What is an OT-manifold?
March 20, 2014
Non-K¨ahler geometry is a mysterious field and new examples were built recently. The class of OT-manifolds form such an example. Their construction
was introduced by Oeljeklaus and Toma in [5]. Our goal in this series of talks is
to describe this construction and give recent developments about their study.
Contents
1 Number-theoretic preliminaries
1.1 Field extensions . . . . . . . . . . . .
1.2 Algebraic and transcendental numbers .
1.3 Number fields . . . . . . . . . . . . .
1.4 Galois group of a field extension . . . .
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2 Construction of OT-manifolds
7
2.1 A special class of Cousin groups . . . . . . . . . . . . . . . . . . 8
2.2 OT-manifolds of simple type . . . . . . . . . . . . . . . . . . . . 10
3 Geometrical properties of OT-manifolds
11
3.1 Non-K¨ahler structure . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Locally conformally K¨ahler structure . . . . . . . . . . . . . . . . 13
1
1
Number-theoretic preliminaries
We start by recalling some facts from number theory, mainly without proof. A
classical references is [2].
In the following, unless specified otherwise, K always denotes a field and K[X]
is the ring of polynomials with coefficients in K.
1.1
Field extensions
Definition 1.1. A field extension of a field K is a pair (L, i) where L is a field
and i : K → L is a morphism. It is denoted by K ⊂ L.
Notice that i is injective because a field only has trivial ideals.
Definition 1.2. When one sees L as a K-vector space, its dimension dimK (L)
is called the degree of L over K and is denoted by [L : K]. An extension is
said to be finite if its degree [L : K] is finite.
Proposition 1.3. Let K ⊂ L ⊂ M be two finite field extensions, then one has
[M : K] = [M : L] × [L : K].
Proof : Choose an L-basis (bj )j∈J of M and a K-basis (ai )i∈I of L. Then,
(ai bj )i∈I,j∈J is a K-basis for M .
Let K ⊂ L be a field extension and A be a subset of K (not necessarily
finite). One notes K(A) (resp. K[A]) the smallest subfield (resp. ring) of L
containing A. It is clear that K[A] ⊂ K(A) hence K(A) is the quotient field of
K[A].
Definition 1.4. Let K ⊂ L be a field extension and let x ∈ L. The extension
K ⊂ K(x) is called simple and x is said to be a primitive element for K(x).
Proposition 1.5. Let K ⊂ L be a field extension and A, B be two subsets of
L. Then one has K(A ∪ B) = K(A)(B) = K(B)(A).
Proof : K(A) ⊂ K(A ∪ B) and B ⊂ K(A ∪ B) imply K(A)(B) ⊂ K(A ∪ B)
while K ⊂ K(A)(B) and A ∪ B ⊂ K(A)(B) imply K(A ∪ B) ⊂ K(A)(B). 2
1.2
Algebraic and transcendental numbers
Let K ⊂ L be a field extension and x ∈ L. Define the following map
θx : K[X] −→ K[x]
P
7−→ P (x).
The map θx may be injective or not. We study these two cases.
Case 1 : θx is not injective. Since K[X] is principal, one has ker θx = (P )
for some polynomial P . Up to choosing P irreducible and monic, it is unique
and called the minimal polynomial of x, denoted µx . Moreover, x is said to
be algebraic over K of degree degK (x) := deg µx . Because µx is irreducible,
the ideal (µx ) is a prime ideal of K[X] hence maximal1 so θx is an isomorphism
between the fields K[X]/(µx ) and K[x] = K(x).
Case 2 : θx is injective: for all P ∈ K[X], P (X) 6= 0 and x is called transcendental over K.
√
Example 1.6. The real number 2 is algebraic over Q while π is transcendental.
Proposition 1.7. Let K ⊂ L be a field extension and x ∈ L be algebraic over
K. Then, [K(x) : K] = degK (x).
Proof : One shows that 1, x, x2 , ..., xdegK (x)−1 is a K-basis of K(x).
Definition 1.8. Let P ∈ K[X] be a polynomial. A splitting field of P is an
extension L of K in which splits, i.e. decomposes as a product of linear factors,
and such that P doesn’t split on any proper subfield of L.
Proposition 1.9. Every polynomial has a splitting field.
