Information AboutSupertrace |
| CATEGORIES ABOUT SUPERTRACE | |
| super linear algebra | |
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: More concretely, if we write out ''T'' is Block Matrix form after the decomposition into even and odd subspaces as follows, : then the supertrace :Tr(''T'') = the ordinary Trace of ''T''0 0 − the ordinary trace of ''T''11. Suppose e1, ..., ep are the even basis vectors and e''p''+, ..., e''p''+''q'' are the odd basis vectors. Then, the components of ''T'', which are elements of ''A'', are defined as : The grading of ''T''''i''''j'' 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 : and the inverse supermatrix : where of course, ''AA''−1 = ''A''−1''A'' = 1 (the identity). We can now check explicitly that the supertrace is Basis Independent |
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