SECANTS, TANGENTS AND THE HOMOGENEITY OF

FREUDENTHAL VARIETIES OF CERTAIN TYPE

Hajime KAJI^{⋆} and Osami YASUKURA^{⋆⋆}

Dedicated to Professor Hiroshi Asano on the occasion of his 68th birthday

Abstract. For a given simple, graded Lie algebra 𝔤 = Σ𝔤_{i} of contact type, we define a certain projective variety V ⊆ ℙ_{∗}(𝔤_{1}) in terms of the Lie bracket, and call V the Freudenthal variety associated to 𝔤. In this article, we introduce a certain stratification of the ambient space ℙ_{∗}(𝔤_{1}), and discuss about projective geometry of V, especially, their secant and tangent loci. We discuss also about the homogeneity of V.

0. Introduction

H. Freudenthal constructed, in a series of his papers (see [F] and its references), the exceptional Lie algebras of type E_{8}, E_{7}, E_{6} and F_{4}, with defining various projective varieties. The purpose of our work is to study projective geometry for his varieties of certain type, which are called varieties of planes in the symplectic geometry of Freudenthal (see [F, 4.11], [O, 2.3]). Precisely speaking, for a given simple, graded Lie algebra 𝔤 = Σ𝔤_{i} of contact type over the complex number field ℂ (see [T]), set

V := {x ∈ 𝔤_{1}|x ≠ 0, (ad x)^{2}𝔤_{−2} = 0},

and define an algebraic set V in ℙ_{∗}(𝔤_{1}) to be the projectivization of V:

V := πV,

where π : 𝔤_{1} \ {0} → ℙ_{∗}(𝔤_{1}) is the natural projection. Then we call V ⊆ ℙ_{∗}(𝔤_{1}) (with the reduced structure) the Freudenthal variety associated to the graded Lie algebra 𝔤, which is a natural generalization of Freudenthal’s varieties mentioned above: Note that V is not necessarily connected in this general setting.

Moreover set

S := {x ∈ 𝔤_{1}|(ad x)^{4}𝔤_{−2} ≠ 0},
T := {x ∈ 𝔤_{1}|(ad x)^{3}𝔤_{−2} ≠ 0, (ad x)^{4}𝔤_{−2} = 0},
U := {x ∈ 𝔤_{1}|(ad x)^{2}𝔤_{−2} ≠ 0, (ad x)^{3}𝔤_{−2} = 0}.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. Primary 14N05, 17B70; Secondary 14M17, 17A40.

Key words and phrases. Freudenthal varieties, secant loci, tangent loci, homogeneity.

—2001/03/28; version 1.6—

Then we have the following stratification of 𝔤_{1}:

𝔤_{1} = S ⊔ T ⊔ U ⊔ V ⊔ {0}.

In the literature, several results have been known about the structure of 𝔤_{1} as a 𝔤_{0}-space, case-by-case for each exceptional Lie algebra of types E_{8}, E_{7}, E_{6} and F_{4}, from the viewpoint of the invariant theory of prehomogeneous vector spaces (see [H], [I], [K], [Mr]): By virtue of those results, it can be shown that the stratification gives the orbit decomposition of the 𝔤_{0}-space 𝔤_{1} for those exceptional Lie algebras.

In this article we study the Freudenthal varieties V with the stratification of the ambient space ℙ_{∗}(𝔤_{1}), from the view-point of projective geometry, not individually but systematically in terms of abstract Lie algebras, without depending on the classification of simple Lie algebras as well as on the known results for each case of types E_{8}, E_{7}, E_{6} and F_{4}.

As a consequence of our work, it turns out that the stratification of 𝔤_{1} is closely related to the secant and tangent loci of V ⊆ ℙ_{∗}(𝔤_{1}). Here, the secant locus Σ_{P} as well as the tangent locus Θ_{P} of V with respect to a given point P ∈ ℙ_{∗}(𝔤_{1}) are defined by

Σ_{P}^{∘} :=\{Q ∈ V |∃R ∈ V \ \{Q\}, P ∈ Q ∗ R\}, Σ_{P} := \overline{Σ_{P}^{∘}} ,
Θ_{P} :={Q ∈ V |P ∈ T_{Q}V },

where we denote by Q ∗ R the projective line in ℙ_{∗}(𝔤_{1}) through Q and R, namely, the secant line of V determined by Q and R, and by T_{Q}V the embedded tangent space to V at Q in ℙ_{∗}(𝔤_{1}) (see, for example, [FR]).

To state the consequence, we need some notation: Let H be the characteristic element of the gradation 𝔤 = ∑ 𝔤_{i}, take E_{+} ∈ 𝔤_{2} and E_{−} ∈ 𝔤_{−2} such that (E_{+}, H, E_{−}) form an 𝔰𝔩_{2}-triple, and define a homogeneous quartic polynomial q over 𝔤_{1} as follows:

2q(x)E_{+} = (ad x)^{4}E_{−}.

Then, the consequence is

Theorem A. We have:

(1) If s ∈ S, then

Σ_{πs} = Σ_{πs}^{∘} = \{π ((ad s)^{3}E_{−} ± \sqrt{3q(s)}s)\} , Θ_{πs} = ∅.

(2) If t ∈ T , then

Σ_{πt} = ∅, Θ_{πt} = {π ((ad t)^{3}E_{−})}.

(3) Assume that V is irreducible. If u ∈ U, then dim Σ_{πu} ≥ 1 and Θ_{πu} ≠ ∅.

In particular, it turns out that non-empty Freudenthal varieties V ⊆ ℙ_{∗}(𝔤_{1}) are socalled varieties with one apparent double point (Corollary A2), that is, for a general point P ∈ ℙ_{∗}(𝔤_{1}) there exists a unique secant line of V through P (see [SR, IX]). We also show that if V is neither empty nor irreducible, then V is a disjoint union of two linear subspaces of ℙ^{2n−1} of dimension n − 1, where 2n = dim 𝔤_{1} (Corollary B2).

On the other hand, we discuss about the homogeneity of V as well. Consider a subalgebra 𝔇_{0} of 𝔤_{0} as follows:

𝔇_{0} := Ker(ad E_{+}|_{𝔤_{0}}),

and denote by G the connected, closed subgroup of Int 𝔤, the inner automorphism group of 𝔤, with Lie algebra 𝔇_{0}. Then, we show Theorem B (Cf. [Mk, (5.8), (2)]). G acts transitively on each irreducible components of V.

Besides the results above, we discuss about the dimension of V (Corollary B1 (1)) and the projective lines contained in V (Corollary B3). Moreover, for the tangent variety of V, we discuss about its defining equation (Corollary A5), its singular locus (Corollary A6) and a certain duality (Corollary B1 (4)), where the tangent variety of V is by definition the union of embedded tangent spaces to V. On the other hand, we discuss also about relationship among the strata of 𝔤_{1} (Corollaries A1, A3 and A4).

The contents of this article are organized as follows: In §1 we define a certain ternary product in 𝔤_{1} and a natural skew-symmetric form on 𝔤_{1}. Then we prove some results on 𝔤_{1} and the stratification above, which are used in §2. In §2 we prove firstly the assertion for secant loci in Theorem A (1), and then we give some corollaries to this result. Using those results, we next prove Theorem B and its corollaries, before continuing the proof of Theorem A. Then, using the results obtained so far, we prove the remaining part of Theorem A. In our study, the ternary product plays a key role, which was originally introduced by K. Yamaguti and H. Asano ([A1], [A3] and [YA]).

