1 Introduction

The aim of this work is to investigate when the variation of Hodge structures associated to a family of Calabi-Yaus is decomposable. We show that assuming the conjecture ”Hodge implies Absolutely Hodge”, the Hodge structure being decomposable can be translated into a condition in Galois theory. Finally we look at some numerical data to try and ascertain whether there is sufficient evidence for a conjecture that the variation of Hodge structures is ”rarely” decomposable.

1.1 Hodge Theory and Galois Theory

Definition 1.1 The total cohomology of a variety X over an algebraically closed field k is H_{tot}^{i}(X) = H_{DR}^{i}(X) × ∏_{l} H_{et}^{i}(X, ℚ_{l}) where H_{DR}^{i} is the de Rham cohomology and H_{ét}^{i} is the etale cohomology.

Let σ : k ↪ ℂ and set σX = X×_{k}Specℂ. Recall that we have comparision isomorphisms:

H_{B}^{i}(σX) ⊗_{ℚ} k →^{≅} H_{DR}^{i}(X)

H_{B}^{i}(σX) ⊗_{ℚ} ℚ_{l} →^{≅} H_{éet}^{i}(X, ℚ_{l})

where H_{B}^{i} is the Betti cohomology.

Definition 1.2 Let \underline{t} = (t_{DR}, (t_{l})_{l}) ∈ H_{tot}^{2i}(X). Then t is a Hodge cycle with respect to σ : k ↪ ℂ if there exists t_{B,σ} ∈ H_{B}^{2i}(X) such that:

1. (rationality) t_{DR} =comp_{B,DR}(t_{B,σ} ⊗ 1) and t_{l} =comp_{B,l}(t_{B,σ} ⊗ 1)

2. (Hodge condition) (t_{B,σ}) ⊗ 1 ∈ H^{i,i}(σX) ⊂ H_{B}^{2i}(σX) ⊗ ℂ

Definition 1.3 We call a Hodge cycle an absolute Hodge cycle if it is a Hodge cycle relative to every embedding σ : k ↪ ℂ.

We will assume the following:

Conjecture 1.4 Every Hodge cycle is an absolute Hodge cycle.

Let k_{0} be a field (not necessarily algebraically closed) contained in k. Then we have the following commutative diagram:

X_{ℂ} ⟶ X ⟶ X_{0}
↓  ↓  ↓
Spec(ℂ) ⟶ Spec(k) ⟶ Spec(k_{0})

There is a natural action of Gal(k/k_{0}) on H_{ét}^{i}(X, ℚ_{l}) and the comparision isomorphisms can be used to get an action on V = H_{B}^{i}(X_{ℂ}).

Definition 1.5 One says that W is a tensor construction from V if it satisfies the following two properties:

1. W = V^{⊗n_{1}} ⊗ V̌^{⊗n_{2}} ⊗ ℚ(1)^{⊗n_{3}}

2. W_{Hdg} = W ⋂(W ⊗ ℂ)^{0,0}

Using Deligne’s article (Art 1, LNM 900, Prop 2.9), the above conjecture implies that the image W_{Hdg} ↪ W ⊗ ℚ_{l} is stabilized by Gal(k/k_{0}).

Definition 1.6 The Mumford-Tate group MT(V) is the largest ℚ-algebraic subgroup of GL(V) × GL(ℚ(1)) whose rational points fix all the elements of W_{Hdg}. MT ′(V) is the projection onto GL(V).

When the associated Hodge structure is polarizable, it can be shown (loc. cit.) that this implies that the Mumford Tate group is reductive. In particular this means that a ℚ-linear transformation is in MT(V) iff it stabilizes all Hodge cycles; i.e. this property uniquely characterizes MT(V).

Proposition 1.7 The action of Gal(k/k_{0}) normalizes MT ′(V) ⊂ GL(V ⊗ ℚ_{l}).

Proof. Let σ ∈ Gal(k/k_{0}), m ∈ MT ′, and w ∈ W_{Hdg}. Then σw = ẃ for

some ẃ ∈ W_{Hdg}. We also observe that mẃ = ẃ. Then (σ^{−1}mσ)w = w By the fact mentioned above (i.e. stabilizing all Hodge cycles implies membership of MT(V)) we have that (σ^{−1}mσ) ∈ MT′(V).

