In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A-supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
T = ( T 00 T 01 T 10 T 11 ) {\displaystyle T={\begin{pmatrix}T_{00}&T_{01}\\T_{10}&T_{11}\end{pmatrix}}}
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, ..., ep are the even basis vectors and ep+1, ..., ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
T ( e j ) = e i T j i . {\displaystyle T(\mathbf {e} _{j})=\mathbf {e} _{i}T_{j}^{i}.\,}
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', ..., ep', e(p+1)', ..., e(p+q)' is given by the supermatrix
e i ′ = e i A i ′ i {\displaystyle \mathbf {e} _{i'}=\mathbf {e} _{i}A_{i'}^{i}}
and the inverse supermatrix
e i = e i ′ ( A − 1 ) i i ′ , {\displaystyle \mathbf {e} _{i}=\mathbf {e} _{i'}(A^{-1})_{i}^{i'},\,}
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
str ( A − 1 T A ) = ( − 1 ) | i ′ | ( A − 1 ) j i ′ T k j A i ′ k = ( − 1 ) | i ′ | ( − 1 ) ( | i ′ | + | j | ) ( | i ′ | + | j | ) T k j A i ′ k ( A − 1 ) j i ′ = ( − 1 ) | j | T j j = str ( T ) . {\displaystyle \operatorname {str} (A^{-1}TA)=(-1)^{|i'|}(A^{-1})_{j}^{i'}T_{k}^{j}A_{i'}^{k}=(-1)^{|i'|}(-1)^{(|i'|+|j|)(|i'|+|j|)}T_{k}^{j}A_{i'}^{k}(A^{-1})_{j}^{i'}=(-1)^{|j|}T_{j}^{j}=\operatorname {str} (T).}
In the case where T is odd, we have
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