Finally we should mention that S. Mukai announced a theorem [Mk, (5.8)] on cubic Veronese varieties without proofs. Our work was originated by looking for proofs of the corresponding statements for Freudenthal varieties (Theorem B, Corollaries A2, A5, B1 (3) and (4)). In fact, we see from his list [Mk, (5.10)] of cubic Veronese varieties that the notion of our Freudenthal varieties seems to coincide with that of cubic Veronese varieties, although we do not know any a priori proof for this coincidence.

1. Preliminaries

Let 𝔤 be a simple Lie algebra of rank ≥ 2, let

𝔤 = 𝔤_{−2} + 𝔤_{−1} + 𝔤_{0} + 𝔤_{1} + 𝔤_{2}

be a graded decomposition of contact type, and let H be the characteristic element: we have

𝔤_{i} = {x ∈ 𝔤|(ad H)x = ix}

and

[𝔤_{i}, 𝔤_{j}] ⊆ 𝔤_{i+j}, 𝔤_{1} ≠ 0, dim 𝔤_{±2} = 1

(see [T]). Taking E_{+} ∈ 𝔤_{2} and E_{−} ∈ 𝔤_{−2} such that (E_{+}, H, E_{−}) form an 𝔰𝔩_{2}-triple, we define a tri-linear map, [, ,] : 𝔤_{1} × 𝔤_{1} × 𝔤_{1} → 𝔤_{1} and a skew-symmetric form, ⟨, ⟩ : 𝔤_{1} × 𝔤_{1} → ℂ by

2[a, b, c] =[c, [b, [a, E_{−}]]] + [c, [a, [b, E_{−}]]],
2⟨a, b⟩E_{+} =[a, b]

with a, b, c ∈ 𝔤_{1}. It is easily shown that the skew-symmetric form ⟨, ⟩ is non-degenerate since 𝔤 is simple (see [T, Lemma 3.2 (2)]). Moreover, we have the following Fact (Yamaguti-Asano). For any v, w, x, y, z ∈ 𝔤_{1}, the following formulas hold:

(S1) [xyz] = [yxz].

(S2) [xyz] − [xzy] = ⟨x, z⟩y − ⟨x, y⟩z + 2⟨y, z⟩x.

(S3) [vw[xyz]] = [[vwx]yz] + [x[vwy]z] + [xy[vwz]].

We should mention that the formulas above with the ternary product [, ,] were originally introduced by Yamaguti and Asano, as the axioms of certain algebras, which they call symplectic triple systems (see [A1], [A3] and [YA]). They also proved the formulas hold for 𝔤_{1} just as in our situation, in order to show that every simple, graded Lie algebra of contact type is systematically constructed from some simple symplectic triple system. Although (S1) is obvious from the definition and (S2) is not difficult to show, some complicated computations are necessary to show (S3). Since the only reference for the proof of this fact is [A3] written in Japanese, we give a proof for (S3) (including (S2)) in Appendix for the convenience of the readers.

Now, for the subalgebra 𝔇_{0} = Ker(ad E_{+}|𝔤_{0}) of 𝔤_{0}, it is easily shown that

𝔤_{0} = 𝔇_{0} ⊕ ℂH.

On the other hand, for a, b ∈ 𝔤_{1} we define a linear map L(a, b) : 𝔤_{1} → 𝔤_{1} by

L(a, b)c := [abc]

with c ∈ 𝔤_{1}. If we define a × b ∈ 𝔤_{0} for a, b ∈ 𝔤_{1} by

−2a × b = [b, [a, E_{−}]] + [a, [b, E_{−}]],

then [a × b, c] = [abc] = L(a, b)c and a × b ∈ 𝔇_{0} (see [KY, Proposition 2 (a)]). Since the adjoint action of 𝔤_{0} on 𝔤_{1} is faithful (see [T, Lemma 3.2 (1)]), we may identify L(a, b) with a × b, so that we may consider L(a, b) ∈ 𝔇_{0}. Note that with the notation above, we have

S = {s ∈ 𝔤_{1}|⟨s, [sss]⟩ ≠ 0},
T = {t ∈ 𝔤_{1}|[ttt] ≠ 0, ⟨t, [ttt]⟩ = 0},
U = {u ∈ 𝔤_{1}|L(u, u) ≠ 0, [uuu] = 0},
V = {v ∈ 𝔤_{1}|v ≠ 0, L(v, v) = 0},

and q(a) = ⟨a, [aaa]⟩. Furthermore, we have

Lemma 1. The subalgebra 𝔇_{0} is linearly spanned by the set {L(a, b)|a, b ∈ 𝔤_{1}}.

Proof. Since 𝔤 is simple, we have [[𝔤_{1}, E_{−}], 𝔤_{1}] = 𝔤_{0} = 𝔇_{0} ⊕ ℂH (see [T, Lemma 3.1]). On the other hand, according to [KY, Proposition 2 (b)], we have

[[a, E_{−}], b] = a × b + ⟨a, b⟩H

for a, b ∈ 𝔤_{1} (see also the proof of (S2) in Appendix). Thus the result follows. □

Lemma 2. For any a, b ∈ 𝔤_{1} and D ∈ 𝔇_{0}, we have ⟨Da, b⟩ + ⟨a, Db⟩ = 0.

Proof. From the definition of 𝔇_{0} and the Jacobi identity we obtain

0 = 2⟨a, b⟩D(E_{+}) = D([a, b]) = [Da, b] + [a, Db]. □

Remark. It is known that Lemmas 1 and 2 above hold for any simple, symplectic triple systems (see [A1, Lemma 2.1 and Theorem 2.9]). On the other hand, all the results below in this section are deduced from Lemmas 1, 2, the non-degeneracy of ⟨, ⟩, and the axioms of symplectic triple systems, (S1), (S2) and (S3). According to [A1, Theorem 2.3] (see also [YA, Theorem 2]), a symplectic triple system is simple if and only if ⟨, ⟩ is nondegenerate. Therefore it turns out that all the results in this section hold for arbitrary simple, symplectic triple systems.

Lemma 3. For any a, b, c ∈ 𝔤_{1} and D ∈ 𝔇_{0}, we have D[abc] = [Dabc] + [aDbc] + [abDc].

Proof. By virtue of Lemma 1, we may assume that D is of the form L(d, e) with d, e ∈ 𝔤_{1}. Then the claim is nothing but (S3). □

Proposition 1. For any x ∈ V, we have:

(1) (Asano [A2]) L(a, x)x = 3⟨a, x⟩x for any a ∈ 𝔤_{1}. In particular, if L(a, x) = 0, then ⟨a, x⟩ = 0.

(2) x ∈ 𝔇_{0}x.

Proof. (1) It follows from (S1), (S2) and x ∈ V that

L(a, x)x = [xax] = [xxa] + ⟨x, x⟩a − ⟨x, a⟩x + 2⟨a, x⟩x = 3⟨a, x⟩x.

(2) This follows from (1) and the non-degeneracy of ⟨, ⟩. □

Proposition 2. For any a, b, c, d ∈ 𝔤_{1}, we have ⟨L(a, b)c, d⟩ = ⟨L(c, d)a, b⟩.

