THE TITS CONSTRUCTION AND THE EXCEPTIONAL

SIMPLE CLASSICAL LIE SUPERALGEBRAS

Georgia Benkart^{1} Alberto Elduque^{2}

July 6, 2001

Abstract. We construct the exceptional simple classical Lie superalgebras D(2, 1; α)

(α ≠ 0, −1), G(3), and F(4) by means of the Tits construction.

Introduction

The well-known Tits construction (see for example, [T], [S], or [J]) builds a simple Lie algebra from a unital composition algebra, a unital Jordan algebra of degree 3, and their Lie algebras of inner derivations. All the exceptional Lie algebras E_{6}, E_{7}, E_{8}, F_{4}, and G_{2} can be realized by this process. In this paper we extend the Tits construction to the exceptional simple classical Lie superalgebras D(2, 1; α) (α ≠ 0, 1), G(3), and F(4). (A description of these simple Lie superalgebras can be found in Kac’s seminal paper [K1] and in [K2] (see also [HK]), which explores connections of them with Jordan superalgebras.) We expand on earlier work of Benkart and Zelmanov [BZ, Sec. 4] in which the Lie superalgebras G(3) and F(4) were described using the Tits construction, and the Freudenthal-Tits magic square was extended to include G(3) and F(4). Here we further enlarge the magic square to encompass the exceptional superalgebras D(2, 1; α) and several of the Lie superalgebras in the series B(m, n) = 𝔬𝔰𝔭(2m + 1, 2n) for small values of m and n. The results of our investigation set the stage for [E], which determines all the forms of the exceptional simple Lie superalgebras. Many (but not all) of the forms of D(2, 1; α) (α ≠ 0, −1), G(3), and F(4) arise by the constructions presented in this paper.

Typeset by AMS-TEX

2000 Mathematics Subject Classification. Primary 17B25, 17A70, 17B60.

1Support from National Science Foundation Grant #DMS–9970119 is gratefully acknowledged.

2Supported by the Spanish DGES (Pb 97-1291-C03-03) and by a grant from the Spanish Dirección General de Enseñanza Superior e Investigación Científica (Programa de Estancias de Investigadores Españoles en Centros de Investigación Extranjeros), while visiting the University of Wisconsin at Madison.

In the (generalized) Tits construction of [BZ, Sec. 3], two Jordan algebras of nondegenerate symmetric bilinear forms are the building blocks of Lie algebras. All orthogonal Lie algebras can be built in this way. This construction has an orthosymplectic counterpart which is described in the final section of the paper. The Tits construction expresses an orthosymplectic Lie superalgebra 𝔬𝔰𝔭(V ⊕ W) of an orthogonal sum V ⊕ W of two superspaces in terms of the orthosymplectic Lie superalgebras of the summands. In particular, this unifies the above mentioned construction of the orthogonal Lie algebras and the construction of the symplectic Lie algebras in [BZ, 3.33].

By a superalgebra we always mean a ℤ_{2}-graded algebra over a field F. An important example is the Grassmann or exterior algebra G, which is a unital associative algebra over F with generators e_{1}, e_{2}, … and defining relations e_{i}e_{j} + e_{j}e_{i} = 0 for i ≠ j and e_{i}^{2} = 0. It has a ℤ_{2}-grading G = G¯0 ⊕ G¯1, where e_{i_{1}} ⋯ e_{i_{m}} belongs to G¯0 if m is even and to G¯1 if m is odd. Let A¯0 ⊕ A¯1 be a superalgebra over F. The subalgebra G(A) = (A¯0 ⊗ G¯0) + (A¯1 ⊗ G¯1) of A ⊗ G (with tensor products over F) is called the Grassmann envelope of A. When V is a particular variety of algebras such as Lie or Jordan, then the superalgebra A is said to be a V-superalgebra if G(A) ∈ V. Thus, A is a Lie (resp. Jordan) superalgebra if and only if G(A) is a Lie (resp. Jordan) algebra.

§1. The Tits Construction

Throughout F will denote a field of characteristic not 2 or 3. Let C be a unital composition algebra over F. Thus, C is a finite-dimensional (not necessarily associative) F -algebra with a nondegenerate quadratic form 𝔫 : C → F such that 𝔫(ab) = 𝔫(a)𝔫(b) for all a, b ∈ C and 𝔫(1) = 1. The map D_{a,b} : C → C given by

(1.1)  D_{a,b}(c) = [[a, b], c] − 3(a, b, c)

is the inner derivation determined by a, b ∈ C. In this expression, [a, b] = ab − ba and (a, b, c) = (ab)c − a(bc). The algebra C has an involution a ↦ \overline{a}, and each element a ∈ C satisfies the degree 2 equation

a^{2} − 𝔱𝔯(a)a + 𝔫(a)1 = 0

where 𝔱𝔯(a) = a + \overline{a}. We denote the trace 0 elements of C by C^{0} = \{a ∈ C | 𝔱𝔯(a) = 0}.

Let J = J¯0 ⊕ J¯1 be a unital Jordan superalgebra with a normalized trace 𝔱 : J → F. That is to say, 𝔱 is a linear map such that 𝔱(1) = 1, 𝔱(J¯1) = 0, and 𝔱((J, J, J)) = 0. (The conditions 𝔱((J¯0, J¯1, J¯1)) = 0 = 𝔱((J¯1, J¯0, J¯1)) do not appear in [BZ, (4.4)], but are needed to extend 𝔱 to a normalized trace 𝔱̃ on the Grassmann envelope.) Then J = F 1 ⊕ J^{0}, where J^{0} = {x ∈ J | 𝔱(x) = 0} ⊇ J¯1. For x, y ∈ J^{0},

(1.2)  xy = 𝔱(xy)1 + x ∗ y,

where x ∗ y = xy − 𝔱(xy)1 is a supercommutative product on J^{0}. For x, y ∈ J^{0}, the map d_{x,y} : J → J defined by

d_{x,y}(z) = x(yz) − (−1)^{x̄ȳ}y(xz)

is the inner superderivation of J determined by x, y. (Our convention here is that x̄ = i (viewed as an element of ℤ_{2}) whenever x ∈ J_{i}.) This map preserves J^{0}, and so gives a superderivation of (J^{0}, ∗).

Given C and J as above, consider the space

(1.3)  T (C, J) := D_{C,C} ⊕ (C^{0} ⊗ J^{0}) ⊕ d_{J,J}

with the superanticommutative product [ , ] specified by

(1.4)

• D_{C,C} is a Lie subalgebra and d_{J,J} is a Lie subsuperalgebra • [D_{C,C}, d_{J,J}] = 0

• [D, a ⊗ x] = D(a) ⊗ x, [d, a ⊗ x] = a ⊗ d(x)

• [a ⊗ x, b ⊗ y] = 𝔱(xy)D_{a,b} + [a, b] ⊗ x ∗ y + 2𝔱𝔯(ab)d_{x,y}

for all D ∈ D_{C,C}, d ∈ d_{J,J}, a, b ∈ C^{0}, x, y ∈ J^{0}.