1.2 The Mumford-Tate Group of a certain 3-dimensional Hodge structure

In this section we consider a Hodge structure V of weight 3 and rank 4 such that V ⊗ ℂ = V^{0,3} ⊕ V^{1,2} ⊕ V^{2,1} ⊕ V^{3,0} with Hodge numbers all equal to 1. We will assume V = V_{1} ⊕ V_{2}, V_{1} ⊗ ℂ = V^{0,3} ⊕ V^{3,0} ,and V_{2} ⊗ ℂ = V^{1,2} ⊕ V^{2,1}. Working over ℂ, we can pick a basis for V_{1} ⊗ ℂ of the form ( \begin{matrix}1 \\ τ_{1} \end{matrix} ) , ( 1 ; τ̄_{1} ) and similarly ( \begin{matrix}1 \\ τ_{2} \end{matrix} ) , ( 1 ; τ̄_{2} ) for V_{2}⊗ℂ. Using the polarization one gets that Im(τ_{1}) < 0 and Im(τ_{2}) > 0. For a Hodge structure W, we define a map μ : 𝔾_{m} → GL(W) by μ(λ)(w^{p,q}) = λ^{−p}w^{p,q} for w^{p,q} ∈ W^{p,q}.

Proposition 1.8 The Mumford-Tate group of a Hodge structure W is the smallest algebraic subgroup G of GL(V)×𝔾_{m} such that μ(𝔾_{m}) ⊂ G_{ℂ}.

Proof. See the article by Deligne referenced above.

For W = V_{1} with respect to the basis given above we have that μ_{1}(λ) = ( \begin{matrix}1 & 0 \\ 0 & λ^{−3} \end{matrix} ) ,and on V_{2} with respect to the given basis μ_{2}(λ) = ( \begin{matrix}λ^{−1} & 0 \\ 0 & λ^{−2} \end{matrix} ) . Now with respect to the standard basis ( \begin{matrix}0 \\ 1 \end{matrix} ) ( \begin{matrix}1 \\ 0 \end{matrix} ), we have

μ_{1}(λ) = \frac{1}{τ̄ − τ} ( \begin{matrix}τ̄λ^{−3}−τ & −λ^{−3}+1 \\ τ̄λ^{−3}τ−ττ̄ & −λ^{−3}τ+τ̄ \end{matrix} )

μ_{2}(λ) = \frac{1}{τ̄ − τ} ( \begin{matrix}τ̄λ^{−2}−τλ^{−1} & −λ^{−2}+λ^{−1} \\ τ̄λ^{−2}τ−λ^{−1}ττ̄ & λ^{−2}τ+λ^{−1}τ̄ \end{matrix} )

and that detμ_{1}(λ) =detμ_{2}(λ) = λ^{−3}.

Now, using the calculations for λ, we can remove most of the possibilities for MT′(V_{i}). The only possibilities are MT′(V_{i}) = GL_{2} or T^{2}. Now we figure out what the possibilities for MT′(V) are for dimension 4. One can show that MT′(V) ⊂ MT′(V_{1}) ⊕ MT′(V_{2}). Hence, MT′(V) ⊂ G where G is one of the following:

a. GL_{2}×GL_{2} b. GL_{2} × T^{2} c. T^{2}× GL_{2} d. T^{2} × T^{2}

Proposition 1.9 MT ′(V) is one of the following groups:

a. \{( \begin{matrix}g_{1} & 0 \\ 0 & g_{2} \end{matrix} ) |g_{i} ∈ GL_{2}, detg_{1} = detg_{2}\}

b. \{( \begin{matrix}g_{1} & 0 \\ 0 & g_{2} \end{matrix} ) |g_{1} ∈ GL_{2}, g_{2} ∈ T^{2}, detg_{1} = detg_{2}\}^{o}

c. \{( \begin{matrix}g_{1} & 0 \\ 0 & g_{2} \end{matrix} ) |g_{1} ∈ T^{2}, g_{2} ∈ GL_{2}, detg_{1} = detg_{2}\}^{o}

d. \{( \begin{matrix}g_{1} & 0 \\ 0 & g_{2} \end{matrix} ) |g_{1} ∈ T^{2}, g_{2} ∈ T^{2}, detg_{1} = detg_{2}\}

e. \{( \begin{matrix}t_{1} & 0 & 0 & 0 \\ 0 & t_{2} & 0 & 0 \\ 0 & 0 & t_{3} & 0 \\ 0 & 0 & 0 & t_{4} \end{matrix} ) |t_{1}t_{2} = t_{3}t_{4} and \frac{t_{1}}{t_{2}} = (\frac{t_{3}}{t_{4}} )^{3}\} = MT ′(V)_{ℂ}

