0.
We can apply similar considerations to the cohomology with coefficients in Z. If H is a Z[G]-module of finite rank over Z, the cohomology of G with coefficients in H is computed as the cohomology of the complex: 0
/H
τ
/H
σ
/H
τ
/ ... .
We have the following proposition. Proposition 2.5. Let X be a compact complex manifold and G an automorphism group of prime order p acting on X. Assume that H ∗ (X, Z) is torsion-free and 3 ≤ p ≤ 19. Then for 0 ≤ k ≤ 2 dim X we have: • H 0 (G, H k (X, Z)) = H k (X, Z)G , k
• H 2i−1 (G, H k (X, Z)) = (Z /p Z)lp−1 (X) , k
• H 2i (G, H k (X, Z)) = (Z /p Z)l1 (X) , for all i ∈ N∗ . Proof.
• By definition, H 0 (G, H k (X, Z)) = Ker τ .
• In odd degrees, H 2i−1 (G, H k (X, Z)) = Ker σ/ Im τ . In the proof of Theorem 74.3 of [8], it is shown that: Ker σ = OK b1 ⊕ ... ⊕ OK br+s−1 ⊕ Abr+s , Im τ = E1 b1 ⊕ ... ⊕ En−1 br+s−1 ⊕ Er+s Abr+s , with b1 , ..., bn , OK -free elements in Ker σ, A an OK -ideal of K and E1 = ... = Er = OK ,
Er+1 = ... = Er+s = (ξp − 1)OK .
And the r and the s in the last equalities are the same as in the proof of Proposition 1.4 (Proposition 5.1 of [6]). Moreover, we find in the proof of Theorem 74.3 of [8] that OK /(ξp − 1)OK = A/(ξp − 1)A = Z /p Z. Hence, k we get Ker σ/ Im τ = (Z /p Z)lp−1 (X) . • For i ≥ 1, H 2i (G, H k (X, Z)) = Ker τ / Im σ. We have seen in the proof of Proposition 1.4 that: H k (X, Z)G ≃ ⊕ri=1 (OK , ai )G ⊕ Z⊕t . By Lemma 2.3 all the elements in ⊕ri=1 (OK , ai )G can be written y +ϕ(y)+ ... + ϕp−1 (y) with y ∈ H k (X, Z). The result follows.
25
Now we state a similar result in the case p = 2. Proposition 2.6. Let X be a compact complex manifold and G an automorphism group of order 2 acting on X. Assume that H ∗ (X, Z) is torsion-free. Then for 0 ≤ k ≤ 2 dim X we have: • H 0 (G, H k (X, Z)) = H k (X, Z)G , k
• H 2i−1 (G, H k (X, Z)) = (Z /p Z)l1,− (X) , k
• H 2i (G, H k (X, Z)) = (Z /p Z)l1,+ (X) , for all i ∈ N∗ . Proof. The same proof as in the last proposition. We can give more precise results on the cohomology of the quotient in imposing additional hypothesis on the degeneration of the spectral sequence. Definition 2.7. Let G be a group of prime order p acting by automorphisms on a complex manifold X. We will say that (X, G) is E2 -degenerate if the spectral sequence of equivariant cohomology with coefficients in Fp degenerates at the E2 term. We will say that (X, G) is E2 -degenerate over Z if the spectral sequence of equivariant cohomology with coefficients in Z degenerates at the E2 -term. 2.2.2
Case where G acts freely
We can use the equivariant cohomology to calculate the integral cohomology of a quotient when the action of the group is free. Proposition 2.8. Let X be a compact complex manifold and G a group of prime order, acting freely on X in such a way that (X, G) is E2 -degenerate over Z. Assume that H ∗ (X, Z) is torsion-free and 3 ≤ p ≤ 19. Then for 0 ≤ 2k ≤ 2 dim X, we have H 2k (X/G, Z) ≃ H 2k (X, Z)G
k−1 k−1 M M 2i+1 2i (Z /p Z)l1 (X) , (Z /p Z)lp−1 (X) i=0
i=0
and for 0 ≤ 2k + 1 ≤ 2 dim X, H 2k+1 (X/G, Z) ≃ H 2k+1 (X, Z)G
k M i=0
2i
(Z /p Z)lp−1 (X)
k−1 M
2i+1
(Z /p Z)l1
(X)
.
i=0
2k Proof. We have H 2k (X/G, Z) ≃ HG (X) by Section 1.3. Moreover E2p,q := p+q p q H (G; H (X)) ⇒ HG (X). Since the spectral sequence of equivariant cohomology is degenerate at the E2 -page,
H 2k (X/G, Z) ≃
2k M
H i (G; H 2k−i (X, Z)),
i=0
26
and by Proposition 2.5, H 2k (X/G, Z) ≃ H 2k (V, Z)G
k−1 M
2i+1
(Z /p Z)lp−1 (X)
k−1 M 2i (Z /p Z)l1 (X) . i=0
i=0
The same formula holds for p = 2. Proposition 2.9. Let X be a compact complex manifold and G a group order of order 2 acting freely on X in such a way that (X, G) is E2 -degenerate over Z. Assume that H ∗ (X, Z) is torsion-free. Then H 2k (X/G, Z) ≃ H 2k (X, Z)G
k−1 k−1 M M 2i+1 2i (Z /p Z)l1,+ (X) (Z /p Z)l1,− (X) i=0
i=0
for 0 ≤ 2k ≤ 2 dim X, and
H 2k+1 (X/G, Z) ≃ H 2k+1 (X, Z)G
k−1 k M M 2i+1 2i (Z /p Z)l1,+ (X) (Z /p Z)l1,− (X) i=0
i=0
for 0 ≤ 2k + 1 ≤ 2 dim X. We can replace the condition of E2 -degeneration over Z by conditions on the lij (X). Proposition 2.10. Let X be a manifold and G a group of prime order acting freely on X. Assume that H ∗ (X, Z) is torsion-free and 3 ≤ p ≤ 19. For 0 ≤ 2k ≤ 2 dim X, assume: 2i i) lp−1 (X) = 0 for all 1 ≤ i ≤ k,
ii) l12i+1 (X) = 0 for all 0 ≤ i ≤ k − 1 when k > 1. Then we have:
1) H 2k (X/G, Z) ≃ H 2k (X, Z)G
k−1 k−1 M M 2i+1 2i (Z /p Z)l1 (X) , (Z /p Z)lp−1 (X) i=0
i=0
2) H 2k−1 (X/G, Z) ≃ H 2k−1 (X, Z)G and H 2k+1 (X/G, Z) ≃ H 2k+1 (X, Z)G . Proof. It is enough to check that all the groups H i (G, H 2k+1−i (X, Z)), 1 ≤ i ≤ 2k and H i (G, H 2k−i−1 (X, Z)), 1 ≤ i ≤ 2k − 2 are trivial. By Proposition 2.5, this is exactly the condition on the lij (X).
27
We have a similar result for p = 2. Proposition 2.11. Let X be a manifold and G a group of order 2 acting freely on X. Assume that H ∗ (X, Z) is torsion-free. For 0 ≤ 2k ≤ 2 dim X, assume: 2i i) l1,− (X) = 0 for all 1 ≤ i ≤ k, 2i+1 ii) l1,+ (X) = 0 for all 0 ≤ i ≤ k − 1, k > 1.
then we have:
1) H 2k (X/G, Z) ≃ H 2k (X, Z)G
k−1 M i=0
2i+1
(Z /2 Z)l1,−
(X)
k−1 M
2i
(Z /2 Z)l1,+ (X) ,
i=0
2) H 2k−1 (X/G, Z) ≃ H 2k−1 (X, Z)G and H 2k+1 (X/G, Z) ≃ H 2k+1 (X, Z)G . Remark: It is also possible to calculate H k (X/G, Fp ) by the spectral sequence of equivariant cohomology with coefficients in Fp when (X, G) is E2 degenerate and the action of G is free. We get similar formulas using Proposition 2.4. Then one can deduce H k (X/G, Z) by the universal coefficient theorem.
2.3 2.3.1
H ∗ -normality Definition
Now we want to calculate the cohomology of X/G when the action of G is not free. A fundamental tool for studying this question is given by the following proposition, which follows from [34]. Proposition 2.12. Let G be a finite group of order d acting on a variety X with the orbit map π : X → X/G, which is a d-fold ramified covering. Then there is a natural homomorphism π∗ : H ∗ (X, Z) → H ∗ (X/G, Z) such that X π∗ ◦ π ∗ = d idH ∗ (X/G,Z) , π ∗ ◦ π∗ = g∗. g∈G
It easily implies the corollary: Corollary 2.13. Let G be a finite group of order d acting on a variety X with the orbit map π : X → X/G, which is a d-fold ramified covering. Then: ∗ 1) π|H ∗ (X/G,Z)/ tors is injective,
2) π∗|H ∗ (X,Z)G ◦ π ∗ = d idH ∗ (X/G,Z) and π ∗ ◦ π∗|H ∗ (X,Z)G = d idH ∗ (X,Z)G , 3) H ∗ (X/G, Q) ≃ H ∗ (X, Q)G . Leaving aside the question of determining the torsion of H ∗ (X/G, Z), we go on to the study of the image of π∗ in H ∗ (X/G, Z)/ tors. 28
Proposition 2.14. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p. Let 0 ≤ k ≤ 2n, and assume that H k (X, Z) is torsion-free. Then there is an exact sequence: 0
/ π∗ (H k (X, Z))
/ H k (X/G, Z)/ tors
/ (Z /p Z)αk
/ 0,
where π : X → X/G is the quotient map and αk is a positive integer. Proof. Let x ∈ H k (X/G, Z)/ tors. Then px = π∗ (π ∗ (x)) with π ∗ (x) ∈ H k (X, Z), which implies the result. It remains to calculate αk . In this section our goal will be to understand, in which cases αk = 0. Definition-Proposition 2.15. Let X be a compact complex manifold of dimension n and G = hϕi an automorphism group of prime order p. Let 0 ≤ k ≤ 2n, and assume that H k (X, Z) is torsion-free. The integer αk from Proposition 2.14 will be called the coefficient of normality. The following statements are equivalent: • αk = 0, • the map π∗ : H k (X, Z) → H k (X/G, Z)/ tors is surjective. • For x ∈ H k (X, Z)G , π∗ (x) is divisible by p if and only if there exists y ∈ H k (X, Z) such that x = y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y). If one of these statements is verified, we will say that (X, G) is H k -normal. If (X, G) is H k -normal for all 0 ≤ k ≤ 2n, we will say that (X, G) is H ∗ normal. Proof. The two first statements are equivalent by Proposition 2.14. We show that the second one is equivalent to the third one. ⇐ Let x ∈ H k (X/G, Z)/ tors. We have π∗ (π ∗ (x)) = px. Hence, π ∗ (x) can be written in the form y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y) for y ∈ π∗ (H k (X, Z)). Then π∗ (π ∗ (x)) = pπ∗ (y) = px, so π∗ (y) = x. ⇒ Let x ∈ H k (X, Z)G such that p divides π∗ (x). Since π∗ : H k (X, Z) → H k (X/G, Z)/ tors is surjective, there is z ∈ H k (X, Z) such that pπ∗ (z) = π∗ (x). We apply π ∗ to this equality. By Proposition 2.12, we get p(z + ϕ∗ (z) + ... + (ϕ∗ )p−1 (z)) = px.
Corollary 2.16. Let X be a compact complex manifold of dimension n and G = hϕi an automorphism group of prime order p. Let 0 ≤ k ≤ 2n. Let Hσk (X, Z) = x + ϕ∗ (x) + ... + (ϕ∗ )p−1 (x)| x ∈ H k (X, Z) . 29
Assume that H k (X, Z) is torsion-free. If the pair (X, G) is H k -normal, then the map 1 π∗ : Hσk (X, Z) → H k (X/G, Z)/ tors p is an isomorphism, and its inverse is π ∗ : H k (X/G, Z)/ tors → Hσk (X, Z). Proof. This map is clearly well defined. Since (X, G) is H k -normal, it is surjective. It remains to show that it is injective. If π∗ (x + ϕ∗ (x) + ... + (ϕ∗ )p−1 (x)) is a torsion element in H k (X/G, Z), then by Corollary 2.13, x + ϕ∗ (x) + ... + (ϕ∗ )p−1 (x) is also a torsion element. Since H k (X, Z) is torsion-free, x + ϕ∗ (x) + ... + (ϕ∗ )p−1 (x) = 0. We will also need the following two lemmas. Lemma 2.17. Let X be a compact complex manifold of dimension n and G = hϕi an automorphism group of prime order p. Let 0 ≤ k ≤ 2n. Assume that H k (X, Z) is torsion-free. Let Kk′ be the overlattice of π∗ (H k (X, Z)G ) obtained by dividing by p all the elements of the form π∗ (y + ϕ(y) + ... + ϕp−1 (y)), y ∈ H k (X, Z). Then: Kk′ = π∗ (H k (X, Z)). Proof. Let y ∈ H k (X, Z), we have π∗ (y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y)) = pπ∗ (y). The result follows. Lemma 2.18. Let X be a compact complex manifold of dimension n and G = hϕi an automorphism group of prime order p. 1) Let 0 ≤ k ≤ 2 dim X, q an integer such that kq ≤ 2 dim X and x ∈ H k (X, Z)G . Then π∗ (x)q = pq−1 π∗ (xq ) + tors . If moreover H kq (X, Z) is torsion-free, then the property that π∗ (x) is divisible by p implies that π∗ (xq ) is divisible by p. 2) Let 0 ≤ k ≤ 2 dim X, q an integer such that kq ≤ 2 dim X, and let (xi )1≤i≤q be elements of H k (X, Z)G . Then π∗ (x1 ) · ... · π∗ (xq ) = pq−1 π∗ (x1 · ... · xq ) + tors . 3) Let 0 ≤ k ≤ 2 dim X, q an integer such that kq = 2n and let (xi )1≤i≤q be elements of H k (X, Z)G . Then π∗ (x1 ) · ... · π∗ (xq ) = pq−1 x1 · ... · xq .
30
Proof.
1) By Corollary 2.13 π ∗ (π∗ (x)q ) = pq xq = π ∗ (pq−1 π∗ (xq )).
The map π ∗ is injective on the torsion-free part, which implies the wanted equality. If moreover π∗ (x) is divisible by p, we can write π∗ (x) = py with y ∈ H kq (X/G, Z). This gives: pq y q = pq−1 π∗ (xq ) + tors . We cannot divide by pq−1 because of the possible torsion of H kq (X/G, Z). We will use the fact that H kq (X, Z) is torsion-free to get round the problem. Applying π ∗ to this equality, we obtain: pq π ∗ (y q ) = pq xq + π ∗ (tors). Since H kq (X, Z) is torsion-free, π ∗ (tors) = 0, and we have π ∗ (y q ) = xq . Pushing down by π∗ , we obtain: py q = π∗ (xq ). 2) The proof is similar 3) By 2), we have: π∗ (x1 ) · ... · π∗ (xq ) = pq−1 π∗ (x1 · ... · xq ) + tors . But H 2n (X, Z) = Z is torsion-free by Poincaré duality. Indentifying H 2n (X, Z) with Z, we can write π∗ (x1 · ... · xq ) = x1 · ... · xq .
2.3.2
H n -normality and cup-product lattice
Under the assumption of the H n -normality, we can determine the cup-product lattice. Proposition 2.19. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p. Assume that H n (X, Z) is torsionfree and (X, G) is H n -normal. Let us denote the sublattice H n (X, Z)G by L. Then: n
1) discr L = paG (X) , with anG (X) ∈ N, 2) H n (X/G, Z)/ tors ≃ L∨ (p), n
3) discr L∨ (p) = prk L−aG (X) . Proof. We need the following lemma. 31
Lemma 2.20. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order. Assume that H n (X, Z) is torsion-free. Then: 1) The projection H n (X, Z) → AH n (X,Z)G H n (X, Z)G ⊕ (H n (X, Z)G )⊥ is an isomorphism. n
2) AH n (X,Z)G ≃ (Z /p Z)aG (X) , with anG (X) ∈ N. 3) Moreover, let x ∈ H n (X, Z)G . Then xp ∈ (H n (X, Z)G )∨ if and only if there is z ∈ H n (X, Z) such that x = z + ϕ(z) + ... + ϕp−1 (z). Proof.
1) The first assertion follows from the unimodularity of H n (X, Z).
2),3) We start by proving that (H n (X, Z)G )⊥ = SG (X). First (H n (X, Z)G )⊥ ⊃ SG (X). Indeed, let y ∈ SG (X) and z ∈ H n (X, Z)G . Then (ϕ∗ )k (y) · z = (ϕ∗ )k (y) · (ϕ∗ )k (z) = y · z for all 0 ≤ k ≤ p. Now we prove (H n (X, Z)G )⊥ ⊂ SG (X). Let y ∈ (H n (X, Z)G )⊥ . Then y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y) ∈ (H n (X, Z)G )⊥ ∩ H n (X, Z)G . Since the cup-product form is non-degenerate, y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y) = 0. Now, let x be a primitive element of H n (X, Z)G and q ∈ N∗ such that x x n G ∨ q ∈ (H (X, Z) ) . Then q ∈ AH n (X,Z)G . By the first assertion, there is ∗ z ∈ H 2 (X, Z) and y ∈ SG (X) such that z = x+y q . Then z + ϕ (z) + ... + ∗
∗ p−1
) (y) (ϕ∗ )p−1 (z) = pq x+ y+ϕ (y)+...+(ϕ . But y+ϕ∗ (y)+...+(ϕ∗ )p−1 (y) = q 0. Hence z + ϕ∗ (z) + ... + (ϕ∗ )p−1 (z) = pq x. Since x is primitive in H n (X, Z)G , q divides p. Hence q = 1 or q = p. If q = p, we get z + ϕ∗ (z) + ... + (ϕ∗ )p−1 (z) = x.
Since (X, G) is H n -normal, from the last lemma and Lemma 2.17, we see that H n (X/G, Z)/ tors = π∗ (L∨ ). Hence by Lemma 2.18 3), H n (X/G, Z)/ tors = L∨ (p). By assertions 1) and 2) of the last lemma and by Proposition 1.3, discr L ⊕ n L⊥ = p2aG (X) . But, since H n (X, Z) is unimodular, discr L = discr L⊥ . Hence n discr L = paG (X) . n Moreover by assertion 2) of the last lemma, L∨ /L = AL ≃ (Z /p Z)aG (X) . ∨ 2an (X) Hence, by Proposition 1.1, we have discr L = (discr L ) · p G . It follows that n n discr L∨ = p−aG (X) and discr L∨ (p) = prk L−aG (X) . 32
Remark: Lemma 2.20 defines anG (X) for all prime numbers p although Proposition-Definition 1.7 defined it just for 2 ≤ p ≤ 19. Corollary 2.21. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p such that H n (X, Z) is torsion-free and (X, G) is H n -normal. Let us denote the lattice H n (X/G, Z)/ tors by N . Then H n (X, Z)G ≃ N ∨ (p). Proof. We denote the sublattice H n (X, Z)G by L. By Proposition 2.19, H n (X/G, Z)/ tors ≃ L∨ (p). The result follows from the equality (L∨ (p))∨ (p) = L. We can also prove an upper bound for the coefficient of normality. We start with the following Proposition. Proposition 2.22. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p acting on X. We assume that H ∗ (X, Z) is torsion-free. Then: 1) discr π∗ (H n (X, Z)) = p(rk H
n
(X,Z)G −an G (X))
, with b ∈ N.
2) If moreover 2 ≤ p ≤ 19, then n
discr π∗ (H n (X, Z)) = pl1 (X) for p 6= 2, and
n
discr π∗ (H n (X, Z)) = 2l1,+ (X) for p = 2. Proof.
1) By Lemma 2.17, π∗ (H n (X, Z)) = Kn′ . Hence Kn′ ⊃ π∗ (H n (X, Z)G ). n
By Proposition 2.19 1), discr H n (X, Z)G = paG , and by Lemma 2.18 3), n
discr π∗ (H n (X, Z)G ) = paG +rk H
n
(X,Z)G
.
And by Lemma 2.20 2),3), we have π∗
Kn′ n (H (X, Z)G )
n
= (Z /p Z)aG .
Hence by Proposition 1.1, discr π∗ (H n (X, Z)) = discr Kn′ = prk H 33
n
(X,Z)G −an G (X)
.
2) It follows from 1), Proposition 1.4 and Proposition 1.6. Corollary 2.23. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p acting on X. We assume that H n (X, Z) is torsion-free. Let αn be the n-th coefficient of normality of (X, G). Then: 1) αn ≤
logp discr π∗ (H n (X,Z)) 2
=
rk H n (X,Z)G −an G (X) . 2
2) If moreover 2 ≤ p ≤ 19, then: αn ≤ αn ≤
ln 1 (X) for p 6= 2, and 2 ln (X) 1,+ for p = 2. 2
Proof. By Proposition 1.1 and Proposition 2.14, discr(H n (X/G, Z)/ tors) = discr(π∗ (H n (X, Z))) · p−2αn . Hence, the result follows from Proposition 2.22. 2.3.3
First general results
We now can state some criteria for the H k -normality. Proposition 2.24. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p acting on X. Assume that H ∗ (X, Z) is torsion-free and 2 ≤ p ≤ 19. Let 0 ≤ k ≤ 2n. k If p = 2 and l1,+ (X) = 0 then (X, G) is H k -normal. If p > 3 and l1k (X) = 0 k then (X, G) is H -normal. In other words, if akG (X) = rk H k (X, Z)G then (X, G) is H k -normal. Proof. By the hypothesis, we can write: ⊕s H k (X, Z) ≃ ⊕ri=1 (OK , ai ) ⊕ OK .
Hence H k (X, Z)G ≃ ⊕ri=1 (OK , ai )G . Let x ∈ H k (X, Z)G , then necessary x ∈ Np . Hence by Lemma 2.3, x = y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y). Therefore, by Proposition-Definition 2.15, (X, G) is H k -normal. Proposition 2.25. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p acting on X. Assume that H ∗ (X, Z) is torsion-free. n If p = 2 and l1,+ (X) = 1 then (X, G) is H n -normal. If p ≥ 3 and l1n (X) = 1 then (X, G) is H n -normal. In other words, if anG (X) = rk H n (X, Z)G − 1 then (X, G) is H n -normal. 34
Proof. By Proposition 2.22 1), we know that discr π∗ (H n (X, Z)) = prk H
n
(X,Z)G −an G (X)
.
Hence, our case: discr π∗ (H n (X, Z)) = p. So by Corollary 2.23 1), the coefficient of normality has to be 0. Now consider the case in which X/G is smooth. Proposition 2.26. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order acting on X. Assume that H ∗ (X, Z) is torsion-free and X/G is smooth. Then (X, G) is H n -normal if and only if rk H n (X, Z)G = anG (X). Proof. Assume that (X, G) is H n -normal, then by Proposition 2.19, H n (X/G, Z)/ tors ≃ L∨ (p) n
G
n
with discr L∨ (p) = prk H (X,Z) −aG (X) . Since X/G is smooth, H n (X/G, Z)/ tors is unimodular. The result follows. It is also possible to calculate the n-th coefficient of normality αn in this case. Proposition 2.27. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p acting on X. Assume that H ∗ (X, Z) is torsion-free, 2 ≤ p ≤ 19 and X/G is smooth. Then: n 1) l1n (X) is even when p > 2, and l1,+ (X) is even when p = 2.
2) αn =
ln 1 (X) 2
when p > 2, and αn =
ln 1,+ (X) 2
when p = 2.