When T (C, J) results in a Lie algebra or Lie superalgebra (see Prop. 1.5 below), we say that it is the Tits construction applied to C and J. Tits [T] was the first to apply this method to obtain an explicit realization of the exceptional simple Lie algebras.

Remark. The last equation in (1.4) is not exactly the multiplication given in [BZ, (3.7) and 3.18], where the product of a ⊗ x and b ⊗ y is defined as

\frac{1}{4}𝔱(xy)D_{a,b} + \frac{1}{2}[a, b] ⊗ x ∗ y + \frac{1}{2}𝔱𝔯(ab)d_{x,y}.

However, the map ϕ with ϕ(D_{a,b}) = D_{a,b}, ϕ(d_{x,y}) = d_{x,y}, and ϕ(a ⊗ x) = 2a ⊗ x provides an isomorphism between the two.

The superversion of [BZ, Prop. 3.9] (with the same proof but taking care of parity signs or alternatively by passing to the Grassmann envelope) states Proposition 1.5. T (C, J) is a Lie superalgebra if and only if the following relations hold:

(i) 0 = ∑_{⟲}(−1)^{x̄_{1}x̄_{3}}𝔱𝔯([a_{1}, a_{2}]a_{3})d_{(x_{1}∗x_{2}),x_{3}}

(ii) 0 = ∑_{⟲}(−1)^{x̄_{1}x̄_{3}}𝔱((x_{1} ∗ x_{2})x_{3})D_{[a_{1},a_{2}],a_{3}}

(iii) 0 = ∑_{⟲}(−1)^{x̄_{1}x̄_{3}}(D_{a_{1},a_{2}}(a_{3}) ⊗ 𝔱(x_{1}x_{2})x_{3}

+[[a_{1}, a_{2}], a_{3}] ⊗ (x_{1} ∗ x_{2}) ∗ x_{3}
+2𝔱𝔯(a_{1}a_{2})a_{3} ⊗ d_{x_{1},x_{2}}(x_{3}))

for any a_{1}, a_{2}, a_{3} ∈ C^{0} and any homogeneous x_{1}, x_{2}, x_{3} ∈ J^{0}.

The notation “∑_{⟲}” signifies summation over the cyclic permutation of the vari-

ables. For example, what is meant by equation (ii) is

\begin{matrix}0 = & (−1)^{x̄_{1}x̄_{3}}𝔱((x_{1} ∗ x_{2})x_{3})D_{[a_{1},a_{2}],a_{3}} + (−1)^{x̄_{2}x̄_{1}}𝔱((x_{2} ∗ x_{3})x_{1})D_{[a_{2},a_{3}],a_{1}} \\ + (−1)^{x̄_{3}x̄_{2}}𝔱((x_{3} ∗ x_{1})x_{2})D_{[a_{3},a_{1}],a_{2}}. \end{matrix}

Remark. The conditions of Proposition 1.5 are satisfied whenever the normalized trace 𝔱̃ on the Grassmann envelope G(J) of J satisfies the Cayley-Hamilton equation 𝔠𝔥_{3}(x) = 0 of 3 × 3 matrices, when it is expressed in terms of the trace:

(1.6) 𝔠𝔥_{3}(x) = x^{3} − 3𝔱̃(x)x^{2} + (\frac{9}{2}𝔱̃(x)^{2} − \frac{3}{2}𝔱̃(x^{2}))x − (𝔱̃(x^{3}) − \frac{9}{2}𝔱̃(x^{2})𝔱̃(x) + \frac{9}{2}𝔱̃(x)^{3})1 = 0.

(A discussion of this can be found in [BZ, Secs. 3 and 4].)

§2. The Jordan Superalgebras D_{μ}

Among the simple Jordan superalgebras, there is a one-parameter family of 4dimensional algebras D_{μ} (μ ≠ 0), having (D_{μ})¯0 = F e ⊕ F f and (D_{μ})¯1 = F x ⊕ F y where

(2.1)  \begin{matrix}e^{2} = e,  f^{2} = f,  ef = 0 \\ xy = e + μf = −yx,  ex = \frac{1}{2}x = fx,  ey = \frac{1}{2}y = fy. \end{matrix}

Observe that e + f = 1 and D_{μ} ≅ D_{μ^{−1}}. The superalgebra D_{1} is the Jordan superalgebra of a supersymmetric bilinear form.

Lemma 2.2.

(1) The superalgebra D_{−1} does not admit a normalized trace.

(2) If μ ≠ 0, −1, then D_{μ} admits a normalized trace given by

𝔱(e) = \frac{μ}{1+μ},  𝔱(f) = \frac{1}{1+μ},  𝔱((D_{μ})¯1) = 0.

Proof. (1) Suppose J = D_{−1}. Then

(x, y, e) = (e − f)e − \frac{1}{2}xy = e − \frac{1}{2}(e − f) = \frac{1}{2}1

so that 1 ∈ (J, J, J). But any normalized trace must satisfy 𝔱(1) = 1 and 𝔱((J, J, J)) = 0. So no normalized trace can exist for J = D_{−1}.

(2) If μ ≠ 0, −1, then it follows that (J, J, J) = F (e − μf) ⊕ J¯1. We must have 𝔱(e + f) = 𝔱(1) = 1 and 𝔱(e − μf) = 0, which imply that 𝔱(e) = μ/(1 + μ) and 𝔱(f) = 1/(1 + μ). □

Lemma 2.3. Suppose μ ≠ 0, −1 and 𝔱 is as in (2) of Lemma 2.2. Let 𝔱̃ be the natural extension of 𝔱 to the Grassmann envelope G(D_{μ}) of D_{μ}. Then all elements v of G(D_{μ}) satisfy the trace identity 𝔠𝔥_{3}(v) = 0 relative to 𝔱̃ if and only if μ = 2 or \frac{1}{2}.

Proof. If μ ≠ 2, \frac{1}{2}, then

𝔠𝔥_{3}(f) = \frac{(2μ−1)(μ−2)}{2(1+μ)^{3}}((1 + μ)f + 1) ≠ 0.

It suffices to consider the superalgebra D_{2} because D_{\frac{1}{2}} ≅ D_{2}. That all elements of G(D_{2}) satisfy 𝔠𝔥_{3}(v) = 0 can be shown by a straightforward computation using the fact that for v = γx + δy, where γ, δ ∈ G¯1 (note we omit tensor symbols), v^{2} = 2γδ(e + 2f) and v^{3} = 0, as γδγ = −γ^{2}δ = 0. □

Lemma 2.4. If μ ≠ 0, −1, then Der D_{μ} = Inder D_{μ} ≅ 𝔬𝔰𝔭(1, 2). The Lie superalgebra Der D_{−1} is not simple.

Proof. Let L = Der D_{μ} = L¯0 ⊕ L¯1. Since (D_{μ})¯0 ≅ F ⊕ F, L¯0 must act trivially on (D_{μ})¯0. For d ∈ L¯0, if d(x) = αx + βy and d(y) = γx + δy, then a straightforward computation shows that the only requirement for d to belong to L is that δ = −α. Thus, L¯0 ≅ 𝔰𝔩_{2}. Any d ∈ L¯1 satisfies d(1) = 0. From x = 2ex and y = 2ey, it follows that d is determined by the value d(e). Hence dim L¯1 ≤ 2. But because d_{D_{μ},D_{μ}} can be seen to have dimension 5, Der D_{μ} = d_{D_{μ},D_{μ}} = InderD_{μ}.

