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When we transpose a (1,1) tensor, shall we simply switch the two indices while keeping their upper/lower positions or switch them and also switch their upper/lower positions? In general, would the left/right order matter for a tensor? Is it true that in contracting indices between two tensors, we want the contracted index to be right close to each other?

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  1. If there is only Grassmann-even variables, then the transposed tensor is $$ \tag{1} (A^{T})_j{}^i ~:=~ A^i{}_j $$ in local coordinates.

  2. Note that for tensors in supervector spaces and supermanifolds, the supertransposition carries additional Grassmann sign factors, see e.g. Ref. 1 for details.

References:

  1. Bryce De Witt, Supermanifolds, Cambridge Univ. Press, 1992.
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  • $\begingroup$ Are you sure the tranpose of a tensor is as (1)? If A is lorentz transformation and B is its inverse, then $B_j{}^i=A^i{}_j$ , which is certainly not A's transpose. $\endgroup$ – Shadumu Mar 13 '15 at 1:09
  • $\begingroup$ That seems to be a separate issue caused by raising and lowering indices with the metric tensor, see e.g. this related Phys.SE post. $\endgroup$ – Qmechanic Mar 13 '15 at 1:36
  • $\begingroup$ A Lorentz matrix satisfies $\Lambda^T\eta\Lambda=\eta$. Or equivalently $\Lambda^T=\eta\Lambda^{-1}\eta^{-1}$. Or equivalently with indices: $\Lambda^{\sigma}{}_{\mu}$$=(\Lambda^T)_{\mu}{}^{\sigma}$$=\eta_{\mu\nu}(\Lambda^{-1})^{\nu}{}_{\rho}\eta^{\rho\sigma}$$=(\Lambda^{-1})_{\mu}{}^{\sigma}$. We stress that the last eq. does not imply that $\Lambda^T$ and $\Lambda^{-1}$ are the same matrix, cf. e.g. this Phys.SE post. $\endgroup$ – Qmechanic Mar 13 '15 at 19:37
  • $\begingroup$ But its transpose and inverse clearly are no the same tensor, just consider a boost in x direction you will see.@Qmechanic♦ $\endgroup$ – Shadumu Mar 13 '15 at 20:59
  • $\begingroup$ @Qmechanic♦ I know that it has to be wrong but...Why your last equation does not imply that $\Lambda^T=\Lambda^{-1}$ $\endgroup$ – MattG88 Jul 20 '18 at 23:30

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