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.
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 $$A_{\,k\,,\,i}\,B_{\,k\,,\,j}=A_{\,k\,,\,(\,i}\,B_{\,k\,)\,,\,j}=A_{\,k\,,\,[\,i}\,B_{\,k\,]\,,\,j}=A_{\,k\,,\,k}\,B_{\,i\,,\,j}=0\,.$$