Now if μ = −1, then direct calculation shows that [L¯1, L¯1] = 0. Thus, L¯1 is an ideal of L = Der D_{μ} in that case. If μ ≠ 0, −1, then 𝔱(Der D_{μ}(D_{μ})) = 𝔱((D_{μ}, D_{μ}, D_{μ})) = 0 so that ker 𝔱 = D_{μ}^{0} is Der D_{μ}-invariant. Likewise so is the supersymmetric form (u, v) = 𝔱(uv) restricted to D_{μ}^{0}. Hence, by a dimension count, Der D_{μ} = 𝔬𝔰𝔭(D_{μ}^{0}) ≅ 𝔬𝔰𝔭(1, 2). □

Theorem 2.5. Let C be a Cayley-Dickson algebra, and assume μ ≠ 0, −1. Then T (C, D_{μ}) is a Lie superalgebra if and only if μ = 2 or \frac{1}{2}. The superalgebra T (C, D_{2}) is a form of the exceptional simple classical Lie superalgebra F(4).

Proof. If μ = 2 or \frac{1}{2}, then G(D_{μ}) satisfies 𝔠𝔥_{3}(v) = 0, so T (C, D_{μ}) is a Lie superalgebra.

Assume μ ≠ 0, −1, and set x_{1} = x, x_{2} = y, and x_{3} = e − μf(=: g) in D_{μ}^{0}. Then 𝔱(x_{1}x_{2}) = 2μ/(1 + μ), while 𝔱(x_{2}x_{3}) = 0 = 𝔱(x_{3}x_{1}). Moreover,

(x_{1} ∗ x_{2}) ∗ x_{3} = \frac{(1−μ)^{2}}{1+μ} g, (x_{2} ∗ x_{3}) ∗ x_{1} = \frac{−(1−μ)^{2}}{2(1+μ)} g, (x_{3} ∗ x_{1}) ∗ x_{2} = \frac{(1−μ)^{2}}{2(1+μ)}g

d_{x_{1},x_{2}}(x_{3}) = x(yg) + y(xg) = \frac{1−μ}{2}(xy + yx) = 0
d_{x_{2},x_{3}}(x_{1}) = y(gx) − g(yx) = \frac{1−μ}{2}yx − g(yx)
= −\frac{1−μ}{2}(e + μf) + e − μ^{2}f = \frac{1+μ}{2}g
d_{x_{3},x_{1}}(x_{2}) = g(xy) − x(gy) = e − μ^{2}f − \frac{1−μ}{2}(e + μf) = \frac{1+μ}{2}g

Thus, condition (iii) in Proposition 1.5 reads

(2.6)  \begin{aligned}\frac{2μ}{1+μ}D_{a,b}(c) & + \frac{(1−μ)^{2}}{2(1+μ)}(2[[a, b], c] + [[b, c], a] + [[c, a], b]) \\ & + (1 + μ)(𝔱𝔯(ca)b − 𝔱𝔯(bc)a) = 0 \end{aligned}

for all a, b, c ∈ C^{0}. Because 𝔱𝔯(ca) = ca + \overline{ca} = ca + ac,

𝔱𝔯(ca)b − 𝔱𝔯(bc)a = (ac + ca)b − a(bc + cb)
= (a, c, b) + (ca)b − a(bc)
= (a, c, b) + (c, a, b) + (a, b, c) + c(ab) − (ab)c
= (a, b, c) − [ab, c] = (a, b, c) − \frac{1}{2}[[a, b], c].

Now in any alternative algebra,

[[a, b], c] + [[b, c], a] + [[c, a], b] = 6(a, b, c).

Hence (2.6) gives

(2.7)  \begin{aligned}0 = & \frac{2μ}{1+μ}([[a, b], c] − 3(a, b, c)) + \frac{(1−μ)^{2}}{2(1+μ)}([[a, b], c] + 6(a, b, c)) \\ & + \frac{1+μ}{2}(2(a, b, c) − [[a, b], c]) \\ = & \frac{4μ^{2}−10μ+4}{1+μ}(a, b, c) = \frac{2(2μ−1)(μ−2)}{1+μ}(a, b, c). \end{aligned}

Since we can find a, b, c ∈ C^{0} such that (a, b, c) ≠ 0, this forces μ = 2 or μ = \frac{1}{2}, as required.

Finally, when μ = 2, a digression on Cayley-Dickson algebras is in order. Let C be such an algebra. Then any D ∈ Der C satisfies D(1) = 0 and D(C^{0}) ⊆ C^{0}, and thus Der C embeds as a subalgebra of the orthogonal Lie algebra 𝔬(C^{0}, 𝔫). Then 𝔬(C^{0}, 𝔫) = Der C ⊕ ad C^{0}, where [D, ad a] = ad D(a) and [ad a, ad b] = 2D_{a,b} − ad [a, b] (see, for instance, [EM, Sec. 2]). Now the spin module for 𝔬(C^{0}, 𝔫) is C with the action given by D·x = D(x) and (ad a)·x = ax+2xa, for any D ∈ Der C, a ∈ C^{0} and x ∈ C. The space Hom_{𝔬(C^{0},𝔫)}(C ⊗ C, F) is spanned by the polar of the norm 𝔫 (that is, the map x⊗y ↦ 𝔱𝔯(\overline{x}y)) [EM, Prop. 10]. The vector space Hom_{𝔬(C^{0},𝔫)}(C ⊗ C, 𝔬(C^{0}, 𝔫)) is one-dimensional too. To see this, first notice that both the adjoint module 𝔬(C^{0}, 𝔫) and the spin module C are selfdual. Hence Hom_{𝔬(C^{0},𝔫)}(C ⊗ C, 𝔬(C^{0}, 𝔫)) ≅ Hom_{𝔬(C^{0},𝔫)}(C ⊗ C ⊗ 𝔬(C^{0}, 𝔫), F) ≅ Hom_{𝔬(C^{0},𝔫)}(𝔬(C^{0}, 𝔫) ⊗ C, C) which contains, at least, the action of 𝔬(C^{0}, 𝔫) on C. Then as an 𝔬(C^{0}, 𝔫)-module, C ⊗ C is generated by F ⊗ C, so that Hom_{𝔬(C^{0},𝔫)}(C ⊗ C, 𝔬(C^{0}, 𝔫)) is embedded in Hom_{Der C}(F ⊗C, 𝔬(C^{0}, 𝔫)) ≅ Hom_{Der C}(F ⊕C^{0}, Der C⊕ad C^{0}) ≅ Hom_{Der C}(C^{0}, C^{0}), which is one-dimensional by Schur’s Lemma.

