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Suppose we have the following equation \begin{equation} \omega_{\rho\sigma}[A^\mu,J^{\rho\sigma}]=B^ \mu \end{equation} where $[A^\mu,J^{\rho\sigma}]=A^\mu J^{\rho\sigma}- J^{\rho\sigma}A^\mu$.

I am starting with topics related to manipulating indices equations representing tensors and I am a little bit lost with the notation. How can one isolate the tensor $A^\mu$ from this equation in terms of the other tensors? My attempt was to multiply both sides by the metric $g^{\rho\sigma}$ so that we can obtain something like \begin{equation} [A^\mu,J^{\rho\sigma}]=\dfrac{1}{\omega}g^{\rho\sigma}B^ \mu \end{equation} since $\omega=g^{\rho\sigma}\omega_{\rho\sigma}$ is now a scalar (number) and it can go dividing on the RHS. After this I am still stuck since the commutator has $A^\mu$ on two terms.

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    $\begingroup$ You can't multiply by the metric here, as then you'd have more than two of each of the dummy indices $\rho$ and \sigma$. It would be more helpful of you gave more context on the problem at hand. $\endgroup$ – Triatticus Apr 19 at 9:29
  • $\begingroup$ Oops. That's true, if I multiply by the metric I am adding free indices to the equation, right? Then I am completely lost in how to proceed. $\endgroup$ – superkingdra Apr 19 at 9:31
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    $\begingroup$ Your $A$'s, $B$'s and $J$'s seem to be operators, not simple tensors. Are your $\omega$'s numbers? Any symmetry properties to be applied to $J^{\rho\sigma}$ and $\omega_{\rho\sigma}$? What is the scalar $\omega$? More context is needed. $\endgroup$ – joigus Apr 19 at 9:48
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    $\begingroup$ Your LHS is 0 if $A,J$ are real numbers. You need to give us more information. $\endgroup$ – jacob1729 Apr 19 at 10:43
  • $\begingroup$ @joigus It's all about the following step, from C.24 to C.25 in page 6 of this notes theorie.ikp.physik.tu-darmstadt.de/qcd/moore/appendixC.pdf, I don't see how to isolate that commutator. $\endgroup$ – superkingdra Apr 19 at 14:20
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In the reference that you provided, you have, $$ \frac{i}{2}\omega_{\mu\nu}\xi_{\alpha}\left[J^{\mu\nu},P^{\alpha}\right]=\omega_{\mu\nu}\xi_{\alpha}\eta^{\nu\alpha}P^{\mu} $$ where $\xi_{\alpha}$ are arbitrary numbers, $\omega_{\mu\nu}$ are antisymmetric, but otherwise arbitrary and numeric, and $J^{\mu\nu}$ are antisymmetric in $\mu$ and $\nu$, never mind they're operators. The $\xi_{\alpha}$ are arbitrary, so they can go. The next step comes from the fact that rank-2 tensors can be expanded in an unambiguous form as the sum of a symmetric and an antisymmetric tensor: $$ T_{\mu\nu}=\frac{1}{2}\left(T_{\mu\nu}+T_{\nu\mu}\right)+\frac{1}{2}\left(T_{\mu\nu}-T_{\nu\mu}\right) $$ The space being a direct sum of symmetric and anti-symmetric subspaces: $$ T_{2}=S_{2}\oplus A_{2} $$ Because you have the contraction of arbitrary antisymmetric constants with operators with the same antisymmetric character in the given indices, you can identify with the LHS by dropping the $\omega_{\mu\nu}$, which is antisymmetric, but only after you anti-symmetrize the LHS on the corresponding indices. This accounts for the $1/2$ disappearing too.

You should always pay heed to the symmetry character of indices, they allow you all kinds of expansions in direct sums that are useful to identify components. Under that light, expressions like the one you were considering, $$ [A^\mu,J^{\rho\sigma}]=\dfrac{1}{\omega}g^{\rho\sigma}B^ \mu $$ would be very suspect from the get go, as you have inconsistent index symmetry on both sides. And $\omega$ being antisymmetric would give zero trace, which is the only scalar you can form from it to first order. I hope that was helpful.

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  • $\begingroup$ I see, it's like we have antisymmetric times antisymmetric in the LHS, which equals symmetric, and we have antisymmetric times something in the RHS, which means that something has to be antisymmetrized in order to be consistent with the symmetry, right? Thanks. $\endgroup$ – superkingdra Apr 20 at 7:45

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