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Instead of focusing on the basis of the rational form of a vector space, we focus on the defining property of the rational form as being ‘symmetric’ under the action of the Galois group. For this we first introduce the action of the Galois group on a vector space and define for each representation a rational subspace.

Consider the natural right action of Gal(E, Q) on the field E, given by ∀σ ∈ Gal(E, Q), ∀x ∈ E : xσ_{= σ}−1_(x).

By defining the action component-wise, there is also a natural right action on the vector space Em_{. Note that the relations}

(σ(λ)x)σ= λxσ and (x + y)σ= xσ+ yσ (9.1) hold for all x, y ∈ Em

, σ ∈ Gal(E, Q) and λ ∈ E. Let ρ : Gal(E, Q) → GL(m, E) be a representation, then it follows immediately from equation (9.1) that the subset defined as

VQ

ρ = {v ∈ E m

| ∀σ ∈ Gal(E, Q), ρσ(v) = vσ} (9.2)

is a rational subspace of Em. This subspace is already close to being a rational form of Em in the sense of the following lemma:

Lemma 9.4. If a set of vectors v1, . . . , vk of VρQ is linearly independent over

Q, then this set is also linearly independent over E as vectors of Em.

Proof. Assume that the lemma does not hold and take vectors v1, . . . , vk of

VQ

ρ with k minimal which are linearly independent over Q but contradict the

statement. This means that there exists xi∈ E such that k

X

i=1

xivi= 0. (9.3)

From the minimality of k it follows that x16= 0 and thus by multiplying this

equation by x−1₁ we can assume that x1= 1. Since k is minimal, it follows that

the xi are the unique elements of E such that x1= 1 and equation (9.3) is true.

If we apply the map ρσ to the equation, we get that

0 = ρσ k X i=1 xivi ! = k X i=1 xiρσ(vi) = k X i=1 xiviσ= k X i=1 σ(xi)vi !σ

because of (9.1). We also have that

k

X

i=1

σ(xi)vi= 0.

Minimality of k and the fact that σ(x1) = σ(1) = 1 imply that σ(xi) = xi for

all i. Because this statement holds for all σ ∈ Gal(E, Q), we conclude that the coefficients xi lie in Q and thus we get a contradiction since vi was a set of

A NEW METHOD FOR CONSTRUCTING ANOSOV LIE ALGEBRAS 165

The lemma shows that the rational subspace VQ

ρ is a rational form of Emif its

dimension is maximal. This motivates the following definition:

Definition 9.5. A representation ρ : Gal(E, Q) → GL(m, E) is called Galois compatible (abbreviated as GC) if and only if dim_Q(VQ

ρ ) = m. Equivalently,

the representation ρ is GC if and only if E ⊗ VQ

ρ = E

m_{or if V}Q

ρ is a rational

form of Em.

The trivial representation is the easiest example of a GC representation, since in that case VQ

ρ = Q

m_{. The next examples we consider are the regular}

representation and more generally, representations given by permutations matrices.

Example 9.6. Let ρ be the regular representation of Gal(E, Q), i.e. take

the vector space over E spanned by the basis {vσ | σ ∈ Gal(E, Q)} and the

representation ρ induced by the relations ρτ(vσ) = vτ σfor all τ, σ ∈ Gal(E, Q).

Every element v of the rational vector space VQ

ρ is given by

v = X

σ∈Gal(E,Q)

σ(x)vσ

for some x ∈ E. This is a rational vector space of dimension [E : Q], which is also the order of the group Gal(E, Q) and thus the dimension of the regular representation ρ. This shows that ρ is indeed GC.

Example 9.7. First we recall what a permutation matrix is. For every permutation π ∈ Sn, there exists a matrix Kπ∈ GL(n, Q) defined as

(Kπ)ij=

1 j = π(i)

0 otherwise

Every matrix of the form Kπ for some π ∈ Sn is called a permutation matrix.

Let ρ : Gal(E, Q) → GL(n, Q) be given by permutation matrices, meaning that

ρσ is a permutation matrix for every σ ∈ Gal(E, Q). Write the standard basis

of En as e1, . . . , en and note that ρ also acts on the basis α = {ei| 1 ≤ i ≤ n}.

By decomposing the basis α into its orbits, we can assume without loss of generality that the action is transitive, i.e. for every i ∈ {1, . . . , n}, there exists a σ ∈ Gal(E, Q) such that ρσ(e1) = ei. In the Appendix we study the relation

between actions of groups on finite sets and subgroups of finite index.