Proof. By proposition 2.14, (H n (X/G, Z)/ tors)/π∗ (H n (X, Z)) = (Z /p Z)αn . Since H n (X/G, Z)/ tors is unimodular, the result follows by Proposition 1.1 and Proposition 2.22. We can also deduce the H k -normality from H kt -normality. Proposition 2.28. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order 2 ≤ p ≤ 19 acting on X. Let 0 ≤ k ≤ 2 dim X, t an integer such that kt ≤ 2 dim X and H ∗ (X, Z) torsion-free (we have H ∗ (X, Z) ⊗ Fp = H ∗ (X, Fp )). Assume that (X, G) is H kt -normal. If S : Symt H k (X, Fp ) → H kt (X, Fp ) x1 ⊗ ... ⊗ xt 7→ x1 · ... · xt is injective and S (Symt H k (X, Fp )) admits a complementary vector space, stable by the action of G, then (X, G) is H k -normal. 35
Proof. We use the same notation for both Jordan decompositions of H k (X, Z) and of H kt (X, Z): H k (X, Fp ) =
p X
⊕lk q (X)
Nq
q=1
=
p X
Nq .
q=1
Let x ∈ H k (X, Z)G . We assume that there is no y ∈ H k (X, Z) such that x = y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y) and we show that π∗ (x) is not divisible by p. Then, by Lemma 2.3, x ∈ / Np . Since S is injective and N1⊗t = N1 , we have xt ∈ / Np . By Lemma 2.3 there is no z ∈ H k (X, Z) such that xt = z +ϕ∗ (z)+...+ ∗ p−1 (ϕ ) (z). Since (X, G) is H kt -normal π∗ (xt ) is not divisible by p. Now, since H kt (X, Z) is torsion-free, by Lemma 2.18 1), π∗ (x) is not divisible by p. In particular, when S is an isomorphism, S (Symt H k (X, Fp )) admits a complementary vector space, stable by the action of G. Moreover, we can calculate the lqkt (X) in terms of lqk (X). Proposition 2.29. Let X be a topological space and G a group of prime order acting on X. Let t and k be integers. Assume that H ∗ (X, Z) is torsion-free. If S : Symt H k (X, Fp ) → H kt (X, Fp ) is an isomorphism, then: lq (Symt H k (X, Fp )) = lqkt (X), where 1 ≤ q ≤ p. Under the hypotheses of the previous Proposition, we can use the following lemma (Lemma 6.14 from [6]): Lemma 2.30. Assume that 3 ≤ p ≤ 19, G = Z /p Z and let M be a finite Fp [G]-module. Then: l1 (M ) · (l1 (M ) + 1) lp−1 (M ) · (lp−1 (M ) − 1) + , 2 2 lp−1 (Sym2 M ) = lp−1 (M ) · l1 (M ), l1 (Sym2 M ) =
p+1 lp (M ) · (lp (M ) − 1) p − 1 · lp (M ) + p · + · lp−1 (M ) 2 2 2 + (p − 1) · lp (M ) · lp−1 (M ) + lp (M ) · l1 (M )
lp (Sym2 M ) =
+ (p − 2) ·
lp−1 (M ) · (lp−1 (M ) − 1) , 2
and li (Sym2 M ) = 0 for 2 ≤ i ≤ p − 2. In some cases, one can guarantee the bijectivity of S .
36
Proposition 2.31. Let X be a topological space. Let t and k be integers and p a prime number. Assume that H ∗ (X, Z) is torsion-free. If the cup product map Symt H k (X, Q) → H kt (X, Q) is an isomorphism and kt H (X, Z)/ Symt H k (X, Z) is p-torsion-free, then: S : Symt H k (X, Fp ) → H kt (X, Fp ) is an isomorphism. Proof.
We prove the injectivity.
Let x1 ⊗ ... ⊗ xt ∈ Symt H k (X, Fp ) such that x1 · ... · xt = 0. Then there exists y ∈ H kt (X, Z) such that x1 · ... · xt = py. Hence y˙ ∈ H kt (X, Z)/ Symt H k (X, Z) is a p-torsion element (here y˙ is the class of y modulo Symt H k (X, Z)). Hence by the hypothesis y˙ = 0. It follows that y ∈ Symt H k (X, Z), so y = y1 · ... · yt with yi ∈ H k (X, Z). Since Symt H k (X, Q) → H kt (X, Q) is injective, x1 ⊗ ... ⊗ xt = py1 ⊗ ... ⊗ yt . So x1 ⊗ ... ⊗ xt = 0. We prove the surjectivity. Let y ∈ H kt (X, Fp ), with y ∈ H kt (X, Z). Since Symt H k (X, Q) → H kt (X, Q) is an isomorphism, there is q ∈ N and x1 ⊗...⊗xt ∈ Symt H k (X, Z) such that q1 x1 · ... · xt = y. Hence y˙ ∈ H kt (X, Z)/ Symt H k (X, Z) is a q-torsion element. But since H kt (X, Z)/ Symt H k (X, Z) is p-torsion-free, p does not divide q. And S ( 1q x1 ⊗ ... ⊗ xt ) = y. 2.3.4
H ∗ -normality and commutative diagrams
Let X be a compact complex manifold of dimension n and G = hϕi an aue → X be a morphism from a tomorphism group of prime order p. Let s : X e compact complex manifold X of dimension n such that ϕ can be extended to an e It means that there exists an automorphism ϕ automorphism of X. e of order p e e = hϕi. of X such that s ◦ ϕ e = ϕ ◦ s. We denote G e We can consider the quotients f := X/ e G. e We get a Cartesian diagram M := X/G and M f M O
r
e X
s
/M O π
π e
/ X.
It induces a commutative diagram on cohomology: H k (M, Z)
q
π∗ π
∗
r∗
k f H (M , Z) m
k 1 H (X, Z) s∗
π e∗ π e∗
- k e Z). H (X, 37
(∗)
e G) e whose H k -normality descends to that of (X, G). The idea is to find (X,
Definition 2.32. Let X be a compact complex manifold of dimension n and e → X be a G = hϕi an automorphism group of prime order p. Let s : X e is a compact complex manifold of dimension n and there morphism such that X e which verifies s ◦ ϕ is an automorphism ϕ e of order p of X e = ϕ ◦ s. We denote e and the induced map X/ e G e → X/G by r. hϕi e by G, e G, e r, s) will be called a pullback of (X, G). • The quadruple (X,
e Fp ) is • If moreover s is a bimeromorphic map, s∗ : H k (X, Fp ) → H k (X, ∗ k injective and s (H (X, Fp )) admits a complementary vector space, stable e the quadruple (X, e G, e r, s) will be called a k-split under the action of G, pullback of (X, G). e G, e r, s) is a k-split pullback of (X, G) and M f = X/ e G e is smooth, • If (X, e e then (X, G, r, s) will be called a regular k-split pullback of (X, G).
e G, e r, s) is a k-split pullback (resp. a regular k-split pullback) for all • If (X, e G, e r, s) is a split pullback (resp. a 0 ≤ k ≤ 2n of (X, G), we say that (X, regular split pullback) of (X, G). e G) e for short, reserving the symbols r, s to denote • We will also write (X, e G, e r, s). the maps in the pullback (X, We have the following lemma.
Lemma 2.33. Let X be a compact complex manifold of dimension n and G = e G) e be a pullback of (X, G). hϕi an automorphism group of prime order p. Let (X, Let 0 ≤ k ≤ 2 dim X and x ∈ H k (X, Z)G . Then: π e∗ (s∗ (x)) = r∗ (π∗ (x)) + tors .
e Z) is torsion-free, then the property that r∗ (π∗ (x)) is divisible If moreover H k (X, by p implies that π e∗ (s∗ (x)) is divisible by p.
Proof. By Diagram (*), we have:
π e∗ (r∗ (π∗ (x))) = s∗ (π ∗ (π∗ (x))) = p · s∗ (x) = π e∗ (e π∗ (s∗ (x))).
The map π e∗ is injective on the torsion-free part, so we get the equality. If moreover r∗ (π∗ (x)) is divisible by p, we can write r∗ (π∗ (x)) = py with y ∈ f, Z). This gives: H k (M π e∗ (s∗ (x)) + tors = py. Applying π e∗ to this equality, we get:
ps∗ (x) = pe π ∗ (y).
38
e Z) is torsion-free, this is also the case for the group s∗ (H k (X, Z)), Since H k (X, hence: π e∗ (y) = s∗ (x). Hence by applying π e∗ , we get:
π e∗ (s∗ (x)) = py.
Lemma 2.34. Let X be a compact complex manifold of dimension n and G = e G) e be a n-split pullback hϕi an automorphism group of prime order p. Let (X, ∗ n G of (X, G). Let K be the overlattice of π e∗ (s (H (X, Z) )) obtained by dividing by p all the elements of the form π e∗ (s∗ (y + ϕ(y) + ... + ϕp−1 (y))), y ∈ H n (X, Z). e Z) are torsion-free. Then: We assume that H ∗ (X, Z) and H ∗ (X, 1) discr K = discr π∗ (H n (X, Z)).
2) If moreover 2 ≤ p ≤ 19, then 2
discr π∗ (H n (X, Z)) = discr K = pl1 (X) for p 6= 2, and 2
discr π∗ (H n (X, Z)) = discr K = 2l1,+ (X) for p = 2. 3) If K is primitive, then (X, G) is H n -normal. Proof.
1) By the last lemma, we have r∗ (Kn′ ) + tors = K. Hence discr r∗ (Kn′ ) = discr K.
Since r is a bimeromorphic map, we get discr Kn′ = discr K. The result then follows from Lemma 2.17. 2) This follows by 1) and Proposition 2.22. 3) We use the same notation for the Jordan decomposition of H n (X, Fp ) and e Fp ). H n (X,
Let x ∈ H n (X, Z)G . We assume that there is no y ∈ H n (X, Z) such that x = y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y) and we show that π∗ (x) is not divisible / Np . by p. By Lemma 2.3 x ∈ ∗ n e Fp ) is injective and s∗ (H n (X, Fp )) admits Since s : H (X, Fp ) → H n (X, e s∗ (x) ∈ a complementary vector space stable under the action of G, / Np . n e ∗ Hence by Lemma 2.3, there is no z ∈ H (X, Z) such that s (x) = z + ϕ(z) + ... + ϕp−1 (z). Since K is primitive, π e∗ (s∗ (x)) is not divisible by ∗ p. Hence by Lemma 2.33, r (π∗ (x)) is not divisible by p. It follows that π∗ (x) is not divisible by p.
39
Proposition 2.35. Let X be a compact complex manifold of dimension n and e G) e be a k-split G = hϕi an automorphism group of prime order p. Let (X, e Z) are torsion-free. If (X, e G) e pullback of (X, G). Assume H ∗ (X, Z) and H ∗ (X, is H k -normal then (X, G) is H k -normal.
Proof. The proof is almost the same as the proof of 3) of Lemma 2.34. We use e Fp ). the same notation for the Jordan decomposition of H k (X, Fp ) and H k (X, k G k Let x ∈ H (X, Z) . We assume that there is no y ∈ H (X, Z) such that x = y + ϕ∗ (y) + ... + (ϕ∗ )p−1 (y) and we show that π∗ (x) is not divisible by p. / Np . By Lemma 2.3 x ∈ e Fp ) is injective and s∗ (H k (X, Fp )) admits a Since s∗ : H k (X, Fp ) → H k (X, e s∗ (x) ∈ complementary vector space, stable under the action of G, / Np . Hence by k e ∗ Lemma 2.3, there is no z ∈ H (X, Z) such that s (x) = z + ϕ(z) + ... + ϕp−1 (z). e G) e is H k -normal, π Since (X, e∗ (s∗ (x)) is not divisible by p. Hence by Lemma ∗ 2.33, r (π∗ (x)) is not divisible by p. It follows that π∗ (x) is not divisible by p. The relation of being a pullback is transitive. Proposition 2.36. Let X be a compact complex manifold of dimension n and G = hϕi an automorphism group of prime order p. Let 0 ≤ k ≤ 2n. Let (X1 , G1 , r1 , s1 ) be a pullback (resp. a k-split pullback, a regular k-split pullback) of (X, G) and (X2 , G2 , r2 , s2 ) be a pullback (resp. a k-split pullback, a regular k-split pullback) of (X1 , G1 ). Then (X2 , G2 , r1 ◦ r2 , s1 ◦ s2 ) is a pullback (resp. a k-split pullback, a regular k-split pullback) of (X, G). We give an example of a split pullback.
Proposition 2.37. Let X be a Kähler manifold of dimension n and G = hϕi an automorphism group of prime order p. Let F ⊂ Fix G be a connected component. Assume that H ∗ (X, Z) is torsion-free. e → X be the blow-up of X in F . Then G extends naturally to X. e Let s : X e this extension. Then (X, e G) e is a split pullback of (X, G). Denote by G Proof. This follows from Theorem 7.31 of [36] (Theorem 1.11).
2.4 2.4.1
Resolution of the quotient By classical blow-ups
Let X be a Kähler manifold of dimension n and G = hϕi an automorphism group of prime order p. In the last section, we have seen that blow-ups of X in a connected component of Fix G provide split pullbacks. In this section we will find all the regular split pullbacks obtained from the blow-up in connected components of Fix G. Then in Section 2.5 and Section 2.6 we will use these regular split pullbacks and Lemma 2.34 3) to get some general theorems.
40
At each fixed point of G, by Cartan Lemma 1 of [7] we can locally linearize the action of G. Thus at a fixed point x ∈ X, the action of G on X is locally equivalent to the action of G = hgi on Cn via g = diag(ξpk1 , ..., ξpkn ), where ξp is a p-th root of unity. Without loss of generality, we can assume that k1 ≤ ... ≤ kn ≤ p − 1. Definition 2.38. kn−1 = 0.
• We will say that a fixed point is of type 0 if k1 = ... =
• We will say that a fixed point is of type 1 if there is i ∈ {1, ..., n} such that k1 = ... = ki = 0 and ki+1 = ... = kn . • We will say that a fixed point is of type 2 if p = 3 and if it is not a point of type 0 or 1. • The fixed points which are not of type 0,1 or 2 will be just called points of other type. When it is defined, we will denote o(x) the type of a fixed point x. Proposition 2.39. Let X be a complex manifold of dimension n and G an automorphism group of prime order p. Let x ∈ Fix G. 1) The variety M = X/G is smooth in π(x) if and only if x is a point of type 0. e be the blow-up of X in the connected components of Fix G of codimen2) Let X f the quotient of X e by the natural action of G on X. e The variety sion ≥ 2 and M f M is smooth if and only if the points of Fix G are of type 0 or 1. Proof. By Lemma 1 of [7], we can assume that X = Cn and
G = diag(ξpk1 , ..., ξpkn ) .
1) By the proof of Proposition 6 of [32], Cn /G is smooth if and only if rk g − id = 1. Hence we get the result. f is smooth at 2) If 0 is a point of type 0, then s∗ (0) is also of type 0 and M π e (s∗ (0)).
Now assume that 0 is of type different from 0. Let G = diag(ξpk1 , ..., ξpkn ) acting on Cn . If k1 = ... = ki = 0, then
Cn /G ≃ Ci × Cn−i / diag(ξpki+1 , ..., ξpkn ) , so without loss of generality, we can assume that all ki are different from fn be the blow-up of Cn in 0 and G e the automorphism group of 0. Let C n f e on C induced by G. We will describe the action of G fn = {((x1 , ..., xn ), (a1 : ... : an )) ∈ Cn × Pn−1 | C rk ((x1 , ..., xn ), (a1 , ..., an )) = 1} . 41
We denote by Oi = ((x1 , ..., xn ), (a1 : ... : an )) ∈ Cn × Pn−1 | ai 6= 0
the chart ai 6= 0. We have
fn ∩ Oi = {((x1 , ..., xn ), (a1 : ... : an )) ∈ Cn × Pn−1 | C
xj = xi aj , j ∈ {1, ..., n}} .
Hence we have an isomorphism: fn ∩ Oi → Cn f:C
((x1 , ..., xn ), (a1 : ... : an )) → (a1 , ..., ai−1 , xi , ai+1 , ..., an )
e on C fn ∩ Oi is an action on Cn given by the diagonal Thus the action of G matrix diag(ξpk1 −ki , ..., ξpki−1 −ki , ξpki , ξpki+1 −ki , ..., ξpkn −ki ). f is smooth if and only if k1 = ... = kn . By assertion 1), M Let X be a complex manifold and G an automorphism group of prime order acting on X. Another idea to get regular split pullbacks of (X, G) is to consider a sequence of blow-ups: MO k
rk
/ ···
r2
πk
XkX
/ M1 O
r1
π1 sk
/ ···
s2
Gk
/ X1 X G1
/M O π
s1
/ X, X G
where each si+1 is the blow-up of Xi in Fix Gi (M = M0 , X = X0 and G = G0 ). We can state the following proposition. Proposition 2.40. There exists k ∈ N such that Mk is smooth if and only if all the fixed points of G have type 0,1 or 2. Moreover in the case where Fix G has points of type 2, M2 is smooth. Proof. By Lemma 1 of [7], we can assume that X = Cn and
G = diag(ξpk1 , ..., ξpkn ) .
1) If 0 is a point of type 0 or 1, by Proposition 2.39, M1 is smooth. If 0D is a point of type 2, we will show that M2 is smooth. Let G = E diag(ξ3k1 , ..., ξ3kn ) acting on Cn . Without loss of generality, we can asfn be the blow-up of Cn in 0. sume that all ki are different from 0. Let C 42
By the proof of Proposition 2.39, on the chart ai 6= 0 the action of G1 is given by the diagonal matrix k
diag(ξ3k1 −ki , ..., ξ3 i−1
−ki
k
, ξ3ki , ξ3 i+1
−ki
, ..., ξ3kn −ki ).
As p = 3, there is a j such that k1 = ... = kj = 1 and kj+1 = ... = kn = 2. So if i ≤ j, by permuting the i-th and the j-th coordinates of the chart Oi , we reduce the action to the form diag(1, ..., 1, ξ3 , ..., ξ3 ). If i > j, by a permutation of coordinates we obtain diag(ξ32 , ..., ξ32 , 1, ..., 1). In both cases, these are points of type 1. Hence all the points of Fix G1 are points of type 0 or 1. Moreover, as there are no fixed points with both eigenvalues ξ3 , ξ32 present in the diagonal matrix of the action, we can conclude that the components of Fix G1 with spectra (1, ..., 1, ξ3 , ..., ξ3 ) and (ξ32 , ..., ξ32 , 1, ..., 1) are disjoint and can be blown up independently. Hence by Proposition 2.39 2), M2 is smooth. 2) Now we will show that in the case of a point of type different from 0, 1 or 2, Mk will never be smooth. We start with
dim X = 2. By Lemma 1 of [7], we can assume that X = C2 and G = diag(ξp , ξpα ) . Since 0 is of type different from 0, 1 or 2, p > 3 and α is not equal to 0 or 1. For x ∈ Fix Gi , we can write:
(Xi , Gi , x) ∼ (C2 , diag(ξp , ξpβ ) , 0),
where ξp is a non-trivial p-th root of the unity. Hence, we can define a sequence as follows:
ui = β ∈ Z /p Z | ∃x ∈ Xi : (Xi , Gi , x) ∼ (C2 , diag(ξp , ξpβ ) , 0) .
o n α 1 ,... Now assume , 1−α , 1−α For instance, u0 = α, α1 , u1 = α − 1, α−1 α that there is i ∈ N such that Mi is smooth. Let i be the smallest integer such that Mi is smooth. Hence by Proposition 2.39 1), ui = {0} and we can write ui−1 = {α1 , ..., αk }. Let x ∈ Fix Gi−1 such that (Xi−1 , Gi−1 , x) ∼
α f2 be the blow-up of (C2 , diag(ξp , ξp j ) , 0), with αj ∈ ui−1 \ {0}. Let C
α j f2 has 2 fixed points a1 and a2 C2 in 0. The action of diag(ξp , ξp ) on C with (see proof of Proposition 2.39):
and
f2 , diag(ξp , ξ αj ) , a′ ) ∼ (C2 , diag(ξp , ξ αj −1 ), 0) (C 1 p p
f2 , diag(ξp , ξ αj ) , a′ ) ∼ (C2 , diag(ξ αj , ξ 1−αj ), 0). (C p p p 2 43
Hence αj − 1 ∈ ui , but ui = {0}. Hence necessarily, αj = 1. Then ui−1 = {1}. We do the same calculation with ui−2 . We can write ui−2 = {α′1 , ..., α′k }. Let x ∈ Fix Gi−2 such that: E D α′ (Xi−2 , Gi−2 , x) ∼ (C2 , diag(ξp , ξp j ) , 0).
with α′j ∈ ui−2 \ {0, 1}. We remark that ui−2 \ {0, 1} is not empty because 2 f2 Mi−1 is not smooth D by definition E of i. Let C be the blow-up of C in αj′ f2 has 2 fixed points a′ and a′ with 0. The action of diag(ξp , ξp ) on C 1
2
(see proof of Proposition 2.39): D ′ ′ E f2 , diag(ξp , ξpαj ) , a1 ) ∼ (C2 , diag(ξp , ξpαj −1 ), 0) (C
and
But
D ′ ′ ′ E f2 , diag(ξp , ξpαj ) , a1 ) ∼ (C2 , diag(ξpαj , ξp1−αj ), 0). (C (C
2
1−α′ α′ , diag(ξp j , ξp j ), 0)
Hence necessarily, done.
α′j
= 2 and
α′j
1−α′j α′ j
2
∼ (C , diag(ξp , ξp
=
1 2
=
p−1 2 .
), 0).
Hence p = 3 and we are
Now, we assume
n > 2. By Lemma 1 of [7], we can assume that X = Cn and G = diag(ξpk1 , ξpk2 , ..., ξpkn ) . Without loss of generality, we can assume that all the ki are different from 0. Since 0 is of type different from 0,1 or 2, p > 3 and the ki are not all equal. Without loss of generality, we can assume that k1 = 1. Since not all the ki are equal, there is j ∈ {1, ..., n} such that kj 6= 1. We also denote α = kj . We denote X ′ = C2 and G′ = diag(ξp , ξpα ) . And we define as before the sequence:
u′i = β ∈ Z /p Z | ∃x ∈ Xi′ : (Xi′ , G′i , x) ∼ (C2 , diag(ξp , ξpα ) , 0) . We define also the following sequence:
Ui = {β ∈ Z /p Z | ∃x ∈ Xi :
(Xi , Gi , x) ∼ (Cn , diag(ξp , ξpβ , ξpt3 , ..., ξptn ) , 0) .
We have to show that Ui 6= {0} for all i ∈ N. But Ui ⊃ u′i . We have seen that u′i 6= {0} for all i ∈ N. The result follows.
Corollary 2.41. Let X be a Kähler manifold and G an automorphism group of prime order acting on X. There exists a regular split pullback of (X, G) obtained as a sequence of blow-ups in connected components of fixed loci if and only if the points of Fix G are of types 0, 1 or 2. 44
2.4.2
By toric blow-ups
Hence, we need another tool to solve all kinds of singularities of the quotient: the toric blowup. Let X be a complex manifold of dimension n and G = hϕi be an automorphism group of prime order p such that Fix G contains only isolated points. We can also solve the singularities of X/G by toric blow-up. As before, at each fixed point of G, we can locally linearize the action of G. Thus, at a fixed point x ∈ X, the action of G on X is locally equivalent to the action of G = hgi on Cn via g = diag(ξpk1 , ..., ξpkn ), where ξp is a p-th root of unity. And we can assume that 1 ≤ k1 ≤ ... ≤ kn ≤ p − 1. We can solve the singularity of Cn /G by toric blow-up. The variety Cn is a toric variety given by the lattice M := Zn and the cone σ = (Q+ )n . And Cn /G is also an affine toric variety given by the lattice M G and the cone σ. The variety Cn /G is also the toric variety associated to M G and the fan Σ in ∨ (M G )∨ and all its faces. To solve the singularity of Q containing the cone σ n C /G, by Theorem 1.30, it is enough to find a new fan Σ′ which is a subdivision of the fan Σ such that all cones of Σ′ will be generated by a subset of a basis of (M G )∨ . Hence, there is a natural map f : XΣ′ → XΣ = Cn /G which is a resolution of Cn /G. Moreover, by Remark 8.3 of [9], the subdivision Σ′ can always be chosen such that f is an isomorphism over Cn /G r {0}. We call this resolution the toric blow-up in 0 of Cn /G. See Section 8 of [9] for more details. Now, we come back to our complex variety X. Since all the fixed points of G are isolated points and by Remark 8.3, we can solve all the singularities of X/G ] → X/G using a toric blow-up on each point. We then get a resolution f : X/G which is an isomorphism over X/G r Sing X/G. We call this resolution the toric blow-up in the singularities of X/G. We will need the following lemma:
Lemma 2.42. Let G = diag(ξpk1 , ..., ξpkn ) be a group acting on Cn . We denote ^ n /G → Cn /G the toric blow-up in 0 of Cn /G. Then, there exists a toric by f : C fn which induce a commutative diagram: variety denoted by C ^ n /G C O
f
/ Cn /G O
(3)
π
π e
fn C Z
s
/ Cn , X G
e G
fn denoted by G e and, with the action of G on Cn extending to an action on C n n ^ f e C /G = C /G. 45
^ n /G is the toric variety associated to Σ′ and M G , where Proof. We know that C ′ G fn be the toric variety associated to the fan Σ and M are defined above. Let C ′ ′ Σ and the lattice M . Since Σ is a subdivision of Σ, there is a natural map fn → Cn . And the action of G on Cn also naturally extends to an action on s:C ^ n /G given on each open toric affine fn . There is also the natural map π fn → C C e:C ′ subset associated to a cone τ ∈ Σ by the inclusion C τ ∨ ∩ M G → C [τ ∨ ∩ M ] which corresponds to the quotient map. Looking at the open toric affine subsets, we see that these maps give a commutative diagram. fn is usually not smooth. However, the singularities Remark: The variety C fn can also be solved by a toric blow-up, finding a subdivision of Σ′ such of C that all cones will be generated by a subset of a basis of M ∨ . Corollary 2.43. Let X be a complex manifold of dimension n and G = hϕi an automorphism group of prime order p such that Fix G contains only isolated ] → X/G be the toric blow-up in the singularities of X/G. points. Let f : X/G e which induces a commutative diagram: Then there exists a variety X ] X/G O
f
/ X/G O π
π e
e X Z
s
/ X, X G
e G
e denoted by G e and, with the action of G on X extending to an action on X ] e e X/G = X/G.
e using the same construct as in Lemma 2.42 on each Proof. We construct X point of Fix G. There is no problem with gluing since all the points of Fix G are fn → Cn of diagram 3 is an isomorphism over isolated points and the map s : C n C r {0}.
e is usually not Remark: We have the same remark as before. The variety X e can also be solved by a toric blow-up. smooth. However, the singularities of X
2.5
The case of fixed points of type 1
During all this section, we will use the following notation. Let X be a compact complex manifold of dimension n and G = hϕi an automorphism group of e → X be the blow-up of X in Fix G. We denote G e prime order p. Let s : X e We can consider the quotient the automorphism group induced by G on X. f := X/ e G. e In this section the fixed points will be of type M := X/G and M 46
f will be always smooth. Hence (X, e G, e r, s) will be a regular split 1, hence M pullback of (X, G), and the following diagram is Cartesian: f M O
r
e X
s
/M O
(4)
π
π e
/ X,
We also denote V = X \ Fix G, U = π(V ), F = s−1 (Fix G), and we use the same symbol F for its image by π e. 2.5.1
The codimension of Fix G
The technique of the proof of the main theorem of this section will be to use f, Z). For Lemma 2.34 3). To do this, we will have to understand K inside H n (M this, we will need the following exact sequence: 0
/ H n (M f, U, Z)
g
/ H n (M f, Z)
/ H n (U, Z)
/ 0.