Definition 1.10. A polynomial P ∈ K[X] is separable if it only has simple
roots in its splitting field. A field extension K ⊂ L is separable if every x ∈ L
is separable over K, i.e. transcendental over K or algebraic over K with its
minimal polynomial µx ∈ K[X] being separable.
Theorem 1.11 (Primitive Element Theorem). Let K ⊂ L be a finite and separable field extension. Then, there is a primitive element x ∈ L for L over K, i.e.
L = K(x).
1
In a principal ring, every non-trivial prime ideal is maximal. This is not true in general,
while the implication maximal⇒prime is always true.
3
1.3
Number fields
Definition 1.12. A number field is a finite extension Q ⊂ K over the rationals.
Proposition 1.13. A number field is always a separable extension.
Proof : A polynomial is separable if and only if it is prime with its formal derivative. Because Q has characteristic 0, for every irreducible polynomial P ∈ Q[X]
of degree > 1, one has P 0 6= 0 hence P has simple roots.
By the primitive element theorem, a number field K is always of the form
Q(α) for α ∈ K an algebraic number over Q.
√
√
√ √
Example 1.14. Q( 2, 3) = Q( 2 + 3).
Definition 1.15. An element x ∈ C is called an algebraic number if it is
algebraic over Q. Moreover, it is an algebraic integer if its minimal polynomial
µx only has integer coefficients, i.e. µx ∈ Z[X]. The set of algebraic integers of
a number field K is denoted OK .
Proposition 1.16. One has OK ∼
= Zn . It is a ring and its quotient field is K.
Definition 1.17. An algebraic integer is called an algebraic unit if it is a divisor
of 1, i.e. if its inverse is also an algebraic integer. If K is a number field, the
∗
.
group of units of the ring OK is denoted by OK
Theorem 1.18. Let K be a number field of degree n. Then there are exactly
n distinct injective homorphisms from K to C.
Proof : Call α a primitive element for K and let µα be its minimal polynomial,
which is of degree n. Name α1 = α, α2 , ..., αn the n distinct roots of µα . Every
element of K = Q(α) can be written uniquely in the form a0 +a1 α+...+an−1 αn−1
where the ai ’s are rational numbers. For k ∈ {1, ..., n}, define σk : K → C by
σk (a0 + a1 α + ... + an−1 αn−1 ) = a0 + a1 αk + ... + an−1 αkn−1 .
The fact that the σk is a field homomorphism is left as an exercise.
Now, if σ : K → C is a field homomorphism, then µα (σ(α)) = σ(µα (α)) = 0
so σ(α) is a root of µα , say αk0 for k0 ∈ {1, ..., n}. We have σ(α) = αk0
so σ(a0 + a1 α + ... + an−1 αn−1 ) = σk0 (a0 + a1 α + ... + an−1 αn−1 ) for all
a0 , ..., an−1 ∈ Q, which shows that σ = σk0 .
4
Definition 1.19. The αi ’s are called the conjugates of α over Q and the fields
Q(αi ) are the conjugate fields of K.
Remark 1.20. Let K be a number field and call σ1 , ..., σs its real embeddings
and σs+1 , ..., σs+2t the complex ones. The norm of an element k ∈ K is the
product σ1 (k)×...×σs+2t (k). In particular, the norm of a unit is ±1. Conversely,
if the norm of an element of OK is ±1 then it is a unit.
Definition 1.21. Let K be a number field. If σ : K → C is an injective
homomorphism which satisfies σ(K) ⊂ R, it is called a real embedding of K,
otherwise it is called a complex embedding.
√
√
3
embedding sending
α
to
2
Example 1.22. Let K = Q(α) with α = 3 2. The √
√
2iπ
4iπ
is real, while the two embeddings sending α to e 3 3 2 and e 3 3 2 are complex
ones.
Let σ : K → C be a complex embedding, σ : K −→ C
is another
x 7−→ σ(x)
complex embedding of K, called conjugate to σ. One has σ 6= σ and σ = σ,
hence the set of complex embeddings of K contains an even number of elements,
which we denote 2t.
On the other hand, note s the number of real embeddings of K. One has
[K : Q] = s + 2t.
The following theorem gives the structure of the group of units of a number
field. One can find a proof in chapter 2, section 4 of [2].