With this in mind, when μ = 2, T := T (C, D_{2}) = Der C ⊕ (C^{0} ⊗ D_{2}^{0}) ⊕ Der D_{2} has even part

T¯0 = Der C ⊕ (C^{0} ⊗ g) ⊕ (Der D_{2})¯0,

where g = e−2f. Thus, T¯0 is a direct sum of two simple Lie algebras, Der C⊕(C^{0}⊗ g) and (Der D_{2})¯0 ≅ 𝔰𝔩_{2}. We argue that the first is isomorphic to the orthogonal Lie algebra 𝔬(C^{0}, 𝔫) in the following way: Since 𝔱(g^{2}) = 2, we have g ∗ g = −g and the map ϕ : Der C ⊕ (C^{0} ⊗ g) → Der C ⊕ ad C^{0} = 𝔬(C^{0}, 𝔫) with ϕ(D_{a,b}) = D_{a,b} and ϕ(a ⊗ g) = ad a is an isomorphism of Lie algebras. Also the odd part of T is

T¯1 = (C^{0} ⊗ (D_{2})¯1) ⊕ (Der D_{2})¯1 = (C^{0} ⊗ (D_{2})¯1) ⊕ d_{e,(D_{2})¯1} ,

and the map ψ : T¯1 ↦ C ⊗(D_{2})¯1 such that ψ(a⊗u) = a⊗u and ψ(d_{e,u}) = \frac{1}{4}(1⊗u), for any a ∈ C^{0} and u ∈ (D_{2})¯1, is a linear isomorphism. Moreover, using ϕ and ψ, we see that as a T¯0-module, T¯1 ≅ C ⊗ (D_{2})¯1 is the tensor product of the spin module for 𝔬(C^{0}, 𝔫) and the natural module for 𝔰𝔩_{2} ≅ (Der D_{2})¯0. This is because for any a, b ∈ C^{0} and u ∈ (D_{2})¯1

ψ([a ⊗ g, 4d_{e,u}]) = −4a ⊗ d_{e,u}(g) = 3a ⊗ u
= (a1 + 2(1a)) ⊗ u and
ψ([a ⊗ g, b ⊗ u]) = ψ([a, b] ⊗ g ∗ u + 2𝔱𝔯(ab)d_{g,u})
= ψ( − \frac{1}{2}[a, b] ⊗ u + 2(ab + ba)(3d_{g,u}))
= −\frac{1}{2}(ab − ba) ⊗ u + \frac{3}{2}(ab + ba) ⊗ u
= (ab + 2ba) ⊗ u .

It follows easily that T is a form of F(4), because the map T¯1 ⊗ T¯1 → T¯0, is uniquely determined up to scalars due to the fact that Hom_{𝔬(C^{0},𝔫)}(C ⊗ C, 𝔬(C^{0}, 𝔫)) and Hom_{𝔬(C^{0},𝔫)}(C ⊗ C, F) are one-dimensional, as are both Hom_{𝔰𝔩_{2}}(V ⊗ V, 𝔰𝔩_{2}) and Hom_{𝔰𝔩_{2}}(V ⊗ V, F), where V is the natural two-dimensional module for 𝔰𝔩_{2}. □

Remarks. (1) When C is a split octonion (Cayley-Dickson) algebra, this construction appears in [BZ, Example 4.7].

(2) The Lie superalgebras T (C, D_{2}), with C a Cayley-Dickson algebra, do not exhaust the forms of F(4). (See [E] for a complete discussion of the forms.)

(3) Equation (2.7) above suggests that if C is a unital associative composition algebra, then T (C, D_{μ}) might still be a Lie superalgebra for any value of μ ≠ 0, −1. That this is indeed true is the content of our next series of results.

§3. The Tits Construction T (C, D_{μ}) with C associative.

Lemma 3.1. Let Q be a quaternion algebra over F (here it is sufficient to assume charF ≠ 2) with canonical involution x ↦ \overline{x}. Let Q^{0} = \{x ∈ Q | 𝔱𝔯(x) = x + \overline{x} = 0\} (a three-dimensional simple Lie algebra under the product [x, y]). Assume V is a 2-dimensional space with a nonzero skew-symmetric form, φ : V × V → F. For u, v ∈ V, let σ_{u,v} ∈ 𝔰𝔭(V, φ) ≅ 𝔰𝔩_{2} (the symplectic Lie algebra) be given by

σ_{u,v}(w) = φ(v, w)u + φ(u, w)v.

For nonzero α, β, γ ∈ F, let 𝔤 = 𝔤¯0 ⊕ 𝔤¯1 be the superalgebra with

\begin{matrix}𝔤¯0 = 𝔰𝔭(V, φ) ⊕ Q^{0} ⊕ Q^{0} \\ 𝔤¯1 = V ⊗ Q \end{matrix}

and with multiplication given by

• the usual Lie bracket in 𝔤¯0

• [(f, p, q), u ⊗ x] = f(u) ⊗ x + u ⊗ (px − xq)

• [u ⊗ x, v ⊗ y] = (α𝔱𝔯(\overline{x}y)σ_{u,v}, βφ(u, v)(x\overline{y} − y\overline{x}), γφ(u, v)(\overline{x}y − \overline{y}x)) for u, v ∈ V, x, y ∈ Q, f ∈ 𝔰𝔭(V, φ), p, q ∈ Q^{0}. Then 𝔤 is a Lie superalgebra if and only if α = β + γ, and in this case, 𝔤 is a form of the exceptional classical simple Lie superalgebra D(2, 1; \frac{γ}{β}).

Proof. We can extend the scalars and assume that F is algebraically closed, and hence that Q = 𝔐_{2}(F) (2 × 2 matrices over F) and Q^{0} = 𝔰𝔩_{2}. Now we may identify Q with End_{F} (V) and x ↦ \overline{x} with the symplectic involution associated to φ so that Q^{0} = 𝔰𝔭(V, φ) too. Consider the linear isomorphism that comes from identifying V^{∗} with V via the form φ:

Φ : V ⊗ V → End_{F} (V), u ⊗ v ↦ uφ(v, ).

Then \overline{Φ(u ⊗ v)} = −Φ(v ⊗ u) and Φ(u_{1} ⊗ v_{1})Φ(u_{2} ⊗ v_{2}) = φ(v_{1}, u_{2})Φ(u_{1} ⊗ v_{2}). Moreover, End_{F} (V) is a module for Q^{0} ⊕ Q^{0} ≅ 𝔰𝔭(V, φ) ⊕ 𝔰𝔭(V, φ) by means of

(p, q).Φ(u ⊗ v) = pΦ(u ⊗ v) − Φ(u ⊗ v)q.

