automorphisms
In this section, we establish a strong relation between gradings on Lie algebras and the existence of certain automorphisms on this Lie algebra. This result is true for any field of characteristic 0, although the proof we give here is given for the rational numbers.
Theorem 8.7. Let nE a Lie algebra over the field E ⊆ C. The Lie algebra nE
has a positive grading if and only if it has an expanding automorphism.
Note that this result for E = Q is exactly the equivalence of Theorem 6.13 that wasn’t proved in Chapter 6.
For complex Lie algebras, Theorem 8.7 was already described in [36] so this section extends these results to arbitrary subfields of C. This translation between positive gradings and expanding automorphisms allows us to use the theory of linear algebraic groups to study positive gradings.
One implication of Theorem 8.7 is immediate, namely that a positive grading implies the existence of expanding automorphisms. Given a positive grading of
nE =M
i>0
nEi ,
take any µ ∈ E such that |µ| > 1 and consider the automorphism ϕµ: nE→ nE
defined as
ϕµ(X) = µiX ∀X ∈ nEi .
The automorphism ϕ only has eigenvalues µi with i > 0 and since |µ| > 1, the automorphism ϕµ is expanding. The hard part is to show that this construction
also works the other way around. We will prove the following, slightly stronger theorem.
Theorem 8.8. Let nE be a Lie algebra over the field E ⊆ C. If ϕ ∈ Aut(nE)
is an expanding automorphism, then there exists a positive grading on nE such
that every automorphism commuting with ϕ preserves this positive grading.
In particular, this positive grading is preserved by ϕ itself. We only give a proof of this theorem in the case E = Q, but this proof generalizes immediately to more general fields. The proof of Theorem 8.8 is based on some facts about algebraic number fields as discussed in Chapter 2.
Take any diagonalizable matrix A ∈ GL(n, Q) with characteristic polynomial
p(X) ∈ Q[X]. Denote by E ⊆ C the minimal splitting field of p(X), which is of
course an algebraic number field. Let v1, . . . , vnbe a basis of eigenvectors for the
matrix A, considered as a linear map Cn→ Cn, with eigenvalues respectively
λ1, . . . , λn. We define the matrix N (A) ∈ GL(n, C) by
N (A)(vi) = NE(λi)vi,
so it has the same eigenvectors as A but with eigenvalues NE(λi) rather than λi.
This definition does not depend on the choice of the basis vi. Let P ∈ GL(n, Q)
be an invertible matrix over Q and consider the matrix P AP−1. If v ∈ Cnis an eigenvector of A corresponding to the eigenvalue λ, then P (v) is an eigenvector of P AP−1 also corresponding to the eigenvalue λ, because
RELATION BETWEEN GRADINGS AND AUTOMORPHISMS 143
This implies that N (P AP−1) = P N (A)P−1, showing that N (A) is invariant under change of basis. Note that if B commutes with A, then B also commutes with N (A).
A priori we only know that N (A) ∈ GL(n, C), but we show that N (A) ∈ GL(n, Q) for every diagonalizable matrix A ∈ GL(n, Q). If p(X) ∈ Q[X] is a polynomial over Q, we denote by L(p) the companion matrix of the polynomial
p. Let L be the primary rational canonical form of A as introduced in Theorem
5.11. This means that
L = L(pm1 1 ) 0 . . . 0 0 L(pm2 2 ) . . . 0 .. . ... . .. ... 0 0 . . . L(pmk k )
with mi> 0 and the polynomials pi(X) irreducible over Q. Denote the degree
of the polynomials pi as ni. Since A is similar over Q to its primary rational
canonical form L, there exists P ∈ GL(n, Q) such that P AP−1= L.
Now take λi a root of the polynomial pi. Note that NE(λi) doesn’t depend on
the choice of the root λi, since all roots of pi are E-conjugate. It’s easy to see
then that N (L) is equal to
N (L) = NE(λ1)1m1n1 0 . . . 0 0 NE(λ2)1m2n2 . . . 0 .. . ... . .. 0 0 0 . . . NE(λk)1mknk ∈ GL(n, Q).
From this it follows that N (A) = P−1N (L)P ∈ GL(n, Q).
Now assume that nQis a rational Lie algebra and ϕ ∈ Aut(nQ) an automorphism.
Then N (ϕ) is a linear map nQ→ nQand we claim that it is also a Lie algebra
automorphism. Consider the complexification nCof nQ, then it suffices to show
that N (ϕ)C= 1
C⊗ N (ϕ) : nC→ nC is a Lie algebra automorphism. By the
linearity of the bracket, it is sufficient to show that N (ϕ)Cpreserves the brackets
between basis vectors for a basis of nC. Since there exists a basis of eigenvectors
for ϕC, we show that N (ϕ)C preserves the bracket of all eigenvectors. Let vλ
and vµ be two eigenvectors of ϕC, then [vλ, vµ] is an eigenvector of eigenvalue
λµ. So [N (ϕ)C(v λ), N (ϕ)C(vµ)] = NE(λ)NE(µ)[vλ, vµ] = NE(λµ)[vλ, vµ] = N (ϕ)C([v λ, vµ]),
with E the minimal splitting field for the characteristic polynomial of ϕ. To define N (ϕ) we assumed that ϕ was semisimple. We now extend the definition of N to arbitrary automorphisms of nQ. Let ϕ : nQ → nQ be an
automorphism, then ϕ = ϕsϕufor some semisimple automorphism ϕs∈ Aut(nQ)
and a unipotent automorphism ϕu ∈ Aut(nQ) because of the ‘Multiplicative
Jordan decomposition’ of Theorem 5.15. Every automorphism commuting with
ϕ also commutes with ϕs. We define N (ϕ) as N (ϕs) of its semisimple part in
the Jordan decomposition.