So we need some conditions on Fix G which will guarantee that this sequence is exact. Definition 2.44. Let X be a compact complex manifold of dimension n and G an automorphism group of prime order p. 1) We will say that Fix G is negligible if the following conditions are verified: • H ∗ (Fix G, Z) is torsion-free. • Codim Fix G ≥
n 2
+ 1.
2) We will say that Fix G is almost negligible if the following conditions are verified: • H ∗ (Fix G, Z) is torsion-free. • n is even and n ≥ 4. • Codim Fix G = n2 , and the purely n2 -dimensional part of Fix G is connected and simply connected. We denote the n2 -dimensional component by Σ. • The cocycle [Σ] associated to Σ is primitive in H n (X, Z). Remark: We might just assume that [Σ] is not divisible by p, but this would imply technical complications.
47
2.5.2
The main theorem
Theorem 2.45. Let G = hϕi be a group of prime order p acting by automorphisms on a Kähler manifold X of dimension n. We assume: i) H ∗ (X, Z) is torsion-free, ii) Fix G is negligible or almost negligible, iii) all the points of Fix G are of type 1. Then: 1) logp (discr π∗ (H n (X, Z))) − h2∗+ǫ (Fix G, Z) is divisible by 2, 2) The following inequalities hold: logp (discr π∗ (H n (X, Z))) + 2 rktor H n (U, Z) f, Z) ≥ h2∗+ǫ (Fix G, Z) + 2 rktor H n (M
≥ 2 rktor H n (U, Z). 3) If, moreover,
logp (discr π∗ (H n (X, Z))) + 2 rktor H n (U, Z) f, Z), = h2∗+ǫ (Fix G, Z) + 2 rktor H n (M
then (X, G) is H n -normal.
Proof. The idea of the proof is to compare K from Lemma 2.34 to its orthogonal f, Z). complement in the unimodular lattice H n (M The proof is a little different if Fix G is negligible or almost negligible. Hence, we give a proof in both cases. The case when Fix G is negligible We consider the following commutative diagram: f, U, Z) H n (NM − 0, Z) = H n (M f/F , NM/F f
g
de π∗
e V, Z) H n (NX/F , NX/F − 0, Z) = H n (X, e e
/ H n (M f, Z)
(5)
π e∗
h
/ H n (X, e Z),
where NX/F − 0 and NM/F − 0 are vector bundles minus the zero section. e f n e We denote T := h(H (X, V, Z)). We will need the following lemmas about properties of T . Lemma 2.46. We have
e Z) = s∗ (H n (X, Z)) ⊕⊥ T. H n (X, 48
Proof. The proof follows from Theorem 7.31 of [36] and its proof (Theorem 1.11). e V, Z) = H n−2 (F, Z), and the map h can be By Thom isomorphism H n (X, e Z), where j is the identified with the morphism j∗ : H n−2 (F, Z) → H n (X, e inclusion in X. As in the proof of Theorem 7.31 of [36], the map e Z) (s∗ , j∗ ) : H n (X, Z) ⊕ H n−2 (F, Z) → H n (X,
is surjective, and its kernel coincides with the image of the map M H n−2rm (Sm , Z) → H n (X, Z) ⊕ H n−2 (F, Z), Sm ⊂Fix G
whereLrm is the codimension of the component Sm of Fix G. But in our case Sm ⊂Fix G H n−2rm (Sm , Z) = 0.
Moreover, the sum is orthogonal with respect to the cup product. Indeed, let x ∈ H n (X, Z) and y ∈ T , then s∗ (s∗ (x)·y) = x·s∗ (y) by the projection formula. Since s∗ (y) = 0, we have s∗ (s∗ (x) · y) = 0, then s∗ (x) · y = 0.
e Z) is unimodular. Lemma 2.47. The sublattice T of H n (X,
Proof. By Lemma 2.46, we have:
e Z) = s∗ (H n (X, Z)) ⊕⊥ T. H n (X,
e Z) and H n (X, Z) endowed with the cup product are unimodSince H n (X, ular, T is also unimodular. By the property of the Thom isomorphism, n de π ∗ (H n (NM , NX/F − 0, Z). e e f/F , NM f/F − 0, Z)) = pH (NX/F
Then by commutativity of the diagram and Proposition 2.12, we have f, U, Z)) = π g(H n (M e∗ (T ).
We deduce the following lemma. Lemma 2.48. 0
1) We have the exact sequence: /π e∗ (T )
/ H n (M f, Z)
/ H n (U, Z)
/ 0.
f, Z) are powers of Fp . 2) The torsion subgroups of H n (U, Z) and H n (M
Proof.
1) We have the following exact sequence:
0
/π e∗ (T )
/ H n (M f, Z) 49
/ H n (U, Z)
/ H n+1 (M f, U, Z).
f, Since H ∗ (Fix G, Z) is torsion-free, by Thom’s isomorphism H n+1 (M U, Z) is torsion-free. Hence it is enough to show: / H n (M f, C)
/π e∗ (T ⊗ C)
0
/ H n (U, C)
/ 0.
f, C) = dim H n (U, C) + Hence, it is enough to show that dim H n (M dim π e∗ (T ⊗ C). By Lemma 2.46 e C) = s∗ (H n (X, C)) ⊕ T ⊗ C. H n (X,
Hence:
e C)G = s∗ (H n (X, C)G ) ⊕ T ⊗ C. H n (X,
Since Codim Fix G ≥
It follows:
n 2
+ 1, H n (V, Z) = H n (X, Z). Hence:
e C)G = s∗ (H n (V, C)G ) ⊕ T ⊗ C. H n (X,
f, C) = dim H n (U, C) ⊕ dim π dim H n (M e∗ (T ⊗ C).
2) Since Codim Fix G ≥ n2 + 1, H n (V, Z) = H n (X, Z). Since H n (X, Z) is torsion-free, H n (V, Z) is torsion-free. Hence by Corollary 2.13, the torsion subgroup of H n (U, Z) is power of Fp . f, Z). Indeed, H n (X, Z) and H ∗ (Fix G, The proof is the same for H n (M Z) are torsion-free. Hence by Theorem 7.31 of [36] (Theorem 1.11), e Z) is torsion-free. Hence the result follows from Corollary H n (X, 2.13. f, Z). We Let Te be the minimal primitive overlattice of π e∗ (T ) in H n (M ⊥ e have T = K by Lemma 2.46 (we recall that K is defined in Lemma 2.34). We will compare the discriminant of K and Te. Then by Proposition 1.3, f, Z). We can state we will be able to know whether K is primitive in H n (M the following result. Lemma 2.49. We have:
n n f 1) Te/T = (Fp )rktor H (U,Z)−rktor H (M ,Z) ,
2∗+ǫ f,Z)−rktor H n (U,Z)) (X)+2(rktor H n (M 2) discr Te = ph .
Proof. By 3) of Lemma 2.18, we have discr π∗ (T ) = prk T . But by Theorem 7.31 of [36], rk T = rk
M
rM m −2
Sm ⊂Fix G k=0
50
H n−2k−2 (Sm , Z).
And since Codim Fix G ≥
n 2
+ 1, we get rk T = h2∗+ǫ (Fix G, Z). So
discr π∗ (T ) = ph
2∗+ǫ
(Fix G,Z)
.
Moreover, by the exact sequence of Lemma 2.48, we have: n n f Te/T = (Fp )rktor H (U,Z)−rktor H (M ,Z) .
Hence, by Proposition 1.1, discr Te =
ph p2(rktor H
2∗+ǫ
(Fix G,Z)
n (U,Z)−rktor H n (M f,Z)
)
.
Conclusion f, Z) will allow us to conclude. Let K be the The unimodularity of H n (M f, Z). We have K ⊥ = Te. Hence by primitive overlattice of K in H n (M Proposition 1.3, 2∗+ǫ f (Fix G,Z)+2(rktor H n (M,Z)−rktor H n (U,Z)) discr K = discr Te = ph .
By Lemma 2.34, we know that discr K = discr π∗ (H n (X, Z)). Then discr K = discr π∗ (H n (X, Z)) ≥ discr K
and
and we get part 2) of the Theorem. By Proposition 1.1, K/K = (Z /p Z)
discr Te ≥ 1
f,Z)−rktor H n (U,Z)) log p (discr π∗ (H n (X,Z)))−h2∗+ǫ (Fix G,Z)−2(rktor H n (M 2
.
We have proved statement 1) of the Theorem. Now if logp (discr π∗ (H n (X, Z))) − h2∗+ǫ (Fix G, Z) f, Z) − rktor H n (U, Z) = 0, − 2 rktor H n (M
f, Z). And we finish the proof by K = K. Hence, K is primitive in H n (M an application of Lemma 2.34 3). The case when Fix G is almost negligible
In this case, n is even, so we can write n = 2m. We consider the following commutative diagram: 2m f (M , U, Z) H 2m (NM f/F − 0, Z) = H f/F , NM
g
de π∗
e V, Z) H 2m (NX/F , NX/F − 0, Z) = H 2m (X, e e 51
/ H 2m (M f, Z) π e∗
h
/ H 2m (X, e Z),
(6)
where NX/F − 0 and NM/F − 0 are vector bundles minus the zero section. e f 2m e We denote R := h(H (X, V, Z)). The following lemma follows Theorem 7.31 of [36] (Theorem 1.11) and its proof. Lemma 2.50. We can write:
with R = T ⊕ Z Σ.
e Z) = s∗ (H 2m (X, Z)) ⊕⊥ T, H 2m (X,
Proof. The proof is very similar to that of Lemma 2.46. e V, Z) = H 2m−2 (F, Z), and the map h can By Thom isomorphism H 2m (X, e Z), where j be identified with the morphism j∗ : H 2m−2 (F, Z) → H 2m (X, e is the inclusion in X. As in the proof of Theorem 7.31 of [36], the map e Z) (s∗ , j∗ ) : H 2m (X, Z) ⊕ H 2m−2 (F, Z) → H 2m (X,
is surjective and its kernel is the image of the map M H 2m−2rk (Sk , Z) → H 2m (X, Z) ⊕ H 2m−2 (F, Z), Sk ⊂Fix G
whereLrk is the codimension of the component Sk of Fix G. But in our case Sk ⊂Fix G H 2m−2rk (Sk , Z) = H 0 (Σ, Z).
The proof of the orthogonality of the sum is the same as in Lemma 2.46. e Z) is unimodular. Lemma 2.51. The sublattice T of H 2m (X,
Proof. The same proof as in Lemma 2.47.
By the property of the Thom isomorphism, de π ∗ (H 2m (NM − f/F , NM/F f 2m 0, Z)) = pH (NX/F , N − 0, Z). Then by the commutativity of the e e X/F f, U, Z)) = π diagram and Proposition 2.12, we have g(H 2m (M e∗ (R).
Lemma 2.52.
1) H 2m−1 (X, Z) = H 2m−1 (V, Z).
2) H 2m (V, Z) = H 2m (X, Z)/ Z Σ.
3) H 2m (V, Z) is torsion-free, and the torsion subgroups of H 2m (U, Z), f, Z) are powers of Fp . H 2m (M 4) We have the exact sequence 0
/π e∗ (R)
/ H 2m (M f, Z)
52
/ H 2m (U, Z)
/ 0.
Proof.
1) We have the following exact sequence:
H 2m−1 (X, V, Z) H
2m
(X, V, Z)
ρ
/ H 2m−1 (X, Z) / H 2m (X, Z)
f
/ H 2m−1 (V, Z) / H 2m (V, Z)
/ /H
2m+1
(X, V, Z).
By Thom’s isomorphism, H 2m−1 (X, V, Z) = 0, H 2m+1 (X, V, Z) = H 1 (Σ, Z) = 0 and H 2m (X, V, Z) = H 0 (Σ, Z). The image of ρ is not trivial in H 2m (X, Z) (see Section 11.1.2 of [36]). Hence the cokernel of f is a torsion group, but H 0 (Σ, Z) is torsion-free. Hence, f is an isomorphism and H 2m−1 (X, Z) = H 2m−1 (V, Z). 2) In view of 1), the exact sequence becomes: / H 0 (Σ, Z)
0
/ H 2m (X, Z)
/ H 2m (V, Z)
/ 0,
which implies the result. 3) The group H 2m (V, Z) is torsion-free, because H 2m (V, Z) = H 2m (X, Z)/ Z Σ and Z Σ is primitive inside H 2m (X, Z). Hence by Corollary 2.13, the torsion subgroup of H 2m (U, Z) is a power of Fp . f, Z). Indeed, H 2m (X, Z) and The proof is the same for H 2m (M ∗ H (Fix G, Z) are torsion-free. Hence by Theorem 7.31 of [36] (Thee Z) is torsion-free. Hence the result follows from orem 1.11), H 2m (X, Corollary 2.13.
4) We have the following exact sequence: /π e∗ (R)
0
/ H 2m (M f, Z)
/ H 2m (U, Z)
/ H 2m+1 (M f, U, Z).
f, Since H ∗ (Fix G, Z) is torsion-free, by Thom’s isomorphism H 2m+1 (M U, Z), is torsion-free. Hence it is enough to show: 0
/π e∗ (R ⊗ C)
/ H 2m (M f, C)
/ H 2m (U, C)
/ 0.
f, C) = dim H n (U, C) + Hence, it is enough to show that dim H n (M dim π e∗ (R ⊗ C). By Lemma 2.47, Hence
e C) = s∗ (H 2m (X, C)) ⊕ T ⊗ C. H 2m (X,
e C)G = s∗ (H 2m (X, C)G ) ⊕ T ⊗ C. H 2m (X, 53
By 2), e C)G = s∗ (H 2m (V, C)G ) ⊕ CΣ ⊕ T ⊗ C. H 2m (X,
Then, by Lemma 2.47,
It follows:
e C)G = s∗ (H 2m (V, C)G ) ⊕ R ⊗ C. H 2m (X,
f, C) = dim H 2m (U, C) + dim π dim H 2m (M e∗ (R ⊗ C). e be the minimal primitive overlattice of π f, Z) and Te Let R e∗ (R) in H 2m (M 2m f the minimal primitive overlattice of π e∗ (T ) in H (M , Z). As before, we need to calculate its discriminant. We start with the following lemma.
Lemma 2.53. There exists x ∈ π e∗ (T ) such that 2m f H (M , Z).
x+(−1)n−1 π e∗ (s∗ (Σ)) p
∈
Proof. Let s1 : Y → X be the blow-up of X in Σ and Σ1 the exceptional e → Y the blow-up in the other components of F such divisor, and s2 : X that s = s2 ◦ s1 . We denote Σ2 = s∗2 (Σ1 ). Consider the following diagram: Σ2
l2
g2
Σ1
e /X
s2
l1
/Y
l0
/ X,
g1
Σ
s1
where l0 , l1 and l2 are the inclusions and gi := si|Σi , i ∈ {1, 2}. P p e , fk − Σ We have π e∗ (OXe ) = OM S f ⊕ L, with L = OM f − Sk ⊂F \Σ
fk is the exceptional divisor associated to the irreducible comwhere each S ponent Sk 6= Σ of F . Thus P f e Sk ⊂F \Σ Sk + Σ f, Z). ∈ H 2 (M p
It follows that
P
Sk ⊂F \Σ
p
fk + Σ e S 54
!m
f, Z). ∈ H 2m (M
By Lemma 2.18 1), we get x+π e∗ (Σm 2 ) f, Z), ∈ H 2m (M p
with x ∈ π e∗ (T ).
(7)
Now, it remains to calculate Σm 2 . By Proposition 6.7 of [11], we have s∗1 l0∗ (Σ) = l1∗ (cm−1 (E)),
where E := g1∗ (NΣ/X )/NΣ1 /Y . Calculating, we find: s∗1 l0∗ (Σ)
m−1 X
m−1−i
(−1)
= l1∗
ci (g1∗ (NΣ/X ))
i=0 m−1 X
m−1−i
· c1 (NΣ1 /Y )
(−1)m−1−i ci (g1∗ (NΣ/X )) · c1 (NΣ1 /Y )m−1−i
= l1∗
i=1
+ (−1)m−1 l1∗ c1 (NΣ1 /Y )m−1 m−1 X
m−1−i
(−1)
= l1∗
i=1 m−1
+ (−1)
ci (g1∗ (NΣ/X ))
m−1−i
· c1 (NΣ1 /Y )
!
! !
Σm 1 .
By applying s∗2 , we get: m−1 Σm (s∗ (Σ) − s∗2 l1∗ (a)) , 2 = (−1)
P m−1−i where a = m−1 ci (g1∗ (NΣ/X )) · c1 (NΣ1 /Y )m−1−i ∈ T . And i=1 (−1) pushing forward via π e∗ , we get: m−1 π e∗ (Σm (e π∗ (s∗ (Σ)) − π e∗ (s∗2 l1∗ (a))) . 2 ) = (−1)
The result follows from (7). Lemma 2.54. We have:
2m 2m f 1) Te/e π∗ (T ) = (Z /p Z)rktor H (U,Z)−rktor H (M,Z)−1 , 2∗+ǫ f ] (Fix G,Z)−2[rktor H 2m (U,Z)−rktor H 2m (M,Z) 2) discr Te = ph .
Proof. By 3) of Lemma 2.18, discr π e∗ (T ) = prk T , By Theorem 7.31 of [36], rk T = rk
M
rM k −2
H 2m−2i−2 (Sk , Z)
Sk ⊂Fix G i=0
=h
2∗+ǫ
(Fix G, Z) − rk H 0 (Σ, Z) − rk H 2m (Σ, Z)
= h2∗+ǫ (Fix G, Z) − 2, 55
where rk is the codimension of the component Sk . So discr π e∗ (T ) = ph
2∗+ǫ
(Fix G,Z)−2
.
Moreover, by the exact sequence of Lemma 2.52, we have: f e π∗ (R) = (Fp )rktor H 2m (U,Z)−rktor H 2m (M,Z) R/e .
But by Lemma 2.53, we already know that there exists x ∈ π e∗ (T ) such x+(−1)m−1 π e∗ (s∗ (Σ)) 2m f that ∈ H (M , Z). We are going to deduce Te/e π∗ (T ). p
f, Z), then If π e∗ (s∗ (Σ)) is divisible by p in H 2m (M (Te/e π∗ (T )), if not both cases
x+(−1)m−1 π e∗ (s∗ (Σ)) p
π e∗ (s∗ (Σ)) p
e π∗ (R)) \ ∈ (R/e
e π∗ (R)) \ (Te/e ∈ (R/e π∗ (T )). Then in
2m 2m f Te/e π∗ (T ) = (Z /p Z)rktor H (U,Z)−rktor H (M,Z)−1 .
Hence by Proposition 1.1,
2∗+ǫ f (Fix G,Z)−2−2[rktor H 2n (U,Z)−rktor H 2n (M,Z)−1 ]. discr Te = ph
Conclusion f, Z). Let K be the primitive overlatWe use the unimodularity of H 2m (M 2m f tice of K in H (M , Z) (We recall that discr K = discr π∗ (H 2m (X, Z)) by Lemma 2.34). We have K ⊥ = Te. Hence by Proposition 1.3, 2∗+ǫ f ] (Fix G,Z)−2[rktor H 2m (U,Z)−rktor H 2n (M,Z) discr K = discr Te = ph .
Since
discr K = discr π∗ (H 2m (X, Z)) ≥ discr K
and
discr Te ≥ 1,
we get the statement 2) of the Theorem. By Proposition 1.1 K/K = (Z /p Z)
[
]
f,Z) logp (discr π∗ (H 2m (X,Z)))−h2∗+ǫ (Fix G,Z)+2 rktor H 2m (U,Z)−rktor H 2m (M 2
Hence, we get the statement 1) of the Theorem. Now if logp (discr π∗ (H 2m (X, Z))) + 2 rktor H 2m (U, Z) f, Z), = h2∗+ǫ (Fix G, Z) + 2 rktor H 2m (M
f, Z). And we finish the proof by K = K. Hence, K is primitive in H 2m (M Lemma 2.34 3).
56
.
2.5.3
f, Z) Calculation of rktor H n (M
In the applications of the above theorem, we will almost never calculate rktor f, Z). We will have: H n (M logp (discr π∗ (H n (X, Z))) + 2 rktor H n (U, Z) = h2∗+ǫ (Fix G, Z).
We give a corollary in the case in which the above equality is satisfied. Corollary 2.55. Let G = hϕi be a group of prime order acting by automorphisms on a Kähler manifold X of dimension n. We assume that: i) H ∗ (X, Z) is torsion-free, ii) Fix G is negligible or almost negligible, iii) all the points of Fix G are of type 1, and iv) logp (discr π∗ (H n (X, Z))) + 2 rktor H n (U, Z) = h2∗+ǫ (Fix G, Z). So: f, Z) is torsion-free, and 1) H n (M
2) (X, G) is H n -normal.
Proof. Statement 1) follows from 2) of Theorem 2.45. So we conclude by 3) of Theorem 2.45. f, Z) is torsion-free is in the following: Another situation where H n (M
Definition 2.56. Let G = hϕi be a group of prime order acting by automorphisms on a complex manifold X of dimension n. Let U := X/G r Sing X/G. We say that (X, G) is U -trivial if H n+1 (U, Z) is torsion-free. Proposition 2.57. Let G = hϕi be a group of prime order acting by automorf be a resolution phisms on a compact complex manifold X of dimension n. Let M of singularities of X/G and U := X/G r Sing X/G. We assume that: i) H ∗ (X, Z) is torsion-free,
f, U, Z) is torsion-free, ii) H n+1 (M
f, U, Z) → H n+1 (M f, Z) is injective, iii) H n+1 (M iv) (X, G) is U -trivial.
f, Z) and H n (M f, Z) are torsion-free. So H n+1 (M
Proof. We have the following exact sequence: 0
/ H n+1 (M f, U, Z)
/ H n+1 (M f, Z)
/ H n+1 (U, Z).
f, U, Z) and H n+1 (U, Z) are torsion-free. Hence H n+1 (M f, Z) is But H n+1 (M f torsion-free. Then by the universal coefficient exact sequence, Hn (M ) is torsionf, Z) is torsion-free. free. Hence, by Poincaré duality, H n (M 57
2.5.4
Calculation of rktor H n (U, Z)
It is possible to calculate rktor H n (U, Z) with the spectral sequence of equivariant cohomology (see Section 2.2.2). Proposition 2.58. Let X be a compact complex manifold of dimension 2m and G an automorphism group of prime order acting on X. Assume that H ∗ (X, Z) is torsion-free, 3 ≤ p ≤ 19 and Fix G is negligible or almost negligible. Assume that (X, G) is E2 -degenerate over Z, or that 2i i) lp−1 (X) = 0 for all 1 ≤ i ≤ m, and
ii) l12i+1 (X) = 0 for all 0 ≤ i ≤ m − 1 when m > 1. Then we have: 1) rktor H 2m (U, Z) = 2) (X, G) is U -trivial.
