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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 trace diagram: More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows, then the supertrace str(T) = the ordinary trace of T00 − the ordinary trace of T11. 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 and the inverse supermatrix

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  • Supertraccia (it)
  • Supertrace (en)
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  • Nella teoria delle superalgebre, se T è una supermatrice quadrata (oppure una matrice a blocchi decomponibile in parti pari e dispari) del tipo: la supertraccia della matrice T è data da: str(T) = l'ordinaria traccia di T0 0 − l'ordinaria traccia di T11. Si può dimostrare che la supertraccia non dipende dalla base scelta per esprimere la supermatrice. (it)
  • 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, then the supertrace str(T) = the ordinary trace of T00 − the ordinary trace of T11. 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 and the inverse supermatrix (en)
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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 trace diagram: More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows, 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 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 and the inverse supermatrix where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is . In the case where T is even, we have In the case where T is odd, we have 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 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. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A. (en)
  • Nella teoria delle superalgebre, se T è una supermatrice quadrata (oppure una matrice a blocchi decomponibile in parti pari e dispari) del tipo: la supertraccia della matrice T è data da: str(T) = l'ordinaria traccia di T0 0 − l'ordinaria traccia di T11. Si può dimostrare che la supertraccia non dipende dalla base scelta per esprimere la supermatrice. (it)
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