Indeed, for any f ∈ 𝔰𝔭(V, φ):

\begin{matrix}Φ(f(u) ⊗ v) = f(u)φ(v, ) = fΦ(u ⊗ v) \\ Φ(u ⊗ f(v)) = uφ(f(v), ) = −uφ(v, f( )) = −Φ(u ⊗ v)f \end{matrix}

from which we can see that Φ is an isomorphism of 𝔰𝔭(V, φ) ⊕ 𝔰𝔭(V, φ)-modules. Now

tr(\overline{Φ(u_{1} ⊗ v_{1})}Φ(u_{2} ⊗ v_{2})) = −tr(Φ(v_{1} ⊗ u_{1})Φ(u_{2} ⊗ v_{2}))
= −φ(u_{1}, u_{2})tr(Φ(v_{1} ⊗ v_{2})) = −φ(u_{1}, u_{2})φ(v_{2}, v_{1})
= φ(u_{1}, u_{2})φ(v_{1}, v_{2}),

Φ(u_{1} ⊗ v_{1})\overline{Φ(u_{2} ⊗ v_{2})}−Φ(u_{2} ⊗ v_{2})\overline{Φ(u_{1} ⊗ v_{1})}
= −φ(v_{1}, v_{2})u_{1}φ(u_{2}, ) + φ(v_{2}, v_{1})u_{2}φ(u_{1}, )
= −φ(v_{1}, v_{2})σ_{u_{1},u_{2}} and
\overline{Φ(u_{1} ⊗ v_{1})}Φ(u_{2} ⊗ v_{2})−\overline{Φ(u_{2} ⊗ v_{2})}Φ(u_{1} ⊗ v_{1})
= −φ(u_{1}, u_{2})σ_{v_{1},v_{2}}.

Therefore, after making the identifications

\begin{aligned}𝔤¯0 & = 𝔰𝔭(V, φ) ⊕ Q^{0} ⊕ Q^{0} = 𝔰𝔭(V, φ) ⊕ 𝔰𝔭(V, φ) ⊕ 𝔰𝔭(V, φ) \\ 𝔤¯1 & = V ⊗ Q ≅ V ⊗ V ⊗ V \end{aligned}

we have that the multiplication in 𝔤 = 𝔤¯0 ⊕ 𝔤¯1 is given by

[ (f_{1}, f_{2}, f_{3}), u_{1} ⊗ u_{2} ⊗ u_{3}]
= f_{1}(u_{1}) ⊗ u_{2} ⊗ u_{3} + u_{1} ⊗ f_{2}(u_{2}) ⊗ u_{3} + u_{1} ⊗ u_{2} ⊗ f_{3}(u_{3})
[u_{1} ⊗ u_{2} ⊗ u_{3}, v_{1} ⊗ v_{2} ⊗ v_{3}]
= (αφ(u_{2}, v_{2})φ(u_{3}, v_{3})σ_{u_{1},v_{1}}, −βφ(u_{1}, v_{1})φ(u_{3}, v_{3})σ_{u_{2},v_{2}},

− γφ(u_{1}, v_{1})φ(u_{2}, v_{2})σ_{u_{3},v_{3}})

so that 𝔤 = Γ(α, −β, −γ) in the notation of Scheunert [Sc, §1, Example 5]. Thus, 𝔤 is a Lie superalgebra if and only if α − β − γ = 0. In this case

𝔤 ≅ Γ( − \frac{α}{β}, 1, \frac{γ}{β} ) ≅ Γ(1, \frac{γ}{β} , −(1 + \frac{γ}{β})) = D(2, 1; \frac{γ}{β}),

as required. □

Theorem 3.2. Let Q be a quaternion algebra and assume μ ≠ 0, −1. Then T (Q, D_{μ}) is a Lie superalgebra which is a form of D(2, 1; μ).

Proof. Let J = D_{μ} for μ ≠ 0, −1. Then Der J ≅ 𝔬𝔰𝔭(J^{0}, 𝔱) ≅ 𝔬𝔰𝔭(1, 2) (Lemma 2.4). For a, b ∈ Q^{0}, D_{a,b} = ad [a, b], so D_{Q,Q} = ad Q^{0} ≅ Q^{0}. Hence we may regard T = T (Q, D_{μ}) as T = Q^{0}⊕(Q^{0}⊗J^{0})⊕𝔬𝔰𝔭(J^{0}, 𝔱), where for a, b ∈ Q^{0}, and u, v ∈ J^{0} we have

\begin{matrix}[a, b ⊗ u] = [a, b] ⊗ u \\ [a ⊗ u, b ⊗ v] = 𝔱(uv)[a, b] + [a, b] ⊗ u ∗ v + 2𝔱𝔯(ab)d_{u,v}. \end{matrix}

Since uv = 𝔱(uv)1 + u ∗ v for all u, v ∈ J, we may make the identifications: Q^{0} ⊕ (Q^{0} ⊗ J^{0}) = Q^{0} ⊗ J and T = (Q^{0} ⊗ J) ⊕ 𝔬𝔰𝔭(J^{0}, 𝔱), where

(3.3)  [a ⊗ u, b ⊗ v] = [ab] ⊗ uv + 2𝔱𝔯(ab)d_{u,v}

for all a, b ∈ Q^{0}, and u, v ∈ J.

Let V = J¯1 and φ(u, v) = 𝔱(uv) for all u, v ∈ V. Then φ is a nonzero skewsymmetric bilinear form on V and (Der J)¯0 ≅ 𝔰𝔭(V, φ). Then for u, v, w ∈ V,

(3.4)  \begin{aligned}uv & = 𝔱(uv)(\frac{1+μ}{2μ}e + \frac{1+μ}{2}f) \\ d_{u,v}(w) & = u(vw) + v(uw) \\ & = 𝔱(vw)(\frac{1+μ}{2μ}e + \frac{1+μ}{2}f)u + 𝔱(uw)(\frac{1+μ}{2μ}e + \frac{1+μ}{2}f)v \\ & = \frac{(1+μ)^{2}}{4μ} (𝔱(vw)u + 𝔱(uw)v) \\ & = \frac{(1+μ)^{2}}{4μ} σ_{u,v}(w) \end{aligned}

Set 𝔤 = 𝔤¯0⊕𝔤¯1, where 𝔤¯0 = 𝔰𝔭(V, φ)⊕Q^{0}⊕Q^{0} and 𝔤¯1 = V ⊗Q, and let ψ : T → 𝔤 be the linear isomorphism given by

(3.5)  \begin{matrix}ψ(a ⊗ e + b ⊗ f + d_{u,v}) = (\frac{(1+μ)^{2}}{4μ} σ_{u,v}, a , b) \\ ψ(a ⊗ u) = u ⊗ a \\ ψ(d_{e,u}) = \frac{1}{4}u ⊗ 1 \end{matrix}

for all a, b ∈ Q^{0}, u, v ∈ J¯1 = V. Note that (Der J)¯1 = d_{e,J¯1}.

