ν − 1}{ν}.

\underline{Remark:} In the cases u = 0 or char k = 0 one gets :

h^{q}(Y, Ω_{Y}^{r} (t)) = δ_{q,r} · δ_{t,0} for 0 < q < d, q + r ≠ d

and the cohomology class of ω^{(q)} is the basis element of H^{q}(Y, Ω_{Y}^{q}). Now we can prove a precision of Theorem 1. For s ≥ 0 and a ∈ \{0, 1, …, [\frac{s}{2}]\} we define the following subset M_{a}^{s} of G_{a}^{s} (cp. Lemma 2): M_{a}^{s} := {λ_{1}m_{1} + … + λ_{u}m_{u} | λ_{i} ∈ ℕ_{0}, ∑_{i=1}^{u} λ_{i} = s − a and ∑_{i=1}^{u} sgnλ_{i} ≥ s − 2a }. Let M_{a}^{s} := ∅ if s < 0 or a < 0.

Theorem 6.

a) Let be 0 < q < d − r. Then H^{q}(Y, Ω_{Y}^{r} (t)) = 0 iff t ∉ M_{a}^{r−q} ∀a ∈ \{0, 1, …, [\frac{r−q}{2}]\}.

b) Let be d − r < q < d. Then H^{q}(Y, Ω_{Y}^{r} (t)) = 0 iff −t ∉ M_{a}^{q−r} ∀a ∈ \{0, 1, …, [\frac{q−r}{2}]\}.

Proof. Let be t ∈ M_{a}^{r−q} for an a ∈ \{0, 1, …, [\frac{r−q}{2}]\}. So we have t = λ_{1}m_{1} + … + λ_{u}m_{u} with λ_{ν} ≥ 0, ∑_{ν=1}^{u} λ_{ν} = r − q − a and ∑_{ν=1}^{u} sgnλ_{ν} ≥ r − q − 2a, i.e., at least r − q − 2a of the coefficients λ_{ν} are positive. Let K = (j_{1}, j_{2}, …, j_{r−q−2a}) be any combitation with 1 ≤ j_{1} < j_{2} < … < j_{r−q−2a} ≤ u and λ_{ν} > 0 ∀ν ∈ K. We set t_{ν} = 2λ_{ν} − 1 ∀ν ∈ K and t_{ν} = 2λ_{ν} ∀ν ∈ {1, 2, …, u} \ K. Then one has t_{ν} ≥ 0, [\frac{t_{ν}+1}{2}] = λ_{ν} ∀ν ∈ \{1, 2, …, u\} and ∑_{ν=1}^{u} t_{ν} = 2·∑_{ν=1}^{u} λ_{ν}−(r−q−2a) = r−q. Thus Z(u, 𝔪, r − q, 0, t) ≠ 0 and H^{q}(Y, Ω_{Y}^{r} (t)) ≠ 0 because of Theorem 5. Let be reversely H^{q}(Y, Ω_{Y}^{r} (t)) ≠ 0, i.e. Z(u, 𝔪, r − q, 0, t) ≠ 0. Then there exists a 𝔱 ∈ ℤ^{u} with t_{ν} ≥ 0 ∀ν, ∑_{ν=1}^{u} t_{ν} = r − q and t = ∑_{ν=1}^{u} [\frac{t_{ν}+1}{2}] · m_{ν}. We set λ_{ν} = [\frac{t_{ν}+1}{2}] and t_{ν} = 2λ_{ν} − ε_{ν} ∀ν ∈ \{1, 2, …, u\}, ε_{ν} ∈ \{0, 1\}. Then one has ε_{ν} ≤ t_{ν} ∀ν, ∑_{ν=1}^{u} ε_{ν} ≤ ∑_{ν=1}^{u} t_{ν} = r − q and ε_{ν} ≡ t_{ν} mod 2 ∀ν, ∑_{ν=1}^{u} t_{ν} = r − q ≡ ∑_{ν=1}^{u} ε_{ν} mod 2. With a = (r − q − ∑_{ν=1}^{u} ε_{ν})/2 follows ∑_{ν=1}^{u} λ_{ν} = \frac{1}{2} · ∑_{ν=1}^{u}(t_{ν} + ε_{ν}) = r − q − a and ∑_{ν=1}^{u} sgnλ_{ν} ≥ ∑_{ν=1}^{u} ε_{ν} = r − q − 2a, i.e. t ∈ M_{a}^{r−q}. Using the Serre duality one gets statement b) . □ Finally we calculate H^{0}(Y, Ω_{Y}^{r} (t)) and H^{d}(Y, Ω_{Y}^{r} (t)) :

Lemma 10. Let be char k > 0, u > 0, 0 < r ≤ d and 0 ≤ s ≤ r. Then one has

h^{0}(Y, Im γ_{s}(t)) = ∑_{ν=s}^{r}(−1)^{ν−s} · ∑_{V_{ν}} h^{0}(Y, A^{r−ν}(t − m_{V_{ν}})

− ∑_{ν=s+1}^{r−1}(−1)^{ν−s}Z(u, 𝔪, r + ν − 1, ν, t),

where V_{ν} runs through the combinations of {1, 2, …, c}, ν at a time, with repetitions.