Pm−1 i=0
2i+1 lp−1 (X) +
Pm−1 i=0
l12i (X),
Proof. 1) We use the equivariant cohomology. When Fix G is negligible, we have H k (V, Z) = H k (X, Z) for all k ≤ 2m. Hence we can exchange V by X in the calculation of H 2m (U, Z), so we get the result by Proposition 2.8 and Proposition 2.10. When Fix G is almost negligible, we have H k (V, Z) = H k (X, Z) for all k ≤ 2m − 2. Moreover, by Lemma 2.52, H 2m−1 (V, Z) = H 2m−1 (X, Z) and H 2m (V, Z) = H 2m (X, Z)/ Z Σ, where Σ is the component of codimension m in Fix G. Since Σ is primitive in H 2m (X, Z), H 2m (V, Z) is torsion-free. Hence, we can replace V by X in the calculation. Then we get the result by Proposition 2.8 and Proposition 2.10. 2) By Proposition 2.10, we have just to check that H 2m+1 (V, Z) is torsionfree. We have the following exact sequence: H 2m+1 (X, V, Z)
/ H 2m+1 (X, Z)
/ H 2m+1 (V, Z)
/ H 2m+2 (X, V, Z).
But by Thom’s isomorphism H 2m+1 (X, V, Z) and H 2m+2 (X, V, Z) are torsion-free. Since H 2m+1 (X, Z) is torsion-free, H 2m+1 (V, Z) cannot have torsion. We have a similar proposition for p = 2. Proposition 2.59. Let X be a compact complex manifold of dimension 2m and G an automorphism group of order 2 acting on X. Assume that H ∗ (X, Z) is torsion-free and Fix G is negligible or almost negligible. Assume that (X, G) is E2 -degenerate over Z, or that 2i i) l1,− (X) = 0 for all 1 ≤ i ≤ m, and
58
2i+1 ii) l1,+ (X) = 0 for all 0 ≤ i ≤ m − 1.
Then we have: 1) rktor H 2m (U, Z) = 2) (X, G) is U -trivial.
Pm−1 i=0
2i+1 l1,− (X) +
Pm−1 i=0
2i l1,+ (X),
We can also give a similar result when n is odd. Proposition 2.60. Let X be a compact complex manifold of dimension 2m + 1 and G an automorphism group of prime order acting on X. Assume that H ∗ (X, Z) is torsion-free, 3 ≤ p ≤ 19 and Fix G is negligible. Assume (X, G) is E2 -degenerate over Z. Then we have: rktor H 2m+1 (U, Z) =
m X
2i lp−1 (X) +
m−1 X
l12i+1 (X).
i=0
i=0
Proof. The same proof using Proposition 2.8. Proposition 2.61. Let X be a compact complex manifold of dimension 2m + 1 and G a group of order 2 acting on X. Assume that H ∗ (X, Z) is torsion-free and Fix G is negligible. Assume (X, G) is E2 -degenerate over Z. Then we have: rktor H 2m+1 (U, Z) =
m X
2i l1,− (X) +
i=0
m−1 X
2i+1 l1,+ (X).
i=0
Remark: Similar results hold over Fp when (X, G) is E2 -degenerate.
2.5.5
Corollaries
The calculation of rktor H n (U, Z) with the spectral sequence of equivariant cohomology and Section 2.3 implies a lot of corollaries. We will just give one example using Lemma 2.22 2). Corollary 2.62. Let G = hϕi be a group of prime order 3 ≤ p ≤ 19 acting by automorphisms on a Kähler manifold X of dimension 2n. We assume: i) H ∗ (X, Z) is torsion-free, ii) Fix G is negligible or almost negligible, iii) all the points of Fix G are of type 1, 2k iv) lp−1 (X) = 0 for all 1 ≤ k ≤ n, and
v) l12k+1 (X) = 0 for all 0 ≤ k ≤ n − 1, when n > 1. Then: 59
1) l12n (X) − h2∗ (Fix G, Z) is divisible by 2, and 2) we have: l12n (X)
+2
"n−1 X
2i+1 lp−1 (X)
+
#
n−1 X
l12i (X)
i=0
i=0
≥ h2∗ (Fix G, Z) # "n−1 n−1 X X 2i+1 2i l1 (X) . lp−1 (X) + ≥2 i=0
i=0
3) If moreover l12n (X)
+2
"n−1 X
2i+1 lp−1 (X)
+
= h (Fix G, Z), then (X, G) is H
2n
l12i (X)
i=0
i=0
2∗
#
n−1 X
-normal.
Pm−1 2i+1 Pm−1 Proof. In Theorem 2.45, we replace rktor H 2m (U, Z) by i=0 lp−1 (X)+ i=0 l12i (X) with Proposition 2.58 and logp (discr π∗ (H n (X, Z))) by l12n with Propof, Z) = 0 by Proposition 2.58 2), sition 2.22 2). Moreover, we have rktor H 2n (M Proposition 2.57, Lemma 2.48 1) and Lemma 2.52 4). There is the same corollary when p = 2. Moreover, when p = 2 all the fixed points are of type 1.
Corollary 2.63. Let G = hϕi be a group of prime order p = 2 acting by automorphisms on a Kähler manifold X of dimension 2n. We assume: i) H ∗ (X, Z) is torsion-free, ii) Fix G is negligible or almost negligible, 2k iii) l1,− (X) = 0 for all 1 ≤ k ≤ n, and 2k+1 iv) l1,+ (X) = 0 for all 0 ≤ k ≤ n − 1, when n > 1.
Then: 2n 1) l1,+ (X) − h2∗ (Fix G, Z) is divisible by 2.
2) We have: 2n (X) l1,+
+2
"n−1 X
2i+1 l1,− (X)
+
n−1 X i=0
i=0
≥ h2∗ (Fix G, Z) # "n−1 n−1 X X 2i+1 2i l1,+ (X) . l1,− (X) + ≥2 i=0
i=0
60
#
2i l1,+ (X)
3) If moreover 2n l1,+ (X)
+2
"n−1 X
2i+1 l1,− (X)
+
n−1 X
#
2i l1,+ (X)
i=0
i=0
= h2∗ (Fix G, Z), then (X, G) is H 2n -normal.
2.6
The case of fixed points of type 2
Let X be a Kähler manifold of dimension 2n and G an automorphism group of order 3. We assume that Codim Fix G ≥ 2 (so that we have no points of type 0 in Fix G). We consider the diagram: MO 2
r2
π2
X2X G2
r1
/ M1 O π1
s2
/M O
(8)
π s1
/ X1 X
/X Z
,
G
G1
where s1 is the blow-up of X in the fixed points of type 2 and s2 is the blow-up of X1 in Fix G1 . By Proposition 2.40 and its proof, Fix G1 only has points of type 1 and M2 is smooth. Definition 2.64. Let X be a Kähler manifold of dimension 2n and G an automorphism group of order 3. • We will say that Fix G is stable if Fix G1 is almost negligible. • We say that a point of Fix G is a stable isolated fixed point of type 2 if it is a point of type 1 (1, ..., 1, 2, ..., 2). p | {z } | {z } n
n
• If n ≥ 4, we say that a point of Fix G is an almost stable isolated fixed point of type 2 if it is a point of type 1 (1, ..., 1, 2, ..., 2) p | {z } | {z } n+1
or
n−1
1 (1, ..., 1, 2, ..., 2) p | {z } | {z } n−1
61
n+1
• if n ≥ 4, we say that a curve in Fix G is an almost stable fixed curve of type 2 if it is a rational curve of type 1 (0, 1, ..., 1, 2, ..., 2) p | {z } | {z } n
n−1
or
1 (0, 1, ..., 1, 2, ..., 2). p | {z } | {z } n
n−1
Proposition 2.65. Let X be a Kähler manifold of dimension 2n and let G be an automorphism group of order 3. Let denote by F1 := {x ∈ Fix G| o(x) = 1} the set of point of type 1. 1) The fixed locus Fix G is stable if and only if F1 is almost negligible and Fix G contains only stable isolated points of type 2; or if F1 is negligible, Fix G contains stable isolated points of type 2 and, at most, one almost stable isolated point of type 2 or one almost stable curve of type 2. 2) We denote by n2 (G) the number of stable points of type 2. o) If Fix G contains only points of type 1 and stable isolated points of type 2, then 2n2 (G) G Hi ), Fix G1 = F1 ⊔ ( i=1
with Hi ≃ Pn−1 .
i) If Fix G contains points of type 1, stable isolated points of type 2, and one almost stable isolated point of type 2, then 2n2 (G)
Fix G1 = F1 ⊔ (
G
Hi ) ⊔ A ⊔ B,
i=1
with Hi ≃ Pn−1 , A ≃ Pn−2 and B ≃ Pn . ii) If Fix G contains points of type 1, stable isolated points of type 2 and one almost stable curve of type 2, C, then 2n2 (G)
Fix G1 = F1 ⊔ (
G
Hi ) ⊔ A ⊔ B,
i=1
with Hi ≃ Pn−1 , A → C fibration of C with fiber Pn−2 and B → C fibration of C with fiber Pn−1 .
62
Proof. Let x ∈ Fix G be a point of type 2. We have: * + (X, G, x) ∼ C2n , diag(1, ..., 1, ξ3 , ..., ξ3 , ξ32 , ..., ξ32 ) , 0 . | {z } | {z } | {z } α
β
2n−α−β
g 2n be the blow-up of C2n in Fix G, Let C
2n g 2n = {((x , ..., x ), (a C × P2n−α−1 | 1 2n α+1 : ... : a2n )) ∈ C
rk ((xα+1 , ..., x2n ), (aα+1 , ..., a2n )) = 1} .
g 2n , On C
*
+
diag(1, ..., 1, ξ3 , ..., ξ3 , ξ32 , ..., ξ32 ) | {z } | {z } | {z } α
acts as follows:
β
2n−α−β
diag(1, ..., 1, ξ3 , ..., ξ3 , ξ32 , ..., ξ32 ) · ((x1 , ..., x2n ), (aα+1 : ... : a2n )) | {z } | {z } | {z } α
β
2n−α−β
= (x1 , ..., xα , ξ3 xα+1 , ..., ξ3 xα+β , ξ32 xα+β+1 , ..., ξ32 x2n ), (aα+1 : ... : aα+β : ξ3 aα+β+1 : ... : ξ3 a2n )) .
Hence, the fixed locus is given by Cα × Pβ−1 and Cα × P2n−α−β−1 .
(9)
Assuming that Fix G1 is almost negligible, then: α + β − 1 ≤ n and 2n − β − 1 ≤ n. That is n − 1 ≤ β and α ≤ 2. If α = 2 then β has to be equal to n − 1. But in this case, we have two fixed components of codimension n which is not allowed. Hence α ≤ 1. If α = 1, then β ≤ n. Hence, there are two cases, β = n or β = n − 1, which correspond to the case of the almost stable fixed curves of type 2. In this case, we will have one component of Fix G1 of codimension n: A. Hence, just one almost stable fixed curve of type 2 is allowed, and then F1 must be negligible. Moreover, A is always primitive as a cocycle by Theorem 7.31 of [36] and A is simply connected if and only if C is simply connected by Lemma 7.32 of [36]. If α = 0 then β ≤ n+1. Therefore, there are three cases: β = n, which is the case of a stable fixed point of type 2, and the cases β = n − 1, β = n + 1, which are the cases of almost stable fixed points of type 2. In the case of almost stable fixed points of type 2, we will have one component of Fix G1 of codimension n. Hence, just one almost stable fixed point of type 2 is allowed, and then F1 must be negligible. The sufficient condition follows from (9). Moreover, we get 2). Theorem 2.66. Let X be a Kähler manifold of dimension 2n and G an automorphism group of order 3. We assume that: 63
i) H ∗ (X, Z) is torsion-free, ii) Fix G is stable. 2k iii) lp−1 (X) = 0 for all 1 ≤ k ≤ n, and
iv) l12k+1 (X) = 0 for all 0 ≤ k ≤ n − 1, when n > 1. We denote by n2 (G) the number of stable isolated points of type 2, by ǫ, the number of almost stable isolated points of type 2 and by η, the number of almost stable curve of type 2 (ǫ, η ∈ {0, 1}). So: 1) l12n (X) − h2∗ (Fix G, Z) is divisible by 2, and 2) we have: l12n (X)
+2
"n−1 X
2i+1 lp−1 (X)
#
l12i (X)
i=0
i=0
2∗
+
n−1 X
≥ h (Fix G, Z) # "n−1 n−1 X X 2i+1 2i l1 (X) − (n2 (G) + ǫ + 2η). lp−1 (X) + ≥2 i=0
i=0
3) If, moreover, l12n (X) + 2
"n−1 X
2i+1 lp−1 (X) +
i=0
= h2∗ (Fix G, Z),
n−1 X
#
l12i (X)
i=0
then (X, G) is H 2n -normal. Proof. We will apply Corollary 2.62 to (X1 , G1 ). The group H ∗ (X1 , Z) is torsion-free because H ∗ (X, Z) is; and by Proposition 2.65 and Theorem 7.31 of [36], the fixed point of type 2 cannot induce torsion on H ∗ (X1 , Z). Hence, we can provide the following lemma: Lemma 2.67. We have: 1) lij (X1 ) = lij (X) for all i 6= 1 and for all 0 ≤ j ≤ 2n, 2) l12j (X1 ) = l12j (X) + n2 (G) + ǫ + 2η for all 2 ≤ j ≤ 2n − 2, 3) l12 (X1 ) = l12 (X) + n2 (G) + ǫ + η and l14n−2 (X1 ) = l14n−2 (X) + n2 (G) + ǫ + η, 4) l12j+1 (X1 ) = l12j+1 (X) for all 0 ≤ j ≤ 2n − 1, 5) h2∗ (Fix G1 , Z) = h2∗ (Fix G, Z) + (2n − 1)n2 (G) + (2n − 1)ǫ + 4(n − 1)η.
64
Proof. Since the exceptional divisors are invariant by the action of G, 1), 2), 3) and 4) follow from Theorem 7.31 of [36] and Definition-Proposition 1.5. It remains to prove 5). By Proposition 2.65 and Lemma 7.32 of [36], we have: h2∗ (Fix G1 , Z) = h2∗ (Fix G, Z) − n2 (G) + 2n2 (G) · n − ǫ + (n − 1)ǫ + (n + 1)ǫ − 2η + 2(n − 1)η + 2nη = h2∗ (Fix G, Z) + (2n − 1)n2 (G) + (2n − 1)ǫ + 4(n − 1)η.
By Lemma 2.67, conditions iv) and v) of Corollary 2.62 are verified for (X1 , G1 ). And since Fix G1 is almost negligible, we can apply Corollary 2.62 to (X1 , G1 ). Hence, we have: 1) l12n (X1 ) − h2∗ (Fix G1 , Z) is divisible by 2. But l12n (X1 ) − h2∗ (Fix G1 , Z) = l12n (X) + n2 (G) + ǫ + 2η + h2∗ (Fix G, Z) + (2n − 1)n2 (G) + (2n − 1)ǫ + 4(n − 1)η Therefore, h2∗ (Fix G, Z) − l12 (X) is divisible by 2. 2) We have: l12n (X1 )
+2
"n−1 X
2i+1 lp−1 (X1 )
+
n−1 X
#
l12i (X1 )
i=0
i=0
≥ h2∗ (Fix G, Z) # "n−1 n−1 X X 2i+1 l12i (X1 ) . lp−1 (X1 ) + ≥2 i=0
i=0
It follows that l12n (X)
+ n2 (G) + ǫ + 2η + 2
"n−1 X
2i+1 lp−1 (X)
i=0
+
n−1 X
(l12i (X)
#
+ n2 (G) + ǫ + 2η) − 2η
i=0
≥ h2∗ (Fix G, Z) + (2n − 1)n2 (G) + (2n − 1)ǫ + 4(n − 1)η # "n−1 n−1 X X 2i+1 2i (l1 (X) + n2 (G) + ǫ + 2η) − 2η. lp−1 (X) + ≥2 i=0
i=0
Therefore, l12n (X)
+2
"n−1 X
2i+1 lp−1 (X)
+
n−1 X
#
l12i (X)
i=0
i=0
+ (2n − 1)(n2 (G) + ǫ + 2η) − 2η
≥ h2∗ (Fix G, Z) + (2n − 1)n2 (G) + (2n − 1)ǫ + 4(n − 1)η # "n−1 n−1 X X 2i+1 2i l1 (X) + (2n − 2)(n2 (G) + ǫ + 2η) − 2η. lp−1 (X) + ≥2 i=0
i=0
65
We subtract (2n − 1)(n2 (G) + ǫ + 2η) − 2η to get # "n−1 n−1 X X 2i+1 2i 2n l1 (X) lp−1 (X) + l1 (X) + 2 i=0
i=0
≥ h2∗ (Fix G, Z) # "n−1 n−1 X X 2i+1 2i l1 (X) − (n2 (G) + ǫ + 2η). lp−1 (X) + ≥2 i=0
i=0
3) If, moreover,
l12n (X)
+2
"n−1 X i=0
2i+1 lp−1 (X)
+
n−1 X
#
l12i (X)
i=0
= h2∗ (Fix G, Z),
by adding (2n − 1)(n2 (G) + ǫ + 2η) − 2η to both sides of the equality, it implies # "n−1 n−1 X X 2i+1 2i 2n l1 (X1 ) = h2∗ (Fix G1 , Z). lp−1 (X1 ) + l1 (X1 ) + 2 i=0
Hence, (X1 , G1 ) is H H 2n -normal.
2.7 2.7.1
2n
i=0
-normal. So by Propositions 2.35 and 2.37, (X, G) is
The case of fixed points of other types The weight of a fixed point
Let X be a compact complex manifold of even dimension n and G = hϕi, an automorphism group of prime order p such that Fix G contain only isolated points. The goal of this section will be to define an object which will take the same place as h2∗+ǫ (Fix G, Z) in Theorem 2.45. Here, the situation is a bit more complicated. It will not be possible to use h2∗+ǫ (Fix G, Z) as in Theorem 2.45. Instead of h2∗+ǫ (Fix G, Z), we will define the weight of an isolated fixed point. To have a good definition of a weight of an isolated fixed point, we need the weight to depend only on the local action of G around the isolated point. The two mains steps of this section are given by Lemma 2.68 and Lemma 2.72. Lemma 2.68 will prove that the future weight of an isolated point depends only on the local action of G on this point, but could also depend on the choice of the toric blow-up (which is not unique). With Lemma 2.72, we finish the section by avoiding the dependence on the choice of the toric blow-up. f → M the toric blow-up in the As usual, we denote M := X/G, r1 : M singularities of M , V := X r Fix G and U := V /G = M r Sing M . And, we have the following diagram: f r1 / M M O π
X, 66
and also the following exact sequence: H n−1 (U, Z)
We can write:
f, U, Z) / H n (M
g
f, U, Z) / H n+1 (M f, U, Z) = H k (M
f, Z) / H n (M
i∗
/ H n (U, Z)
(10)
f, Z) / H n+1 (U, Z). / H n+1 (M
⊥ M
x∈Sing M
f, Ux , Z), H k (M
for all 0 ≤ k ≤ 2n. Let x ∈ Fix G, working locally around x. Assume that the action of G on X at x is locally equivalent to the action of n /G → Cn /G be the toric blow-up ^ G = hgi on Cn . As in Lemma 2.42, let f : C n of C /G in 0. We denote by he1 i , ..., hen i the cones of dimension 1 in the fan ^ n /G Σ of Cn /G. Let hf1 i , ..., hfk i be the cone of dimension 1 of the fan Σ′ of C added to Σ. The fi are chosen inside the cone he1 , ..., en i. And let hu1 i , ..., hum i be the cone of dimension 1 added to Σ′ to get a smooth toric compactification n /G of C n /G. We can chose the u out of the cone he , ..., e i. And we denote ^ ^ C i 1 n ′ ^ n /G. We denote the inclusion by j : C ^ n /G ֒→ C ^ n /G. And we Σ the fan of C ^ n /G r j(f −1 (0)). ^ n /G r f −1 (0) = Cn /G r {0} and U ′′ := C denote U ′ := C
f, Z) generated Lemma 2.68. Let x ∈ Sing M . Let Γx be the subgroup of H n (M −1 −1 f \ r (x) and by the cocycles supported on r1 (x). Let Ux := M 1 f, Ux , Z) → H n (M f, Z). gx : H n (M
So:
1) Γx ⊂ Im gx and Im gx /Γx is a torsion group; 2) Im gx /Γx depends only on the local action of G on x and of the choice on the toric blow-up; 3) gx is injective. Proof. Let x ∈ Fix G. Assume that the action of G on X at x is locally ^ n /G equivalent to the action of G = hgi on Cn . We start by working on C ^ n /G. Let Γ′ be the subgroup of H n (C ^ n /G, Z) generated by the cocycles and C −1 supported on j(f (0)). We have the following exact sequence: g ^ n /G, U ′′ , Z) H n (C
′
^ n /G, Z) / H n (C
i∗
/ H n (U ′′ , Z)
We have Γ′ ⊂ Im g ′ ; indeed, a cocycle supported on j(f −1 (0)) will be sent to 0 ^ n /G is the inclusion. by the i∗ where i : U ′′ ֒→ C 67
^ n /G, Z) generated by the cocycles supAlso let Γ be the subgroup of H n (C n /G r j(C n /G). The groups Γ′ and Γ are orthogonal. A cocycle in ^ ^ ported on C ^ n /G r j(C ^ n /G) does not meet a cocycle in j(f −1 (0)). The proof of the lemma C will be in 6 steps. First, we prove that ^ n /G, Z) H n (C Γ′ ⊕ Γ is a torsion group. Then we will prove that Im g ′ /Γ′ is a torsion group. In the third step, we give an expression of Im g ′ /Γ′ . In Step 4, we deduce that Im gx /Γx is a torsion group. In step 5, we prove that Γx ≃ Γ′ and finally, we finish the proof in step 6. • Step 1:
^ n /G,Z) H n (C Γ′ ⊕Γ
is a torsion group
By Section 5.7, Section 10.7, and Theorem 10.8 of [9], the cocycles sup′ ported on j(f −1 (0)) correspond to the cones in Σ containing at least one of ^ n /Gr j(C ^ n /G) also correspond to the the fi . The cocycles supported on C ′
cones in Σ containing at least one of the ui . By Theorem 10.8 of [9], the ^ n /G, Z) are generated by all the cocycles corresponding elements in H n (C n ^ n of H to the cones in Σ of dimension n2 . Let Γ0 be the subgroup E (C /G, Z) D ′
generated by cocycles corresponding to the cones ei1 , ..., ei n . Hence, we 2 have: ^ n /G, Z) H n (C Γ0 = . Γ′ ⊕ Γ Γ0 ∩ (Γ′ ⊕ Γ) We prove that
Γ0 Γ0 ∩(Γ′ ⊕Γ) G ∨
is a torsion group. Let (x1 , ..., xn ) be a basis of
∨ the lattice (M ) . And let x∨ 1 , ..., xn be the dual basis. By Section 10.7 of [9], we have: n X j=1
x∨ i (ej ) [hej i] +
k X
x∨ i (fj ) [hfj i] +
m X
x∨ i (uj ) [huj i] = 0,
j=1
j=1
for all i ∈ {1, ..., n}, and with [C] as the cocycle corresponding to the cone C. Let E be the matrix of (e1 , ..., en ) in the basis (x1 , ..., xn ). Then we have the following matricial equation: t
E · ([he1 i] , ..., [hen i]) = B, Pm ∨ Pk with Bi = − j=1 x∨ i (fj ) [hfj i] − j=1 xi (uj ) [huj i].
