The Fundamental Group of Some Siegel Modular Threefolds

9 Pages·0.1 MB·English

by  K. Hulek

The Fundamental Group of Some Siegel
Modular Threefolds
Klaus Hulek and G. K. Sankaran
4
9
9
1 The moduli space A1,p of abelian surfaces with a polarisation of type (1,p) and a level
structure, for p an odd prime, is a singular quasi-projective variety. It has been studied
n
a in detail in [HKW1]. In the present note we prove the following.
J

8
Theorem Let X be any desingularisation of an algebraic compactification of A for p
1 1,p
an odd prime. Then X is simply connected.

1
v In fact it is enough to prove this for one such X, since any two X have the same funda-
3
0 mental group.
0 If p = 5 or 7 then X is known to be rational and therefore simply connected. For p = 5
1
this goes back to [HM]; for p = 7 it has recently been shown in [MS] that X is birationally
0
4 equivalent to a Fano variety of type V , which is known to be rational. It does not seem
22
9
to be known whether X is rational if p = 3, but it is shown in [HS] that if p is large then
/
m
X is of general type.
o
e
g
1 Generalities about fundamental groups
-
g
l
a
The general facts in this section are all well known.
:
v
i
X Lemma 1.1 Let M be a connected simply connected real manifold and G a group acting
r discontinuouslyon M. Let x ∈ M be a base point. Then the quotient map ϕ : M −→ M/G
a
induces a map ψ : G −→ π (M/G,ϕ(x)) which is surjective.
1
Note that we do not require the action of G to be free or even effective. The map ψ is
defined as follows. If g ∈ G let θ : [0,1] −→ M be any path in M from x to g(x) : that
g
is, θ (0) = x and θ (1) = g(x). Then ϕ ◦ θ : [0,1] −→ M/G is a closed loop in M/G,
g g g
with ϕ◦θ (0) = ϕ◦θ (1) = ϕ(x). We put ψ(g) = [ϕ◦θ ] ∈ π (M/G,ϕ(x)). One checks
g g g 1
easily that ψ is well defined and that it is a group homomorphism. It is shown in [G] that
ψ is surjective.
Lemma 1.2 Let M be a connected complex manifold and M an analytic subvariety.
0
Then the inclusion induces a map π (M \M ,x) −→ π (M,x) for any base point x ∈ M
1 0 1
which is surjective for codimCM0 ≥ 1 and an isomorphism for codimCM0 ≥ 2.
Lemma 1.3 If X and X are birationally equivalent smooth projective varieties then
1 2
∼
π (X ) = π (X ).
1 1 1 2
Proofs of both these lemmas can be found in [HK].
1
2 Applications to the case of A
1,p
We fix some notation. S is the Siegel upper half-plane of degree 2. We denote by A
2 1,p
the moduli space of abelian surfaces with a (1,p)-polarisation and a level structure, and
by A(p2) the moduli space of abelian surfaces of level p2. We let Γ and Γ(p2) be the
1,p
corresponding arithmetic subgroups of Sp(4,Z), so
Z Z Z pZ
  pZ pZ pZ p2Z 
Γ = γ ∈ Sp(4,Z)|γ −1 ∈
1,p  Z Z Z pZ 
 
 
 Z Z Z pZ 
 
 
 
 
(see [HKW1]) and Γ(p2) is the principal congruence subgroup of level p2. We denote by
Γ0 the arithmetic subgroup of Sp(4,Q)
1,p
Z Z Z pZ
  pZ Z pZ pZ 
Γ0 = γ ∈ Sp(4,Q)|γ ∈
1,p  Z Z Z pZ 
 