Theorem 1.23 (Dirichlet’s Unit Theorem). Let K be a number field of degree
s + 2t where s is its number of real embeddings and 2t the number of complex
∗ ∼
embeddings. One has OK
= C × Zs+t−1 where C is a finite group of roots of
unity of K.
∗ ∼
Remark 1.24. If s > 0, then OK
= {±1} × Zs+t−1 . Indeed, if ε ∈ K is a
k
root of unity, one has σ1 (ε ) = 1 for some integer k > 0. Because σ1 is a
homomorphism, this implies that σ1 (ε)k = 1, i.e. σ1 (ε) = ±1. The injectivity
of σ1 (and the fact that σ1 (1) = 1) forces ε = ±1.
1.4
Galois group of a field extension
Definition 1.25. Let K ⊂ L be a field extension. An automorphism σ : L → L
is called a K-automorphism if σ|K = idK .
5
Let P ∈ K[X] be a polynomial, L its splitting field over K and σ be a
K-automorphism of L. If α ∈ L is a root of P then σ(α) is also a root of P .
Definition 1.26. Let K ⊂ L be a field extension. The Galois group of this
extension, denoted AutK (L), is the group of K-automorphisms of L. If L is the
splitting field of a polynomial P ∈ K[X], AutK (L) is called the Galois group
of P .
The following should be clear:
Theorem 1.27. Let P [X] be a polynomial of degree m and A = {α1 , ..., αn } ⊂
L the set of the n 6 m distinct roots of P in its splitting field L. Then AutK (L)
is isomorphic to a subgroup of the symmetric group Sn .
Proof : This isomorphism is given by AutK (L) −→ Sn .
σ
7−→ σ|A
Theorem 1.28. If P ∈ K[X] is a separable polynomial and L is its splitting
field, one has # AutK (L) = [L : K].
Definition 1.29. Let K ⊂ L be an algebraic field extension. It is called normal
or quasi-Galois if every irreducible polynomial P ∈ K[X] having a root in L
has all its other roots in L.
Example 1.30. The extension K ⊂ K is normal. If K ⊂ Ω is an algebraic
closure of K, it is also normal.
Definition 1.31. Let K ⊂ L be a finite field extension. It is called a Galois
extension if it is normal and separable.
Proposition 1.32. For a finite field extension K ⊂ L, the following properties
are equivalent:
• K ⊂ L is a Galois extension;
• # AutK (L) = [L : K];
• Every irreducible polynomial P ∈ K[X] which admits a root in L is separable and splits in L;
• L is the splitting field of a separable polynomial P ∈ K[X].
6
2
Construction of OT-manifolds
Let K be an algebraic number field and n := [K : Q] be its degree over Q. The
field K admits exactly s real embeddings and 2t complex ones (with n = s + 2t).
We call the real embeddings σ1 , ..., σs and σs+1 , ..., σs+2t the complex ones, ordered in such a way that σs+i = σ s+i+t for 1 6 i 6 t. For the construction we
assume that s > 0 and t > 0.
Call m := s + t and consider the map σK : K −→ Rs × Ct ⊂ Cm .
a 7−→ (σ1 (a), ..., σm (a))
The set σK (OK ) is a lattice of real rank n in Cm , so OK acts properly discontinuously on Cm . Moreover, there is an action of the multiplicative group
K ∗ = K \ {0} on Cm given by az := (σ1 (a)z1 , ..., σm (a)zm ) for a ∈ K and
z = (z1 , ..., zm ) ∈ Cm . In particular, σK (OK ) is stable under this action.
∗
:
Define the following subgroup of OK
∗,+
∗
OK
:= {u ∈ OK
| σi (u) > 0 for all 1 6 i 6 s}.
Let H be the complex upper half-plane. One has a free action of the semi∗,+
direct product OK
n OK on Hs × Ct . This action is defined by
(u, a).z := (σ1 (u)z1 + σ1 (a), ..., σm (u)zm + σm (a))
∗,+
for (u, a) ∈ OK
n OK and z = (z1 , ..., zs , zs+1 , ..., zm ) ∈ Hs × Ct .
∗,+
Define ` : OK
→ Rm by
`(a) = (log |σ1 (a)|, ..., log |σs (a)|, 2 log |σs+1 (a)|, ..., 2 log |σm (a)|).