Proof. First note that the equality of determinants is an immediate consequence of our calculations following Proposition 1.8. In case a one can see that the Mumford-Tate group is either the one indicated in the statement of the theorem or it’s a subgroup thereof with an additional defining condition g_{1} = g_{2} or g_{2} = g_{0}g_{1}g_{0}^{−1} for some 2 × 2 invertible matrix g_{0}. Suppose the first possibility occurs, i.e. g_{1} = g_{2} = g. Then the matrix ( \begin{matrix}0 & I \\ I & 0 \end{matrix} ) commutes with all the elements of MT′(V). Hence the corresponding element in (V̌ ⊗ V)_{ℂ} is fixed by the complexification of the Mumford-Tate group. It follows that ( \begin{matrix}0 & I \\ I & 0 \end{matrix} ) corresponds to an element of bidegree (0,0) in the weight 0 Hodge structure (V̌ ⊗ V)_{ℚ}. Thus it gives an endomorphism of Hodge structures of V, which maps V_{1} isomorphically onto V_{2}, which is absurd as those Hodge structures are not isomorphic. This rules out the case g_{1} = g_{2}. The argument that allows one to exclude the second case is similar.

For case b note that MT′(V) contains both SL_{2} and T which intersect trivially, hence the claim follows from a dimension argument.

For d and e we have 2 ≤ dim MT′(V) ≤ 3. If dim MT′(V) = 3 we obtain case d. If dim MT′(V) = 2, the two tori are isogenous and the associated quadratic imaginary fields are the same. Consideration of the action of the Galois group forces the additional relation occuring in case e.

Observation 1.10 In cases b-e in the above proposition, an element of the normalizer of MT ′(V) in GL(V) doesn’t interchange the pieces of V, but in case a it can happen. However, the matrix that flips the pieces has trace 0.

1.3 A Criterion for Irreducibility

Now we consider the family Y_{s} of hypersurfaces in ℙ^{4} given by:

Y_{s} : s(y_{0} + y_{1} + y_{2} + y_{3} + y_{4})^{5} = y_{0}y_{1}y_{2}y_{3}y_{4}.

We want to apply the results of the previous sections to determine the s ∈ ℂ for which H^{3}(Y_{s}, ℚ) is reducible. We will give a sufficient condition for the irreducibility of the Hodge structure. Assume the Hodge structure is reducible for some s. Then from de Jong’s lecture we know that s ∈ ℚ̄ if we assume Conjecture 1.4. Let k_{0} = ℚ[s], and let O_{k_{0}} be the ring of integers of k_{0}. Pick a prime ℘ ⊂ O_{k_{0}} and let F rob_{℘} be a Frobenius element in Gal(ℚ̄/k_{0}) for ℘. For simplicity, assume that ℘ is a degree one prime. Also assume that ℘ lies above (p) ⊂ ℤ. We can then compute the characteristic polynomial P_{℘} of F rob_{℘} acting on

H_{et}^{3}(Y_{s} ⊗ ℚ̄, ℚ_{l}) →^{≅} H_{B}^{3}(Y_{s} ⊗ ℂ, ℚ) ⊗ ℚ_{l} for l ≠ p. (Note that P_{℘} depends on

s, but for simplicity we omit the s here.) We can compute P_{℘} by reducing the equation for Y_{s} mod ℘ and by counting points over 𝔽_{p}, 𝔽_{p^{2}}. (We know the contributions of the pieces H_{et}^{i}(Y_{s} ⊗ ℚ̄, ℚ_{l}) for i ≠ 3, because the Betti numbers for those pieces are all 0 or 1.) Then there are two cases to consider: Case 1: F rob_{℘} flips the two factors V_{1} and V_{2}. Then, as observed above (See Observation 1.10), we know that trace(F rob_{℘}) = 0. This occured only once in our numerical data, so we will ignore this case. Case 2: F rob_{℘} does not flip the pieces. In this case, the characteristic polynomial P_{℘} which has degree 4 must factor into two quadratic factors over ℚ_{l} for every l ≠ p, because F rob_{℘} preserves the two two-dimensional spaces V_{1} and V_{2}. We know the shape of the polynomial P_{℘}:

