Supertrace information

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,

${\displaystyle T={\begin{pmatrix}T_{00}&T_{01}\\T_{10}&T_{11}\end{pmatrix}}}$

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, ..., e_p are the even basis vectors and e_p+1, ..., e_p+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as

${\displaystyle T(\mathbf {e} _{j})=\mathbf {e} _{i}T_{j}^{i}.\,}$

The grading of Tⁱ_j is the sum of the gradings of T, e_i, e_j mod 2.

A change of basis to e_1', ..., e_p', e_(p+1)', ..., e_(p+q)' is given by the supermatrix

${\displaystyle \mathbf {e} _{i'}=\mathbf {e} _{i}A_{i'}^{i}}$

and the inverse supermatrix

${\displaystyle \mathbf {e} _{i}=\mathbf {e} _{i'}(A^{-1})_{i}^{i'},\,}$

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

${\displaystyle \operatorname {str} (A^{-1}TA)=(-1)^{|i'|}(A^{-1})_{j}^{i'}T_{k}^{j}A_{i'}^{k}=(-1)^{|i'|}(-1)^{(|i'|+|j|)(|i'|+|j|)}T_{k}^{j}A_{i'}^{k}(A^{-1})_{j}^{i'}=(-1)^{|j|}T_{j}^{j}=\operatorname {str} (T).}$

In the case where T is odd, we have

${\displaystyle \operatorname {str} (A^{-1}TA)=(-1)^{|i'|}(A^{-1})_{j}^{i'}T_{k}^{j}A_{i'}^{k}=(-1)^{|i'|}(-1)^{(1+|j|+|k|)(|i'|+|j|)}T_{k}^{j}(A^{-1})_{j}^{i'}A_{i'}^{k}=(-1)^{|j|}T_{j}^{j}=\operatorname {str} (T).}$

The ordinary trace is not basis independent, so the appropriate trace to use in the Z₂-graded setting is the supertrace.

The supertrace satisfies the property

${\displaystyle \operatorname {str} (T_{1}T_{2})=(-1)^{|T_{1}||T_{2}|}\operatorname {str} (T_{2}T_{1})}$

for all T₁, T₂ in End(V). In particular, the supertrace of a supercommutator is zero.

In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E -> A which vanishes on supercommutators.^[1] Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.