∗,+
By Dirichlet’s units theorem, `(OK
) is a lattice of
rank m − 1 in the real
Preal
m
m
(m−1)-dimensional vector space L := {x ∈ R | i=1 xi = 0}. The projection
pr : L → Rs given by the s first coordinates is surjective so there are subgroups
∗,+
U of rank s of OK
such that pr(`(U )) is a lattice of rank s of Rs . Such an U
is called admissible.
Lemma 2.1. The set (Hs ×Ct )/σK (OK ) is diffeomorphic to (R>0 )s ×(S1 )n and
∗,+
the action of a subgroup U of OK
on it is properly discontinuous and cocompact
if and only if U is admissible.
7
Proof :
the map
are real,
Hs × C t
Write zj = xj + iyj for 1 6 j 6 s. One has Hs ∼
= (R>0 )s × Rs via
(z1 , ..., zs ) 7→ (y1 , ..., ys , x1 , ..., xs ). Since the s first embeddings of K
an element a ∈ OK sends a point (y1 , ..., ys , x1 , ..., xs , zs+1 , ..., zm ) ∈
to the point
(y1 , ..., ys , x1 + σ1 (a), ..., xs + σs (a), zs+1 + σs+1 (a), ..., zm + σm (a)) ∈ Hs × Ct .
This shows that (Hs ×Ct )/σK (OK ) ∼
= (R>0 )s ×((Rs ×Ct )/σK (OK )), which
s
1 n
is in turn diffeomorphic to (R>0 ) × (S ) .
∗,+
Since the quotient of Hs × Ct by OK
n OK is the same as the quotient of
∗,+
s
t
(H × C )/σK (OK ) by OK and since the projection map (R>0 )s × (S1 )n →
∗,+
(R>0 )s is proper, the action of a subgroup U of OK
on (Hs × Ct )/σK (OK ) is
proper if and only if it is proper on (R>0 )s .
∗,+
Indeed, the action of OK
on (Hs ×Ct )/σK (OK ) send an element (y1 , ..., ys ) ∈
∗,+
s
(R>0 ) to (σ1 (u)y1 , ..., σs (u)ys ), for u ∈ OK
.
This and the cocompactness of the action is now equivalent to the definition
of an admissible subgroup.
∗,+
The previous lemma shows that the action of OK
n OK on Hs × Ct is not
properly discontinuous when t > 1, while it is always the case for every admissible
∗,+
subgroup of OK
.
∗,+
Definition 2.2. Choose an admissible subgroup U of OK
, the quotient X :=
s
t
X(K, U ) = (H × C )/(U n OK ) is a compact complex manifold 2 of dimension
m. This manifold is called an Oeljeklaus-Toma manifold, or an OT-manifold
for short.
2.1
A special class of Cousin groups
Lemma 2.3 ([5], lemma 2.4). Every holomorphic function on Hs × Ct /σK (OK )
is constant.
Proof : The proof is divided into two steps:
Step 1: For every v ∈ Hs , one sees that ({v} × Ct )/σK (OK ) is dense in
(v + Rs × Ct )/σK (OK ).
2
When one has a fixed-point free and properly discontinuous action of a group G on a
complex manifold W , then the quotient W/G has a complex structure induced from W . See
[4], Theorem 2.2 for a proof.
8
0
Let V be the connected component of the closure of σK (OK ) in Rs containing
0
0. Let M := σK−1 (V ) ⊂ OK . Assume that V 6= Rs . This implies that the rank
of M is strictly smaller than n. Now let α ∈ OK be a primitive element for K.
There is a multiplicative action on OK of α, and the induced linear action of α
on Rs leaves V invariant. This implies that M is also invariant under the action
of α, which implies that the characteristic polynomial of α has a factor of same
degree as the rank of M , a contradiction.
Step 2: For every holomorphic function f on Hs × Ct /σK (OK ) and v ∈ Hs , the
map f is bounded on (v + Rs × Ct )/σK (OK ) ∼
= (S1 )n . In particular the lift fe of
f to Hs × Ct is bounded on (v + Rs ) × Ct , hence constant on {v} × Ct thanks to
Liouville’s theorem. The observation in Step 1 implies that the function fe is constant on (v+Rs )×Ct , and hence constant on Hs ×Ct by the identity principle. Remark 2.4. This implies that the complex Lie group Cm /σK (OK ) is a Cousin
group, i.e. a connected complex Lie group having no non-constant holomorphic
functions.