P_{℘} = x^{4} + a_{1}x^{3} + a_{2}x^{2} + p^{3}a_{1}x + p^{6}

and the roots λ_{1}, λ_{2}, λ_{3}, λ_{4} can be numbered so that they satisfy the equations λ_{1}λ_{2} = p^{3}, λ_{3}λ_{4} = p^{3}. Now we can ask: When is there an l such that P_{℘} is irreducible over ℚ_{l}? It is certainly sufficient that

1. P_{℘} is irreducible over ℚ and

2. If K is the splitting field of P_{℘}, then Gal(K/ℚ) contains a 4-cycle.

However, because of the shape of the polynomial and relations on the roots, the Galois group Gal(K/ℚ) can only be the Klein 4-group V_{4} or it contains a 4-cycle. Therefore, the group we wish to eliminate is V_{4}. We can check if Gal(K/ℚ) is V_{4} by checking whether the discriminant of P_{℘} is a square in ℚ^{∗}.

With notation as above, we can summarize our results as follows:

Summary 1.11 Given an s in ℚ̄, if there exists a prime ℘ of O_{k_{0}} such that P_{℘} satisfies:

1. trace(P_{℘}) ≠ 0

2. P_{℘} is irreducible over ℚ

3. disc(P_{℘}) is not a square in ℚ^{∗} then H^{3}(Y_{s}, ℚ) is irreducible.

1.4 Some Numerical Data

Now we present some numerical data to see if the P_{℘} are ever reducible for any s. There certainly is not enough evidence to even conjecture whether the Hodge structure is ever decomposable or not because we were only able to count the points of Y_{s} over 𝔽_{p} and 𝔽_{p^{2}} for very small primes p. We are excluding the case where p = 5, and for each p we look at s ∈ {1, ⋯, p − 1}, s ≠ 1/5^{5} mod p. In the following we compute the characteristic polynomial P_{℘} = P_{℘,s} for a given prime p and all allowed values of s.

s  P_{℘} for p = 3  irredcible ? disc a square ?
1 x^{4} + 5x^{3} + 45x^{2} + 135x + 729  yes  no
s  P_{℘} for p = 7  irreducible ? disc a square ?
1 x^{4} + 5x^{3} + 385x^{2} + 7^{3} · 5x + 7^{6}  yes  no
2 x^{4} + 25x^{3} + 350x^{2} + 7^{3} · 25x + 7^{6}  yes  no
3 x^{4} + 10x^{3} + 420x^{2} + 7^{3} · 10x + 7^{6}  yes  no
4 x^{4} − 5x^{3} − 210x^{2} − 7^{3} · 5x + 7^{6}  yes  no
6 x^{4} − 35x^{3} + 805x^{2} − 35 · 7^{3}x + 7^{6}  yes  no

For p = 11, again, for each allowed value of s, the corresponding characteristic polynomial has nonzero trace, is irreducible and its discriminant is not a square in ℚ^{∗}.

s  P_{℘} for p = 13  irreducible ? disc a square ?
4  x^{4} + 10x^{3} − 910x^{2} + 13^{3} · 10x + 13^{6}  no !  no
6 x^{4} − 120x^{3} + 7670x^{2} + 13^{3} · (−120)x + 13^{6}  no !  no

All other allowed values for s are ok when p is 13. For p = 17 all allowed values of s are ok.

s  P_{℘} for p = 19  irreducible ? disc a square ?
13 (x^{2} − 95x + 19^{3})(x^{2} + 100x + 19^{3})  no !  no

All other allowed values for s are ok when s is 19. When p is 23 and s = 18, the corresponding characteristic polynomial is x^{4} + 8050x^{2} + 23^{6} which has trace 0 and whose discriminant is a square. When p = 31 and s = 5, the corresponding charcateristic polynomial factors as (x^{2} − 217x + 31^{3}) · (x^{2} + 108x + 31^{3}). When p = 41 and s = 18, the corresponding characteristic polynomial factors as (x^{2} − 372x + 41^{3}) · (x^{2} + 328x + 41^{3}).