Now using (3.3) and (3.4), we have for a, b ∈ Q^{0}, u ∈ J¯1 that

[a ⊗ e, b ⊗ u] = [a, b] ⊗ \frac{1}{2}u + 2𝔱𝔯(ab)d_{e,u}

↦^{ψ} u ⊗ \frac{1}{2}([a, b] + 𝔱𝔯(ab)1) = u ⊗ ab

[a ⊗ f, b ⊗ u] = [a, b] ⊗ \frac{1}{2}u + 2𝔱𝔯(ab)d_{f,u}

= [a, b] ⊗ \frac{1}{2}u − 2𝔱𝔯(ab)d_{e,u}

↦^{ψ} u ⊗ \frac{1}{2}([a, b] − 𝔱𝔯(ab)1) = −u ⊗ ba

[a ⊗ e, 4d_{e,u}] = −4a ⊗ d_{e,u}(e) = a ⊗ u ↦^{ψ} u ⊗ a
[a ⊗ f, 4d_{e,u}] = −4a ⊗ d_{e,u}(f) = −a ⊗ u ↦^{ψ} −u ⊗ a
[a ⊗ u, 4d_{e,v}] = 4a ⊗ d_{e,v}(u)  (for u, v ∈ J¯1)

= 4a ⊗ (e(vu) − \frac{1}{2}vu)

= 4𝔱(vu)a ⊗ (e(\frac{1+μ}{2μ}e + \frac{1+μ}{2}f) − \frac{1}{2}(\frac{1+μ}{2μ}e + \frac{1+μ}{2}f))

= −4𝔱(uv)a ⊗ (\frac{1+μ}{4μ}e − \frac{1+μ}{4}f)

= 𝔱(uv)( − \frac{1+μ}{2μ}(a\overline{1} − 1\overline{a}) ⊗ e − \frac{1+μ}{2}(\overline{a}1 − \overline{1}a) ⊗ f)

↦^{ψ} φ(u, v)(0, −\frac{1+μ}{2μ}(a\overline{1} − 1\overline{a}), −\frac{1+μ}{2μ}(\overline{a}1 − \overline{1}a))

[a ⊗ u, b ⊗ v] = [a, b] ⊗ uv + 2𝔱𝔯(ab)d_{u,v}

= 𝔱(uv)(\frac{1+μ}{2μ}[a, b] ⊗ e + \frac{1+μ}{2}[a, b] ⊗ f) + 2𝔱𝔯(ab)d_{u,v}

↦^{ψ} ( − \frac{(1+μ)^{2}}{2μ} tr(\overline{a}b)σ_{u,v}, −\frac{1+μ}{2μ}φ(u, v)(a\overline{b} − b\overline{a}), −\frac{1+μ}{2}φ(u, v)(\overline{a}b − \overline{b}a))

[4d_{e,u}, 4d_{e,v}] = 16(d_{d_{e,u}(e),v} + d_{e,d_{e,u}(v)}) = 16(−\frac{1}{4}d_{u,v})

= −4d_{u,v} ↦^{ψ} ( − 𝔱𝔯(\overline{1}1)\frac{(1+μ)^{2}}{2μ} σ_{u,v}, 0, 0).

These computations demonstrate that ψ is an isomorphism from T = T (Q, D_{μ}) into the superalgebra of the last lemma with the parameters α = −(1 + μ)^{2}/(2μ), β = −(1 + μ)/(2μ), and γ = −(1 + μ)/2. Because β + γ = α, this is a Lie superalgebra, and because μ = γ/β, it is a form of D(2, 1; μ). □

Remark. If K is a 2-dimensional unital composition algebra and μ ≠ 0, −1, then

(3.6)  T = T (K, D_{μ}) = (K^{0} ⊗ (D_{μ})^{0}) ⊕ Der D_{μ} ,

where Der D_{μ} ≅ 𝔬𝔰𝔭(1, 2) is a subsuperalgebra of T and K^{0} ⊗ (D_{μ})^{0} is the natural module V for 𝔬𝔰𝔭(1, 2). The space Hom_{𝔬𝔰𝔭(1,2)}(V ⊗ V, 𝔬𝔰𝔭(1, 2)) is one-dimensional, because as an 𝔬𝔰𝔭(1, 2)-module, V ⊗ V ≅ End(V), which decomposes as the direct sum of a trivial module spanned by the identity transformation, a copy of 𝔬𝔰𝔭(V) = 𝔬𝔰𝔭(1, 2), and the irreducible module of super symmetric transformations on V with zero supertrace, which is isomorphic to V. Thus, the multiplication (K^{0} ⊗ (D_{μ})^{0}) × (K^{0} ⊗ (D_{μ})^{0}) → Der D_{μ} is uniquely determined up to scalars. From this it follows that T is a form of 𝔰𝔩(1, 2) = 𝔰𝔩(V) ≅ 𝔬𝔰𝔭(V) ⊕ V. The superalgebra T (F, D_{μ}) = Der D_{μ} ≅ 𝔬𝔰𝔭(1, 2) (μ ≠ 0, −1).

§4. The Jordan Superalgebra J(V, ϑ) and the Freudenthal-Tits Magic Square

In this section we will focus on another simple Jordan superalgebra, J = J(V, ϑ), the Jordan superalgebra of a nondegenerate supersymmetric bilinear form ϑ on the superspace V = V¯0 ⊕ V¯1 having V¯0 = 0 and dim V¯1 = 2. Thus, J¯0 = F 1, J¯1 = F x ⊕ F y with xy = −yx = 1 and ϑ(x, y) = 1. The only normalized trace on J is given by 𝔱(1) = 1 and 𝔱(x) = 𝔱(y) = 0 (compare [BZ, Example 4.6]). If u = ϵ_{0} +ϵ_{1}x+ϵ_{2}y in the Grassmann envelope G(J), where ϵ_{0} ∈ G¯0, and ϵ_{1}, ϵ_{2} ∈ G¯1, then

\begin{aligned}u^{2} & = (ϵ_{0}^{2} + 2ϵ_{1}ϵ_{2}) + 2ϵ_{0}ϵ_{1}x + 2ϵ_{0}ϵ_{2}y \\ u^{3} & = ϵ_{0}(ϵ_{0}^{2} + 6ϵ_{1}ϵ_{2}) + 3ϵ_{0}^{2}(ϵ_{1}x + ϵ_{2}y) \end{aligned}

Using these expressions, we see that both 𝔠𝔥_{2}(u) = 0 and 𝔠𝔥_{3}(u) = 0. As a result, T (C, J) is a Lie superalgebra for any unital composition algebra C. It is not difficult to verify that T (C, J) is a form of the 31-dimensional exceptional simple Lie superalgebra G(3) if dim C = 8. (In [BZ, Example 4.6], the choice C = 𝕆, the split octonion algebra, was used to obtain G(3).) In fact all forms of G(3) can be obtained in this way - see [E] for the details. When dim C = 4, then T (C, J) is a form of 𝔬𝔰𝔭(3, 2) = B(1, 1); when dim C = 2, it is a form of 𝔬𝔰𝔭(1, 2) = B(0, 1); and when dim C = 1, T (C, J) ≅ Der J ≅ 𝔰𝔩_{2}. Thus, the Freudenthal-Tits magic square can be augmented with two columns corresponding to J(V, ϑ) and D_{μ} for μ ≠ 0, −1 to give the following table which displays T (C, J).