In [7], Vogt exhibits a special class of Cousin groups by showing 3 :
Theorem 2.5. Let C = Cn /Λ be a Cousin group. Then the following conditions
are equivalent:
1. The space H 1 (C, OC ) is finite-dimensional.
2. Let P = (In S) be a period basis of Λ. Then there exist constants C > 0
and a > 0 such that kt σS + t τ k > C exp(−a|σ|) for all σ ∈ Zn \ {0} and
all τ ∈ Zm , where n + m is the rank of Λ.
3. Every topologically trivial line bundle over C comes from a representation
of Λ.
In fact, thanks to the following generalization of Liouville’s theorem on approximation of algebraic numbers, one can show that the Cousin groups Cm /σK (OK )
arising in the construction of OT-manifolds belong to this special class.
3
The theorem given in [7], p. 208 has 8 equivalent assertions, here we only recall three of
them.
9
Theorem 2.6. Let α1 , ..., αm be algebraic numbers, of respective degrees nk ,
with deg Q(α1 , ..., αm ) = n, and let
P (z1 , ..., zm ) =
N1
X
k1 =0
···
Nm
X
km
∈ Z[z1 , ..., zm ].
ak1 ,...,km z1k1 · · · zm
km =0
If P (α1 , ..., αm ) is non-zero, then one has the inequality
1−δn
|P (α1 , ..., αm )| > L(P )
m
Y
L(αk )−δNk n/nk ,
k=1
with δ = 1 if all the αi are real, δ = 1/2 otherwise and where L(P ) is the sum
of the absolute values of the coefficients of P (and L(α) is the quantity L(µ),
with µ being the minimal polynomial of α).
Theorem 2.7. Let Λ ⊂ Cn be a lattice such that C = Cn /Λ is a Cousin group
with a period basis whose coefficients are all algebraic numbers, then C satisfies
the equivalent conditions of theorem 2.5. In particular, Cousin groups arising in
the construction of OT-manifolds belong to this family.
2.2
OT-manifolds of simple type
We recall two lemmas and a definition (see [5]) for further use:
∗,+
Lemma 2.8. Let U be a subgroup of OK
which is not contained in Z. Then
the two following conditions are equivalent:
1. The action of U on OK admits a (non-trivial) proper invariant sub-module
of lower rank.
2. There exists a proper intermediate field extension Q ⊂ K 0 ⊂ K with
∗,+
U ⊂ OK
0 .
Definition 2.9. We say that X(K, U ) is of simple type if U does not satisfy
one of the equivalent conditions of the previous lemma.
∗,+
Lemma 2.10. Let Q ⊂ K 0 ⊂ K an intermediate field extension with U ⊂ OK
0
0
0
an admissible subgroup for K. Let s , 2t be the number of real and complex
embeddings of K 0 respectively. Then s = s0 , t0 > 0 and U is admissible for K 0 .
For more details, see [5].
10
3
3.1
Geometrical properties of OT-manifolds
Non-K¨
ahler structure
Definition 3.1. Let M be a complex manifold with complex structure J. A
Riemannian metric g = h · , · i is compatible with J if hJX, JY i = hX, Y i for
all vector fields X, Y ∈ T X. A complex manifold with a compatible Riemannian
metric is called a Hermitian manifold. The 2-form ω(X, Y ) := g(JX, Y ) is
the associated fundamental form. One recovers g by g(X, Y ) = ω(X, JY ).
Finally, g is a K¨
ahler metric if ω is closed i.e. dω = 0 and ω is called the
K¨
ahler form. A complex manifold is called a K¨
ahler manifold if it admits a
K¨ahler metric and a non-K¨
ahler manifold otherwise.
Example 3.2. Cn , Pn (C) (ω = i∂∂ log |z|2 ), complex tori, Riemann surfaces
are K¨ahler manifolds.
Proposition 3.3. Let (X, g) be a compact K¨ahler manifold. Then there exists
a decomposition
M
H p,q (X).
H k (X, C) =
p+q=k
Moreover, one has H p,q (X) = H q,p (X).