But (e1 , ..., en ) is a basis of (M G )∨ Q . Therefore, E ∈ GLn (Q). Hence, all the [hei i] can be written as linear combinations of the [hfj i] and of the 68
[huj i] with rational coefficients. So by Section 10.7 of [9], all the elements of Γ0 can be written as a linear combination of elements of Γ′ and of Γ with rational coefficients. This proves that ^ n /G, Z) H n (C Γ0 = ′ Γ ⊕Γ Γ0 ∩ (Γ′ ⊕ Γ) is a torsion group. • Step 2: Im g ′ /Γ′ is a torsion group Now, we look at the following diagram: n /G, C n /G, Q) ^ ^ H n (C g⊗Q g ⊗Q n /G, U ′′ , Q) ^ / H n (C
n n /G, Q) ^ / H (C
H n−1 (U ′ , Q)
^ n /G, U ′ , Q) / H n (C O
^ n /G, Q) / H n (C O
/ H n (U ′ , Q)
H n−1 (U ′ , Q)
/ H n (Cn /G, U ′ , Q)
/ H n (Cn /G, Q)
/ H n (U ′ , Q), (11)
H n−1 (U ′′ , Q)
′
/ H n (U ′′ , Q)
ϕ
n /G, C n /G, Z) → H n (C n /G, Z). We know that H n (U ′ , Q) = ^ ^ ^ with g : H n (C n−1 ′ H (U , Q) = 0, and so by commutativity of the diagram:
^ n /G, Q) ϕ : Im g ′ ⊗ Q → H n (C is an isomorphism. Moreover, ^ n /G, C ^ n /G, Q) H n (C
g⊗Q
^ n /G, Q) / H n (C
^ n /G, Q) / H n (C
is an exact sequence, therefore, ^ n /G, Q) ≥ dim H n (C ^ n /G, Q). dim Im g ⊗ Q + dim H n (C Hence ^ n /G, Q). dim Im g ⊗ Q + dim Im g ′ ⊗ Q ≥ dim H n (C
(12)
But Im g ⊗ Q ∩ Im g ⊗ Q = 0. Indeed, let y ∈ Im g ⊗ Q ∩ Im g ⊗ Q. Since y ∈ Im g ⊗ Q, we have ϕ(y) = 0. But ϕ gives an isomorphism between ^ n /G, Q). Since y ∈ Im g ′ ⊗ Q, we have y = 0. Since Im g ′ ⊗ Q and H n (C ′
′
^ n /G, Q), Im g ⊗ Q ⊕ Im g ′ ⊗ Q ⊂ H n (C 69
we have, by (12), ^ n /G, Q). Im g ⊗ Q ⊕ Im g ′ ⊗ Q = H n (C n
(13)
^ n
/G,Z) We also have Γ ⊂ Im g. But we have H (ΓC′ ⊕Γ which is a torsion group. Hence n /G, Q) = Γ′ ⊗ Q ⊕Γ ⊗ Q . ^ (14) H n (C
Then, necessarily, Im g ′ ⊗ Q = Γ′ ⊗ Q . So
Im g′ Γ′
• Step 3:
is a torsion group. Im g′ Γ′
= coker ψ ′
^ n /G associated to the We define ψ ′ . Let Fi be the subtoric variety of C Tk ′′ n /G r F . We have U ′′ = ^ cone hfi i. We denote Ui′′ := C i i=1 Ui . We look at the following diagram: Lk
i=1
^ n /G, U ′′ , Z) H n (C ❘❘❘i ❘❘❘ ρ ❘❘❘ ❘❘❘ ψ′ ❘(
^ n /G, U ′′ , Z) H n (C
g′
^ n /G, Z). / H n (C
coker ψ ′ The map g ′ is injective. Indeed, if we consider the diagram (11) with integral coefficients, we see that the kernel of g ′ is given by H n−1 (U ′ , Z). But by equivariant cohomology, we can show that H n−1 (U ′ , Z) = 0 (it is the same technique as used in Section 2.5.4). ^ n /G, U ′′ , Z) = H n−2 (F , Moreover, by Thom’s isomorphism, we have H n (C i i ^ n Z). We denote the inclusion by ji : Fi ֒→ C /G. Then ρ is given by the sum j1∗ + ... + jk∗ . Hence, Im ρ = Γ′ . Now we prove that
Im g ′ . Γ′ We have by commutativity of the diagram: coker ψ ′ =
Im g ′ Im g ′ Im g ′ = . = Γ′ Im ρ Im g ′ ◦ ψ ′ 70
And since g ′ is injective, we have: ^ n /G, U ′′ , Z) Im g ′ H n (C = = coker ψ ′ . Im g ′ ◦ ψ ′ Im ψ ′ • Step 4:
Im gx Γx
is a torsion group
f and we consider the following diaNow, we come back to the variety M gram: Lk
i=1
Lk
^ n /G, U ′′ , Z) H n (C i
f, Ux,i , Z) H n (M PPP PPP ρ PPPx ψx PPP PP( gx n f / H n (M f, Z), H (M , Ux , Z)
i=1
ψ′
n ^ n H (C /G, U ′′ , Z) coker ψ ′
coker ψx (15)
f \ Fi . We have the following exact sequence: where Ux,i := M H n (M, Ux , Z)
gx
/ H n (M f, Z)
∗ jx
/ H n (Ux , Z),
f the inclusion. Since Γx is sent on 0 by jx∗ , we have with jx : Ux ֒→ M Γx ⊂ Im gx . The group Im gx Γx is a torsion group. It is due to the fact that coker ψ ′ = coker ψx is a torsion f, Ux , Z) group. Indeed, let z ∈ ImΓxgx , with z ∈ Im gx . There is y ∈ H n (M Lk f, Ux,i , Z) and t ∈ N such that z = gx (y). Hence, there is u ∈ i=1 H n (M such that tψx (u) = y. And so tz ∈ Γx , so tz = 0.
• Step 5: Γ′ ≃ Γx We denote by VΓ′ the group generated by the ^ n /G supported on j(f eties of C
−1
n 2
-dimensional subvari-
(0)) and by VΓx the group generated ^ n /G supported on r−1 (x). Since by the n2 -dimensional subvarieties of C 1 −1 −1 r1 (x) = j(f (0)), there is an isomorphism f : VΓ′ → VΓx . Moreover, f is an isometry for the intersection pairing because there is an isomorf containing r−1 (x) and an open subset phism between an open subset of M 1 ^ n /G containing j(f −1 (0)). We will prove that, in fact, f induces an of C isometry between Γ′ and Γx . Letting F ′ ∈ VΓ′ and Fx ∈ VΓx , we denote 71
by [F ′ ] the cocycle in Γ′ associated to F ′ and by [Fx ] the cocycle in Γx associated to Fx . We must prove that [f(F ′ )] = 0 if and only if [F ′ ] = 0. First, assume that [f(F ′ )] = 0. Hence, the intersections of f(F ′ ) with all the elements in VΓx are equal to 0. Since f is an isometry, the intersection of F ′ with all the elements in VΓ′ are equal to 0. But since [F ′ ] ∈ Γ′ the intersections of [F ′ ] with all the elements in Γ are also equal to 0. Hence ^ n /G, Z). Since the cup product by step 1, [F ′ ] is orthogonal to all H n (C is a non-degenerate form, we have [F ′ ] = 0. To prove the reciprocal we need the two following lemmas: f, U, Z)) and r1∗ (π∗ ( Lemma 2.69. When n is even, the lattices g(H n (M n H (X, Z))) are orthogonal for the cup product. Proof. Here we use Corollary 2.43. We consider the following commutative diagram given by this corollary: MO ′
r2
π′
/M f O
r1
s2
/X e
s1
G′
(16)
π
π e
X ′Z
/M O
Z
/ X, X G
e G
e and M ′ its quotient where X ′ is the toric blow-up in the singularities of X, ′ by the induced action by G on X . It induces the following commutative diagram on cohomology groups: H n (U, Z) O
H n (U, Z) O
H n (U, Z) O
(17)
i∗
H n (M ′ , Z) o
r2∗
(π ′ )∗
H n (X ′ , Z) o j ′∗
H n (V, Z)
f, Z) o H n (M
r1∗
π e∗
s∗ 2
e Z) o H n (X, e j∗
H n (V, Z)
H n (M, Z) π∗
s∗ 1
π0∗
H n (X, Z) j∗
{ H n (V, Z).
The proof will be a bit technical because there are many cohomology groups involved. But in fact, the proof is just based on the projection formula and get around the cohomology groups on singular varieties. f, U, Z)) and x ∈ r1∗ (π∗ (H n (X, Z)), we show that Letting y ∈ g(H n (M x · y = 0. We consider x0 ∈ H n (X, Z) such that x = r1∗ (π∗ (x0 ). And we 72
consider π e∗ (x · y) = π e∗ (x) · π e∗ (y). By commutativity of our diagram, we have: π e∗ (x) = s∗1 (π ∗ (π∗ (x0 ))) = s∗1 (px0 ).
Then,
s∗2 (e π ∗ (x · y)) = s∗ (px0 ) · s∗2 (e π ∗ (y)). Hence, by projection formula, π ∗ (y))). π ∗ (x · y))) = px0 · s∗ (s∗2 (e s∗ (s∗2 (e But y is a linear combination of cocycles supported on r1−1 (x), and so s∗ (s∗2 (e π ∗ (y))) will be points. Therefore, s∗ (s∗2 (e π ∗ (x · y))) = 0. Hence, s∗2 (e π ∗ (x · y)) = 0.
(18)
f, Z) → H 2n (X ′ , Z) is injective. So, we It is not clear that s∗2 ◦ π e∗ : H 2n (M have to prove that (18) implies that x · y = 0. The map r1∗ : H 2n (M, Z) → f, Z) is surjective. So we can write x·y = r∗ (z) with z ∈ H 2n (M, Z). H 2n (M 1 Then, by commutativity of the diagram, π e∗ (x · y) = s∗1 (π ∗ (z)). It follows by (18) that s∗ ((π ∗ (z)) = 0. Then π ∗ (z) = 0, and z = 0, so x · y = 0.
f, Z) is injective Lemma 2.70. 1) The map r1∗ : H n (M, Z) → H n (M modulo torsion. 2) The group
f,Z) H n (M
∗ n f g(H n (M,U,Z))⊕r 1 (π∗ (H (X,Z))
is a torsion group.
f, Z) is injective modulo torProof. 1) Since r1∗ : H 2n (M, Z) → H 2n (M sion and the cup-product is a non-degenerate form on H n (M, Z), the f, Z) is injective modulo torsion. map r1∗ : H n (M, Z) → H n (M 2) We just have to prove that
We have:
f, Q). f, U, Q)) ⊕ r1∗ (π∗ (H n (X, Q)) = H n (M g(H n (M f, U, Q)) ⊕ r1∗ (π∗ (H n (X, Q)) ⊂ H n (M f, Q). g(H n (M
Hence, it is enough to prove that
f, U, Q)) + dim r∗ (π∗ (H n (X, Q)) ≥ dim H n (M f, Q). dim g(H n (M 1
By Corollary 2.13, we have:
dim H n (V, Q)G = dim H n (U, Q).
73
Since Fix G is just isolated points, we have: dim H n (V, Q)G = dim H n (X, Q)G . And one more time by Corollary 2.13: dim π∗ (H n (X, Q) = dim H n (X, Q)G . Hence, dim π∗ (H n (X, Q) = dim H n (U, Q). So it follows by 1) that dim H n (U, Q) = dim r1∗ (π∗ (H n (X, Q))). Hence, the result follows from exact sequence (10).
Now, assume that [F ′ ] = 0. Hence, the intersections of F ′ with all the elements in VΓ′ are equal to 0. Since f is an isometry, the intersection of f(F ′ ) with all the elements in VΓx is equal to 0. But since [f(F ′ )] ∈ Γx , by Lemma 2.69, the intersections of [f(F ′ )] with all the elements in r1∗ (π∗ ( H n (X, Z))) are also equal to 0. Hence by Lemma 2.70 2), [f(F ′ )] f, Z). Since the cup product is a non-degenerate is orthogonal to all H n (M ′ form, we have [f(F )] = 0.
• Step 6: End of the proof: Hence by Step 4:
dim Im gx ⊗ Q = dim Γx ⊗ Q .
(19)
dim Γx ⊗ Q = dim Γ′ ⊗ Q .
(20)
dim Γ′ ⊗ Q = dim g ′ ⊗ Q .
(21)
By Step 5: By Step 2: ′
In Step 3, we see that g is injective. Hence, ^ n /G, U ′′ , Q). Im g ′ ⊗ Q = H n (C
(22)
It follows by (19), (20), (21) and (22) that ^ n /G, U ′′ , Q) = dim Im g ⊗ Q. dim H n (C x ^ n /G, U ′′ , Q) = H n (M f, Ux , Q), we have: dim Im gx ⊗ And since H n (C n f Q = dim H (M , Ux , Q). Therefore, gx is injective modulo torsion. But 74
^ n /G, U ′′ , Z) is without torsion. Indeed, g ′ is injecf, Ux , Z) = H n (C H n (M n /G, Z) is without torsion by Theorem 10.8 of [9]. Hence ^ tive and H n (C gx is injective. Now, by commutativity of (15), we have: Im gx Im gx Im gx = = . Γx Im ρx Im gx ◦ ψx And since gx is injective, we have: f, Ux , Z) H n (M Im gx = = coker ψx . Im gx ◦ ψx Im ψx
And since coker ψx = coker ψ ′ , ImΓxgx depend only on the local action of G in x and of the choice of the toric blow-up.
f, Lemma 2.71. Assume that H n (X, Z) is torsion-free. We have: discr g(H n (M l U, Z)) = p with l ∈ N.
f, U, Z)). By Proof. Let Te be the minimal primitive over-lattice of g(H n (M exact sequence (10) and Lemma 2.68 3), we deduce the exact sequence g
f, U, Z)) / g(H n (M
0
/ Te
i∗
/ 0,
/ TU
(23)
with TU being a sub-torsion-group of H n (U, Z). Since H n (X, Z) is torsion-free and Fix G contains only isolated points, by Corollary 2.13, H n (U, Z) has just Z /p Z-torsion. Hence, (23) becomes: 0
f, U, Z)) / g(H n (M
g
/ Te
i∗
/ (Z /p Z)a
/ 0,
(24)
′
So by Proposition 1.1, it remains only to prove that discr Te = pl , with l′ ∈ N. f, Z). Let K be the minimal primitive over-lattice of r1∗ (π∗ (H n (X, Z)) in H n (M We know by Lemma 2.18 3) that discr r1∗ (π∗ (H n (X, Z))) = pd , with d ∈ N. f, U, Z))⊥ by Lemma 2.69 and 2.70 2). Since H n (M f, Z) is Moreover K = g(H n (M l′ ′ e unimodular, by Proposition 1.3, we have discr T = discr K = p with l ∈ N. Lemma 2.72. Consider the following exact sequence: g ^ n /G, U ′′ , Z) H n (C
′
^ n /G, Z) / H n (C
/ H n (U ′′ , Z)
^ n /G, U ′′ , Z). / H n+1 (C
The integer ^ n /G, U ′′ , Z) logp discr Im g ′ + 2 rktor H n+1 (C does not depend of the toric blow-up. Moreover, we have one of the five following cases: 75
^ n /G, U ′′ , Z) = Z /p Z, tors H n (U ′′ , Z) = i) logp discr Im g ′ = 0, tors H n+1 (C ^ n /G, Z) = 0. Z /p Z and tors H n (C ^ n /G, U ′′ , Z) = 0, tors H n (U ′′ , Z) = 0 and ii) logp discr Im g ′ = 0, tors H n+1 (C ^ n /G, Z) = Z /p Z. tors H n (C ^ n /G, U ′′ , Z) = 0, tors H n (U ′′ , Z) = Z /p Z iii) logp discr Im g ′ = 2, tors H n+1 (C ^ n /G, Z) = 0. and tors H n (C n /G, U ′′ , Z) = 0, tors H n (U ′′ , Z) = 0 and ^ iv) logp discr Im g ′ = 0, tors H n+1 (C ^ n /G, Z) = Z /p Z. tors H n (C n /G, U ′′ , Z) = 0, tors H n (U ′′ , Z) = 0 and ^ v) logp discr Im g ′ = 1, tors H n+1 (C ^ n /G, Z) = 0. tors H n (C
Proof. The following open inclusion ^ n /G ^ n /G ֒→ C C induces the following commutative diagram on cohomology: 0
0
▼▼▼ ▼▼▼ ▼▼▼ ▼▼&
^ n /G, Z) H n−1 (C
^ n /G, C ^ n /G, Z) o H n (C O PPP PPP PPP g PPP ' g′ n /G, U ′′ , Z) n /G, Z) n /G, U ′′ , Z) / H n (C^ /0 / H n (C^ / H n (U ′′ , Z) δe / H n+1 (C^ ◆◆◆ ◆◆◆ρ ◆◆◆ ◆◆◆ & e d n /G, U ′ , Z) n /G, Z) n /G, U ′ , Z) / H n (C^ / H n (U ′ , Z) δe / H n+1 (C^ / H n (C^ ▼▼▼ ▼▼▼ ▼▼▼ ▼& n+1 n+1 ^ n ^ n ^ n / H (C /G, C /G, Z) H (C /G, U ′ , Z) ❍❍ ❍❍ ❍❍ ❍❍ ❍❍ $ ^ n /G, U ′ , Z) H n (C
H n−1 (U ′′ , Z)
H n−1 (U ′ , Z)
0
0
76
^ n /G, Z) = 0 for all The zeros in the diagram come from the fact that H 2k+1 (C n /G is a smooth complete toric variety (see[9] Theorem ^ 0 ≤ k ≤ n − 1 because C 10.8). By equivariant cohomology, we can show that H n (U ′ , Z) = Z /p Z and H n−1 (U ′ , Z) = 0 (same calculation as in Section 2.5.4). Moreover, we have ^ n /G, U ′ , Z) = H n (C ^ n /G, U ′′ , Z) ⊕ H n (C ^ n /G, C ^ n /G, Z), H n (C and ^ n /G, U ′ , Z) = H n+1 (C ^ n /G, U ′′ , Z) ⊕ H n+1 (C ^ n /G, C ^ n /G, Z). H n+1 (C We will also need the following lemma. Lemma 2.73.
1)
Im g Γ
is a torsion group.
2) Im g and Im g ′ are orthogonal. Proof.
1) By (13) and (14), we also have: Im g ⊗ Q = Γ ⊗ Q . ′
2) By step 2) of the proof of Lemma 2.68, ImΓ′g is a torsion group. Moreover, the lattices Γ and Γ′ are orthogonal. Hence by 1), the result follows. ^ n /G) • First case: Im g ⊕ Im g ′ is primitive in H n (C In this case, ρ is equal to 0. And ^ n /G) = Im g ⊕⊥ Im g ′ . H n (C Hence, discr Im g = discr Im g ′ = 1. Moreover, ^ n /G, U ′ , Z) = Z /p Z . H n+1 (C Hence we have two sub-cases: ^ n /G, U ′′ , Z) = Z /p Z and H n+1 (C ^ n /G, C ^ n /G, Z) = 0, H n+1 (C or ^ n /G, C ^ n /G, Z) = Z /p Z and H n+1 (C ^ n /G, U ′′ , Z) = 0. H n+1 (C
77
^ n /G, U ′′ , Z) = Z /p Z * First sub-case: H n+1 (C ^ n /G) and H n+1 (C ^ n /G, U ′′ , Z) In this case, Im g ′ is primitive in H n (C = Z /p Z, hence by the diagram: tors H n (U ′′ , Z) = Z /p Z . n /G) and H n+1 (C n /G, C n /G, ^ ^ ^ Furthermore, Im g is primitive in H n (C Z) = 0, hence by the diagram:
^ n /G, Z) = 0. tors H n (C It is the case i). ^ n /G, C ^ n /G, Z) = Z /p Z * Second sub-case: H n+1 (C For the same reason as before, we get tors H n (U ′′ , Z) = 0, and
n /G, Z) = Z /p Z . ^ tors H n (C
It is the case ii). ^ n /G) • Second case: Im g ⊕ Im g ′ is not primitive in H n (C In this case, ρ is not equal to 0. Hence: discr(Im g ⊕ Im g ′ ) = p2 , and by the diagram: ^ n /G, U ′ , Z) = H n+1 (C ^ n /G, U ′′ , Z) = H n+1 (C ^ n /G, C ^ n /G, Z) = 0. H n+1 (C Therefore, there are three sub-cases: discr Im g = p2 and discr Im g ′ = 1, or discr Im g = 1 and discr Im g ′ = p2 , or discr Im g = p and discr Im g ′ = p. * discr Im g = p2 and discr Im g ′ = 1
78
^ n /G, Z) and H n+1 (C ^ n /G, In this case, Im g is not primitive in H n (C ^ n /G, Z) = 0. Hence by the diagram: C ^ n /G, Z) = Z /p Z . tors H n (C n /G, Z) and H n+1 (C n /G, U ′′ , ^ ^ We have also Im g ′ primitive in H n (C Z) = 0. Hence by the diagram:
tors H n (U ′′ , Z) = 0. It is the case iv). * discr Im g = 1 and discr Im g ′ = p2 For the same reason as before, ^ n /G, Z) = 0, tors H n (C and tors H n (U ′′ , Z) = Z /p Z . It is the case iii). * discr Im g = p and discr Im g ′ = p ^ n /G, Z) and H n+1 ( In this case, Im g ′ and Im g are primitive in H n (C n /G, U ′′ , Z) = H n+1 (C n /G, C n /G, Z) = 0. Hence by the diagram: ^ ^ ^ C n /G, Z) = 0, ^ tors H n (C
and tors H n (U ′′ , Z) = 0. It is the case v). ^ n /G, U ′′ , Z) does not Now we prove that w := logp discr Im g ′ + 2 rktor H n+1 (C depend on the toric blow-up. We will need the following lemma. Lemma 2.74. The elements discr Im g and U ′′ do not depend on the toric blowup. Proof. Changing the toric blow-up will just change f −1 (0), hence U ′′ is not n /G, C n /G, Z) is also not modified. By the same ^ ^ modified. The group H n (C technique used in Lemma 2.68, we can prove that discr Im g is also not changed. To do that, we need the following lemma. n /G, Z) = 0. ^ Lemma 2.75. We have H n−1 (C
79
Proof. We have the following commutative diagram:
n /G, U ′′ , Z) ^ H n−1 (C
^ n /G, U ′ , Z) H n−1 (C
q
0
0
n /G, Z) ^ / H n−1 (C
/ H n−1 (U ′′ , Z)
^ n /G, Z) / H n−1 (C
/ H n−1 (U ′ , Z)
0
^ n /G, Z) = 0 by Theorem 10.8 of [9]. The group H n−1 (U ′′ , Z) The group H n−1 (C ′ = 0 because g is injective by Step 3 of the proof of Lemma 2.68. And H n−1 (U ′ , Z) = 0 by equivariant cohomology. Hence, by commutativity of the diagram, ^ n /G, Z) = 0. q = 0. It follows H n−1 (C ^ n /G associated to the cone hu i. We Let Gi be the subtoric variety of C i T k ^ n /G r G . We have C ^ n /G = denote Ui := C i i=1 Ui . As in Lemma 2.68, we look at the following diagram: Lk
i=1
^ n /G, U , Z) H n (C ❘❘❘i ❘❘❘ ρ ❘❘❘ ❘❘❘ ψ ❘(
^ n /G, C ^ n /G, Z) H n (C
g
^ n /G, Z). / H n (C
coker ψ By Lemma 2.75, g is injective. Hence, as in Lemma 2.68, we deduce that Im g = coker ψ which does not depend on the toric blow-up. We have discr Γ Γ which does not depend on the toric blow-up. It follows by Lemma 2.73 1) that discr Im g does not depend on the toric blow-up. When we change the toric blow-up, discr Im g ′ could change but discr Im g and H n (U ′′ , Z) does not change. We sum up all the values of discr Im g and tors H n (U ′′ , Z) in all the cases. i) discr Im g = 1 and tors H n (U ′′ , Z) = Z /p Z, ii) discr Im g = 1 and tors H n (U ′′ , Z) = 0, iii) discr Im g = 1 and tors H n (U ′′ , Z) = Z /p Z, iv) discr Im g = p2 and tors H n (U ′′ , Z) = 0, 80
v) discr Im g = p and tors H n (U ′′ , Z) = 0. The unique situation that could happen in changing the toric blow-up is moving from case i) to case iii) or the contrary. But in both cases, w = 2. This completes the proof. We need also the following lemma. Lemma 2.76. We assume that n is even. Let x ∈ Fix G, we denote Ux := M \ {π(x)}. We have: n f f, U, Z)) = L⊥ 1) We have g(H n (M x∈Sing M g(H (M , Ux , Z)).
f, Ux , Z)) = pax with ax an integer. 2) discr g(H n (M L n f f, U, Z) = Proof. 1) We have H n (M x∈Sing M H (M , Ux , Z). So by taking the image by g, we have: M f, U, Z)) = f, Ux , Z)). g(H n (M g(H n (M (25) x∈Sing M
Moreover if x and y are two different points in Fix G, the cocycles in f, Ux , Z)) and in g(H n (M f, Uy , Z)) do not meet. Hence, the sum of g(H n (M equation (25) is orthogonal.
2) It follows from 1) and Lemma 2.71.
