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,

[ , ] + [ , ] ○ c_{L,L} = 0

which is the same as

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

Jacobi identity,

[, [ , ]] + [, [ , ]] ○ σ + [, [ , ]] ○ σ^{2} = 0,

where σ ∈ S_{3} 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 ○ c_{A,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 M_{1}, M_{2} A-modules, M_{1} ⊗ M_{2} is taken as the tensor product of M_{1} as a right module with M_{2} as a left module.

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

(A^{p|q})_{0} = A_{0} ⊗ (k^{p|q})_{0} ⊕ A_{1} ⊗ (k^{p|q})_{1}

(A^{p|q})_{1} = A_{1} ⊗ (k^{p|q})_{0} ⊕ A_{0} ⊗ (k^{p|q})_{1}

We say that an A-module M is free if it is isomorphic (in the category of A-modules) to A^{p|q} for some (p, q). This is equivalent to saying that M contains p even elements {e_{1}, …, e_{p}} and q odd elements {ε_{1}, …, ε_{q}} such that

M_{0} = span_{A0}{e_{1}, …, e_{p}} ⊕ span_{A1}{ε_{1}, …, ε_{q}}

M_{1} = span_{A1}{e_{1}, …, e_{p}} ⊕ span_{A0}{ε_{1}, …, ε_{q}}

We shall also say M as the free module generated over A by the even elements e_{1}, …, e_{p }and the odd elements ε_{1}, …, ε_{q}.

Let T: A^{p|q} → A^{r|s }be a morphism of free A-modules and then write e_{p+1}, …., e_{p+q} for the odd basis elements ε_{1}, …, ε_{q}. Then T is defined on the basis elements {e_{1}, …, e_{p+q}} by

T(e_{j}) = ∑_{i=1}^{p+q} e_{i}t^{i}_{j}

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

T = (^{T1 T2} _{T3 T4})

where T_{1} is an r x p matrix consisting of even elements of A, T_{2} is an r x q matrix of odd elements, T_{3} is an s x p matrix of odd elements, and T_{4} 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, T_{1} & T_{4}, which are even and T_{2} & T_{3}, which are odd, is determined. When we define T on the basis elements, the basis elements precedes the coordinates t^{i}_{j}. 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)(e_{j}) = S(∑_{i} e_{i}t^{i}_{j}) = ∑_{i,k}e_{k}s_{i}^{k}t^{i}_{j}

hence for any x ∈ A^{p|q} , we can express x as the column vector x = ∑e_{i}x^{i} and so T(x) is given by the matrix product T x.