Here we suppose that J is given by (I) F, (II) H(𝔐_{3}(F), t), (3 × 3 symmetric matrices relative to the transpose t), (III) 𝔐_{3}(F)^{+} (3 × 3 matrices under the product uv + vu), (IV) H(𝔐_{3}(Q), ∗) (3 × 3 hermitian matrices over Q = 𝔐_{2}(F) relative to the conjugate transpose ∗), (V) H(𝔐_{3}(𝕆), ∗), (3×3 hermitian matrices over a split octonion algebra 𝕆 relative to the conjugate transpose ∗), (VI) J(V, ϑ), (VII) D_{μ} (μ ≠ 0, −1)

Extended Freudenthal-Tits Magic Square

C \ J  I  II  III  IV  V  VI  VII
F  0  A_{1}  A_{2}  C_{3}  F_{4}  A_{1}  B(0, 1)
F ⊕ F  0  A_{2}  A_{2}⊕A_{2}  A_{5}  E_{6}  B(0, 1)  A(1, 0)
Q  A_{1}  C_{3}  A_{5}  D_{6}  E_{7}  B(1, 1)  D(2, 1; μ)
𝕆  G_{2}  F_{4}  E_{6}  E_{7}  E_{8}  G(3)  F(4) (μ=2, \frac{1}{2} )

§5. Simple Jordan Superalgebras With a Normalized Trace

The foregoing constructions which give the superalgebras T (C, J) in the last two columns of the magic square have as one ingredient a unital Jordan superalgebra J with a normalized trace 𝔱 whose extension 𝔱̃ to the Grassman envelope G(J) satisfies the trace identity 𝔠𝔥_{3}(x) = 0. Here we determine which simple Jordan superalgebras admit such a trace.

Proposition 5.1. The only finite-dimensional simple unital Jordan superalgebras J with J¯1 ≠ 0 over a field of characteristic ≠ 2, 3 whose Grassmann envelope G(J) satisfies the trace identity 𝔠𝔥_{3}(x) = 0 relative a normalized trace on J are D_{2} and the Jordan superalgebra J(V, ϑ) = F 1⊕V of a supersymmetric bilinear form ϑ such that V = V¯1 and dim V = 2.

Proof. Assume J = J¯0⊕J¯1 is a unital simple Jordan superalgebra with a normalized trace 𝔱 and J¯1 ≠ 0. We claim:

• If there are nonzero elements a, b, c ∈ J¯1 with ab = 1 and ac = 0 = bc, then G(J) does not satisfy 𝔠𝔥_{3}(x) = 0.

Indeed if such elements exist, then for x = ϵ_{1}a + ϵ_{2}b + ϵ_{3}c with ϵ_{1}, ϵ_{2}, ϵ_{3} ∈ G¯1, we have x^{2} = 2ϵ_{1}ϵ_{2}1 and x^{3} = 2ϵ_{1}ϵ_{2}ϵ_{3}c, so that 𝔠𝔥_{3}(x) = x^{3} − \frac{3}{2}2ϵ_{1}ϵ_{2}x = −ϵ_{1}ϵ_{2}ϵ_{3}c ≠ 0.

Note that this implies:

• If J is a simple Jordan superalgebra J(V, θ) = F 1 ⊕ V of a supersymmetric bilinear form θ and dim V¯1 > 2, then G(J(V, θ)) does not satisfy 𝔠𝔥_{3}(x) = 0.

• If J = J(V, θ) and dim V¯0 ≥ 2, then for all v ∈ V¯0 with v^{2} = α1 ≠ 0, we have v ∈ (J, J, J) so that 𝔱(v) = 0. Then v^{3} = αv and 𝔠𝔥_{3}(v) = αv − \frac{3}{2}αv = −\frac{1}{2}αv ≠ 0.

• If K_{10} is the 10-dimensional Kac superalgebra (see [BE] for a description) then G(K_{10}) does not satisfy 𝔠𝔥_{3}(x) = 0.

This is a consequence of the previous statement because the even part of K_{10} contains a subalgebra of the form J(V, θ) with V = V¯0 and dim V¯0 = 4.

• If J = J(V, θ) and dim V¯0 = 1, dim V¯1 = 2, then after a scalar extension, J is isomorphic to D_{1} and Lemma 2.3 applies.

• If J is a simple Jordan superalgebra and J¯0 is not semisimple, then G(J) does not satisfy 𝔠𝔥_{3}(x) = 0.

Indeed, if J¯0 is not semisimple, then from [MZ] it follows that either J¯0 contains a subquotient J(V, θ) with V = V¯0 and dim V¯0 ≥ 2 or a subquotient F [x]/(x^{p}) where p = char(F) > 3, which evidently does not satisfy 𝔠𝔥_{3}(v) = 0.

• When J is one of the Jordan superalgebras 𝔐_{m,n}(F)^{+}, P_{n}(F), or Q_{n}(F), the envelope G(J) does not satisfy 𝔠𝔥_{3}(v) = 0.

It is a direct consequence of Lemma 2.3, that any Jordan superalgebra J containing a subalgebra isomorphic to D_{−1}, has a Grassmann envelope which fails to satisfy 𝔠𝔥_{3}(v) = 0. We observe that 𝔐_{1,1}(F)^{+} ≅ D_{−1}, so any 𝔐_{m,n}(F)^{+} contains D_{−1} as

a subalgebra. In P_{n}(F) = \{( \begin{matrix}a & b \\ c & a^{t} \end{matrix} ) | a, b, c ∈ 𝔐_{n}(F), b = −b^{t}, c = c^{t}\} (n ≥ 2),

the elements e = E_{1,1} + E_{n+1,n+1}, f = E_{2,2} + E_{n+2,n+2}, x = E_{1,n+2} − E_{2,n+1}, y = E_{n+1,2} + E_{n+2,1} span a subalgebra isomorphic to D_{−1}. This happens in

Q_{n}(F) = \{( \begin{matrix}a & b \\ b & a \end{matrix} ) | a, b ∈ 𝔐_{n}(F)\} (n ≥ 2) with the same e, f, y but with

x = E_{1,n+2} + E_{2,n+1}. Consequently, the assertion holds.

The results in the steps • combine to complete the proof. □

§6. The Tits construction and orthosymplectic Lie superalgebras

The (generalized) Tits construction considered in [BZ, Sec. 3] does not require starting with a unital composition algebra. An interesting case treated in [BZ, 3.28–3.33] begins with two Jordan algebras of symmetric bilinear forms. This construction has its counterpart for superalgebras.