Proof. It follows from (S2) that

⟨[abc], d⟩ = ⟨[acb], d⟩ + ⟨a, c⟩⟨b, d⟩ − ⟨a, b⟩⟨c, d⟩ + 2⟨b, c⟩⟨a, d⟩,
⟨[cda], b⟩ = ⟨[cad], b⟩ + ⟨c, a⟩⟨d, b⟩ − ⟨c, d⟩⟨a, b⟩ + 2⟨d, a⟩⟨c, b⟩.

Therefore, using (S1) and Lemma 2, we have

⟨[abc], d⟩ − ⟨[cda], b⟩ = ⟨[acb], d⟩ − ⟨[cad], b⟩ = ⟨[acb], d⟩ + ⟨b, [acd]⟩ = 0. □

Proposition 3. For any x ∈ V and D, E ∈ 𝔇_{0}, we have:

(1) (Asano [A2]) L(Dx, x) = 0.

(2) ⟨Dx, x⟩ = 0.

(3) ⟨Dx, Ex⟩ = 0.

(4) L(Dx, Ex)x = 0.

Proof. By virtue of Lemma 1, it suffices to show these formulas for D, E ∈ 𝔇_{0} of the form L(a, b) with a, b ∈ 𝔤_{1}.

(1) It follows from (S3), (S1) and x ∈ V that

0 = [ab[xxc]] = [[abx]xc] + [x[abx]c] + [xx[abc]] = 2[[abx]xc].

(2) This follows from (1) and Proposition 1 (1). This follows from Proposition 2 as well: Indeed, we have ⟨L(a, b)x, x⟩ = ⟨L(x, x)a, b⟩ = 0.

(3) It follows from Proposition 2 and (1) that ⟨L(a, b)x, L(c, d)x⟩ = ⟨L(x, L(c, d)x)a, b⟩ = 0.

(4) It follows from (S2) that

L([abx], [cdx])x = [[abx]x[cdx]] + ⟨[abx], x⟩[cdx] − ⟨[abx], [cdx]⟩x + 2⟨[cdx], x⟩[abx],

which is equal to zero: Indeed, L([abx], x) = 0 follows from (1), ⟨[abx], x⟩ = ⟨[cdx], x⟩ = 0 from (2), and ⟨[abx], [cdx]⟩ = 0 from (3). □

Proposition 4. If S = ∅, then T = ∅ and V = ∅.

Proof (Asano [A4]). For any a, b ∈ 𝔤_{1} and λ, μ ∈ ℂ, we have

0 = q(λa + μb) =λ^{4}q(a) + λ^{3}μ(⟨b, [aaa]⟩ + ⟨a, [baa]⟩ + ⟨a, [aba]⟩ + ⟨a, [aab]⟩)
+ λ^{2}μ^{2}(⋯) + ⋯ + μ^{4}q(b).

In particular, we have ⟨b, [aaa]⟩+⟨a, [baa]⟩+⟨a, [aba]⟩+⟨a, [aab]⟩ = 0. Since it follows from (S1) and (S2) that

(†)  [baa] = [aba] = [aab] − 3⟨a, b⟩a,

and it follows from Lemma 2 that ⟨a, [aab]⟩ = ⟨b, [aaa]⟩, we have

(‡)  ⟨a, [baa]⟩ = ⟨a, [aba]⟩ = ⟨a, [aab]⟩ = ⟨b, [aaa]⟩.

Therefore we have 4⟨b, [aaa]⟩ = 0 for any b ∈ 𝔤_{1}, so that [aaa] = 0 since ⟨, ⟩ is nondegenerate.

Similarly to the argument above, we obtain from [aaa] = 0 that [baa]+[aba]+[aab] = 0. Using (†), we have

3[aab] = 6⟨a, b⟩a.

Therefore, if L(a, a) = 0, then ⟨a, b⟩a = 0 for any b ∈ 𝔤_{1}, so that a = 0 since ⟨, ⟩ is non-degenerate. □

Proposition 5. For any a ∈ 𝔤_{1}, we have:

(1) L(a, [aaa]) = 0.

(2) L(a, a)^{2}a = 3q(a)a.

(3) L([aaa], [aaa]) = −3q(a)L(a, a). In particular, a ∈ T if and only if [aaa] ∈ V.

(4) [[aaa][aaa][aaa]] = −9q(a)^{2}a.

(5) q([aaa]) = 9q(a)^{3}.

Proof. (1) Setting v = w = x = y := a in (S3), using (S1), we have 2[[aaa]az] = 0. (2) It follows from (S2) and (1) that

[aa[aaa]] = [a[aaa]a] + ⟨a, [aaa]⟩a − ⟨a, a⟩[aaa] + 2⟨a, [aaa]⟩a = 3⟨a, [aaa]⟩a.

(3) It follows from (S3) that

[aa[[aaa]az]] = [[aa[aaa]]az] + [[aaa][aaa]z] + [[aaa]a[aaz]].

Then it follows from (1) and (2) that 0 = 3q(a)[aaz] + [[aaa][aaa]z].

(4) It follows from (3) and (2) that

L([aaa], [aaa])[aaa] = −3q(a)L(a, a)[aaa] = −9q(a)^{2}a.

(5) This follows from (4). □

Proposition 6. For any D ∈ 𝔇_{0} and x ∈ V, we have q(Dx) = 0.

Proof. It follows from Proposition 3 (4) and Lemma 3 that

0 = D([DxDxx]) = [D^{2}xDxx] + [DxD^{2}xx] + [DxDxDx],

so that [DxDxDx] = −2[D^{2}xDxx]. Hence it follows from Proposition 3 (3) that

⟨Dx, [DxDxDx]⟩ = −2⟨Dx, [D^{2}xDxx]⟩ = 0. □

Proposition 7. For a = u + v with [uuu] = 0 and v ∈ V, we have:

(1) L(a, a) = L(u, u) + 2L(u, v).

(2) [aaa] = 3[uuv] − 6⟨u, v⟩(u − v).

(3) q(a) = 12⟨u, v⟩^{2}.

Proof. (1) This follows from (S1) and v ∈ V.

(2) It follows from (1), [uuu] = 0, (S2) and Proposition 1 (1) that

[aaa] = [uuu] + [uuv] + 2[uvu] + 2[uvv]
= [uuv] + 2([uuv] + ⟨u, u⟩v − ⟨u, v⟩u + 2⟨v, u⟩u) + 2(3⟨u, v⟩v)
= 3[uuv] − 6⟨u, v⟩u + 6⟨u, v⟩v.

(3) It follows from (2), Lemma 2, Proposition 3 (2) and [uuu] = 0 that

q(a) = 3⟨u, [uuv]⟩ + 3⟨v, [uuv]⟩ − 6⟨u, v⟩⟨u + v, u − v⟩
= −3⟨[uuu], v⟩ + 12⟨u, v⟩^{2}
= 12⟨u, v⟩^{2}. □

Remark. It is easily shown that V ⊆ ℙ_{∗}(𝔤_{1}) has no trisecant line. Recall that a trisecant line of V ⊆ ℙ_{∗}(𝔤_{1}) is by definition a projective line L in ℙ_{∗}(𝔤_{1}) such that 3 ≤ #(L∩V) < ∞. Suppose that there exist x, y ∈ V such that πx ≠ πy and {πx, πy} ⊊ πx ∗ πy ∩ V, so that we may assume that x + y ∈ V. Then if follows from Proposition 7 (1) that λx + μy ∈ V for any λ, μ ∈ ℂ, that is, πx ∗ πy ⊆ V.