Alternatively, we could use the generalized eigenspaces of ϕ to define N (ϕ), see Theorem 5.3. If nQ= ⊕
λ∈CVλis the decomposition into generalized eigenspaces
of ϕ, then N (ϕ) is given by X 7→ NE(λ)X for every X ∈ Vλ.
We conclude this discussion with the following proposition:
Proposition 8.9. Let ϕ ∈ Aut(nQ) be an automorphism of a rational Lie
algebra nQ. Then N(ϕ) ∈ Aut(nQ) is an automorphism with only rational
eigenvalues and commuting with every automorphism which commutes with ϕ. Moreover, if ϕ is expanding, then N(ϕ) is also expanding.
Proof. The only thing left to show is the last statement, namely that N(ϕ) is
expanding if ϕ is expanding. The eigenvalues of N (ϕ) are equal to NE(λ) with
λ an eigenvalue of ϕ and E the splitting field of the characteristic polynomial of ϕ. So it suffices to show that if λ and all its E-conjugates are > 1 in absolute
value, then also NE(λ) > 1. Since NE(λ) is just the product of the E-conjugates
of λ, this follows immediately.
From this proposition, the proof of Theorem 8.8 with E = Q now follows directly.
Proof of Theorem 8.8. Let ϕ : nQ→ nQ be an expanding automorphism. Then
N (ϕ) is an expanding automorphism with only eigenvalues in Q. By squaring ϕ if necessary we can assume that all eigenvalues of N (ϕ) are positive. Since
all eigenvalues are rational, the corresponding eigenspaces are rational subspace of n. Take the R-grading n =L
r∈RVr where Vr is the eigenspace of eigenvalue
er. It’s easy to see that this is indeed an R-grading. Since N (ϕ) is expanding,
it follows that the grading is positive. The existence of a positive R-grading implies the existence of a positive grading by just renaming the subspaces Vr,
as explained above.
If ψ commutes with ϕ, it also commutes with N (ϕ) and thus it preserves the eigenspaces of N (ϕ). Since the grading is given by eigenspaces of N (ϕ), the last statement follows.
RELATION BETWEEN GRADINGS AND AUTOMORPHISMS 145
Remark 8.10. The proof for general fields E is almost identical. If E ⊆ F is a
field extension of finite degree n, then there also exist exactly n monomorphisms
E → C which fix the field E pointwise. For general matrices in A ∈ GL(k, E),
we can then define the norm NE(A) which is a matrix with only eigenvalues in
E and which commutes with every matrix commuting with A. If ϕ ∈ Aut(nE),
then NE(ϕ) will again be an automorphism of nE.
The only difference is that E is in general not a subfield of R. This means that we do not find an R-grading directly from the automorphism NE(ϕ). But the
linear map ˜ϕ which maps an eigenvector X corresponding to an eigenvalue λ to
˜
ϕ(X) = |λ|2X
has eigenvalues in E ∩ R since |λ|2 ∈ E ∩ R. Moreover, this map is also an automorphism of the Lie algebra.
We now generalize these results to cohopfian F -groups. The translation of expanding maps to the Lie algebra is an expanding automorphism. We start by showing that the translation of a non-trivial self-cover is a partially expanding map.
Assume that N is an F -group which is not cohopfian. This means there exists an injective group morphism ϕ : N → N which is not surjective. From Section 6.2 we know that ϕ has characteristic polynomial p(X) ∈ Z[X]. Since ϕ is not surjective, Proposition 2.30 implies that | det(ϕ)| > 1. If we decompose p(X) into its Q-irreducible components
p(X) = p1(X) . . . pl(X),
then pi(X) ∈ Z[X] and at least one pi(X) satisfies |pi(0)| > 1. This means that
N (ϕ) is an automorphism with only eigenvalues in Z and at least one eigenvalue λ with |λ| > 1. By squaring N (ϕ) if necessary we get that nQ has a partially
expanding automorphism.
Theorem 8.11. Let N be an F-group with corresponding Lie algebra nQ. If
ϕ : N → N is an injective group morphism which is not surjective, then nQ has
a partially expanding automorphism which commutes with every automorphism commuting with ϕ.
So the translation of Theorem 8.8 from expanding maps to non-trivial self-covers is the following.
Theorem 8.12. Let E be a field of characteristic 0 and nE a Lie algebra over
E. The Lie algebra nE has a non-trivial non-negative grading if and only if it
Proof. Let ϕ : nE→ nE be a partially expanding automorphism and consider
the automorphism NE(ϕ) ∈ Aut(nE). We can assume that all eigenvalues of
NE(ϕ) are positive real numbers, just as in the case of expanding automorphisms.
The eigenspaces nE
r for the eigenvalues er forms an R grading of nE which is
non-negative and non-trival. Every automorphism commuting with ϕ also commutes with NE(ϕ) and thus preserves the eigenspaces.