Now we can define the weight of a fixed point. Definition 2.77. (Weight of a fixed point) Let X be a compact complex manifold of even dimension n and G = hϕi, an automorphism group of prime order p such that Fix G contains only isolated points. Let x ∈ Fix G and Ux := M \ {π(x)}. We define the weight of x by: f, Ux , Z)) + 2 rktor H n+1 (M f, Ux , Z), w(x) = logp discr g(H n (M
where g is given in the exact sequence 10. Then, we have the following lemma. Lemma 2.78. We have
f, U, Z)) + 2 rktor H n+1 (M f, U, Z) = logp discr g(H n (M
X
w(x).
x∈Fix G
Proof. It follows from the definition of the weight and Lemma 2.76 1). Now we provide the following fundamental proposition on the weight of a fixed point.
81
Proposition 2.79. Let X be a compact complex manifold of even dimension n and G = hϕi an automorphism group of prime order p such that Fix G contains only isolated points. The weight of a fixed point x ∈ Fix G depends only on the local action of G on it. Proof. Let x ∈ Fix G. Assume that the action of G on X at x is locally equivalent to the action of G = hgi on Cn . We keep the same notation as ^ n /G, U ′′ , Z). f, Ux , Z) = H n+1 (C before. We have H n+1 (M Moreover, we have proved that discr Im g ′ = discr Im gx .
Indeed, by Step 5 of the proof of Lemma 2.68, we know that the lattices Γx and ′ Γ′ are isometric. Moreover, by Step 6 of this same proof, we know that ImΓ′g and ImΓxgx are isomorphic. So discr Im g ′ = discr Im gx . Therefore, ^ n /G, U ′′ , Z). f, Ux , Z) = discr Im g ′ + 2 rktor H n+1 (C discr Im gx + 2 rktor H n+1 (M (26) And the result follows by Lemma 2.72. The following notation follows. Notation 2.80. Let X be a complex manifold of dimension n and G = hϕi an automorphism group of prime order p such that Fix G contains only isolated points. Let x ∈ Fix G. Assuming that the action of G on X is locally equivalent to the action of G = hgi on Cn via g = diag(ξpk1 , ..., ξpkn ), then instead of w(x) we write w( 1p (k1 , ..., kn )). Moreover, Lemma 2.72 and (26) provide the following proposition. Proposition 2.81. The weight of a fixed point can only be equal to 0, 1 or 2. 2.7.2
The main theorem
Theorem 2.82. Let G = hϕi be a group of prime order p acting by automorphisms on a compact complex manifold X of even dimension n. We assume that: i) H ∗ (X, Z) is torsion-free, ii) Fix G is finite without a point of weight 2. Then: 1) logp (discr π∗ (H n (X, Z))) −
P
x∈Fix G w(x)
82
is divisible by 2.
2) The following inequalities hold: logp (discr π∗ (H n (X, Z))) + 2 rktorp H n (U, Z) X f, Z) ≥ w(x) + 2 rktorp H n (M x∈Fix G
≥ 2 rktorp H n (U, Z).
3) If, moreover, X
logp (discr π∗ (H n (X, Z)))+2 rktorp H n (U, Z) =
x∈Fix G
f, Z), w(x)+2 rktorp H n (M
then (X, G) is H n -normal. Proof. The idea of the proof is the same as that of Theorem 2.45. f, U, Z) = Since there is no point of weight 2, by Lemma 2.72, we have H n+1 (M 0. The exact sequence (10) becomes: / H n (M f, U, Z)
0
g
i∗
/ H n (M f, Z)
/ H n (U, Z)
/ 0.
f, U, Z)) in H n (M f, Z). Then, Let Te be the primitive over-lattice of g(H n (M we have the exact sequence (24) of the proof of Lemma 2.71: 0
/ g(H n (M f, U, Z))
g
/ Te
i∗
/ (Z /p Z)a
/ 0.
Now the proof is very similar to the one in Theorem 2.45. By Proposition 1.1: f, U, Z)) discr g(H n (M discr Te = 2a p f, U, Z)) discr g(H n (M = . n n f,Z)) p2(rktorp H (U,Z)−rktor H (M f, Z). Let K be the primitive overlattice of We use the unimodularity of H n (M n f in H (M , Z). By Lemma 2.69 and 2.70 2), we have K ⊥ = Te. Hence by Proposition 1.3,
r1∗ (π∗ (H n (X, Z))
discr K = discr Te
= plogp discr g(H
n f f,Z)−rktorp H n (U,Z) (M ,U,Z))+2 rktor H n (M
(
).
Then discr r1∗ (π∗ (H n (X, Z))) = discr π∗ (H n (X, Z)) ≥ discr K
and
discr Te ≥ 1
and by Lemma 2.78, we get part 2) of the Theorem. By Proposition 1.1, K/K = (Z /p Z)
logp (discr π∗ (H n (X,Z)))−
P
n n f x Fix G w(x)+2 rktorp H (U,Z)−2 rktor p H (M ,Z) 2
83
.
We have proved statement 1) of the Theorem. Now if X
log p (discr π∗ (H n (X, Z))) −
x Fix G
f, Z) = 0, w(x) + 2 rktorp H n (U, Z) − 2 rktorp H n (M
f, Z). By K = r1∗ (π∗ (H n (X, Z))). Hence, r1∗ (π∗ (H n (X, Z))) is primitive in H n (M n Lemma 2.70 1), it follows that π∗ (H (X, Z)) is primitive. And this completes the proof. 2.7.3
Corollary
As in Section 2.5.5, we give a corollary using Proposition 2.22 2) and the calculation of the torsion of H n (U, Z). Corollary 2.83. Let G = hϕi be a group of prime order 3 ≤ p ≤ 19 acting by automorphisms on a compact complex manifold X of dimension 2n. We assume: i) H ∗ (X, Z) is torsion-free, ii) Fix G is finite without a point of weight 2, 2k iv) lp−1 (X) = 0 for all 1 ≤ k ≤ n, and
v) l12k+1 (X) = 0 for all 0 ≤ k ≤ n − 1, when n > 1. Then: 1) l12n (X) − 2) we have:
P
x∈Fix G
w(x) is divisible by 2, and
l12n (X) + 2
2i+1 lp−1 (X) +
w(x)
x∈Fix G
≥2
"n−1 X
#
n−1 X
l12i (X)
i=0
i=0
X
≥
"n−1 X
2i+1 lp−1 (X)
+
n−1 X
#
l12i (X)
.
#
=
i=0
i=0
3) If, moreover, l12n (X)
+2
"n−1 X
2i+1 lp−1 (X)
+
n−1 X i=0
i=0
then (X, G) is H 2n -normal.
84
l12i (X)
X
x∈Fix G
w(x),
Proof. By Proposition 2.59, (X, G) is U -trivial. Moreover, since there is no f, U, Z) = 0. Then by Proposition point of weight 2, by Lemma 2.72, H 2n+1 (M 2n f 2.57, rktor H (M , Z) = 0. In Theorem 2.82, we replace rktor H 2m (U, Z) by Pm−1 2i+1 Pm−1 2i n i=0 lp−1 (X)+ i=0 l1 (X) with Proposition 2.58 and logp (discr π∗ (H (X, Z ))) by l12n with Proposition 2.22 2). i Remark: Corollary 2.83 is also true for p = 2, exchanging l1i with l1+ and i i lp−1 with l1− . In this case, it has only one advantage compared to Corollary 2.63: X can be a complex but non-Kähler manifold.
2.7.4
Calculation of the weight of fixed points
When the dimension of X is bigger than 2, it will be very hard to calculate the weight of fixed points by hand. Hence, in the applications, we will never do a direct calculation. Since the weight of the fixed points only depends on the local action of G on it, we will calculate the weight in applying Theorem 2.84 in simple examples. We give a first example in this section. We will see another example in Section 3. In Theorem 2.84, we do not assume that Fix G is without a point of weight 2; the consequence is a proof much more technical than the proof of Theorem 2.82. Theorem 2.84. Let G = hϕi be a group of prime order p ≤ 19 acting by automorphisms on a compact complex manifold X of even dimension n. We assume that: i) H ∗ (X, Z) is torsion-free, ii) Fix G is finite, iii) (X, G) is U -trivial. Then: 1) logp (discr π∗ (H n (X, Z))) −
P
x∈Fix G w(x)
is divisible by 2,
2) The following inequalities hold: logp (discr π∗ (H n (X, Z)))+2 rktorp H n (U, Z) ≥
X
w(x) ≥ 2 rktorp H n (U, Z).
x∈Fix G
Proof. Since (X, G) is U -trivial, the exact sequence (10) becomes: ∗
f, U, Z) g / H n (M f, Z) i / H n (U, Z) / fH n+1 (M f, U, Z) h / H n+1 (M f, Z) / 0. 0 / H n (M
f, U, Z)) in H n (M f, Z). Then Let Te be the primitive over-lattice of g(H n (M we have the exact sequence (24): 0
f, U, Z)) / g(H n (M
g
85
/ Te
i∗
/ (Z /p Z)a
/ 0.
f, U, Z) is torsion-free, we have: Now, since g is injective and H n (M f, Z) ⊕ (Z /p Z)a . tors Im i∗ = tors H n (M
(27)
The difficulty which did not appear in the proof of Theorem 2.82 is the calculation of a. After this calculation, the proof will be the same. We define the integer b ∈ N as follows: i∗
f, Z)/ tors H n (M
We have:
/ H n (U, Z)/ tors
f
/ (Z /p Z)b
/ 0.
H n (U, Z)/ tors = (Z /p Z)b . Im i∗ / tors
It follows that: tors H n (U, Z) ⊕ (Z /p Z)b = tors Im i∗ ⊕ tors Im f. Hence, by (27), f, Z) ⊕ (Z /p Z)a ⊕ tors Im f. tors H n (U, Z) ⊕ (Z /p Z)b = tors H n (M
(28)
f, U, Z) is a torsion group, we also have: Since h is surjective and H n+1 (M f, U, Z) = tors Im f ⊕ tors H n+1 (M f, Z). H n+1 (M
f, Z) = But by universal coefficient formula and Pointcaré duality, tors H n+1 (M n f tors H (M , Z). Therefore, f, U, Z) = tors Im f ⊕ tors H n (M f, Z). H n+1 (M
(29)
And so by (28) and (29):
f, Z)−(rktor H n+1 (M, U, Z)−rktor H n (M f, Z)) a = rktor H n (U, Z)+b−rktor H n (M = rktor H n (U, Z) + b − rktor H n+1 (M, U, Z).
(30)
Working as in Theorem 2.82, the following lemma will allow us to simplify b in the equations. f, e be the primitive over-lattice of r∗ (π∗ (H n (X, Z)) in H n (M Lemma 2.85. Let H 1 Z)/ tors. Let ρ ∈ N such that: e H
r1∗ (π∗ (H n (X, Z)) Then, we have b ≤ ρ.
86
= (Z /p Z)ρ .
Proof. Since p ≤ 19, we have by Definition-Proposition 1.5 1): ⊕s H n (X, Z) ≃ ⊕ri=1 (OK , ai ) ⊕ OK ⊕ Z⊕t .
Hence, by the proof of Proposition 1.4: H n (X, Z)G ≃ ⊕ri=1 (OK , ai )G ⊕ Z⊕t . We write H n (X, Z)G = P ⊕ L with P ≃ ⊕ri=1 (OK , ai )G and L ≃ Z⊕t . Hence we have (31) r1∗ (π∗ (H n (X, Z)G )) = r1∗ (π∗ (P)) ⊕ r1∗ (π∗ (L )). And by Lemma 2.3 and the proof of Proposition 1.4, we have: r1∗ (π∗ (H n (X, Z))) =
1 ∗ r (π∗ (P)) ⊕ r1∗ (π∗ (L )). p 1
(32)
We will compare H n (U, Z)/ tors and i∗ (r1∗ (π∗ (H n (X, Z)G ))) in two different ways to get the inequality of Lemma 2.87 and then deduce the inequality of Lemma 2.85. Lemma 2.86. We have H n (U, Z)/ tors i∗ (r1∗ (π∗ (H n (X, Z)G )))
= (Z /p Z)b+ρ+α+rk P ,
where α = logp discr Te.
Proof. We have by Proposition 1.1: f, Z)/ tors H n (M = (Z /p Z)α . e Te ⊕ H
(33)
Moreover, by definition of ρ: Te ⊕
e Te ⊕ H
r1∗ (π∗ (H n (X, Z)))
= (Z /p Z)ρ .
(34)
And by (31), (32) and Lemma 2.70 1): Te ⊕ r1∗ (π∗ (H n (X, Z))) = (Z /p Z)rk P . Te ⊕ r∗ (π∗ (H n (X, Z)G ))
(35)
1
Hence, by (33), (34) and (35):
f, Z)/ tors H n (M
Te ⊕ r1∗ (π∗ (H n (X, Z)G ))
= (Z /p Z)α+ρ+rk P .
f, Z)/ tors it follows that: Since Te is primitive in H n (M f, Z)/ tors) i∗ (H n (M
i∗ (r1∗ (π∗ (H n (X, Z)G )))
And since
H n (U,Z)/ tors f i∗ (H n (M,Z)/ tors)
= (Z /p Z)α+ρ+rk P .
= (Z /p Z)b , we get our result. 87
(36)
(37)
Lemma 2.87. We have: b + ρ + α ≤ rk L. Proof. In fact, we will prove that: b + ρ + α + rk P ≤ rk H n (X, Z)G . To do this, we must first prove that: 1 ∗ ∗ i (r1 (π∗ (H n (X, Z)G ))). p
H n (U, Z)/ tors ⊂
(38)
We consider the following diagram: i∗
f, Z) o H n (M π e
r1∗
H n (M, Z)
∗
e Z) o H n (X,
i∗ 0
& / H n (U, Z) π0∗
π∗
s∗ 1
H n (X, Z)
j∗
/ H n (V, Z)
By Corollary 2.13, we have:
H n (U, Z)/ tors ⊂
1 π0∗ (H n (V, Z)G )/ tors . p
(39)
It remains to prove that: i∗ (r1∗ (π∗ (H n (X, Z)G ))) = π0∗ (H n (V, Z)G )/ tors .
(40)
Let y ∈ H n (X, Z)G , by commutativity of the diagram we have i∗ (r1∗ (π∗ (y)) = i∗0 (π∗ (y)).
(41)
And we also have: π0∗ (i∗0 (π∗ (y))) = j ∗ (π ∗ (π∗ (y))) = pj ∗ (y) = π0∗ (π0∗ (j ∗ (y))). But by Corollary 2.13, π0∗ is injective modulo torsion, hence i∗0 (π∗ (y)) = π0∗ (j ∗ (y)) + tors .
(42)
So by (41) and (42), we have: π0∗ (j ∗ (H n (X, Z)G ))/ tors = i∗0 (π∗ (H n (X, Z)G ))/ tors = i∗ (r1∗ (π∗ (H n (X, Z)G )))/ tors .
Since i∗ (r1∗ (π∗ (H n (X, Z)G ))) is torsion-free, we have: π0∗ (j ∗ (H n (X, Z)G ))/ tors = i∗ (r1∗ (π∗ (H n (X, Z)G ))). And since j ∗ (H n (X, Z)G ) = H n (V, Z)G , we get (40). So by (39) we have (38). Now, we have 1 ∗ ∗ n G p i (r1 (π∗ (H (X, Z) ))) H n (U, Z)/ tors
88
= (Z /p Z)γ ,
with γ ∈ N. By Lemma 2.70 and Corollary 2.13, i∗ ◦ r1∗ ◦ π∗ is injective, hence: 1 ∗ ∗ n G p i (r1 (π∗ (H (X, Z) ))) i∗ (r1∗ (π∗ (H n (X, Z)G )))
= (Z /p Z)rk H
n
(X,Z)G
.
From Lemma 2.86 and the two last equalities, it follows that: b + ρ + α + rk P + γ = rk H n (X, Z)G = rk L + rk P. Since γ ≥ 0, this completes the proof. Now we finish the proof of Lemma 2.85. Since discr Te = pα , by Proposition 1.1 and (36), we have: discr r1∗ (π∗ (H n (X, Z)G )) = p2ρ+2 rk P+α .
(43)
But we also know by that Definition-Proposition 1.5 1) and Definition-Proposition 1.7 2), 3) that: discr H n (X, Z)G = prk P . And by Lemma 2.18 3), discr r1∗ (π∗ (H n (X, Z)G )) = prk P+rk H
n
(X,Z)G
= prk L+2 rk P .
(44)
Hence, by (43) and (44), we have: rk L = 2ρ + α. And Lemma 2.87 concludes the proof. Now the proof is very similar to that of Theorem 2.82. By Proposition 1.1 and (24), f, U, Z)) discr g(H n (M . discr Te = 2a p And by (30),
discr Te =
f, U, Z)) discr g(H n (M
p2(rktorp H
n (U,Z)−b−rktor H n+1 (M f,U,Z))
.
f, Z). By Lemma 2.69 and 2.70 2), K ⊥ = We use the unimodularity of H n (M e T . Hence, by Proposition 1.3, e = discr Te discr H
= plogp discr g(H
n f f,U,Z)−b−rktorp H n (U,Z) (M ,U,Z))+2 rktor H n+1 (M
(
And by definition of ρ and Proposition 1.1, e · p2ρ . discr r1∗ (π∗ (H n (X, Z))) = (discr H) 89
).
Then, discr r1∗ (π∗ (H n (X, Z))) e = discr H p2ρ = plogp discr g(H
n f f,U,Z)−b−rktorp H n (U,Z) (M ,U,Z))+2 rktor H n+1 (M
(
)
≥1
and by Lemma 2.78 and Lemma 2.85, we get part 2) of the theorem. Moreover, f, U, Z)) logp discr π∗ (H n (X, Z)) − 2ρ = logp discr g(H n (M f, U, Z) − b − rktorp H n (U, Z) + 2 rktor H n+1 (M
Statement 1) of the theorem follows.
Corollary 2.88. Let G = hϕi be a group of prime order 3 ≤ p ≤ 19 acting by automorphisms on a compact complex manifold X of dimension 2n. We assume that: i) H ∗ (X, Z) is torsion-free, ii) Fix G is finite, 2k iv) lp−1 (X) = 0 for all 1 ≤ k ≤ n, and
v) l12k+1 (X) = 0 for all 0 ≤ k ≤ n − 1, when n > 1. Then: 1) l12n (X) − 2) we have:
P
x∈Fix G
w(x) is divisible by 2, and
l12n (X) + 2 ≥
X
"n−1 X
2i+1 lp−1 (X) +
w(x)
≥2
2i+1 lp−1 (X)
+
n−1 X i=0
i=0
l12i (X)
i=0
i=0
x∈Fix G
"n−1 X
#
n−1 X
#
l12i (X)
.
Proof. The U -triviality is obtained by Proposition In Theorem 2.84, P Pm−12.59. 2i+1 2i we replace rktor H 2m (U, Z) by m−1 i=0 lp−1 (X) + i=0 l1 (X) with Proposition 2.58 and logp (discr π∗ (H n (X, Z))) by l12n with Proposition 2.22 2). Remark: There is no conclusion about the H n -normality in Corollary 2.88 and Theorem 2.84. Because when the fixed locus contains points of weight 2 the statement 3) of Corollary 2.88 and Theorem 2.84 could be false because b could be > 0. 90
Proposition 2.89. In dimension 2 the weight of an isolated fixed point is always 1.
Proof. Let G = diag(ξp , ξpα ) acting on C2 with α ∈ {1, ..., (p − 1)}. It enough to show that we are in the case v) of Lemma 2.72. EThe toric variety C2 is given D− → → − by the lattice M = Z2 and the cone σ = i , j . The invariant lattice M G → − − → → − − → → − → − is generated by p i , p j , (p − α) i + j and i + k j with k ∈ {1, ..., (p − 1)} such that kα = p − 1 in Z /p Z. → − → − − → Since k(p − α) ≡ 1 mod p, (p i , (p − α) i + j ) is a basis of M G . We denote → − → − → − → − → − → − → u := p i and − v := (p − α) i + j . Let ( i ∨ , j ∨ ) be the canonical basis of the → → ∨ dual N := M . We write elements of the canonical basis (− u ∨, − v ∨ ) of (M G )∨ → − − → →∨ − →∨ − → − → − → − in function of i and j . We have u ∨ ( i ) = p1 and u ∨ ((p − α) i + j ) = 0. → − → Hence − u ∨ ( j ) = − p−α p . And so 1− (p − α) − → →∨ − → u∨ = i ∨ − j . p p → − → − − → − − And we have → v ∨ ( i ) = 0 and → v ∨ ((p − α) i + j ) = 1. So → − − → v ∨ = j ∨. The toric variety C2 /G is given by the lattice (M G )∨ and the fan nD− → − → Eo → E D− →∨ E D− , j ∨ , i ∨, j ∨ . Σ= i
Hence, in the canonical basis of (M G )∨ , it provides the fan:
→ → → → → → Σ = {hp− u ∨ + (p − α)− v ∨ i , h− v ∨ i , hp− u ∨ + (p − α)− v ∨, − v ∨ i} . → − → − We denote by f1 ∨ ,...,fk ∨ the vectors which generate the new cones of dimension → ∨ ^ 2 /G. And we denote by − 1 of the fan of C g1 ∨ ,...,− g→ m the vectors which generate ^ 2 /G. We are free to choose, for the new cones of dimension 1 to get the fan of C example, →∨ − − → → → f1 = → u∨+− v ∨ and − g1 ∨ = −− u ∨. → → → → → We also denote − e ∨ = p− u ∨ + (p − α)− v ∨ and − e ∨=− v ∨ . By Theorem 10.8 of 1
2
n /G, Z) is isomorphic to the group ^ [9], H (C i hD− h hD− → Ei →∨ → Ei → → ∨ g1 i] , ..., [h− g→ Z [h− e1 ∨ i] , [h− e2 ∨ i] , f1 ∨ , ..., fk ∨ , [h− m i] n
quotiented by
→ p [h− e1 ∨ i] +
m k hD− X − →∨ − → Ei X − → → → gj ∨ (− u ) [h− gj ∨ i] fi (→ u ) fi ∨ + j=1
i=1
91
and → → [h− e2 ∨ i] + (p − α) [h− e1 ∨ i] +
m k hD− X →∨ − − → Ei X − → → → gj ∨ (− v ) [h− gj ∨ i] , fi (→ v ) fi ∨ + j=1
i=1
where the brackets indicate the cocycle associated to the cone. As inhDLemma →∨ Ei − 2.68, we denote by Γ′ and Γ the lattice generated by the cocycles , fi → − → − i ∈ {1, ..., k} and [h g ∨ i], j ∈ {1, ..., m} respectively. Then, the class [h e ∨ i] of i
j
n ^ → → − n /G,Z) − → → [h→ ei ∨ i] generates H (ΓC′ ⊕Γ . Moreover, since f1 ∨ (− u ) = 1 and − g1 ∨ (− u ) = −1, x+y → − ′ ∨ with x ∈ Γ and y ∈ Γ not divisible by p. It follows that we have [h e i] =
1
′
p
′
Γ = Im g and Γ = Im g and we are necessarily in situation v) of Lemma 2.72. Therefore, the weight is equal to 1. We give one example of the use of Corollary 2.88. Proposition 2.90. Letting p be a prime number, we have w( 1p (1, 2, 3, ..., p − 1)) = 1. Proof. We will apply Corollary 2.88 to the following action on Pp−1 . ϕ : Pp−1 −→ Pp−1 (a0 : a1 : · · · : ap−1 ) → (a0 : a1 ξp : · · · : ap−1 ξpp−1 ), where ξp is a p-th root of unity. We consider G = hϕi. We have p fixed points with all the same type: p1 (1, ..., (p − 1)). We have l1p−1 (Pp−1 ) = 1. Hence, by 2) of Corollary 2.88, we have: 1 (1, ..., (p − 1)) ≥ p − 1. p ≥ pw p Hence, w
1 (1, ..., (p − 1)) = 1. p
Proposition 2.91. When X is Kähler, the isolated fixed points of type 1 are of weight 1. Proof. Let G = hϕi be a group of prime order p acting by automorphisms on a complex manifold X of dimension n such that Fix G contains only isolated point of type 1. We consider the diagram (4) from the beginning of Section 2.5: f M O
r
e X
s
/M O π
π e
92
/ X.
