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Consider the following contraction between two vector fields

$$ A_{k,i}B_{k,j} $$ Summation over $k$ is implied. I want to decompose this into parts that are symmetric/antisymmetric w.r.t the permutation of $i$ and the $k$th component of $B$. My attempt is to write these parts as

\begin{align} A_{k,(i}B_{k),j} &= \tfrac{1}{2} (A_{k,i}B_{k,j} + A_{k,k}B_{i,j}) \\ A_{k,[i}B_{k],j} &= \tfrac{1}{2} (A_{k,i}B_{k,j} - A_{k,k}B_{i,j}) = \tfrac{1}{2} \delta_{ik}^{ab} A_{k,a}B_{b,j} \end{align}

I would like to know whether this procedure is valid even in non-Euclidean spaces? In general, what are the restrictions one should consider before applying such tricks to tensors?

Also please help me to properly place the $($ or $[$ for the case $A_{i,j}B_{k,\ell}$ where now $i$th component of $A$ is to permute the $k$th component of $B$. If I write this as $A_{(i,j}B_{k),\ell}$ then it seems that three indices are to be symmetrized, but actually symmetrization of just the two is desired.

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  • $\begingroup$ If I may, I also ask a side question here: It seems that if the divergence of $A$ vanishes, then the symmetric and antisymmetric part of the tensor $A_{k,i}B_{k,j}$ are equal. What does this imply? Does this mean $A_{k,i}B_{k,j}$ is identically zero? $\endgroup$
    – Bjaam
    Jun 10, 2023 at 13:13

1 Answer 1

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  • Index (anti-) symmetrization is just an operation on tensor components. It does not matter on which space those tensors are defined.

  • A convention for symmetrisation of $i$ and $\ell$ in $A_{i,k}\,B_{j,\ell}$ is to use bars: $$ A_{\,(\,i\,\color{red}|\,,\,k}\,B_{\,j\,,\,\color{red}|\,\ell)}=\frac{A_{\,i\,,\,k}\,B_{\,j\,,\,\ell}-A_{\,\ell\,,\,k}\,B_{\,j\,,\,i}}{2}\,. $$ Whatever is between the bars is unchanged. See: Sean Carroll, Spacetime and Geometry. This also works for anti-symmetrization and it does not matter how many indices are between the bars or in the compartments $(\;\;|$ and $|\;\;)\,.$

  • What if $A$ is divergence free? $$ A=\begin{pmatrix}y\\x\end{pmatrix} $$ is obviously divergence free. Its Jacobian is $$ J_A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\,. $$ Now take $B=A$ then $$ A_{k,i}\,B_{k,j}=J_A^\top J_A=\begin{pmatrix}1&0\\0&1\end{pmatrix}\not=0\,. $$ If you are puzzled now please note that $A_{\,k\,,\,[\,i\,}\,B_{\,k\,]\,,\,j}$ is not the anti-symmetric part of $A_{k,i}\,B_{k,j}\,.$ It is a contraction of $A_{k\,,\,[\,i\,}\,B_{\ell\,]\,,\,j}$ which is anti-symmetric only in $i$ and $\ell\,.$

  • Claim. The following are equivalent: \begin{align} &&(i) &&A_{\,k\,,\,(\,i}\,B_{\,k\,)\,,\,j}&=A_{\,k\,,\,[\,i}\,B_{\,k\,]\,,\,j}\\[2mm] &&(ii) &&A_{\,k\,,\,i}\,B_{\,k\,,\,j}&=\frac{1}{2}A_{\,k\,,\, [\,i}\,B_{\,k\,]\,,\,j}\\[2mm] &&(iii) &&A_{\,k\,,\,i}\,B_{\,k\,,\,j}&=\frac{1}{2}A_{\,k\,,\,(\,i}\,B_{\,k\,)\,,\,j}\\[2mm] &&(iv) &&A_{\,k\,,\,k}\,B_{\,i\,,\,j}&=0\\[2mm] &&(v) &&\text{ $A$ is divergence free,}&\text{ or $B$ is constant } \end{align} Further, when $B$ is constant then $$A_{\,k\,,\,i}\,B_{\,k\,,\,j}=A_{\,k\,,\,(\,i}\,B_{\,k\,)\,,\,j}=A_{\,k\,,\,[\,i}\,B_{\,k\,]\,,\,j}=A_{\,k\,,\,k}\,B_{\,i\,,\,j}=0\,.$$

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  • $\begingroup$ Thanks. My initial guess was that (in general and not only for a specific example like yours) it might be proportional to $\delta_{ij}$ or some symmetric tensor in $i$ and $j$ (although I don't have proof or disproof yet), but I got confused exactly the way you cautioned. What does $A_{k,i}B_{k,j} =A_{k,(i}B_{k),j} = A_{k,[i}B_{k],j}$ mean truly? $\endgroup$
    – Bjaam
    Jun 12, 2023 at 8:21
  • $\begingroup$ Now for using bars: what if $i$ and $j$ were to be contracted in $A_{i,k}B_{j,\ell}$? I will check out Caroll's. $\endgroup$
    – Bjaam
    Jun 12, 2023 at 8:21
  • $\begingroup$ $A_{k,[i}B_{k],j}$ means truly (as I wrote in my last sentence) that it is the contraction of $A_{k,[i}B_{\ell],j}\,.$ Same applies to $A_{k,(i}B_{k),j}\,.$ I think you can make your life easer when you first (anti-)symmetrize and then contract. This also should answer the question in your last comment. $\endgroup$
    – Kurt G.
    Jun 12, 2023 at 8:30
  • $\begingroup$ Yes, I got that. The last comment was regarding the placement of bars. $\endgroup$
    – Bjaam
    Jun 12, 2023 at 9:14
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    $\begingroup$ A.: $A_{(i,|k|}B_{j),\ell}\,.$ $\endgroup$
    – Kurt G.
    Jun 12, 2023 at 9:22

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