Proposition 8. For u ∈ 𝔤_{1} with [uuu] = 0 and v ∈ V, we have:

(1) L(u, [uuv]) = 2⟨u, v⟩L(u, u).

(2) [uuv] ∈ V ∪ {0}. Moreover, ℂv + ℂ[uuv] ⊆ V ∪ {0}.

Proof. (1) Setting a := u + v, we have

0 = L(a, [aaa]) (∵ Proposition 5 (1))
= L(u + v, 3[uuv] − 6⟨u, v⟩(u − v)) (∵ Proposition 7 (2))
= 3L(u, [uuv]) + 3L(v, [uuv]) − 6⟨u, v⟩L(u + v, u − v)
= 3L(u, [uuv]) − 6⟨u, v⟩L(u, u) (∵ Proposition 3 (1), v ∈ V).

(2) Similarly to the above, we have

0 = L([aaa], [aaa]) + 3q(a)L(a, a) (∵ Proposition 5 (3))

= 9{L([uuv], [uuv]) − 4⟨u, v⟩L(u, [uuv]) + 4⟨u, v⟩^{2}(L(u, u) − 2L(u, v))}

+ 36⟨u, v⟩^{2}(L(u, u) + 2L(u, v)) (∵ Proposition 7, (S1))

= 9{L([uuv], [uuv]) − 4⟨u, v⟩L(u, [uuv]) + 8⟨u, v⟩^{2}L(u, u)}.

Then, using (1), we obtain L([uuv], [uuv]) = 0. It follows from Proposition 3 (1) that L(v, [uuv]) = 0, so that the latter part follows as well. □

Proposition 9. For a ∈ 𝔤_{1}, we have:

(1) 3L(a, a)^{2}b = 8⟨b, [aaa]⟩a + 8⟨a, b⟩[aaa] + ⟨a, [aaa]⟩b for any b ∈ 𝔤_{1}.

(2) If a ∈ S, then the linear map L(a, a) has full rank.

Proof. (1) It follows from (S2) that

L(a, a)^{2}b = [aa[aba]] + 3⟨a, b⟩[aaa].

Moreover it follows from (S2) that [aa[aba]] = [a[aba]a] + 3⟨a, [aba]⟩a. Since ⟨a, [aba]⟩ = ⟨b, [aaa]⟩ by Proposition 2, we have

(⋆)  [aa[aba]] = [a[aba]a] + 3⟨b, [aaa]⟩a.

On the other hand, it follows from (S1) and (S3) that [ab[aaa]] = 2[a[aba]a] + [aa[aba]], while it follows from (S2) and Proposition 5 (1) that [ab[aaa]] = ⟨a, [aaa]⟩b − ⟨a, b⟩[aaa] + 2⟨b, [aaa]⟩a. Therefore, we have

(⋆⋆)  2[a[aba]a] + [aa[aba]] = ⟨a, [aaa]⟩b − ⟨a, b⟩[aaa] + 2⟨b, [aaa]⟩a.

Thus it follows from (⋆) and (⋆⋆) that

3[aa[aba]] = ⟨a, [aaa]⟩b − ⟨a, b⟩[aaa] + 8⟨b, [aaa]⟩a.

Therefore, combining this with the first formula, we obtain the required result.

(2) (Asano [A5]) Note that 𝔤_{1} = ℂa ⊕ ℂ[aaa] ⊕ (ℂa ⊕ ℂ[aaa])^{⊥}: Indeed, it follows from ⟨a, [aaa]⟩ ≠ 0 that ℂa ∩ ℂ[aaa] = (ℂa + ℂ[aaa]) ∩ (ℂa + ℂ[aaa])^{⊥} = {0}, and from the non-degeneracy of ⟨, ⟩ that dim(ℂa + ℂ[aaa]) + dim(ℂa + ℂ[aaa])^{⊥} = dim 𝔤_{1}.

Now, it is clear from (1) that

L(a, a)^{2}|_{ℂa+ℂ[aaa]} = 3⟨a, [aaa]⟩1_{ℂa+ℂ[aaa]},
L(a, a)^{2}|_{(ℂa+ℂ[aaa])^{⊥}} = \frac{1}{3}⟨a, [aaa]⟩1_{(ℂa+ℂ[aaa])^{⊥}}.

Therefore L(a, a)^{2} has full rank if ⟨a, [aaa]⟩ ≠ 0, so does L(a, a). □

2. Proofs and Corollaries

Proof of Theorem A. (1) We first show the assertion for secant loci: we prove the following

Theorem A’. For any s ∈ S there exists a unique pair {x, y} ⊆ V such that s = x + y, and we have

\{x, y\} = \{\frac{1}{2}s + \frac{1}{2\sqrt{3q(s)}}[sss], \frac{1}{2}s − \frac{1}{2\sqrt{3q(s)}}[sss]\} .

Proof. If z = λs + μ[sss] with λ, μ ∈ ℂ, then it follows from (S1), Proposition 5 (1) and (3) that

L(z, z) = λ^{2}L(s, s) + 2λμL(s, [sss]) + μ^{2}L([sss], [sss])
= (λ^{2} − 3μ^{2}q(s))L(s, s).

Taking λ := 1/2 and μ := ±1/2\sqrt{3q(s)}, we see that λ^{2} − 3μ^{2}q(s) = 0, so that x, y ∈ V with s = x + y.

For the uniqueness, it follows from Proposition 7 (2) and (3) that

x − y = ±\frac{1}{\sqrt{3q(s)}}[sss]

for such a pair {x, y}. Therefore {x, y} is uniquely determined by s. □

Before proving the remaining part of Theorem A, we give some corollaries to Theorem A’, and show Theorem B and its corollaries.

Corollary A1. S = {x + y ∈ 𝔤_{1}|x, y ∈ V, ⟨x, y⟩ ≠ 0}.

Proof. This follows from Proposition 7 (3) and Theorem A’. □

Corollary A2 (Cf. [Mk, (5.8), (3)]). If V ≠ ∅, then V is non-degenerate in ℙ_{∗}(𝔤_{1}), and moreover V is a variety with one apparent double point. In particular, for any a ∈ 𝔤_{1} \{0}, there exists x ∈ V such that ⟨a, x⟩ ≠ 0 if V ≠ ∅.

Proof. Recall that V is said to be non-degenerate if there is no hyperplane in ℙ_{∗}(𝔤_{1}) containing V. To prove the non-degeneracy of V, suppose that there exists a proper subspace 𝔰 of 𝔤_{1} such that V ⊆ ℙ_{∗}(𝔰). Since πS is a non-empty (Proposition 4), Zariski open subset of an irreducible space ℙ_{∗}(𝔤_{1}), πS is dense in ℙ_{∗}(𝔤_{1}). Therefore, there exists a point R ∈ πS \ ℙ_{∗}(𝔰), and, according to Theorem A’, we have R ∈ P ∗ Q for some P, Q ∈ V. Then we have P ∗ Q ⊆ ℙ_{∗}(𝔰), so that R ∈ ℙ_{∗}(𝔰). This is a contradiction, and V is non-degenerate. Moreover it follows from Theorem A’ that for any point R in the dense open πS, there exists a unique secant line through R, that is, P ∗ Q above, so that V is a variety with one apparent double point.

For the last part, since ⟨, ⟩ is non-degenerate, a^{⊥} := {b ∈ 𝔤_{1}|⟨a, b⟩ = 0} is a proper subspace of 𝔤_{1}. Therefore we have V \ a^{⊥} ≠ ∅ since V is non-degenerate. □

Corollary A3. S = ∅ if and only if V = ∅, that is, V = ∅, and in that case T = ∅.

