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

Image:Trace.png

More concretely, if we write out T in 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

str(T) = the ordinary trace of T0 0 − 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(\mathbf{e}_j)=\mathbf{e}_i T^i_j.\,

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

\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. In the case where T is even, we have

\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 the case where T is odd, we have

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

\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.

[edit] See also

[edit] Reference

  1. ^ N. Berline, E. Getzler, M. Vergne, Heat Kernels and Dirac Operators, Springer-Verlag, 1992, ISBN 0-387-53340-0, p. 39.