 
 Z 1Z Z Z 
 p 
 
 
(in terms of moduli, this corresponds to keeping the (1,p)-polarisation but forgetting the
level structure). We denote by A∗ the toroidal compactification of A described in
1,p 1,p
[HKW1], and by A∗(p2) the usual toroidal (or Igusa) compactification of A(p2). Since
Γ(p2) is neat, A∗(p2) is nonsingular: indeed, any nonsingular compactification of A(p2)
will serve our purpose.
Let A′ be the smooth part of A . The space A is a quotient of S by Γ and A′
1,p 1,p 1,p 2 1,p 1,p
is the quotient by Γ of an open part S′ ⊂ S . Since the singular locus of A has
1,p 2 2 1,p
codimension 2,S′ is the complement of an analytic subset of codimension 2 in S . In
2 2
particular, by Lemma 1.2, S′ is simply connected.
2
Remark. The quotient map S′ −→ A′ = S′/Γ is not unramified. The ramification
2 1,p 2 1,p
is in codimension 1 but the isotropy groups are generated by reflections so the quotient
is smooth. If we restrict to the complement of the ramification locus our covering space
will not necessarily be simply connected.
Proposition 2.1 Let f : X −→ A∗ be a resolution of singularities (so X is smooth and
1,p
f is a birational morphism which is an isomorphism away from f−1(SingA∗ )). Then
1,p
there are surjective homomorphisms Γ −→ π (A′ ) and π (A′ ) −→ π (X).
1,p 1 1,p 1 1,p 1
Proof. The map Γ −→ π (A′ ) exists and is surjective by Lemma 1.1. The map
1,p 1 1,p
π (A′ ) −→ π (X) comes from Lemma 1.2: we may consider A′ as a subset of X via
1 1,p 1 1,p
f−1, which exists on A′ . 2
1,p
We now choose a convenient resolution of singularities to work with. Γ is a normal
1,p
subgroup of Γ0 and the quotient (which is isomorphic to SL(2,F ) as an abstract group)
1,p p
acts on A∗ (this is shown in [HKW1]). By the general results of Hironaka ([H]) or by an
1,p
easy explicit construction we can choose f : X −→ A∗ to be an equivariant resolution,
1,p
so that Γ0 /Γ acts on X. Henceforth X will always be such a resolution.
1,p 1,p
Let ψ : Γ −→ π (X) be the composite of the two maps in Proposition 2.1. Then ψ is
1,p 1
surjective. We shall show that Kerψ = Γ and hence π (X) = 1.
1,p 1
2
3 Level p2
In this section we use results of Kn¨oller ([K]) to prove that Γ(p2), which is a subgroup of
Γ , lies in the kernel of ψ.
1,p
Since Γ(p2) is a normal subgroup of Γ , there is an action of Γ /Γ(p2) on A(p2) and the
1,p 1,p
quotient is A . The quotient map induces a rational map h : A∗(p2)−− → X, which is
1,p
a morphism over A′ . We can resolve the singularities of this map so as to get a diagram
1,p
Y
σ HHH h˜
H
H
? Hj
h
A∗(p2) -X
where Y is smooth, σ is an isomorphism over A′(p2) = h−1(A′ ) and h˜ is a morphism.
1,p
In particular we can think of A′(p2) as a subset of Y and A′ as a subset of X, and then
1,p
h˜|A′(p2) is the quotient map A′(p2) −→ A′1,p.
Theorem 3.1 (Kn¨oller [K]) Y is simply connected.
Corollary 3.2 Γ(p2) is contained in Kerψ.
Proof. The quotient map q : S′ −→ A′(p2) induces a surjection Γ(p2) −→ π (Y). Let
2 1
M ∈ Γ(p2), let x ∈ S′ be a base point and let θ be a path from x to M(x) in S′.
2 M 2
Then there is a null homotopy H : [0,1]×[0,1] −→ Y such that H (0,t) = qθ (t) and
M M M
˜ ˜ ˜
H (1,t) = q(x) ∈ Y. So hH is a null homotopy of hqθ in X. But [hqθ ] is the class
M M M M
ψ(M) ∈ π (X), since h˜q is the same as the quotient map S′ −→ A′ . 2
1 2 1,p
4 Another element of the kernel
Proposition 4.1 The element
1 0 1 0
 0 1 0 0 
M =
0 0 0 1 0
 