Proof. This follows from Proposition 4 and Theorem A’. □

Proof of Theorem B. We prove

Theorem B’. We have:

(1) G acts transitively on each of irreducible components of V, and t_{x}V = 𝔇_{0}x for any x ∈ V, where t_{x}V is the Zariski tangent space to V at x.

(2) 𝔇_{0}x = (𝔇_{0}x)^{⊥} with 2 dim 𝔇_{0}x = dim 𝔤_{1} for any x ∈ V, and 𝔤_{1} = 𝔇_{0}x ⊕ 𝔇_{0}y for any x, y ∈ V with ⟨x, y⟩ ≠ 0.

Proof. (1) Let G_{0} be the connected, closed subgroup of Int 𝔤 with Lie algebras 𝔤_{0}. We first show

Claim. G_{0} acts transitively on S.

Proof of Claim. We may assume that S ≠ ∅, so that S is irreducible since it is a Zariski open subset of an irreducible space 𝔤_{1}. Moreover, since S is stable under the action of G_{0}, it suffices to show that

[𝔤_{0}, s] = 𝔤_{1}

for any s ∈ S. Take an arbitrary a ∈ 𝔤_{1}. It follows from Proposition 9 (2) that there exists b ∈ 𝔤_{1} such that a = L(s, s)b. On the other hand, it follows from (S2) that L(s, s)b = (L(s, b) + 3⟨s, b⟩ ad H)s ∈ [𝔤_{0}, s]. Thus, we have a ∈ [𝔤_{0}, s]. □

Now, it follows from 𝔤_{0} = 𝔇_{0} ⊕ ℂH that G_{0} = G × ℂ^{×}. Taking account of Proposition 1 (2), we see that to show (1) of Theorem B’ it suffices to show that G_{0} acts transitively on each of irreducible components of V. Take an arbitrary x ∈ V. There exists y ∈ V such that ⟨x, y⟩ ≠ 0 by Corollary A2. Then we show that for any x′ in a Zariski open neighborhood V \ y^{⊥} of x in V, there exists g ∈ G_{0} such that gx = x′, which implies the required result. Since it follows from Proposition 7 (3) that x + y, x′ + y ∈ S, there exists g ∈ G_{0} such that g(x + y) = x′ + y by Claim, where we note that gx, gy ∈ V since V is G_{0}-stable. Therefore, it follows from Theorem A’ that

{gx, gy} = {x′, y}.

If gx = x′, then there is nothing to prove: Otherwise, we have gx = y and gy = x′, so that g^{2}x = x′. This completes the proof.

(2) We note that

(♣)  dim 𝔇_{0}x ≤ \frac{1}{2} dim 𝔤_{1}

holds for any x ∈ V: Indeed, it follows from Proposition 3 (3) that 𝔇_{0}x ⊆ (𝔇_{0}x)^{⊥}, and from the non-degeneracy of ⟨, ⟩ that dim(𝔇_{0}x)^{⊥} = dim 𝔤_{1} − dim 𝔇_{0}x.

Now, we show that

(♠)  𝔤_{1} = 𝔇_{0}x + 𝔇_{0}y.

Take an arbitrary a ∈ 𝔤_{1}, and set s := x + y. Since ⟨x, y⟩ ≠ 0, it follows from Proposition 7 (3) that s ∈ S. Then, it follows from Proposition 9 (2) that there exists b ∈ 𝔤_{1} such that a = L(s, s)b. On the other hand, it follows from x, y ∈ V and (S2) that

L(s, s)b = [xyb] + [yxb]
= [xby] + ⟨x, b⟩y − ⟨x, y⟩b + 2⟨y, b⟩x
+ [ybx] + ⟨y, b⟩x − ⟨y, x⟩b + 2⟨x, b⟩y
= ([byx] + 3⟨y, b⟩x) + ([bxy] + 3⟨x, b⟩y),

which is contained in 𝔇_{0}x + 𝔇_{0}y by Proposition 1 (2). Thus we have a ∈ 𝔇_{0}x + 𝔇_{0}y.

Combining (♣) and (♠), we obtain that 𝔇_{0}x ∩ 𝔇_{0}y = 0 for any y ∈ V with ⟨x, y⟩ ≠ 0: We note that for any x ∈ V there exists y ∈ V such that ⟨x, y⟩ ≠ 0 (Corollary A2). Thus the equality holds in (♣), and the required results follow. □

Remark 1. It follows from Claim above that 𝔇_{0}s = [sss]^{⊥} for any s ∈ S: Indeed, note that it follows from Lemma 1, Propositions 2 and 5 (1) that 𝔇_{0}a ⊆ [aaa]^{⊥} for any a ∈ 𝔤_{1} since ⟨L(b, c)a, [aaa]⟩ = ⟨L(a, [aaa])b, c⟩ = 0. Since [sss]^{⊥} has codimension 1 in 𝔤_{1}, the assertion follows from 𝔇_{0}s + ℂs = 𝔤_{0}s = 𝔤_{1}.

Remark 2. One can deduce Theorem B’ (1) from Linear Section Theorem [KY, Theorem B], using a generalization of a theorem of Richardson [V, Lemma, p. 469], as well as from Theorem B’ (2), using the finiteness theorem for the number of nilpotent orbits [V, Proposition 2, p. 469]: Note that both of those proofs depend essentially on the argument [V, Lemma, p. 469]. On the other hand, (2) follows from (1) in Theorem B’ by Theorem A’, as follows:

Proof of (1) ⇒ (2) in Theorem B’. Similarly to the proof of (2) above, it suffices to show (♠) for x, y ∈ V with ⟨x, y⟩ ≠ 0. Take an arbitrary a ∈ 𝔤_{1}, and consider a line in 𝔤_{1} as follows:

σ : ℂ → 𝔤_{1}; λ ↦ (x + y) + λa.

Since σ(0) = x+y ∈ S by Proposition 7 (3), for a sufficiently small ε > 0 we have σ(λ) ∈ S for any λ ∈ Δ, where we set Δ := {λ ∈ ℂ||λ| < ε}. Then it follows from Theorem A’ that there exist curves ξ, η : Δ → V such that ξ(0) = x, η(0) = y and

σ(λ) = ξ(λ) + η(λ)

for any λ ∈ Δ. Then we have

a = \frac{d}{dλ}|_{λ=0} σ(λ) = \frac{d}{dλ}|_{λ=0} ξ(λ) + \frac{d}{dλ}|_{λ=0} η(λ) ∈ t_{x}V + t_{y}V.

According to (1), we have t_{x}V = 𝔇_{0}x and t_{y}V = 𝔇_{0}y, so that a ∈ 𝔇_{0}x + 𝔇_{0}y. □

Recall that the tangent variety of V, denoted by Tan V, is the union of embedded tangent spaces to V, and the projective dual of V, denoted by V^{∨}, is the set of hyperplanes tangent to V (see, for example, [FR, §3]): In other words, we set

Tan V := ⋃_{P ∈V} T_{P} V ⊆ ℙ_{∗}(𝔤_{1}), V^{∨} := ⋃_{P ∈V} (T_{P} V)^{∨} ⊆ ℙ_{∗}(𝔤_{1})^{∗},

where we define ℙ_{∗}(𝔤_{1})^{∗} to be ℙ_{∗}(𝔤_{1}^{∗}), that is, the set of hyperplanes in ℙ_{∗}(𝔤_{1}), and for a linear subspace 𝔰 ⊆ 𝔤_{1}, we define ℙ_{∗}(𝔰)^{∨} to be ℙ_{∗}(Ker(𝔤_{1}^{∗} ↠ 𝔰^{∗})) ⊆ ℙ_{∗}(𝔤_{1})^{∗}, that is, the set of hyperplanes containing ℙ_{∗}(𝔰). From Theorem B’ we immediately obtain

Corollary B1. Assume that V ≠ ∅. Then we have:

(1) V is equi-dimensional with 2 dim V + 1 = dim ℙ_{∗}(𝔤_{1}).

