Super Lie Algebra

JacobiatorIdentity

A super Lie algebra L is an object in the category of super vector spaces together with a morphism [ , ] : L ⊗ L → L, often called the super bracket, or simply, the bracket, which satisfies the following conditions

Anti-symmetry,

[ , ] + [ , ] ○ cL,L = 0

which is the same as

[x, y] + (-1)|x||y|[y, x] = 0 for x, y ∈ L homogenous.

Jacobi identity,

[, [ , ]] + [, [ , ]] ○ σ + [, [ , ]] ○ σ2 = 0,

where σ ∈ S3 is a three-cycle, i.e. taking the first entity of [, [ , ]] to the second, and the second to the third, and then the third to the first. So, for x, y, z ∈ L homogenous, this reads

[x + [y, z]] + (-1)|x||y| + |x||z|[y, [z, x]] + (-1)|y||z| + |x||z|[z, [x, y]] = 0

It is important to note that in the super category, these conditions are modifications of the properties of the bracket in a Lie algebra, designed to accommodate the odd variables. We can immediately extend this definition to the case where L is an A-module for A a commutative superalgebra, thus defining a Lie superalgebra in the category of A-modules. In fact, we can make any associative superalgebra A into a Lie superalgebra by taking the bracket to be

[a, b] = ab – (-1)|a||b|ba,

i.e., we take the bracket to be the difference τ – τ ○ cA,A, where τ is the multiplication morphism on A.

A left A-module is a super vector space M with a morphism A ⊗ M → M, a ⊗ m ↦ am, of super vector spaces obeying the usual identities; that is, ∀ a, b ∈ A and x, y ∈ M, we have

a (x + y) = ax + ay

(a + b)x = ax + bx

(ab)x  = a(bx)

1x = x

A right A-module is defined similarly. Note that if A is commutative, a left A-module is also a right A-module if we define (the sign rule)

m . a = (-1)|m||a|a . m

for m ∈ M and a ∈ A. Morphisms of A-modules are defined in the obvious manner: super vector space morphisms φ: M → N such that φ(am) = aφ(m) ∀ a ∈ A and m ∈ M. So, we have the category of A-modules. For A commutative, the category of A-modules admits tensor products: for M1, M2 A-modules, M1 ⊗ M2 is taken as the tensor product of M1 as a right module with M2 as a left module.

Turning our attention to free A-modules, we have the notion of super vector kp|q over k, and so we define Ap|q := A ⊗ kp|q where

(Ap|q)0 = A0 ⊗ (kp|q)0 ⊕ A1 ⊗ (kp|q)1

(Ap|q)1 = A1 ⊗ (kp|q)0 ⊕ A0 ⊗ (kp|q)1

We say that an A-module M is free if it is isomorphic (in the category of A-modules) to Ap|q for some (p, q). This is equivalent to saying that M contains p even elements {e1, …, ep} and q odd elements {ε1, …, εq} such that

M0 = spanA0{e1, …, ep} ⊕ spanA11, …, εq}

M1 = spanA1{e1, …, ep} ⊕ spanA01, …, εq}

We shall also say M as the free module generated over A by the even elements e1, …, eand the odd elements ε1, …, εq.

Let T: Ap|q → Ar|s be a morphism of free A-modules and then write ep+1, …., ep+q for the odd basis elements ε1, …, εq. Then T is defined on the basis elements {e1, …, ep+q} by

T(ej) = ∑i=1p+q eitij

Hence T can be represented as a matrix of size (r + s) x (p + q)

T = (T1 T2 T3 T4)

where T1 is an r x p matrix consisting of even elements of A, T2 is an r x q matrix of odd elements, T3 is an s x p matrix of odd elements, and T4 is an s x q matrix of even elements. When we say that T is a morphism of super A-modules, it means that it must preserve parity, and therefore the parity of the blocks, T1 & T4, which are even and T2 & T3, which are odd, is determined. When we define T on the basis elements, the basis elements precedes the coordinates tij. This is important to keep the signs in order and comes naturally from composing morphisms. In other words if the module is written as a right module with T acting from the left, composition becomes matrix product in the usual manner:

(S . T)(ej) = S(∑i eitij) = ∑i,keksiktij

hence for any x ∈ Ap|q , we can express x as the column vector x = ∑eixi and so T(x) is given by the matrix product T x.