Proof. For 0 ≤ s < r we have the exact sequences (cp. (2))

0 \overline{}H^{0}(Y, Im γ_{s+1}(t)) \overline{}⨁_{V_{s}} H^{0}(Y, A^{r−s}(t − m_{V_{s}})) \overline{}H^{0}(Y, Im γ_{s}(t)) ^{\overline{}^{d}}

\overline{}^{d} H^{1}(Y, Im γ_{s+1}(t)) ^{\overline{}^{α}} ⨁_{V_{s}} H^{1}(Y, A^{r−s}(t − m_{V_{s}})).

Now we look at the homomorphism

α_{s+1}^{(1)} : Ψ^{1,r−s−1} \overline{}⨁_{V_{s}} Θ^{1,r−s}.

From Lemma 9, 8., 10. and 6. (with q = 0) we know its kernel: Ker α_{s+1}^{(1)} = Im d_{s}^{(0)} = B(1, s + 1) ⊕ C(1, s + 1) (cp. (13) and (20)). Therefore the kernel of the above α is Ker α = (B(1, s + 1) ⊕ C(1, s + 1)) ∩ H^{1}(Y, Im γ_{s+1}(t)) and this is a k-vector space of dimension Z(u, 𝔪, r + s, s + 1, t) (cp. Corollary 6). In particular it is easy to see Z(u, 𝔪, 2r − 1, r, t) = 0 (cp. the definition of Z). □ Because of Im γ_{0} = Ω_{Y}^{r} one gets Theorem 7. Let Y ⊆ ℙ_{k}^{n} be a nonsingular d-dimensional complete intersection of multi-degree 𝔪 = (m_{1}, m_{2}, …, m_{c}). If char k = p divides the integers m_{1}, m_{2}, …, m_{u} and only these (0 < u ≤ c), then one has for 0 < r ≤ dim Y :

h^{0}(Y, Ω_{Y}^{r} (t)) = ∑_{ν=0}^{r}(−1)^{ν} · ∑_{V_{ν}} h^{0}(Y, A^{r−ν}(t − m_{V_{ν}})

− ∑_{ν=1}^{r−1}(−1)^{ν} Z(u, 𝔪, r + ν − 1, ν, t)

where V_{ν} runs through the combinations of {1, 2, …, c}, ν at a time, with repetitions. Using the Serre duality one gets : h^{d}(Y, Ω_{Y}^{r} (t)) = h^{0}(Y, Ω_{Y}^{d−r}(−t)). Corollary 8. h^{0}(Y, Ω_{Y}^{1} (t)) and h^{d}(Y, Ω_{Y}^{d−1}(t)) = h^{0}(Y, Ω_{Y}^{1} (−t)) are independent of char k ∀t ∈ ℤ. Corollary 9. If dim Y ≤ 2 then h^{q}(Y, Ω_{Y}^{r} (t)) is independent of char k ∀q, r, t. Corollary 10. If g.c.d.(m_{1}, m_{2}, …, m_{u}) ∤ t then h^{q}(Y, Ω_{Y}^{r} (t)) is given by the same formula like in the complex case. Corollary 11. In the simple case u = 0 or char k = 0 one has the formula

h^{0}(Y, Ω_{Y}^{r} (t)) = ∑_{ν=0}^{r}(−1)^{ν} · ∑_{V_{ν}} h^{0}(Y, A^{r−ν}(t − m_{V_{ν}}).

\underline{Remark:} From (1) and (2) we know

χ(Y, Ω_{Y}^{r} (t)) = ∑_{ν=0}^{r} ∑_{μ=0}^{c}(−1)^{ν+μ} ∑_{V_{ν}} ∑_{K_{μ}} χ (ℙ^{n}, Ω_{ℙ^{n}}^{r−ν}(t − m_{V_{ν}} − m_{K_{μ}}))

where V_{ν} runs through the combinations of {1, 2, …, c}, ν at a time, with repetitions and K_{μ} runs through the combinations of {1, 2, …, c}, μ at a time, without repetition. The theorems 5 and 7 and Corollary 7 give h^{q}(Y, Ω_{Y}^{r} (t)) with q + r ≠ d. Finally we have

h^{d−r}(Y, Ω_{Y}^{r} (t)) = (−1)^{d−r} · χ(Y, Ω_{Y}^{r} (t)) + ∑_{{q=0 \atop q≠d−r}}^{d}(−1)^{q+d−r+1} h^{q}(Y, Ω_{Y}^{r} (t)).

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