Let V and W be two superspaces endowed with even supersymmetric nondegenerate bilinear forms ( |)_{V} and ( |)_{W}, and let J(V) = F 1 ⊕ V and J(W) = F 1 ⊕ W be the associated Jordan superalgebras. On the vector space

T (J(V), J(W)) = 𝔬𝔰𝔭(V) ⊕ (V ⊗ W) ⊕ 𝔬𝔰𝔭(W)

we specify a superanticommutative multiplication by

(6.1)

• 𝔬𝔰𝔭(V) and 𝔬𝔰𝔭(W) are subsuperalgebras,

• [𝔬𝔰𝔭(V), 𝔬𝔰𝔭(W)] = 0,

• [D, v ⊗ w] = D(v) ⊗ w , [d, v ⊗ w] = (−1)d̄ ; v̄v ⊗ d(w)

• [v ⊗ w, v′ ⊗ w′] = (−1)^{w̄v̄′}(D_{v,v′}(w | w′)_{W} + (v | v′)_{V} d_{w,w′}) for any homogeneous elements D ∈ 𝔬𝔰𝔭(V), d ∈ 𝔬𝔰𝔭(W), v, v′ ∈ V, w, w′ ∈ W, where

(6.2)  \begin{aligned}D_{v,v′}(v′′) & = (−1)^{v̄′v̄′′}(v | v′′)_{V} v′ − (−1)^{v̄(v̄′+v̄′′)}(v′ | v′′)_{V} v \\ d_{w,w′}(w′′) & = (−1)^{w̄′w̄′′}(w | w′′)_{W} w′ − (−1)^{w̄(w̄′+w̄′′)}(w′ | w′′)_{W} w. \end{aligned}

The argument in [BZ, 3.30], with appropriate parity signs inserted, shows that T (J(V), J(W)) ≅ 𝔬𝔰𝔭(V ⊕ W), where the supersymmetric bilinear form on V ⊕ W is the orthogonal sum of the ones on V and W. Thus, the result of the Tits construction is

(6.3)  𝔬𝔰𝔭(V ⊕ W) = 𝔬𝔰𝔭(V) ⊕ (V ⊗ W) ⊕ 𝔬𝔰𝔭(W).

When V = V¯0, and W = W¯0, (6.3) gives

(6.4)  𝔬(V¯0 ⊕ W¯0) = 𝔬(V¯0) ⊕ (V¯0 ⊕ W¯0) ⊕ 𝔬(V¯0)

which is (3.31) of [BZ] and comes from the Tits construction there.

Another interesting instance of the above occurs when V = V¯1 and W = W¯1. In this case 𝔬𝔰𝔭(V) = 𝔰𝔭(V¯1) and 𝔬𝔰𝔭(W) = 𝔰𝔭(W¯1), so that (6.3) says

(6.5)  𝔰𝔭(V¯1 ⊕ W¯1) = 𝔰𝔭(V¯1) ⊕ (V¯1 ⊗ W¯1) ⊕ 𝔰𝔭(V¯1).

Hence, the Lie algebra 𝔰𝔭(V¯1 ⊕W¯1) is obtained from the Jordan superalgebras J(V¯1) and J(W¯1) (see [BZ, 3.33]).

Finally, when V = V¯0 and W = V¯1, then by (6.3)

(6.6)  \begin{aligned}𝔬𝔰𝔭(V¯0 ⊕ V¯1) & = 𝔬𝔰𝔭(V¯0) ⊕ (V¯0 ⊗ V¯1) ⊕ 𝔬𝔰𝔭(V¯1) \\ & = 𝔬(V¯0) ⊕ (V¯0 ⊗ V¯1) ⊕ 𝔰𝔭(V¯1), \end{aligned}

which is the standard realization of 𝔬𝔰𝔭(V¯0 ⊕ V¯1) as a module for its even part 𝔬(V¯0) ⊕ 𝔰𝔭(V¯1).

We discuss these Tits constructions from two different perspectives. For the first, we assume F is algebraically closed of characteristic 0. Recall that the maximal regular semisimple subalgebras of a simple Lie algebra can be read from the corresponding affine Dynkin diagram by deleting a node (as in [D]). Thus, from the affine diagrams of (i) B_{r}^{(1)}, (ii) D_{r}^{(1)}, (iii) C_{r}^{(1)}, (see [K4, p. 54]), we obtain that (i) 𝔬_{2m+1} ⊕ 𝔬_{2n} (m + n = r) is maximal in 𝔬_{2r+1}; (ii) 𝔬_{2m} ⊕ 𝔬_{2n} (m + n = r) is maximal in 𝔬_{2r}; (iii) 𝔰𝔭_{2m} ⊕ 𝔰𝔭_{2n} (m + n = r) is maximal in 𝔰𝔭_{2r}. When the Lie algebra is viewed as a module for the maximal semisimple subalgebra under the adjoint action, the resulting decomposition corresponds to one in (6.4) or (6.5). Note that 𝔬_{2m+1} ⊕ 𝔬_{2n+1} is a maximal semisimple subalgebra of 𝔬_{2r} for r = m + n + 1 that is not regular, but there is a decomposition as in (6.4) in this case as well.

Our second interpretation of the orthosymplectic Tits construction comes from the identification of 𝔬𝔰𝔭(V) with ∧^{2}V as modules for 𝔬𝔰𝔭(V). Here ∧^{2}V = (V ⊗ V)/S, where S is the subspace of V ⊗ V spanned by the elements u ⊗ v + (−1)^{ūv̄}v ⊗ u for all homogeneous u, v ∈ V. The map u ∧ v ↦ D_{u,v}, where u ∧ v = (u ⊗ v) + S and D_{u,v} is as in (6.2), provides an 𝔬𝔰𝔭(V)-module isomorphism. Now we have the usual identity for vector spaces [FH, Exer. 1.2],

∧^{2}(V ⊕ W) = ⨁_{j=0}^{2} ( ∧^{j} (V) ⊗ ∧^{2−j}(W)),

which can be viewed as a decomposition into modules for 𝔬𝔰𝔭(V)⊕𝔬𝔰𝔭(W) (compare (6.3)). The fact that the Lie superalgebra structure is respected is the content of the Tits construction.

One further comment on these constructions – when V = V¯0 ⊕ V¯1, then ∧^{2}V ≅ ∧^{2}V¯0⊕(V¯0⊗V¯1)⊕S^{2}V¯1. If dim V¯1 is even, this decomposition has an interpretation as a splitting of ∧^{2}V into 𝔬(V¯0)⊕𝔰𝔭(V¯1)-modules as in (6.6). If one further decomposes V¯1 into isotropic subspaces V¯1 = V′ ⊕V′′ as in [K3, Sec. 2], the result is a 5-grading of 𝔬𝔰𝔭(V) given by

𝔬𝔰𝔭(V) = S^{2}V′ ⊕ (V¯0 ⊗ V′) ⊕ ( ∧^{2} V¯0 ⊕ (V′ ⊗ V′′)) ⊕ (V¯0 ⊗ V′′) ⊕ S^{2}V′′,

whose zero component ∧^{2}V¯0 ⊕ (V′ ⊗ V′′) is isomorphic to 𝔬(V¯0) ⊕ 𝔤𝔩(V′).

References

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[K3] V.G. Kac, A sketch of Lie superalgebra theory, Comm. in Math. Physics 53 (1977), 31–64.

[K4] V.G. Kac, Infinite Dimensional Lie Algebras, Third Ed., Cambridge U. Press, Cambridge, 1990.

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[T] J. Tits, Algèbres alternatives, algèbres de Jordan, et algèbres de Lie exceptionelle, Nederl. Akad. Weten. Proc. Ser. A 69 (1966), 223-237.

Department of Mathematics, University of Wisconsin, Madison, Wisconsin 53706

   E-mail address: benkart@math.wisc.edu

Departamento de Matemáticas, Universidad de Zaragoza, 50009 Zaragoza, Spain

   E-mail address: elduque@posta.unizar.es