For every 1 ≤ i ≤ n, fix an element σi∈ Gal(E, Q) with σi(e1) = ei. Note that

every element σ ∈ Gal(E, Q) such that σ(e1) = ei is of the form σ = σih for

some h ∈ H. Let H be the subgroup of Gal(E, Q) which fixes the vector e1 and

of index n in the group Gal(E, Q) as follows from the Appendix and thus the field E0 _{has degree n over Q. Let x ∈ E}0, then every element

v = n X i=1 σi(x)ei is an element of VQ

ρ . Since E0 has degree n over Q, this is a vector space of

dimension n over Q and thus ρ is GC.

Of course, there are also examples of representations which are not GC.

Example 9.8. Consider the field E = Q(i) with complex conjugation σ : E → E as a field automorphism, meaning that σ is defined by σ(i) = −i. The Galois

group of E over Q is equal to

Gal(E, Q) = {1E, σ} ' Z2.

Now consider the representation ρ : Gal(E, Q) → GL(2, E) defined as

ρ(σ) =0 −i i 0



.

The subspace VQ

ρ is given by the condition

x y  ∈ VQ ρ ⇐⇒ 0 −i i 0  x y  =σ −1_(x) σ−1(y)  =σ(x) σ(y)  ⇐⇒ ( −iy = σ(x) ix = σ(y) ⇐⇒ ( iσ(y) = x ix = σ(y) ⇐⇒ ( −x = i2_{x = x} ix = σ(y) ⇐⇒ x y  =0 0  and thus VQ

ρ is the trivial subspace. This shows that the representation ρ is

not GC, although it is conjugate over GL(n, E) to the representation

σ 7→1 0

0 −1 

∈ GL(2, Q) which is GC.

A NEW METHOD FOR CONSTRUCTING ANOSOV LIE ALGEBRAS 167

The goal of the remaining part of this section is to show that every rational representation, so every representation into GL(m, Q), is Galois compatible.

Proposition 9.9. Let E be a Galois extension of Q. Every representation

Gal(E, Q) → GL(m, Q) is Galois compatible.

To prove this statement we will first show that it holds for irreducible representations and then apply the following lemma.

Lemma 9.10. Let ρi : Gal(E, Q) → GL(ni, E) with i ∈ {1, 2} be

representations, then the following are equivalent: 1. ρ1 and ρ2 are Galois compatible.

2. ρ1⊕ ρ2: Gal(E, Q) → GL(n1+ n2, E) is Galois compatible.

Proof. Let VQ

ρ1 and Vρ2Q be the rational subspaces corresponding to ρ1and ρ2,

then by definition it follows that

VQ ρ1⊕ρ2 = V Q ρ1⊕ V Q ρ2.

Indeed, we have the following equivalences (v, w) ∈ VQ ρ1⊕ρ2 ⇐⇒ (v, w) σ_{= (ρ} 1⊕ ρ2)σ(v, w) ∀σ ∈ Gal(E, Q) ⇐⇒ (vσ_{, w}σ_{) = ((ρ} 1)_σ(v), (ρ2)_σ(w)) ∀σ ∈ Gal(E, Q) ⇐⇒ v ∈ VQ ρ1, w ∈ V Q ρ2

and thus the claim holds. The statement of the lemma then easily follows by using Lemma 9.4.

Example 9.8 showed that being GC is not preserved under E-equivalence for general fields E. For the field Q this is true though.

Lemma 9.11. Let ρi: Gal(E, Q) → GL(n, E) with i ∈ {1, 2} be representations

which are Q-equivalent. Then ρ1 is GC if and only if ρ2 is GC.

Proof. Let P ∈ GL(n, Q) such that ρ1(σ) = P ρ2(σ)P−1for every σ ∈ Gal(E, Q).

Then the vector spaces VQ

ρ1 and V Q ρ2 satisfy VQ ρ1 = P V Q ρ2P −1

The proof of the proposition is now immediate by using the previous lemmas.

Proof of Proposition 9.9. From Lemma 9.10 and Lemma 9.11, it follows that

it is sufficient to prove the proposition for the Q-irreducible representations of Gal(E, Q). Since every Q-irreducible representation is a subrepresentation of the regular representation (see e.g. [68, Corollary 9.5.]), the statement follows from Example 9.6 above.