 0 0 0 1 
 
of Γ is in Kerψ.
1,p
Proof. We follow the procedure of [K], and choose a path θ : [0,1] −→ S , namely
M0 2
θ (t) = ic(1−t)1+ictM (1)
M0 0
where c ∈ R is a positive constant to be specified later. θ gives rise to a closed loop
M0
in A , starting and finishing at ic1 (so when we specify c we are just choosing a base
1,p
point).
3
The map e : S −→ C∗ ×C×S, corresponding to the central boundary component D
l0 2 l0
∗
of A (see [HKW1]) is given by
1,p
τ τ
e : 1 2 7−→ (e2πiτ1,τ ,τ ).
l0 (cid:18) τ τ (cid:19) 2 3
2 3
So e θ (t) = (e−2πce−2πit,0,ic). The homotopy
l0 M0
H(s,t) = ((se2πit +1−s)e−2πc,0,ic)
is evidently a null homotopy of e θ in C×C×S . As in [K], we let V be the interior of
l0 M0 1
the closure of S /P in C×C×S : then the image of H lies in V. But by [HKW2], A∗
2 l0 1 1,p
is nonsingular near D : the map V −→ A∗ is branched but in some open set U ⊂ V
l0 1,p
the branching comes from reflections. If we choose c sufficiently large then the image of
H will lie entirely in U, and H will therefore give rise to a null homotopy which does
∗
not meet the singular locus of A . In particular, the loop in A corresponding to θ
1,p 1,p M0
actually lies in A′ , and θ is actually a path in S′. So ψ(M ) is trivial (it is even trivial
1,p M0 2 0
as an element of π (A∗ \Sing A∗ )). 2
1 1,p 1,p
Now we know that Kerψ contains M and all of Γ(p2). It is also, of course, a normal
0
subgroupofΓ :butbecausewecouldchooseX tohaveanactionofΓ0 /Γ itisactually
1,p 1,p 1,p
normal as a subgroup of Γ0 , since Γ0 acts on π (X) = Γ /Kerψ by conjugation in
1,p 1,p 1 1,p
Sp(4,Q).
5 Calculations in Γ
1,p
The remark at the end of section 4, above, shows that to prove the theorem it is enough
to check that any normal subgroup of Γ0 containing Γ(p2) and M must contain Γ .
1,p 0 1,p
We shall give a set of generators for Γ and show how to produce each one, starting with
1,p
M and Γ(p2) and using multiplication, inversion, and conjugation by elements of Γ0 . A
0 1,p
similar procedure is carried out by Mennicke [M] for principal congruence subgroups of
Sp(2n,Z).
Proposition 5.1 Γ is generated by the elements M ,M ,M and M together with the
1,p 1 2 3 4
subgroups j (SL(2,Z)) and j (Γ (p)), where
1 2 1
1 0 0 0 1 0 0 p
 0 1 0 0   0 1 p 0 
M = M =
1 0 1 1 0 2 0 0 1 0
   
 1 0 0 0   0 0 0 1 
   
1 1 0 0 1 0 0 0
 0 1 0 0   −p 1 0 0 
M = M =
3 0 0 1 0 4 0 0 1 p
   
 0 0 −1 1   0 0 0 1 
   
a 0 b 0
 0 1 0 0  a b 
j (SL(2,Z)) = | ∈ SL(2,Z)
1  c 0 d 0 (cid:18) c d (cid:19) 
 
 
 0 0 0 1 
 
 
 
 
4
1 0 0 0
 0 a 0 pb  a b 
j (Γ (p)) = | ∈ Γ (p) .
2 1  0 0 1 0 (cid:18) c d (cid:19) 1 
 
 
 0 c/p 0 d 
 
 
 
Remark. Of course one could find a finite set of generations for Γ by doing so for
1,p
SL(2,Z) and for Γ (p).
1
Proof. It is easier to work, as in Chapter 1 of [HKW1], with
∗ ∗ ∗ ∗
Γ˜ = γ ∈ Sp(Λ,Z)|γ ≡  0 1 0 0 mod p
1,p  ∗ ∗ ∗ ∗ 
 
