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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2 Answers
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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.
answered Mar 11, 2015 at 19:57
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$\begingroup$ 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. $\endgroup$– ShadumuCommented Mar 13, 2015 at 1:09
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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}$$=(\Lambda^{-1})_{\mu}{}^{\sigma}$. We stress that the last eq. does not imply that $\Lambda^T$ and $\Lambda^{-1}$ are the same matrix, cf. e.g. this Phys.SE post. $\endgroup$– Qmechanic ♦Commented Mar 13, 2015 at 19:37
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2$\begingroup$ $\Lambda^T$ is a matrix for a linear map in ${\cal L}(V^{\ast};V^{\ast})$ while $\Lambda^{-1}$ is a matrix for a linear map in ${\cal L}(V;V)$, so they are like apples and oranges. $\endgroup$– Qmechanic ♦Commented Jul 21, 2018 at 4:36
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1$\begingroup$ Actually, another question, why is it ok to write that equation? It seems to me based on the positions of the indices that $(A^T)^i{}_j$ is in $V\otimes V^*$ while $A^i{}_j$ is in $V^*\otimes V$, so we shouldn't be able to write that equation. $\endgroup$ Commented Aug 18, 2021 at 5:03
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To answer your question, we first need to look up the definition of the transpose. After reformulating your question, it will be very straightforward to answer it.
Definition. Let $V$ and $W$ be vector spaces over a field $F$ and $A\colon V\to W$ a linear map. Then the transpose of $A$ is the linear map $$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}$$
