5
$\begingroup$

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?

$\endgroup$

2 Answers 2

3
$\begingroup$
  1. 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.

  2. 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.

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

  4. 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.
$\endgroup$
9
  • $\begingroup$ Are you sure the tranpose of a tensor is as (3)? 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, 2015 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, 2015 at 1:36
  • 1
    $\begingroup$ 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, 2015 at 19:37
  • 2
    $\begingroup$ $\Lambda^T$ is a matrix for a linear map in ${\cal L}(V^{\ast};V^{\ast})$ while $\Lambda^{-1}$ is a matrix for a linear map in ${\cal L}(V;V)$, so they are like apples and oranges. $\endgroup$
    – Qmechanic
    Jul 21, 2018 at 4:36
  • 1
    $\begingroup$ Actually, another question, why is it ok to write that equation? It seems to me based on the positions of the indices that $(A^T)^i{}_j$ is in $V\otimes V^*$ while $A^i{}_j$ is in $V^*\otimes V$, so we shouldn't be able to write that equation. $\endgroup$
    – user56834
    Aug 18, 2021 at 5:03
0
$\begingroup$

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: $$(\beta\circ \Phi\circ\alpha^{-1})(M)=M^\mathrm{T}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.