As a consequence, the first Betti number b1 (X) = dimC H 1 (X, C) of a
compact K¨ahler manifold X is equal to 2h1,0 = 2 dim H 1 (X, OX ).
Example 3.4. Hopf surfaces.
Lemma 3.5. Let u1 , . . . , us be some s ≥ 1 multiplicatively independent units in
UK . Then either Q(u1 , . . . , us ) is a proper subfield of K or UK contains s multiplicatively independent units v1 , . . . , vs ∈ S(u1 , . . . , us ) := {ak11 · · · aks s | ki ∈ N}
of degree d each.
Definition 3.6. A reciprocal unit is an algebraic unit x ∈ C such that the
inverse 1/x is one of its conjugate over Q.
Proposition 3.7. Thefirst
two Betti numbers of an OT-manifold X = X(K, U )
s
are b1 = s and b2 =
. For b2 we need to assume that U contains a non2
reciprocal unit.
11
Proof : The manifold X is an Einlenberg-MacLane space of type K(G, 1),
that is, it has a fundamental group isomorphic to G and its universal covering is
contractible. By theorem I in [3] (pp. 482-483), the cohomology groups of X
with coefficients in Q are the same as those of its fundamental group. So we are
left with the computation of H 1 (U n OK , Q) and H 2 (U n OK , Q).
As in [5], we use the Lyndon-Hochschild-Serre spectral sequence associated
to the short exact sequence
0 → OK → U n OK → U → 0.
One has E2p,q = H p (U, H q (OK , Q)) ⇒ H p+q (U n OK , Q) and the following
exact sequence:
E20,1
}|
{
z
1
U
1
OK
1
0 → H (U, Q ) → H (U n OK , Q) → H (OK , Q)
→ H 2 (U, QOK ) → H 2 (U n OK , Q)1 → H 1 (U, H 1 (OK , Q))
|
{z
}
1,1
E2
where H 2 (U n OK , Q)1 is defined by the exact sequence
0 → H 2 (U n OK , Q)1 → H 2 (U n OK , Q) → H 2 (OK , Q)U .
{z
}
|
0,2
E2
See for instance [6].
If one proves that E20,1 = E20,2 = E21,1 = 0, the result follows. For the fact
that E20,1 = E21,1 = 0, the proofs are in [5], proposition 2.3, so we will not repeat
it here (the existence of a non-reciprocal unit in U is not needed).
∼
As for the group E20,2 = H 2 (OK , Q)U P
= Alt2 (OK , Q)U , recall that an element of Alt2 (OK , Q)U is of the form γ = i<j ai,j σi ∧ σj with ai,j ∈ C. Moreover, the U -invariance of γ means that for every pair (i, j) such that ai,j 6= 0,
one has σi (η)σj (η) = 1 for all η ∈ U . Now when U contains a non-reciprocal
unit η0 and therefore the relation σi (η0 )σj (η0 ) = 1 can never hold for any choice
of i < j. Hence γ is trivial and so is the group E20,2 .
Proposition 3.8. An OT -manifold X = X(K, U ) is non-K¨ahler.
12
Proof : One has dimC H 1 (X, C) = s and dim H 1,0 = dim H 0,1 > s.
Let ρ : U → C be a group homomorphism. We will associate to this homomorphism a principal C-bundle above X and show that if this bundle is trivial
then ρ is also trivial, which will lead to the desired inequality. Consider the action
of U on the product ((Hs × Ct )/σK (OK )) × C given by
u.(x, z) := (u.x, z + ρ(u))
for all u ∈ U , x ∈ (Hs × Ct )/σK (OK ) and z ∈ C. The quotient F of
((Hs × Ct )/σK (OK )) × C under this action of U is a principal C-bundle above
X. Indeed, the action of ξ ∈ C on an element [x, z] := U.(x, z) ∈ F is given
by ξ.[x, z] := [x, z + ξ]. If F is trivial, it must have a global section, i.e. there
is a holomorphic function f : (Hs × Ct )/σK (OK ) → C satisfying the equality
f (u(x)) = f (x) − ρ(u) for all u ∈ U and u ∈ U . By lemma 2.3, the function f
is constant and this implies ρ ≡ 0. This proves that h1,0 > s. If an OT manifold
was K¨ahler we would have s > 2s, a contradiction.
