Transpose of (1,1) tensor 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?
 A: *

*Recall that (1,1) tensors can be identified with linear operators $$\begin{align}A~=~&\sum_{ij} e_i~A^i{}_j~e^{\ast j}\cr
~\in~& {\cal L}(V;V)~\cong~V\otimes V^{\ast},\end{align}\tag{1}$$ where $V$ is the underlying vector space.


*The transposed element is of the form $$
\begin{align}A^T~=~&\sum_{ij} e^{\ast j}~(A^{T})_j{}^i~e_i\cr
~\in~&{\cal L}(V^{\ast};V^{\ast})~\cong~V^{\ast}\otimes V,\end{align}\tag{2}$$ where $V^{\ast}$ is the dual vector space.


*If there is only Grassmann-even variables, then the transposed tensor is
$$   (A^{T})_j{}^i ~:=~ A^i{}_j \tag{3}$$
in local coordinates.


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

*

*Bryce De Witt, Supermanifolds, Cambridge Univ. Press, 1992.

A: To answer your question, we first need to look up the definition of the transpose. After reformulating your question, it will be very straightforward to answer it.

Definition. Let $V$ and $W$ be vector spaces over a field $F$ and $A\colon V\to W$ a linear map. Then the transpose of $A$ is the
linear map $$A^\mathrm{T}\colon W^*\to V^*$$ satisfying
$A^\mathrm{T}(x)=x\circ A$ for all $x\in W^*$. We also consider the following function:
\begin{align}
\Phi\colon L(V,W)&\to L(W^*,V^*)\\
A&\mapsto A^\mathrm{T}
\end{align}

We are considering the case where $V$ is $n$-dimensional and $V=W$. Let $\displaystyle{F^{n\times n}}$ be the set of $n\times n$-matrices with entries in $F$. Let $v_1,\ldots,v_n$ be a basis of $V$, then the isomorphisms
\begin{align}
\alpha\colon L(V,V)&\to F^{n\times n}
\end{align}
and
\begin{align}
\beta\colon L(V^*,V^*)&\to F^{n\times n}
\end{align}
defined by
$$\alpha(A)_{ij}:=A^i{}_j:=v^i(Av_j)$$
and
$$\beta(B)_{mn}:=B_m{}^n:=v_mBv^n:=(Bv^n)(v_m)$$
allow us to identify tensors/linear maps with matrices. What we can do is ask how the matrix $M$ assigned to $A\in L(V,V)$ is related to the matrix $N$ assigned to $A^\mathrm{T}\in L(V^*,V^*)$.
By unwrapping the definitions, one easily sees that $N$ is the transpose of $M$, i.e. $N_{ij}=M_{ji}$. In other words:
$$(\beta\circ \Phi\circ\alpha^{-1})(M)=M^\mathrm{T}$$
