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- Supertrace abstract "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:File:Trace.pngMore concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,then the supertracestr(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 asThe 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 supermatrixand the inverse supermatrixwhere 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 haveIn the case where T is odd, we haveThe ordinary trace is not basis independent, so the appropriate trace to use in the Z2-graded setting is the supertrace.The supertrace satisfies the propertyfor 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.".
- Supertrace thumbnail Trace.png?width=300.
- Supertrace wikiPageID "1249738".
- Supertrace wikiPageRevisionID "543902030".
- Supertrace hasPhotoCollection Supertrace.
- Supertrace subject Category:Super_linear_algebra.
- Supertrace comment "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:File:Trace.pngMore concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,then the supertracestr(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. ".
- Supertrace label "Supertraccia".
- Supertrace label "Supertrace".
- Supertrace sameAs Supertraccia.
- Supertrace sameAs m.04m2ff.
- Supertrace sameAs Q3978058.
- Supertrace sameAs Q3978058.
- Supertrace wasDerivedFrom Supertrace?oldid=543902030.
- Supertrace depiction Trace.png.
- Supertrace isPrimaryTopicOf Supertrace.
