Supertrace

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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, Tr(T) is defined by the following tangle diagram:

Image:Trace.png

More concretely, if we write out T is block matrix form after the decomposition into even and odd subspaces as follows,

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

then the supertrace

Tr(T) = the ordinary trace of T0 0 − the ordinary trace of T11.

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(\mathbf{e}_j)=\mathbf{e}_i T^i_j.\,

The grading of Tij is the sum of the grading of i and the grading of j mod 2.

A change of basis to e1', ..., ep', e(p+1)', ..., e(p+q)' is given by the supermatrix

\mathbf{e}_{i'}=\mathbf{e}_i A^i_{i'}

and the inverse supermatrix

\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

\operatorname{str}(A^{-1} T A)=(-1)^{|i'|} (A^{-1})^{i'}_j T^j_k A^k_{i'}=(-1)^{|i'|}(-1)^{(|i'|+|j|)(|i'|+|j|)}T^j_k A^k_{i'} (A^{-1})^{i'}_j=(-1)^{|j|} T^j_j =\operatorname{str}(T).\,

In particular, if we have a Z2-graded Hilbert space and T is an operator over it, the appropriate trace to use is the supertrace.

See also: Berezinian.

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