 
 0 0 0 1 
   
 
 
where Λ is the symplectic form
1 0
0
 0 p 
.
−1 0
 0 
 0 −p 
 
Γ˜ is an Sp(4,Q)-conjugate of Γ . More precisely
1,p 1,p
Γ˜ = RΓ R−1
1,p 1,p
where
1
 1 
R = .
1
 
 p 
 
Thus we wish to show that Γ˜ is generated by
1,p
1 0 0 0 1 0 0 1
˜  0 1 0 0  ˜  0 1 p 0 
M = , M = ,
1 0 1 1 0 2 0 0 1 0
   
 p 0 0 1   0 0 0 1 
   
1 1 0 0 1 0 0 0
M˜ =  0 1 0 0 , M˜ =  −p 1 0 0 ,
3 0 0 1 0 4 0 0 1 1
   
 0 0 −p 1   0 0 0 1 
   
a 0 b 0
 0 1 0 0  a b 
˜ (SL(2,Z)) = | ∈ SL(2,Z)
1  c 0 d 0 (cid:18) c d (cid:19) 
 
 0 0 0 1 
 
 
 
and  
1 0 0 0
 0 a 0 b  a b 
˜ (Γ (p)) = | ∈ Γ (p) .
2 1  0 0 1 0 (cid:18) c d (cid:19) 1 
 
 
 0 c 0 d 
 
 
 
 
5
Recall that a vector v ∈ Z4 is short if and only if there exists w ∈ Z4 such that vΛtw = 1 :
otherwise it is long. If K ∈ Γ˜ then the second row of K is certainly long. Suppose
1,p
the first row is long also. Then, since the first two rows of K span a Λ-isotropic plane,
K ≡ 0 mod p. But then
14
0 ∗ 0 0
 0 1 0 0 
K ≡ mod p
∗ ∗ ∗ ∗
 
 ∗ ∗ ∗ ∗ 
 
so K is not invertible mod p. But K ∈ Sp(4,Z) and in particular K is invertible.
So the first row of K must be short. It is shown in the course of the proof of Proposition
3.38 in [HKW1] that, by multiplying K on the right by a suitable product of M˜ and
i
elements of ˜ (SL(2,Z)), we may assume that the first row of K is (1,0,0,0). Then the
1
symplectic condition gives
1 0 0 0
 ∗ a 0 c 
K =
∗ ∗ 1 ∗
 
 ∗ b 0 d 
 
−1
a b a b
for some ∈ Γ (p). Multiplying by ˜ transforms this into
(cid:18) c d (cid:19) 1 2(cid:18) c d (cid:19)
1 0 0 0
 −np 1 0 0 
∗ m 1 n
 
 mp 0 0 1 
 
for some integers m,n. Multiplying on the right by M˜−nM˜−m reduces to
4 1
1 0 0 0
 0 1 0 0 
∗ 0 1 0
 
 0 0 0 1 
 
which is in ˜ (SL(2,Z)). This proves the proposition. 2
1
Now we produce all the generators using the allowable means described above.
Proposition 5.2 M ,M ,M ,M ,j (SL(2,Z)) and j (Γ (p)) can all be generated from
1 2 3 4 1 2 1
M and Γ(p2) by multiplication, inversion, and conjugation in Γ0 .
0 1,p
1 1
Proof. (i) M = j and the smallest normal subgroup of SL(2,Z) containing
0 1(cid:18) 0 1 (cid:19)
M is the whole of SL(2,Z), e.g. [B]. Hence we can make j (SL(2,Z)).
0 1
0 p 0 0
1 1
(ii) M−1M M M−1 =  p p2  and L =  0 p2  ∈ Γ(p2), so
4 0 4 0 1
 0 1   0 1 
   
   
M = M−1M M M−1L−1 and we have made M .
2 4 0 4 0 1 2
6
1 0 2 0
(iii) M M M−1M−1M M =  0 1 0 0  and since we got j (SL(2,Z)) in (i) we
1 0 1 1 0 1 0 0 1 0 1
 