(2) T_{P} V = ℙ_{∗}(𝔇_{0}x) for any P ∈ V with x ∈ π^{−1}P.

(3) (Cf. [Mk, (5.8), (4)]) T_{P} V = (T_{P} V)^{∨} for any P ∈ V.

(4) (Cf. [Mk, (5.8), (1)]) Tan V = V^{∨}. Here we identify ℙ_{∗}(𝔤_{1}) with its dual space ℙ_{∗}(𝔤_{1})^{∗} by means of the non-degenerate, skewsymmetric form ⟨, ⟩.

Proof. (1) follows directly from Theorem B’. For (2), it follows from Theorem B’ (1) that T_{P} V = ℙ_{∗}(ℂx + 𝔇_{0}x) (see, for example, [KOY, Lemma 2.1]), which is equal to ℙ_{∗}(𝔇_{0}x) by Proposition 1 (2). Now, (3) follows from Theorem B’ (2) since we have (𝔇_{0}x)^{⊥} = Ker(𝔤_{1}^{∗} ↠ (𝔇_{0}x)^{∗}), and (4) follows as well. □

Remark. Since V is not necessarily irreducible, Corollary B1 (1) does not follow directly from Corollary A2.

Corollary B2. If V is neither empty nor irreducible, then V ⊆ ℙ_{∗}(𝔤_{1}) is a disjoint union of two linear varieties of dimension n − 1 in ℙ^{2n−1}:

V ≃ ℙ^{n−1} ⊔ ℙ^{n−1},

where 2n = dim 𝔤_{1}.

Proof. Let {V_{i}}_{1≤i≤k} be the set of irreducible components of V with k ≥ 2, and let 𝔰_{i} be the linear subspace of 𝔤_{1} spanned by π^{−1}V_{i}. It follows from Theorem B that each V_{i} is an orbit of G, so that 𝔰_{i} is G-stable. Moreover, by virtue of an argument by Zak [Z, pp. 49–50] we see that 𝔰_{i} is an irreducible G-module. Since ℙ_{∗}(𝔰_{i}) has a unique closed orbit of G (see, for example, [S, Ch. 1, §4.6.1, Lemma]), we see that 𝔰_{i} ∩ 𝔰_{j} = 0 if i ≠ j. Taking account of the non-degeneracy of V (Corollary A2), we obtain

𝔤_{1} = 𝔰_{1} ⊕ ⋯ ⊕ 𝔰_{k}.

On the other hand, we have dim 𝔰_{i} ≥ dim π^{−1}V_{i} = n by Corollary B1 (1). Therefore, it follows that 𝔤_{1} = 𝔰_{1} ⊕ 𝔰_{2} with k = 2 and V_{i} = ℙ_{∗}(𝔰_{i}) ≃ ℙ^{n−1}. □

Corollary B3. If V ≠ ∅ and U ≠ ∅, then V is covered by the lines in ℙ_{∗}(𝔤_{1}) contained in V.

Proof. We may assume that V is irreducible: Otherwise the claim is obvious from Corollary B2. Take an arbitrary u ∈ U. It suffices to show that

V \ (L(u, u)V ∪ Ker L(u, u)) ≠ ∅.

Indeed, if v ∈ V \ (L(u, u)V ∪ Ker L(u, u)), then πv ≠ π[uuv] ∈ ℙ_{∗}(𝔤_{1}), and it follows from Proposition 8 (2) that the secant line πv ∗ π[uuv] is contained in V. Taking account of the homogeneity of V (Theorem B’ (1)), we see that this holds for any v ∈ V.

Now, we have L(u, u)V ⊆ L(u, u)𝔤_{1} ⊆ u^{⊥} since ⟨[uua], u⟩ = −⟨a, [uuu]⟩ = 0 for any a ∈ 𝔤_{1} by Lemma 2. Therefore we have

V \ (L(u, u)V ∪ Ker L(u, u)) ⊇ V \ (u^{⊥} ∪ Ker L(u, u)).

On the other hand, we have V \ (u^{⊥} ∪ Ker L(u, u)) ≠ ∅: Indeed, u^{⊥} as well as Ker L(u, u) are proper subspaces of 𝔤_{1}, and V is irreducible and non-degenerate (Corollary A2). Thus the claim follows. □

Corollary B4. If V is irreducible, then the linear map L(u, u) has rank at least 2 for any u ∈ U.

Proof. We have rk L(u, u) ≥ 1 since L(u, u) ≠ 0. Suppose rk L(u, u) = 1: we set

Q := ℙ_{∗}(L(u, u)𝔤_{1}) ∈ ℙ_{∗}(𝔤_{1}).

Since π[uuv] = Q for any v ∈ V \ Ker L(u, u), it follows from Proposition 8 (2) that P ∗ Q ⊆ V for any P ∈ V \ ℙ_{∗}(Ker L(u, u)) with P ≠ Q. Since V is irreducible, V is a cone with vertex Q. On the other hand, it follows from Theorem B that V is smooth. Therefore V is a linear variety. This contradicts to Corollaries A2 and B1 (1). □

Now we return to proving Theorem A.

Proof of Theorem A (continued). (1) For the tangent locus, suppose that πs ∈ T_{πx}V for some x ∈ V. Then it follows from Corollary B1 (2) that s = Dx for some D ∈ 𝔇_{0}, and from Proposition 6 that q(s) = q(Dx) = 0. This contradicts to s ∈ S.

(2) For the secant locus, the assertion easily follows from Proposition 7: Indeed, suppose that πt ∈ πx ∗ πy for some x, y ∈ V. Then we may assume that t = x + y. We see from Proposition 7 (2) and (3) that [ttt] ≠ 0 implies ⟨x, y⟩ ≠ 0, which implies q(t) = 12⟨x, y⟩^{2} ≠ 0. This contradicts to t ∈ T .

Next for the tangent locus, it follows from Proposition 5 (3) that [ttt] ∈ V. Moreover, we have t ∈ 𝔇_{0}[ttt]: Indeed, for any a, b ∈ 𝔤_{1}, it follows from Propositions 2 and 5 (1) that ⟨L(a, b)[ttt], t⟩ = ⟨L([ttt], t)a, b⟩ = 0, so that ⟨𝔇_{0}[ttt], t⟩ = 0 by Lemma 1. Now, our claim follows from Theorem B’ (2). Therefore it follows from Corollary B1 (2) that π[ttt] ∈ Θ_{πt}.