In theoretical physics, string theory, non-K¨ahler examples are needed to work
with.
3.2
Locally conformally K¨
ahler structure
f its universal covering.
Definition 3.9. Let M be a complex manifold and M
Then M is called a locally conformally K¨
ahler manifold ( LcK manifold, for
short) if there exists a representation ρ : π1 (M ) → R>0 and a positive closed
f such that g ∗ ω = ρ(g)ω for all g ∈ π1 (M ).
(1, 1)-form ω on M
Example 3.10. A K¨ahler manifold is of course LcK. A Hopf manifold is also
LcK.
P
Proof : Let ω := i dzj ∧ dzj be the K¨ahler metric on Cn \ {0} and
ρ : Z −→ R>0 .
λ 7−→ |λ|2
Example 3.11. When t = 1, an OT-manifold is LcK.
13
Theorem 3.12. An OT-manifold X(K, U ) admits an LcK metric if and only if
the following holds:
for all u ∈ A,
|σs+1 (u)| = ... = |σs+t (u)|.
(1)
Proof : Assume that ω is a K¨ahler metric on Hs × Ct upon which U n OK
acts by homotheties. Then ω can be written as
X
ω=
hij dzi ∧ dz j .
i,j=1,s+t
Since ω descends to (Hs × Ct )/σK (OK ) ∼
= (R>0 )s × (S1 )n , by averaging ω
on the (S1 )n part we can assume that all the coefficients of ω depend only on
z1 , . . . , zs . Now we show that all hii , i = s + 1, . . . , s + t are constant. Assume
this is not the case, i.e.:
∂hii
6= 0 for some 1 ≤ k ≤ s.
∂zk
From the assumption that ω is a K¨ahler form, hence closed, we get
∂hki
6= 0,
∂zi
a contradiction with the fact that hki does not depend on zi for i > s + 1.
Since ω is an LcK metric on X, there is a map ρ such that g ∗ ω = ρ(g) for
all g ∈ U n OK . Because there are constant coefficients in ω, ρ is equal to
ρ(u, a) = |σs+1 (u)|2 = ... = |σs+t (u)|2 for all (u, a) ∈ U n OK .
We are left with checking that this condition is sufficient. Let X(K, U ) be
an OT-manifold with U satisfying condition (1) and define the following real
function on Hs × Ct :
ϕ(z) :=
s
Y
i
z − zj
j=1 j
! 1t
+
t
X
|zs+k |2 .
k=1
This definition of ϕ is very natural: when t = 1, this function is the same as
the function F defined in [5], example p. 169, to show that when t = 1 an
OT-manifold admits an LcK metric.
It is enough to prove that it is a K¨ahler potential on Hs × Ct . For this, we
will see that the matrix (∂zp ∂zq ϕ1 ) is positive definite, where we set ϕ1 (z) =
14
Q
s
i
j=1 zj −zj
1t
. For all q ∈ {1, ..., s}, one has:
∂zq ϕ1 (z) =
1 1
ϕ1 ,
t zq − zq
and for all p ∈ {1, ..., s}, one has:
∂zp ∂zq ϕ1 (z) =
 1
−1

ϕ1 if p 6= q


2
 t (zp − zp )(zq − zq )


1
−1


(1 + t)
ϕ1 if p = q.
2
t
(zp − zp )2
1
ϕ1 B where the matrix B is
t2
 (1+t)

1
1
·
·
·
2
4y1 y2
4y1 ys
4y
 11 (1+t)

 4y2 y1 4y2 · · · 4y12 ys 
,
2
B=
 ..

 .

(1+t)
1
1
· · · 4y2
4ys y1
4ys y2
Hence (∂zp ∂zq ϕ1 ) =
s
and zj = xj + iyj for all j ∈ {1, ..., s + t}. Notice that B is the sum between
a diagonal positive definite matrix and a positive semidefinite one, hence B is
positive definite.
Now, let ω0 := i∂∂ϕ and for all g = (u, a) ∈ A n OK set ρ(g) := |σs+1 (u)|2 .
First, notice that because u is a unit we have
(σ1 (u)...σs (u))(|σs+1 (u)|2 ...|σs+t (u)|2 ) = 1.