 0 −2 0 1 
 
1 −2
may multiply by j to get
1(cid:18) 0 1 (cid:19)
1 0 0 0
 0 1 0 0 
0 0 1 0
 
 0 −2 0 1 
 
Call this matrix L . Since p is odd we can find integers λ,µ such that −2λ + p2µ = 1.
2
1 0
Consider the matrix L = j , so L ∈ Γ(p2). We then have
3 2(cid:18) p3 1 (cid:19) 3
1 0 0 0
 0 1 0 0  1 0
LλLµ = = j .
2 3 0 0 1 0 2(cid:18) p 1 (cid:19)
 
 0 1 0 1 
 
1 p
Call this element L . We also have j ∈ Γ(p2), but we have not got all of
4 2(cid:18) 0 1 (cid:19)
j (Γ (p)) yet. However we need L as an auxiliary.
2 1 4
(iv) L M M M−1M−1 = M−1 so we can now make M .
4 1 0 1 0 3 3
1 0
(v) Put L = j . Then
5 1(cid:18) 1 1 (cid:19)
L−1M−1L M L−1 = M
5 2 5 2 1 4
and we have made M .
4
(vi) M L L M−1L−1L = M−1, so we can make M .
3 5 4 3 5 2 1 1
(vii) It remains to show how to get all of j (Γ (p)). So far we have not used the freedom
2 1
to conjugate by Γ0 rather than just by Γ . Define
1,p 1,p
p2Z pZ
Γ′(p2) = Q ∈ Γ (p)|Q−1 = .
1 (cid:26) 1 (cid:18) p3Z p2Z (cid:19)(cid:27)
Thus Q ∈ Γ′(p2) if and only if j (Q) ∈ Γ(p2). If instead Q ∈ SL(2,Z) then j (Q) ∈ Γ0 ,
1 2 2 1,p
1 0
and we have already generated j (P), where P = , so the problem is now to
2 (cid:18) p 1 (cid:19)
generate Γ (p) using P, elements of Γ′(p2), and conjugation by elements of SL(2,Z).
1 1
λp+1 αp
Ageneralelement ofΓ (p)isoftheform andsince ithasdeterminant
1 (cid:18) βp µp+1 (cid:19)
equal to 1
(λ+µ)p+(λµ−αβ)p2 = 0,
and in particular λ+µ ≡ 0 mod p. Suppose that λ ≡ 0 mod p, so that µ ≡ 0 mod p also.
Then
λp+1 αp λp+1 αp
P−β =
(cid:18) βp µp+1 (cid:19) (cid:18) −βλp2 −αβp2 +µp+1 (cid:19)
7
which is in Γ′(p2). So we can generate any element of Γ (p) for which λ ≡ 0 mod p.
1 1
Suppose then that (λ,p) = 1. If (α,p) = 1 also then p|(λ−kα) for some integer k. But
1 0
now we can conjugate by ∈ SL(2,Z):
(cid:18) k 1 (cid:19)
1 0 λp+1 αp 1 0 (λ−kα)p+1 αp
=
(cid:18) k 1 (cid:19)(cid:18) βp µp+1 (cid:19)(cid:18) −k 1 (cid:19) (cid:18) β′p (µ+kα)p+1 (cid:19)
and so we can generate all elements of Γ (p) for which λ ≡ 0 mod p or α 6≡ 0 mod p. The
1
remaining elements are of the form
λp+1 α′p2
(cid:18) ∗ µp+1 (cid:19)
1 p
and we multiply by ∈ Γ′(p2):
(cid:18) 0 1 (cid:19) 1
1 p λp+1 α′p2 ∗ (α′ +µ)p2 +p
=
(cid:18) 0 1 (cid:19)(cid:18) ∗ µp+1 (cid:19) (cid:18) ∗ µp+1 (cid:19)
and (α′ +µ)p+1 6≡ 0 mod p so this is something we have already generated. 2
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[M] J.Mennicke, ZurTheorie der Siegelschen Modulgruppe, Math. Annalen 159(1965),
115-129.
K. Hulek G.K. Sankaran
Institut fu¨r Mathematik Department of Pure Mathematics
Universit¨at Hannover and Mathematical Statistics
Postfach 6009 University of Cambridge
D-30060 Hannover Cambridge CB2 1SB
Germany England
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