The blow-up of Cn in 0 remains a toric variety because 0 is fixed by the action of f is a toric blow-up of M . And by Lemma 2.46, Lemma the torus. Therefore, M 2.48 1), and Lemma 2.18 3), we have all the isolated fixed points at weight 1. Proposition 2.92. If X is a Kähler manifold of dimension 2n, the isolated fixed points of type 2 are of weight 1. Proof. Let
*
G=
+
diag(ξ3 , ..., ξ3 , ξ32 , ..., ξ32 ) | {z } | {z } α
2n−α
g 2n be the blow-up of C2n in 0, acting on C2n . Let C
g 2n = {((x , ..., x ), (a : ... : a )) ∈ C2n × P2n−1 | C 1 2n 1 2n
rk ((x+1 , ..., x2n ), (a1 , ..., a2n )) = 1} .
g 2n , On C
*
+
diag(ξ3 , ..., ξ3 , ξ32 , ..., ξ32 ) | {z } | {z } α
acts as follows:
2n−α
diag(ξ3 , ..., ξ3 , ξ32 , ..., ξ32 ) · ((x1 , ..., x2n ), (a1 : ... : a2n )) | {z } | {z } α
=
2n−α (ξ3 x1 , ..., ξ3 xα , ξ32 xα+1 , ..., ξ32 x2n ), (a1
: ... : aα : ξ3 aα+1 : ... : ξ3 a2n ) .
Hence, the fixed locus is given by {0} × Pα−1 and {0} × P2n−α−1 . The action of the torus (C∗ )2n on (C)2n extends to the following action on g 2n : C
(y1 , ..., y2n )·((x1 , ..., x2n ), (a1 : ... : a2n )) = ((y1 x1 , ..., y2n x2n ), (y1 a1 : ... : y2n a2n )) . g 2n is fixed by the action of torus So the fixed locus of the action G on C g 2n in the fixed locus of G (C ) . And so the blow-up (classical blow-up) of C remains a toric variety. Now let X be a Kähler manifold of dimension 2n and G an automorphism group of order 3 acting on X such that Fix G is finite. Then, looking the diagram (8): ∗ 2n
MO 2
r2
π2
X2X G2
/ M1 O
r1
π1 s2
/ X1 X
/M O π
s1
/X Z G
G1
it follows that M2 is a toric blow-up of M . 93
,
Now we will calculate the weight of fixed points using M2 . Let x ∈ Fix G a point of type 2 and Ux := M r {π(x)}. By Theorem 7.31 of [36], we have: H 2n (X1 , Z) = s∗1 (H 2n (X, Z))
⊥ M
(⊕⊥ x∈Fix G Tx ),
with Tx ≃ (−1) as a lattice. And once more by Theorem 7.31 of [36], we have: H 2n (X2 , Z) = s∗ (H 2n (X, Z)) with s = s2 ◦ s1 and Tx′ = (⊕2n−1−α H 2n−2i−2 (Pα−1 , Z)) i=0
⊥ M
∗ ′ (⊕⊥ x∈Fix G s2 (Tx ) ⊕ Tx ),
M α−1 2n−2i−2 2n−1−α (⊕i=0 H (P , Z)).
Without loss of generality, we can assume that α ≤ n. It follows that: rk Tx′ = 2α.
(45)
fx = s∗2 (Tx ) ⊕ Tx′ . Since X2 is smooth and X is smooth, T fx is a We denote T unimodular lattice of rank 2α + 1. fx ) is generated by cocycles supported on r−1 (r−1 (π(x))), Moreover, since π2∗ (T 2 1 fx ) is a sublattice of g(H n (M2 , Ux , Z)). Furby the exact sequence (10), π2∗ (T thermore, fx ) ⊗ Q = dim g(H n (M2 , Ux , Q)). dim π2∗ (T (46)
Indeed, by Lemma 2.72, H n+1 (M2 , Ux , Z) is a torsion group. Hence, by the exact sequence (10): ! X n n dim H (M2 , Q) = dim g(H (M2 , Ux , Q)) ⊕ dim H n (U, Q)), (47) x∈Fix G
with U = M r Sing M and V = X r Fix G as usual. And we have: ! X fx ⊗ Q ⊕ dim H n (X, Q)). dim H n (X2 , Q) = dim T x∈Fix G
Since the exceptional divisors are fixed by G, by Corollary 2.13 we have: ! X n f dim H (M2 , Q) = dim Tx ⊗ Q ⊕ dim H n (X, Q)G ). x∈Fix G
fx ) ⊂ g(H n ( Since dim H n (X, Q)G = dim H n (V, Q)G = dim H n (U, Q) and π2∗ (T M2 , Ux , Z)), by (47) it follows (46).
94
g(H n (M2 ,Ux ,Z)) is a torsion group. By f π2∗ (T x) n ax 2) discr g(H (M2 , Ux , Z)) = p with ax ∈ N. By Lemma 2.18 3) fx ) = p2α+1 . Hence, by Proposition 1.1, have discr π2∗ (T
By (46), it follows that
Lemma 2.76 and (45) we
2α+1−ax g(H n (M2 , Ux , Z)) = (Z /p Z) 2 . fx ) π2∗ (T
Hence, ax is odd. And we have:
w(x) = ax + 2 rktor H n+1 (M2 , Ux , Z). But by Proposition 2.81, w(x) can be only equal to 0,1 or 2. It follows that w(x) = 1. 2.7.5
Returning to the case of fixed points of type 2
From Corollary 2.83 and Proposition 2.92, we deduce the following Corollary. Corollary 2.93. Let G = hϕi be a group of order 3 acting by automorphisms on a compact Kähler manifold X of dimension 2n. We assume that: i) H ∗ (X, Z) is torsion-free, ii) Fix G is finite, 2k iv) lp−1 (X) = 0 for all 1 ≤ k ≤ n, and
v) l12k+1 (X) = 0 for all 0 ≤ k ≤ n − 1, when n > 1. Then: 1) l12n (X) − # Fix G is divisible by 2, and 2) we have: l12n (X)
+2
"n−1 X
2i+1 lp−1 (X)
+
n−1 X
#
l12i (X)
i=0
i=0
≥ # Fix G # "n−1 n−1 X X 2i+1 l12i (X) . lp−1 (X) + ≥2 i=0
i=0
3) If, moreover, l12n (X)
+2
"n−1 X
2i+1 lp−1 (X)
+
n−1 X i=0
i=0
then (X, G) is H 2n -normal. 95
#
l12i (X)
= # Fix G,
2.7.6
The case of simply connected surfaces
We can also give a practical application in the case of surfaces. Corollary 2.94. Let G = hϕi be a group of prime order 2 ≤ p ≤ 19 acting by automorphisms on a simply connected complex surface X. We assume that: i) Fix G is finite and non-empty. 2 2 ii) lp−1 (X) = 0 when p > 2 and l1,− (X) = 0 when p = 2.
So (X, G) is H 2 -normal. Proof. The cohomology graded group H ∗ (X, Z) is torsion-free. Since X is simply connected, H1 (X, Z) = 0. And so by Poincaré duality, H 3 (X, Z) = 0. Moreover, by the universal coefficient theorem H 1 (X, Z) = 0 and H 2 (X, Z) is torsion-free. By Proposition 2.89, the weight of all the fixed points is 1. Hence, by Corollary 2.83 and Corollary 2.63, it is enough to prove that l12 (X)+2 = # Fix G 2 when p > 2 and l1,+ (X) + 2 = # Fix G when p = 2. By Proposition 4.5 and Corollary 4.4 of [6], we have: 2 (X), # Fix G = 2 + l12 (X) + lp−1
(48)
# Fix G = 2 + l12 (X),
(49)
if p > 2 and 2 2 if p = 2. Since we have assumed that lp−1 (X) = 0 when p > 2 and l1,− (X) = 0 when p = 2, the equalities (48) and (49) become
# Fix = 2 + l12 (X),
(50)
2 # Fix = 2 + l1,+ (X),
(51)
if p > 2 and which is what we wanted to prove.
3
Applications
In this section, we give some examples of applications of the results from the last section.
3.1
Quotient of a K3 surface by symplectic groups
Corollary 3.1. Let S be a K3 surface and G a symplectic group of prime order p acting on S. Then (S, G) is H 2 -normal. Proof. The surface S is simply connected. Since the action is symplectic, the fixed locus can only be isolated points. From Nikulin [28], we have p ≤ 7 and Fix G 6= ∅. Hence, by Corollary 2.94, it remains to prove that for all 3 ≤ p ≤ 7, 2 2 lp−1 (X) = 0, l1,− (X) = 0 for p = 2. 96
• p=2 We know that H 2 (S, Z)i ≃ U 3 ⊕ E8 (−2) (see for instance [14]). So by Proposition 2.19 1) and Definition-Proposition 1.7 3), l22 (S) = 8. Since rk H 2 (S, Z) = 22 and rk H 2 (S, Z)i = 14, by Proposition 1.6, ∗
2 rk H 2 (S, Z) − rk H 2 (S, Z)i = l22 (S) + l1,− (S) = 22 − 14 = 8. 2 Hence, l1,− (S) = 0.
• p=3 By Theorem 4.1 of [13], H 2 (S, Z)G ≃ U ⊕ U (3)2 ⊕ A22 . So by Proposition 2.19 1) and Definition-Proposition 1.7 3), l32 (S) = 6. Since rk H 2 (S, Z) = 22 and rk H 2 (S, Z)G = 10, by Proposition 1.4, rk H 2 (S, Z) − rk H 2 (S, Z)G = 2l32 (S) + l22 (S) = 22 − 10 = 12. Hence l22 (S) = 0. • p=5 By Theorem 4.1 of [13], H 2 (S, Z)G ≃ U ⊕ U (5)2 . So by Proposition 2.19 1) and Definition-Proposition 1.7 3), l52 (S) = 4. Since rk H 2 (S, Z) = 22 and rk H 2 (S, Z)G = 6, by Proposition 1.4, rk H 2 (S, Z) − rk H 2 (S, Z)G = 4l52 (S) + l42 (S) = 22 − 6 = 16. Hence, l42 (S) = 0. • p=7
4 1 By Theorem 4.1 of [13], H (S, Z) ≃ U (7) ⊕ . So by Proposition 1 2 2 2.19 1) and Definition-Proposition 1.7 3), l7 (S) = 3. Since rk H 2 (S, Z) = 22 and rk H 2 (S, Z)G = 4, by Proposition 1.4, 2
G
rk H 2 (S, Z) − rk H 2 (S, Z)G = 6l72 (S) + l62 (S) = 22 − 4 = 18. Hence, l62 (S) = 0. In this case (p = 7), we can also apply Proposition 2.25, since l12 (S) = 1. Corollary 3.2. Let S be a K3 surface and G a symplectic group of prime order p acting on S. The following table summarizes our results: p 2 3 5 7
(H 2 (S/G, Z), ·) E8 (−1) ⊕ U (2)3 U (3) ⊕ U 2 ⊕ A22 U (5) ⊕ U 2 4 −3 U⊕ −3 4
Proof. It follows from Corollary 3.1, Theorem 4.1 of [13], and Proposition 2.19 2).
97
3.2
Quotient of a K3 surface by non-symplectic groups
Corollary 3.3. Let S be a K3 surface and G a non-symplectic group of automorphisms on S of order p ≥ 3. We assume that Fix G is finite (it is always the case when p = 17 or 19, and never when p = 13). Then (S, G) is H 2 -normal. Proof. The surface S is simply connected. By [1] and [2], Fix G 6= ∅ and p ≤ 19. Hence, by Corollary 2.94, it remains to prove that for all 3 ≤ p ≤ 19, 2 lp−1 (X) = 0. • p=3 By Table 2 of [1], H 2 (S, Z)G ≃ U (3)⊕E6∨ (3). We have discr E6 = 3, hence discr E6∨ (3) = 53 . So by Proposition 2.19 1) and Definition-Proposition 1.7 3), l32 (S) = 7. Since rk H 2 (S, Z) = 22 and rk H 2 (S, Z)G = 8, by Proposition 1.4, rk H 2 (S, Z) − rk H 2 (S, Z)G = 2l32 (S) + l22 (S) = 22 − 8 = 14. Hence, l22 (S) = 0. • p=5 By Table 2 of [2], H 2 (S, Z)G ≃ H5 ⊕ A∨ 4 (5), with 2 1 H5 = , 1 −2 and
−4 1 1 1 1 −4 1 1 . A4 = 1 1 −4 1 1 1 1 −4
5 We have discr A4 = 5, hence discr A∨ 4 (5) = 3 . So by Proposition 2.19 1) 2 and Definition-Proposition 1.7 3), l5 (S) = 4. Since rk H 2 (S, Z) = 22 and rk H 2 (S, Z)G = 6, by Proposition 1.4,
rk H 2 (S, Z) − rk H 2 (S, Z)G = 4l52 (S) + l42 (S) = 22 − 6 = 16. Hence, l42 (S) = 0. • p=7 By Table 3 of [2], H 2 (S, Z)G ≃ U (7) ⊕ K7 , with −4 1 K7 = . 1 −2 So by Proposition 2.19 1) and Definition-Proposition 1.7 3), l72 (S) = 3. Since rk H 2 (S, Z) = 22 and rk H 2 (S, Z)G = 4, by Proposition 1.4, rk H 2 (S, Z) − rk H 2 (S, Z)G = 6l72 (S) + l62 (S) = 22 − 4 = 18. Hence, l62 (S) = 0. In this case, we can also apply Proposition 2.25 since l12 (S) = 1. 98
• p=11 By Table 4 of [2], H 2 (S, Z)G ≃ U (11). So by Proposition 2.19 1) and 2 Definition-Proposition 1.7 3), l11 (S) = 2. Since rk H 2 (S, Z) = 22 and 2 G rk H (S, Z) = 2, by Proposition 1.4, 2 2 rk H 2 (S, Z) − rk H 2 (S, Z)G = 10l11 (S) + l10 (S) = 22 − 2 = 20. 2 Hence, l10 (S) = 0.
In this case, we could also apply Proposition 2.24 since l12 (S) = 0. • p=17 By Table 6 of [2], H 2 (S, Z)G ≃ U ⊕ L17 , with −2 1 0 1 1 −2 0 0 , L17 = 0 0 −2 1 1 0 1 −4
and discr L17 = 17. So by Proposition 2.19 1) and Definition-Proposition 2 1.7 3), l17 (S) = 1. Since rk H 2 (S, Z) = 22 and rk H 2 (S, Z)G = 6, by Proposition 1.4, 2 2 rk H 2 (S, Z) − rk H 2 (S, Z)G = 16l17 (S) + l16 (S) = 22 − 6 = 16.
2 (S) = 0. Hence, l16
• p=19 By Table 7 of [2], H 2 (S, Z)G ≃ U ⊕ K19 , with −10 1 K19 = . 1 −2 2 (S) = 1. So by Proposition 2.19 1) and Definition-Proposition 1.7 3), l19 2 2 G Since rk H (S, Z) = 22 and rk H (S, Z) = 4, by Proposition 1.4, 2 2 rk H 2 (S, Z) − rk H 2 (S, Z)G = 18l19 (S) + l18 (S) = 22 − 4 = 18. 2 Hence, l16 (S) = 0.
Corollary 3.4. Let S be a K3 surface and G a non-symplectic group of prime order p ≥ 3, acting on S. We assume that Fix G is finite. The following table summarizes our results:
99
p 3 5 7 11 17 19
With L17
−2 1 0 1 −2 0 = 0 0 −2 1 0 1
(H 2 (S/G, Z), ·) U ⊕ E6 −2 −5 ⊕ A4 −5 −10 −4 3 U⊕ 3 −4 U U (17)⊕ L∨ 17 (17) −10 9 U (19) ⊕ 9 −10 2 −1 0 0 1 0 and A4 = −1 2 −1 0 . 0 −1 2 −1 1 0 0 −1 2 −4
Proof. It follows from Corollary 3.1, [2], [1] and Proposition 2.19 2). From proof of Corollary 3.1 and Corollary 3.3, we note the following proposition: Proposition 3.5. Let S be a K3 surface and G a group of automorphisms on S 2 of prime order. We assume that Fix G is finite but non-empty, then lp−1 (S) = 0 2 for p ≥ 3 and l1− (S) = 0 for p = 2. In other words, m2G (S) = a2G (S). We recall that mkG (S) is defined in Definition-Proposition 1.7.
3.3
Quotient of a complex torus of dimension 2 by − id
Corollary 3.6. Let A be a complex torus of dimension 2. We denote A = A/ − id. Then H 2 (A, Z) endowed with the cup product is isometric to U (2)3 . Proof. The ring H ∗ (A, Z) is torsion-free and Fix G contains 16 isolated points. The map (− id)∗ acts on H 1 (A, Z) as − id and acts on H 2 (A, Z) as id. Hence 1 2 2 l1,− (A) = 4, l1,+ (A) = 6 and l1,− (A) = 0. Therefore the equality of Corollary 2.63 3) is verified: 6 + 2 + 2 × 4 = 16. Hence (A, − id) is H 2 -normal. We finish the proof by Proposition 2.19 2).
3.4
Recall on the quotient of a K3[2] -type manifold by a symplectic involution
From [20] we have the following Theorem. Theorem 3.7. Let X be deformation equivalent to S [2] and G a symplectic automorphism group of order 2. Then (X, G) is H 4 -normal and H 2 -normal. 100
Proof. This follows from Section 2.2, Lemma 2.28 and Lemma 2.29 of [20]. Remark: The H 4 -normality can be find again with Corollary 2.63. We will give an example of a non H 2 -normal pair. Let S be a K3 surface and i a symplectic involution on S. We denote by ι the involution induced by i on S [2] . By Theorem 4.1 of [24], the fixed locus of ι is the union of 28 points and a K3 surface Σ. Consider the partial resolution of singularities r′ : M ′ → M obtained by blowing up Σ := π(Σ), where π : S [2] → M is the quotient map. ′ Denote by Σ the exceptional divisor. Let s′ : X ′ → S [2] be the blow-up of S [2] in Σ, and denote by Σ′ the exceptional divisor in X ′ . Denote by ι′ the involution on X ′ induced by ι. We have M ′ ≃ X ′ /ι′ , and we denote by π ′ : X ′ → M ′ the quotient map. We sum up the notation in the diagram: MO ′
r′
π′
π
X ′Z
s′
/ S [2] . Z ι
ι′
We have
/M O
H 2 (X ′ , Z) ∼ = H 2 (S [2] , Z) ⊕ Z Σ′ ,
We also write δ ′ = s′∗ (δ), where δ is the half diagonal of S [2] . Counter-example 3.8. The element π1∗ (Σ′ +δ ′ ) is divisible by 2 in H 2 (M ′ , Z). Thus (X ′ , ι′ ) is not H 2 -normal, because there is no y ∈ H 2 (X ′ , Z) such that Σ′ + δ ′ = y + ι′ (y). And we find the coefficient of normality: α2 (X ′ ) = 1. Proof. The element π∗′ (Σ′ + δ ′ ) is divisible by 2 in H 2 (M ′ , Z) by Lemma 3.36 of [20]. Moreover Σ′ + δ ′ ∈ N1 , then by Lemma 2.3, Σ′ + δ ′ cannot be written in the form y + ι′ (y) for some y ∈ H 2 (X ′ , Z). Since (X, G) is H 2 -normal, we get α2 (X ′ ) = 1. We also recall the theorem from [20] about the Beauville–Bogomolov form of M ′ . Let X be an irreducible symplectic manifold of K3[2]-type and σ a symplectic involution on X. By Theorem 4.1 of [24], the fixed locus of σ is also the union of 28 points and a K3 surface Σ. We still denote M ′ the partial resolution of M in Σ. Theorem 3.9. The Beauville–Bogomolov lattice H 2 (M ′ , Z) is isomorphic to E8 (−1) ⊕ U (2)3 ⊕ (−2)2 , and the Fujiki constant of M ′ is 6.
101
3.5 3.5.1
Quotient of a K3[2] -type manifold by a symplectic automorphism of order 3 Symplectic groups and Beauville–Bogomolov forms
Here, we study the quotient of a manifold of K3[2] -type X by a symplectic group G of prime order. Hence M = X/G will be a singular irreducible symplectic variety. Moreover, in the case where the non-free locus of G is finite, the codimension of the singular locus of M = X/G will be 4. Hence, in this situation, we can use Theorem 1.23 and also the Torelli Theorem of Namikawa (Theorem 1.22). Therefore, we can calculate the Beauville–Bogomolov form in this case. We will need the following proposition. Proposition 3.10. Let X be a manifold of K3[2] -type, G a symplectic group of prime order p with Fix G finite. Then s 3p3 BX (α, β), BX/G (π∗ (α), π∗ (β)) = CX/G where CX/G is the Fujiki constant of X/G and α, β are in H 2 (X, Z)G . Proof. By (1) of Theorem 1.23, we have (π∗ (α))4 = CX/G BX/G (π∗ (α), π∗ (α))2 . By Theorem 1.12, the Fujiki constant of X is 3. Hence we also have: α4 = 3BX (α, α)2 . Moreover, by Lemma 2.18, 3), (π∗ (α))4 = p3 α4 . By (2) of Theorem 1.23, we get the result. 3.5.2
Beauville–Bogomolov lattice
We study the case p = 3. We have the following corollary. Corollary 3.11. Let X be a manifold of K3[2] -type. Let G be an order 3 group of numerically standard symplectic automorphisms of X. Then (X, G) is H 2 -normal and H 4 -normal. Proof. By Proposition 2.28, Proposition 2.31 and Theorem 1.16, if (X, G) is H 4 -normal then (X, G) is H 2 -normal. Hence we have just to show the H 4 normality. We will apply Theorem 2.66 (we could also apply Corollary 2.93). By Theorem 2.5 of [22] and Example 4.2.1 of [21], we know that Fix G consists of 27 isolated points. Moreover the action of G is symplectic on X, hence all the fixed points are of type 13 (1, 1, 2, 2). Hence there are stable fixed 102
points. By Theorem 1.16, H ∗ (X, Z) is torsion-free. It remains to show that l22 (X) = l24 (X) = 0. By definition of numerically standard and Theorem 4.1 of [13], we know that H 2 (X, Z)G ≃ U ⊕ U (3)2 ⊕ A22 ⊕ (−2), where H 2 (X, Z) is endowed with the Beauville–Bogomolov form. Hence by Lemma 1.10 and Definition-Proposition 1.7, l32 (X) = 6. Since rk H 2 (X, Z) = 23 and rk H 2 (X, Z)G = 11, by Proposition 1.4, rk H 2 (X, Z) − rk H 2 (X, Z)G = 2l32 (X) + l22 (X) = 23 − 11 = 12. Hence l22 (X) = 0. Then, by Proposition 2.29, Proposition 2.31, Theorem 1.16 and Lemma 2.30, l24 (X) = 0. Now, we show that l14 (X) + 2 1 + l21 (X) + l23 (X) + l12 (X) = h2∗ (Fix G, Z).
Since H odd (X, Z) = 0 (Theorem 1.16) and Fix G consists of 27 isolated points, we have just to show: l14 (X) + 2(1 + l12 (X)) = 27.
