Divisor class groups of affine complete intersections
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1 Divisor class groups of affine complete intersections Lambrecht, 2015
2 Introduction Let X be a normal complex algebraic variety. Then we can look at the group of divisor classes, more precisely Weil divisor classes: Cl X, Cartier divisor classes, i.e. Picard group Pic X. If X is smooth they coincide. The case X affine is particularly interesting: Let O(X ) be the coordinate ring of X. O(X ) is factorial if and only if Cl X is trivial, O(X ) is almost factorial if and only if Cl X is a torsion group. Definition An integral domain R is called almost factorial if for each x R \ R {0} there is an n > 0 such that x n can be written as a product of prime elements.
3 Concrete results: in the case of affine complete intersections. We shall use topological results from singularity theory, and line bundles with connection, mixed Hodge structure or Deligne-Beilinson cohomology. Recall Pic X H 1 (X, OX ). Start with X smooth and compact. GAGA allows to switch to analytic context: exponential sequence H 1 (X an ; Z) H 1 (X, O X ) H 1 (X, OX ) Also Hodge theory is useful. Recall Deligne cohomology: Let Z(p) := (2πi) p Z and Z(p) D be the complex Z(p) X an Ω 0 X... Ω p 1 an X 0... an Then HD k (X an ; Z(p)) := H k (X an ; Z(p) D ).
4 H 2 D (X an ; Z(1)) Pic X, H 2 D (X an ; Z(p)) Pic ci X for p > 1. Here Pic ci X refers to line bundles with integrable connection. This leads to an exact sequence H 0 (X, Ω 1 X ) Pic ci X Pic X H 1 (X, Ω 1 X ). Note: Pic ci (X ) H 1 (X an ; C ).
5 Noncompact case Now drop the condition that X is compact. Use good compactification X (i.e. X smooth, D := X \ X is a divisor with normal crossings). Similarly as above, we can use (generalized) Hodge theory, Deligne-Beilinson cohomology, or comparison with Pic cir X, where Pic cir refers to regular integrable connections. More precisely: we have an exact sequence H 0 (X, Ω 1 X (log D)) Pic cir (X ) Pic X H 1 (X, Ω 1 X (log D)) cf. H.-Lê (in preparation).
6 The Deligne-Beilinson cohomology can be defined as follows: HDB k (X, Z(p)) := Hk ( X an, Z(p) DB ), where Z(p) DB := cone(rj an Z(p) X an Ω p 1 (log D))[ 1]. X an Here j : X X is the inclusion. Theorem Pic X = ker(h 2 DB (X, Z(1)) H3 ( X an, X an ; Z)) (H., 2015). This is plausible because of the exact sequence H 2 ( X an, X an ; Z) HDB 2 ( X, Z(1)) HDB 2 (X, Z(1)) H 3 ( X an, X an ; Z) compared with Z r Pic X Pic X 0 where r = number of irreducible components of D.
7 Main results Theorem Let X be a smooth complex algebraic variety. a) b 1 (X ) = 0 Pic 0 X = 0, b) b 1 (X ) = b 2 (X ) = 0 Pic X H 2 (X an ; Z), c) If H 1 (X an ; Z) = 0, Pic X is free of rank b 2 (X ), d) H 1 (X an ; Z) = 0, b 2 (X ) = 0 Pic X = 0.
8 Proof: a) b 1 (X ) = 0 Pic 0 X = 0: different possibilities: Let X be as above. Then we have an exact sequence, by H.-Lê: H 0 ( X, Ω 1 X (log D)) Pic cir (X ) Pic(X ) c 1 H 2 (X an ; C) The group on the left vanishes. Suppose that [L] Pic 0 (X ). By the sequence above it has an inverse image of the form [L, ] in Pic cir (X ). But Pic cir (X ) Pic ci (X an ) H 1 (X an ; C ). Look at H 1 (X an ; C) H 1 (X an ; C ) c 1 H 2 (X an ; Z). So [L, ] has an inverse image in H 1 (X an ; C) but this group vanishes: b 1 (X ) = 0. So [L, ] = 0, hence [L] = 0.
9 Other way to prove b 1 (X ) = 0 Pic 0 X = 0: apply a result of H.-Lê (2005), namely Pic 0 X = 0 Gr W 1 H1 (X an ; Q) = 0. Or: use Deligne-Beilinson cohomology. We have an exact sequence H 1 ( X, O X ) H2 DB (X, Z(1)) H2 (X an ; Z) The left group vanishes: b 1 (X ) = 0. Finally Pic 0 (X ) = ker(h 2 DB (X, Z(1)) H2 (X an ; Z)). b) Analogously we have three different possibilities to prove this. c), d) straightforward.
10 Application: Let f 1,..., f k C[z 1,..., z n+k ] be polynomials which are weighted homogeneous of degree d 1,..., d k with respect to weights w 1,..., w n+k. We assume that X := {f 1 =... = f k 1 = 0} is a complete intersection with isolated singularity and that f k X has an isolated singularity at 0, too. Put X t := X f 1 k ({t}). Theorem Assume t 0. O(X t ) is factorial if n 3, Pic X t is free of finite rank if n = 2. Proof: X t has the homotopy type of the Milnor fibre of f k X, so of a bouquet of n-spheres.
11 Now let t = 0. Assume n 2, then X 0 is normal. Let K be the link of X 0 (with respect to 0), it is a deformation retract of X an 0 \ {0}. Theorem O(X 0 ) is factorial if n 4. Cl X 0 is free of rank b 2 (K) if n = 3. In particular, O(X 0 ) is factorial if n = 3 and b 2 (K) = 0, i.e. K is a 5-dimensional rational homology sphere. n = 2, b 1 (K) = 0 implies that Cl(X 0 ) H 1 (K; Z). In particular, O(X 0 ) is factorial (resp. almost factorial) if K is an integral (resp. rational) homology sphere. Proof: This follows because K is n 2-connected. For n = 2, b 1 (K) = b 2 (K) because of Poincaré duality.
12 Example Let f j be a linear combination of z a 1 1,..., za n+k n+k, j = 1,..., k, with general coefficients. So we have weighted homogeneous polynomials with same degree. Following Brieskorn, associate a graph with vertices 1,..., n + k: Join i and j by an edge if a i, a j are not coprime. For any component C let (B) be the following condition: either C consists of a singe vertex, or C has an odd number of vertices, and for any two vertices i j in C we have gcd(a i, a j ) = 2. By a topological theorem (H., 1972) we deduce: If n = 3 and there are k components with condition (B) the ring O(X 0 ) is factorial. If n = 2 and there are k (resp. k + 1) components with condition (B) the ring O(X 0 ) is almost factorial (resp. factorial).
13 Homogeneous case From now on assume that the polynomials are even homogeneous. This case is easier: Then the closure X t of X t in P n+k is smooth, and the part at infinity is a connected smooth divisor (n 2). So we have an exact sequence 0 Z Cl X t Cl X t 0. So apply known results about divisor class groups in the compact case. This yields another way to find factorial coordinate rings:
14 Example Let p be prime 5, f (z) := z 1 z p z 2 z p z 3 z p 1 1. Put X t := {f = t}. Then for t 0: O(X t ) is factorial. This is because the Picard number of X t is 1 according to Shioda (1981). Note that we may not apply our previous result: Milnor number 0.
15 Picard group of X 0 Note that X 0 is not smooth, so we cannot expect that Weil divisors are Cartier divisors. Then Theorem Pic X 0 = 0 Cf. Fossum (1973). THANK YOU FOR YOUR ATTENTION!
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