To complete the proof, it suffices to show that for any t ∈ T , if t ∈ t_{x}V for some x ∈ V, then πx = π[ttt]. We may assume that there is a curve ξ : Δ → V such that ξ(0) = x and

lim_{λ→0} \frac{1}{λ}\{ξ(λ) − ξ(0)\} = t,

where Δ := {λ ∈ ℂ||λ| < ε} with a sufficiently small ε > 0. Setting

τ(λ) := \frac{1}{λ}\{ξ(λ) − ξ(0)\}

with λ ∈ Δ \ {0}, we have lim_{λ→0} τ(λ) = t. Moreover, by the assumption [ttt] ≠ 0, taking ε smaller if necessary, we may assume that [τ(λ)τ(λ)τ(λ)] ≠ 0 for any λ ∈ Δ \ {0}. On the other hand, it follows from Proposition 7 (2) that

λ^{3}[τ(λ)τ(λ)τ(λ)] = −6⟨ξ(λ), −ξ(0)⟩(ξ(λ) + ξ(0)).

Thus it follows that ⟨ξ(λ), −ξ(0)⟩ ≠ 0, so that τ(λ) ∈ S by Proposition 7 (3). Now, it follows from Theorem A (1) that

\{πξ(λ), πξ(0)\} = Σ_{πτ(λ)}^{∘} = \{π ([τ(λ)τ(λ)τ(λ)] ± \sqrt{3q(τ(λ))}τ(λ))\} .

Therefore, using lim_{λ→0} q(τ(λ)) = q(t) = 0, we have

πx = lim_{λ→0} πξ(λ) = lim_{λ→0} π ([τ(λ)τ(λ)τ(λ)] ± \sqrt{3q(τ(λ))}τ(λ))

= lim_{λ→0} π ([τ(λ)τ(λ)τ(λ)])

= π[ttt]. □

(3) It follows from Corollary A2 that there exists v ∈ V such that ⟨u, v⟩ ≠ 0. If z = λu + μ[uuv] with λ, μ ∈ ℂ, then it follows from Proposition 8 that

L(z, z) = λ^{2}L(u, u) + 2λμL(u, [uuv]) + μ^{2}L([uuv], [uuv])
= λ(λ + 4μ⟨u, v⟩)L(u, u).

Taking (λ, μ) = (1, −1/4⟨u, v⟩), (0, 1/4⟨u, v⟩) with λ(λ + 4μ⟨u, v⟩) = 0, we set

\{x, y\} := \{u − \frac{1}{4⟨u, v⟩}[uuv], \frac{1}{4⟨u, v⟩}[uuv]\} .

Then we have that u = x + y with x, y ∈ V: Note that it follows from Proposition 8 (1) that [uuv] ≠ 0 since L(u, u) ≠ 0 and ⟨u, v⟩ ≠ 0.

Thus we see that π[uuv] ∈ Σ_{πu}^{∘} for any v ∈ V with ⟨u, v⟩ ≠ 0, so that

π(L(u, u)(V \ u^{⊥}) \ {0}) ⊆ Σ_{πu}^{∘}.

Since V is irreducible, we obtain π(L(u, u)V \ {0}) ⊆ Σ_{πu}. On the other hand, it follows from Corollary A2 that π(L(u, u)V \ {0}) is non-degenerate in ℙ_{∗}(L(u, u)𝔤_{1}), so that dim π(L(u, u)V\{0}) ≥ 1 since dim ℙ_{∗}(L(u, u)𝔤_{1}) ≥ 1 by Corollary B4 and π(L(u, u)V\{0}) is irreducible. Thus we obtain dim Σ_{πu} ≥ 1.

Next for the tangent locus, we note that for u ∈ U and v ∈ V it follows

(⋆)  ⟨u, v⟩ = 0, [uuv] ≠ 0 ⇒ u ∈ 𝔇_{0}[uuv].

Indeed, for any a, b ∈ 𝔤_{1} it follows from Propositions 2 and 8 (1) that

⟨L(a, b)[uuv], u⟩ = ⟨L([uuv], u)a, b⟩ = 2⟨u, v⟩⟨L(u, u)a, b⟩ = 0.

Therefore, we have ⟨𝔇_{0}[uuv], u⟩ = 0 by Lemma 1, so that u ∈ 𝔇_{0}[uuv] by Theorem B’ (2), where one should note that [uuv] ∈ V ∪ {0} by Proposition 8 (2).

Now, we see from (⋆) that

π(L(u, u)(V ∩ u^{⊥}) \ {0}) ⊆ Θ_{πu}.

On the other hand, since V is irreducible and non-degenerate in ℙ_{∗}(𝔤_{1}) as above, V ∩ℙ_{∗}(u^{⊥}) is also non-degenerate in ℙ_{∗}(u^{⊥}). Therefore, it follows from Corollary B4 that V ∩ℙ_{∗}(u^{⊥}) ⊈ ℙ_{∗}(Ker L(u, u)), so that π(L(u, u)(V ∩ u^{⊥}) \ {0}) ≠ ∅. Thus we obtain Θ_{πu} ≠ ∅. □

Corollary A4. If V ≠ ∅, then U = {x + y ∈ 𝔤_{1}|x, y ∈ V, ⟨x, y⟩ = 0, L(x, y) ≠ 0}.

Proof. This follows from Proposition 7 (1), (2) and the first paragraph of the proof of Theorem A (3). □

Corollary A5 (Cf. [Mk, (5.8), (1)]). If V is irreducible, then Tan V is a quartic hypersurface in ℙ_{∗}(𝔤_{1}) defined by q.

Proof. Denote by T the quartic hypersurface:

T := π({t ∈ 𝔤_{1}|q(t) = 0, t ≠ 0}) = πT ⊔ πU ⊔ V.

Note that it follows from Proposition 4 that q is not identically zero since V ≠ ∅. Then, we see that Tan V ⊆ T follows from Proposition 6 and Corollary B1 (2), and Tan V ⊇ T follows from Theorem A (2) and (3). □

Corollary A6. If V is irreducible, then the singular locus of Tan V is an algebraic set in ℙ_{∗}(𝔤_{1}) defined by the system of cubic equations, [uuu] = 0, with variables in u over 𝔤_{1}.

Proof. By virtue of Corollary A5, it suffices to show that the hypersurface defined by q = 0 has a singularity at a ∈ 𝔤_{1} if and only if [aaa] = 0, that is,

dq(a) = 0 ⇔ [aaa] = 0,

for a ∈ 𝔤_{1}. Using (‡) in the proof of Proposition 4, for λ ∈ ℂ and b ∈ 𝔤_{1} we have

\frac{1}{λ} \{q(a + λb) − q(a)\} = ⟨b, [aaa]⟩ + ⟨a, [baa]⟩ + ⟨a, [aba]⟩ + ⟨a, [aab]⟩ + λ(⋯)
= 4⟨b, [aaa]⟩ + λ(⋯)
→ 4⟨b, [aaa]⟩ (λ → 0).

Therefore dq(a) = 0 if and only if ⟨b, [aaa]⟩ = 0 for any b ∈ 𝔤_{1}, which is equivalent to [aaa] = 0 since ⟨, ⟩ is non-degenerate. □

Remark. It follows from Proposition 5 (4) that the rational map

γ : ℙ_{∗}(𝔤_{1}) −−→ ℙ_{∗}(𝔤_{1})

given by a ↦ [aaa] is a Cremona transformation of ℙ_{∗}(𝔤_{1}) of order 2 (see, for example, [SR, III]), that is, a birational automorphism of ℙ_{∗}(𝔤_{1}) with γ^{2} = 1, and the base locus is πU. It follows from Proposition 5 that γπT ⊆ πV and πT = γ^{−1}πV, and that γ gives an automorphism of πS. Moreover it follows from Proposition 5 (3) and Theorem A (2) that γ^{−1}P = T_{P} V ∩ πT for any P ∈ V. From the proof of Corollary A6 we see that γ is explicitly given by the partial differentials of q.