(52)
It remains to calculate l14 (X) and l12 (X). By Proposition 1.4, rk H 2 (X, Z)G = l12 (X) + l32 (X). Since rk H 2 (X, Z)G = 11 and l32 (X) = 6, we get l12 (X) = 5. Now by Proposition 2.29, Proposition 2.31, Theorem 1.16 and Lemma 2.30, l14 (X) = 5×6 2 = 15. Hence we get (52). We conclude by Theorem 2.66. We deduce the following theorem. Theorem 3.12. Let X be a manifold of S [2] -type. Let G be an order 3 group of numerically standard symplectic automorphisms of X. We denote M3 = X/G. Then the Beauville–Bogomolov lattice H 2 (M3 , Z) is isomorphic to U (3) ⊕ U 2 ⊕ A22 ⊕ (−6), and the Fujiki constant of M3 is 9. Proof. By definition of numerically standard and Theorem 4.1 of [13], there is an isometry of lattices H 2 (S [2] , Z)G ≃ U ⊕ U (3) ⊕ U (3) ⊕ A2 ⊕ A2 ⊕ (−2). Now, we need a lemma. Lemma 3.13. Let X be an irreducible symplectic manifold of K3[2] -type and G a symplectic automorphism group of order 3 ≤ p ≤ 19. We have AH 2 (X,Z)G = (Z /2 Z) ⊕ (Z /p Z)aG (X) . We denote AH 2 (X,Z)G ,p := (Z /p Z)aG (X) . Then the projection H 2 (X, Z) 2 (X) → AH 2 (X,Z)G ,p H 2 (X, Z)G ⊕ SG is an isomorphism. Moreover, let x ∈ H 2 (X, Z)G and assume xp ∈ (H 2 (X, Z)G )∨ . If xp ∈ AH 2 (X,Z)G ,p then there is z ∈ H 2 (X, Z) such that x = z + ϕ(z) + ... + ϕp−1 (z). 103
Proof. The first assertion follows from Lemma 1.10 and its proof. Now let x ∈ H 2 (X, Z)G such that xp ∈ AH 2 (X,Z)G ,p . By the first assertion, there is z ∈ H 2 (X, Z) and y ∈ SG (X) such that z = x+y p . Then z + ϕ(z) + ... + ϕp−1 (z) = x + y+ϕ(y)+...+ϕ p z + ϕ(z) + ... + ϕp−1 (z) = x.
p−1
(y)
. But y + ϕ(y) + ... + ϕp−1 (y) = 0, so
Here we have AH 2 (X,Z)G ,3 = AU(3) ⊕ AU(3) ⊕ AA2 ⊕ AA2 . Hence by Lemma f2 the minimal 3.13, we have 31 π∗ (U (3)) ⊂ H 2 (M3 , Z). And if we denote by A 2 f primitive overgroup of π∗ (A2 ) in H (M3 , Z), we will have A2 /π∗ (A2 ) = Z /3 Z. We denote (a, b) an integral basis of A2 , with BX (a, a) = BX (b, b) = −2 ∈ AA2 . Then by Lemma 3.13 and Corollary and BX (a, b) = 1. Hence a−b 3 f2 /π∗ (A2 ) is gener3.11, π∗ (a) − π∗ (b) is divisible by 3 in H 2 (M3 , Z), and A ∗ (b) ∗ (b) choose π∗ (a)−π and π∗ (a)−π + π∗ (b) = 3 3 matrix of the sublattice generated by a − b −6 3 A2 (3) = . 3 −6
π∗ (a)−π∗ (b) . Hence we can 3 π∗ (a)+2π∗ (b) f2 . The as a basis of A 3 and a + 2b in A2 is
ated by
By Corollary 3.11 and Lemma 3.13, we have 1 1 H 2 (M3 , Z) = π∗ (U ) ⊕ π∗ (U (3))2 ⊕ π∗ (A2 (3))2 ⊕ π∗ (−2). 3 3 Then by Proposition 3.10, the Beauville–Bogomolov form of H 2 (M3 , Z) gives the lattice s s s s ! ! ! ! 81 81 81 81 1 2 1 2 3 ⊕ U ⊕ A2 3 ⊕ −2 U CM3 3 CM3 3 CM3 CM3 s s s s ! ! ! ! 9 9 9 9 2 2 ⊕U ⊕ A2 ⊕ −6 . =U 3 CM3 CM3 CM3 CM3 It follows that CM3 = 9 and we get the lattice U (3) ⊕ U 2 ⊕ A22 ⊕ (−6).
3.6
Quotient of a K3[2] -type manifold by a symplectic automorphism of order 5
First we prove the H 4 -normality. Corollary 3.14. Let X be a manifold of S [2] -type. Let G be an order 5 group of symplectic automorphisms of X. Then (X, G) is H 4 -normal.
104
Proof. By Theorem 6.2.9 of [21], Fix G consists of 14 isolated points. The local action of G around an isolated fixed point could be of three different types: 1 1 1 5 (1, 1, 4, 4), 5 (1, 1, 1, 2) or 5 (1, 2, 3, 4). By the proof of Proposition 6.2.14 of [21], we have 12 points of type 51 (1, 1, 4, 4), 1 of type 15 (1, 1, 1, 2) and 1 of type 1 5 (1, 2, 3, 4). We will apply Corollary 2.88 to calculate the weight of these points. By Theorem 1.16, H ∗ (X, Z) is torsion-free. Hence, it remains to show that l42 (X) = l44 (X) = 0. By Corollary 7.2.8 of [21], we know that (X, G) can be deformed to a pair (S [2] , ψ [2] ) with ψ, a symplectic automorphism on S. Hence, by Theorem 4.1 of [13], we know that H 2 (X, Z)G ≃ U ⊕ U (5)2 ⊕ (−2), where H 2 (X, Z) is endowed with the Beauville–Bogomolov form. By Lemma 1.10 and Definition-Proposition 1.7, l52 (X) = 4. Since rk H 2 (X, Z) = 23 and rk H 2 (X, Z)G = 7, by Proposition 1.4, rk H 2 (X, Z) − rk H 2 (X, Z)G = 4l52 (X) + l42 (X) = 23 − 7 = 16. So l42 (X) = 0. But we cannot apply Proposition 2.29, Proposition 2.31 and 4 (X,Z) Lemma 2.30 to show that l44 (X) = 0 because by Theorem 1.16, SymH2 H 2 (X,Z) has Z /5 Z-torsion. Therefore, we will have to use the basis of Theorem 1.17 to prove the following Lemma: Lemma 3.15. We have: 1) l14 (X) = 6, 2) l54 (X) = 54, 3) l44 (X) = 0,
Proof. Since (X, G) can be deform to a pair (S [2] , ψ [2] ) with ϕ a symplectic automorphism on S, we calculate l14 (S [2] ), l54 (S [2] ) and l44 (S [2] ). We denote ϕ = ψ [2] . We use notation of Section 1.5.3 and Section 1.5.1. First note that ϕ∗ (q2 (αk ) |0i) = q2 (ψ ∗ αk ) |0i , ϕ∗ (q1 (αk )q1 (αj ) |0i) = q1 (ψ ∗ αk )q1 (ψ ∗ αj ) |0i , ϕ∗ (m1,1 (αk ) |0i) = m1,1 (ψ ∗ αk ) |0i ,
ϕ∗ (q1 (1)q1 (x) |0i) = q1 (1)q1 (x) |0i .
Moreover, ϕ∗ (γk ) = j(ψ ∗ (αk )). Hence if γk ∈ Ni then q2 (αk ) |0i and m1,1 (αk ) |0i are also in Ni , for all i equal to 1, 4 or 5. And q1 (αk )q1 (αj ) |0i are in N1 if and only if γk and γj are in N1 else in N5 . Moreover q1 (1)q1 (x) |0i is in N1 . Hence l44 (X) = 0. Hence, by Theorem 1.17 we have: l14 (X) = 1+(l12 (X)−1)+(l12 (X)−1)+
l2 (X)(l12 (X) + 1) (l12(X) − 1)(l12 (X) − 2) = 1 . 2 2 105
We have seen that l52 (X) = 4 and rk H 2 (X, Z)G = 7. Hence by Proposition 1.4 l12 (X) = 3. Hence we get l14 (X) = 6. Moreover, by Theorem 1.16, we have rk H 4 (S [2] , Z) = 276. Hence by Proposition 1.4, we have l54 (X) = 54. Now, by Corollary 2.88, we have: 6 − 12w( 51 (1, 1, 4, 4)) − w( 15 (1, 1, 1, 2)) − divisible by 2. Hence w( 15 (1, 1, 1, 2)) + w( 51 (1, 2, 3, 4)) is divisible by 2. And w( 15 (1, 2, 3, 4))
1 1 1 6 + 2(3 + 1) ≥ 12w( (1, 1, 4, 4)) + w( (1, 1, 1, 2)) + w( (1, 2, 3, 4)) ≥ 2(3 + 1), 5 5 5 1 1 1 14 ≥ 12w( (1, 1, 4, 4)) + w( (1, 1, 1, 2)) + w( (1, 2, 3, 4)) ≥ 8. 5 5 5 Moreover, by Proposition 2.90, we know that w( 15 (1, 2, 3, 4)) = 1. It follows that w( 15 (1, 1, 4, 4)) = w( 15 (1, 1, 1, 2)) = w( 15 (1, 2, 3, 4)) = 1. And by Corollary 2.83, we conclude that (X, G) is H 4 -normal. The last proof prove also the following Proposition. Proposition 3.16. We have w( 15 (1, 1, 4, 4)) = w( 15 (1, 1, 1, 2)) = 1. Now, we prove the H 2 -normality. Corollary 3.17. Let X be a manifold of S [2] -type. Let G be an order 5 group of symplectic automorphisms of X. Then (X, G) is H 2 -normal.
Proof. By Corollary 7.2.8 of [21], we can assume that (X, G) = (S [2] , ψ [2] ) with ψ a symplectic automorphism on S. Here we cannot apply Proposition 2.28, Proposition 2.31 to prove the H 2 -normal from the H 4 -normal because by H 4 (X,Z) Theorem 1.16, Sym 2 H 2 (X,Z) has Z /5 Z-torsion. Hence, we have to use one more time the basis of Theorem 1.17. By Theorem 4.1 of [13], we know that there is an isometry: H 2 (S [2] , Z)G = U ⊕ U (5)2 ⊕ (−2). By Lemma 3.13, π∗ (U (5)2 ) is divisible by 5. Hence by Definition-Proposition 2.15, to finish the proof, we have to show that π∗ (U ⊕ (−2)) is primitive in H 2 (M5 , Z), where M5 := X/G. To show this, we consider the sublattice of H 4 (S [2] , Z)G : S = Sym2 (U ⊕ (−2)) ⊕ Z q1 (1)q1 (x) |0i . By Proposition 1.24 and Proposition 1.19, the matrix of this lattice is 12 −2 1 −2 2 1 ⊕ U (2)2 . 1 1 1
Hence its discriminant is 48. Since 5 does not divide 48, the only elements in S of the form y + ϕ(y) + ... + ϕ4 (y) are the elements divisible by 5. 106
Now let u ∈ U ⊕ (−2) be such that π∗ (u) is divisible by 5. We will show that u is divisible by 5. Hence by Lemma 2.18 1), π∗ (u2 ) is divisible by 5. But since (X, G) is H 4 -normal, by Definition-Proposition 2.15, there is y ∈ H 4 (X, Z) such that u2 = y + ϕ(y) + ... + ϕ4 (y). But we have u2 ∈ S, we have seen that in this case, u2 is divisible by 5. We will see that this implies that u is divisible by 5. We denote (u1 , u2 , δ) a basis of U ⊕(−2), with BS [2] (u1 , u1 ) = BS [2] (u2 , u2 ) = BS [2] (u1 , δ) = BS [2] (u2 , δ) = 0, BS [2] (u1 , u2 ) = 1 and BS [2] (δ, δ) = −2. We can write u = au1 + bu2 + cδ, then u2 = a2 u21 + 2abu1 · u2 + b2 u22 + c2 δ 2 + 2acu1 · δ + 2bcu2 · δ. By taking the cup product of u2 with u21 and u22 , we see by Proposition 1.24 that a2 and b2 have to be divisible by 5. Hence a and b are divisible by 5. And taking the cup product of u2 by δ 2 , we conclude by Proposition 1.24 that c2 has to be divisible by 5. So u is divisible by 5. This proves the corollary. Theorem 3.18. Let X be a manifold of S [2] -type. Let G be an order 5 group of symplectic automorphisms of X. We denote M5 = X/G. Then the Beauville– Bogomolov lattice H 2 (M5 , Z) is isomorphic to U (5)⊕ U 2 ⊕ (−10), and the Fujiki constant of M5 is 15. Proof. By Corollary 7.2.8 of [21], we can assume that there is a K3 surface S such that X = S [2] and G is a natural automorphism group. By Theorem 4.1 of [13], there is an isometry of lattices H 2 (S [2] , Z)G ≃ U ⊕ U (5)2 ⊕ (−2). Here we have ATG (X),5 = A2U(5) . Hence by Lemma 3.13, we have 15 π∗ (U (5)) ⊂ H 2 (M5 , Z). It follows by Corollary 3.17 that 1 H 2 (M5 , Z) = π∗ (U ) ⊕ π∗ (U (5))2 ⊕ π∗ (−2). 5 Then by Proposition 3.10, the Beauville–Bogomolov form on H 2 (M5 , Z) gives the lattice s s s 1 3 · 53 1 3 · 53 3 · 53 2 U⊕ U (5) ⊕ (−2) 25 CM5 25 CM5 CM5 s s s 15 15 2 15 = U (5) ⊕ U ⊕ (−10). CM5 CM5 CM5 It follows that CM5 = 15 and we get the lattice U (5) ⊕ U 2 ⊕ (−10).
3.7
Quotient of a K3[2] -type manifold by an automorphism of order 11
There are two different examples of automorphisms of order 11 on a manifold of K3[2] -type (Example 4.5.1 and Example 4.5.2 in [21]). 107
In both examples the fixed locus is a set of 5 isolated points. But as it 2 G is explained in Section 7.4.4 of [21], the lattice H (X, Z) will be different in 6 2 2 these cases. In the first case the lattice is 2 8 −3, and in the second 2 −3 8 2 1 0 1 6 0 . Moreover, it is proved in Section 7.4.4 of [21] that these are 0 0 22 the only possible invariant lattices for p = 11. Hence we obtain the following corollary. Corollary 3.19. Let X be a manifold of S [2] -type. Let G be a symplectic automorphism group of X of order 11. Then (X, G) is H 2 -normal and H 4 normal. Proof. In the both cases, we have discr H n (X, Z)G = 112 . Hence by Proposition 2.19 1), anG (X) = 2. Hence anG (X) = rk H n (X, Z)G − 1. It follows by Proposition 2.25 that (X, G) is H 4 -normal. Then by Proposition 2.28, Proposition 2.31 and Theorem 1.16, (X, G) is H 2 -normal. Let X be a manifold of S [2] -type and G a symplectic 6 2 2 G of X of order 11. If H (X, Z) is isomorphic to 2 8 2 −3
automorphism group 2 −3, we will denote 8 2 1 0 1 the quotient X/G by M11 , and if H 2 (X, Z)G is isomorphic to 1 6 0 , we 0 0 22 2 will denote the quotient X/G by M11 . We have the following theorem.
Theorem 3.20. Let X be a manifold of S [2] -type. Let G be a symplectic automorphism group of order 11, acting on X, of one of the above two types. Then the Fujiki constant is 33 in both cases and the Beauville–Bogomolov lattices are: 2 3 −8 2 −9 0 1 2 6 −16 , H 2 (M11 H 2 (M11 , Z) ≃ 3 , Z) ≃ −9 46 0 . −8 −16 50 0 0 2
Proof. Let
Λ111
6 := 2 2
2 2 8 −3 , −3 8
g 1 the minimal primitive overlattice of π (Λ1 ) in H 2 (M 1 , and denote by Λ ∗ 11 11 11 1 Z) (here it is H 2 (M11 , Z)/ tors). Then by Lemma 3.13 and Corollary 3.19, g 1 /π (Λ1 ) = (Z /11 Z)2 . We denote by (a, b, c) an integral basis of Λ1 with Λ ∗ 11 11 11 BX (a, a) = 6, BX (b, b) = 8, BX (c, c) = 8, BX (a, b) = BX (a, c) = 2 and g 1 we can choose BX (b, c) = −3. By Lemma 3.13, for an integral basis of Λ 11 108
∗ (c) π∗ (a)−3π∗ (c) ∗ (c) ( π∗ (b)−π , , π∗ (a)+8π ). The matrix of the sublattice of Λ111 gen11 11 11 erated by b − c, a − 3c and a + 8c is 2 × 11 3 × 11 −8 × 11 3 × 11 6 × 11 −16 × 11 . −8 × 11 −16 × 11 50 × 11
1 Then by Proposition 3.10, the Beauville–Bogomolov form on H 2 (M11 , Z) gives the lattice s 2 3 −8 1 3 · 113 3 6 −16 . 1 11 CM11 −8 −16 50
1 = 33, and we get the lattice It follows that CM11
2 3 −8 3 6 −16 . −8 −16 50
Now let
Λ211
2 := 1 0
1 0 6 0 , 0 22
g 2 the minimal primitive overgroup of π (Λ2 ) in H 2 (M 2 , and denote by Λ ∗ 11 11 11 2 2 Z) (here it is H (M11 , Z)/ tors). Then by Lemma 3.13 and Corollary 3.19, g 2 /π (Λ2 ) = (Z /11 Z)2 . We denote by (a, b, c) an integral basis of Λ2 with Λ ∗ 11 11 11 BX (a, a) = 2, BX (b, b) = 6, BX (c, c) = 22, BX (b, c) = BX (a, c) = 0 and g 2 we can choose BX (a, b) = 1. Then by Lemma 3.13, for an integral basis of Λ 11 π∗ (a)−2π∗ (b) π∗ (a)+9π∗ (b) π∗ (c) ( , , ). The matrix of the sublattice Λ211 generated 11 11 11 by a − 2b, a + 9b and c is 2 × 11 −9 × 11 0 −9 × 11 46 × 11 0 . 0 0 2 × 11 2 Then by Proposition 3.10, the Beauville–Bogomolov form on H 2 (M11 , Z) gives the lattice s 2 −9 0 1 3 · 113 −9 46 0 . 2 11 CM11 0 0 2
1 = 33, and we get the lattice It follows that CM11
2 −9 0 −9 46 0 . 0 0 2
109
3.8
Quotient of K3[2] by a natural non-symplectic automorphism of order 3
Lemma 3.21. Let S be a K3 surface and G = hψi a non-symplectic automorphisms group of order 3 on S with a finite fixed locus. The group G induces a natural non-symplectic automorphism group of order 3 on S [2] , G with a finite fixed locus. We have: # Fix G = 9. Proof. By Theorem 2.2 of [1], we have # Fix G = 3. We deduce # Fix G. Let ξ ∈ S [2] with Supp ξ = {x1 , x2 }. The point ξ is fixed by G if and only if {x1 , x2 } = {ψ(x1 ), ψ(x2 )}. This holds if and only if x1 = ψ(x1 ) and x2 = ψ(x2 ) or x1 = ψ(x2 ) and x2 = ψ(x1 ). Since ψ has order 3, the second case is possible only if x1 = x2 . It follows that ξ is fixed by G if and only if x1 = ψ(x1 ) and x2 = ψ(x 2 ). If x1 6= x2 , there are 3 such points. Let x ∈ Fix G, the group G acts on ξ ∈ S [2] | Supp ξ = {x} ≃ P1 and necessarily has 2 fixed points. It follows that # Fix G = 3 + 2 × 3 = 9. Corollary 3.22. Let S be a K3 surface and G a natural non-symplectic group of automorphisms of X = S [2] of order 3. We assume that Fix G is finite. Then (X, G) is H 4 -normal and H 2 -normal. Proof. By Proposition 2.28, Proposition 2.31, and Theorem 1.16, if (X, G) is H 4 -normal, then (X, G) is H 2 -normal. Hence we have just to show the H 4 normality. We will apply Corollary 2.93. By Theorem 1.16, H ∗ (X, Z) is torsion-free. It remains to show that l22 (X) = l24 (X) = 0. Since G is natural, by Table 2 of [1] we know that H 2 (X, Z)G ≃ U (3) ⊕ ∨ E6 (3) ⊕ (−2) where H 2 (X, Z) is endowed with the Beauville–Bogomolov form. Hence, by Lemma 1.10 and Definition-Proposition 1.7, l32 (X) = 7. Since rk H 2 (X, Z) = 23 and rk H 2 (X, Z)G = 9, by Proposition 1.4, rk H 2 (X, Z) − rk H 2 (X, Z)G = 2l32 (X) + l22 (X) = 23 − 9 = 14. Therefore, l22 (X) = 0. Then, by Proposition 2.29, Proposition 2.31, Theorem 1.16 and Lemma 2.30, l24 (X) = 0. Now, we show that l14 (X) + 2 1 + l21 (X) + l23 (X) + l12 (X) = h2∗ (Fix G, Z).
Since H odd (X, Z) = 0 (Theorem 1.16) and by Lemma 3.21, Fix G consists of 9 isolated points, we just have to show that l14 (X) + 2(1 + l12 (X)) = 9.
(53)
It remains to calculate l14 (X) and l12 (X). By Proposition 1.4, rk H 2 (X, Z)G = l12 (X) + l32 (X). Since rk H 2 (X, Z)G = 9 and l32 (X) = 7, we get l12 (X) = 2. Now by Proposition 2.29, Proposition 2.31, Theorem 1.16, and Lemma 2.30, l14 (X) = 2×3 2 = 3. Hence we get (53). We conclude with Corollary 2.93. 110
3.9 3.9.1
Summary Singular symplectic surfaces
Let Yp be the quotient of a K3 surface by a symplectic automorphism of order p. The surfaces Y p are simply connected by Lemma 1.2 of [10]. Hence they are singular irreducible symplectic surfaces. Proposition 3.23. We have H 1 (A, C) = H 3 (A, C) = 0. Proof. Indeed (− id)∗ acts as − id on H 1 (X, C) and on H 3 (X, C). The following table summarizes our results:
3.9.2
X/G Y2 Y3 Y5
b2 14 10 6
χ 16 12 8
Y7
4
6
A
6
8
(H 2 (X/G, Z), ·) E8 (−1) ⊕ U (2)3 U (3) ⊕ U 2 ⊕ A22 U (5) ⊕ U 2 4 −3 U⊕ −3 4 U (2)3
Non-symplectic surfaces
Let Zp be the quotient of a K3 surface by a non-symplectic automorphism of order p with only isolated fixed points. The surfaces Z p are simply connected by Lemma 1.2 of [10]. The following table summarizes our results:
with L17
3.9.3
X/G Z3
b2 8
χ 10
Z5
6
8
Z7
4
6
Z 11 Z 17
2 6
4 8
Z 19
4
6
−2 1 0 1 1 −2 0 0 = 0 0 −2 1 1 0 1 −4
(H 2 (S/G, Z), ·) U ⊕ E6 −2 −5 ⊕ A4 −5 −10 −4 3 U⊕ 3 −4 U U (17)⊕ L∨ 17 (17) −10 9 U (19) ⊕ 9 −10 2 −1 0 0 −1 2 −1 0 and A4 = 0 −1 2 −1. 0 0 −1 2
Singular irreducible symplectic fourfold
For the sake of completeness, we will also give the fourth Betti number b4 and the topological Euler number χ for all the varieties that we have studied.
111
Lemma 3.24. Let X be an irreducible symplectic manifold deformation equivalent to S [2] and let G be a group of automorphisms of X of prime order with 3 ≤ p ≤ 19, p 6= 5 and m2G (X) = a2G (X). Let r = rk H 2 (X, Z)G . Then: • b2 (X/G) = r, • b3 (X/G) = 0, • b4 (X/G) = • χ(X/G) =
r(r+1) 2
(23−r)2 2(p−1) ,
+
(r+4)(r+1) 2
+
(23−r)2 2(p−1) .
Proof. • By Proposition 5.1 of [6], we know that r = lp2 (X) + l12 (X) and 4 rk H (X, Z)G = lp4 (X) + l14 (X). And by Corollary 5.10 of [6], we have lp2 (X) = 23−r p−1 . Hence by Proposition 6.6 and Lemma 6.14 of [6], we can calculate lp4 (X) and l14 (X). We get: lp4 (X)
+
l14 (X)
2 2 lp (X) r(r + 1) + (p − 1) . = 2 2
Then: rk H 4 (X, Z)G =
r(r + 1) (23 − r)2 + . 2 2(p − 1)
• By Poincaré duality, the Euler number of X/G is equal to χ(X/G) = 2 + 2 rk H 2 (X, Z)G + rk H 4 (X, Z)G . Hence χ(X/G) = 2 + 2r +
r(r + 1) (23 − r)2 + . 2 2(p − 1)
Lemma 3.25. Let X be an irreducible symplectic manifold of S [2] -type. Let G be a group of symplectic automorphisms of X of order 5. Then b4 (X/G) = 60 and χ(X/G) = 76. Proof. Follows from Lemma 3.15. The following table summarizes our results: X/G M′ M3 M5
b2 16 11 7
b4 178 102 60
χ 212 126 76
CX/G 6 9 15
1 M11
3
26
34
33
2 M11
3
26
34
33
112
BX/G E8 (−1) ⊕ U (2)3 ⊕ (−2)2 U (3) ⊕ U 2 ⊕ A22 ⊕ (−6) U (5) ⊕ U 2 ⊕ (−10) 2 3 −8 3 6 −16 −8 −16 50 2 −9 ⊕ (2) −9 46
Glossary akG (X): Definition-Proposition 1.7 and Lemma 2.20 Coefficient of normality: Definition-Proposition 2.15 E2 -degenerate, E2 -degenerate over Z: Definition 2.7 h2*+ǫ (X, Z ): Notation 2.1 H k -normal, H ∗ -normal: Definition-Proposition 2.15 lji (X), lk± (X): Definition-Proposition 1.5 mkG (X): Definition-Proposition 1.7 Natural automorphism: Definition 1.13 Negligible, almost negligible (fixed locus): Definition 2.44 Numerically standard pair: Definition 1.14 Pull back, k-split pullback, regular k-split pullback (of (X, G)): Definition 2.32 Stable fixed locus, (almost) stable fixed (point, curve): Definition 2.64 Standard pair: Definition 1.13 Toric blow-up: Section 2.4.2 Type (of a fixed point): Definition 2.38 U -trivial: Definition 2.56 Weight (of a fixed point): Definition 2.77
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