Then, write
∂∂(ϕ ◦ g)(z) =
1
(σ1 (u)...σs (u))
1
t
∂∂ϕ1 (z) + ∂∂
t
X
k=1
= ρ(g)∂∂ϕ1 (z) + ρ(g)∂∂
t
X
k=1
= ρ(g)∂∂ϕ(z)
15
|zs+k |2
|σs+k (u)zs+k + σs+k (a)|2
We now obtain the following equalities:
g ∗ ω0 = g ∗ (i∂∂ϕ) = i∂∂(ϕ ◦ g) = iρ(g)∂∂(ϕ) = ρ(g)ω0 .
This concludes the proof.
Theorem 3.13. If a number field K of degree d = s + 2t over Q with s real and
2t complex embeddings, where t ≥ 2, contains a nonreciprocal unit u ∈ UK of
degree d whose 2t algebraic conjugates satisfy (1) then for some integers m ≥ 0
and q ≥ 2 we have
s = (2t + 2m)q − 2t.
(2)
On the other hand, if s and t ≥ 2 satisfy (2) then there is a number field K with
s real and 2t complex embeddings that contains a nonreciprocal unit u ∈ UK of
degree d satisfying (1).
Theorem 3.14. Let K be a number field of degree d = s + 2t over Q with s
real and 2t complex embeddings, where t ≥ 2 and s is not of the form (2). Then
the rank of the subgroup U of UK of units satisfying (1) is smaller than s and,
therefore, for any choice of an admissible subgroup U for K the Oeljeklaus-Toma
manifold X(K, U ) has no LcK metric.
Proof : Recall that no OT-manifold admits a K¨ahler structure (this is Proposition 2.5, loc. cit.).
Assume that K is a number field of degree d = s + 2t with t > 2 and with
s not being of the form (2). We now suppose that the rank of the subgroup U
of UK of units satisfying equation (1) is at least (therefore, equal to) s and we
want to show that it leads to a contradiction.
First, notice that l(U ) has a trivial intersection with the kernel of the projecting map P : S → Rs where l and P are defined in the introduction of the
main article. Thus, U is an admissible subgroup of UK . Now, consider the OTmanifold X(K, U ); it admits an LcK metric by Theorem 3.12. Call u1 , ..., us a
family multiplicatively independent units generating U and K 0 := Q(u1 , ..., us ).
The same argument as the main article shows that K 0 can not be a proper subfield of K. As a consequence of Lemma 3.5 and Theorem 3.13, every element u
of U satisfies |σs+j (u)| = 1, for all j ∈ {1, ..., t}.
Let ω be a K¨ahler form on Hs × Ct giving rise to an LcK metric on X(K, U ).
For all g = (u, a) ∈ U n OK , one has g ∗ ω = |σs+1 (u)|2 ω (see the paragraph
before Theorem 3.12), which simplifies as g ∗ ω = ω. The form ω being invariant
16
under the action of A n OK , it descends to a K¨ahler form on X(K, U ). This
implies that X(K, U ) is a K¨ahler manifold, which is the desired contradiction. 17
References
[1] L. Battisti and K. Oeljeklaus, Holomorphic line bundles over domains in
Cousin groups and the algebraic dimension of OT-manifolds, preprint at
arXiv:1306.3047v1, 2013, 13 p.
[2] Borevich, A. I. and Shafarevich, I. R.,Number theory, Translated from the
Russian by Newcomb Greenleaf. Pure and Applied Mathematics, Vol. 20,
Academic Press, 1966,
[3] S. Eilenberg, and S. MacLane,Relations between homology and homotopy
groups of spaces,Ann. of Math. (2) 46,(1945), 480–509.
[4] K. Kodaira, Complex manifolds and deformation of complex structures,
Translated from the 1981 Japanese original by K. Akao, Springer-Verlag,
2005, x+465
[5] K. Oeljeklaus and M. Toma, Non-K¨ahler compact complex manifolds associated to number fields, Ann. Inst. Fourier 55 (2005), 161–171.
[6] Chih Han Sah, Cohomology of split group extensions, J. Algebra 29 (1974),
255–302.
[7] C. Vogt, Line bundles on toroidal groups, J. Reine Angew. Math. 335
(1982), 197–215.
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