In this section we describe the structure of the module of logarithmic vector fields for a linear free divisor, see [28] and [25], and we give a bound on the number of semisimple vector fields in a basis of such module.
Let D ⊂ Cn be a linear free divisor defined by the homogeneous polynomial
f = det((δi(xj))i,j)∈C[x1, . . . , xn] of degreen, whereδ1, . . . , δn is a basis of weight
zero vector fields of Der(−logD). Thenδi(f)∈C·f and there is the standard Euler vector fieldχ=Pn
i=1xi∂/∂xi∈ hδ1, . . . , δniC.
Since χ(f)/f =n6= 0, we can assume that δ1 =χ and δi(f) = 0 fori= 2, . . . , n.
Soδ2, . . . , δnis a degree zero basis of the annihilator Ann(D) ofD. Sinceχvanishes
only at the origin, the origin of the affine coordinate systemx1, . . . , xn is uniquely
determined. A coordinate change between two degree zero bases of Der(−logD) can always be chosen linear. Among all possible linear coordinate changes, lets+ 1 be the maximal number of linearly independent diagonal weight zero logarithmic vector fields.
Forδ ∈DerCn a weight zero vector field, we write δS for its semisimple part and
δN for its nilpotent part. Then we have the following:
Theorem 1.4.1. ([25], Theorem 6.1) Let D =V(f)⊂Cn be a linear free divisor.
Then there exists a global degree zero basis χ, σ1, . . . , σs, ν1, . . . , νn−s−1 such that
1. [χ, σi] = 0and [χ, νj] = 0for all i= 1, . . . , s and j= 1, . . . , n−s−1;
2. the σi are simultaneously diagonalizable with eigenvalues in Q and σi(f) = 0;
3. the νj are nilpotent and νj(f) = 0;
5. ifδ∈Γ(Cn,Der(−logD))is a weight zero vector field such that[χ, δ] = 0and [σi, δ] = 0for i= 1, . . . , s, then δS ∈ hχ, σ1, . . . , σsiC.
Moreover, s ≥ 1 and if s = n−1 then f = x1· · ·xn defines a normal crossing
divisor.
In Theorem 1.4.1, one can perform the Gauss algorithm on the diagonalsσ1, . . . , σs.
Thenσi≡xi∂/∂xi mod Pjn=s+1C·xj∂/∂xj.
The following Lemma is useful in many examples:
Lemma 1.4.2. ([25], Lemma 6.3) Let σ = Pn
i=1wixi∂/∂xi. Then xi∂/∂xj is an
eigenvector of adσ for the eigenvalue wi−wj.
We now improve this description of Der(−logD), putting a lower bound on s
and noticing that each logarithmic vector field that is nilpotent annihilates each irreducible component ofD.
Proposition 1.4.3. Let D ⊂ Cn be a linear free divisor, let f = Qk
i=1fi be a
reduced defining equation forD written as a product of irreducible polynomials and lets+ 1 be the number of semisimple vector fields in the basis ofDer(−logD) given by Theorem 1.4.1. Thens+ 1≥k.
Proof. We can suppose that f1, . . . , fk are the semi-invariant polynomials given by
Proposition 1.3.9 and letχ1, . . . , χk be the corresponding independent characters.
We want to show that there exist σ1, . . . , σk ∈Der(−logD) such that dfj(σi) =
δijfj. This will conclude the proof because if they are not semisimple, we can take
their semisimple part and if they exist, they are automatically linearly independent. Becauseχ1, . . . , χkare independent, so aredeχ1, . . . , deχkas elements of the dual
space of gD. Hence there exist ν1, . . . , νk ∈ gD such that deχi(νj) = δij. Let
now σ1, . . . , σk be the corresponding vector fields on Cn. Because fi is a semi-
invariant with character χi, we have that fi(gx) = χi(g)fi(x) for all x ∈ Cn and
g ∈ G◦D. Differentiating the previous expression with respect to g, we have that
dxfi(σj) =deχi(νj)fi =δijfi.
Notice that if in the previous Proposition we consider the case k = 1, then the inequality is strict because we always have s≥1, as proved in Theorem 1.4.1.
Example 1.4.4. 1. Consider the linear free divisor D=V((yz+xw)zw)⊂C4
with Saito matrix
x 3x x z y 0 2y −w z z −z 0 w −2w 0 0
This linear free divisor has3 components and3semisimple vector fields in the basis from Theorem 1.4.1.
2. Consider the linear free divisor D = V(y2z2 −4xz3 −4y3w + 18xyzw −
27x2w2)⊂
C4 with Saito matrix
x 0 x y y 3x y 2z z 2y −z 3w w z −3w 0
This linear free divisor has just1 component but 2 semisimple vector fields in the basis from Theorem 1.4.1.
Proposition 1.4.5. ([6], Proposition 11.8) Let G be a connected algebraic group with Lie algebrag and ν ∈g. Then ν is semisimple if and only if it is tangent to a torus in G.
Proposition 1.4.6. Let D ⊂ Cn be a linear free divisor, let f = Qki=1fi be a
reduced defining equation forD written as a product of irreducible polynomials and let ν ∈ Der(−logD) be a nilpotent vector field. Then ν ∈ Ann(V(fi)) for all
i= 1, . . . , k.
Proof. To prove the statement it is enough to show that for everyg ∈SI(G◦D,Cn) we havedg(ν) = 0.
Consider g∈ SI(G◦D,Cn) with character χ and let v ∈gD be the corresponding
nilpotent element toν. Then we have thatdg(ν) =deχ(v)g, so it is enough to prove
thatdeχ(v) = 0.
We can assume that the character χ:G◦D −→ C∗ is non-trivial, then χ induces an isomorphism of a one dimensional quotient ofG◦D, a torus, onto the image. The corresponding one dimensional quotient of the Lie algebragD is then isomorphic to the Lie algebra of the 1-torus and thus consists of semisimple elements by Proposition 1.4.5. In particular, all nilpotent elements ofgD must lie in the kernel of deχ. Remark 1.4.7. The conclusion of Proposition 1.4.6 does not hold ifν is semisimple, also if ν ∈Ann(D).
Proof. Consider the linear free divisor D = V(f) = V((yz +xw)zw) ⊂ C4 and
σ = x∂/∂x+ 2y∂/∂y−z∂/∂z ∈Der(−logD). Then σ(f) = 0 but σ(yz+xw) =