Appendix. Proof of (S3)

We firstly show (S2), since it is partly used in showing (S3) below.

Proof of (S2) (Asano [A3]). It follows from the Jacobi identity that [[x, y]E_{−}] = [[x, E_{−}]y]+ [x[y, E_{−}]], so that [x[y, E_{−}]] = [y[x, E_{−}]] + 2⟨x, y⟩H. Adding [y[x, E_{−}]] on the both sides, applying \frac{1}{2} ad z, we get

(*)  [xyz] = [z[y[x, E_{−}]]] − ⟨x, y⟩z.

Therefore it follows from the Jacobi identity that

[xyz] − [xzy] = {[z[y[x, E_{−}]]] − ⟨x, y⟩z} + {[y[z[x, E_{−}]]] − ⟨x, z⟩y}
= ⟨x, z⟩y − ⟨x, y⟩z + [[x, E_{−}], [y, z]]
= ⟨x, z⟩y − ⟨x, y⟩z + 2⟨y, z⟩[[x, E_{−}]E_{+}].

Thus, we get the formula since we have [[x, E_{−}]E_{+}] = x. □

Proof of (S3) (Asano [A3]). It suffices to show that

(♡)  [L(x, x), L(y, y)] = 2L(L(x, x)y, y).

Indeed, setting x := v + w in (♡), and using (♡) again, we obtain

[L(v, w), L(y, y)] = 2L(L(v, w)y, y).

Moreover setting y := x + y in this formula, and using (♡) again, we obtain

[L(v, w), L(x, y)] = L(L(v, w)x, y) + L(x, L(v, w)y).

Then we have

[L(v, w), L(x, y)] = L(L(v, w)x, y) + L(x, L(v, w)y)

⇔ L(v, w)L(x, y) = L(L(v, w)x, y) + L(x, L(v, w)y) + L(x, y)L(v, w),

and the latter is nothing but (S3).

To prove (♡), we show the following

Lemma. Denote by xy^{∗} the map z ↦ ⟨z, y⟩x, and set x̃ := [x[x, E_{−}]]. Then: (1) L(x, y) = − ad y ∘ ad x ∘ ad E_{−} − 2xy^{∗} − 2yx^{∗} − ⟨x, y⟩1.

(2) y(L(x, x)y)^{∗} = −yy^{∗} ∘ L(x, x).

(3) ad L(x, x)y = [ad y, ad x̃].

(4) ad x̃ ∘ ad E_{−} = ad E_{−} ∘ ad x̃.

Proof. (1) This follows from (*) and the Jacobi identity: Indeed, we have

L(x, y)z = [z[y[x, E_{−}]]] − ⟨x, y⟩z
= [[z, y][x, E_{−}]] + [y[z[x, E_{−}]]] − ⟨x, y⟩z

= 2⟨z, y⟩[E_{+}[x, E_{−}]] + [y[[z, x]E_{−}]] + [y[x[z, E_{−}]]] − ⟨x, y⟩z = −2⟨z, y⟩x + 2⟨z, x⟩[y[E_{+}, E_{−}]] + [y[x[z, E_{−}]]] − ⟨x, y⟩z

= −2⟨z, y⟩x − 2⟨z, x⟩y − [y[x[E_{−}, z]]] − ⟨x, y⟩z.

(2) Using Lemma 2 in §1, we have

y(L(x, x)y)^{∗}z = ⟨z, L(x, x)y⟩y = −⟨L(x, x)z, y⟩y = −yy^{∗}(L(x, x)z).

(3) Since it follows from (*) that L(x, x) = −(ad x̃)|𝔤_{1}, we have

[L(x, x)y, z] = −[[x̃, y]z] = [[y, z]x̃] + [[z, x̃]y] = [ad y, ad x̃]z.

(4) Since [[x, E_{−}]E_{−}] = 0, we have [x̃, E_{−}] = 0, so that [x̃[E_{−}, z]] = [E_{−}[x̃, z]]. □

Proof of (♡).

[L(x, x), L(y, y)] − {L(L(x, x)y, y) + L(y, L(x, x)y)}
= [L(x, x), −(ad y)^{2} ∘ ad E_{−} − 4yy^{∗}]
− {− ad y ∘ ad L(x, x)y ∘ ad E_{−}
− 2(L(x, x)y)y^{∗} − 2y(L(x, x)y)^{∗} − ⟨L(x, x)y, y⟩1}
− {− ad L(x, x)y ∘ ad y ∘ ad E_{−}
− 2y(L(x, x)y)^{∗} − 2(L(x, x)y)y^{∗} − ⟨y, L(x, x)y⟩1}
(∵ Lemma (1))
= [ad x̃, (ad y)^{2} ∘ ad E_{−}] − [L(x, x), 4yy^{∗}]
+ ad y ∘ [ad y, ad x̃] ∘ ad E_{−} + [ad y, ad x̃] ∘ ad y ∘ ad E_{−}
+ 4y(L(x, x)y)^{∗} + 4(L(x, x)y)y^{∗}
(∵ L(x, x) = − ad x̃|𝔤_{1} and Lemma (3))
= ad x̃ ∘ (ad y)^{2} ∘ ad E_{−} − (ad y)^{2} ∘ ad E_{−} ∘ ad x̃
− 4L(x, x) ∘ yy^{∗} + 4yy^{∗} ∘ L(x, x)
+ (ad y)^{2} ∘ ad x̃ ∘ ad E_{−} − ad y ∘ ad x̃ ∘ ad y ∘ ad E_{−}
+ ad y ∘ ad x̃ ∘ ad y ∘ ad E_{−} − ad x̃ ∘ (ad y)^{2} ∘ ad E_{−}
+ 4y(L(x, x)y)^{∗} + 4(L(x, x)y)y^{∗}
= −4L(x, x) ∘ yy^{∗} + 4yy^{∗} ∘ L(x, x) + 4y(L(x, x)y)^{∗} + 4(L(x, x)y)y^{∗}
(∵ Lemma (4))
= −4L(x, x) ∘ yy^{∗} + 4(L(x, x)y)y^{∗} (∵ Lemma (2)).

The last term is equal to zero, since we have

(L(x, x) ∘ yy^{∗})(z) = L(x, x)(⟨z, y⟩y) = ⟨z, y⟩L(x, x)y = (L(x, x)y)y^{∗}(z). □

Acknowledgements. Both of the authors are supported in part by Grant-in-Aid for Scientific Research (C), No. 10640046, Japan Society for the Promotion of Science. They would like to thank Prof. Hiroshi Asano and Prof. Soji Kaneyuki for their invaluable advice and stimulating discussion. In particular, the proof of Theorem B’ (1) was inspired by Prof. Asano [A5], where the discussion with him is essential.

The first author is supported also in part by Waseda University Grant for Special Research Projects, Individual Research, No. 99A-187, Waseda University. He would like to thank Prof. Eduardo Esteves and Prof. Arnaldo Garcia for their invaluable comments. This work was completed while the first author stayed in IMPA (=Instituto de Matemática Pura e Aplicada). He wishes to thank the people at IMPA, especially Prof. Esteves and Prof. Garcia for their warm hospitality. Muito obrigado!

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